• Author / Uploaded
• Neil Handa

Citation preview

JEE

MAIN

COMPLETE

CHEMISTRY

JEE

MAIN

COMPLETE

CHEMISTRY K.L. Kapoor Formerly Associate Professor, Hindu College, University of Delhi, Delhi

McGraw Hill Education (India) Private Limited CHENNAI

McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San  Juan Santiago  Singapore Sydney Tokyo Toronto

Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai-600116 Complete Chemistry—JEE Main Copyright © 2018, McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. Price: `875/1  2 3 4 5 6 7 8 9   7085462   22 21 20 19 18

Printed and bound in India ISBN (13):  978-93-87572-55-3 ISBN (10):  93-87572-55-2 Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Sri Krishna Graphics, Delhi and printed at Cover Designer: Neeraj Dayal

visit us at: www.mheducation.co.in

To Our Readers... How to Crack the JEE

T

o help students preparing for the JEE Main, there was need for a book which included a variety of Multiple Choice Questions (MCQs) designed on the basis of the prescribed syllabus for this examination. This book is an attempt in this direction and will help students in developing a strong foundation and enough confidence to take the JEE Main. The various topics of chemistry may be classified into three branches—Physical, Inorganic and Organic. The book covers these three branches in 29 chapters. Each chapter starts with the synopsis of theory followed by MCQs along with answers and hints and solutions to arrive at correct answers. Wherever needed, the chapter is divided into sections to cover the subject in easily understandable portions to help in better grasping of the subject matter. Each section/chapter ends with MCQs from the previous years' AIEEE and JEE Main. This will help students in getting an idea about the types and levels of questions asked in this competitive examination. The answers and solutions to these questions are provided separately, immediately after the questions. The analyses of these papers (provided on the next page) reveals that in most of the cases, one question is asked from each chapter and the entire syllabus is covered in the examination paper. This book provides extensive coverage of the theory as well as the associated MCQs. The contents of each chapter are covered in various sections. At the end of all the sections, more extensive MCQs based on the Entire Chapter along with their solutions are also included. It will be beneficial for the students to adopt the present book as the reference book along with their main text book. The MCQs included in this book should be attempted along with the class-room teaching of the subject matter. A regular and periodical review of the theory and MCQs from this book will help students in gaining enough confidence to appear in the JEE Main and enable them to face the challenge of successfully clearing this examination. From the analyses of previous years’ question papers, a pattern of predominant topics emerges on which students should pay more attention while preparing for the examination. These are: Physical Chemistry—Entire portion Inorganic Chemistry has been thoroughly revised and updated—Chemical families—perioidic properties, structures of compounds containing Si, N, P, S, halogens and inert gases, d-block elements and coordination chemistry Organic Chemistry—Stereoisomerism, SN1 and S22 Reactions, Reactions involving rearrangement, Chemistry of typical reactions shown by phenols, aldehydes and ketones and amines, relative acidity/basicity of phenolic, Carboxylic acids and amines, polymers, carbohydrates, stereochemistry involved in halogenation of alkenes and dehalogenation of halogenated compounds to give alkene, reactions involving Grignard reagent and diazonium salt.

K.L. Kapoor

Trend Analysis Physical Chemistry   Chapter

1. 2.

3. 4. 5. 6. 7.

8.

9. 10.

Some Basic Concepts States of Matter Gaseous State Liquid State Solid State Atomic Structure Chemical Bonding Solutions Chemical Thermodynamics Chemical and Ionic Equilibria Chemical Equilibria Ionic Equilibria Redox Reactions and Electrochemistry Redox Reactions Electrolysis Conduction Galvanic Cells Chemical Kinetics Surface Chemistry Total

2014 (offline) 1

I 1

2014 (online) II III 1 1

2015 2016 (offline) (offline) IV —

1

1 — 1 1

2 — 1 1

1 — 1 2

— — 1 3

1 1 2 —

1 1 1

1 1 1

2 1 1

3 1 —

— — 2

2 2 —

2

1 —

1 —

— 2

1 1

1 1

1 — 2 1 1 — 13

— 2 — — 1 — 13

— — — 1 2 — 14

1 1 — — 1 — 12

3 — — — 1 — 14

1 — 1 — — — — 1 2 12

1

2017 (offline) 2

1

1

1

1 2

1 1 2

1 1 2

1 2 1

1 1 1 11

1 1 1 15

Inorganic Chemistry Chapter

11. 12. 13. 14.

Chemical Families–Periodic properties General Principles and Processes of Isolation of Metals Hydrogen s-Block Elements

2014 (offline) —

I —

2014 (online) II III 1 2

IV 1

1

1

1 — —

1 — —

— — —

— — —

1 1 1

1

— — —

2015 2016 (offline) (offline)

2017 (offline) 1

1 (Contd...)

viii  Trend Analysis

  Chapter 15. 16. 17. 18. 19.

Study of p-Block Elements (Groups 13, 14 and 15) Study of p-Block Elements (Groups 16, 17 and 18) d– and f– Block Elements Coordination Chemistry and Organometallics Nuclear Chemistry Total

2014

2014 (online)

2015

1

1

2

1

2

2

2016 (offline) 1

2017

1

—

1

1

—

1

2

2

1

2

—

2

1

1

2

2 — 6

2 — 6

1 — 5

2 — 8

2 — 6

1 — 9

2

1

10

4

Organic Chemistry   Chapter

2014 (offline) I

20. 21. 22.

23. 24.

25. 26. 27. 28. 29.

Purification and Characterization of Organic Compounds Some Basic Principles Hydrocarbons Alkanes Alkenes Alkynes Benzene Organic Compounds Containing Halogens Organic Compounds Containing Oxygen Alcohols Phenols Ethers Aldehydes & Ketones Carboxylic Acids Organic Compounds Containing Nitrogen Synthetic & Natural Polymers Biomolecules & Biological Processes Chemistry in Action Principles Related to Practical Chemistry Total

2014 (online) II III

2015 2016 2017 (offline) (offline) (offline) IV

1 —

— 1

— 2

— 1

— 1

1 2

— — — —

— — — —

— 1 1 —

1 1 — —

— — 1

— — — 1

2

2

1

—

1

— 2 — 1 1 2

1 — 1 2 — 1

— 1 — 2 — 1

2 — — — 2 1

1

2

—

1 —

— —

— 11

1 11

1

2

1

2

1

1

1

— 2 1 — 1 1

— — — — — 1

1

1

1 1 1

—

1

1

1

1

— 1

1 1

1 —

— 2

1 2

1

1 11

— 10

— 10

— 9

9

1 11

About JEE Main 1.  Introduction and Scheme of Examination The Joint Entrance Examination from the year 2013 for admission to the undergraduate programmes in Engineering is being held in two parts, JEE-Main and JEE-Advanced. Only the top 1,50,000 candidates (including all categories) based on performance in JEE Main will qualify to appear in the JEE Advanced examination. Admissions to IITs will be based only on category-wise All India Rank (AIR) in JEE Advanced, subject to condition that such candidates are in the top 20 percentile categories. Admission to NITs will be based on 40% weightage for performance in Class XII board marks (normalized) and the remainder 60% weightage would be given to performance in JEE Main and a combined All India Rank (AIR) would be decided accordingly. In case any State opts to admit students in the engineering Colleges affiliated to state Universities where States require separate merit list to be provided based on relative weightages adopted by the states, then the merit list shall be prepared with such relative weightages as may be indicated by States.

2.  Eligibility Criteria and List of Qualifying Examinations for JEE(Main) Exam The minimum academic qualification for appearing in JEE(Main) is that the candidate must have passed in final examination of 10+2 (Class XII) or its equivalent referred to as the qualifying examination (see below). However, admission criteria in the concerned institution/university will be followed as prescribed by concerned university/institution and as per the guidelines & criteria prescribed by AICTE.

Qualifying Examinations List of Qualifying Examinations (i) The +2 level examination in the 10+2 pattern of examination of any recognized Central/State Board of Secondary Examination, such as Central Board of Secondary Education, New Delhi, and Council for Indian School Certificate Examination, New Delhi (ii) Intermediate or two-year Pre-University Examination conducted by a recognized Board/University. (iii) Final Examination of the two-year course of the Joint Services Wing of the National Defence Academy. (iv) Any Public School/Board/University Examination in India or in foreign countries recognized by the Association of Indian Universities as equivalent to 10+2 system. (v) H.S.C. Vocational Examination. (vi) A pass grade in the Senior Secondary School Examination conducted by the National Open School with a minimum of five subjects. (vii) 3 or 4-year diploma recognized by AICTE or a State Board of Technical Education.

x  About JEE Main

3.  Pattern of Examination Subject combination for each paper and type of questions in each paper are given below: Paper 1

Paper 2

Subjects Physics, Chemistry & Mathematics

Mathematics – Part I Aptitude Test – Part II & Drawing Test – Part III

Type of Questions Objective type questions with equal weightage to Physics, Chemistry & Mathematics Objective type questions Objective type questions questions to test Drawing Aptitude

Duration 3 Hours

3 Hours

Requirement of papers for different courses is given in the table below: Course B.E/B.TECH B.ARCH/B. PLANNING

Papers Paper – 1 Paper – 2

Scoring and Negative Marking There will be objective type questions with four options having single correct answer. For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted. No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet.

Syllabus SECTION —A Physical Chemistry Unit 1  Some Basic Concepts in Chemistry Matter and its nature, Dalton’s atomic theory; Concept of atom, molecule, element and compound; Physical quantities and their measurements in Chemistry, precision and accuracy, significant figures, S.I. Units, dimensional analysis; Laws of chemical combination; Atomic and molecular masses, mole concept, molar mass, percentage composition, empirical and molecular formulae; Chemical equations and stoichiometry.

Unit 2  States of Matter Classification of matter into solid, liquid and gaseous states. Gaseous State: Measurable properties of gases; Gas laws—Boyle’s law, Charles’ law, Graham’s law of diffusion, Avogadro’s law, Dalton’s law of partial pressure; Concept of Absolute scale of temperature; Ideal gas equation, Kinetic theory of gases (only postulates); Concept of average, root mean square and most probable velocities; Real gases, deviation from Ideal behaviour, compressibility factor, van der Waals equation, liquefaction of gases, critical constants. Liquid State: Properties of liquids—vapour pressure, viscosity and surface tension and effect of temperature on them (qualitative treatment only). Solid State: Classification of solids: molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea); Bragg’s Law and its applications, Unit cell and lattices, packing in solids (fcc, bcc and hcp lattices), voids, calculations involving unit cell parameters, imperfection in solids; electrical, magnetic and dielectric properties.

Unit 3  Atomic Structure Discovery of sub-atomic particles (electron, proton and neutron); Thomson and Rutherford atomic models and their limitations; Nature of electromagnetic radiation, photoelectric effect; spectrum of hydrogen atom, Bohr model of hydrogen atom—its postulates, derivation of the relations for energy of the electron and radii of the different orbits, limitations of Bohr’s model; dual nature of matter, de-­Broglie’s relationship, Heisenberg uncertainty principle. Elementary ideas of quantum mechanics, quantum mechanical model of atom, its important features, y and y2, concept of atomic orbitals as one electron wave functions; Variation of y and y2 with r for 1s and 2s orbitals; various quantum numbers (principal, angular momentum and magnetic quantum numbers) and their significance; shapes of s, p and d—orbitals, electron spin and spin quantum number; rules for filling electrons in orbitals—aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of elements, extra stability of half-filled and completely filled orbitals.

xii  Syllabus

Unit 4  Chemical Bonding and Molecular Structure Kossel: Lewis approach to chemical bond formation, concept of ionic and covalent bonds. Ionic Bonding: Formation of ionic bonds, factors affecting the formation of ionic bonds; calculation of lattice enthalpy. Covalent Bonding: Concept of electronegativity, Fajan’s rule, dipole moment; Valence Shell Electron Pair Repulsion (VSEPR) theory and shapes of simple molecules. Quantum mechanical approach to covalent bonding: Valence bond theory—Its important features, concept of hybridization involving s, p and d orbitals; Resonance. Molecular Orbital Theory: Its important features, LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, concept of bond order, bond length and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications.

Unit 5  Chemical Thermodynamics Fundamentals of Thermodynamics: System and surroundings, extensive and intensive properties, state functions, types of processes. First Law of Thermodynamics: Concept of work, heat, internal energy and enthalpy, heat capacity, molar heat capacity, Hess’s law of constant heat summation; Enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization and solution. Second Law of Thermodynamics: Spontaneity of processes; DS of the universe and DG of the system as criteria for spontaneity, DGo (Standard Gibbs energy change) and equilibrium constant.

Unit 6  Solutions Different methods for expressing concentration of solution—molality, molarity, mole fraction, percentage (by volume and mass both), vapour pressure of solutions and Raoult’s Law—Ideal and non-ideal solutions, vapour pressure— composition plots for ideal and non-ideal solutions; Colligative properties of dilute solutions—relative lowering of vapour pressure, depression of freezing point, elevation of boiling point and osmotic pressure; Determination of molecular mass using colligative properties; Abnormal value of molar mass, van’t Hoff factor and its significance.

Unit 7  Equilibrium Meaning of equilibrium, concept of dynamic equilibrium. Equilibria Involving Physical Processes: Solid–liquid, liquid–gas and solid–gas equilibria, Henry’s law, general characteristics of equilibrium involving physical processes. Equilibria Involving Chemical Processes: Law of chemical equilibrium, equilibrium constants (Kp and Kc) and their significance, significance of DG and DGo in chemical equilibria, factors affecting equilibrium concentration, pressure, temperature, effect of catalyst; Le­Chatelier’s principle. Ionic Equilibrium: Weak and strong electrolytes, ionization of electrolytes, various concepts of acids and bases (Arrhenius, Bronsted—Lowry and Lewis) and their ionization, acid—base equilibria (including multistage ionization) and ionization constants, ionization of water, pH scale, common ion effect, hydrolysis of salts and pH of their solutions, solubility of sparingly soluble salts and solubility products, buffer solutions.

Syllabus  xiii

Unit 8  Redox Reactions and Electrochemistry Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number, balancing of redox reactions. Eectrolytic and metallic conduction, conductance in electrolytic solutions, specific and molar conductivities and their variation with concentration: Kohlrausch’s law and its applications. Electrochemical cells—Electrolytic and Galvanic cells, different types of electrodes, electrode potentials including standard electrode potential, half-cell and cell reactions, emf of a Galvanic cell and its measurement; Nemst equation and its applications; Relationship between cell potential and Gibbs’ energy change; Dry cell and lead accumulator; Fuel cells; Corrosion and its prevention.

Unit 9  Chemical Kinetics Rate of a chemical reaction, factors affecting the rate of reactions c­ oncentration, temperature, pressure and catalyst; elementary and complex reactions, order and molecularity of reactions, rate law, rate constant and its units, differential and integral forms of zero and first order reactions, their characteristics and half-lives, effect of temperature on rate of reactions—Arrhenius theory, activation energy and its calculation, collision theory of bimolecular gaseous reactions (no derivation).

Unit 10  Surface Chemistry Adsorption: Physisorption and chemisorption and their characteristics, factors affecting adsorption of gases on solids—Freundlich and Langmuir adsorption isotherms, adsorption from solutions. Catalysis: Homogeneous and heterogeneous, activity and selectivity of solid catalysts, enzyme catalysis and its mechanism. Colloidal state: Distinction among true solutions, colloids and suspensions, classification of colloids—lyophilic, lyophobic; multi molecular, macro-molecular and associated colloids (micelles), preparation and properties of colloids—Tyndall effect, Brownian movement, electrophoresis, dialysis, coagulation and flocculation; Emulsions and their characteristics.

SECTION — B Inorganic Chemistry Unit 11  Classificaton of Elements and Periodicity in Properties Modem periodic law and present form of the periodic table, s, p, d and f block elements, periodic trends in properties of elements­atomic and ionic radii, ionization enthalpy, electron gain enthalpy, valence, oxidation states and chemical reactivity.

Unit 12  General Principles and Processes of Isolation of Metals Modes of occurrence of elements in nature, minerals, ores; steps involved in the extraction of metals — concentration, reduction (chemical. and electrolytic methods) and refining with special reference to the extraction of Al, Cu, Zn and Fe; Thermodynamic and electrochemical principles involved in the extraction of metals.

Unit 13  Hydrogen Position of hydrogen in periodic table, isotopes, preparation, properties and uses of hydrogen; physical and chemical properties of water and heavy water; Structure, preparation, reactions and uses of hydrogen peroxide; Classification of hydrides—ionic, covalent and interstitial; Hydrogen as a fuel.

xiv  Syllabus

Unit 14  s-Block Elements (Alkali and Alkaline Earth Metals) Group - 1 and 2 Elements General introduction, electronic configuration and general trends in physical and chemical properties of elements, anomalous properties of the first element of each group, diagonal relationships. Preparation and properties of some important compounds—sodium carbonate, sodium chloride, sodium hydroxide and sodium hydrogen carbonate; Industrial uses of lime, limestone, Plaster of Paris and cement; Biological significance of Na, K, Mg and Ca.

Unit 15  p-Block Elements Group-13 to Group-18 Elements General Introduction: Electronic configuration and general trends in physical and chemical properties of elements across the periods and down the groups; unique behaviour of the first element in each group. Groupwise study of the p-block elements Group-13 Preparation, properties and uses of boron and aluminium; structure, properties and uses of borax, boric acid, diborane, boron trifluoride, aluminium chloride and alums. Group-14 Tendency for catenation; Structure, properties and uses of allotropes and oxides of carbon, silicon tetrachloride, silicates, zeolites and silicones. Group-15 Properties and uses of nitrogen and phosphorus; Allotrophic forms of phosphorus; Preparation, properties, structure and uses of ammonia nitric acid, phosphine and phosphorus halides, (PCl3, PCl5); Structures of oxides and oxoacids of nitrogen and phosphorus. Group-16 Preparation, properties, structures and uses of dioxygen and ozone; Allotropic forms of sulphur; Preparation, properties, structures and uses of sulphur dioxide, sulphuric acid (including its industrial preparation); Structures of oxoacids of sulphur. Group-17 Preparation, properties and uses of chlorine and hydrochloric acid; Trends in the acidic nature of hydrogen halides; Structures of Interhalogen compounds and oxides and oxoacids of halogens. Group-18 Occurrence and uses of noble gases; Structures of fluorides and oxides of xenon.

Unit 16  d – and f – Block Elements Transition Elements General introduction, electronic configuration, occurrence and characteristics, general trends in properties of the first row transition elements—physical properties, ionization enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation, interstitial compounds, alloy formation; Preparation, properties and uses of K2 Cr2 O7 and KMnO4.

Syllabus  xv

Inner Transition Elements Lanthanoids — Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction. Actinoids — Electronic configuration and oxidation states.

Unit 17  Co-ordination Compounds Introduction to co-ordination compounds, Werner’s theory; ligands, co-ordination number, denticity, chelation; IUPAC nomenclature of mononuclear co-ordination compounds, isomerism; Bonding ­Valence bond approach and basic ideas of Crystal field theory, colour and magnetic properties; importance of co-ordination compounds (in qualitative analysis, extraction of metals and in biological systems).

Unit 18  Environmental Chemistry Environmental Pollution: Atmospheric, water and soil. Atmospheric Pollution: Tropospheric and stratospheric. . Tropospheric Pollutants: Gaseous pollutants: Oxides of carbon, nitrogen and sulphur, hydrocarbons; their sources, harmful effects and prevention; Green house effect and Global warming; Acid rain; Particulate Pollutants: Smoke, dust, smog, fumes, mist; their sources, harmful effects and prevention. Stratospheric Pollution: Formation and breakdown of ozone, depletion of ozone layer—its mechanism and effects. Water Pollution: Major pollutants such as pathogens, organic wastes and chemical pollutants; their harmful effects and prevention. Soil Pollution: Major pollutants such as Pesticides (insecticides,. herbicides and fungicides), their harmful effects and prevention. Strategies to control environmental pollution.

SECTION—C Organic Chemistry Unit 19  Purification and Characterisation of Organic Compounds Purification: Crystallization, sublimation, distillation, differential extraction and chromatography—principles and their applications. Qualitative Analysis: Detection of nitrogen, sulphur, phosphorus and halogens. Quantitative Analysis (basic principles only) Estimation of carbon, hydrogen, nitrogen, halogens, sulphur, phosphorus. Calculations of empirical formulae and molecular formulae; Numerical problems in organic quantitative analysis.

Unit 20  Some Basic Principles of Organic Chemistry Tetravalency of carbon; Shapes of simple molecules—hybridization (s and p); Classification of organic compounds based on functional groups: - C = C - , - C = C - and those containing halogens, oxygen, nitrogen and sulphur, Homologous series; Isomerism - structural and stereoisomerism. Nomenclature (Trivial and IUPAC)

xvi  Syllabus

Covalent Bond Fission—Homolytic and heterolytic: free radicals, carbocations and carbanions; stability of carbocations and free radicals, electrophiles and nucleophiles. Electronic Displacement in a Covalent Bond: Inductive effect, electromeric effect, resonance and hyperconjugation. Common Types of Organic Reactions: Substitution, addition, elimination and rearrangement.

Unit 21  Hydrocarbons Classification, isomerism, IUPAC nomenclature, general methods of preparation, properties and reactions. Alkanes: Conformations: Sawhorse and Newman projections (of ethane); Mechanism of halogenation of alkanes. Alkenes: Geometrical isomerism; Mechanism of electrophilic addition: addition of hydrogen, halogens, water, hydrogen halides (Markownikoff’s and peroxide effect); Ozonolysis, oxidation, and polymerization. Alkynes: acidic character; addition of hydrogen, halogens, water and hydrogen halides; polymerization. Aromatic hydrocarbons: Nomenclature, benzene—structure and aromaticity; Mechanism of electrophilic substitution: halogenation, nitration, Friedel–Craft’s alkylation and acylation, directive influence of functional group in mono-substituted benzene.

Unit 22  Organic Compounds Containing Halogens General methods of preparation, properties and reactions; Nature of C-X bond; Mechanisms of substitution reactions. Uses/environmental effects of chloroform, iodoform, freons and DDT.

Unit 23  Organic Compounds Containing Oxygen General methods of preparation, properties, reactions and uses. Alcohols, Phenols and Ethers Alcohols: Identification of primary, secondary and tertiary alcohols; mechanism of dehydration. Phenols: Acidic nature, electrophilic substitution reactions: halogenation, nitration and sulphonation, Reimer Tiemann reaction. Ethers: Structure. Aldehyde and Ketones: Nature of carbonyl group; Nucleophilic addition to >C=O group, relative reactivities of aldehydes and ketones; Important reactions such as—Nucleophilic addition reactions (addition of HCN, NH3 and its derivatives), Grignard reagent; oxidation; reduction (Wolff Kishner and Clemmensen); acidity of a - hydrogen, aldol condensation, Cannizzaro reaction, Haloform reaction; Chemical tests to distinguish between aldehydes and Ketones. Carboxylic Acids: Acidic strength and factors affecting it.

Unit 24  Organic Compounds Containing Nitrogen General methods of preparation, properties, reactions and uses. Amines: Nomenclature, classification, structure basic character and identi-fication of primary, secondary and tertiary amines and their basic character. Diazonium Salts: Importance in synthetic organic chemistry.

Syllabus  xvii

Unit 25  Polymers General introduction and classification of polymers, general methods of polymerization - addition and condensation, copolymerization; Natural and synthetic rubber and vulcanization; some important polymers with emphasis on their monomers and uses - polythene, nylon, polyester and bakelite.

Unit 26  Biomolecules General introduction and importance of biomolecules. Carbohydrates: Classification: aldoses and ketoses; monosaccharides (glucose and fructose), constituent monosaccharides of oligosacchorides (sucrose, lactose, maltose) and polysaccharides (starch, cellulose, glycogen). Proteins: Elementary Idea of a - amino acids, peptide bond, polypeptides; proteins: primary, secondary, tertiary and quaternary structure (qualitative idea only), denaturation of proteins, enzymes. Vitamins: Classification and functions. Nucleic Acids: Chemical constitution of DNA and RNA. Biological functions of Nucleic acids.

Unit 27  Chemistry in Everyday Life Chemicals in Medicines: Analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamins-their meaning and common examples. Chemicals in Food: Preservatives, artificial sweetening agents-common examples. Cleansing Agents: Soaps and detergents, cleansing action.

Unit 28  Principles Related to Practical Chemistry ∑ Detection of extra elements (N,S, halogens) in organic compounds; Detection of the following functional groups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl and amino groups in organic compounds. ∑ Chemistry involved in the preparation of the following: Inorganic compounds: Mohr’s salt, potash alum. Organic compounds: Acetanilide, p-nitroacetanilide, aniline yellow, iodoform. ∑ Chemistry involved in the titrimetric excercises - Acids bases and the use of indicators, oxalic-acid vs KMnO4, Mohr’s salt vs KMnO4. ∑ Chemical principles involved in the qualitative salt analysis: Cations: Pb2+, Cu2+, AI3+, Fe3+, Zn2+, Ni2+, Ca2+, Ba2+, Mg2+, NH4+. Anions: CO32–, S2–, SO42–, NO2– , NO3– , CI – , Br–, I– .

(Insoluble salts excluded).

∑ Chemical principles involved in the following experiments:

1. Enthalpy of solution of CuSO4



2. Enthalpy of neutralization of strong acid and strong base.



3. Preparation of lyophilic and lyophobic sols.



4. Kinetic study of reaction of iodide ion with hydrogen peroxide at room temperature.

Contents To Our Readers... Trend Analysis About JEE Main Syllabus 1. Some Basic Concepts of Chemistry Section-1 Physical Quantities  1.1

v vii ix xi 1.1–1.35

Section-2 Significant Figures (or Digits)  1.7 Section-3 Atomic and Molecular Masses  1.9 Section-4 Laws of Chemical Combination  1.14 Section-5 Composition of a Solution  1.19 2. States of Matter Gaseous State  2.1 Unit-1

2.1–2.90

Section-1 Ideal Gases  2.1 Section-2 Kinetic-Molecular Theory of Gases  2.7 Section-3 Real Gases  2.12 Unit-2

Liquid State  2.37

Unit-3

Solid State  2.43

Section-1 Crystal Systems  2.43 Section-2 Closest Packings of Atoms  2.54 Section-3 Structures of Ionic Compounds  2.64 Section-4 Imperfection in Solids  2.74 3. Atomic Structure Section-1 Development of Structure of Atom  3.1

3.1–3.39

Section-2 Quantum-Mechanical Model of Atom  3.8 4. Chemical Bonding and Molecular Structure Section-1 Bond Formation and VSEPR Theory  4.1

4.1–4.42

Section-2 VB and MO Theories 4.13 5. Solutions Section-1 Composition of a Solution  5.1

5.1–5.42

Section-2 Liquid Solutions  5.7 Section-3 Colligative Properties  5.16 6. Chemical Thermodynamics Section-1 Basic Definitions and First Law of Thermodynamics  6.1

6.1–6.45

xx  Contents

Section-2 Thermochemistry 6.9 Section-3 Criteria of Spontaneity  6.17 7. Chemical and Ionic Equilibria Chemical Equilibrium  7.1 Unit-1 unit-2

7.1–7.75

Ionic Equilibrium  7.27

Section-1 Concepts of Acids and Bases  7.27 Section-2 The pH Scale and pH of Acid and Base Solutions  7.32 Section-3 Hydrolysis of Salts  7.39 Section-4 Buffer Solutions  7.44 Section-5 Solubility Product  7.48 Section-6 Acid-Base Indicators  7.53 8. Redox Reactions and Electrochemistry Redox Reactions and Electrolysis  8.1 Unit-1

8.1–8.58

Section-1 Redox Reactions  8.1 Section-2 Electrolytic Cell  8.8 Unit-2

Electrolytic Conduction  8.18

Unit-3

Galvanic Cells  8.30

9. Chemical Kinetics

9.1–9.33

10. Surface Chemistry

10.1–10.9

11. Chemical Families–Periodic Properties

11.1–11.19

12. General Principles and Processes of Isolation of Metals

12.1–12.8

13. Hydrogen

13.1–13.7

14. S-Block Elements (Alkali and Alkaline Earth Metals)

14.1–14.25

The Group 1 Elements – Alkali Metals  14.1 The Group 2 Element – Alkaline Earth Metals  14.12 15. Study of the p-Block Elements (Groups 13, 14 and 15)

15.1–15.60

The Group 13 Elements – Boron Family  15.1 The Group 14 Elements – Carbon Family  15.16 The Group 15 Elements  15.35 16. Study of the p-Block Elements (Groups 16, 17 and 18)

16.1–16.48

The Group 16 Elements  16.1 The Group 17 Elements  16.22 The Group 18 Elements  16.41 17. d– and f–Block Elements

17.1–17.28

18. Coordination Chemistry and Organometallics

18.1–18.26

19. Nuclear Chemistry

19.1–19.8

20. Purification and Characterization of Organic Compounds

20.1–20.9

21. Some Basic Principles

21.1–21.46

Contents  xxi

22. Hydrocarbons Section-1 Alkanes   22.1

22.1–22.61

Section-2 Alkenes   22.13 Section-3 Alkynes   22.31 Section-4 Benzene 22.43 Section-5 Sources of Hydrocarbons  22.58 23. Organic Compounds Containing Halogens (Haloalkanes and Haloarenes) Section-1 Haloalkanes 23.1 Section-2 Haloarenes 23.5 24. Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) Section-1 Alcohols 24.1

23.1–23.14

24.1–24.73

Section-2 Phenols 24.16 Section-3 Ethers 24.31 Section-4 Aldehydes and Ketones  24.34 Section-5 Carboxylic Acids  24.55 25. Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.1–25.24

Section-1 Cyanides and Isocyanides  25.1 Section-2 Nitro Compounds  25.2 Section-3 Amines 25.4 26. Synthetic and Natural Polymers

26.1–26.14

27. Biomolecules and Biological Processes

27.1–27.24

28. Chemistry in Action

28.1–28.13

29. Principles Related to Practical Chemistry

29.1–29.15

Annexure Practice Test Paper–I Practice Test Paper–II Practice Test Paper–III Practice Test Paper–IV Practice Test Paper–V JEE (Main) ChemistryOffline Solved Paper—2017  JEE (Main) ChemistryOnline Solved Paper 1—2017  JEE (Main) ChemistryOnline Solved Paper 2—2017

AN.1–AN.15 PI.1–PI.3 PII.1–PII.4 PIII.1–PIII.5 PIV.1–PIV.4 PV.1–PV.4 JEEC.1-JEEC.8 JEEC.1-JEEC.8 JEEC.1-JEEC.9

1 Some Basic Concepts of Chemistry Section 1

Physical Quantities

In physical sciences, we commonly deal with quantities such as pressure, volume, mass, temperature, current, etc. These quantities are known as physical quantities. A physical quantity has two components, namely, numerical value and its unit, and is written as Physical quantity = (Numerical value) (Unit) The International Union of Pure and Applied Chemistry (IUPAC) has recommended the use of seven physical quantities having their own dimensions. Their dimensions are completely independent of one another and it is for this reason, these are known as dimensionally independent physical quantities. These physical quantities along with their recommended symbols are given in Table 1. Seven Base Physical Quantities Physical quantity

Symbol

Length

l

Mass

m

Time

t

Electric current

I

Thermodynamic temperature

T

Amount of substance

n

Luminous intensity

Iv

Of the seven physical quantities, luminous intensity is not needed in physical chemistry. It is used in optical photometry and is, therefore, included here, only for the sake of completeness.

International System of Units, commonly abbreviated as SI. The SI units of seven base physical quantities are listed in Table 2.

Complete Chemistry—JEE Main

Seven Base Physical Quantities Physical quantity Length Mass Time Electric current Thermodynamic temperature Amount of substance Luminous intensity

Name of SI unit metre kilogram second ampere kelvin mole candela

Symbol for SI unit m kg s A K mol cd

The SI base units stated in Table 2 have been precisely The metre is the length of path travelled by light in vacuum during a time interval of 1/299 792 458 of a second. The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram. This prototype is a polished cylinder of platinum-iridium alloy which was chosen for its durability and resistance to corrosion. The cylinder is kept at the International Bureau of Weights and Measures in a suburb of Paris, France. The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between

of negligible cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 ¥ 10–7 newton per metre of length. The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. The candela is the luminous intensity, in a give direction, of a source that emits monochromatic radiation of frequency 540 ¥ 1012 hertz and that has a radiant intensity in that direction of (1/683) watt per steradian. Of the seven base physical quantities, it is worth highlighting the physical quantity ‘amount of substance’ along with its unit ‘mol’. This is because a basic change in our conventional way of referring to this quantity is required. Suppose we have 0.5 mol of a substance (say, hydrogen). Conventionally, we speak or write it as Number of moles of hydrogen = 0.5 (or moles of hydrogen = 0.5 mol) Now suppose we have 1 kg of hydrogen. We never speak or write it as Number of kg of hydrogen = 1 since we are not accustomed to speak or write it like this. Instead, we say Mass of hydrogen = 1 kg Many such examples can be cited. For example, we never say Number of kg of apples = 1 Number of litre of milk = 1 Number of dozen of apples = 1 Number of metre of a line = 1 A mole of a substance is a collection of 6.022 ¥ 1023 particles of that substance. It is like a dozen which is a collection of 12 articles. When we do not refer to a collection of 12 articles as Number of dozen of articles = 1 why should we refer to a collection of 6.022 ¥ 1023 articles as Number of moles of articles =1? So, some inconsistency has arisen in referring to these two quantities which basically belong to the same category. To avoid this, IUPAC has recommended the use of the phrase ‘amount of substance’ for a physical quantity whose unit is ‘mol’. More precisely, the word ‘amount’ is exclusively reserved whenever the quantity is to be expressed in terms of moles. It is like using the words mass for kg, volume for m3 (or L) and length for metre. Thus, the use of the phrase ‘number of moles equal to 0.5’ should be completely replaced by ‘amount of substance equal to 0.5 mol’. It may be mentioned that to write Amount of substance = 0.5 g is not acceptable as the word ‘amount’ is exclusively reserved for the unit ‘mol’ and not for ‘g’.

Some Basic Concepts of Chemistry

entities of that substance. The proportionality factor is the same for all substances and is equal to l/NA, where NA is Avogadro constant which has a value of NA = 6.022 ¥ 1023 mol–1 ( Earlier, NA was commonly referred to as Avogadro number. But it is not a pure number as it has both a numerical value as well as a unit. So, It is a physical quantity and is called Avogadro constant.) Mathematically, the amount of substance is given as n = N / NA NA is Avogadro constant. For example, 3.011.5 ¥ 1023

where N molecules of dihydrogen will contain

Amount of dihydrogen = 3.011 ¥ 1023 / (6.022 ¥ 1023 mol–1) = 0.5 mol

Physical quantities other than base physical quantities are known as derived physical quantities. These may be expressed in terms of base physical quantities by multiplication and division. Table 3 records some of the derived physical quantities. SI derived units of a few physical quantities Physical quantity

SI unit

Symbol for SI unit

area

(length)2

square metre

m2

volume

(length)3

cubic metre

m3

density

mass/volume

kilogram per cubic metre

speed

distance/time

metre per second

m s–1

acceleration

speed/time

metre per square second

m s–2

heat capacity

Dq/Dt

—

J K–1

(l/m)(Dq/Dt)

—

J kg–1 K–1

molar heat capacity

(l/n)(Dq/Dt)

—

J K–1 mol–1

amount concentration

amount of substance/volume of solution

mole per cubic metre

mol m–3

molality

amount of substance/mass of solvent

mole per kilogram

mol kg–1

kg m–3

Some physical quantities have been assigned special names and symbols. These are described in Table 4. Special names and symbols for certain SI derived units Physical quantity frequency force

Name of SI unit Symbol for SI unit — mass ¥ accleration

Hertz

Hz

s–1

Newton

N

kg m s–2 kg m–1 s–2 (=N m–2)

pressure

force/(length)2

Pascal

Pa

energy

force ¥ length

Joule

J

kg m2 s–2 (= N m)

power

energy/time

Watt

W

kg m2 s–3 (= J s–1)

Coulomb

C

As

electric charge

current ¥ time

electric potential difference

—

Volt

V

kg m2 s–3 A–1 (= J A–1 s–1)

electric resistance

—

Ohm

W

kg m2 s–3 A–2 (= V A–1)

electric conductance

—

Siemens

S

kg–1 m–2 s3 A2 (= A V–1 = W–1)

—

Tesla

T

kg s–2 A–1

Complete Chemistry—JEE Main

–2

Fraction

Symbol

Multiple

Symbol

d

10

deca

da

hecto

h

kilo

k

10

-1

10

-2

centi

c

102

10-3

milli

m

103

micro

m

10

6

mega

M

n

10

9

giga

G

pico

p

1012

tera

T

femto

f

1015

peta

P

a

18

exa

E

10

deci

-6

10-9 10

nano

-12

10-15 10

-18

atto

10

g is known as

unit is treated the same way as any other quantity in an algebraic operation is treated. For example, in algebra, each of the following expressions represents one and the same thing. =6

2

=3

= 1.5 ¥ 4

=2¥3

and so on. Similarly, the following expressions represent one and the same thing, V V = 25.0 cm3 = 25.0 cm3 and so on.

4

=1.5

1.5

=4

V = 1 cm3 25.0

Note that writing an expression of the type V/cm3 = 25.0 is very convenient while writing the headings in tables and as labels on the axes of graphs. A physical expression should also be dimensionally correct. For example, the conversion expression of Celsius temperature to kelvin temperature may be written as T = qC + 273.15 This expression is numerically correct but not dimensionally as the unit of qc is °C and that of T is K. One can add or subtract two physical quantities if they have the same unit. Thus, a correct conversion equation would be T/K = qC/°C + 273.15 For example, for 25 °C, we would have T/K= 25 °C/°C + 273.15 = 25 + 273.15 = 298.15

or

T = 298.15 K

A few other examples are Ea k A log ÊÁ ˆ˜ = log ÊÁ ˆ˜ ; Ë k∞ ¯ Ë k ∞ ¯ 2.303RT

Êhˆ Ê Aˆ E ln Á ˜ = ln Á ˜ + ; Ë h∞ ¯ Ë h∞ ¯ RT

D H Ê pˆ ln Á ˜ = - vap + C Ë p∞ ¯ RT

Note that the division by k°, h° and p° (which stand for the corresponding unit physical quantities) make the expression within the logarithm brackets unitless.

Some Basic Concepts of Chemistry

Other examples are pH = –log {[H+]/mol dm–3};

pK°w = –log [Kw/(mol dm–3)2};

pKa° = –log {Ka/mol dm–3}

The use of standard equilibrium constant K° (= K/(mol dm–3)Ân ) would avoid the division by units. The conversion of one unit to another may be carried out by

. It involves

the following steps. Desired unit =1 Given unit Multiply the given value by the above identity and simplify the expression Arrange the identity so as to have

Convert 57.8 m into cm unit. Identify expression

102 cm =1 1m

Hence,

Ê 102 cm ˆ (57.8 m) Á = 57.8 ¥ 102 cm Ë 1 m ˜¯

Convert 1.5 g cm–3 in terms of kg m–3. Identity expressions

unit itself. For example,

1 kg 1m = 1 and 2 =1 3 10 g 10 cm

Hence,

1.5 g cm–3 =

(1.5 g) (1 kg/103 g) t = 1.5 ¥ 103 kg m–3 (1 cm3 ) (1 m/102 cm)3

1 kg m–3 = 1 (103 g) (102 cm)–3 = 10–3 g cm–3

It is adviseable to use the complete value of a physical quantity (= numerical value ¥ unit) while evaluating its value from the given expression. If SI Units are used for every physical quantity, the result of the expression will also come out in terms of SI base unit. Volume of one mole of an ideal gas at 27 °C and 1 atm pressure. Expression to be used Here

V = nRT/p T = (27 + 273.15) K = 300.15 K p = 1 atm = 101. 325 ¥ 103 Pa R = 8.314 J K–1 mol–1

Hence

V=

(1 mol) (8.314 J K -1 mol-1 ) (300.15 K) = 2.463 ¥ 10–2 J Pa–1 (101.325 ¥ 103 Pa)

J Pa–1 = (kg m2 s–2) (kg m–1 s–2)–1 = m3

and thus

V = 2.463 ¥ 10–2 m3 = 2.463 ¥ 10–2 (10 dm)3 = 24.63 dm3

MULTIPLE CHOICE QUESTIONS ON SECTION 1 Identify the correct choice in the following questions. 1. The symbol for SI unit kg m–1 s–2 is (a) J (b) N (c) Pa 2. Which one of the following is the base physical quantity?

(d) C

Complete Chemistry—JEE Main

3. Which of the following unit represent the joule unit? 2 s–2 (b) N m (c) Pa m–3 (a) kg m 4. The value of Avogardo constant is (b) 6.022 ¥ 1023 atoms (c) 6.022 ¥ 1023 mol–1 (a) 6.022 ¥ 1022 5. Which of the following conversion units is correct? (b) 1 J = 1 kg m2 s–2 (c) 1 C = 1 A s–1 (a) 1 Pa = 1 kg m s–2 18 is (a) tera (b) peta (c) exa –15 is (a) nano (b) pico (c) femto 8. One gigametre stands for (b) 109 m (c) 1012 m (a) 106 9. Which of the following expressions is dimensionally correct? t T (b) pH = –log [H+] (a) = c + 273.15 K ∞C (c)

T Ê tF ˆ 5 = Á - 32˜ ÊÁ ˆ˜ + 273.15 Ë ¯ Ë 9¯ K ∞F

(d)

(d) C V (d) 6.022 ¥ 1022 mol–1 (d) 1 N = 1 kg m–1 s2 (d) atto (d) giga (d) 1015 m

TF Ê 9 ˆ Ê tc ˆ = Á ˜ Á ˜ + 32 °F Ë 5 ¯ Ë ∞C ¯

10. Which one of the followings is the correct conversion expression of 1 J? 2 s–2 (b) 1 N m (c) 1 N m–2 (a) 1 kg m

(d) 1 Pa m3

ANSWERS 1. (c) 7. (c)

2. (b) 8. (b)

3. (c) 9. (b)

4. (c) 10. (c)

5. (b)

6. (d)

HINTS AND SOLUTIONS 1. kg m–1 s–2 = kg m s–2/m2 =

mass × acceleration force = = pressure. The SI symbol of pressure is pascal (Pa) area area

2. Electric current is the base physical quantity. 3. The unit Pa m3 represents joule. 4. Avogadro constant is 6.022 ¥ 1023 mol–1. ¥ acceleration. Its unit will be N = (kg) (m s–2). Pa = (kg m s–2) (m–2) = kg m–1 s–2. ¥ time. Its unit will be C = A s ¥ distance = N m = (kg m s–2)m = kg m2 s–2. 10. See Q.5

Some Basic Concepts of Chemistry

Section 2

Significant Figures (or Digits)

The numerical value of a physical quantity is determined experimentally. Due to experimental error and the limitation include all those digits which are known with certainty, and the last digit indicates the uncertain value. The total number .

cant. 3. Zeros at the end of a number without a decimal point are ambiguous. For example, 7500 may have two or

digit and ending with the digit that has uncertain value.

Number

Number

0.0002

1

3.800

4

0.0038

2

3.080

4

0.0380

3

3.0805

5

0.3800

4

0.3805

4

The counting of discrete variables (such as peas, pencils, erasers, and so on) leads to an exact

An algebraic operation may involve numerical quantities of

1. In an arithmetic operation involving addition and/or subtraction, the answer should include a factor of uncertainty equal to the maximum uncertainty present in the numbers being added and/or subtracted.

3. In an arithmetic operation involving mixed manipulations (addition, subtraction, multiplication and division),

If the digit following the last digit to be retained is more than 5, the last digit to be retained is increased by one. If the digit following the last digit to be retained is less than 5, the last digit is left unchanged. For

Complete Chemistry—JEE Main

digit by 1 if it is odd and leave it unchanged if it is even. For

A number may be a small or large. To express the number in a compact manner, N ¥ 10 n where N is a number with a single non-zero digit to the left of the decimal point and n is an integer. For example, 150 is written as 1.50 ¥ 102 Avogadro constant as 6.022 ¥ 1023 mol–1 Planck’s constant as 6.626 ¥ 10–34 J s

16 0 + 0 16

.0 . 094 . 13 . 224

¨ involves maximum uncertainty

67 . 69 – 2 . 113 65 . 577

Æ rounded off to 16.2

0 . 0 8 3 ¨ minimum significant digits ¥1 0 . 1 0 . 8 3 8 3 Æ rounded off to 0.84 0.0154 = 0.000017440 Æ rounded off to 0.0000174. 883 In scientific notation, this number is written as 1.74 ¥ 10–5

Let

=

94 678 È 1 - 1 ˘ (2.303)(8.314) ÍÎ 298 308 ˙˚

94 678 È ˘ È 308 - 298 ˘ =Í ˙ ÍÎ 298 ¥ 308 ˙˚ = 0.538739 ( 2 . 303 )( 8 . 314 ) Î ˚ The second calculation involves division and subtraction. The actual operation involves 1 1 = 0.003 25 ; = 0.003 36 and 308 298 Since the answer involves two sig digits. Hence, the answer will be 0.54.

0.003 36–0.003 25 = 0.000 11

¨ involves maximum uncertainty Æ rounded off to 65.58

Some Basic Concepts of Chemistry

MULTIPLE CHOICE QUESTIONS ON SECTION 2 Identify the correct choice in the following questions. (a) 1 (b) 2 (c) 3 (d) 4 –11 m and that of a proton is 1.5 ¥10–15 m. The ratio of the radius of 2. The radius of a hydrogen atom is 5.29 ¥ 10 (b) 35.266 ¥ 103 (c) 3.5 ¥ 104 (a) 3.526 ¥ 104 3. If p = 3.141 59, then the value of 4p (c) 1.963 5 ¥ 10–3 (a) 0.001 964 (b) 1.96 ¥ 10–3 4. The answer to the expression 1.561 ¥ 103 – 1.80 ¥ 102 + 2.02 ¥ 104 (b) 2.158 ¥ 104 (c) 2.158 1 ¥ 104 (a) 2.16 ¥ 104 (a) 3

(b) 4

(d) 3.526 666 7 ¥ 103 (d) 3.526 666 7 ¥ 10–2 (d) 21.59 ¥ 104

(c) 5

(d) 6

ANSWERS 1. (c)

2. (c)

3. (b)

4. (a)

5. (a)

HINTS AND SOLUTIONS 123

300 -11

2.

5.29 ¥ 10 m = 3.5 ¥ 104 -15 1.5 ¥ 10 m

4. The result will carry 2 digits after decimal.

Section 3

Atomic and Molecular Masses

The IUPAC has recommended the following terms while dealing with atomic and molecular masses. The relative atomic mass of an element is the ratio of the average mass per of an element to 1/12 of the mass of an atom of the nuclide carbon-12, i.e. mass of an atom (1) A r= (1/12) mass of an atom of 12 C The relative molecular mass of a compound is the ratio of the average carbon-12, i.e. Mr =

mass of a molecule (1/12) mass of an atom of 12 C

(2)

Complete Chemistry—JEE Main

The quantities Ar and Mr are formerly known as atomic weight and molecular weight, respectively. It may be noted that Ar and Mr carry no units as these are simply the ratio of two masses. The quantity (1/12) mass of an atom mu, unit; u, also commonly abbreviated as amu). Hence 1 amu = 1 u = = 1.661 ¥ 10–24 g

12

ma (12 C) (0.012 kg mol-1 ) /(6.022 ¥ 1023 mol-1 ) = = 1.661 ¥ 10–27 kg 12 12

ma). It is simply a mass and thus has the unit of mass (i.e. g or kg). From Eq. (1), it follows that mass of an atom, ma = Ar mu

molecular mass. It is simply a mass and thus carries the unit of mass (i.e. g or kg). For Eq. (2), it follows that mass of a molecule, mf = Mr mu

composition is known as a molar mass. Mathematically, we write it as M=

m n

Since the unit of mass (m) is kg or g, and that of amount of substance (n) is mol, it follows that the unit of molar mass is kg mol–1 or g mol–1. If a system has N entities (atoms or molecules), we have For elements

For Compounds

m = N ma

m = N mf

n = N/NA

n = N/NA m N mf = N A mf = N A ( M r mu ) = M r ( N A mu ) Mm = = n N/N A

Mm =

m n

=

N ma N /N A

= N A ma = N A ( Ar mu ) = Ar ( N A mu )

= Ar ( I g mol–1)

= Mr ( I g mol–1)

In words, the relative atomic (or molecular) mass is the numerical value of the molar mass expressed in g mol–1. We will have Relative atomic mass of sodium = 23 Atomic mass of sodium = 23 u = 23 (1.66 ¥ 10–27 kg) = 3.82 ¥ 10–26 kg Molar mass of sodium = 23 g mol–1 Relative molecular mass of carbon dioxide = 44 Molecular mass of carbon dioxide = 44 u = 44 (1.66 ¥ 10–27 kg) = 7.30 ¥ 10–26 kg Molar mass of carbon dioxide = 44 g mol–1 Quite often, we do not specify the units of atomic, molecular and molar masses. Hence, our statements are not exact and precise. In other words, a mere replacement of the term weight by mass without carrying the associated unit does not complete our adoption of IUPAC recommendations. So, whenever we write or speak of atomic, molecular and molar masses, we must state the associated units. It may be noted that the terms gram atomic weight, gram molecular weight, gram formular weight, gram atom, gram molecule, etc., are obsolete terms and should thus be abandoned.

Some Basic Concepts of Chemistry

An atom or a molecule in nature exists in its natural isotopic composition. The terms mentioned above for atomic or molecular masses refer to an average value of this natural isotopic composition. For example, natural magnesium consists of three isotopes as mentioned in the following. 24Mg

ma = 23.985 u

25Mg

78.99%

ma = 24.986 u

10.00%

26

Mg ma = 25.983 u 11.01% The average atomic mass of Mg is ma(Mg) = (0.789 9 ¥ 23.985 + 0.100 0 ¥ 24.986 + 0.110 1 ¥ 25.983) u = (18.946 + 2.498 6 + 2.861) u = 24.306 u As mentioned earlier, 1 mol of a substance contains 6. 022 ¥ 1023 constituent particles (atoms or molecules or ions). This fact is expressed by Avogadrao constant which has value of 6.022 ¥ 1023 mol–1 It is expressed by the symbol of NA. For a given amount of substance, say n, the number of constituent particles will be given as N = n NA For example, The number of molecules in 0.5 mol glucose is N = (0.5 mol) (6.022 ¥ 1023 mol–1) = 3.011 ¥ 1023 The amount (n) of a substance in a given mass (m) may be calculated from the expression m M where M is the molar mass of the substance. n=

The molar mass of a compound is sum of molar masses of its constituent atoms. For example, the molar mass of H2SO4 is MH2SO4 = 2 MH + MS + 4 MO = (2 ¥ 1 + 32 + 4 ¥ 6) g mol–1 = 98 g mol–1 The per cent of H, S and O is H2SO4 may be calculated by the expression Mass per cent of element =

No. of its atom × Molar mass of atom ¥ 100 Molar mass of the compoound

For example, in H2SO4 we have Mass per cent of H =

2 ¥ 1 g mol-1 ¥ 100 = 2.04 % 98 g mol-1

Mass per cent of S =

1 ¥ 32 g mol-1 ¥ 100 = 32.65 % 98 g mol-1

Mass per cent of O =

4 ¥ 16 g mol-1 ¥ 100 = 65.31 % 98 g mol-1

The chemical formula of a compound may be determined from the mass percentages of atoms present in it. The latter are ascertained experimentally by employing appropriate procedures. With these percentages known, one can proceed to determine the molecular formula as indicated in the following. 1. Take the mass of each element equal to its mass percentage and divide this by the corresponding molar mass (if the mass is taken in grams) or molecular mass (if the mass is taken in terms of atomic mass unit). This gives the relative amounts (if mass is taken in gram) or number of atoms (if mass is taken in atomic mass unit) of different elements present in the compound.

Complete Chemistry—JEE Main

2. Divide the amounts (or number of atoms) of elements by the smallest amount (or number of atoms) to give a simple relative ratio of atoms present in the compound. 3. If the relative ratio of atoms involves noninteger(s), then multiply all the simple ratios by a suitable number to get a whole number ratio for each element. 4. An empirical formula is written by taking as many atoms as given by the whole number ratio. The molecular formula represents the actual number of atoms of each element present in a molecule of the compound. The molecular formula is either the same as the empirical formula or a simple multiple of the empirical formula, i.e., Molecular formula = (Empirical formula)n where n = 1, 2, 3 .... The value of n is equal to the ratio of molar mass of the compound and molar empirical formula mass, i.e. Molar mass n= Molar empirical formula mass –1.

To determine empirical formula from the given data, we proceed as follows. Element

Per cent

Mass of element taken

C

47.5

47.5 g

H

2.54

2.54 g

Cl

50.0

50.0 g

Amount of atom 47.5 g 12 g mol -1 2.54 g 1 g mol

-1

Simple relative ratio of atoms 3.95 mol

= 3.95 mol

1.41 mol 2.54 mol

= 2.54 mol

50.0 g 35.5 g mol

-1

1.41 mol

= 1.41 mol

1.41 mol

Relative whole number of ratio of atoms

= 2.80

2.8 ¥ 5 = 14

= 1.80

1.8 ¥ 5 = 9

=1

1.41 mol

1¥5=5

Hence, the required empirical formula is C14H9Cl5. The above calculations in short may be carried out as follows. 47.5 2.54 50.0 : : 12 1 35.5

3.95 : 2.54 : 1.41 ::

3.95 2.54 1.41 : : :: 2.80 : 1.80 : 1 1.41 1.41 1.41

¥ ¥ ¥ Empirical Formula molar mass = (14 ¥ 12 + 9 ¥ 1 + 5 ¥ 35.5) g mol–1 = 354.5 g mol–1 Since, the empirical formula molar mass is the same as the molar mass, the molecular formula of DDT is C14H9Cl5. A gaseous hydrocarbon gives upon combustion 3.08 g CO2 and 0.72 g H2O. Determine its empirical formula. mCO2 Ê MC ˆ 1 MC Mass of C in the mass mCO2 = ¥ mCO2 ¥ mCO2 ˜ = Amount of C in mCO2 = Á M CO2 Ë M CO2 ¯ M C M CO2 Similarly, 2mH 2O Ê 2M H ˆ 1 Amount of H in the mass mH2O = Á ¥ mH 2O ˜ = Ë M H2O ¯ M H M H2O For the given data, we have Amount of C =

3.08 g = 0.07 mol 44 g mol-1

Amount of H =

2 ¥ 0.72 g = 0.08 mol 18 g mol-1

Empirical Formula C7H8.

Some Basic Concepts of Chemistry

MULTIPLE CHOICE QUESTIONS ON SECTION 3 Identify the correct choice in the following questions.

2.

3.

4.

5. 6.

7. 8.

9. 10.

(b) 1.661 ¥ 10–27 kg (c) 1.661 ¥ 10–25 g (d) 1.661 ¥ 10–25 kg (a) 1.661 ¥ 10–27 g The relative atomic mass of sodium is 23. Which of the following statements is correct about sodium? (a) Atomic mass of sodium is 23 u. (b) Atomic mass of sodium is 3.82 ¥ 10–26 kg. (c) Molar mass of sodium is 23 g mol–1 (d) The number of atoms in 24 kg of sodium is 6.022 ¥ 1023. Chlorine exists as 35Cl (atomic mass = 34.9688 u; 75.77%) and 37Cl (atomic mass = 36.965 9 u; 24.23%). The atomic mass of natural existing chlorine is (a) 35.452 7 u (b) 36.423 6 u (c) 34.123 5 u (d) 32.345 8 u 3.9 g of an organic compound on combustion gives 13.2 g of CO2 and 2.7 g of H2O. The empirical formula of the compound is (c) C2H4 (d) C6H6 (a) CH (b) CH2 Which one of the following has maximum number of atoms? (a) 16 g of C (b) 20 g of Na (c) 45 g of S (d) 15 g of N An organic compound contains 20.0% C, 6.66% H, 47.33% N and the rest was oxygen. Its molar mass is 60 g mol–1. The molecular formula of the compound is (b) CH2NO (c) C2H6NO (d) CH18NO (a) CH4N2O Mass per cent of Na in Na2CO3 is (a) 21.52% (b) 31.20% (c) 38.20% (d) 43.40% –1) and 30.1 mass% of oxygen, its molecular If an iron oxide has 69.9 mass% of Fe (molar mass = 56. 0 g mol formula will be (c) Fe3O4 (d) Fe2O6 (a) FeO (b) Fe2O3 The number of oxygen atoms in 24.9 g of CuSO4◊5H2O is (molar mass of Cu = 63 g mol–1) (a) 2.41 ¥ 1024 (b) 3.01 ¥ 1024 (c) 5.42 ¥ 1023 (d) 5.42 ¥ 1024 A compound contains 11.99% N, 13.70% O, 9.25% B and 65.06% F. Its empirical formula is (molar mass of B is 10.8 g mol–1) (b) NOBF4 (c) N2OF2 (d) NO2F2 (a) NOBF2

ANSWERS 1. (b) 7. (d)

2. (d) 8. (b)

3. (a) 9. (c)

4. (a) 10. (b)

5. (c)

HINTS AND SOLUTIONS 1. mu =

ma (12 C) M (12 C) / N A (0.012 kg mol-1 ) /(6.022 ¥ 1023 mol-1 ) = = 1.661 ¥ 10–27 kg = 12 12 12

2. The number of atoms in 23 g of Na will be equal to 6.022 ¥ 1023.

6. (a)

Complete Chemistry—JEE Main

3. ma = (0.757 7 ¥ 34.968 8 + 0.242 3 ¥ 36.965 9) u = 35.452 7 u 13.2 2 ¥ 7.2 44 15 5. Larger the amount of an element, larger the number of atoms. n(C) = m(C) / M(C) = 16 g/12 g mol–1 = 1. 333 mol

n(Na) = m(Na) / M(Na) = 20 g/ 23 g mol–1 = 0.870 mol

n(S) = m(S) / M(S) = 45 g/ 32 g mol–1 = 1.406 mol

n(N) = m(N)/ M(N) = 15 g/14 g mol–1 = 1.071 mol

20 47.33 26.01 6.66 : : : 12 14 16 1 Empirical formula CN2OH4 Molar empirical mass is 60 g mol–1 (same as the given molar mass). Hence, Molecular formula is CN2OH4. 2M Na 46 ¥ 100 = ¥ 100% = 43.40% 7. Mass per cent of Na = M Na 2CO3 106 69.9 30.1 : 56.0 16.0

1.881 1.248

Molecular formula Fe2O3

9. Molar mass of CuSO4·5H2O = (63 + 32 + 4 ¥ 16 + 5 ¥ 18) g mol–1 = 249 g mol–1 Amount of CuSO4·5H2O in the given mass is n =

24.9 g m = = 0.1 mol M 249 g mol-1

Number of oxygen atoms = 9 ¥ (0.1 mol) (6.022 ¥ 1023 mol–1 ) = 5.42 ¥ 1023 11.99 13.7 9.25 65.06 : : : 14 16 10.8 19 4

Section 4

Laws of Chemical Combination

Based on the study of chemical reactions the following laws have been established. ( ) This law states that the mass is conserved in a chemical reaction. This law states that all pure samples (drawn from different sources) of the same compound contain the same elements combined in the same proportion by mass. This law states that the two elements, A and B versa) are in the ratio of small whole numbers. This law was established by Gay Lussac. According to this law, the volume of reactants and products involved in gaseous reactions are related to each other by small integers, provided the volumes are measured at the same temperature and pressure conditions. For example, in the reaction 2H2(g) + O2(g) Æ 2H2O(g), we have 2 Volume of hydrogen combines with 1 volume of oxygen to give 2 volume of water. This law follows from the fact that under similar conditions of temperature and pressure, equal volumes of all gases contain equal number of particle (Avogadro’s hypothesis). The volume occupied by 1 mol gas at standard condition of temperature (0 °C) and pressure (1 atm), abbreviated as STP, is 22.4 L. Thus, in the above example, 2 ¥ 22.4 L of H2 combines of 22.4 L of O2 to give 2 ¥ 22.4 L of H2O(g).

Some Basic Concepts of Chemistry

The laws of chemical combination may be explained on the basis of Dalton 1. Matter is composed of atoms which cannot be created or destroyed. 2. All the atoms of one element are alike (i.e. they have the same size, shape and mass) but are different from those of other element. 3. In compound, atoms combine together in the ratio of small whole numbers and are held by chemical forces. The smallest entity of a compound is known as a molecule.

The branch of chemistry which deals with mass relationships in chemical reactions is called stoichiometry. This branch 1. Conservation of mass 2. The relative masses of atoms 3. The concept of the mole According to the law of conservation of mass, the total mass of the products formed in a chemical equation is equal to the total mass of reactants that are consumed during the progress of the equation. The law of conservation of mass implies that the number of atoms of each kind must be the same on both sides of a chemical equation. An equation satisfying this criterion is known as a balanced chemical equation. A balanced chemical equation provides quantitative information regarding the consumption of reactants and creation of products. The numbers which appear before the chemical symbols and which balance the equation (with the understanding proportional to the number of molecules or the amounts of the constituents that change during the reaction. For a general case (1) nAA+nBB Æ nCC + nDD where nA,nB,nC and nD The progress of a reaction is described in terms of a physical quantity known as (symbol, x, pronounced xi)’. It is expressed as -

Dn Dn Dn DnA =- B = C = D =x nA nB nC nD

(2)

where Dn represents the change in the amount of the substance. The negative and positive signs in the above expressions are due to the fact that DnA and DnB are negative (i.e. their amounts decrease with the progress of reaction) and DnC and DnD are postive (i.e. their amounts increase with the progress of reaction). The unit of extent of reaction (x) is mol. – DnA = nA mol

Decrease in the amount of A is

Increase in the amount of C is

DnC = nC mol

Decrease in the amount of B is – DnB = nB mol Increase in the amount of D is DnD = nD mol In other words, we say that nA mol of A on reacting with nB mol of B gives nC mol of C and nD mol of D. For example, for the reaction Pb(NO3)2 + 2KI Æ PbI2 + 2KNO3, we have 1 mol of Pb(NO3)2 on reacting with 2 mol of KI gives 1 mol of PbI2 and 2 mol of KNO3. In terms of masses consumed/produced, in the reaction Pb(NO3)2 Molar mass

we have

331.2 g

mol–1

+

Æ PbI2

2KI 166 g

mol–1

461.0 g

+ mol–1

2KNO3 101.1 g mol–1

331.2 g of Pb(NO3)2 on reacting with 2 ¥ 166 g of KI gives 461.0 g of PbI2 and 2 ¥ 101.1 g of KNO3.

In terms of molecules, we have 1 molecule of Pb(NO3)2 on reacting 2 molecules of KI gives 1 molecule of PbI2 and 2 molecules of KNO3 In terms of molecular masses, we have 331.2 u of Pb(NO3)2 on reacting with 2 ¥ 166 u of KI gives 461.0 u of PbI2 and 2 ¥ 101.1 u of KNO3. A given reaction may be initiated with any amounts of reactants, but the consumption of reactants and production of products will be governed by the relation similar to that given by Eq. (2), that is, the relative amounts

Complete Chemistry—JEE Main

appeared in the balanced chemical equation. As the reaction proceeds, the amounts of reactants continue to decrease. The reaction continues to proceed till the amount of one of the reactants is exhausted. This reactant is known as . To determine the limiting reagent, we proceed as follows. Determine the maximum value of extent of reaction with the initial amount of each of the reactants. Initial amount of the reactant xmax = Corresponding stoichiomeetric coefficient . The amounts of products formed will be governed by this limited reagent. For example, Let the reaction 2NaOH + H2SO4 Æ Na2SO4 + 2H2O be started with 100 g each of NaOH and H2SO4. We will have Reactants

NaOH

H2SO4

Initial mass, m0

100 g

100 g

Molar mass, M

40 g mol–1

98 g mol–1

100 mol = 2.5 mol 40

100 mol = 1.02 mol 98

Initial amounts, n0 =

m0 M

2.5 mol 1.02 mol n0 = 1.25 mol = 1.02 mol 2 1 n Since, the extent of reaction is smaller for H2SO4, this reagent will act as limiting reagent. The amounts of products, Na2SO4 and H2O, will be decided by this reagent as its amount is exhausted earlier. Thus, we will have Amount of NaOH unreacted = (n0 – nxmax) = (2.50 – 2 ¥ 1.02)mol = 0.46 mol Amount of Na2SO4 = nxmax = (1) (1.02 mol) = 1.02 mol Amount of H2O = nxmax = (2) (1.02 mol) = 2.04 mol xmax =

MULTIPLE CHOICE QUESTIONS ON SECTION 4 Identify the correct choice in the following questions. (a) Lavoisier (b) Proust (c) Dalton (b) Gay Lussac 2. For the chemical reaction nAA + nBB ÆnCC + nDD, the extent of reaction is given by the expression Dn Dn Dn Dn Dn Dn Dn Dn (a) x = A = C (b) x = A = - C (c) x = - A = C (d) x = - A = - C nA nC n n nA nC nA nC A C 2A(g) + 3B(g) Æ 4C(g) + 5D(g)

3. For the reaction n0

1.5 mol

2.0 mol

the amounts of A, B, C and D when the reaction has proceeded to the extent = 0.15 mol, respectively, are (a) 1.0 mol, 1.25 mol, 1.0 mol and 1.25 mol (b) 1.2 mol, 1.55 mol, 0.60 mol and 0.75 mol (c) 0.5 mol, 0.5 mol, 2.0 mol and 2.5 mol (d) 0.9 mol, 1.1 mol, 1.2 mol and 1.5 mol 4. Consider the reaction N2(g) + 3H2(g) Æ 2NH3(g) n0

1.5 mol

1.5 mol

With the progress of reaction it is found that there is a formation of 0.8 mol of NH3. At this stage, the extent of reaction is (a) 0.2 mol (b) 0.3 mol (c) 0.4 mol (d) 0.5 mol 5. For the reaction N2(g) + 3H2(g) Æ 2NH3(g) m0

1400 g

250 g

Some Basic Concepts of Chemistry

6. 7. 8.

9.

10.

which of the following statement is correct? (a) N2(g) acts as a limiting reagent (b) H2(g) acts as a limiting reagent (c) The extent of reaction when the reaction is over is 41.67 mol (d) At the end of reaction, the mass of NH3 formed is 1416.78 g In the reaction N2H4 + 3O2 Æ 2NO2 + 2H2O, the mass of O2 required to combine with 745 g of N2H4 will be (a) 2120 g (b) 2235 g (c) 2436 g (d) 2510 g A mixture of 50.0 g of S and 100.0 g Cl2 reacts of form S2Cl2. The mass of S2Cl2 formed will be (a) 150.0 g (b) 105.5 g (c) 121.0 g (d) 135.1 g In the reaction Fe2O3(s) + 3C(s) Æ 2 Fe(s) + 3CO(g), 453 kg of iron was obtained from 752 kg of a sample –1) of Fe2O3. The perentage of Fe2O3 (a) 75% (b) 80% (c) 86% (d) 92% In the reaction 2NH4Cl(s) + Ca(OH)2(s) Æ CaCl2(s) + 2NH3(g) + 2H2O(g), the mass of NH3 formed by heating –1) a mixture containing 10.0 g each of NH4Cl and Ca(OH)2 (a) 1.5 g (b) 2.5 g (c) 3.2 g (d) 4.6 g 1.8 g of Mg is burnt in a closed vessel which contains 0.8 g of oxygen. Which of the following statements is correct? Given ; M(Mg) = 24.3 g mol–1. (a) 0.05 mol of MgO is formed (b) 0.8 g of Mg is left behind (c) Oxygen is completely used in the reaction (d) The extent of reaction at the completion of reaction is 0.025 mol.

ANSWERS 1. (a) 7. (b)

2. (c) 8. (c)

3. (b) 9. (c)

4. (c) 10. (b)

5. (a)

6. (b)

HINTS AND SOLUTIONS 1. Law of conservation of mass was established by Lavoisier. Dn Dn 2. The extent of reaction is x=- A = C nA nC 3. We have Dn - A = x fi –DnA = nAx = 2 ¥ 0.15 mol = 0.30 mol. Hence nA = n0,A + DnA = 1.5 mol – 0.30 mol = 1.20 mol nA Dn - B = x fi –DnB = nBx = 3 ¥ 0.15 mol = 0.45 mol. nB = n0,B + DnB = 2.0 mol – 0.45 mol = 1.55 mol nB nC = DnC = nCx = 4 ¥ 0.15 mol = 0.60 mol and nD = DnD = nDx = 5 ¥ 0.15 mol = 0.75 mol x=

4. We have 5. n0(N2) =

DnNH3 n NH3

=

0.8 mol = 0.4 mol 2

m0 ( N 2 ) 1400 g = = 50.0 mol M ( N 2 ) 28 g mol-1

xmax (N2) =

n0 ( N 2 ) 50.0 mol = = 50.0 mol n (N2 ) 1

n0(H2) =

m0 (H 2 ) 250 g = = 125.0 mol M (H 2 ) 2 g mol-1

xmax(H2) =

Since xmax(H2) < xmax(N2), H2 acts as a limiting agent.

n0 (H 2 ) 125.0 mol = = 41.67 mol n (H 2 ) 3

Complete Chemistry—JEE Main

The extent of reaction when the reaction is over will be xmax(H2), i.e. 41.67 mol DnNH3 = nNH3 xmax = 2 ¥41.67 mol = 83.34 mol mNH3 = (DnNH3) (MNH3) = (83.34 mol) (17 g mol–1) = 1416.78 g 6. We have

N2H4 + 3O2 Æ 2NO2 + 2H2O 32 g

96 g

Ê 96 g ˆ Mass of O2 required = Á (745 g) = 2235 g Ë 32 g ˜¯ 7. The reaction is

2S n0

n0 (S) =

xmax (S) =

+

50.0 g

Cl2

Æ

S2Cl2

100.0 g

50.0 g = 1.5625 mol 32 g mol-1

n0 (Cl2 ) =

n0 (S) 1.5625 mol = = 0.7812 mol n (S) 2

xmax (Cl2 ) =

100.0 g = 1.4084 mol 71.0 g mol-1 n0 (Cl2 ) 1.4084 mol = = 1.4084 mol 1 n (Cl2 )

S acts as limiting reagent. Hence Dn(S2Cl2) = n(S2Cl2) xmax(S) = (1)(0.7812 mol) = 0.7812 mol Dm(S2Cl2) = Dn(S2Cl2) M(S2Cl2) = (0.7812 mol)(135 g mol–1) = 105.5 g Fe2O3(s) + 3C(s) Æ 2Fe(s) + 3CO(g)

8. We have Molar mass

160 g mol–1

56 g mol–1

2 ¥ 56 g of Fe(s) will be obtained from 160 g of Fe2O3. Mass of Fe2O3 from which 453 kg of Fe is obtained will be 160 g m= ¥ 453 kg = 647.14 kg 2 ¥ 56 g Hence,

Per cent purity of sample = xmax ( NH 4 Cl) =

9. We have

xmax (Ca (OH)2 ) =

647.14 kg ¥ 100 = 86% 752 kg

m NH 4Cl / M NH 4Cl n NH 4Cl

=

10.0 g / 53.5 g mol-1 = 0.0935 mol 2

mCa(OH)2 / M Ca(OH)2 n Ca(OH)2

=

10 g / 74 g mol-1 = 0.1351 mol 1

Since xmax(NH4Cl) < xmax (Ca(OH)2), the limiting reagent is NH4Cl. DnNH3 = nNH3 xmax(NH4Cl) = 2 ¥ 0.0935 mol = 0.1870 mol mNH3 = DnNH3 MNH3 = (0.1870 mol) (17 g mol–1) = 3.18 g 10. The reaction is xmax (Mg) = xmax (O 2 ) =

2Mg + O2 Æ 2MgO mMg / M Mg n Mg

=

1.8 g / 24.3 g mol-1 = 0.037 mol 2

0.8 g / 32 g mol-1 = 0.025 mol 1

Since xmax(O2) < xmax(Mg), the limiting reagent is oxygen. DnMgO = nMgOxmax(O2) = 2(0.025 mol) = 0.05 mol

Some Basic Concepts of Chemistry

Ê 2M Mg ˆ 48.6 Mass of Mg consumed = Á ˜ mO2 = 32 ¥ 0.8 g = 1.22 g M Ë O2 ¯ Mass of Mg left behind = 1.8 g – 1.22 g = 0.58 g

Section 5

Composition of a Solution

Most of Chemical reactions occur in solution. It is important to know their concentrations. Some of the terms used in

Mass percentage of solute =

Mass of solute ¥ 100 Mass of solution

A solution contains 15 g of sucrose in 100 g water. Calculate the mass percentage of sucrose. Mass of solvent, m1 = 100 g Mass of solute, m2 = 15 g Mass of solution, m = m1 + m2 = 115 g m2 Ê 15 g ˆ ¥ 100 = 13.04% ¥ 100 = Á Mass percentage of sucrose = Ë 115 g ˜¯ m1 + m2 Throughout, the subscripts 1 and 2 represent solvent and solute, respectively.

Amount fraction of solute = Amount fraction of solvent =

n2 Amount of solute = Amount of (solvent + solute) n1 + n2 Amount of solvent n1 = Amount of (solvent + solute) n1 + n2

The sum of amount fractions of solute and solvent is equal to 1. Calculate the amount of fractions of solute and solvent containing 18.0 g of glucose (molar mass = 180 g mol–1) and 90.0 g of water.

Amount of solute, n2 =

m2 18.0 g = = 0.10 mol M 2 180 g mol-1

Amount fraction of glucose, Amount fraction of water,

Molarity =

1

2=

Amount of solvent, n1 =

m1 90.0 g = = 5.00 mol M1 18.0 g mol-1

n2 0.10 mol = = 0.020 n1 + n2 (5.00 + 0.10) mol

=1–

2

= 1 – 0.020 = 0.980

Amount of solute Volume of solution in dm3

i.e.

M=

n2 V

Since the unit of amount of solute is mol, the unit of molarity is mol dm–3, i.e. mol L–1. This unit is commonly abbreviated by the symbol M (roman style) and is spelled as molar. Molarity is temperature dependent as the volume of solution varies with temperature.

Complete Chemistry—JEE Main

Calculate the molarity of glucose containing 18 g of glucose (molar mass = 180 g mol–1) in 90 mL of an aqueous solution. m2 18 g = = 0.1 mol Volume of solution, V = 90 mL = 0. 090 L Amount of glucose, n2 = M 2 180 g mol-1 Molarity of glucose, M =

Molality =

n2 0.1 mol = = 1.1 mol L–1 V 0.090 L

Amount of solute Mass of solvent in kg

Since the unit of amount of solute is mol, the unit of molality is mol kg–1. This unit is spelled as molal. The molality is temperature independent quantity. 10.6 g of sodium carbonate (molar mass = 106 g mol–1) is present in 100 mL of aqueous solution. If density of solution is 1.018 g mL–1, calculate the molality of sodium carbonate in solution. n 10.6 g = 0.10 mol Amount of solute, n2 = 2 = M 2 106 g mol-1 Mass of solution, m = rV = (1.018 g mL–1)(100 mL) = 101.8 g Mass of solvent, m1 = mass of (solution – solute) = 101.8 g –10.6 g = 91.2 g = 91.2 ¥ 10–3 kg Molality of sodium carbonate, m =

n2 0.1 mol = = 1.096 mol kg–1 m1 91.2 ¥ 10-3 kg

MULTIPLE CHOICE QUESTIONS ON SECTION 5 Identify the correct choice in the following questions. 1. The mass of sodium carbonate (Na2CO3) to prepare 400 mL of 0.275 molar aqueous solution is (a) 5.83 g (b) 11.66 g (c) 17.49 g (d) 18.20 g 2. If the density of CH3OH is 0.80 kg L–1, the volume of methanol to prepare 2.5 L of 0.25 M aqueous solution is (a) 25.0 mL (b) 32.0 mL (c) 45.0 mL (d) 56.0 mL 3. The mass of CaCO3(s) that reacts completely with 50 mL of 0.75 M HCl is (a) 3.750 g (b) 2.788 g (c) 2.020 g (d) 1.875 g 4. The mass of ethanol (molar mass = 46 g mol–1) to be added to 1.0 kg of water so as to have its amount fraction equal to 0.2 is (a) 319.5 g (b) 432.1g (c) 638.9 g (d) 719.3 g 5. The density of a 2.0 M solution of acetic acid in water is 1.01 g cm–3. The molality of acetic acid in solution is (b) 2.25 mol kg–1 (c) 2.50 mol kg–1 (d) 3.0 mol kg–1 (a) 2.12 mol kg–1 6. A solution of acetic acid has molarity equal to 1.35 M and molality equal to 1.45 mol kg–1. The density of solution will be (b) 1.125 g mL–1 (c) 1.012 g mL–1 (d) 0.994 g mL–1 (a) 1.251 g mL–1 7. The density of a 3 molal solution of sodium thiosulphate (Na2S2O3·5H2O) solution is 1.25 g cm–3. The percentage by mass of sodium thiosulphate is (a) 35.25% (b) 39.20% (c) 42.67% (d) 46.67% –3. If the density of solution is 1.14 g cm–3, the molality 8. The molarity of a sulphuric acid solution is 2.32 mol dm of the solution will be (b) 2.25 mol kg–1 (c) 2.62 mol kg–1 (c) 1.98 mol kg–1 (a) 2.54 mol kg–1

Some Basic Concepts of Chemistry

9. If 20.0 cm3 of 1.0 M CaCl2 and 60.0 cm3 of 0.20 M CaCl2 are mixed, the molarity of the resultant solution is (a) 0.80 M (b) 0.60 M (c) 0.40 M (d) 0.20 M 10. The concentration of ethanol in the solution called 86-proof vodka is 6.5 M. If the density of the solution is 0.95 g cm–3, the amount fraction of ethanol in vodka is (a) 0.304 (b) 0.252 (c) 0.205 (d) 0.152

ANSWERS 1. (b) 7. (c)

2. (a) 8. (a)

3. (d) 9. (c)

4. (c) 10. (d)

5. (b)

6. (c)

HINTS AND SOLUTIONS 1. Since molarity, M = n2/V, we have Amount of Na2CO3 required, n2 = MV = (0.275 mol L–1) (0.40 L) = 0.11 mol Mass of Na2CO3 required, m = n2 M2 = (0.11 mol) (106 g mol–1) = 11.66 g 2. Since molarity, M = n2/V, we have Amount of CH3OH required, n2 = MV = (0.25 mol L–1) (2.5 L) = 0.625 mol Mass of CH3OH required, m2 = n2M2 = (0.625 mol) (32 g mol–1) = 20.0 g = 20.0 ¥ 10–3 kg Volume of CH3OH required, V =

m2 20.0 ¥ 10-3 kg = = 0.025 L = 25.0 mL r 0.80 kg L-1

3. The reaction is CaCO3 + 2HCl Æ CaCl2 + CO2 + H2O The amount of HCl in 50 mL (= 0.050 L) of 0.75 M solution is n = MV = (0.75 mol L–1) (0.050 L) = 0.0375 mol 3 will be 0.0375 mol/2 = 0.01875 mol m = nM = (0.01875 mol) (100 g mol–1) = 1.875 g. Mass CaCO3 required will be 2 = n2/(n1 + n2) Hence, Now,

2

(n1 + n2) = n2 or

n1 =

2n 1 =

(1 –

2)n2

m1 1000 g = = 55.56 mol . M1 18 g mol-1

or

2n 1

=

1n 2

or

n2 =

2n 1

1

Hence, n2 = (0.2) (55.56 mol)/(0.8) = 13.89 mol

Mass of ethanol required is m2 = n2M2 = (13.89 mol) (46 g mol–1) = 638.9 g 5. For 2.0 M solution, we will have Volume of solution, V = 1000 cm3 Amount of acetic acid, n2 = 2.0 mol –3 Mass of solution, m = rV = (1.01 g cm ) (1000 cm3) = 1010 g Mass of acetic acid, m2 = n2M2 = (2.0 mol) (60 g mol–1) = 120 g Mass of water in 1010 g of solution m1 = 1010 g – 120 g = 890 g = 0.890 kg n2 2.0 mol = = 2.25 mol kg–1 Molality of acetic acid = m1 0.890 kg 6. For 1.35 M solution, we will have n2 = 1.35 mol and V = 1000 mL –1 The solution is also 1.45 mol kg . The mass of solvent to have 1.35 mol of solute will be Ê 1.35 mol ˆ = 0.93103 kg = 931.03 g m1 = Á Ë 1.45 mol kg -1 ˜¯ The mass of solution will be

m = m1 + n2M2 = 931.03 g + (1.35 mol) (60 g mol–1) = 1012.03 g

The density of the solution will be

r=

m 1012.03 g = = 1.012 g mL-1 V 1000 mL

Complete Chemistry—JEE Main

7. Molar mass of Na2S2O3·5H2O is M = (2 ¥ 23 + 2 ¥ 32 + 3 ¥ 16 + 5 ¥ 18) g mol–1 = 248 g mol –1 The mass of 3 mol of Na2S2O3·5H2O is m2 = (3 mol) (248 g mol–1) = 744 g The mass of solution will be m = m1 + m2 = 1000 g + 744 g = 1744 g Mass percentage of sodium thiosulphate =

m2 Ê 744 g ˆ ¥ 100 = 42.67 % ¥ 100 = Á Ë 1744 g ˜¯ m

8. For 1 L of solution, we have n2 = 2.32 mol and V = 1000 cm3 Mass of solution, m = Vr = (1000 cm3) (1.14 g cm–3) = 1140 g Mass of sulphuric acid, m2 = n2M2 = (2.32 mol) (98 g mol–1) = 227.36 g Mass of solvent, m1 = m – m2 = 1140 g – 227.36 g = 912.64 g = 0.913 kg n2 2.32 mol = = 2.54 mol kg–1 Molality of solution = m1 0.913 kg 9. The molarity of the resultant solution will be M1V1 + M 2V2 (1.0 M )(20 cm3 ) + (0.20 M )(60.0 cm3 ) 20 + 12 = M = 0.40 M = M= 3 3 V1 + V2 (20 cm + 60 cm ) 80 10. For 1 L of vodka, we have n2 = 6.5 mol and V = 1000 cm3 Mass of solution, m = Vr = (1000 cm3) (0.95 g cm–3) = 950 g Mass of ethanol, m2 = n2M2 = (6.5 mol) (46 g mol–1) = 299 g Mass of water, m1 = m – m2 = 950 g – 299 g = 651 g m 651 g Amount of water, n1 = 1 = = 36.167 mol M1 18 g mol-1 n2 6.5 = = 0.152 Amount fraction of ethanol, 2 = n1 + n2 (36.167 + 6.5)

Some Basic Concepts of Chemistry

MULTIPLE CHOICE QUESTIONS FOR THE ENTIRE CHAPTER Identify the correct choice in the following questions. 1. The number of atoms present in 0.05 g of water is (b) 1.67 ¥ 1022 (c) 5.02 ¥ 1021 (d) 1.67 ¥ 1021 (a) 1.67 ¥ 1023 2. The value of Avogadro constant is (b) 6.022 ¥ 1023 mol–1 (c) 6.022 ¥ 1023 mol (d) 6.022 ¥ 10–23 mol-1 (a) 6.022 ¥ 1023 3. A certain compound has the molecular formula X4O6. If 10.0 g of the compound contains 5.62 g of X, the atomic mass of X is (a) 62.0 amu (b) 48.0 amu (c) 32.0 amu (d) 30.8 amu 4. Select the quantity of NO2 (c) 7.0 ¥ 1022 molecules (d) 8.0 ¥ 10–1 mol (a) 100 amu (b) 1.0 ¥ 10–3 g 5. The atomic mass unit is equal to (b) 1.66 ¥ 10–25 kg (c) 1.66 ¥ 10–26 kg (d) 1.66 ¥ 10–27 kg (a) 1.66 ¥ 10–24 kg 11 6. Boron occurs in two varieties, namely, 10 atomic mass of naturally occurring element is reported as 10.82 amu. The per cent of 10B in this naturally occurring boron is (a) 10 (b) 19 (c) 29 (d) 35 7 6 7. Lithium occurs in two isotopes, namely, exists 7.4% of 6Li in naturally occurring lithium, then its atomic mass will be (a) 6.2 amu (b) 6.5 amu (c) 6.94 amu (d) 7.2 amu 8. In SI units, the atomic mass unit is represented by the symbol (a) g (b) kg (c) u (d) mg 9. If ml is the mass of 2 neutrons + 2 protons + 2 electrons and m2 is the mass of an a-particle + 2 electrons, then (b) m1 < m2 (a) m1 > m2 (c) ml = m2 (d) m1> may > or < m2 depending of its physical state. 10. The numerical values of molar mass and relative molar mass are identical when the former is expressed in (a) kg mol–1 (b) g mol–1 (c) mg mol–1 (d) cg mol–1 11. Which of the following expressions is dimensionally correct? (a) T = t + 273.15 (b) T/K = t/°C + 273.15 (c) T = t + 273.15 K (d) T = t + 273.15 °C 12. The Celsius and Fahrenheit temperatures corresponding to 64.15 K respectively are (a) –209 °C, –334.2 °F (b) –20.9 °C, –34.4 °F (c) 209 °C, 344.2 °F (d) 20.9 °C, 34.4 °F 13. Which of the following expressions is not dimensionally correct? (b) m = ItM/F |ne | (a) h = pr4pt/8lV 2 2 2 2 2 (c) E = –2p m(e /4pe0) /n h (d) Kc = Kp(RT)Dvg where the various symbols have their usual meanings. 14. Which of the following conversion units is correct? (a) J Pa–1 ∫ m3 (b) (J s2 kg–1)1/2 ∫ m (c) J kg–1 m–1 ∫ m s–1 (d) J1/2 kg–1/2 m–1 = Hz 15. In the IUPAC recommendations, the number of dimensionally independent physical quantities is (a) 1 (b) 3 (c) 5 (d) 7 16. The unit of ‘amount of substance’ is (c) kg (d) mol (a) g (b) cm3

Complete Chemistry—JEE Main

17. The length 109 m may be represented as (a) 10 km (b) 100 Mm (c) 1 Gm (d) 1 Tm 18. One micrometer stands for (b) 10–6 m (c) 10–9 m (d) 10–12 m (a) 10–3 m 19. Which of the following is the unit of energy? (b) N m (c) Pa m3 (d) Pa m–1 (a) kg m2 s–2 20. The unit Siemens is of physical quantity (a) Resistance (b) Conductance (c) Molar conductivity (d) Conductivity 18 is (a) tera (b) peta (c) exa (d) atto 22. Which of the following physical quantities is a base physical quantity? (a) Mass (b) Time (c) Electric charge (d) Thermodynamic temperature 23. Which of the following physical quantities is a derived physical quantity? (a) Electric current (b) Speed (c) Molar heat capacity (d) Electric resistance 24. Which of the following physical quantities has SI unit of kg m2 s–3 A–1? (a) Electric resistance (b) Electric potential (c) Power (d) Energy 2 s–3 A–2? 25. Which of the following physical quantities has SI unit of kg m (a) Electric resistance (b) Electric potential (c) Power (d) Energy (a) J A–1 s–1

(a) nano, n

(b) kg–1 m–2 s3 A2 (c) kg m2 s–2 12 and its symbol respectively are (b) giga, G (c) tera, T 15 and its symbol respectively are (b) Peta, P (c) Tera, T –12 and its symbol respectively are (b) atto, a (c) femto, f

(a) 10-9

(b) 10–12

(a) mega, M (a) Giga,G

(b) 1012 (a) 109 32. One femtometer stands for (b) 10–15 m (a) 10–3 m

(d) kg s–2 A–1 (d) peta, P (d) Exa, E (d) pico, p

(c) 10–15

(d) 10–18

(c) 1015

(d) 1018

(c) 10–9 m

(d) 10–18 m

33. The volume of concentrated sulphuric acid (98 mass % H2SO4, density 1.84 g cm–3) required to prepare 5 dm3 of 0.5 mol dm–3 solution of sulphuric acid is (b) 136 cm3 (c) 204 cm3 (d) 272 cm3 (a) 68 cm3 –3 34. The molarity of concentrated sulphuric acid (r = 1.834 g cm ) containing 95% of H2SO4 by mass is (a) 4.44 M (b) 8.88 M (c) 13.32 M (d) 17.78 M 35. Molarity of liquid water at 4 °C is (b) 55.56 mol L–1 (c) 0.556 mol L–1 (d) 18 mol L–1 (a) 5.56 mol L–1 36. The molarity of 20.0 mass % H2SO4 solution of density 1.14 g cm–3 is (b) 2.02 mol dm–3 (c) 2.32 mol dm–3 (d) 2.82 mol dm–3 (a) 1.52 mol dm–3 –1 ) containing 36.5 mass per cent of HCl required for preparing 2 L of 0.1 M hydrochloric acid solution is about (a) 15.4 mL (b) 16.7 mL (c) 17.5 mL (d) 22.2 mL

Some Basic Concepts of Chemistry –1) containing 70 mass percent

39.

40. 41.

42.

of acid to prepare 1.0 M solution would be (a) 120.6 mL (b) 130.6 mL (c) 135.6 mL (d) 145.6 mL The volume of 0.25 M NaOH to be added to 250 mL of 0.15 M NaOH so that the resultant solution is 0.2 M would be (a) 250 mL (b) 350 mL (c) 450 mL (d) 550 mL Which of the concentration units is temperature dependent? (a) Molality (b) Amount fraction (c) Molarity (d) Parts per million Medicinal alcohol is an aqueous solution of C2H6O that contains half alcohol and half water by volume. If this contains 44.29 % alcohol by mass, the density of pure alcohol is about (a) 0.795 g/mL (b) 0.895 g/mL (c) 0.443 g/mL (d) 0.557 g/mL The expression converting molality (m) into molarity (M) is (b) m = M/(r – MM2) (c) m = (r + MM2)/M (d) m = (r – MM2)/M (a) m = M(r + MM2) where r and M2 are density of solution and molar mass of the solute, respectively.

43. If in a reaction HNO3 is reduced to NO, the mass of HNO3 absorbing one mole of electrons would be (a) 12.6 g (b) 21.0 g (c) 31.5 g (d) 63.0 g – is reduced to NH +, the mass of NO– absorbing one mole of electrons would be 44. If in a given reaction NO3 4 3 (a) 31.0 g (b) 12.4 g (c) 6.29 g (d) 7.75 g –3 and an atomic mass of 55.85 amu. The volume occupied by 1 mol of Fe is 45. Iron has a density of 7.86 g cm (b) 7.11 cm3 mol–1 (c) 4.28 ¥ 104 cm3 mol–1 (d) 22.8 cm3 mol–1 (a) 0.141 cm3 mol–1 – + 46. The stoichiometry numbers of Cu2O, NO 3 and H in the balanced chemical equation of the reaction Cu2O + NO–3 + H+ æÆ Cu2+ + NO + H2O, respectively, are (a) 2, 4, 4 (b) 3, 2, 14 (c) 2, 3, 6 (d) 4, 2, 6 2– + 47. The stoichiometry numbers of Cr2O 7 , CH3CHO, H in the balanced chemical equation of the reaction + 3+ Cr2O 2– 7 + CH3CHO + H æÆ Cr + CH3COOH + H2O, respectively, are (a) 1, 2, 4 (b) 1, 4, 2 (c) 1, 3, 8 (d) 1, 4, 6 48. The stoichiometry numbers of MnO–4, Pb2+ and H2O in the balanced chemical equation of the reaction Mn2+ + PbO2 + H+ æÆ MnO4– + Pb2+ + H2O, respectively, are (a) 2, 5, 4 (b) 1, 4, 2 (c) 1, 2, 6 (d) 1, 2, 3 49. The mass of H2O2 that is completely oxidised by 30.2 g of KMnO4 (molar mass = 158 g mol–1) in acidic medium is (a) 12 g (b) 15 g (c) 17 g (d) 1 g –1 50. The volume V of K2Cr2O7 (molar mass 294 g mol ) solution (mass concentration 0.005 g per mL) is equivalent to 35.0 mL of 0.02 M KMnO4 (molar mass 158 g mol–1), when these solutions are used in the titration in acid solution. The value of V would be (a) 46.75 mL (b) 34.3 mL (c) 29.05 mL (d) 22.5 mL –1 ), is converted into ferric oxide and sulphur dioxide. The mass of 51. Iron pyrite, FeS2 residue left behind from 1.08 g iron pyrite ignited to a constant mass in air would be (a) 0.72 g (b) 0.85 g (c) 0.54 g (d) 0.46 g 52. 0.5 g of sample of an alloy gave 0.30 g of Mg2P2O7 (molar mass of Mg and P are 24 g mol–1 and 31 g mol–1, respectively). The percentage of Mg in the alloy would be (a) 10.5% (b) 12.97% (c) 15.23% (d) 18.31% 53. A sample of NaClO3 is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in –1 ) obtained will be (a) 0.35 g (b) 0.41 g (c) 0.48 g (d) 0.54 g –1 ) and AlCl3 54. A 0.50 g sample containing only anhydrous FeCl3 –1 ). The mass of FeCl3 in the sample is mol–1 (a) 0.3126 g (b) 0.4157 g (c) 0.2345 g (d) 0.1567 g

Complete Chemistry—JEE Main

55. One gram of a mixture of Fe3O4 and Fe2O3 is treated with H2 3Fe2O3 + H2 Æ 2 Fe3O4 + H2O The sample weighs 0.97 g after the above reaction is completed. The per cent of Fe2O3 in the mixture is about (a) 16 (b) 14 (c) 12 (d) 90 56. The combustion of 4.24 mg of an organic compound produces 8.45 mg of CO2 and 3.46 mg of water. The mass percentages of C and H in the compound, respectively, are (a) 54.4, 9.1 (b) 9.1, 54.4 (c) 27.2, 18.2 (d) 18.2, 27.2 57. A 2.0 g mixtute of Na2CO3 and NaHCO3 suffered a loss of 0.12 g on heating. The percentage of Na2CO3 in the mixture is (a) 83.8 (b) 16.2 (c) 38.8 (d) 61.2 58. A compound contains 48.98% C, 2.72% H and 48.32% Cl by mass. Its empirical formula is (b) C3H2Cl (c) C12H10Cl5 (d) C15H12Cl6 (a) C13H8Cl5 59. An organic compound weighing 0.778 g was subjected to Kjeldahl’s method to determine its nitrogen content. The evolved ammonia was absorbed in 100 cm3 of 1 M HC1. The excess acid required 147.4 cm3 of 0.5 M KOH for complete neutralisation. The mass per cent of nitrogen in the compound was (a) 23.67 (b) 32.67 (c) 47.33 (d) 74.33 60. An organic compound contains 20.0% C, 6.66% H, 47.33% N and the rest was oxygen. Its molar mass is 60 g mol–1. The molecular formula of the compound is (b) CH2NO2 (c) C2H6NO (d) CH18NO (a) CH4N2O 61. The simplest formula of the compound containing 32.5% K, 0.839% H, 26.7% S and 39.9% O by mass is (b) KHSO3 (c) KHSO4 (d) K2H2S2O7 (a) KHSO2 62. The simplest formula of a compound which contains 85.6% C and 14.4% H by mass is (c) C2H3 (d) CH3 (a) CH (b) CH2 63. The simplest formula of a compound containing 21.9% Mg, 27.8% P and 50.3% O by mass is (b) MgP2O4 (c) Mg2P2O7 (d) Mg3PO4 (a) Mg2P3O5 64. The mass of Na2CO3 (91.2% purity) required for the neutralisation of 45.6 mL of 0.25 M HC1 solution is (a) 0.6625 g (b) 0.4265 g (c) 0.5765 g (d) 0.8473 g 65. The sulphur in 1.0 g sample of steel is burned to sulphur dioxide and absorbed in 50.0 mL of 0.1 M sodium hydroxide solution. The excess sodium hydroxide requires 24.0 mL of 0.15 M hydrochloric acid solution for neutralization. The percentage of sulphur in the sample is (a) 1.50% (b) 2.24% (c) 3.25% (d) 4.5% 66. A sample of organic material weighing 1.058 g is decomposed by the Kjeldahl method. The ammonia is distilled into 50.0 mL of 0.100 M HCl solution. The excess remaining acid required 10.10 mL of 0.098 M NaOH solution. The percentage of nitrogen in the sample would be (a) 5.45% (b) 10.90% (c) 12.4% (d) 16.7%

ANSWERS 1. 7. 13. 19. 25. 31. 37.

(c) (c) (d) (d) (a) (a) (b)

2. 8. 14. 20. 26. 32. 38.

(b) (c) (c) (b) (d) (b) (d)

3. 9. 15. 21. 27. 33. 39.

(d) (a) (d) (c) (c) (b) (a)

4. 10. 16. 22. 28. 34. 40.

(d) (b) (d) (c) (b) (d) (c)

5. 11. 17. 23. 29. 35. 41.

(d) (b) (c) (a) (d) (b) (a)

6. 12. 18. 24. 30. 36. 42.

(b) (a) (b) (b) (d) (c) (b)

Some Basic Concepts of Chemistry

43. 49. 55. 61.

(b) (c) (d) (b)

44. 50. 56. 62.

(d) (b) (a) (b)

45. 51. 57. 63.

(b) (a) (a) (c)

46. 52. 58. 64.

(b) (b) (b) (a)

47. 53. 59. 65.

(c) (c) (c) (b)

48. 54. 60. 66.

(a) (a) (a) (a)

HINTS AND SOLUTIONS Ê 0.05 g ˆ 1. Number of atoms = 3 Á (6.022 ¥ 1023 mol–1) = 5.02 ¥ 1021 Ë 18 g mol-1 ˜¯ 3. If is atomic mass, then

4 ¥ 10 g = 5.62 g 4 + 96 u

which gives

= 30.8 u

Ê ˆ 1g –22 g 4. 100 amu = (100) Á 23 ˜ = 1.66 ¥ 10 Ë 6.022 ¥ 10 ¯ Mass of 7.0 ¥ 1022 molecules =

6. 7. 9. 12.

7.0 ¥ 1022 ¥ 46 g = 5.35 g 6.022 ¥ 1023

Mass of 8.0 ¥ 10–1 mol = 0.8 ¥ 46 g = 36.8 g If is the fraction of 10B, we have (10.01 amu) + (1 – ) (11.01 amu) = 10.82 amu This gives = 0.19. Hence, percentage is 19%. We have, M = [(7.4) (6.00 amu) + (100 – 7.4) (7.00 amu)]/100 = 6.93 amu In the formation of 42He from its constituents, there occurs the loss of mass. This is known as mass defect. and tF = {(–209 ¥ 1.8) + 32}°F = –344.2 °F tC = (64.15 – 273.15)°C = –209 °C; Unit of work Unit of (force ¥ distance) kg m s -2 (m) = = kg m 2 s -3 A -1 = As Unit of charge Unit of (current ¥ time) I = E/R. Hence,

Unit of potential =

Unit of potential kg m 2 s -3 A -1 = = kg m 2 s -3 A -2 Unit of current A F = qBu. Hence,

Unit of resistance =

Unit of B = 33. V =

Unit of force kg m s -2 = kg s -2 A -1 = (Unit of charge) (Unit of velocity) (A s) (m s -1 )

m(100 / 98) (n2 M 2 )(100 / 98) ( MV ) M 2 (100 / 98) = = r r r

= {(0.5 mol dm–3) (5 dm3) (98 g mol–1) (100/98)}/(1.84 g cm–3) = 136 cm3 95 g / 98 g mol-1 = 0.01778 mol cm–3 = 17.78 mol dm–3 100 g / 1.834 g cm -3 35. One litre of water weighs 1000 g (assuming density equal to 1.00 g/mL). 34. Molarity =

m / M 1000 g / 18 g mol-1 = = 55.56 mol L-1 V 1L m 100 g 36. Volume of 100 g solution is V= = = 87.72 cm3 -1 r 1.14 g cm 20.0 g m = = 0.207 mol n= Amount of sulphuric acid in 100 g solution is M 98 g mol-1 Hence,

Molarity =

Complete Chemistry—JEE Main

0.207 mol n = = 2.32 mol dm -3 -3 3 V 87.72 ¥ 10 dm 36.5 g m = = 1.0 mol 37. In 100 g of HCl solution, the amount of HCl is n = M 36.5 g mol-1 m 100 g Volume of 100 g of HCl solution, V= = = 83.3 mL r 1.20 g mL-1 The amount of HCl in 2 L of 0.1 M HCl solution, n2 = MV = (0.1 mol L–1) (2 L) = 0.2 mol Ê 83.3 mL ˆ Hence, the volume of solution containing 0.2 mol HC1 is V= Á (0.2 mol) = 16.66 mL Ë 1.0 mol ˜¯ Molarity of sulphuric acid is

M=

38. Mass of 10.0 mL of nitric acid, m = Vr = (10.0 mL)(1.40 g mL–1) = 14.0 g Mass of nitric acid in this solution = (14.0 g) (70/100) = 9.8 g Amount of nitric acid, n = m/M = 9.8 g/63 g mol–1 = 0.1556 mol Ê 0.1556 mol ˆ n = Á = 15.56 M Molarity of solution M = Ë 0.01 L ˜¯ V This solution is to be diluted to prepare 1.0 M solution. (10 mL) (15.56 M ) Volume of the resultant solution = = 155.6 mL (1.0 M ) Volume of water to be added = 155.6 mL – 10.0 mL = 145.6 mL 39. Let V be the volume of 0.25 M NaOH solution. Total amount of NaOH after mixing the two solutions is n = V (0.25 mol L–1) + (0.25 L) (0.15 mol L–1) Total volume of the solution = V + 0.25 L Molarity of the resultant solution

M=

V (0.25 mol L-1 ) + (0.25 L)(0.15 mol L-1 ) V + 0.25 L

V (0.25 mol L-1 ) + (0.25 L)(0.15 mol L-1 ) = 0.2 mol L-1 V + 0.25 L Solving for V, we get V = 0.25 L = 250 mL 41. The mass of water per 100 g of medicinal alcohol is 55.71 g. Its volume will be 55.71 cm3. Thus, density of alcohol = 44.29 g / 55.71 cm3 = 0.795 g cm–3 Equating this to 0.2 M, we get

+5

M=

+2

ÆNO 43. We have H N O3 ææ 62 g mol-1 = 21 g eq -1 Change in oxidation number = –3. Equivalent mass of HNO3 = -1 3 eq mol -3 +5 44. We have N O3- ææ Æ N H 4+ 62 g mol-1 = 7.75 g eq -1 Change in oxidation number = –8. Equivalent mass of NO–3 = -1 8 eq mol Molar mass 55.85 g mol-1 = = 7.11 cm3 mol-1 45. Molar volume = density 7.86 g cm -3 49. The reaction is 2KMnO4 + 5H2O2 + 6H+ æÆ 2Mn2+ + 5O2 + 8H2O + 2K+ 2 ¥ 151 g of KMnO4 reacts completely with 5 ¥ 34 g of H2O2 30.2 g of KMnO4 reacts completely with

5 ¥ 34 g H 2 O 2 (30.2 g KMnO 4 ) = 17 g H 2 O 2 2 ¥ 151 g KMnO 4

50. The reactions are 1 14 1 7 Cr2 O72- + H + + e- ææ Æ Cr 3+ + H 2 O 6 6 3 6 Thus,

and

1 8 1 4 MnO -4 + H + + e - ææ Æ Mn 2+ + H 2 O 5 5 5 5

1 1 n ÊÁ Cr2 O72- ˆ˜ ∫ n ÊÁ MnO -4 ˆ˜ i.e. 6n(Cr2 O72- ) = 5n(MnO 4- ) ¯ Ë5 ¯ Ë6

Some Basic Concepts of Chemistry

Now

n(KMnO4) = VM = (0.035 L) (0.02 mol L–1) = 0.0007 mol Ê 0.005 g mL-1 ˆ 0.005 = V ÊÁ mol mL-1 ˆ˜ n(K2CrO7) = VM = V Á -1 ˜ ¯ Ë 294 Ë 294 g mol ¯

Thus

Ê 0.005 mol mL-1 ˆ = 5(0.0007 mol) 6V Á ˜¯ Ë 294

or

Ê 5 ˆ Ê 0.0007 ¥ 294 ˆ mL = 34.3 mL V = ÁË ˜¯ ÁË ˜¯ 6 0.005

51. Molar mass of FeS2 = (56 + 2 ¥ 32) g mol–1 = 120 g mol–1 Molar mass of Fe2O3 = (2 ¥ 56 + 3 ¥ 16) g mol–1 = 160 g mol–1 11 The reaction is 2FeS2 + O æÆ Fe2O3 + 4SO2 2 2 160 Hence, the mass of residue will be ¥ 1.08 g = 0.72 g 2 ¥ 120 52. Molar mass of Mg2P2O7 = (2 ¥ 24 + 2 ¥ 31 + 7 ¥ 16) g mol–1 = 222 g mol–1 2 ¥ 24 ¥ 0.30 g = 0.06486 g Mass of Mg in the given Mg2P2O7 = 222 0.064 86 g ¥ 100 = 12.97% Mass percentage of Mg in alloy = 0.5 g 3 NaClO3 ææ Æ NaCl + O 2 53. The reaction is 2 0.16 g = 0.005 mol Amount of O2 released = 32 g mol-1 2 Amount of NaCl formed or AgCl precipitated = ¥ 0.005 mol 3 2 ˆ Ê Mass of AgCl precipitated = Á ¥ 0.005 mol˜ (108 + 35.5) g mol–1 = 0.48 g ¯ Ë3 m 1.435 g = = 0.01 mol 54. Amount of AgCl obtained, n = M 143.5 g mol-1 Let be the mass of FeCl3, then 0.50 g Amount of FeCl3 = and Amount of A1Cl3 = -1 133.5 g mol-1 162 g mol È ˘ 0.50 g 3Í + = 0.01 mol -1 -1 ˙ 133.5 g mol ˚ Î162 g mol Solving for , we get 1 È 0.01 ¥ 162 ¥ 133.5 = - 81˘˙ g = 0.3126 g (-28.5) ÍÎ 3 ˚ Hence,

55. Molar mass of Fe2O3 = (2 ¥ 56 + 3 ¥ 16) g mol–1 = 160 g mol–1 Molar mass of Fe3O4 = (3 ¥ 56 + 4 ¥ 16) g mol–1 = 232 g mol–1 Loss of mass in the reaction due to conversion of Fe2O3 to Fe3O4 is (2 ¥ 232 – 3 ¥ 160)g mol–1 = –16 g mol–1 16 0.03 ¥ 480 = g = 0.9 g If is the mass of Fe2O3 in the given mixture, then = (1 – 0.97)g or 3 ¥ 160 16 Per cent of Fe2O3 in the mixture = 0.9 ¥ 100 = 90% Ê 12 ˆ Ê 8.45 ˆ 100 = 54.4 56. Mass per cent of C = ÁË ˜¯ ÁË % ˜ 44 4.24 ¯

Ê 2 ˆ Ê 3.46 ˆ 100 = 9.1 Mass per cent of H = Á ˜ Á % Ë 18 ¯ Ë 4.24 ˜¯

57. Loss is due to the reaction 2 NaHCO3 Æ Na2CO3 + H2O + CO2 2 ¥ 84 g

106 g

which involves the loss of 62 g (= 2 ¥ 84 g – 106 g). If is the mass of NaHCO3 in the mixture, then

Complete Chemistry—JEE Main

Ê 168 g ˆ =Á (0.12 g) = 0.325 g Ë 62 g ˜¯

Per cent Na2CO3 =

(2.0 - 0.325)g ¥ 100 = 83.8 2.0 g

48.98 2.72 48.32 : : 12 1 35.5 Empirical formula = C3H2Cl 59. Amount of HCl to start with n1 = VM = (100 ¥ 10–3 L) (1.0 mol L–1) = 0.100 mol Amount of excess of HCl

n2 = VM = (147.4 ¥ 10–3 L) (0.5 mol L–1) = 0.0737 mol

Amount of N in the compound = Amount of NH3 evolved = Amount of acid consumed, n = n1 – n2 = (0.100 – 0.0737) mol = 0.0263 mol Mass of N is the compound mN = nM = (0.0263 mol) (14 g mol–1) = 0.3682 g Per cent of N =

mN morganic

¥ 100 =

0.3682 g ¥ 100 = 47.33% 0.778 g

20.0 6.66 47.33 26.01 : : : 12 1 14 16 Empirical formula = CH4N2O; Molar empirical formula mass = 60 g mol–1 Molecular formula = CH4N2O 32.5 0.039 26.7 39.9 : : : 39 1 32 16 Empirical formula = KHSO3 85.6 14.4 : 12 1

2

21.9 27.8 50.3 : : 24.3 31.0 16 2P 2O 7 64. The reaction is

Na2CO3 + 2HC1 Æ 2NaCl + H2CO3

Amount of HCl to be neutralised = VM = (45.6 ¥ 10–3 L) (0.25 mol L–1) = 11.4 ¥ 10–3 mol Amount of Na2CO3 required = 0.5 ¥ 11.4 ¥ 10–3 mol = 5.7 ¥ 10–3 mol Mass of Na2CO3 required = (5.7 ¥ 10–3 mol) (106 g mol–1) (100/91.2) = 0.6625 g 65. The reaction of NaOH and SO2 is 2NaOH + SO2 Æ Na2SO3 + H2O Amount of NaOH to start with = VM = (50 ¥ 10–3 L) (0.1 mol L–1) = 5 ¥ 10–3 mol Amount of NaOH after the absorption of SO2 = (24.0 ¥ 10–3 L) (0.15 mol L–1) = 3.6 ¥ 10–3 mol Amount of NaOH used in absorbing SO2 = 5 ¥ 10–3 mol – 3.6 ¥ 10–3 mol = 1.4 ¥ 10–3 mol Amount of SO2 formed = (0.5) (1.4 ¥ 10–3 mol) = 7.0 ¥ 10–4 mol Amount of S in the steel = 7.0 ¥ 10–4 mol (7.0 ¥ 10-4 mol) (32 g mol-1 ) ¥ 100 = 2.24% Mass percent of S = 1.0 g 66. Amount of HCl to start with = (50.0 ¥ 10–3 L) (0.10 mol L–1) = 50.0 ¥ 10–4 mol Amount of excess HCl = (10.0 ¥ 10–3 L) (0.098 mol L–1) = 9.8 ¥ 10–4 mol Amount of NH3 evolved or amount of N in the compound = (50.0 – 9.8) ¥ 10–4 mol = 41.2 ¥ 10–4 mol Mass of N in the compound = (41.2 ¥ 10–4 mol) (14 g mol–1) = 0.05768 g Percentage of N in the compound =

0.05768 ¥ 100 = 5.45% 1.058

Some Basic Concepts of Chemistry

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will (a) Remain unchanged (b) Be a function of the molecular mass of the substance (c) Decrease twice (d) Increase twice

3. 4. 5.

6.

7.

8. 9.

+ 520 mL of 1.2 M (a) 1,344 M M (c) 1.20 M (d) 1.50 M How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? (a) 2.5 ¥ 10–2 (b) 3.125 ¥ 10–2 (c) 0.02 (d) 1.25 ¥ 10–2 Density of a 2.05 M solution of acetic acid water is 1.02 g/mL. The molality of the solution is (a) 0.44 mol kg–1 (b) 1.14 mol kg–1 (c) 3.28 mol kg–1 (d) 2.28 mol kg–1 In the reaction, 2Al (s) + 6HCl(aq) Æ 2Al3+(aq) + 6Cl–(aq) + 3H2(g) (a) 6 L HC1(aq) is consumed for every 3 L H2(g) produced (b) 33.6 L H2(g) is produced regardless of temperature and pressure for every mole Al that reacts (c) 67.2 L H2(g) at STP is produced for every mole Al that reacts (d) 11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will be (a) 1.64 (b) 1.88 (c) 1.22 (d) 1.45 The density of a solution prepared by dissolving 120 g urea (molecular mass = 60 u) in 1000 g of water is 1.15 g mL–1. The molarity of the solution is (a) 0.50 M (b) 1.78 M (c) 1.02 M (d) 2.05 M The molarity of a solution obtained by mixing 750 mL of 0.5 M HC1 and 250 mL of 2 M HC1 will be (a) 0.875 M (b) 1.00 M (c) 1.75 M (d) 0.975 M A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is (a) C2H4 (b) C3H4 (c) C6H5 (d) C7H8 is

11. Dissolving 120 g of a compound of relative molar mass 60 in 1000 g of water gave a solution of density 1.12 g/mL. The molarity of the solution is (a) 1.00 M (b) 2.00 M (c) 2.50 M (d) 4.00 M 12. The mass of oxygen in 3.6 mol of water is (a) 115.2 g (b) 57.6 g (c) 28.8 g (d) 18.4 g 13. A gaseous compound of nitrogen and hydrogen contains 12.5 mass% of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is (a) NH2 (b) N3H (c) NH3 (d) N2H4 14. The mass of BaSO4 formed upon mixing 100 mL of 20.8% BaCl2 solution with 50 mL of 9.8% H2SO4 solution (a) 23.3 g (b) 11.65 g (c) 30.6 g (d) 33.2 g 15. A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. (a) BaCl ◊ H2O

(b) BaCl2 ◊ 2H2O

(c) BaCl2 ◊ 3H2O

(d) BaCl2 ◊ 4H2O

Complete Chemistry—JEE Main

16. A + 2B + 3C AB2C3. Reaction of 6.0 g of A, 6.0 ¥ 1023 atoms of B, and 0.036 mol of C yields 4.8 g of compound AB2C3. If the atomic masses of A and C are 60 and 80 amu, respectively, the atomic mass B of is (Avogadro constant = 6 ¥ 1023 mol–1) (a) 70 amu (b) 60 amu (c) 30 amu (d) 40 amu 17. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (molecular 2+ ions by the resin when expressed in mole per gram resin? (a) 1/103 (b) 1/206 (c) 2/309 (d) 1/412 18. 5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant (a) Isobutane (b) Ethane (c) Butane (d) Propane 19. An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% (a) 600 (b) 200 (c) 400 (d) 300 20. The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S in (a) 0.25 mol

(b) 0.50 mol

(c) 0.333 mol

(d) 0.125 mol –

in aqueous

(a) 400 mL (b) 600 mL (c) 200 mL (d) 800 mL 22. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is (b) C4H8 (c) C4H10 (d) C3H6 (a) C3H8

ANSWERS 1. 7. 13. 19.

(a) (d) (d) (c)

2. 8. 14. 20.

(a) (a) (b) (d)

3. 9. 15. 21.

(b) (d) (b) (a)

4. 10. 16. 22.

(d) (a) (c) None

5. (d) 11. (b) 17. (d)

6. (c) 12. (b) 18. (d)

HINTS AND SOLUTIONS 1. The atomic mass unit will become twice and the relative molar mass will become half such that the product of these two remains unchanged. 2. Total amount of solute = (0.480 ¥ 1.5 + 0.520 ¥ 1.2) mol = (0.720 + 0.624) mol = 1.344 mol Total volume of solution = (480 + 520) mL = 1000 mL = 1 L Molarity of solution = 1.344 mol/l L = 1.344 M 3. In Mg3(PO4)2, there are 8 atoms of oxygen. Hence, 0.25 mol of oxygen atoms will be contained in (1 mol/8 mol) (0.25 mol) = 3.125 ¥ 10–2 mol of Mg3(PO4)2. 4. In 1 L of solution, we have r(acetic acid soln) = 1.02 g/mL = 1020 g/L m(acetic acid) = (2.05 mol/L) (60 g mol–1) = 123 g/L m(water) = ( 1020 – 123) g/L = 897 g/L Molality of acetic acid solution = (2.05 mol)/(0.897 kg) = 2.28 mol kg–1

Some Basic Concepts of Chemistry

2Al(s) + 6HCl(aq) Æ 2Al3+(aq) + 6Cl–(aq) + 3H2(g)

5. We have

2 mol

6 mol

3 mol 3 ¥ 22.4 L at STP

3 ¥ 22.4 L = 11.2 L 6 6. In 100 g solution, 29 g H2SO4 is present. The volume of this solution as determined from the given molarity is For 1 mol of HCl, the volume of H2 gas evolved will be

n (29 g / 98 g mol-1 ) = = 0.0822 L ∫ 82.2 mL M 3.60 mol L-1 m 100 g = 1.22 g mL-1 Hence, density of the solution is r = = V 82.2 mL V=

7. Amount of urea dissolved, n =

120 g m = = 2 mol M m 60 g mol-1

Total mass of solution, m = 120 g + 1000 g = 1120 g Volume of solution, V =

m 1120 g = = 973.9 mL = 0.9739 L r 1.15 g mL-1

n 2 mol = = 2.05 mol L-1 V 0.9739 L 8. Amount of HCl in the resultant solution is n = MlVl + M2V2 = (0.5 M) (750 ¥ 10–3 L) + (2 M) (250 ¥ 10–3 L) = 875 ¥ 10–3 mol Total volume of the resultant solution is V = Vl + V2 = (750 + 250) ¥ 10–3 L = 1 L Molarity of solution, M =

Molarity of the resultant solution is M =

n 875 ¥ 10-3 mol = = 0.875 mol L-1 = 0.875 M V 1L

Ê 2M H ˆ 2 ÁM ˜ mH 2O = 18 ¥ 0.72 g = 0.08 g Ë H2O ¯ m 0.08 g = 0.08 mol Amount of H in the molecule is nH = H = M H 1 g mol-1

9. Mass of H in the obtained water is

Mass of C in the obtained CO2 is

Ê MC ˆ 12 ÁM ˜ mCO2 = 44 ¥ 3.08 g = 0.84 g Ë CO2 ¯

Amount of C in the molecule is

nC =

mC 0.84 g = = 0.07 mol M C 12 g mol-1

The ratio of amounts of C and H in the molecule is nC nH Hence, the empirical formula of the compound is C7H8. 10. Let there, be 1 g of O2, then mass of N2 will be 4 g. 1g = (1 / 32) mol 32 g mol-1 1 1 : :: 7 : 32 Ratio of their amounts is 32 7 Amount of O2 =

Amount of N2 =

The same ratio holds good for the number of molecules.

11. Total mass of the solution = 120 g + 1000 g = 1120 g Volume of the solution, V = Amount of compound, n =

4g = (1 / 7) mol 28 g mol-1

m 1120 g = = 1000 mL = 1.0 L r 1.12 g mL-1

120 g m = = 2.0 mol M 60 g mol-1

Complete Chemistry—JEE Main

Molarity of solution, M =

n 2.0 mol = = 2.0 M V 1.0 L

12. Mass of oxygen, m = nM = (3.6 mol) (16 g mol–1) = 57.6 g 87.5 14

12.5 1

Molar empirical formula mass = 16 g mol–1 Molar mass = ( 2 ¥ vapour density) g mol–1 = 2 ¥ 16 g mol–1 Molecular formula (NH2), i.e. N2H4. 2

14. Amount of BaCl2 in 100 mL of 20.8% BaCl2 = Amount of H2SO4 in 50 mL of 9.8% H2SO4 =

m 20.8 g = = 0.1 mol M (137 + 2 ¥ 35.5) g mol-1

m (9.8 / 2) g = = 0.05 mol M 98 g mol-1

From the reaction BaCl2 + H2SO4 Æ BaSO4 + 2HCl 0.1 mol

0.05 mol

We conclude that 0.05 mol of BaSO4 will be formed Hence m(BaSO4) = (0.05 mol) (233 g mol–1) = 11.65 g 15. Amount of anhydrous BaCl2, n =

m 52 g = = 0.25 mol M (137 + 2 ¥ 35.5) g mol-1

Let the hydrated barium chloride be BaCl2 ◊ H2O. The mass of 0.25 mol of this salt will be equal to 61 g (given) Hence, (0.25 mol) (208 + ¥ 18)g mol–1 = 61 g Solving for , we get = 2. Hence, the formula of hydrated barium chloride is BaCl2 ◊ 2H2O A 6.0 g 6 = mol 60

+

2B

+

6.0 ¥ 1023 atoms

Maximum extent 0.1 mol of reaction

3C 0.036 mol

AB2C3 4.8 g

1 mol

0.5 mol

0.012 mol

The limiting reagent is the atom C. The amount of AB2C3 formed will also be 0.012 mol. Mass of AB2C3 formed is 4.8 g. Hence (0.012 ¥ 60 + (0.024 mol) MB + 036 ¥ 80) g = 4.8 g Thus, atomic mass of B is 50 amu. This gives MB = 50 g mol–1. 17. The exchanging reaction is

2 C8H7SO3Na + Ca2+Æ(C8H7SO3)2Ca + 2Na+

For one mole of Ca2+ ions, two moles of resin is required. Thus, the maximum uptake of Ca2+ ions will be 1 mol Ca2+/2 mol resin ∫1 mole Ca2+/(2 × 206) g resin ∫ (1/412) mol Ca2+/g of resin 18. For the alkane CnH2n+2, the combustion reaction is Ê 3n + 1ˆ O Æ n CO2 + (n + 1) H2O CnH2n + 2 + Á Ë 2 ˜¯ 2 For 5 L of CnH2n+2, the volume of oxygen required is (5 L) {(3n + 1)/2}. Equating this to 25 L, we get

Some Basic Concepts of Chemistry

Ê 3n + 1ˆ (5 L) Á = 25 L Ë 2 ˜¯ This gives n = 3. Hence, the alkane is propane. 19. 8% sulphur implies that 8 g of sulphur is present in 100 g of the compound. The minimum molecular weight of the compound is obtained when 32 g of sulphur (i.e. one mole of sulphur) is present. Thus minimum molecules Ê 100 ˆ (32) = 400 weight = Á Ë 8 ˜¯ 20. Molar mass of arsenic acid, H3AsO4, is M = (3 + 74.92 + 4 × 16) g mol–1 = 143.42 g mol–1 Arsenic pentasulphide is As2S5. 2 mol of H3AsO4 will provide 1 mol of As2S5. 35.5 g (= (35.5/143.92) mol) of H3AsO4 will provide

1 Ê 35.5 ˆ Á ˜ mol of As2O5 which is 0.123 mol. 2 Ë 143.92 ¯

21. 0.04 mol of OH– implies 0.04 eq of OH–. Hence

N1V1 = 0.04 eq V1 =

Hence,

i.e.

(0.1 eq L–1) V1 = 0.04 eq

0.04 L = 0.4 L ∫ 400 mL 0.1

Ê 20 ˆ (375 mL) = 75 mL 22. Volume of O2 present in the given volume of air = Á Ë 100 ˜¯ Volume of air without O2 = 375 mL – 75 mL = 300 mL The combustion reaction may be represented as y C Hy + ÊÁ x + ˆ˜ O2 Ë 4¯ 15 mL

Æ

75 mL

CO2

+

y HO 2 2

(330 – 300) mL = 30 mL

Obviously, the reaction is y HO 2 2 y/4 = 5 – = 5 – 2 = 3 i.e. y = 12

C Hy + 5O2 Æ 2CO2 + Thus

=2

and

The hydrocarbon is C2H12, which theoretically, is not possible. Had the volume of gas after the combustion of hydrocarbon been 345 mL, the hydrocarbon would be C3H8 as seen from the following calculations. yˆ y Ê CO2 + HO C Hy + ÁË x + ˜¯ O2 Æ 4 2 2 15 mL

75 mL

(345 – 300) mL = 45 mL

The value of x will be 45 mL/15 mL = 3 and that of y will be +

75 mL y = =5 15 mL 4 y = 4(5 – ) = 4( 5 – 3) = 8.

The net reaction will be

C3H8 + 5O2 Æ 3 CO2 + 4 H2O

2 States of Matter

UNIT 1

GASEOUS STATE

SECTION 1

Ideal Gases

Experimentally Derived Gaseous Laws Boyle’s Law Vμ

1 p

V=

K p

or

pV = K

(1)

where K p lV l = p 2V 2 where Vl and V2 are volumes at pressures p1 and p2

Fig. 1

(2)

Fig. 2

Charles Law Vt = a + bt

(3)

2.2 Complete Chemistry—JEE Main

where t

a and b a and b are b=

V0 / 273.15 1 ∞C

and

a=V

t / ∞C ˆ Ê V0 / 273.15 ˆ Ê 273.15 + t / ∞C ˆ t = V0 ÊÁ1 + Vt = V + Á ˜¯ = V0 ÁË ˜ ˜ Ë Ë 1 ∞C ¯ 273.15 273.15 ¯

(4)

t ∞

T K

ˆ Ê T / K ˆ = Ê V0 T = KT VT = V0 Á ˜ Á Ë 273.15 ¯ Ë 273.15 K ˜¯

(6)

where K

Fig. 3

Fig. 4

Gay-Lussac’s Law pt = a + bt

and

pT = KT

(7)

pT = KT

Fig. 5

Fig. 6

States of Matter 2.3

Equation of State p1 , V1 , T1 æStep æææ Æ p2 ,V ¢, T1 æStep æææ Æ p2 ,V2 , T2 (1) (2) p 1V 1 = p 2V ¢ fi V ¢/T1 = V2/T2 fi V¢

Step 1 Step 2

p1V1 V2T1 = p2 T2

or

V ¢ = p1V 1/p2 V ¢ = V2T1/T2

p1V1 p2V2 = T1 T2

or

pV =K T

(8)

where K †

K

Universal Gas Constant R

K = nR where n

pV = nRT

(9)

(pressure) (volume) (amount of gas) (kelvin temperature) R are (force)(llength ) (force/length 2 ) (length)3 work (or energy) = R= = (amount of gas) (kelvin) (amount of gas)(kelvin) (amount of gas) (kelvin) R=

R

Concept of an Ideal Gas and Value of Gas Constant

When H2

T = 273.15 K

He 22.414

Ideal gas

ideal gas. pV/atm L

pV 2, N 2

p

2

p same value of pV

O2 p/atm

Fig. 7

R=

22.414 atm L pV = nT (1 mol) (273.15 K )

p

†

–1

V

mol–1

3

3

p and T

V K

, we have

N2

2.4 Complete Chemistry—JEE Main 3)

R 3 3,

K–1 mol–1

3

K–1 mol–1

3

K–1 mol–1

K–1 mol–1 –1

R

we also have

mol–1

R –1

R

mol

–1

Dalton’s Law of Partial Pressures

p = p1 + p2 pV = (n1 + n2 + · · ·) RT ;

where

p1V = n1RT ;

p1 n1 n = 1 = x1 , = p n1 + n2 +  ntotal

also

p2V = n2RT · · ·

p2 n2 n = 2 = x2 = p n1 + n2 +  ntotal

(11)

where x1, x2, …

such that ptotal = p

+ pwater

(12)

Graham’s Law of Diffusion

T and p, we have r2 = r1

r1 r2

r2 = r1

or

r2 Ê p2 ˆ Ê M1 ˆ = Á ˜Á Ë p1 ¯ Ë M 2 ˜¯ r1

M1 M2

(13)

1/ 2

(14) p1 and p2

Illustration M

–1

M 4 4 +1 1 p 4=x 4p= 4 +1

p

Ê p ˆ Ê M CH 4 ˆ nHe r = He = Á He ˜ Á ˜ nCH 4 rCH 4 Ë pCH 4 ¯ Ë M He ¯

1/ 2

=x

p=

Ê 16 bar ˆ Ê 16 g mol -1 ˆ = 16 = Á Ë 4 bar ˜¯ ÁË 4 g mol -1 ˜¯

–1 4

States of Matter 2.5

Avogadro’s Law ¥

23

as ¥

NA

23

Density of a Gas

mol–1 pV = nRT

pV = nRT = (m/M) RT where m

M r=

m pM = V RT

MULTIPLE CHOICE QUESTIONS ON SECTION 1

pV = K

K does not depend upon

pV versus p

p

(a) p where p

pV)p = ¥

not correct? 2

dm3

3

R ¥

7

–1 –1

p = a + bt, where b (d) a/273 K

p /273 K

¥

3

¥

23

cm3

not correct?

mol–1

–1

mol–1 –1 mol–1

mol–1 3

¥

¥

23

¥

22

22

–1 –3

–3

–3 2

2

(a) p

2

> pN > p 2

2

2

2

p

2

–3 –1

2

>p

2

> pN

2

(c) pN > p 2

2

>p

2

(d) pN > p 2

2

3 3

4

3

3

3

3

3

>p

2

2.6 Complete Chemistry—JEE Main –1 –1

–1

–1

–1

3

¥

¥

8

3

¥

7

¥

7

6

ANSWERS

HINTS AND SOLUTIONS K pV = constant at all pressures, the plot of pV versus p b=p b = (∂p/∂t)V = (p pV)p ∫

¥

3

¥

3

–1

¥

dm3

–1

mol–1

m)3 –1

n=

mol–1

–2

–1

pM (1 atm)(124 g mol -1 ) = r= RT (0.082 dm3 atm K -1 mol -1 ) (620 K ) Mav =

mol–1

pV (6.0 atm)(500 ¥ 10 dm ) = RT (0.082 dm3 atm K -1 mol-1 )(300 K )

N = nNA

If x

3

3

-3

r = pM/RT, we have

2

cm3

3

R

2

¥

23

3

mol–1

¥

22

–3

rRT (0.50 g L-1 )(0.082 L atm K -1mol-1 )(300 K ) = (1 atm) p

–1

M av = x M H 2 + (1 - x) M O2 = M O2 + ( M H 2 - M O2 ) x x=

or

n n

M av - M O2 M H 2 - M O2

=

12.3 g mol -1 - 32 g mol -1 2 g mol -1 - 32 g mol -1

2)

=

m (O 2 ) 25 g = = 0.78 mol ; M (O 2 ) 32 g mol-1

2)

=

m(CO 2 ) 40 g = = 0.909 mol M (CO 2 ) 44 g mol-1

n(N2) =

m( N 2 ) 35 g = = 1.25 mol M ( N 2 ) 28 g mol-1

States of Matter 2.7 3

rNH3 rHCl or

x

4

ÊM ˆ = Á HCl ˜ Ë M NH3 ¯

1/ 2

i.e.

x/t 36.5 ˆ 1 / 2 = ÊÁ ˜ = 1.465 (3 m - x) / t Ë 17 ¯

x = (3 m – x

x

x p2 = p

– pwater

p2V2/T2 = p1V1/T1 Ê p1 ˆ Ê T2 ˆ Ê 760 ˆ Ê 300 ˆ (125 cm3 ) = 144.0 cm3 V2 = Á ˜ Á ˜ V1 = ÁË ˜ ˜Á Ë p2 ¯ Ë T1 ¯ 725 ¯ Ë 273 ¯ Ê p ˆÊT ˆ Ê 700 ˆ Ê 273 ˆ (1.43 g L-1 ) r2 = Á 2 ˜ Á 1 ˜ r1 = Á Ë 760 ˜¯ ÁË 290 ˜¯ Ë p 1 ¯ Ë T2 ¯

r = pM/RT 3)

p n=

3

= (p

–1

p

pV {(10-10 / 760) atm} (10-3 L) = = 5.35 ¥ 10-18 mol -1 -1 RT (0.082 L atm K mol )(300 K )

N = n NA

¥

–18

¥

23

mol–1

¥

6

(2 p ) (0.85 V ) pV = T 348 K fi

T

SECTION 2

t

Kinetic-Molecular Theory of Gases

Kinetic Theory of Gases

1 N where N

È 1 m(u 2 + u 2 +  + u 2 ) ˘ μ T 1 2 N ˙ ÍÎ 2 ˚ m

u 1, u 2

2.8 Complete Chemistry—JEE Main

1 mu 2 = KT 2

(16) u2

where K u2 =

u12 + u22 +  + u N2 N

(17)

Kinetic Gas Equation pV =

1 m N u2 3

(18)

Derivation of Gaseous Laws

pV =

2 Ê1 2 N Á m u 2 ˆ˜ = NKT ¯ Ë 3 2 3

(19) T, we have

Boyle’s Law pV = constant

p, we have

Charles Law Ê 2 NK ˆ T V= Á Ë 3 p ˜¯

or

V = (constant) T V, we have

Gay-Lussac’s Law Ê 2 NK ˆ T p= Á Ë 3 V ˜¯

or

p = (constant) T

p and T, we have

Avogadro’s Law Ê 2 KT ˆ N V= Á Ë 3 p ˜¯

or

V = (constant) N pVm = RT and

Expression of Root Mean Square Speed

pVm = u2 =

1 (mN A )u 2 3

or

RT =

1 M u2 3

3RT M

or

urms =

u2 =

M = mNA

3RT M

Average Kinetic Energy of Gaseous Molecules 1 KE = mu 2 2

KE =

where k

1 Ê 3RT ˆ 1 Ê 3RT ˆ 3 Ê R ˆ 3 mÁ = T = kBT ˜= m 2 Ë M ¯ 2 ÁË mN A ˜¯ 2 ÁË N A ˜¯ 2 k =

R 8.314 J K -1 mol -1 = = 1.38 ¥ 10-23 J K -1 N A 6.022 ¥ 1023 mol -1

(21)

States of Matter 2.9

N A (KE) =

3 3 ( N A kB )T = RT 2 2

(22)

Graham’s Law of Diffusion rμ

u2

r2 p2 = r1 p1

μ

1/ M

M1 M2

Average Speed of Molecules u + u2 +  + u N u= 1 N u=

8kBT 8RT = pM pm ump, u and urms

Comparison of Different Speeds

2 RT 8RT 3RT 8 : : :: 2 : : 3 :: 1.414 : 1.595 : 1.732 :: 1 : 1.128 : 1.224 M pM M p

Distribution of Molecular Speeds

dNu M ˆ3/ 2 = 4p ÊÁ Ë 2pRT ˜¯ N Nu/N M

+ du

Mu2/2RT)*u2 du

(23) u and u

N) (dNu/du) T2 > T1 T1 and T2

u

Fig. 8

General Characteristics of Fig. 8

ump

represented as ump ump =

2 RT = M

Nu/du)/du

2kBT m

(24)

Mu2/2RT ump upon the value of M/T 2 2

molecules at temperature 2T

M at temperature 2T molecules at temperature T

2.10 Complete Chemistry—JEE Main

2 RT 8RT 3RT : : M pM M

ump u urms

2 : 8/ p : 3

ump < u < urms

MULTIPLE CHOICE QUESTIONS ON SECTION 2

1 2 (a) pV = mNu 3

1 pV = mNu 2 3

1 2 (c) pV = mNump 3

(d) pV =

(a) T

T2

(c)

T

(d) T 3/2

(c)

3 pr

(d)

3p / r

(a) 3r / p where p, r and m

2 mNu 3

3RT / m

f 1, f 2 and f 3 (a) f1 > f2 > f3

f1 > f3 > f2

(c) f2 > f3 > f1 E

(a) E (N2) > E (c) E (N2) = E

2) > E 2) = E

2

2 2) 2

–1

2

(d) f3 > f2 > f1 2

E (N2) < E 2) < E (d) E 2) > E (N2) > E not correct?

–1

(a)

–1 –1

(a) f1 > f2 > f3 f1 and f2

f1 < f2 < f3

f2 f1 and f2

2)

–1

f1, f2 and f3

(a) f1

2)

(c) f1 > f3 > f2

f1 decreases and f2 f1 and f2 decrease

–1

(d) f2 > f1 > f3

–1

,

States of Matter 2.11

not correct? (a)

8 RT / pM

8 kBT / pmf

(c)

8 p / pr

(d)

8r / pp ¥

(a)

ANSWERS

HINTS AND SOLUTIONS pV =

1 mNu 2 3

RT/M urms =

3 pVm 3RT = = M M

3p 3p = r M / Vm

3RT2 3RT1 3R = ( T2 - T1 ) M M M T2 - T1 373 K - 273 K = 2.79 K ;

473 K - 373 K = 2.44 K

573 K - 473 K = 2.19 K ;

673 K - 573 K = 2.00 K

3RT2 3RT1 =2 M M

fi

T2

t2

3RT2 2 RT1 = M M

fi

T2 = 2 T1

3T2 = 2T1

fi

fi

T2 = 4T1 = 4(300 K )

T2 = (2 / 3)(300 K ) = 200 K

3 n ÊÁ RT ˆ˜ ¯ Ë2 Amount of N2 =

2

8g = 0.286 mol 28 g mol-1 =

2

=

8g = 0.182 mol 44 g mol-1

3 3 RT = (8.314 J K -1 mol-1 )(300 K ) = 3741.3 J mol-1 2 2 3 3 -1 -1 -1 E = RT = (1.987 cal K mol )(300 K ) = 894.15 cal mol 2 2 E =

8g = 0.25 mol 32 g mol-1

¥ 32)

2.12 Complete Chemistry—JEE Main

3 3 RT = (0.082 L atm K -1 mol-1 )(300 K ) = 36.9 L atm mol-1 2 2 3 ¥ E = RT = 3741.3 J mol-1 ∫ 3741.3 kPa dm3 mol-1 2 E =

ump =

2 RT Ê 2(8.314 J K -1 mol -1 ) (300 K ) ˆ =Á ˜¯ M Ë 0.032 kg mol -1

–1

–1

1/ 2

= 394.8 m s -1

f2

f1

2 RT1 / M 2 RT2 / M

8 pVm 8RT = = pM pM

u=

T1 300 K = = 0.866 T2 400 K

=

8p 8p = p( M / Vm ) pr

Also

8kBT p mf

u=

(3 / 2) RT1 T1 300 K = = = 0.75 (3 / 2) RT2 T2 400 K

SECTION 3

Real Gases

Deviation from Ideal Behaviour

Z=

Vm, real gas Vm, ideal gas

=

Vm pVm = RT/p RT

Z Z versus p (1) Z

2 2

2

†

temperature, the pV

†

2

2

States of Matter 2.13 T4 T3

T1 > T2 > T3 > T4

T2 T1 1.0 Z

Ideal gas

0

Fig. 9

200 400 p/101.325 kPa

600

Fig. 10

pV = nRT

Van Der Waals Equation of State for a Real Gas Causes of Deviations from Ideal Behaviour

Evidence for Molecular Attractions and Molecular Volume

Derivation of Van der Waals Equation p V = nRT.

Correction for Volume b

p V = nRT, V

V = Vr – nb

(26)

b r, then Ê 4 3ˆ b = 4NA Á p r ˜ ¯ Ë3 Correction for Molecular Attraction

Decrease in velocity of a single molecule Fig. 11

2.14 Complete Chemistry—JEE Main

n V ( ) Number of molecules colliding per second with the side of the vessel μ

n/V n n -Dp μ ÊÁ ˆ˜ ÊÁ ˆ˜ ËV ¯ ËV ¯ -Dp = a

- Dp μ

i.e

n2 V2

n2 V2

(27)

where a p = pr

p = pr +

n2 a V2

(28)

Ê n2 a ˆ ÁË p + V 2 ˜¯ (V - nb) = nRT

(29) a and b

Units of van der Waals Constants a and b 6 mol–2 nb dm3 mol–1 or cm3 mol–1

n2a/V2 6 mol–2 3 mol–1 b

a

Comment on the values of a and b

b as a

For example, 2 2 2

3

mol–1; mol–1; 3 mol–1;

b b b

3

6

a a a

mol–2 mol–2 6 mol–2 6

Illustration

3

2

a 3

We have n

mol–1 3;

V

p= Now

T

a

nRT n2 a - 2 V - nb V

(2.0 mol)(8.314 kPa dm3 K -1 mol-1 )(300 K ) nRT = (12.0 dm3 ) - (2.0 mol) (0.0427 dm3 mol-1 ) V - nb n2 a (2.0 mol) 2 (364 kPa dm6 mol-2 ) = V2 (12.0 dm3 ) 2

6

6

2

mol–2

b

mol–2 and b

¥

–3

2)

dm3 mol–1

=

States of Matter 2.15

P

nRT (2.0 mol)(8.314 kPa dm6 K -1 mol -1 )(300 K ) = p= V 12.0 dm3

Applicability of van der Waals Equation Ê a ˆ ÁË p + V 2 ˜¯ (Vm - b) = RT m

b Ê a ˆ ÁË p + V 2 ˜¯ Vm = RT

or

pVm +

m

a = RT Vm

Vm

Z=1–

or

a Vm RT

(31)

Z a/VmRT

V

p

Z

p When p

b

Vm

2 a/V m

Vm

p p(Vm – b) = RT

pb RT

Z=1+

or

(32)

Z Vm

a/Vm2

b

Vm

a p(Vm – b) = RT

a/V m2

Z

p

Virial Equation of State Z=

pVm B C =1+ + + RT Vm Vm2

(33)

where B, C

Ê a ˆ ÁË p + V 2 ˜¯ (Vm - b) = RT

or

p=

m

or

Z=

RT a - 2 Vm - b Vm

pVm Vm a b ˆ Ê = = Á1 RT Vm - b Vm RT Ë Vm ˜¯

-1

-

2 ˘ È a a b Ê b ˆ = Í1 + + Á ˜ + ˙ Vm RT ÎÍ Vm Ë Vm ¯ ˙˚ Vm RT

2

a ˆ 1 Ê b ˆ Ê + + = 1 + ÁË b ˜ RT ¯ Vm ÁË Vm ˜¯

(34)

2.16 Complete Chemistry—JEE Main

Boyle Temperature B=b-

a RT T =

a Rb

Conditions for a Real Gas to Behave Ideally p=

p=

n2 a nRT - 2 V - nb V b

(36) a

nRT V

(37)

nRT nRT n2 a = - 2 V V - nb V

n2 a nRT nRT = 2 V nb V V

or

(38) a

b to the constant a

b T and V a=

a and b

RTb 1 - nb / V

and

b=

a RT + na / V

T

(39) a and b

nab V= a - RTb V

a and b

a na T= Rb RV

(41) V

na/RV

a/Rb T =

a Rb

(42)

Comparison of van der Waals Pressure of a Gas and the Corresponding Ideal Gas Pressure preal < p preal > p preal < p

a>

RTb 1 - nb / V

or

b
p

a


a RT + na / V

States of Matter 2.17

Critical Constants Andrews Isotherms p-V

T4

Fig. 12

constant pressure p V

p

vapour pressure

Tc

(4) At temperature Tc

Tc pc the temperature Tc Vc temperature Tc

pc

Characteristics of Critical Isotherm critical constants the temperature Tc Fig. 13

Continuity of State surface of discontinuity

2.18 Complete Chemistry—JEE Main

Critical Constants and van der Waals Constants RT ˆ 2 a ab Ê Vm3 - Á b + Vm + Vm =0 ˜ Ë p ¯ p p

Ê a ˆ ÁË p + V 2 ˜¯ (Vm - b) = RT m

(Vm – Vc ) 3

Vm3 - Vc3 - 3VcVm2 + 3Vc2Vm = 0

(43)

(44)

RT ˆ a ab Ê Vm3 - Á b + c ˜ Vm2 + Vm =0 Ë pc ¯ pc pc Vm pc = a/27b2

Vc = 3b;

and

Tc = 8a/27Rb

(46)

pV/RT pcVc 3 = RTc 8

(47)

Inversion Temperature

T

T =

2a Rb

(48)

MULTIPLE CHOICE QUESTIONS ON SECTION 3

Z (a) Z (c) Z factor Z, (a) a

T and p T and p

Z > 1 and b a and b Z

Z

Z (d) Z

T and p T and p and Z < 1 at lower T and p a and b a

Z < 1 and b a and b Z1

States of Matter 2.19

Z (a) Z (c) Z = 1

Z 1 or Z 3 mol–1

b 6

a

(a) a = 27 pc

b2

mol–2 and b

a = 3pcVc

(a) T > Tc and p > pc where Tc and pc a

2

3

a not correct? (a) a = 27(RTc)2/64pc (d) a = 27 RVcTc/8 6 mol–2, and b 3 mol–1 a= T) and pressure (p) are T < Tc and p > pc

T = Tc and p < pc 6

mol–1

mol–2 and b

3

(d) T > Tc and p < pc

mol–1

preal and p p p

(a) preal preal (c) preal (d) preal

p p T and V

Ê V ˆ (a) a = RTb Á Ë V + nb ˜¯

a V ˆ a = RTb ÊÁ Ë V - nb ˜¯

Ê b ˆ (c) a = RT Á Ë V + nb ˜¯

T and V (a) b =

a RT + na/V n2 a a - nRTb

b=

a RT - na/V

(c) b =

aRT V - na

a na + Rb RV

b ˆ ˜ V - nb ¯ a

(d) b =

a RT - n 2 a / V

a and b V=

n2 a a + nRTb

(c) V =

nba a - RTb

V (a) T =

Ê (d) a = RT ÁË

b

T (a) V =

b

(d) V =

nba a + RTb

a and b T=

a na Rb RV

(c) T =

a n2 a Rb RV

(d) T =

a n2 a Rb RV 2

2.20 Complete Chemistry—JEE Main

preal and p preal > p RTb (a) a < 1 - nb / V

b


a RT + na / V

(d) b =

a RT + na / V

b=

a RT - na / V

preal and p preal < p (a) a
1 Z = 1-

Z - ZB =

a Ê 1 1ˆ Vm R ÁË TB T ˜¯

T>T ,

Z

a Vm RT

pVm a pb ab =1+ + 2 RT Vm RT RT Vm RT

a pb + Vm RT RT and

Z -1 =

or

Z=

ZB = 1 -

a Vm RTB

a Ê 1 1ˆ Vm R ÁË TB T ˜¯

Z>1

Ê 4 3ˆ b = 4 NA Á pr ˜ ¯ Ë3 Ê 3b ˆ r= Á Ë 16 N A p ˜¯

1/ 3

1/ 3

È ˘ 3 ¥ 25.12 cm3 mol -1 =Í ˙ -1 23 Î16(6.022 ¥ 10 mol )(3.14) ˚

Tc =

1/ 3 1 3ˆ -8  ÊÁ cm ˜¯ = 1.36 ¥ 10 cm = 13.6 nm Ë 4 ¥ 1023

8a 8(27.0 kPa dm6 mol-2 ) = 27 Rb 27(8.314 kPa dm3 K -1mol-1 )(0.024 dm3 mol-1 ) 2

a=

3 pcVc2

= 3 pc (3b) = 27 pc b 2

2;

27( RTc )2 Ê 3RTc ˆ = a = 3 pcVc2 = 3 pc Á Ë 8 pc ˜¯ 64 pc

States of Matter 2.21

9 Ê 3RTc ˆ 2 a = 3 pcVc2 = 3 Á (Vc ) = RTcVc ˜ Ë 8Vc ¯ 8 a 364.0 kPa dm6 mol -2 = Rb (8.314 kPa dm3 K -1 mol -1 ) (0.0364 dm3 mol -1 )

TB =

∫ T £ Tc

p ≥ pc

2a 2(127.2 kPa dm6 mol -2 ) = Rb (8.314 kPa dm3 K -1 mol -1 )(0.0318 dm3 mol -1 )

T =

p=

and

n2 a nRT - 2 V - nb V

Now

nRT (2.0 mol)(8.32 kPa dm3 K -1 mol -1 )(300 K ) = V - nb (12.1 dm3 ) - (2.0 moll)(0.05 dm3 mol -1 ) n 2 a (2.0 mol) 2 (366 kPa dm6 mol -2 ) = V2 (12.1 dm3 ) 2 p

MULTIPLE CHOICE QUESTIONS ON ENTIRE UNIT

Gaseous Laws V0 ˆ Ê Ê 273.15 + t / ∞C ˆ T Vt=V ÁË ˜¯ (c) VT = Á Ë 273.15 273.15 K ˜¯

(a) Vt = a + bt

pM = rRT not correct?

(a) pV = nRT –1

(a) R (c) R

mor–1 3 –1 K mol–1

(d) Vt = V t 1 2 (d) pV = mNu 3

(c) p = rRT 3

K–1 mol–1 3 –1 K mol–1

R (d) R

2

–1

–1

–1

–1

2.22 Complete Chemistry—JEE Main

V/T versus T V/T versus T V versus p V versus p pV versus p pV versus p (a) p =

2V E 3N

p=

(a) r2 = r1(T1 p2/T2 p1) ¥ ¥

14

3V E 2N

(c) p =

r2 = r1 (T2T1/p2 p1) R –2) (mm3) K–1 mol–1 –2) (mm3) K–1 mol–1

¥

¥ –3

2N E 3V

(d) p =

(c) r2 = r1 (p2T2/p1T1)

(d) r2 = r1 (p1T1/p2T2)

–2

11

3N E 2V

3

¥ ¥

12

¥

12

–2)

(mm3) K–1 mol–1 –2) (mm3) K–1 mol–1 ¥ –12

16

–3

–3

–3

–3

¥

13

–3

–1 –1

–3

3

–1

–1

–1

–1

–3 2

3

4 –3

2

3

3

3 3

2

3

4

æÆ N2

3

2

2

4

æÆ N2

2 2

4

rA/r (a) (pA /p ) (MA/M )1/2

MA/M ) (pA/p )1/2

r (a) p

p

r p

M p

=4

(c) (pA/p ) (M /MA)1/2

r and M

= 3M (c) p

(d) (MA/M ) (p /pA)1/2

p

=6

(d) p

p

= 1/6

States of Matter 2.23

2

(a)

r1 p1 = r2 p2

2

M1 M2

r1 p1 = r2 p2

4

M2 M1

(c) R

r1 = r2

p1 M1 p2 M 2

(d)

p1 M 2 p2 M1

correct?

3

K–1 mol–1 –1 mol–1

3

¥

2

r1 = r2

2

12

K–1 mol–1 –2) (mm3) K–1 mol–1

2

2

r(A) = 2r( ) and 2M(A) = M( ), the p(A) and p r

r

(a) 1

3

(c) 1/3

31

(d) 3/1

2

2

2

pVm)p Æ 3

3

3

3

Speeds and Kinetic Energy (d) 3k T/m

(a)

3RT/M

3kBT/m

(c) 2RT/M

(a)

3RT/M

2RT/M

(c)

8RT / pM

(d)

(a)

3kBT/m

2kBT/m

(c)

8kBT/pm

(d) 3RT/M

8kBT/pM

2.24 Complete Chemistry—JEE Main

(a)

3 ¥ 8.314 ¥ 273 J mol-1 2

(c) 3 ¥

¥

1 ¥ 8.314 ¥ 273 J mol-1 2 3 8.314 ¥ 273 ˆ J (d) ÊÁ ˆ˜ ÊÁ Ë 2 ¯ Ë 6.022 ¥ 1023 ˜¯

–1

3 (a) ÊÁ RT ˆ˜ mol ¯ Ë2 (a) (urms) (c) (urms)

ˆ Ê3 (c) Á kBT ˜ mol ¯ Ë2

RT) mol

> (urms) < (urms) 4 2

> (urms) > (urms) 4

> (urms) > (urms) 3

4

3

urms) (d) (urms)

2 2

speed (ump (a) ump uav urms (c) uav ump ums

2

(a) (1/2)k T

< (urms) > (urms) 2 urms), 2

< (urms) < (urms) 4

4

< (urms) < (urms) 3 3

2 2

uav

ump urms uav uav urms ump?

2

E) 2) = E 2) = E ) < E 2 2) < E

(a) E (c) E

(d) (3 k T) mol

2

E (d) E

2 2)

k T

–1

>E ) 2 >E 2)

>E ) 2 T(N2)

3 p/r .

(c) urms μ pVm

(d) urms μ p / Vm

(c) T

(d) T

CV, m = (3/2) R

2)

< T(N2)

2)

=

7 T(N2)

States of Matter 2.25

Cp, m = (3/2) R R Cp, m and CV, m Cp, m and CV, m

R 2

Real Gases

b 4 (a) b = ÊÁ p r 3ˆ˜ N A ¯ Ë3

r

Ê 4 3ˆ (c) b = 2 Á p r ˜ N A ¯ Ë3

4 b = ÊÁ p r 3ˆ˜ ¯ Ë3

4 (d) b = 4 ÊÁ p r 3 ˆ˜ N A ¯ Ë3

Ê p + na ˆ (V - nb) = nRT ˜ ÁË V¯

na (a) ÊÁ p + 2 ˆ˜ (V - nb) = nRT Ë V ¯ Ê n2 a ˆ (V - nb) = nRT (c) Á p + V ˜¯ Ë

Ê n2 a ˆ (d) Á p + 2 ˜ (V - nb) = nRT Ë V ¯ a 3

3

mol–1

6 mol–2

Z(= pVm /RT ) pressure? (a) Z = l – a/Vm RT

Z = 1 + a/Vm RT

(c) Z = 1 + pb/RT

(d) Z = 1 – pb/RT

Z(= pVm/RT) pressure? (a) Z = 1 – a/Vm RT

Z = 1 + a/Vm RT

(c) Z = 1 + pb/RT

(d) Z = 1 – pb/RT

(a) T = aR/b

T = ab/R

(c) T = a/Rb

(d) T = aRb

(a) T = 2 a/Rb

T = a/2Rb

(c) T = 2 aR/b

(d) T = 2 aRb

(c) low T and low p

(d) low T

T and low p (a) low T

T

p

(a) low T and low p

p

T and low p

T

p

T and low p

T

T

p

T and low p

p a

2,

N2

3

6

4

mol 2

2

p

3

4

atm

2.26 Complete Chemistry—JEE Main

p + a/V 2m

(a) Vm – b

(c) RT

(d) 1/RT Vm

3

3

3 2

2

Z) versus p At low temperature and low pressure, Z (a) the constant a

b

b

a and b (a) m3

a a and b

3 mol–1

3

a? 2

3

2

b? 2

2

2

p,

2

Z versus pressure p

p,

Z versus pressure p

l = l/( 2 p s 2 N*) where s (a) l μ p/T (c) l μ T/p (a) Vc = b

Vc = 2b

(a) Tc = 8a/(27Rb)

pc = a (3b2)

l μ (p/T)2 (d) l b (c) Vc = 3b (c) pcVc = (3/8) RTc

2

N*

T and p (d) Vc = 4b not correct? (d) Vc = 3b

States of Matter 2.27

f1 and f2

c and c + dc

(a) f1 > f2 (c) f1 = f2

2

at 2T

f1 < f2 (d) f1 and f2

Miscellaneous Questions H 2 (g )

He(g )

N 2 (g )

O 2 (g )

(I)

(II)

(III)

(IV) T (=

a b

decreased to

2

and T,

2.28 Complete Chemistry—JEE Main

g

Cp, m and CV, m Z

p

ANSWERS

(d)

(c)

HINTS AND SOLUTIONS rH 2 rHe

=

M He = 2 = 1.414 M H2

pV = nRT =

m RT M

m=

pVM = RT

Ê 725 mmHg ˆ -1 ÁË 760 mmHg ¥ 1 atm˜¯ (0.75 L)(71 g mol ) (0.082 atm L K -1 mol -1 )(293 K )

= 2.11 g

States of Matter 2.29

pM (1 atm)(28 g mol -1 ) = = 1.25 g L-1 RT (0.082 L atm K -1mol -1 )(273 K )

r=

V= V + at pV = K

V = KT

and

2 N p = ÊÁ ˆ˜ E 3ËV ¯

1 2 1 pV = mNu 2 = N ÊÁ mu 2 ˆ˜ . ¯ 3 3 Ë2 pV = nRT =

m RT M

r=

–2

–2

2 cm)–2

r2 Ê p2 ˆ Ê T1 ˆ = r1 ÁË T2 ˜¯ ÁË p1 ˜¯

pM RT –4

–1 s–2

–3

6

–1 s–2

pV (1.103 ¥ 1010 dyn m -2 )(22.414 ¥ 106 mm3 mol -1 ) = T (273 K )

¥

3

Vm R=

¥

–2

¥

mol–1

6

¥

10

–2

mm3 mol–1

N=

pVN A (101.325 ¥ 10 -12 kPa )(1 dm3 )(6.022 ¥ 1023 mol -1 ) = = 7.244 ¥ 1010 RT (8.314 kPa dm3 K -1 mol -1 )(101.325 K )

r=

pM (3 ¥ 101.325 kPa ) (17 g mol -1 ) = = 2.07 g dm -3 RT (8.314 kPa dm3 K -1 mol -1 )(300 K )

14

–2)

(mm3) K–1 mol–1

Ê p ˆÊT ˆ Ê 700 Torr ˆ Ê 273 K ˆ = 1.20 g L-1 r2 = r1 Á 2 ˜ Á 1 ˜ = (1.43 g L-1 ) Á Ë 300 K ˜¯ ÁË 760 Torr ˜¯ Ë T2 ¯ Ë p1 ¯ M =

r RT (0.001 kg dm -3 )(8.314 kPa dm3 K -1 mol -1 )(300 K ) = = 0.0246 kg mol -1 = 24.6 g mol -1 p (101.325 kPa )

M =

r RT (2.52 g dm -3 )(8.314 kPa dm -3 K -1 mol -1 )(600 K ) = 124 g mol -1 = p (101.325 kPa ) 124 =4 31

M av =

r RT (0.543 g dm -3 )(8.314 kPa dm3 K -1 mol -1 )(300 K ) = 13.37 g mol -1 = p (101.325 kPa )

If x

x M + (1 – x)M

–1

x(4) + (1 – x

2

x 2

2.68 ¥ 100 = 20.2 2.68 + 10.56 300 K ÊT ˆÊ p ˆ ˆ Ê 760 Torr ˆ Ê = 162.9 cm3 V2 = V1 Á 2 ˜ Á 1 ˜ = (136.5 cm3 ) Á ˜ Ë (725 - 25) Torr ¯ ÁË 273 K ˜¯ Ë p2 ¯ Ë T1 ¯ x

2

4

x + 2(1 – x)] mol = (2 + x) mol Now

pV = nRT

2.30 Complete Chemistry—JEE Main

p1 n1T1 = ; p2 n2T2 2

Ê 0.5 atm ˆ = Ê 1 ˆ Ê 300 K ˆ ˜ ˜ Á ÁË 4.5 atm ¯ Ë 2 + x ¯ ÁË 1200 K ˜¯

or

x=

4.5 - 2 = 0.25 0.50 ¥ 4

4

r μ1/ M

r μpA

Also,

where pA rA pA Ê mB ˆ = rB pB ÁË M A ˜¯

1/ 2

pV = nRT = (m/M)RT fi pM = (m/V) RT = rRT r M Y Ê rX ˆ Ê M Y ˆ pX = X = =2¥3=6 pY M X rY ÁË rY ˜¯ ÁË M X ˜¯ n –1

(a) n (c) n

–1

= 1 mol

r1 = (M2/M1)1/2r2

–1

–1)1/2r

–1)/4

n1 = n2 R R R

3

–1

R ¥

14

2

–1

= (1/4)r2 fi

= (3/16) mol and –1 mol–1 –1 mol–1 3 atm K–1 mol–1

mol–1

r1 Ê 1 ˆ n /t 1 n 1 =Á ˜ fi 1 = fi 1 = ¯ Ë r2 n2 / t 4 n2 4 4 –1

m1 3 3

atm

–1 mol–1

K–1

–1 mol–1 –2)

mol–1 –2)

3

mm)3 K–1 mol–1

(mm3) K–1 mol–1

r M B rA M B pA = A = = (2) (2) = 4 rB M A pB M A rB

p = rRT/M

r1 T2 Êr ˆ Ê r ˆ (300 K ) = 600 K = fi T2 = Á 1 ˜ T1 = Á Ë 0.5 r ˜¯ Ë r2 ¯ r2 T1

p = rRT/M

urms = 3 p / r ,

–1

n (d) n

we get

u1 = u2

r2 1 = r1 3 1/ 3

r1 t2 = = r2 t1

M2 M1 t = 32 / 2 (5 s) = 20 s t = 44 / 2 (5 s) = 5 22 s

t = 4 / 2 (5 s) = 5 2 s t = 28 / 2 (5 s) = 5 14 s

States of Matter 2.31 3

pVm)pÆ

3

3

RT Ê 3 RT ˆ ˜¯ = (3 mol) RT ÁË 2 urms = 3RT / M ,

ump = 2 RT / M ,

M H 2 < M CH 4 < M NH3 < M CO2

urms μ 1/ M .

uav = 8 RT / pM ,

urms = 3RT / M

ump : uav : urms :: 2 : 8 / p : 3 :: 1.414 : 1.595 : 1.732 :: 1 : 1.128 : 1.224 urms = 3RT/M ; urms

p or V at

constant temperature as pV

1/ 2

urms =

3RT È 3(8.314 J K -1 mol -1 ) (480 K ) ˘ =Í ˙ M 48 ¥ 10 -3 kg mol -1 Î ˚

2 RT 2 R (300 K ) =2 . M M urms =

= 499.4 m s -1 ª 500 m s -1

T

3RT 3p 3 ¥ 100 ¥ 103 Pa = = = 1732 m s -1 r M 0.1 kg m -3 È 3 (8.314 J mol-1 K -1 ) (300 K ) ˘ ÍÎ 2 ˙˚ uav = 8 RT / pM , we have (uav )927 º C È (927 + 273) K ˘ Ê 1200 ˆ 1 / 2 = =Á ˜ =2 (uav ) 27 º C ÍÎ (27 + 273) K ˙˚ Ë 300 ¯

fi

(uav )927ºC = 2(0.3 m s -1 ) = 0.6 m s -1 urms = 3RT / M we have

p/r

urms μ RT =

pVm

urms = 3RT / M , we have

T (327 + 273) K = 4 16

fi

T (H 2 ) T (N2 ) = ( 7 )2 2 28

(urms )H 2 = 7 (urms ) N 2

∫

T or

T

2)

= T(N2)/2

CV, m = (3/2) RT and Cp, m

RT urms = 3RT / M ,

we have 2

u1 T M = 1 2 u2 T2 M1

64 1 2 ÊM ˆÊu ˆ T2 = Á 2 ˜ Á 2 ˜ T1 = ÊÁ ˆ˜ ÊÁ ˆ˜ (300 K ) = 1200 K Ë 4 ¯ Ë 2¯ Ë M1 ¯ Ë u1 ¯

2.32 Complete Chemistry—JEE Main

b a/V 2m fi

p(Vm – b) = RT

2 ) (V – b) = RT p + a/V m m

fi

pVm – pb = RT

b

Z=

p

pVm pb =1+ RT RT

Vm

a Vm, real,/V

Vm, real > V

Z =1-

3

V

a Vm RT

Z

Z = 1 + pb/RT. b

mol–1

Z

Vm

Ê pVm a ˆ a ÁË p + V 2 ˜¯ (Vm ) = RT or pVm = RT - a / Vm or Z = RT = 1 - V RT m m pV = nRT = l=

N N pN A p RT fi N * = = = NA V RT kT

kT

l μ T/p

2ps 2 p

c and c + dc f= M

2)/2T

=M

dN 1 Ê M ˆ = 4p Á Ë 2 p RT ˜¯ N dc 2)/T,

3/ 2

exp( - Mc 2 / RT )c 2

hence, f

N = nNA = (m/M)NA fi N μ (1/M crms = 3RT / M

fi crms μ 1 / M , fi

n(3/2) RT = (m/M) (3/2) RT a

μ1/M,

b

2

Tc and pc Tc and pc dN 1 M ˆ 3 / 2 - Mu 2 / 2 RT 2 = 4p ÊÁ e u Ë 2p RT ˜¯ N du ump = 2 RT / M , T = a/Rb

dN 1 M ˆ 3 / 2 -1 = 4p ÊÁ e (2 RT / M ) μ M1/2 Ë 2pRT ˜¯ N du T μ a/b

a

r μ1/ M a T = 2a/Rb h= least value of m1/2/b1/3

(mkB t )1 / 2 p3 / 2s 2

, where s =2r μ b1/3

h μ m1/2/b1/3

States of Matter 2.33

–1) =

RT RT

–1

–1

3R ) = 3R

T –1

¥ 32/28) K = 343 K

T

(crms )H 2 = 3R (300 K ) /(0.002 kg mol -1 ) and (cav ) He = 8R (300 K ) /(p ¥ 0.004 kg mol -1 ) (crms )H 2 (caV ) He

=

(3 / 2) >1 8 /(4p) g

g

Tc b a

Z = 1 + pb/ Z = 1 – a/Vm RT

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE-MAIN

(c) 313/293 b

313 / 293

(d) 2

[2004]

[2004] not

[2005]

(a) 2 3

–4

1 273 ¥ 3 298

(c)

dm3

(d)

1 2

[2007]

3

¥

–3

a (a) a b 2> a 2 6 (c) a and b 2< a and b

–1

R

2

¥ b

1 3

2 2

6

¥

–3

2

b

2

6

[2011]

[2011]

2.34 Complete Chemistry—JEE Main

(a) 1 + RT/pb

pb/RT

(d) 1 – pb/RT

c (a) c*: c : c = 1.225 : 1.128 : 1 (c) c*: c : c = 1 : 1.128 : 1.225

[2012]

c

c

c*: c : c = 1.128 : 1.225 : 1 (d) c*: c : c = 1 : 1.225 : 1.128

[2013]

Z (a) Z = 1 – a/VmRT

Z = 1 – pb/RT

(c) Z = 1 + pb/RT

(d) Z = 1 + RT/pb [2014]

[2003]

p=

nRT n 2 - a ÊÁ ˆ˜ ËV ¯ V - nb p = nRT/V,

when

[2014]

[2014, online]

[2014, online] 3 3

2

2

[2014, online] not

[2015, online]

[2015, online] p and temperature T1 are

V pf

T2 T1 pi, V

T1 pi, V

T1 pf, V

T2 pf, V

States of Matter 2.35

Ê T1 ˆ (a) 2p Á Ë T1 + T2 ˜¯

Ê T2 ˆ p Á Ë T1 + T2 ˜¯

Ê TT ˆ (c) 2p Á 1 2 ˜ Ë T1 + T2 ¯

Ê TT ˆ (d) p Á 1 2 ˜ Ë T1 + T2 ¯ u

2

u

(a) 2u (a) 1 + pb/RT

[2016]

pb/RT

(c) 4u

(d) u/2

(c) 1 – pb/RT

(d) 1 – b/VRT [2016, online]

[2016, online]

ANSWERS

HINTS AND SOLUTIONS k (KE) 40 º C (KE) 20 º C

=

(273 + 40) K 313 = (273 + 20) K 293 b

–1

nCH 4 =

m 16 g mol -1

and

nO2 =

m 32 g mol-1

xO2 =

–1

2

nO2 nCH 4 + nO

= 2

m

(m / 32 g) 1 / 32 1 = = (m / 16 g) + (m / 32 g) 3 / 32 3

3

n=

pV (3170 Pa )(10 -3 m3 ) = RT (8.314 J K -1 mol -1 )(300 K )

¥

–3

mol

a a a

a

2 2

2)

6

b b

crms = 3RT / M ;

2

6

cav = 8RT / pM ;

2

cmp = 2 RT / M

Ê a ˆ ÁË p + V 2 ˜¯ (Vm - b ) = RT m

a / Vm2 0.732, hence, CsCl forms body-centred cubic unit cell. 10. For closest-packed rock salt structure where cations occupy octahedral voids, rc/ra = 0.414. Hence, ra = rc/0.414 = 72 pm/0.414 = 173.9 pm. 11. If rc = 100 pm, then ra = 100 pm/0.414 = 241.5 pm. Edge length of cube, a = 2 (rc+ ra) = 683 pm a3 = (683 pm)3 = 318.6 ¥ 106 pm3 = (318.6 ¥ 106) (10–10 cm)3 = 318.6 ¥ 10–24 cm3 = 3.19 ¥ 10–22 cm3. 12. Fe2O3 has corundum unit cell. 13. The coordination number of Cs+ an Cl– are 8 and 8, respectively, as CsCl occurs in body-centre cubic unit cell. 14. CsCl forms body-centred cubic unit cell. Hence, 2(rc + ra) = 3a This gives

a=

2 (rc + ra ) 3

=

2(181 + 167) 3

pm = 401.8 pm

States of Matter 2.73

15. For MgO, rc/ra = 86 pm/126 pm = 0.682. It is more than 0.414 but less than 0.732. Thus MgO occurs as rock-salt structure. The edge length unit cell is 2(ra + rc ) = 2(86 +126) pm = 424 pm 16. In zinc-blende closest unit cell, cations occupy tetrahedral voids for which rc/ra = 0.225. Hence, ra = rc /0.225 = 60 pm/0.225 = 266.67 pm. Anions occupy face centred cubic positions. Hence, 4ra = 2 a or a = 2 2 ra = (2 2 ) (2.667 pm) = 754.3 pm. 17. ra = rc /0.225 = 69.4 pm/0.225 =308.4 pm; r=

a = 2 2 ra = 2 2 (308.4 pm) = 872.4 pm

NM 4(100 g mol-1 ) 400 g cm -3 = 1.0 g cm -3 = = -1 -10 3 23 3 399.84 NA a (6.022 ¥ 10 mol )(872.4 ¥ 10 cm)

Anions form face-centred cubic unit cell.

Hence

Hence, ra = rc /0.225 = 90 pm/0.225 = 400 pm a = 2 2 ra = 2 2 (400 pm) =1131.4 pm =1.13 nm a = 2 2 ra = 2 2 (353.56 pm) =1000 pm

19. ra = rc /0.225 = (79.55 pm)/0.225 =353.56 pm; There are 4 molecules of M2 NM 4(200 g mol-1 ) = Hence r= =1.33 g mol–1 N A a3 (6.022 ¥ 1023 mol-1 )(1000 ¥ 10-10 cm)3 20. A2+ occupy tetrahedral voids and B3+ occupy octahedral voids and r B3+ = 0.414 ¥ 126 pm = 52.2 pm Hence rA2+ = 0.225 ¥ 126 pm = 28.4 pm 21. We have 2 (r++ r–) = 3 (a/2).

Hence a =

4(r+ + r- ) 3

=

4 (59 pm + 126 pm ) 3

= 427.24 pm

22. There are 4 molecules of BeO in a unit cell, Hence r=

NM 4(9 + 16) g mol-1 ) = = 2.13 g cm–3 N A a3 (6.022 ¥ 1023 mol-1 )(427.24 ¥ 10-10 cm)3

23. Volume occupied by ions is 4 4 4 u = 4 ÊÁ p ra3 ˆ˜ + 4 ÊÁ p rb3 ˆ˜ = 4 ÊÁ ˆ˜ (3.14) [ (126 pm)3 + (59 pm)3] = 3.69 ¥ 107 pm3 ¯ Ë3 ¯ Ë3 Ë 3¯ Volume of unit cell is

V = a3 = (427.24 pm)3 = 7.80 ¥ 107 pm3

Fraction of volume occupied is

f=

u 3.69 ¥ 107 pm3 = = 0.47 V 7.80 ¥ 107 pm3

24. In spinel structure, 1/8 of tetrahedral voids are occupied by A2+ and 1/2 of octahedral voids are occupied by B3+. 25. Zinc blende unit cell, carbon atoms will occupy face-centred cubic unit cell with half of tetrahedral voids occupied by carbon atoms. Carbon atoms form face-centred cubic unit cell = 4. Carbon atoms from half of tetrahedral voids = 4 Total number of carbon atoms = 8 26. The fraction of volume occupied by carbon in diamond unit cell is 3p / 16 which is 0.34. See text. 27. We have r=

4( M MX / N A ) [4(ra + rc ) / 3 ]3

=

3 3 ( M MX / N A ) (3 3 )[(120 g mol-1 ) /(6.022 ¥ 1023 mol-1 )] = 16(ra + rc )3 16 [(120 + 80) ¥ 10-10 cm]3

(3 3 )(20 ¥ 10-23 ) g = 8.11 g cm–3 16 ¥ 8.0 ¥ 10-24 cm3 28. The distance between corner and tetrahedral void is 3a / 4 . Hence 2rC = 3a /14. This gives a = 8rC / 3 = 8(77 pm)/1.732 = 355.7 pm =

2.74 Complete Chemistry—JEE Main –10 –10 3 a/4 . Hence, a = 8rM / 3 = 8 ¥ 86.6 ¥10 cm/1.732 = 400 ¥ 10 cm u = a3 = (4.00 ¥10–8 cm)3 = 6.4 ¥ 10–23 cm3

29. 2rM =

8 atoms ˆ (1 cm3 ) Ê 8 atoms ˆ V = Ê N = ÁË = 1.25 ¥ 1023 atoms ˜ ÁË -23 3˜ u ¯ 6.4 ¥ 10 cm ¯ 30. The edge length of unit cell is The volume of unit cell is

a = 2(ra + rc) = 2 (180 + 120) pm = 600 pm = 600 ¥ 10–10 cm u = (600 ¥ 10–10 cm)3 = 2.16 ¥ 10–22 cm3

The number of molecules in V = 1 cm3 is

SECTION 4

Ê 4 molecules ˆ (1 cm)3 ˜ N = ÁË = 1.85 ¥ 1022 molecules 2.16 ¥ 10-22 cm3 ¯

Imperfection in Solids

The term ‘imperfection in solids’ implies the departure from the perfect periodic arrangement of atoms, molecules or ions in the structure of crystalline substance. The perfect arrangement of the species is expected to exist only at the lowest temperature of 0 K. At higher temperatures, the crystalline substances exhibit departure from the ideal mentioned in the following.

1. STOICHIOMETRIC DEFECT This defect does not change the stoichiometry of solids. This includes (i) Vacancy defect—presence of vacancies in the lattice points. (ii) Interstitial defect—dislocation of species from lattice point to interstitial site. In ionic solids, the above two defects do occur with the constraints of electrical neutrality of ionic solids. Schottky Defect This is vacancy defect. To maintain electrical neutrality, both cations and anions are missing from lattice points (Fig. 44). The schottky defects are more common in ionic compounds with high coordination numbers and with similar sizes of cations and anions. The absence of ions leads to decrease in the density of solid as compared to perfect crystals. Examples are NaCl and CsCl Fig. 44 Schottky defect Frenkel Defect This is interstitial defect. In a solid involving large differences in sizes of cations and anions, the smaller-sized ions shift from the regular position to the interstitial positions. (Fig. 45) The density of crystals exhibiting Frenkel defect remains unchanged. Examples are ZnS, AgCl, AgBr and AgI. Alkali halides do not show Frenkel effect as the ions cannot occupy interstitial sites. 2. IMPURITY DEFECTS In impurity defect, some of the lattice points are occupied by ions other than the expected ion. For example, if molten NaCl is crystallized in the presence of SrCl2, some of Na+ lattice sites are occupied by Sr2+ ions. To maintain electrical neutrality, some of the lattice points are unoccupied (Fig. 46). The creation of cationic vacancies imparts electrical conductivity to the ionic crystals.

Fig. 45

Frenkel defect

3. NON-STOICHIOMETRIC DEFECTS quantity as compared to the expected stoichiometric compound. Two categories of non-

Fig. 46

States of Matter 2.75

The preparation of FeO always leads to non-stoichiometric compound Feo.95O. This compound contains both Fe2+ and Fe3+ ions. To maintains electrical neutrality, Fe2+ ions from three lattice points are replaced by two Fe3+ ions at two sites leaving third site vacant (Fig. 47).

Illustration 1

The composition of a sample of wustite is Fe0.93O. What percentage of iron exists as Fe(III).

Fig. 47

For every 1 mol of oxygen atoms, 0.93 mol of iron atoms is present. If x is the amount of Fe(III) present, then Amount of positive charge = x(+3) + (0.93 mol – x)(+2) = 1.86 mol + x Amount of negative charge = (1 mol) (2) To maintain electrical neutrality, we have 1.86 mol + x = (1 mol)2 This gives x = 0.14 mol Percentage of Fe(III) = (0.14 mol/0.93 mol) ¥100 = 15.1%

Illustration 2

A sample of wustite FexO contains 25 mol% of Fe(III). What is the value of x and simple formula

of FexO. Amount of Fe(III) = 0.25x Amount of Fe(II) = 0.75x To maintain electrical neutrality, we have (0.25x)(3) + (0.75x)(2) = 2 This gives x = 2/2.25 = 8/9 Hence, the compound is Fe8/9O.

Metal Exess Defect due to Anionic Vacancies (i) Heating of solid NaCl in the atmosphere of sodium metal vapours causes the deposition of sodium on the surface of solid. The Cl– ions from the bulk of solid moves to the surface and combines with sodium to give sodium chloride. The electrons set free here move to the bulk and occupy anionic vacancies (Fig. 48). Fig. 48 These trapped electrons in the anionic vacancies are referred to as F-centres and is responsible for the interesting properties to the crystal such as photoconductors, n-type semiconductor and characteristic colours. For example, the excess of K+ in KCl imparts violet colour and Li+ in LiCl imparts pink colour. (ii) The strong heating of ZnO leads to the formation of a yellow-coloured non-stiochiometric compound. Zinc oxide loses oxygen reversibly on heating ( ZnO Æ Zn + 12 O 2 + 2e - ) . The excess of Zn2+ ions in crystal move to interstitial sites and electrons occupy the neighbouring interstitial positions. Because of free electrons, the electrical conductivity is enhanced.

Electrical Properties Conductors These are good conductors of electricity. Their conductivity ranges between 104 S m–1 and 107 S m–1 (where S stands for Siemens which is ohm–1). Insulators These are poor conductor of electricity. Their conductivity ranges between 10–20 S m–1 and 10–10 S m–1 Semiconductors These lie in between conductors and insulator and have conductivity ranges from 10–6 S m–1 to 4 10 S m –1.

Band Theory These conductivity of solids depends on the valence electrons in their constituent atoms. The conductivity ability has been explained on the basis of band theory. The later involves the formation of molecular orbitals from the valence atomic orbitals of atoms. Because of very large number of atoms, the energies of one set of molecular orbitals are very close to each and constitute what is called band of molecular orbitals.

2.76 Complete Chemistry—JEE Main

Metals contain one, two or three electrons in the outermost shell of their atoms. Their atomic orbitals form partially-

Nonmetals contain 5,6 or 7 electrons in the outermost shell of their atoms. Their atomic orbitals form atmost fully-

Elements of Group 14 of periodic table having atoms with 4 valence electrons constitute semiconductors. In this case, the gap between lower-energy band and conduction band is small (Fig. 49 c) and electrons can jump from lowerenergy band to conduction band and their number increases with increase in temperature. This causes an increase in conductivity of semiconductors with increase in temperature.

Fig. 49

Enhancement of Conductivity of Semiconductors Silicon and germanium are the examples of semicounductors. Their conductivity is too low for any practical use. However their conductivity can be increased by introducing a very small quantity of atoms having 3 or 5 electrons in the outermost shell. This process is called dopping. n-Type Semiconductors Si and Ge belong to Group 14 of periodic table. When dopping is carried out with P or As (which belong to Group 15 with 5 valence electrons), some of the Si or Ge is replaced by P or As. In the latter atom, available for conduction causing an increase in the conductivity of dopped semiconductor. In n-type semiconductor, n stands for negatively charged free electron. p-Type Semiconductor If dopping is done with the atoms containing 3 valence electrons (i.e. elements from Group 13 of periodic table such as boron, gallium or indium), the resultant semiconductor is known as p-type semiconductor. Here all the three electrons of dopped atom are involved in the bonding with neighbouring atoms and a hole exist in is generated. The latter moves in the opposite direction of the movement of electrons. This movement of electrons or hole enhances the conductivity of semiconductor.

Magnetic Properties of Solids

This alignment produces a net magnetization on the microscopic level. The orbital motion of electrons around nucleus also generates tiny magnets but their contribution to total magnetization The magnetic moment due to the spinning of electrons is given by m = n(n + 2) m B

States of Matter 2.77

where n is the number of unpaired electrons and mB is the base unit of magnetic moment and is known as Bohr magneton. Its value is 9.274 ¥ 10–24 J T–1. Paramagnetic Substnace Examples are O2, Cu2+ and Fe3+.

Diamagnetic Substance Examples are N2, H2O, and NaCl. These substance are repelled by the external

Ferromagnetic Substance The substances which can be permanently magnetized by the substance. Examples are Fe, Co, Ni and CrO2. These substances are strongly attracted towards magnetic

(Fig. 49a) and this alignment remains unaffected Antiferromagnetic Substance

Fig. 2.49

In these substances, half of the magnetic moment vectors align in one direction and

no magnetic properties (Fig. 49b). Example is MnO. Ferrimagnetic Substance In these substances, the alignment of magnetic moment vector in opposite direction is uneven resulting into a net magnetic moment (Fig. 49c). Examples are Fe3O4 and ferrites with general formula MFe2O4 where M is Mg2+, Cu2+, Zn2+, etc.

Dielectric Properties

generates an induced dipole moment in the constituent of the substance. In general, the induce dipole moment is directly pind μ E or pind = a E where a is known as polarizability of the constituent molecules. The induced dipole moments may align in the direction Ferroelectric substance The induced dipole moments are aligned to induce a net dipole moment. Examples are barium titnate (BaTiO3), sodium potassium tartrate (Rochelle salt) and potassium dihydrogen phosphate (KH2PO4) Antiferroelectric substance The induced dipoles are alternatively aligned in the opposite directions resulting into no dipole moment. Example is lead zirconate (PbZrO3). Piezoelectricity The crystal in which alignment of induced dipoles generates net dipole moment show the phenomenon of piezoelectricity. There occurs emission of electrons when a stress is applied to the crystal. Pyroelectricity: Some of the polar crystal emit electrons when these are heated. Superconductivity: A substance offering no resistance to the passage of electrons is said to be superconductor. Most metals become superconducting at temperature of about 2-5 K. At this temperature, the substance becomes diamagnetic. Examples are Ba0.7K0.3BiO3(30 K) and YBa2Cu3O7(90K).

2.78 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS ON SECTION 4 Identify the correct choice in the following questions. 1. Schottky defect is (a) vacancy defect (b) interstitial defect (c) impurity defect 2. Frenkel defect is (a) vacancy defect (b) interstitial defect (c) impurity defect 3. Iron oxide preparation always leads to nonstoichiometric compound. It involves

(d) stoichiometric defect (d) stoichiometric defect

4. In the compound Fe0.85O, the percentage of iron existing as Fe(III) is (a) 25.3% (b) 30.3% (c) 35.3% (d) 40.3% 5. A sample of wustite FexO contanis 15 mol% of Fe(III). The sample has the molecular formula of (a) Fe0.93O (b) Fe0.90O (c) Fe0.86O (d) Fe0.82O 6. Heating of sodium chloride in the presence of sodium vapour creates F-centres which represent (a) cationic vacancies (b) anionic vacancies (c) octahedral voids (d) tetrahedral voids 6 S m–1 7. A substance has conductivity of 10 (a) conductor (b) insulator (c) semiconductor (d) superconductor (a) conductor (b) insulator (c) semiconductor 9. The magnetic moment of an atom or molecule or ion is given by the expression (b) n(n + 2) mB (c) n mB (a) n(n + 1) mB where n is number of unpaired electrons and mB is Bohr magneton. 10. The value of Bohr magneton is (b) 9.274 ¥ 10–22 J T–1 (c) 9.274 ¥ 10–23 J T–1 (a) 9.274 ¥ 10–21 J T–1 11. Which of the following material can be permanently magnetized? (a) Paramagnetic substance (b) Diamagnetic substance (c) Ferromagnetic substance

(d) superconductor (d) (n+1) mB

(d) 9.274 ¥ 10–24 J T–1

12. Which of the following substance does not have magnetic properties? (a) Paramagnetic substance (b) Ferromagnetic substance (c) Antiferromagnetic substance (d) Ferrimagnetic substance 13. Which of the following facts regarding the effect of increasing temperature on the conductivity of a substance is correct? (a) Increases for metals and decreases for semiconductors (b) Decreases for metals and increases of semiconductors (c) Increases for both metals and semiconductors (d) Decreases for both metals and semiconductors 14. The amount of cationic vacancies per mole of Fe0.88O is (a) 0.10 mol (b) 0.11 mol (c) 0.12 mol (d) 0.14 mol 15. The amount of cationic vacancies per mole of a nonstoichiometric iron oxide is 0.14 mol. The formula of the compound is (b) Fe0.82O (c) Fe0.86O (d) Fe0.93O (a) Fe0.75O

States of Matter 2.79

ANSWERS 1. (b) 7. (a) 13. (b)

2. (b) 8. (c) 14. (c)

3. (a) 9. (b) 15. (c)

4. (c) 10. (d)

5. (a) 11. (c)

6. (b) 12. (c)

HINTS AND SOLUTIONS 1. Schottky defect is vacancy defect. 2. Frenkel defect is interstitial defect. 4. For every 1 mol of oxygen atoms, 0.85 mol of iron atoms is present. If x is the amount of Fe(III) present, then the electrical neutrality requires that x(+3) + (0.85 mol – x) (2) = (1 mol) 2 This gives

x = 0.3 mol

Percentage of

Fe(III) = (0.3 mol/0.85 mol) ¥ 100 = 35.3% Amount of positive charge = Amount of negative charge

5. We will have

(0.15 mol) x (+3) + (0.85 mol) x (+2) = (1 mol) (2) 2 This gives x = = 0.93. The compound is Fe0.93O 2.15 6. F–centres are anionic vacancies. 7. Conductors has conductivity in the range 104 – 107 S m–1 8. Conductivity of semiconductor increases with increase in temperature. n(n + 2) mB

9. The expression of magnetic moment is 10. mB = 9.274 ¥ 10–24 J T–1

12. Antiferromagnetic substance do not have magnetic properties. 13. For metals, conductivity decreases while for semiconductors, it increases. 14. In non-stoichiometric iron oxide, 3 mol of Fe2+ ions are replaced by 2 mol of Fe3+ ions and 1 mol of cationic vacancies is created. If x is the amount of Fe(III) present in the given compound, then x(+3) + (0.88 mol –x) (+2) = (1 mol) (2) This gives

x = 0.24 mol. (III)

(II)

Hence, the given compound is Fe0.24 Fe0.64 O . The cationic vacancies will be 0.24 mol/2 = 0.12 mol 15. 0.14 mol of cationic vacancy implies the presence of 0.28 mol of Fe(III). If x is the amount of Fe(II), then electrical neutrality requires that (0.28 mol) (+3) + x (2) = (1 mol) (2) This gives

(III)

(II)

x = 0.58 mol of Fe(II). Hence the formula is Fe0.28 Fe0.58 O , i.e. Fe0.86O

2.80 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS ON ENTIRE UNIT Identify the correct choice in the following questions. Crystal Systems 1. In the primitive unit cell, the points are present at the (a) corners of the unit cell (b) centre of the unit cell (c) centre of each face of the unit cell (d) one set of faces of the unit cell 2. Which of the following statements is not correct? (a) There are seven crystal systems (b) There are fourteen Bravais lattices (c) There are one primitive and two nonprimitive cubic unit cells (d) In the body-centred unit cell, the points besides present at the comers are also present at the centre of all faces 3. Molecular solids are usually (a) good electrical conductors (b) quite hard (c) quite brittle (d) volatile 4. The number of nearest neighbours around each particle in a face-centred cubic lattice is (a) 4 (b) 6 (c) 8 (d) 12 5. Bragg equation for the scattering of X-rays by crystals is (a) nl = d sin q (b) nd= l sin q (c) nl = 2d sin q (d) nd = 2l sin q 6. Which of the following dimensions represents a cubic unit cell? (a) a = b = c; a = b = g = 90° (b) a = b π c; a = b = g = 90° (c) a = b = c; a = b = g π 90° (d) a = b π c; a = b = 90°, g = 120° 7. Which of the following dimensions represents a tetragonal unit cell? (a) a = b = c; a = b = g = 90° (b) a = b π c; a = b = g = 90° (d) a = b π c; a = b = 90°, g = 120° (c) a = b = c; a = b = g π 90° 8. Which of the following dimensions represents a trigonal unit cell? (a) a = b = c; a = b = g = 90° (b) a = b π c; a = b = g = 90° (c) a = b = c; a = b = g π 90° (d) a = b π c; a = b = 90°, g = 120° 9. Which of the following dimensions represents a hexagonal unit cell? (a) a = b = c; a = b = g = 90° (b) a = b π c; a = b = g = 90° (c) a = b = c; a = b = g π 90° (d) a = b π c; a = b = 90°, g = 120° 10. Which of the following dimensions represents a triclinic unit cell? (a) a = b = c; a = b = g = 90° (b) a = b π c; a = b = g = 90° (c) a π b π c; a = b = 90° π g (d) a π b π c; a π b π g π 90° 11. Which of the following dimensions represents a monoclinic unit cell? (a) a = b = c; a = b = g = 90° (b) a = b π c; a = b = g = 90° (d) a π b π c; a π b π g π 90° (c) a π b π c; a = b = 90° π g 12. Which of the following dimensions represents a orthorhomic unit cell? (a) a π b π c; a = b = g = 90° (b) a π b π c; a = b = 90° π g (c) a π b π c; a π b π g = 90° (d) a = b π c; a = b = 90° π g 13. Which of the following units can be used to built the structure of diamond? (a) Tetrahedral (b) Octahedral (c) Hexagonal (d) Cubic 14. Which of the following units can be used to built the structure of graphite? (a) Tetrahedral (b) Octahedral (c) Hexagonal (d) Cubic

States of Matter 2.81

15. The unit cell present in the crystal lattice of diamond is (a) face-centred cube (b) tetragonal (c) hexagonal (d) trigonal 16. Which of the following statements is not correct? (a) The molecular crystals are soft and possess comparatively low melting points. (b) The ionic crystals have high melting points. (c) Covalent crystals are poor conductor of heat and electricity. (d) Metallic crystals are not ductile and malleable. 17. In a molecular crystal the forces holding together the atoms within a given molecule are (a) covalent bonds (b) ionic bonds (c) metallic bonds (d) van der Waals forces 18. Ionic solids tend to be (a) good electrical conductors (b) soft (c) volatile (d) brittle 19. Which of the following Bravais lattices does not exist in the crystal lattice having cubic unit cell? (a) Primitive (b) Body-centred (c) Face-centred (d) End-centred 20. The number of Bravais lattices in the crystal lattice having orthorhombic unit cell is (a) 1 (b) 2 (c) 3 (d) 4 21. The number of Bravais lattices in the crystal lattice having tetragonal unit cell is (a) 1 (b) 2 (c) 3 (d) 4 22. Which of the following lattices having the indicated unit cell has only primitive Bravais lattice? (a) Monoclinic (b) Trigonal (c) Cubic (d) Tetragonal 23. Which of the following lattices does not have only primitive Bravais lattice? (a) Lattice with unit cell triclinic (b) Lattice with unit cell hexagonal (c) Lattice with unit cell monoclinic (d) Lattice with unit cell trigonal 24. The number of nearest neighbours around each particle in a primitive cubic lattice is (a) 4 (b) 6 (c) 8 (d) 12 25. Which of the following groups of crystal systems has the crystallographic axes a = b = c? (a) Trigonal, Hexagonal (b) Trigonal, Cubic (c) Trigonal, Hexagonal, Cubic (d) Orthorhombic, Cubic 26. Which of the following groups of crystal systems has the crystallographic axes a = b π c? (a) Hexagonal, Tetragonal (b) Hexagonal, Monoclinic (c) Hexagonal, Orthorhombic (d) Cubic, Orthorhombic 27. Which of the following groups of crystal systems has the crystallographic axes a π b π c? (a) Trigonal, Triclinic (b) Tetragonal, Monoclinic (c) Trigonal, Tetragonal (d) Triclinic, Monoclinic 28. Which of the following groups of crystal systems has the crystallographic anlges a = b = g = 90°? (a) Tetragonal, Cubic, Orthorhombic (b) Tetragonal, Cubic, Monoclinic (c) Cubic, Orthorhombic, Jrigonal (d) Hexagonal, Cubic, Orthorhombic 29. Which of the following groups of crystal systems has the crystallographic angles a = b = 90° π g ? (a) Trigonal, Tetragonal (b) Hexagonal, Monoclinic (c) Tetragonal, Hexagonal (d) Triclinic, Monoclinic 30. Which of the following crystal systems has crystallographic angles a π b π g ? (a) Monoclinic (b) Triclinic (c) Trigonal (d) Tetragonal 31. Which of the following crystal systems has the largest number of Bravais lattices? (a) Cubic (b) Tetragonal (c) Orthorhomic (d) Hexagonal 32. Which of the following groups of crystal systems has only two Bravais lattices? (a) Tetragonal, Monoclinic (b) Tetragonal, Cubic (c) Tetragonal, Orthorhombic (d) Orthorhombic, Hexagonal

2.82 Complete Chemistry—JEE Main

33. Which of the following crystal systems has three Bravais lattices? (a) Tetragonal (b) Orthorhombic (c) Cubic (d) Monoclinic 34. The number of lattices having only one Bravais lattice is (a) one (b) two (c) three (d) four Of the three cubic Bravais lattices, primitive (P), body-centred (I) and face-centred (F), of elements, answer the following questions. 35. The number of atoms per unit cell are (a) P-one; I-two; F-four (b) P-one; I-four; F-two (c) P-two; I-one; F-four (d) P-two; I-four; one

Packing in Crystalline Solids 36. The unit cell present in ABC ABC... packing of atoms is (a) hexagonal (b) tetragonal (c) face-centred cubic (d) primitive cube 37. The unit cell present in ABAB... packing of atoms is (a) hexagonal (b) tetragonal (c) face-centred cube (d) primitive cube 38. The factor which makes a solid to have a low density is (a) close-packing (b) high atomic mass (c) high occupancy of tetrahedral holes (d) large atomic radius 39. The closest-packed layers in the face-centred cubic unit cell are perpendicular to (a) the face of the cell (b) the face diagonal of the cell (c) edges of the cell (d) the body diagonal of the cell 40. If the same type of atoms are packed in hexagonal closest packing (HCP) and cubical closest packing (CCP) separately, then (a) density of HCP will be greater than CCP (b) density of HCP will be smaller than CCP (c) density of HCP will be equal to CCP (d) density of HCP may be equal or greater than or smaller than CCP depending upon the temperature of the system 41. In the closest packing of atoms, (a) the size of tetrahedral void is greater than that octahedral void (b) the size of tetrahedral void is smaller than that of octahedral void (c) the size of tetrahedral void is equal to that of octahedral void (d) the sizes of tetrahedral and octahedral voids vary depending upon their locations 42. In the closest packing of atoms, there are (a) one tetrahedral void and two octahedral voids per atom (b) two tetrahedral voids and one octahedral void per atom (c) two of each of tetrahedral and octahedral voids (d) three of each of tetrahedral and octahedral voids 43. In the closest packing of atoms A (radius ra) the radius of atom C (radius rc hole in terms of ra is (b) rc = 0.225 ra (c) rc = 0.732 ra (d) rc = 0.155 ra (a) rc = 0.414 ra 44. In the closest packing of atoms A (radius ra) the radius of atom C (radius rc) in terms of ra is (a) rc = 0.732 ra (b) rc = 0.414 ra (c) rc = 0.225 ra (d) rc = 0.155 ra 45. In the primitive cubic unit cell of closest packed atoms, the radius of atom (r) is related to the edge length (a) of unit cell by the expression (c) r = 3 a/4 (d) r = 4a/ 3 (a) r = a/2 (b) r = a/2 2

States of Matter 2.83

46. In the body-centred cubic unit cell of closest packed atoms, the radius of atom (r) is related to the edge length (a) of unit cell by the expression (c) r= 3 a/4 (d) r = 2 a (a) r = a/2 (b) r = a/2 2 47. In the face-centred cubic unit cell of closest packed atoms, the radius of atom (r) is related to the edge length (a) of unit cell by the expression (c) r = a/2 2 (d) 3 a/4 (a) r = a/2 (b) r = a 2 48. The fraction of volume occupied in a primitive cubic unit cell is (a) 0.52 (b) 0.48 (c) 0.55 (d) 0.45 49. The fraction of volume occupied in a body-centred cubic unit cell is (a) 0.32 (b) 0.68 (c) 0.35 (d) 0.65 50. The fraction of volume occupied in a face-centred cubic unit cell is (a) 0.26 (b) 0.30 (c) 0.74 (d) 0.70 51. The coordination number of atoms packed in hexagonal close-packed structure is (a) 4 (b) 6 (c) 8 (d) 12 52. The radius of atom is related to the edge length of unit cell by the relations (b) (P) r = a/2; (I) r = 3 a/4; (F) r = a/2 2 (a) (P) r = a/4; (I) r = 3 a/2; (F) r = a/2 2 (c) (P) r = 3 a/4; (I) r = a/2; (F) r = a / 2 (d) (P) r = a/4; (I) r = 3 a/2; (F) r = a/ 2 53. The coordination number of atoms are (a) P : I: F : : 6 : 6 : 8 (b) P : I: F : : 8 : 6 : 6 (c) P : I: F : : 6 : 8 : 6 (d) P : I: F : : 6 : 6 : 6 54. The per cent void volume per unit cell are (a) P : I : F : : 25.95 : 31.98 : 47.64 (b) P : I : F : : 52.36 : 68.02 : 74.05 (c) P : I : F : : 47.64 : 25.95 : 31.98 (d) P : I: F : : 47.64 : 31.98 : 25.95 55. The portion of edge-length not occupied by atoms are (b) P : I : F : : 0 : a (1 –1/ 2 ) : a(1 – 3 /2) (a) P : I : F : : 0 : a (1 – 3 /2) : a(1 – 1/ 2 ) (d) P : I: F : : a (1 – 3 /2) : a(1 – 1/ 2 ) : 0 (c) P : I : F : : a (1 – 1/ 2 ) : 0 : a(l – 4 3 /2) 56. Diamond has face-centred cubic unit cell. The number of carbon atoms per unit cell is (a) 2 (b) 4 (c) 6 (d) 8 57. An element crystallizes in hexagonal unit cell. The number of atoms per unit cell is (a) 2 (b) 4 (c) 6 (d) 8

Ionic Solids 58. In the normal spinels some of the tetrahedral holes in the close-packed structure are occupied by (a) oxide ions (b) dipositive ions (c) tripositive ions (d) tetrapositive ions 59. Which of the following substances does not crystallize in the rock-salt structure? (a) NaCl (b) KC1 (c) MgO (d) CsCl 60. Which of the following statements is true in the rock-salt structure of an ionic compound? (a) Coordination number of cation is four whereas that of anion is six (b) Coordination number of cation is six whereas that of anion is four (c) Coordination number of each cation and anion is four (d) Coordination number of each cation and anion is six 61. Which of the following statements is true in the body-centred cubic structure of an ionic compound? (a) Coordination number of each cation and anion is eight (b) Coordination number of each cation and anion is six (c) Coordination number of each cation and anion is four (d) Coordination number of cation is six whereas that of anion is four

2.84 Complete Chemistry—JEE Main

62. The general formula, of an ionic compound crystallizing in body-centred cubic structure is (c) A2B (d) AB3 (a) AB (b) AB2 63. The general formula of an ionic compound crystallizing in rock-salt structure is (c) AB3 (d) A3B (a) AB (b) AB2 64. Which of the following statements is true in the zinc-blende type structure of an ionic compound? (a) Coordination number of each cation and anion is eight (b) Coordination number of each cation and anion is six (c) Coordination number of each cation and anion is four (d) Coordination number cation is four whereas that of anion is six 65. The general formula of an ionic compound crystallizing in zinc-blende structure is (c) A2B (d) AB3 (a) AB (b) AB2 (a) (b) (c) (d)

Coordination number of anion is four and that of cation is eight Coordination number of anion is eight and that of cation is four Coordination number of each cation and anion is four Coordination number of each cation and anion is eight

(a) AB (a) (b) (c) (d)

70.

71.

72.

73. 74.

75.

76.

(b) AB2

(c) A2B

(d) AB3

Coordination number of anion is four and that of cation is eight Coordination number of anion is eight and that of cation is four Coordination number of each cation and anion is four Coordination number of each cation and anion is eight

(c) A2B (d) AB3 (a) AB (b) AB2 If the anions (B) form hexagonal closest packing and cations (A) occupy only 2/3 octahedral holes in it, then the general formula of the compound is (c) A2 B3 (d) A3 B2 (a) AB (b) AB2 In spinel structure, oxides ions are cubical-closest packed, whereas l/8th of tetrahedral holes are occupied by cations A2+ and 1/2 of octahedral holes are occupied by cations B3+ ions. The general formula of the compound having spinel structure is (a) A2BO4 (b) AB2O4 (c) A2B4O (d) A4B2O In a cubic unit cell, seven of eight comers are occupied by atom A and centres of faces are occupied by B. The general formula of the substance having this type structure would be (b) A7B24 (c) A7B12 (d) A7 B36 (a) A7B6 An ionic compound is expected to have octahedral structure if rc/ra lies in the range of (a) 0.414 to 0.732 (b) 0.732 to 0.82 (c) 0.225 to 0.414 (d) 0.155 to 0.225 An ionic compound is expected to have tetrahedral structure if rc/ra (a) is more than 0.732 (b) lies in the range of 0.414 to 0.732 (c) lies in the range of 0.225 to 0.414 (d) lies in the range of 0.155 to 0.225 The structure of an ionic compound is expected to be cubic if rc/ra (a) is greater than 0.732 (b) lies in the range of 0.414 to 0.732 (c) lies in the range of 0.225 to 0.414 (d) lies in the range of 0.155 to 0.225 Which of the following expressions is true in case of a sodium chloride unit cell? (a) rc + ra = a (b) rc + ra = a/2 (c) rc + ra = 2a (d) rc + ra = 2 a where rc, ra and a have their usual meanings

States of Matter 2.85

77. Ammonium chloride crystallizes in a body-centred cubic lattice with a unit distance of 387 pm. If the size of Cl– ion is 181 pm, the size of NH4+ ion would be (a) 206 pm (b) 116 pm (c) 174 pm (d) 154 pm 78. The edge length of sodium chloride unit cell is 564 pm. If the size of CI– ion is 181 pm, the size of Na+ ions would be (a) 282 pm (b) 383 pm (c) 101pm (d) 167 pm 79. A compound formed by elements A and B crystallises in a cubic structure where A atoms are at the corners of a cube and B atoms are at the face centres of the cube. The formula of the compound is (b) AB3 (c) AB2 (d) AB (a) AB4 80. The 8:8 type packing is present in (c) CsCl (d) KC1 (a) NaCl (b) CaF2 81. An ionic compound AxBy crystallized in hexagonal unit cell in which the anions B occupy the lattice points and the cations A occupy 2/3rd octahedral voids. The simplest formula of the compound is (b) A2B3 (c) A3B4 (d) A2B (a) AB2 82. An ionic compound AxByCz crystallizes in cubical-closest packing in which the ions A occupy the lattice points, the ions B occupy one-eighth of tetrahedral voids and the ions C occupy one-half of octahedral voids. The simplest formula of the compound is (b) A4B2C (c) A2BC4 (d) A2B4C (a) A4BC2

Density of Crystalline Solids 83. The expression to compute the density of cubic crystal is NÊ M ˆ a3 Ê M ˆ a3 Ê M ˆ a3 Ê N A ˆ (a) r = 3 Á (b) r = (c) r = (d) r = Á ˜ ˜ Á ˜ 6 ÁË N A ˜¯ N Ë NA ¯ a Ë NA ¯ NË M ¯ where the various symbols have their usual meanings. 84. The number of atoms associated with a single primitive cubic unit cell is (a) 1 (b) 2 (c) 4 (d) 8 85. The number of atoms associated with a single face-centred cubic unit cell is (a) 1 (b) 2 (c) 4 (d) 8 86. The number of atoms associated with a single body-centred cubic unit cell is (a) 1 (b) 2 (c) 4 (d) 8 87. The unit cell length of sodium chloride crystal is 564 pm. Its density would be (a) 2.165 g cm–3 (b) 4.330 g cm–3 (c) 1.082 g cm–3 (d) 3.247 g cm–3 –1 88. The cubic unit cell of aluminium (molar mass 27.0 g mol ) has an edge length of 405 pm. Its density is 2.70 g cm–3. The type of unit cell is (a) Primitive (b) Face-centred (c) Body-centred (d) End-centred 89. A substance which has face-centred cubic crystal has a density of 1.984 g cm–3 and the edge length of unit cell equal to 630 pm. The molar mass of the substance is (b) 56.02 g mol–1 (c) 74.70 g mol–1 (d) 65.36 g mol–1 (a) 37.35 g mol–1 90. If the atomic mass of elements are the same in the three cubic lattices, then the order of densities is (b) rP < rI < rF (c) rF < rP < rI (d) rF < rI < rP (a) rP < rF < rI 91. Potassium has a body-centred cubic structure with the nearest neighbour distance 452 pm. Its relative atomic mass is 39. Its density would be (b) 0.81 g cm–3 (c) 0.91 g cm–3 (d) 1.01 g cm–3 (a) 0.71 g cm–3 92. A metal crystallises in two cubic phases, FCC and BCC with unit length equal to 350 pm and 300 pm, respectively. The ratio of densities of FCC and BCC is about (a) 1.26 (b) 1.36 (c) 1.46 (d) 1.66

2.86 Complete Chemistry—JEE Main

93. The pyknometric density of sodium chloride crystal is 2.16 g cm–3 while its X-ray density is 2.18 g cm–3. The fraction of unoccupied site in sodium chloride crystal is (a) 0.0043 (b) 0.0056 (c) 0.0093 (d) 0.0082

Point Defects, Dielectric Properties, etc. 94. Which of the following statements is not correct? (a) The density of the crystal exhibiting Schottky defect is less as compared to that of the perfect crystal (b) The density of the crystal exhibiting Frenkel defect is less as compared to that of the perfect crystal (c) The Schottky defects are more common in ionic compounds with high coordination numbers and where the sizes of ions are small (d) In alkali halides, Frenkel defects are not found. 95. Which of the following statements is not correct? (a) Silver bromide shows both types of Schottky and Frenkel defects (b) The compound ferrous oxide can be prepared in its stoichiometric composition (c) The creation of cationic holes in ionic crystals imparts electrical conductivity to such crystals (d) Heating of zinc oxide leads to the formation of a yellow coloured nonstoichiometric compound. 96. Which of the following is expected to be diamagnetic? (c) Ti2O3 (d) VO (a) TiO (b) TiO2 97. Which of the following is expected to be paramagnetic? (c) NaCl (d) Benzene (a) TiO (b) TiO2 98. An electron trapped in an anionic site in a crystal is called (a) F-centre (b) Frenkel defect (c) Schottky defect (d) Interstitial defect 99. In a ferromagnetic material (a) all the magnetic moment vectors are aligned in one direction. (b) half of the magnetic moment vectors point in one direction and the rest in the opposite direction. (c) all the magnetic moment vectors are randomly oriented. (d) is characterised by small magnetic moment. 100. In an antiferromagnetic material (a) all the magnetic moment vectors are aligned in one direction. (b) half of the magnetic moment vectors point in one direction and the rest in opposite direction. (c) all the magnetic moment vectors are randomly oriented. (d) is characterised by a very large magnetic moment. 101. In a ferrimagnetic material (a) all the magnetic moment vectors point in one direction. (b) all the magnetic moment vectors are randomly oriented. (c) contains equal number of magnetic moment vectors oriented in opposite directions. (d) contains unequal number of magnetic moment vectors in opposite directions.

ANSWERS 1. 7. 13. 19. 25. 31. 37.

(a) (b) (a) (d) (b) (c) (a)

2. 8. 14. 20. 26. 32. 38.

(d) (c) (c) (d) (a) (a) (d)

3. 9. 15. 21. 27. 33. 39.

(d) (d) (a) (b) (d) (c) (d)

4. 10. 16. 22. 28. 34. 40.

(d) (d) (d) (b) (a) (c) (c)

5. 11. 17. 23. 29. 35. 41.

(c) (c) (a) (c) (b) (a) (b)

6. 12. 18. 24. 30. 36. 42.

(a) (a) (d) (b) (b) (c) (b)

States of Matter 2.87

43. 49. 55. 61. 67. 73. 79. 85. 91. 97.

(a) (b) (a) (a) (b) (a) (b) (c) (c) (a)

44. 50. 56. 62. 68. 74. 80. 86. 92. 98.

(c) (c) (d) (a) (b) (c) (c) (b) (a) (a)

45. 51. 57. 63. 69. 75. 81. 87. 93. 99.

(a) (d) (c) (a) (c) (a) (b) (a) (c) (a)

46. 52. 58. 64. 70. 76. 82. 88. 94. 100.

(c) (b) (b) (c) (c) (b) (a) (b) (b) (b)

47. 53. 59. 65. 71. 77. 83. 89. 95. 101.

(c) (c) (d) (a) (b) (d) (a) (c) (b) (d)

48. 54. 60. 66. 72. 78. 84. 90. 96.

(a) (d) (d) (a) (b) (c) (a) (b) (b)

HINTS AND SOLUTIONS 48. f = (4/3) pr3/a3 = (4/3) p(a/2)3 /a3 = p/6 =0.52 77. Body diagonal of cube = 3 a = (1.732) (387 pm) = 670.3 pm. Thus 2(r+ + r–) = 670.3 pm 670.3 which gives r+ = pm – 181 pm = 154 pm 2 564 78. 2(r+ + r–) = 564 pm; r+ = pm –181 pm = 101 pm 2 Ê 58.5 g mol-1 ˆ NÊ M ˆ 4 = 87. r = 3 Á = 2.16 ¥ 106 g m–3 = 2.l6 g cm–3 ˜ a Ë N A ¯ (564 ¥ 10-12)3 ÁË 6.022 ¥ 1023 mol-1 ˜¯ Ê 6.022 ¥ 1023 mol-1 ˆ N 88. N = r a3 ÊÁ A ˆ˜ = (2.70 g cm -3 )(405 ¥ 10-10 cm)3 Á =4 Ë M ¯ Ë 27.0 g mol-1 ˜¯ The unit cell is face-centred cube. 89. M = 91. r =

r a3 N A (1.984 g cm -3 )(630 ¥ 10-10 cm3 )(6.022 ¥ 1023 mol-1 ) = = 74.7 g mol-1 4 N

NM (2)(29 g mol -1 ) = = 9.11 ¥ 105 g m -3 = 0.91 g cm -3 a3 N A (2 ¥ 452 ¥ 10 -12 m / 3 )3 (6.022 ¥ 1023 mol -1 )

92. For FCC, N = 4 and for BCC, N =2. Hence Ê a3 ˆ Ê 4 ˆ Ê (3.0)3 ˆ r (FCC) Ê N ˆ = = Á 3˜ = 1.26 r (BCC) Ë a ¯ FCC ÁË N ˜¯ BCC ÁË (3.5)3 ˜¯ ÁË 2 ˜¯ 93. f =

( M / r1 ) - ( M / r2 ) (1 / r1 ) - (1 / r2 ) r2 - r1 r2 2..18 = = = -1 = - 1 = 1.00926 - 1 = 0.00926 ( M / r2 ) r1 r1 2.16 (1 / r2 )

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. What type of crystal defects is indicated in the diagram below? Na+ Cl– Na+ Cl– Na+ Cl– Cl– Cl– Na+ Na+ Na+ Cl– Cl– Na+ Cl– + – Cl Na Cl Na+ Na+ (a) Frenkel defect (b) Schottky defect (c) Interstital defect (d) Frenkel and Schottky defects

[2004]

2.88 Complete Chemistry—JEE Main

2. An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound will be (c) A3B (d) A2B [2005] (a) AB (b) AB3 3. Total volume of atoms present in a face-centred cubic unit cell of a metal (r is atomic radius) 16 20 3 24 3 12 pr3 (b) pr (c) pr (d) pr3 [2006] (a) 3 3 3 3 4. In a compound, atoms of element Y form cubical-closest packing and those of elements X occupy 2/3 of tetrahedral voids. The formula of the compound will be. (b) X4Y3 (c) X2Y3 (d) X2Y [2008] (a) X3Y 5. Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom? (a) 157 pm (b) 181 pm (c) 108 pm (d) 127 pm [2009], [2011] 6. The edge length of a face-centred cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is (a) 144 pm (b) 288 pm (c) 398 pm (d) 618 pm [2010] 7. Percentage of free space in cubic close-packed structure and in body-centred packed structure are respectively (a) 48% and 26% (b) 30% and 26% (c) 26% and 32% (d) 32% and 48% [2011] 8. In a face-centred lattice, atom A occupies the corner positions and atom B occupies the faced centre position. If one atom of B is missing from one of the face-centred points, the formula of the compound is (b) A2B (a) A2B5 (d) A2B3 [2011], [2014 online] (c) AB2 9. Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of lithium will be (a) 175 pm (b) 300 pm (c) 240 pm (d) 152 pm [2012] 10. Which of the following exists as covalent crystals in the solid state? (a) Iodine (b) Silicon (c) Sulphur (d) Phosphorus [2012] 11. CsCl crystallises in body centred cubic lattice. If a is its edge length then which of the following expression is correct? (b) rCs+ + rCl– = 3 a/2 (c) rCs+ + rCl– = 3 a (d) rCs+ + rCl– = 3a [2014] (a) rCs+ + rCl– =3a/2 12. How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00 g? (b) 2.57 ¥ 1021 (c) 5.14 ¥ 1021 (d) 1.28 ¥ 1021 (a) 1.71 ¥ 1021

[2003]

13. The total number of octahedral void(s) per atom present in a cubical close-packed structure is (a) 2 (b) 4 (c) 1 (d) 3

[2014 online]

14. The appearance of colour in solid alkali metal halides is generally due to (a) Schottky defect (b) Frenkel defect (c) Interstitial position (d) F-centres

[2014 online]

15. In a monoclinic unit cell, the relation of sides and angles are respectively (a) a = b π c and a = b = g = 90° (b) a π b π c and a = b = g = 90° (c) a π b π c and a = g = b = 90° = a (d) a π b π c and a π b π g π 90° [2014 online] 16. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately (a) 1.86 Å (b) 3.22 Å (c) 5.72 Å (d) 0.93 Å [2015]

ANSWERS 1. (b) 7. (c) 13. (b)

2. (b) 8. (a) 14. (d)

3. (a) 9. (d) 15. (c)

4. (b) 10. (b) 16. (a)

5. (d) 11. (b)

6. (a) 12. (b)

States of Matter 2.89

HINTS AND SOLUTIONS 1. Missing of species from their expected location gives rise to Schottky defect. 2. Each corner of a cube is shared amongst eight cubic unit cells. The contribution of atoms A at the eight corners of a cubic unit cell is as follows. 1 Contribution from corners = 8 ÊÁ ˆ˜ = 1 Ë 8¯ Each face of a cube is common between two unit cells. The contribution of atoms B at the centre of six faces of a cubic unit cell is as follows. 1 Contribution from corners = 6 ÊÁ ˆ˜ = 3 Ë 2¯ Hence, the empirical formula of the compound would be AB3. 3. In a face-centred cubic unit cell, there are four atoms. Hence, the volume occupied by atoms is 4(4/3 pr3), i.e. (16/3) pr3. 4. In closest-packed atoms, there are two tetrahedral voids per closest-packed atom. Hence, the formula of the compound will be YX2(2/3) i.e. YX4/3 or X4Y3 Alternatively, there are four Y per unit cell and 8(2/3) X. Hence, the formula is Y4X16/3 i.e. Y12X16 or Y3X4. 5. For a face-centred cubic unit cell, atoms touch each other along the face-diagonal of the cube. Hence 4r=

2a

r = ( 2 ) (361 pm)/4 = 127.6 pm 6. In the face-centred cubic cell of an ionic substance, the cation-anion touch each other along the edges of the cube. The example of such an ionic compound is sodium chloride. Hence, This gives

2r+ + 2r– = a fi 2(110 pm) + 2r– = 508 pm. This gives r_ = 144 pm 7. In cubical-closest packing, atoms touch each other along the face diagonal of the cube. Hence, 4r = 2 a. The number of atoms in a unit cell is four. Hence Volume of a cube = a3 È4 Ê 2 ˆ3˘ 4 a˜ ˙ ; Fraction of volume occupied = 2p = 0.74 Volume occupied by atoms = Í p Á 3 4 Ë ¯ ˙˚ ÍÎ 6 Percentage of void volume = (1– 0.74) (100) = 26% In a body-centred cube, atoms touch each other along the body-diagonal of the cube. Hence, 4r = 3 a The number of atoms in a unit cell is two. Hence Volume of a cube = a3 È 4 Ê 3a ˆ 3 ˘ 3p Volume occupied by atoms = 2 Í p Á = 0.68 ˜¯ ˙ ; Fraction of volume occupied = 3 4 Ë ÍÎ ˙˚ 8 Percentage of void volume = (1 – 0.68) (100) = 32% 8. There are eight comers and six faces in a cube. Each atom at the comer is shared amongst eight unit cells and by the two unit cells at the centre of each face. Thus, contributions of A and B from a unit cell are as follows. Contribution of atom A = 8(1/8) = 1 and Contribution of atom B = 5(1/2) = 5/2 Hence, the formula of the compound is AB5/2 i.e. A2B5. 9. In body centred cubic structure, atoms (radius: r) touch each other along the body diagonal of the cube (edge length : a). Thus 4r = 3 a Ê 3ˆ Ê 1.732 ˆ (351 pm) = 152 pm ˜ r = Á ˜ a = ÁË 4 4 ¯ Ë ¯

2.90 Complete Chemistry—JEE Main

10. Iodine, sulphur and phosphorus exist I2, S6 and P4, respectively. Silicon does exist as covalent crystal in the solid state. 11. In a body-centred cubic arrangements, ions touch each other along the body diagonal. or rCs+ + rCl– = 3 a/ 2 Hence, 2rCs+ + 2 rCl– = 3 a 12. The are four molecules of NaCl per unit cell. Hence, the mass of one unit cell is m = 4(23 + 35.5) g/6.022 ¥ 1023 The number of unit cell in 1.00 g of crystal will be N= 13. 14. 15. 16.

6.022 ¥ 1023 = 2.57 ¥ 1021 4 ¥ (23 + 35.5)

There are 4 octahedral voids. These are situated at the centres of cube and at the centres of edges. The electrons, which occupy the anion vacancy are called F-centres. This imparts colour to the alkali metal halides. For monoclinic unit cell, the structural parameters are a π b π c and a = b = 90° π g In a body centred cubic lattice, atoms touch each other along the body diagonal of the cube. Hence 4r = 3a or r = ( 3 / 4) a ( 3 / 4)(4.29Å) = 1.86 Å

3 Atomic Structure

SECTION 1

Development of Structure of Atom

The word ‘atomic’ has its origin in an adjective in Greek meaning ‘not divisible’. Based on experimental evidence, Dalton (1808) proposed that matter is composed of atoms which cannot be created, divided or destroyed. At the end of nineteenth century, experimental evidence established that an atom can be subdivided. It is made up of three subatomic particles, namely, protons neutrons and electrons. Their main characteristics are as follows. Mass

Charge

Proton

1.672 ¥ 10

Neutron

1.675 ¥

Electron

9.1096 ¥ 10–31 kg

–27

10–27

¥10–19 C

Symbol p

kg

1.6022

kg

No charge

n

–1.6022 ¥ 10–19 C

e

After the discovery of proton, neutron and electron, their arrangement in atom was proposed as described in the following.

Thomson Model (1898) An atom is positively-charged sphere of radius about 10–10 m in which negatively-charged electrons are uniformly embedded so as to have an overall electrically neutral atom. Rutherford Model In 1910, Rutherford carried out an experiment in which a thin sheet of gold foil was bombarded with a beam of a-particles (He2+). He observed that : (i) Most of a (ii) A few of a Based on these observations, Rutherford proposed the following model of an atom: An atom is made up of tiny positively-charged nucleus containing practically whole of mass of the atom. Electrons are revolving around the nucleus with high speeds in circular path, known as orbits. Rutherford also concluded that: (i) The radii of nuclei of atoms are of the order of 10–15 m to 10–14 m, and (ii) The radii of outer region where electrons move is of the order of 10–10 m. Thus, most of space in an atom is emptly and the volume occupied by the nucleus is less than 10–12 of the total volume of an atom. Subsequently, it was discovered that the nucleus of an atom is made up of two particles, namely, positively-charged protons and electrically-neutral neutrons.

3.2 Complete Chemistry—JEE Main

Rutherford model of an atom could not be sustained based on the following fact. In order to maintain circular motion of electron around the nucleus, it is subjected to acceleration. According to Maxwell theory of electromagnetic radiation a charged particle when subjected to acceleration should lose energy in the form of electromagnetic radiation. Hence, the electron will lose energy continuously and eventually will fall into the nucleus. Calculations have shown that only 10–8 s will be required for the electron to fall in the nucleus. Rutherford model also do not explain the discrete spectra shown by the atoms. In 1885, Balmer analyzed the spectra shown by hydrogen atom and established the expression Ê n2 ˆ l /nm = 364.56 Á 2 2 2 ˜ Ë n2 - n1 ¯ for the observed spectra in the visible and ultraviolet region of radiations. The value of n1= 2 and n2 can take values 3,4, . . ., etc.

Bohr Model In order to explain the stability of hydrogen and hydrogen-like species (such as He+, Li2+, . . ., etc.), Niels Bohr, in 1913, proposed the following two assumptions: 1. The electrons in an atom revolve around the nucleus only in certain allowed circular orbits without losing any energy. 2. The electron can jump from one of the orbits to another and can there by gain or lose energy equivalent to the difference in energy of the two involved orbits in the form of electromagnetic radiation such that DE = hn = h (c/l) (1) where DE is the energy difference between the two orbits. n (or l) is the frequency (or wavelength) of the radiation and c is the speed of light ( 3 ¥ 108 m s–1). Criterion for stability of the atom The stability of the atom is explained based on the expression |attractive (centripetal) force| = centrifugal force ( Ze)(e) mu2 = (2) r ( 4p e 0 ) r 2 where Z is the atomic number of the nucleus, e is the elementary charge, u is the speed of electron in the orbit of radius r, and eo is the permittivity of a vacuum and has a value of 8.854 ¥ 10–12 C2 N–1 m–2. i.e.

Postulate for Allowed Stationary Orbits quantum restriction on the angular momentum of the revolving electrons. h mu r = n ÊÁ ˆ˜ Ë 2p ¯

(3)

where n has integral values 1,2,3, . . ., and is known as quantum number. The symbol h is Planck’s constant and has a value of 6.626 ¥ 10–34 J s. Radius of Permitted Orbits Equation (2) gives Z e2 = mu2 ( 4p e 0 ) r Eliminating u by using Eq. (3), we get 2 Z e2 = m Èn ÊÁ h ˆ˜ 1 ˘ Í Ë 2p ¯ m r ˙ ( 4p e 0 ) r Î ˚

This gives

È ˘ h2 r = n2 Í 2 ˙ 2 Î 4p m ( Z e / 4p e 0 ) ˚

(4)

Atomic Structure

3.3

Energy of electron in a orbit We have E = kinetic energy + potential energy =

Z e2 1 mu 2 2 ( 4p e 0 ) r

1 1 1 (mu )2 mu 2 - mu 2 = - mu 2 = 2 2 2 m Eliminating mu by using Eq. (3), we get E=

E= -

(5)

1 È Ê h ˆ 1 ˘2 nÁ ˜ 2m ÍÎ Ë 2p ¯ r ˙˚

Substituting the expression of r from Eq. (4), we get E= -

1 È Ê h ˆ 4p 2 m( Ze2 / 4p e 0 )2 ˘ 1 Ín Á ˜ ˙ =- 2 2 2 2m Î Ë 2p ¯ n h n ˚

Speed of Electron in an orbit

Substituting

È 2p 2 m( Z e2 / 4pe 0 )2 ˘ Í ˙ h2 Î ˚

E = – (1/2)mu2

(6)

from Eq. (5) in Eq. (6), we get

1 È 2p 2 m ( Z e 2 / 4p e 0 ) 2 ˘ 1 - mu 2 = - 2 Í ˙ 2 n Î h2 ˚ u=

This gives

2p ( Ze / 4p e 0 ) nh

(7)

Points to Remember: n2 Z Z2 ; E μ– 2 ; u= Z n n E = –KE, PE = –2KE and PE = 2E where KE and PE are kinetic energy and potential energy, respectively rμ

Discrete spectra of Hydrogen Atom DE = E2 – E1 =

The energy difference between two Bohr orbits is

2p 2 m( Ze2 / 4p e 0 )2 Ê 1 1ˆ - 2˜ Á 2 2 Ë n1 n2 ¯ h

(8)

The frequency and wavenumber of radiation having energy DE are n=

~

n=

where

R• =

DE 2p 2 m( Ze2 / 4p e 0 )2 Ê 1 1ˆ - 2˜ = Á 2 2 h Ë n1 n2 ¯ h Ê 1 1ˆ 1 DE 2p 2 m( Ze2 / 4p e 0 )2 Ê 1 1ˆ = - 2 ˜ = Z 2R • Á 2 - 2 ˜ = Á 3 2 Ë n1 n2 ¯ l hc Ë n1 n2 ¯ hc 2p 2 m( Ze2 / 4pe 0 )2 h3 c

(9)

(10)

and is known as Rydberg constant. Its value is

R• = 1.097 37 ¥107 m–1 The spectra shown by hydrogen atom can be explained by assigning different values of n1 and n2 in Eq. (10) keeping in mind that n2 > n1, These are shown in Table 1 and Fig. 1

3.4 Complete Chemistry—JEE Main

Table.1 Spectral series

Values of

Lyman

Region of radiation

n1

n2

1

2,3,4, . . .

ultraviolet

Balmer

2

3,4,5, . . .

Visible

Paschen

3

4,5,6, . . .

Near infrared

Brackett

4

5,6, . . .

Infrared

Pfund

5

6,7, . . .

Far infrared

Ionization Energy of Hydrogen Atom For determining ionization energy, we set n1 = 1 and n2 = • . This gives 2 p 2 m (e 2 / 4 p e 0 ) 2

= 2.182 ¥10–18 J h2 For one mole of hydrogen atom IE = NA(2.182 ¥10–18 J) = (6.022 ¥1023 mol–1) (2.182 ¥ 10–18 J) = 1314 ¥ 103 J mol–1 = 1314 kJ mol–1 IE =

Bohr theory is inadequate in explaining the experimental facts of multi-electronic

Fig. 1

The Bohr theory was not in agreement with the principles discovered later on. These are wave-particle duality by de-Broglie and the uncertainty principle of Heisenberg. In the light of these facts, Bohr theory was replaced by the current quantum theory.

Illustrative problems 1. Given: R• = 1.0974 ¥ 107 m–1 DE = or

n1= 2 and n2 = 3. Hence

Ê 1 hc 1ˆ = R• hc Á 2 - 2 ˜ l Ë n1 n2 ¯

Ê 1 1 1ˆ Ê 1 1ˆ = R• Á 2 - 2 ˜ = (1.0974 ¥ 107 m -1 ) Á - ˜ = 1.524 ¥ 106 m–1 Ë 4 9¯ l Ë n1 n2 ¯

l = 1/(1.52 ¥ 106 m–1) = 6.56 ¥ 10–7 m = 656 nm 2. What transition in the hydrogen atom will have the same energy as the transition in He+ ion from the level n = 4 to the level n = 2 . 1ˆ 3 Ê 1 1ˆ Ê 1 For helium ion DE = Z2R•hc Á 2 - 2 ˜ = (22)R•hc Á 2 - 2 ˜ = R•hc ¯ Ë 4 2 4 Ë n1 n2 ¯ For hydrogen atom

Ê 1 1ˆ DE= R•hc Á 2 - 2 ˜ Ë n1 n2 ¯

Obviously n1 = 1 and n2 = 2. Hence, the transition n = 2 to n =1 in hydrogen atom will have the same wavelength as the transition n = 4 to n = 2 in He+ ion. 3. Calculate the radius of n = 2 orbit in Li2+ ion. Given : e2/4p e0 = 2.30 ¥ 10–28 N m2.

Atomic Structure

3.5

È ˘ (6.626 ¥ 10-34 J s)2 ˘ È h2 2) r = n2 Í = (2 Í ˙ 2 31 28 2 ˙ 2 2 Î (4)(3.14) (9.1 ¥ 10 kg )(3 ¥ 2.30 ¥ 10 N m ) ˚ Î 4p m( Ze / 4p e 0 ) ˚

= 7.09 ¥ 10–11 m =70.9 pm. p e0 =1.112 ¥ 10–10 C2 N–1 m–2.

4.

2p( Ze / 4p e 0 ) nh 2

The expression of speed is u = For H-atom,

Z = 1. Hence u=

2(3.14)(1)(1.6 ¥ 10-19 C) 2 /(1.112 ¥ 10-10 C2 N -1 m -2 ) = 2.18 ¥ 106 m s–1 (1)(6.626 ¥ 10-34 J s) +

5. Given : 4p e0 =1.112 ¥ 10 The speed of electron is

–10

2

–1

C N m

ion.

–2

2 u = 2p( Ze / 4p e 0 ) nh 2(3.14)(2)(1.6 ¥ 10-19 C) 2 /(1.112 ¥ 10-10 C2 N -1 m -2 ) = (1)(6.626 ¥ 10-34 J s)

= 4.36¥106 m s–1 Ê ˆ h2 r = n2 Á 2 2 Ë 4p m( Ze / 4p e 0 ) ˜¯ The value Ze2/4p e0 is (2)(1.6 ¥ 10-19 C)2 Z e2 = = 4.60 ¥ 10–28 N m2 4p e 0 (1.112 ¥ 10-10 C2 N -1 m -2 ) Now

È ˘ (6.626 ¥ 10-34 J s) 2 –11 m r = (1)2 Í 2 -31 -28 2 ˙ = 2.66 ¥ 10 Î (4)(3.14) (9.1 ¥ 10 kg )(4.60 ¥ 10 N m ) ˚

The number of revolution made is n=

u 4.36 ¥ 106 m s -1 = = 2.61 ¥ 1016 s -1 -11 2pr 2(3.14)(2.66 ¥ 10 m)

MULTIPLE CHOICE QUESTIONS ON SECTION 1 1. Isotopes contain (a) same number of neutrons (b) same number of protons (c) same mass number (d) same number of electrons 2. Isotones contains (a) same number of neutrons (b) same number of protons (c) same number of electrons (d) same mass number 3. The ratio of E2 – E1 and E5 – E4 of orbits in hydrogen-like species is (a) 108/7 (b) 100/3 (c) 152/3 (d) 50/3 4. Which of the following orbit of He+ ion will have the same energy as that of second orbit of hydrogen atom? (a) n = 1 (b) n = 2 (c) n = 3 (d) n = 4

3.6 Complete Chemistry—JEE Main

5. The ratio of kinetic energy and total energy of an electron in Bohr orbit of hydrogen-like species is (a) 1/2 (b) 1/4 (c) –1 (d) –1/2 6. The ratio of potential energy and total energy of an electron in Bohr orbits of hydrogen atom is (a) 2 (b) 1/2 (c) 1 (d) –1/2 7. The wavelength of radiation emitted when an electron jump from n = 4 to n = 2 energy levels of hydrogen atom is (Given: Rydberg constant = 1.097 ¥107 m–1) (a) 342 nm (b) 252 nm (c) 486 nm (d) 890 nm 8. The number of possible spectral lines emitted when electron in n = 4 Bohr orbit reaches to n = 1 Bohr orbit is (a) 1 (b) 2 (c) 4 (d) 6 +, the energy E of an electron in nth orbit of hydrogen atom 9. If E1 2 will be (b) E1/2n2 (c) E1/4n2 (d) E1/8n2 (a) E1/2n 10. For which of the following hydrogen-like species, the wavelength of radiation emitted due to electron shifts from = 2 to n = 1, has a maximum value? (c) Li2+ (d) Be3+ (a) H (b) He+ 11. For which of the following electronic transitions in hydrogen-like species, the frequency of emitted radiation will be maximum? (a) n = 2 to n = 1 (b) n = 3 to n = 2 (c) n = 4 to n = 2 (d) n = 6 to n = 4 + having n = 1 is about (Given : e2/ 4p e = 2.30 ¥ 10–28 N m2) 0 (a) 53.4 pm (b) 106.8 pm (c) 26.7 pm (d) 42.1 pm n =1) of hydrogen atom is 53.4 pm. The Bohr orbit having n = 3 in Li2+ will be (a) 53.4 pm (b) 106.8 pm (c) 120.1 pm (d) 160.2 pm also permitted for the electron in Bohr orbits of hydrogen atom? (a) –6.8 ev (b) –4.53 ev (c) –3.4 ev (d) –2.27 ev ¥ 105 m s–1. The speed of electron in the n = 2 level of Li2+ ion will be (b) 1.09 ¥106 m s–1 (c) 1.45 ¥106 m s–1 (d) 1.82 ¥106 m s–1 (a) 7.27 ¥105 m s–1 16. The ionization energy of hydrogen atom is 1314 kJ mol–1. The energy of n = 3 electronic level in this atom is (a) –2.182 ¥ 10–18 J (b) –4.27 ¥10–18 J (c) –2.42 ¥ 10–19 J (d) –2.182¥10–19 J 17. The ionization energy of hydrogen atom is 1314 kJ mol–1. The ionization energy of He+ ion is expected to be (a) 1314 kJ mol–1 (b) 2628 kJ mol–1 (C) 3952 kJ mol–1 (D) 5256 kJ mol–1

ANSWERS 1. (b) 7. (c) 13. (d)

2. (a) 8. (d) 14. (c)

3. (b) 9. (c) 15. (b)

4. (d) 10. (a) 16. (c)

5. (c) 11. (a) 17. (d)

HINTS AND SOLUTIONS 1. Isotopes contain same number of protons. 2. Isotones contain same number of neutrons. 3. E2 – E1 μ –

9 2 E2 - E1 3 400 100 3 2 Z2 Z2 Z2 Z2 Z ; = ¥ = Z ; E5 – E4 μ – = = + + 400 4 3 9 E5 - E4 4 4 1 25 16

6. (a) 12. (c)

Atomic Structure

3.7

4. Since E μ – Z2/n2, the orbit n = 4 of He+ will have the same energy as that of n = 2 of hydrogen atom 5. The ratio KE/E = –1 6. E = –KE = PE/2 Thus PE/E = 2 7.

Ê 1 1 1ˆ 1ˆ 3 Ê 1 = R• Á 2 - 2 ˜ = (1.097 ¥ 107 m -1 ) Á 2 - 2 ˜ = (1.097 ¥ 107 mol-1 ) ÊÁ ˆ˜ ¯ Ë Ë l Ë n1 n2 ¯ 16 ¯ 2 4

l = 4.86 ¥ 10–7 m = 486 nm 8. The possible spectral emissions correspond to the transfer n = 4 to n = 3, n = 4 to n =2, n = 4 to n = 1, n = 3 to n = 2, n = 3 to n =1 and n = 2 to n = 1. There are 6 emissions. 9. Since En μ – Z2/n2, we have E1(He+) μ – 10.

22 = –4 12

En(H) = -

and

12 1 = - 2 . Hence 2 n n

En(H) =

E1/ 4 n

2

=

E1 4n 2

4 Ê 1 1 1ˆ 1 1 3 l= . = Z 2 R• Á 2 - 2 ˜ = Z 2 R• ÊÁ 2 - 2 ˆ˜ = Z 2 R• . Hence, 3Z 2 R• Ë1 l Ë n1 n2 ¯ 2 ¯ 4 Lesser the value of Z (i.e. hydrogen atom) larger the wavelength.

11. We have

Ê 1 1ˆ n/Z2R•c = Á 2 - 2 ˜ . Ë n1 n2 ¯

Hence

Choice a

1 1 1 1 3 108 - 2 = 2- 2 = = 2 4 144 2 n1 n2 1

Choice b

1 1 1 1 5 20 - 2 = 2- 2 = = 2 36 144 3 n1 n2 2

Choice c

1 1 1 1 3 27 - 2 = 2- 2 = = 2 16 144 4 n1 n2 2

Choice d

1 1 1 1 5 5 - 2 = 2- 2 = = 2 144 6 n1 n2 4 144

Obviously, the choice a is correct. ˘ È h2 2 12. r = n2 Í 2 ˙ = (1) 2 m Ze 4 p ( / 4 p e ) Î 0 ˚ = 2.66 ¥ 10–11 m = 26.6 pm. 13. Since r μ n2/Z, we have rLi2+ (32 / 3) = 2 =3 fi rH 1 /1 14. Since E μ – Z2/n2, we can have

È ˘ (6.626 ¥ 10-34 J s) 2 Í -31 -28 2 2 ˙ Î 4(3.14) (9.1 ¥ 10 kg )(2 ¥ 2.30 ¥ 10 N m ) ˚

rLi2+ = 3rH = 3 ¥ 53.4 pm = 160.2 pm E = –3.4 eV for n = 2

15. Since u μ Z/n, we will have u (Li 2+ ) 3 / 2 = 1/1 u (H)

fi

Ê 3ˆ Ê 3ˆ 5 -1 u (Li+) = ÁË ˜¯ u(H) = ÁË ˜¯ (7.27 ¥ 10 m s ) =1.09 ¥ 106 m s–1 2 2

16. Ionization energy per hydrogen atom is

(1314 ¥ 103 J mol-1 ) IE = = 2.182 ¥10–18 J (6.022 ¥ 1023 mol-1 )

The energy of n = 1 orbit in hydrogen atom will be The energy of n = 3 orbit will be 17. Since E μ – Z2/n2, we have Hence,

E3 =

E1 32

=-

E1 = –2.182 ¥10–18 J

2.182 ¥ 10-18 J = –2.42 ¥ 10–19 J 9

E1 (He+ ) - 4 / 1 = =4 -1 / 1 E1 (H)

fi

E1(He+) = 4 E1(H)

IE(He+) = 4 IE(H) = 4 ¥ 1314 kJ mol–1 = 5256 kJ mol–1

3.8 Complete Chemistry—JEE Main

SECTION 2

Quantum-Mechanical Model of Atom

The quantum-mechanical model of an atom came into existence after the discovery of wave-particle duality and uncertainty principle.

Wave-Particle Duality A subatomic particle can display wave nature as well as particle nature. This dual behaviour is inter-related by deBroglie relation, according to which, we have h h i.e. mu = p= l l where the momentum p ( = mu) represents the particle nature and the wavelength l represents the wave nature of the subatomic particle. Illustration de–Broglie wavelength of an electron travelling with a speed of 1% of the speed of light. h (6.626 ¥ 10-34 J s) = = 2.43 ¥ 10–10 m = 243 pm mu (9.1 ¥ 10-31 kg )(3 ¥ 106 m s -1 ) For particles of larger mass (say, 1 g), l l=

(Note

The uncertainty Principle According to this principle postulated by Heisenberg, we have, It is not possible to design an experiment with the help of which one can determine simultaneously the precise values of both position and momentum of subatomic particles. Mathematically, the uncertainty principle is expressed as h Dp Dx ≥ 4p where D p and D x are the root mean square uncertainties in momentum and position of the particle, respectively and h is Planck’s constant. Illustration Uncertainty in speed of an electron if the uncertainty in position is 5 pm. h h i.e. Dp Dx ≥ (m Du ) Dx ≥ 4p 4p Hence

Du ≥

Ê 6.626 ¥ 10-34 J s ˆ Ê ˆ 1 h Ê 1 ˆ = = 1.16¥107 m s–1 Á Á ˜ -31 -12 ˜ Á 4 (3.14) Ë ¯ Ë (9.1 ¥ 10 kg)(5 ¥ 10 m) ˜¯ 4p Ë m D x ¯

Comment ¥ 106 m s–1. The uncertainty of locating its position by 5 pm leads to much larger uncertainty in its speed. Thus, talking the precise position and speed of electron in Bohr theory has no meaning. (Note For particles of larger mass (say. 1 g), the uncertainties in position and velocity are very small and one can talk about their precise position as well as its speed.)

Dual Nature of Radiation Even the radiations have dual nature of wave as well as that of particle. This was established by Albert Einstein in 1905. In wave nature, a radiation is characterised by its wavelength (l) or frequency (n) or wave number (n~). These are inter-related through the expressions c 1 and n~ = n= l l where c is the speed of light. In particle nature, a radiation is considered to be a stream of photons, each carrying energy equal to hn, i.e. E = hn = hc/l

Atomic Structure

3.9

Photoelectric Effect The photoelectric effect (discovered by Heinrich Hertz) involves the instantaneous emission of electrons when a clean metal plate in vacuum is exposed to a beam of ultraviolet radiation (Fig. 2) Main Observations of the Experiment 1. Electrons are emitted only when the plate is irradiated with radiation of frequency equal to or greater than some minimum frequency, known as threshold frequency. 2. The number of electrons emitted is proportional to the intensity of incident radiation. 3. The kinetic energy of emitted electrons depends on the frequency of the incident radiation and it increases linearly with increase in the frequency of incident radiation (Fig. 3)

Fig. 2

Explanation of Photoelectric Effect When a photon of incident radiation collides with the electron of metal, the electron acquires energy equal to the energy of the photon. Thus, the emitted electron carries energy as given by the expression 1 mu2 + IE 2 where KE, PE and IE stand for kinetic energy, potential energy and ionization hn = KE + PE=

energy, respectively. The latter is IE = hn0 where n0 is known as the threshold frequency of the metal. Hence

Fig. 3

1 mu2 = h(n – n0) 2 that is, the kinetic energy of emitted electrons varies linearly with the frequency of the incident radiation (Fig. 3). The slope of linear plot is Planck’s constant. KE =

Stopping Potential In Fig. 2, if the potential applied to the electrodes is increased, a stage is reached when no electron reaches to the negative electrode. At this stage, kinetic energy of electrons becomes zero and is thus given by the expression KE = |eV0| where V0 is known as stopping potential. Illustration The velocity and stopping potential of electrons when a metal (threshold frequency = 1.5 ¥1016 s–1) is exposed to the radiation of wavelength 15 nm. n=

c 3 ¥ 108 m s -1 = = 2.0 ¥ 106 m s–1 l 15 ¥ 10-9 m

1 mu 2 = h(n – n0) = (6.626 ¥ 10–34 J s) (0.5 ¥ 1016 s–1) = 3.3 ¥ 10–18 J 2 Ê 2 ¥ 3.3 ¥ 10-18 J ˆ = 2.7 ¥ 106 m s–1 u= Á Ë 9.1 ¥ 10-31 kg ˜¯

and

V0 =

KE Ê 3.3 ¥ 10-18 J ˆ =Á = 20.63 V e Ë 1.60 ¥ 10-19 C ˜¯

Quantum-Mechanical Behaviour of Electron in an Atom in terms of probable distribution based on the wave nature of the electron. The wave nature of the electron is described by wave function which are the solutions of Schrödinger equation.

3.10 Complete Chemistry—JEE Main

Ê ∂2 8p 2 me ∂2 ∂2 ˆ + (E - V ) = 0 Y + + ÁË ∂x 2 ∂y 2 ∂z 2 ˜¯ h2 where V is the potential energy of the electron in hydrogen-like species and is given as Ze2 V = ( 4p e ) r 0 The Schrödinger equation can be solved precisely for hydrogen-like species For this, the Schrödinger equation is expressed in spherical polar system of coordinates (Fig. 4) and its solutions have the form Yr,q,j = Rr Qq Fj where R, Q and F are r -, q - and j - dependent mathematical expressions, respectively.

Fig. 4

Solution of r-Dependent Equation The r - dependent equation provides the variations in the function R with the distance r of electron from the nucleus. The solution of R provides more than one r-dependent behaviour involving two constants n and l. These are the quantum numbers and have the following values. n 1, 2, 3, . . . l 0, 1, 2, . . ., (n – 1) ;(a total n values) A few expressions of Rn,l are as follows. ÊZˆ R1, 0 = 2 Á ˜ Ë a0 ¯ R2, 1 =

1/ 2

exp(–Zr/a0)

1 Ê Z ˆ Á ˜ 3 Ë 2a0 ¯

3/ 2

Ê Z ˆ R2, 0 = 2 Á Ë 2a0 ˜¯

3/ 2

Zr ˆ Ê ÁË 2 - a ˜¯ exp(–Zr/2a0) 0

Ê Zr ˆ ÁË a ˜¯ exp(–Zr/2a0) 0

Solution of q - Dependent Equation The q - dependent equation provides the variation in the function Q with the angle q. The solution of Q provides more than one q-dependent solutions involving two constants l and |m|, where the quantum number m takes the value m 0, ±1, ±2, . . ., ±l ;(a total of 2l +1 values) A few expression of Ql, |m| are as follows. Q0,0 = 1 / 2

Q1,0 = ( 6 / 2) cos q

Q1,±1 = ( 3 / 2)sin q

Solution of j-Dependent Equation The j-dependent equation provides variation in the function F with the angle j. The solution of F provides more than one j-dependent solutions involving the constant m. A few expressions of m are as follows: F0 = 1 / 2 p

F+1 = (1 / 2p ) exp(ij )

F–1 = (1 / 2p ) exp(-ij )

Orbitals and Quantum Numbers The solution of Schrödinger equation may be represented as Yn, l, m(r, q, j) = Rn, l (r) Ql, |m|(q)Fm(j) where the constants n, l and m are known as quantum numbers. Their values, as described above, are: n 1, 2, 3, . . . l 0, 1, 2, . . .,(n – 1) ;(a total of n values) m 0, ±1, ±2, . . ., ± l ;(a total of 2l + 1 values) Note The maximum value of l is limited by n and that of m by l. Each solution of Schrödinger equation represents an orbital of an atom. Quantum Number n In hydrogen-like species the energy of electron in an atom is governed by the quantum number n and is known as principal quantum number. The energy expression is

Atomic Structure

E= -

3.11

1 Ê 2p 2 m( Ze2 / 4p e 0 )2 ˆ ˜¯ n 2 ÁË h2

Note This expression is the same as that obtained in Bohr theory. The different energy shells are designated by the symbols K, L, M, . . . depending upon the value of n equal to 1, 2, 3, . . ., respectively. Quantum Number l The quantum number l, known as azimuthal or subsidiary quantum number, characterises the angular momentum of the electron by the expression Ê hˆ ;(h is Planck’s constant) L = l (l + 1) Á ˜ Ë 2p ¯ The value of l is designated by the symbols s, p, d, f, g, . . ., depending upon the values of l equal to 0, 1, 2, 3, 4, . . ., respectively.

Quantum Number m The quantum number m, known as magnetic quantum number, characterises the z-component of angular momentum of the electron by the expression h Lz = m ÊÁ ˆ˜ Ë 2p ¯ The z The relation between l and m is Lz = L cos q m=

This gives

h h m ÊÁ ˆ˜ = ÊÁ l (l + 1) ˆ˜ cos q Ë 2p ¯ Ë 2p ¯

i.e

l (l + 1) cos q

Designation of an Orbital An orbital is designated by writing the value of principal quantum number n followed by the symbol prescribed to the azimuthal quantum number l. The value of m is added as the subscript to the symbol of l. Illustrations n=1

l=0

m=0

1s

1 orbital

n=2

l=0 l=1

m=0 m = +1 m=0 m = –1

2s 2p+1 2p0 2p–1

4 orbitals

In general, the number of orbitals in the give value of principal quantum number n is n2. All these orbitals are degenerate, that is, they have same energy.

Spin Quantum Number An electron in an orbital spin around its own axis. This spinning produces angular momentum whose value is given by the expression L=

h s ( s + 1) ÊÁ ˆ˜ Ë 2p ¯

where s is known as spin quantum number and has a value of 1/2. The z-component of this angular momentum is given by Ê hˆ Lz = ms ÁË 2p ˜¯ where ms, known as magnetic spin quantum number, is either +1/2 or –1/2. The electron having ms = 1/2 is labelled as a-spin and is represented by a vertical arrow pointing upward i.e. by (≠). The electron having ms = –1/2 is labelled as b-spin and is represented by a vertical arrow pointing downward i.e. by (Ø).

3.12 Complete Chemistry—JEE Main

Probable Distribution of Electron Around the Nucleus Fm2 )

Y 2n, l, m(=R2n , l Ql2, |m| evaluated at the given point.

Probable Distribution of s-orbitals For s-orbitals, Y 2n,0,0 μ R2n,0 as the angular dependence has a constant value. 1s orbital: The plots of R1,0 versus r and R21, 0 versus r are shown in Figs 5a and 5b.

Fig. 5a

In Figs 5a and 5b, the symbol a0 represents Bohr radius. Dot-Population Picture In dot-population, the value of relative probability at a given location is shown by the density of dots near that location. This picture gives the most realistic description of the electron’s time average distribution in the atom. Equal-Probability Contours In equal-probability contours, the points having identical probability are joined. The probability contour enclosing 90 to 95% probability is known as shape of orbital. Figures 5c and 5d display dot-population picture and shape of orbital of 1s orbital.

Fig. 5b

Fig. 5c

Fig. 5d

Radial Distribution Function nucleus, the total amount of the dot population near the nucleus is very small owing to the small volume near the nucleus. In order to visualise the total amount of the dot population within a spherical shell placed at radii r and r + dr from the nucleus, the radial distribution function F = (volume of the spherical shell) (probability density) = (4 p r2dr) (R2) Thus, the radial distribution functi the electron in a spherical shell of thickness dr located at the distance r from the nucleus.

Radial Function of 1s Orbital For 1s orbital, radial distribution function is shown in Fig. 5e. A maximum is observed at a0 which is exactly the same as that of Bohr radius. 2s orbital: For 2s orbital, the plots of R2,0, R22,0 and 4pr2R22,0 versus r are shown in Figs 6a, 6b and 6c.

Fig. 5e

Atomic Structure

Fig. 6a

Fig. 6b

3.13

Fig. 6c

Nodal Point The point where the function R has a zero value is known as spherical nodal point. R2,0 has one nodal point at r = 2a0. In general, the spherical nodal points in Rn, l function is equal fo (n – l – 1) Maxima in Radial Distribution Function In the plot of 4pr2R22, 0 versur r, there are two maxima at about 0.04 nm and 0.3 nm. The maxima at 0.04 nm is smaller than that at 0.3 nm. In general, the number of maxima in the radial distribution plot of s orbital is equal to the principal quantum number n. Of these, the maximum value is observed at the largest value of r.†

Dot-Population Picture and Equal-Probability contour For 2s orbitals, these are shown in Figs. 6d and 6e.

Probable Distribution of 2p Orbitals r-Dependence of 2p orbitals For 2p orbital, n = 2 and l = 1. Since r-dependent function Rn, l depends only on n and l, the same function will hold good for all the three values of m, namely, +1, 0 and –1. 2 Figures 7a, 7b and 7c show the plots of R2,1, R2, 1 and 2R2 versur r, respectively. 4pr 2, 1

Fig. 7a

Fig. 7b

Fig. 6d

Fig. 6e

Fig. 7c

† With the increase in the value of n, the largest maximum lies farther away from the nucleus. This implies that the size of orbital increases with increase in value of the principal quantum number n.

3.14 Complete Chemistry—JEE Main

Notable features of these plots are: There is no nodal points. This is in consistent with the general formula of n – l –1. R2, 1 has a zero value at r = 0 in contrast to R2, 0 which has a maximum value at r = 0. 2 at r = 4a There is one maximum in the plot of 4pr2R2, 1 0 2 )/dr equal to zero. d(r2R2.1 We have R2,1 μ r e-r / 2 a0 Hence This gives

Ê r4 ˆ d 2 2 d 4 - r / a0 (r R2, 1 ) = (r e ) = Á 4r 3 - ˜ e - r / a 0 = 0 a0 ¯ dr dr Ë r = 4a0

Angular Dependence of 2p orbitals For 2p orbitals, we have 6 cosq 2

Q1, 0 =

Q1, | ±1 | =

3 sinq 2

The angular dependence of an orbital is governed by the functions Q and F. 1

F0 =

2p

F+1 =

1 2p

e ij

and

F-1 =

1 2p

e - ij

Plot of Q1,0F0 or 2pz orbital There is no j dependence as F0 does not involve angle j. Twodimensional plot of the function Q1,0F0 versus q gives two circles in xz-plane (or yz-plane), one just above the xy-nodal plane and the other just below it (Fig. 8a). The shape of angular probability distribution is given by the plot of (Q1,0F0 )2 versus q. This is shown in Fig. 8b. Note The number of nodal surfaces in the angular distribution function is equal to the value of quantum number l. Shape of 2p 0 orbital The shape (i.e. 90% probability contour diagram) is obtained by combining Fig. 8a Fig. 8b the r-dependence and angular dependence functions. Figure 8c displays the shape of 2p0 orbital. It consists of two spheroidal lobes pointing in z-direction. Thus, 2p0 orbital is also known as 2pz orbital. The nucleus is situated between the two lobes and lie in the xy-nodal plane. Plots of 2px and 2py orbitals Since the function F+1 and F–1 carry the imaginary quantity i = -1 , these are converted into real functions by taking the following linear combinations. Fx = Fy =

1 2 -i 2

( F+1 + F-1 ) = ( F+1 - F-1 ) =

1 2 -i 2

1 2p 1 2p

(eij + e-ij ) =

1

(eij - e-ij ) =

1

p p

cos j sin j

When the functions are Fx and Fy are separately combined with Q1, |±1| function and plotted against q j equal to 0 and p, respectively), one gets the shapes similar to that shown in Fig. 8a, with the directions along x– and y– axes, respectively.

Fig. 8c

Atomic Structure

3.15

Shpes of 2px and 2py The shapes of 2px and 2py orbitals are of the appearance as shown in Fig. 8c, the two lobes point along x– and y– axis, respectively. (Figs 8d and 8e)

Fig. 8d

2px orbital

Fig. 8e

2py orbital

Shapes of 3d orbitals Proceeding similarly as in the case of 2p orbitals, one can draw the shapes of 3d orbitals. These are shown in Figs 9 (a-e). These include 2(=l axes (3dz2 and 3dx2 – y2 ), and (ii) orbitals having maximum probability distribution in between the two axes (3dxy, dxz and 3dyz).

Fig. 9a

3dz2

Fig. 9c

Fig. 9b

3dyz

Fig. 9d

3dxy

3dxz

3.16 Complete Chemistry—JEE Main

Fig. 9e

3dx2 – y2

Summary of Number and Types of Orbitals in an Atom In an atom, we have 1. Various quantum shells are governed by the principal quantum number, n, which can take values of 1,2,3, . . ., These shells are designated as K,L,M, . . ., and so on. 2. Each quantum shell consists of sub-shells which are characterized by the azimuthal quantum number, l. The number of sub-shells is equal to the value of n as l can take values of 0, 1, . . ., (n – 1). These sub-shells are designated as s, p, d, f, g, . . ., and so on. 3. Each sub-shells consists of orbitals which are characterized by the magnetic quantum number m. The number of orbitals is equal to 2l +1, since m can take values of +l through zero up to –l, i.e. +l, + (l – 1), . . ., 0. . ., –( l – 1), –l. 4. The total number of orbitals in a quantum shell is equal to n2. Illustration Permitted and not permitted orbitals in an atom. (i) n = 0; l = 0; m = 0 Not permitted as n cannot be equal to zero (ii) n = 2; l = 1, m = +1 Permissible (iii) n = 1; l = 1, m = 0 Not permitted as l cannot be equal to n. (iv) n = 1; l = 0, m = –1 Not permitted as |m| cannot be greater that l. (v) n = 2; l = 1, m = 1 Permissible General Comments on Quantum Numbers Quantum numbers n decides the size and energy of the orbital Quantum numbers l decides the shape of the orbital Quantum numbers m decides the orientation of the orbital General Comment about the Orbitals in a Multi-electron Atom Though the energy of orbitals in hydrogen-like species is decided by the quantum number n only, but in a multielectron atoms, it is decided by both the quantum numbers n and l. This is primarily due to electronic repulsions in the atom. In general The energy of an orbital increases with increase in the value of n + l. For the two orbitals with the same value of n + l, the orbital having higher value of n has the higher energy. The angular orientation of orbitals in the multi-electron atoms are the same as those in hydrogen-like species. Thus, the scheme of designation of orbitals remains the same, for example 1s, 2s, 2px, 2py, 2pz, 3s, 3px...etc.

1. Aufbau Principle The word aufbau is a German word which means ‘building up’. The principle is: The two rules to predict the relative energies of various orbitals as mentioned earlier are: (i) Energy increases with increase in the value of n + l. (ii) For the same value of n + l, the lower value of n has lower energy. Guided by these, the order of energies of orbitals are shown in Fig. 10

Atomic Structure

3.17

Fig. 10

2. Pauli Exclusion Principle The principle is: No two electrons in an atom have the same values for all the four quantum numbers n, l, m and ms.

According to this principle, only two electrons can be accommodated in an orbital since for the same orbital, the quantum numbers, n, l, and m have the same value and thus ms will differ having the values of +1/2 and –1/2, respectively. 3. Hund’s Rule: The rule is: Electrons with the same spin enter degenerate orbitals (which have identical energies) one by one till all of them are singly occupied. This is followed by pairing of electrons with opposite spins. Illustration Filling of 2p orbitals

Fig. 11

Explanation By occupying different orbitals with the same spin, electron-electron repulsive interactions is minimised because the magnetic dipoles generated by the spinning of electrons will have like poles farther away from each other causing lesser repulsion between them. The exception is

Illustrations: 24Cr 29Cu

(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)2 (3d)4 (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)2 (3d)9

24Cr 29Cu

(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)1 (3d)5 (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)1 (3d)10

Note: each other and their order of energies is reversed once these are occupied by electrons.

3.18 Complete Chemistry—JEE Main

MULTIPLE-CHOICE QUESTIONS ON SECTION 2 1. If the kinetic energy of a sub-atomic particle is increased 8 times, its de-Broglie wavelength becomes x times the original wavelength. The value of x is (d) 1/ 2 (a) 1/8 (b) 1/4 (c) 1/ 2 2 2. The stopping potential of the electrons emitted in a photoelectric experiment is V. The de-Broglie wavelength of the electron when it is emitted from the metal surface will be (b) h / 2emV (c) h 2 / 2emV (d) h / 2emV (a) h 2 / 2emV 3. The de-Broglie wavelength of the particle A is four times the wavelength of the particle B. If mass of A is two times the mass of B, then the ratio of kinetic energy of A and that of B will be (a) 1/16 (b) 1/8 (c) 1/32 (d) 1/ 32 4. The de-Broglie wavelength of electron in Bohr orbits displays the following trend with increase in the value of quantum number n. (a) Increase (b) Decrease (c) No change (d) Cannot be predicted 5. The de-Broglie wavelength of a proton travelling with a speed of 0.1% of the speed of light will be (a) 1.32 pm (b) 1.32 nm (c) 2.62 pm (d) 5.14 nm –27 kg) is 5 mm, the uncertainty in its speed 6. If the uncertainty in locating the position of a proton (m = 1.67 ¥ 10 will be (b) ≥ 6.3 ¥ 10–3 m s–1 (c) ≥ 6.3 ¥ 10–2 m s–1 (d) £ 6.3 ¥ 10–2 m s–1 (a) ≥ 9.3 ¥ 10–3 m s–1 7. The kinetic energy of electrons in a photoelectric effect is 1.3252 ¥ 10–19 J. If the threshold frequency of the metal is 5.50 ¥ 1014 Hz, the frequency of incident radiation is (b) 6.50 ¥ 1014 Hz (c) 7.50 ¥ 1014 Hz (d) 8.50 ¥ 1014 Hz (a) 4.50 ¥ 1014 Hz 15 Hz) is irradiated with the radiation of frequency 10.40 ¥ 1015 Hz. The 8. A metal (threshold frequency 8.0 ¥ 10 stopping potential of the emitted electrons is about (a) 67.6 V (b) 82.1 V (c) 120.1 V (d) 99.4 V 9. The number of spherical nodes in an orbital of hydrogen-like species is (a) n – l (b) n – l – 1 (c) n + l – 1 (d) n – l – 2 10. The number of angular nodes in an orbital of hydrogen-like species is (a) n (b) l (c) n – l (d) n + l 11. The total number of orbitals in M shell of an atom is (a) 1 (b) 4 (c) 9 (d) 16 12. Which of the following orbitals is not permitted in an atom? (a) n = 2, l = 0, m = 0 (b) n = 2, l = 1, m = 0 (c) n = 2, l = 0, m = 1 (d) n = 3, l = 2, m = 1 13. If q is the angle made by orbital angular momentum with the z-axis, then the quantum number l and m of an atom are related to each other through the expression (a) m = l cos q (b) m ={l(l + 1)}cos q (c) l = m cos q (d) m = l (l + 1) cos q 14. The total number of nodes in an orbital is equal to the value of (a) n (b) n – 1 (c) n – 2 (d) n + l – 1 15. In a sub-shell of an atom, the total number of allowed orbitals is equal to the value of (a) l (b) l + 1 (c) l + 2 (d) 2 l + 1 16. Which of the following species travelling with the same speed will have maximum de-Broglie wavelength? (a) H (b) D (c) T (d) a–particle 17. Consider the ground state of Cu atom (Z = 29). The number of electrons with the azimuthal quantum numbers l = 1 and 2 are, respectively, (a) 12, 5 (b) 12, 10 (c) 16, 6 (d) 14, 8

Atomic Structure

same energy? (i) n = 2, l = 1, m = 0 (ii) n = 3, l = 1, m = 1 (iii) n = 4, l = 0, m = 0 (a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) 19. Which of the following arrangement of orbitals having quantum numbers (i) n = 4, l = 2 (ii) n = 5, l = 1 (iii) n = 4, l = 3 is correct regarding their relative energies? (a) (i) < (ii) < (iii) (b) (i) < (iii) < (ii) (c) (ii) < (i) < (iii) 24Cr is (b) [Ar] (3d)4 (4s)2 (c) [Ar] (3d)6 (4s)0 (a) [Ar] (3d)5 (4s)1 47Ag is (a) 3d (b) 4d (c) 5d 22. Which of the following statements is correct? (b) The electron densities in the xy and yz planes are zero in 3dxz orbital. (c) The electron density in the xy plane in 3dz2 orbital is zero. (d) The electron density in the xy plane in 3dxy orbital is zero. 23. The number of unpaired electrons in 27Co is (a) 2 (b) 3 (c) 4 24. The numbers of spherical and angular nodes in 4f orbitals, respectively, are (a) 1, 3 (b) 1, 4 (c) 2, 3 25. Which of the following sets represents isoelectronic species? (b) Na+, F–, V3+ (c) K+, Cl–, Mg2+ (a) K+, Ca2+, Sc3+

3.19

(iv) n = 3, l = 1, m = 0 (d) (ii) and (iv)

(d) (ii) < (iii) < (i) (d) [Ar] (4d)5 (4s)1 (d) 6d

(d) 5 (d) 0, 3 (d) Cr3+, Fe2+, Co3+

ANSWERS 1. 7. 13. 19. 25.

(c) (c) (d) (a) (a)

2. 8. 14. 20.

(b) (d) (b) (a)

3. 9. 15. 21.

(c) (b) (d) (b)

4. 10. 16. 22.

(b) (b) (a) (b)

5. 11. 17. 23.

(a) (c) (b) (b)

6. 12. 18. 24.

(b) (c) (d) (d)

HINTS AND SOLUTIONS 1. de-Broglie relation is

mu = h/l. The kinetic energy becomes

2 8T1 l12 l12 l T2 l1 h2 2 1 fi = fi = fi l2 = 1 l Hence, = 2 mu 2 = 2 2 2 T 8 l2 2 2 T1 l2 1 2 ml 2 2. The kinetic energy of the electron emitted from the metal surface will be h p2 1 fi l= Hence, 2 m eV = = eV fi p 2 = 2 m eV mu 2 = eV fi l 2m 2 TA mB lB2 Ê 1 ˆ Ê 1 ˆ 2 1 = =Á ˜Á ˜ = 3. The ratio will be 32 TB mA lA2 Ë 2 ¯ Ë 4 ¯

T=

h 2 m eV

4. Since in Bohr theory p = n(h/2p), increase in p with increase in the value of n will be associated with decrease in de-Broglie wavelength. In fact, the expression of wavelength is l = 2p/n .

3.20 Complete Chemistry—JEE Main

5. l =

h (6.626 ¥ 10-34 J s) = = 1.32 ¥ 10–12 m = 1.32 pm mu (1.672 ¥ 10-27 kg)(3 ¥ 105 m s -1 )

6. Du ≥ 7. Since

h Ê 1 ˆ (6.626 ¥ 10-34 J s) = = 6.3 ¥ 10–3 m s–1 4p ÁË mp Dx ˜¯ (4)(3.14)(1.67 ¥ 10-27 kg)(5 ¥ 10-6 m) KE = hn – hn0, n=

we have

KE (1.3252 ¥ 10-19 J ) + n0 = + 5.50 ¥ 1014 s -1 = 2.0 ¥ 1014 s–1 + 5.50 ¥ 1014 s–1 = 7.50 ¥ 1014 s–1 h (6.626 ¥ 10-34 J s)

8. |eV| = hn – hn0 h(v - n 0 ) (6.626 ¥ 10-34 J s)(2.40 ¥ 1016 s -1 ) = V= = 99.4 V e (1.6 ¥ 10-19 C) 9. The number of spherical nodes (i.e. nodes in the plot of Rn, l versus r) is n – l – 1. 10. The number of angular node is equal to the value of l (azimuthal quantum number). 11. For M shell of an atom, n = 3. The total number of orbital in a shell is equal to the value of n2. 12. The value of m cannot be greater than the value of l. Lz = m(h/2p) and Lz =L cos q. Hence, m= l (l + 1) cos q L = l (l + 1) (h/2p), Total number of nodes = (spherical + angular) nodes = n – l – 1 + l = n – 1 The total number of orbitals in a sub-shell is equal to the number of allowed values of magnetic quantum number. de-Broglie wave length is inversely proportional to the mass of the species travelling with the same speed. (l = h/mu) 2 2 6 2 6 10 1 29Cu is (1s) (2s) (2p) (3s) (3p) (3d) (4s) l =1 corresponds to p orbital. The number of electrons in p orbitals is 6 + 6 = 12 l =2 corresponds to d orbital. The number of electrons in d orbitals is 10 18. Orbitals having the same value of n + l will have the same energy irrespective of the value of m. 19. Larger the value of n + l, larger the energy. For the same value of n + l, larger the value of n, larger the energy. 5 1 24Cr is [Ar] (3d) (4s) atom. 10(5s)1. 47 22. 3dxz involves xy and yz nodal planes. 7 2 27Co is [Ar] (3d) (4s) . Thus, it contains 3 unpaired electrons in 3d orbitals 13. 14. 15. 16.

. 24. Spherical nodes = n – l – 1 = 4 – 3 –1 = 0 Angular nodes = l = 3 + + 3+ 25. K , Ca , Sc all contain 18 electrons. The electrons in rest of species are: Na+(10), F–(10), V3+(20), Cl–(18), Mg2+(10), Cr5+(21), Fe+(24) and Co3+(24).

Atomic Structure

3.21

MULTIPLE CHOICE QUESTIONS FOR THE ENTIRE CHAPTER General Characteristics 1. The approximate radii of the nuclei of atoms lie in the range of (b) 10 13 m 10 14 m (c) 10 14 m 10 15 m (d) 10 15 m 10 16 m (a) 10 12 m 10 13 m 2. The increasing order for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle (a) is (a) e, p, n, a (b) n, p, e, a (c) n, p, a, e (d) n, a, p, e 3. The fundamental particle responsible for keeping the components of nucleus together is (a) meson (b) neutron (c) positron (d) antiproton Ge is 4. An isotone of 76 32 Ge (b) 77 (c) 77 (d) 76 (a) 77 32 33As 34Se 34Se 5. The number of neutrons in dipositive zinc ion (Z = 30) with mass number 70 is (a) 34 (b) 36 (c) 38 (d) 40 6. The mass of a neutron is (a) equal to the mass of a proton (b) greater than the mass of a proton (c) smaller than the mass of a proton (d) equal to the sum of masses of a proton and an electron 7. The mass of one mole of electrons is (a) 0.548 g (b) 0.548 mg (c) 0.548 cg (d) 0.548 dg 8. The number of electrons carrying a total charge of 1 C is (b) 6.25 ¥ 1018 (c) 6.24 ¥ 1020 (d) 6.023 ¥ 1022 (a) 6.023 ¥ 1023 7 –12 9. The binding energy of 3Li atom is 6.258 ¥ 10 J. If atomic masses of electron, proton and neutron are 0.0005 u, 1.0073 u and 1.0086 u, respectively, the actual mass of Li atom is (a) 6.899 u (b) 6.985 u (c) 7.016 u (d) 7.312 u 10. Naturally occurring copper is assigned atomic mass equal to 63.546 u. If it consists of two isotopes 63Cu (atomic mass = 62.930 u) and 65Cu (atomic mass = 64.928 u), then mass percentage of isotope 65Cu in the naturally occurring copper is (a) 30.83 (b) 41.15 (c) 56.2 (d) 59.1

Bohr Theory 11. The expression for the Bohr radius of hydrogen-like species is ˘ È h2 (a) n2 Í 2 ˙ 2 ÍÎ 4p m Ze 4p e 0 ˙˚

(

(b)

)

1 n2

˘ È h2 Í 2 ˙ 2 ÍÎ 4p m Ze 4pe 0 ˙˚

(

)

È 4p 2 m h 2 ˘ 1 È 4p 2 m h 2 ˘ (c) n2 Í 2 (d) 2 Í 2 ˙ ˙ n Î Ze 4 pe 0 ˚ Î Ze 4 p e 0 ˚ 12. The expression for the energy of electron in the hydrogen-like species is

(

2 È 2 1 Í 2p m Ze 4pe 0 (a) 2 n Í h2 Î

(c)

È 1 Í 2p 2 m h 2 n 2 Í Ze2 4 pe 0 Î

(

)

˘ ˙ 2˙ ˚

)

2

˘ ˙ ˙ ˚

(

2 È 2 1 Í 2p m Ze 4pe 0 (b) - 2 n Í h2 Î

(d) -

È 1 Í 2p 2 m h 2 n 2 Í Ze2 4 pe 0 Î

(

)

˘ ˙ 2˙ ˚

)

2

˘ ˙ ˙ ˚

3.22 Complete Chemistry—JEE Main

13. The expression of Rydberg constant is (a)

2p 2 m h3 c

(e 2

4p e 0 )

2

(b)

2p 2 m (e 2 4p e 0 ) h3 c

2

(c)

(e 2

4p e 0 )

2

2p 2 m h3 c

14. Which of the following atoms has the largest atomic radius? (a) 3Li (b) 5B (c) 7N

(d)

2 p 2 m h3 c e 2 4p e 0

(d) 8F

15. The speed of an electron in the innermost orbit of the hydrogen atom (Bohr radius = 52.9 pm, me = 9.11 10–31 kg) is (b) 2.19 106 m s–1 (c) 2.19 107 m s–1 (d) 2.19 108 m s–1 (a) 2.19 104 m s–1 16. The ionization energy of hydrogen atom in the ground state is 2.17 10–11 erg. The ionization energy of Li2+ in the ground state will be (a) 1.953 10–15 J (b) 1.953 10–16 J (c) 1.953 10–17 J (d) 1.953 10–18 J 17. (a) 52.9 cm (b) 52.9 nm (c) 52.9 pm (d) 52.9 Å 18. With increasing quantum number, the energy difference between adjacent levels of hydrogen atom (a) increases (b) decreases 19. The wavelength of radiation required to remove the electron of hydrogen atom (ionization energy 21.7 10–12 erg) from n = 2 orbit to n = • is (b) 3.664 10–5 cm (c) 3.664 10–6 cm (d) 3.664 10–7 cm (a) 3.664 10–4 cm 20. The angular momentum of an electron in the second Bohr orbit is given by (c) 2(h/2p) (d) 2(2 + 1) (h/2p) (a) 1(h/2p) (b) 2 (h/2p) 21. The energy difference between the two orbits of a hydrogen atom is Ê 1 1ˆ DE = R•hc Á 2 - 2 ˜ Ë n1 n2 ¯ For the Lyman series, the values of n1 and n2 are given as (b) n1 = 2, n2 = 3, 4, 5, … (a) n1 = 1, n2 = 2, 3, 4, … (d) n1 = 4, n2 = 5, 6, 7, … (c) n1 = 3, n2 = 4, 5, 6, … 22. For the Balmer series, the values of n1 and n2 in the expression Ê 1 1ˆ DE = R•hc Á 2 - 2 ˜ Ë n1 n2 ¯

are (a) n1 = 1, n2 = 2, 3, 4, … (b) n1 = 2, n2 = 3, 4, 5, … (d) n1 = 4, n2 = 5, 6, 7, … (c) n1 = 3, n2 = 4, 5, 6, … 23. For the Paschen series, the values of n1 and n2 in the expression Ê 1 1ˆ DE = R•hc Á 2 - 2 ˜ Ë n1 n2 ¯

are (b) n1 = 2, n2 = 3, 4, 5, … (a) n1 = 1, n2 = 2, 3, 4, … (c) n1 = 3, n2 = 4, 5, 6, … (d) n1 = 4, n2 = 5, 6, … 24. For the Brackett series, the values of n1 and n2 in the expression

are (a) n1 = 1, n2 = 2, 3, 4, … (c) n1 = 3, n2 = 4, 5, 6, …

Ê 1 1ˆ DE = R•hc Á 2 - 2 ˜ Ë n1 n2 ¯ (b) n1 = 2, n2 = 3, 4, 5, … (d) n1 = 4, n2 = 5, 6, 7, …

Atomic Structure

3.23

25. For the Pfund series, the values of n1 and n2 in the expression Ê 1 1ˆ DE = R•hc Á 2 - 2 ˜ Ë n1 n2 ¯

26.

27.

28.

29.

30. 31. 32.

33.

are (b) n1 = 3, n2 = 4, 5, 6, … (a) n1 = 2, n2 = 3, 4, 5, … (d) n1 = 5, n2 = 6, 7, … (c) n1 = 4, n2 = 5, 6, 7, … Which of the following statements is not true? (a) Lyman spectral series of hydrogen atom lies in the ultraviolet region of electromagnetic radiation (b) Balmer spectral series of hydrogen atom lies in the visible region of electromagnetic radiation (c) Pashen spectral series of hydrogen atom lies in the visible region of electromagnetic radiation (d) Brackett spectral series of hydrogen atom lies in the infrared region of electromagnetic radiation 106 m s–1. Its velocity in the second orbit would be (b) 4.38 106 m s–1 (c) 5.5 105 m s–1 (d) 8.76 106 m s–1 (a) 1.10 106 m s–1 10–18 J. Its energy in the second orbit would be (a) – 1.09 10–18 J (b) – 4.36 10–18 J (c) – 5.45 10–19 J (d) – 8.72 10–18 J Hydrogen atom in its ground state is excited by a radiation of wavelength 97.26 nm. The longest wavelength it emits is (a) 1250 nm (b) 1400 nm (c) 1875 nm (d) 2050 nm The ground state electronic energy in Be3+ ion is (b) 3.49 ¥ 10–15 J (c) 3.49 ¥ 10–16 J (d) 3.49 ¥ 10–17 J (a) 3.49 ¥ 10–14 J In Bohr theory, the ratio of kinetic energy and potential energy of an electron in any of orbit of hydrogen atom is (a) 1/2 (b) – 1/2 (c) 1/4 (d) – 1/4 Which of the following statements about Bohr theory of hydrogen atom is not correct? (a) Centripetal force acting on the electron is Ze2/(4p e0 r). (b) Centrifugal force acting on the electron is mv2/r (c) Angular momentum of electron is n (h/2p) (d) Bohr radius is equal to 52.9 pm. The expression of speed of electron in Bohr radii is given by (a) v =

n ( h / 2p ) Ze2 /(4pe 0 )

(b) v =

(c) v =

n 2 ( h / 2p ) Ze2 /(4pe 0 )

(d) v =

2p ( Ze2 / 4p e 0 ) nh 4p 2 ( Ze2 / 42p e 0 ) n2 h2

34. Which of the following statements regarding Bohr theory of hydrogen atom is not correct? (a) Kinetic energy of an electron in an orbit is half of magnitude of its potential energy. (b) Kinetic energy of an electron in an orbit is equal to the magnitude of its potential energy. (c) Total energy of an electron is negative of its kinetic energy. (d) Potential energy of an electron in an orbit is equal to – mv2. 35. The ionization energy of one mole of hydrogen atoms is about (a) 1313 J (b) 1313 kJ (c) 1616 J (d) 1616 kJ 36. In Bohr theory, which of the following expression for the energy of an electron in Bohr orbit is correct? (b) E μ – Z/n2 (c) E μ – n2/Z (d) E μ – Z2/n2 (a) E μ Z/n2 37. In Bohr theory, which of the following expression for the speed of an electron in Bohr orbit is correct? (a) v μ Z/n (b) v μ n/Z (c) v μ Z2/n2 (d) v μ n2/Z2

3.24 Complete Chemistry—JEE Main

38. In Bohr theory, which of the following expression for the radii of Bohr orbits is correct? (c) r μ n2/Z (d) r μ Z2/n2 (a) r μ n/Z (b) r μ n2/Z2 39. Which of the following spectral series of hydrogen atom lies in the ultraviolet region of electromagnetic radiation? (a) Lyman (b) Balmer (c) Paschen (d) Brackett 40. Which of the following spectral series of hydrogen atom lies in the far infrared region of electromagnetic radiation? (a) Lyman (b) Balmer (c) Paschen (d) Pfund 41. Which of the following spectral series of hydrogen atom lies in the visible region of electromagnetic radiation? (a) Lyman (b) Balmer (c) Paschen (d) Brackett + will have the same wavelength as that 42. Which of the following electronic transitions in the line spectrum of He (a) 4 ¨æ 2 (b) 2 ¨æ 1 (c) 3 ¨æ 1 (d) 4 ¨æ 1 43. The kinetic energy of an electron in He+ is maximum when it is present in the orbit having (a) n = 1 (b) n = 2 (c) n = 3 (d) n = • 44. The longest wavelength in Balmer series is (a) 656.5 nm (b) 770 nm (c) 360 nm (d) 450 nm (a) Stark effect (b) Zeeman effect (c) Bohr effect (d) Anti-Zeeman effect 46. In Bohr theory of He+, the energy difference between two successive energy levels (a) increases with increase in the value of n (b) decreases with increase in the value of n (c) remains constant with increase in the value of n (d) initially increases followed by a decrease with increase in the value of n 2+ ion will be r (a) 3r (b) r/3 (c) 9r (d) r/9 2+ ion relative that of electron in the third Bohr orbit of H atom is (a) 1 (b) 2 (c) 4 (d) 9

Quantum Numbers and Orbitals 49. The shape of an orbital is governed by the (a) principal quantum number (b) azimuthal quantum number (c) magnetic quantum number (d) spin quantum number 50. The orientation of an orbital in an atom is governed by the (a) principal quantum number (b) azimuthal quantum number (c) magnetic quantum number (d) spin quantum number 51. For a given value of principal quantum number n, the number of allowed values of azimuthal quantum number l, is given by (a) n 2 (b) n (c) n 1 (d) n2 52. For a given value of azimuthal quantum number l, the number of allowed values of magnetic quantum number m, is given by (a) l + 1 (b) l + 2 (c) 2l + 1 (d) 2l + 2 53. Which of the following sets of quantum numbers is not allowed? (a) n = 2, l = 0, m = + 1 (b) n = 2, l = 1, m = + 1 (c) n = 2, l = 0, m = 0 (d) n = 2, l = 1, m = 1 54. The value of Rydberg constant is (b) 1.09678 106 m–1 (a) 1.09678 105 m–1 7 –1 (c) 1.09678 10 m (d) 1.09678 108 m–1

Atomic Structure

55. Which of the following orbitals is symmetric about the y-axis? (b) py (c) dyz (a) px

(d) dxy

(a) n = 3 (b) l = 2 (c) m = 2 (d) s = 1/2 57. Which of the following quantum numbers n, l, m and ms, respectively, is not possible? (a) 2, 1, 0, – 1/2 (b) 2, 2, 0, 1/2 (c) 3, 2, 1, 1/2 (d) 3, 2, 0, 1/2 58. How many unpaired electrons are present in Ni2+ (Z = 28)? (a) 0 (b) 2 (c) 4 (d) 8 not theoretically possible? (b) 3p6 (c) 4s1 (d) 4f12 (a) 2d5 60. The maximum number of electrons that can be accommodated in a quantum shell is equal to (c) 2n2 (d) n(n + 1) (a) n (b) n2 61. The number of unpaired electrons present in the ground state of Cr is (a) 3 (b) 4 (c) 5 (d) 6 62. Which of the electrons is most tightly bound to the nucleus? (a) 4s (b) 4p (c) 4d (d) 4f 63. Which of the following atoms would be expected to be most paramagnetic? (b) 4Be (c) 5B (d) 6C (a) 3Li 64. Tick the correct statement on the aufbau principle? (a) (n – 1)d subshell is always lower in energy than ns subshell (b) (n – 1)d subshell always has energy more than ns subshell (c) 5d is lower in energy than 4f (d) 5f is lower in energy than 7s Z = 58) is 10 (5s)2 (5p)6 (4f)3 (6s)1 (b) (5d)10 (5s)2 (5p)6 (5d)1 (6s)2 (a) (4d) 10 2 6 3 2 (c) (4d) (5s) (5p) (5d) (6s) (d) (4d)10 (5s)2 (5p)6 (4f)2 (5d)0 (6s)2 66. Aufbau order is not violated in

67. Which of the following statements is not correct?

(c) Aufbau order is not obeyed in cases where energy difference between ns and (n – 1)d subshells is larger. 5

(a) 24 (b) 25 (c) 26 69. The number of electrons present in the l = 2 orbital of Cr is (a) 3 (b) 4 (c) 5 70. Which of the isoelectronic species has more electrons than neutrons? (b) Ne (c) Na+ (a) O2– 71. The number of electrons with l = 2 in an atom having atomic number 23 is (a) 2 (b) 3 (c) 4

(4s)1 is (d) 27 (d) 6 (d) Mg2+ (d) 5

3.25

3.26 Complete Chemistry—JEE Main

72. The ion having 18 electrons in the outermost shell is (b) Al 3+ (Z = 13) (c) K+ (Z = 19) (a) Cu+ (Z = 29) 73. The number of unpaired electrons present in an atom with atomic number 23 is (a) 1 (b) 2 (c) 3

(d) As3+ (Z = 33) (d) 5

Quantum Mechanical Approach (a) at the nucleus (b) at the Bohr radius (c) at a large distance from the nucleus 75. The number of nodal points in the plot of Rn, 0 versus r is equal to the value of (a) n (b) l (c) n – l (d) n – (l + 1) 76. For 2s orbital, the nodal surface exists at the distance (b) 1.5 a0 from the nucleus (a) a0 from the nucleus (c) 2 a0 yz plane? (b) py (c) pz (d) dyz (a) px 78. The number of radial nodes in the 3p probability density distribution is (a) 0 (b) 1 (c) 2 (d) 3 79. The angular dependence of an orbital is decided by the (a) principal quantum number only (b) azimuthal quantum number only (c) magnetic quantum number only (d) both azimuthal and magnetic quantum numbers 80. Which of the following symbols correctly represents an orbitals in an atom? (b) n, l, m = Rn, l l m (c) n, l, m = Rn, l l, m m (d) n, l, m = Rn, l l, |m| m (a) n, l, m = Rn l m 81. The angular momentum of an electron in an atomic orbital is governed by the (a) principal quantum number (b) azimuthal quantum number (c) magnetic quantum number (d) spin quantum number 82. The z-component of angular momentum of an electron in an atomic orbital is governed by the (a) azimuthal quantum number (b) principal quantum number (c) magnetic quantum number (d) spin quantum number 83. An orbital with l = 0 is (a) symmetrical about the nucleus (b) symmetrical about the x-axis only (c) symmetrical about the y-axis only (d) symmetrical about the z-axis only 84. The energy of an orbital in a multielectron atom depends on the (a) principal quantum number only (b) principal and azimuthal quantum numbers only (c) principal, azimuthal and magnetic quantum numbers (d) principal, azimuthal, magnetic and spin quantum numbers 85. For silver metal, threshold frequency for the emission of photoelectron is 1.13 1017 s–1. The kinetic energy of electrons emitted when silver is irradiated with wavelength 1.5 nm is (b) 5.76 10–16 J (c) 5.76 10–15 J (d) 5.76 10–14 J (a) 5.76 10–17 J 86. The binding energy of electrons in a metal is 193 kJ mol–1. The threshold frequency of the metal is (b) 4.83 1012 Hz (c) 4.83 1014 Hz (d) 4.83 1016 Hz (a) 4.83 1010 Hz 87. The momentum of a particle having de Broglie wavelength of 0.1 nm is (a) 6.626 10–21 kg m s–1 (b) 6.626 10–22 kg m s–1 (c) 6.626 10–23 kg m s–1 (d) 6.626 10–24 kg m s–1

Atomic Structure

3.27

88. An electron is accelerated through a potential difference of 500 V. Its de Broglie wavelength would be (a) 55 pm (b) 5.5 pm (c) 0.55 pm (d) 55 nm 89. A proton is accelerated to one tenth of the velocity of light. If its velocity can be measured with a precision of 1%, then the uncertainty in its position is equal to or greater than (a) 1.93 nm (b) 19.3 nm (c) 19.3 pm (d) 193 pm 90. The angular momentum of an electron in f orbital is (b) 6 (h / 2p) (c) 2 (h / 2p) (d) 20 (h / 2p) (a) 12 (h / 2p) 91. Which of the following statements on the atomic wave function is not correct? (a) may be a real valued wave function (b) may be in some cases be a complex function (c) (d) 92. Which of the following statements on the square of atomic wave function at a point near to the nucleus is not correct? (a) 2 may be positive, negative or imaginary (b) 2 is proportional to electron density (c) 2 is a normalized wave function (d) 2 93. Which of the following statements on quantum numbers is not correct? (a) Quantum numbers n, l, m and ms are needed to describe an electron in an atom completely. (b) Quantum numbers n, l, m and s are obtained by solving the Schrödinger wave equation. (c) A subshell in an atom can be designated with two quantum numbers n and l. (d) The maximum value of l is equal to n – 1 and that of m is ± l. 94. Which of the following statements is not correct? (a) The wave function depicting the dependence on r involves two quantum numbers n and l. (b) The wave function depicting the angular dependence involves two quantum numbers l and m. (c) The spin quantum number is not the outcome of the Schrödinger equation. (d) The lowest energy state of an atom corresponds to n = 0. 95. In an orbital, the signs of lobes indicate the (a) sign of the wave function (b) sign of the probability distribution (c) presence or absence of electron (d) sign of charge 96. f orbitals are characterised by the quantum number (a) n = 4 (b) l = 3 (c) m = 3 (d) s = 1/2 97. Which of the following sets of quantum numbers is not allowed? (a) n = 2, l = 1, m = 1 (b) n = 2, l = 1, m = 0 (c) n = 2, l = 2, m = 1 (d) n = 3, l = 2, m = – 1 2 2 98. In the plot of r Y versus r for 1s orbital of hydrogen atom, maximum occurs at (c) r = 2a0 (d) r = • (a) r = 0 (b) r = a0 (Bohr radius) 99. The plot of r2Y 2 versus r for 2s orbital of hydrogen atom exhibits (a) one maximum (b) two maxima (c) no maximum (d) three maxima 100. The angular momentum of an electron in p orbitals is equal to (a) zero (b) (h/2p) (c) 2 (h/2p) (d) 6 (h/2p) 101. The angular momentum of an electron in d orbitals is equal to (a) zero (b) (h/2p) (c) 2 (h/2p) (d) 6 (h/2p) 102. The pz orbital corresponds to the magnetic quantum m equal to (a) 0 (b) + 1 (c) – 1 (d) + 2

3.28 Complete Chemistry—JEE Main

103. The magnetic quantum number corresponding to dz2 orbital is (a) + 2 (b) + 1 (c) 0 (d) – 2 104. For which of the following species, the electronic distribution is spherically symmetrical? (d) B (a) H (b) Na (c) Cl – (a) at the nucleus (b) at a distance equal to Bohr radius from the nucleus

106.

107.

108.

109.

110. 111. 112.

(d) at a distance equal to twice of the Bohr radius from the nucleus. A dyz orbital has (a) no nodal plane (b) one nodal plane in the yz plane (c) two nodal planes in the xy and xz planes (d) Three nodal planes in the xy, xz and yz planes. A dx2 – y2 orbital has (a) no nodal plane (b) one nodal plane in the xy plane (c) two nodal planes in the xz and yz planes (d) two nodal planes at angles 45° to x and y axes. A dyz orbital is directed (a) along the y– and z–axes (b) along the x– and y–axes (c) along the centre of y– and z–axes (d) along the centre of x– and y–axes A dxy orbital is directed (a) along the x– and y–axes (b) along the z–axis (c) along the centre of x and z–axes (d) along the centre of x– and y–axes If an electron is to be located within 0.1 nm, the approximate uncertainty in its speed will be about (b) 6 ¥ 105 m s–1 (c) 10–2 m s–1 (d) 102 m s–1 (a) 10–5 m s–1 The number of radial nodes in 4s orbital is (a) 2 (b) 3 (c) 4 (d) 5 The wavelength of an electron accelerated by 100 V of potential difference is about (a) 122.8 pm (b) 61.4 pm (c) 30.7 pm (d) 15.35 pm The r-dependent wave function of hydrogen atom is given by the expression Ê 2r 2r 2 ˆ + exp (– r/3 a0) R = (constant) Á 3 a0 9 a02 ˜¯ Ë

Answer the following four questions. 113. The above wave function depends on the quantum numbers (a) n (b) n and l (c) n, l and m 114. The number of nodes in this wave functions is/are (a) 0 (b) 1 (c) 2 115. The above wave function represents (a) 1s orbital (b) 2s orbital (c) 3s orbital 116. One of the nodes lies at the distance r from the nucleus, where r is about (a) 1.5a0 (b) 2.5a0 (c) 5.5a0 117. One of the maximum value of R2 lies approximately at (b) r = 4.0a0 (c) r = 11.47a0 (a) r = 2.0a0 The r-dependent wave function of hydrogen atom is given by the expression Ê 2r r 2 ˆ R = (constant) Á - 2 ˜ exp (r/3a0) Ë a0 3a0 ¯ Answers the following questions.

(d) n, l, m and s (d) 3 (d) 3p orbital (d) 7.1a0 (d) r = 12.5a0

Atomic Structure

118. The number of nodes (besides at r = 0) in this wave function is (a) 0 (b) 1 (c) 2 119. The above wave function represents (a) 2s (b) 2p (c) 3p 120. The node (besides at r = 0) occurs at (b) 4a0

(a) 2a0

(d) 3 (d) 3s

(c) 6a0

(d) 8a0

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79. 85. 91. 97. 103. 109. 115.

(c) (b) (b) (b) (d) (b) (a) (a) (b) (b) (d) (c) (c) (d) (a) (d) (c) (c) (d) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 86. 92. 98. 104. 110. 116.

(d) (b) (a) (c) (c) (a) (c) (a) (c) (b) (a) (c) (a) (d) (c) (a) (b) (c) (b) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75. 81. 87. 93. 99. 105. 111. 117.

(a) (c) (b) (a) (a) (b) (a) (b) (b) (b) (d) (c) (d) (b) (d) (b) (b) (a) (b) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76. 82. 88. 94. 100. 106. 112. 118.

(b) (a) (c) (b) (c) (b) (d) (b) (c) (b) (b) (a) (c) (c) (a) (d) (c) (c) (a) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77. 83. 89. 95. 101. 107. 113. 119.

(d) (a) (c) (c) (c) (b) (b) (b) (a) (a) (d) (b) (a) (a) (d) (a) (d) (d) (b) (c)

HINTS AND SOLUTIONS 4. Isotones have same number of neutrons (= mass number – atomic number). 7. m = NA me = (6.022 ¥ 1023 mol–1) (9.1 ¥ 10–31 kg) = 5.48 ¥ 10–7 kg ∫∫ 0.548 mg 8. N = 1 C/e = 1 C/(1.6 ¥ 10–19 C) = 6.25 ¥ 1018 9. 73Li contains, 3p, 4n and 3e. Their sum of masses is m¢ = [3(1.0073 u) + 4 (1.0086) + 3(.0005)] u = 7.0578 u Mass corresponding to mass defect is D m = E/c2 = (6.258 ¥ 10–12 J)/(3 ¥ 108 m s–1)2 = 6.953 ¥ 10–29 kg In atomic unit, we have D m¢ =

Dm 6.953 ¥ 10-29 = u = 0.0419 u 1.66 ¥ 10-27 kg / u 1.66 ¥ 10-27

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 90. 96. 102. 108. 114. 120.

(b) (b) (b) (d) (d) (d) (a) (d) (c) (c) (d) (a) (b) (b) (a) (b) (a) (c) (c) (c)

3.29

3.30 Complete Chemistry—JEE Main

Actual mass of Li is m = m¢ – D m¢ = (7.0578 – 0.0419 u) = 7.0159 u 10. We have 63.546 u = x (62.930 u) + (1 – x) (64.928 u) x=

Hence

64.928 - 63.546 = 0.69 64.928 - 62.930

Mass per cent of 65Cu = 0.31 ¥ 100 = 31% 14. Atomic size decreases along the period.

(

)

2 ˘ È 1.602 ¥ 10-19 C 2p Ê e 2 ˆ 2 (3.14) Í ˙ = 2.19 15. d = = h ÁË 4 pe 0 ˜¯ 6.626 ¥ 10-34 J s Í1.112 ¥ 10-10 C2 N -1 m -2 ˙ Î ˚

(

)

16. ELi = Z2EH = (32) (2.17

10–11 erg) = 1.953

È h2 17. r = n Í 2 2 ÍÎ 4 p m e 4 p e 0

˘ Ê e h2 ˆ ˙ = n 2 Á 0 2 ˜ = 12 Ë pme ¯ ˙˚

2

(

= 5.29

)

2

=

(

10–17 J

)(

È 8.854 ¥ 10-12 C2 N -1 m -2 6.626 ¥ 10-34 J s Í 2 Í (3.14) 9.1 ¥ 10-31 kg 1.602 ¥ 10-19 C Î

(

)(

)

)

2

˘ ˙ ˙ ˚

10–11 m = 52.9 pm

È1 1˘ 18. DE = R•hc Í 2 - 2 ˙ ; Î n1 n2 ˚ 19. DE•

10–10 erg = 1.953

106 m s–1

1 DE• 22

1

=

(

Ê 1 1ˆ ÁË n 2 - n 2 ˜¯ decreases with increase in the value of n1 and n2 (= n1 + 1) 1 2 1 (21.7 4

10–12 erg) = 5.425

)(

10–12 erg = 5.425

)

6.626 ¥ 10-34 J s 3 ¥ 108 m s -1 hc = l= = 3.664 DE 5.425 ¥ 10-19 J

(

)

10–7 m

10–19 J

3.664

10–5 cm

26. Paschen spectral series lies in the near infrared region of electromagnetic radiation. 27. The expression of velocity is 1 Ê 1 Ze2 ˆ ; n ÁË 2 h e 0 ˜¯

d=

Thus

v2 n1 = v1 n2

d = (2.19

1 106 m s–1) ÊÁ ˆ˜ = 1.10 Ë 2¯

106 m s–1

28. The expression of energy is E=

29. DE =

1 (constant). Thus n2

Ê n2 ˆ E2 = E1 Á 12 ˜ = (– 2.18 Ë n2 ¯

1 10–18 J) ÊÁ ˆ˜ = – 5.45 Ë 4¯

hc (6.626 ¥ 10-34 J s) (3 ¥ 108 m s -1 ) = = 2.0438 ¥ 10–18 J l (97.26 ¥ 10-9 m)

Also

Ê1 1ˆ DE = (2.18 ¥ 10–18 J) Á 2 - 2 ˜ Ë1 n2 ¯

Hence

Ê 1 DE =1– Á 2 Ë 2.18 ¥ 10-18 n2 n2 =

ˆ 2.0436 ¥ 10-18 =1– = 1 – 0.9374 ˜ 2.18 ¥ 10-18 J¯

1 / (1 - 0.9374)  4

10–19 J

Atomic Structure

3.31

The transition n2 = 4 Æ n1 = 3 will emit the longest wavelength. Hence 1 1 DE = (2.18 ¥ 10–18 J) ÊÁ 2 - 2 ˆ˜ = 1.06 ¥ 10–19 J Ë3 4 ¯ l=

hc (6.626 ¥ 10-34 J s) (3 ¥ 108 m s -1 ) = = 1.875 ¥ 10–6 m = 1875 nm DE (1.06 ¥ 10-19 J)

30. Ground state energy of Be3+ = Z2EH = (16) (2.18 ¥ 10–18 J) = 3.49 ¥ 10–17 J 31. KE =

1 mu 2 2

and

32. Centripetal force is

PE =

Z e2 = – mu2. (4 p e 0 )r

Therefore

KE = 1 PE 2

Ze2/(4 p e0 r2).

35. IE = NAR • hc = (6.022 ¥ 1023 mol–1) (1.09678 ¥ 107 m–1) (6.626 ¥ 10–34 J s) (3 ¥ 10 8 m s–1) = 1.313 ¥ 106 J mol –1 = 1313 kJ mol –1 42. First line in the Lyman series of hydrogen atom corresponds to 1 1 3 DE = R• hc ÊÁ 2 - 2 ˆ˜ = R•hc ¯ Ë1 4 2 For He +, the spectral line occurs at Ê 1 1ˆ 1 1 5 R•hc DE = Z2 R• hc Á 2 - 2 ˜ = (22)R• hc ÊÁ 2 - 2 ˆ˜ = Ë2 36 Ë n1 n 2 ¯ 4 ¯ 43. Velocity of electron decreases with increase in the value of n in Bohr theory. The electron will have maximum kinetic energy in the n = 1 orbit. 44. For Balmer series, the longest wavelength corresponds to the transition 3 ¨ 2. Ê 1 1ˆ 1 1 5 R• hc DE = R• hc Á 2 - 2 ˜ = R• hc ÊÁ 2 - 2 ˆ˜ = Ë2 36 Ë n1 n 2 ¯ 3 ¯ l=

hc 36 36 = = = 6.565 ¥ 10–7 m = 656.6 nm DE 5 R• 1.09678 ¥ 107 m -1

46. The expression of energy is K E = - 2 ; where K is constant n È 1 1˘ (2n + 1) K Hence DE = En + 1 – En = – K Í - 2˙ = 2 n ˚ [n ( n + 1)]2 Î ( n + 1) For

n=1 n=2 n=3

DE = 3K/4 = 0.75 K DE = 5K/36 = 0.14 K DE = 7K/144 = 0.049 K

47. The expression of radius is r 2+ n2 (12 / 3) K 1 K; where K is constant. Hence Li = 2 = Z rH+ (1 /1) K 3 48. The expression of speed is Z u (Li 2 + ) 3 K /1 K; where K is constant. Hence = =9 u= n u (H) (1/ 3) K 51. The allowed values of l are 0, 1, …, (n – 1), a total of n values. rn =

3.32 Complete Chemistry—JEE Main

The allowed values of m are 0, ± 1, ± 2, …, ± l, a total of 2l + 1 values. m > l is not allowed. The value of l cannot exceed n – 1. 2+ (1s)2(2s)2(2p)6(3s)2(3p)6(3d)8. 28Ni (electron = 26) is For 2d orbital, l > n, which is not allowed. The number of orbitals in a shell is equal to n2. Each orbital can accommodate two electrons. 2(2s)2(2p)6(3s)2(3p)6(3d)5 (4s)1. Energy increases with increase in n + l value and thus the electron is less tightly bound. The most paramagnetic atom would contain larger number of unpaired electrons. Energy increases with increase in the value of n + l 1 1 2 58Ce is (4f) (5d) (6s) . But due to very similar energies of 4f and 5d orbitals in all the lanthanides excepting Gd (4f75d16s2) and 71Lu (4f14, 5d1, 6s2), (5d)1 electron is shifted to (4f) 2 0 2 58Ce is (4f) (5d) (6s) . 2 2 6 2 6 5 1 (2s) (2p) (3s) (3p) (3d) (4s) . 70. Each species contains 10 electrons. The number of neutrons are as follows O2– : 16 – 8 = 8 Ne : 20 – 10 = 10 Na+ : 23 – 11 = 12 Mg2+ : 24 – 12 = 12 2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)3 (4s)2 Z = 23) is (1s) For l = 2, we have d orbital.

52. 53. 57. 58. 59. 60. 61. 62. 63. 64.

Cu+ Al3+

(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (1s)2 (2s)2 (2p)6

K+ 3+ 33As

(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (4s)2

2

74. 75. 76. 78. 79. 80. 81.

(2s)2 (2p)6 (3s)2 (3p)6 (3d)3 (4s)2 The number of unpaired electrons will be 3 in (3d)3 The probability implies the value of y 21s. The function y has a maximum value at r = 0. The number of nodal point in the plot of Rn, 0 versus r is n – (l + 1). For example in R1, 0 (i.e. 1s orbital) has no node, and R2, 0 (i.e. 2s orbital) has one node. At r = 2a0, there is a node as the value of function R2, 0 is zero at this distance The number of radial node is equal to n – (l + 1). For 3p orbital, n = 3 and l = 1. n,l,m = Rn,l { l,|m| m}: the terms within the brackets represents angular dependence. The function R depends on the quantum numbers n and l, the function Q depends on l and | m | and the function F depends on m. Ê h ˆ The expression is L = l (l + 1) Á . Ë 2 p ˜¯

h Lz = m ÊÁ ˆ˜ . Ë 2p ¯ 83. An orbital with l = 0 involves no dependence on the angles q and j. Thus, it is symmetrical about the nucleus. 84. The energy of a single electron species such as H, He+, and Li2+ depends only the principal quantum number whereas for a multielectron species, it depends on the principal as well as azimuthal quantum number. 85. Energy carried by one photon of wavelength 1.5 nm is 82. The expression is

E1 =

(

)(

)

6.626 ¥ 10-34 J s 3 ¥ 108 m s -1 hc = 1.325 = l 1.5 ¥ 10-9 m

(

)

10–16 J

Energy carried by one photon of frequency 1.13 1017 s–1 is E2 = h = (6.626 10–34 J s) (1.13 1017 s–1) = 7.49 Kinetic energy of a single electron is DE = E1 – E2 = (13.25 – 7.49)

10–17 J = 5.76

10–17 J

10–17 J

Atomic Structure

3.33

86. Binding energy of electron per atom of the metal Ebind = =

KE =

1 2 p2 mv = 2 2m l=

Now

h = p

= 5.49 89. Dx ≥

10–19 J

Ebind 3.20 ¥ 10-19 J = 4.83 = h 6.626 ¥ 10-34 J s

h 6.626 ¥ 10-34 J s = = 6.626 l 0.1 ¥ 10-9 m

87. p =

88.

193 ¥ 103 J mol-1 = 3.20 6.022 ¥ 1023 mol-1

Thus h

=

10–24 J s m–1 = 6.626 p2 = eV 2m

(

1014 s–1 10–24 kg m s–1

p=

or 6.626 ¥ 10-34 J s

)(

2 m eV

)

1/2

ÈÎ( 2) 9.1 ¥ 10-31kg 1.6 ¥ 10-19 C (500 V )˘˚ 10–11 m = 54.9 pm

2 m eV

6.626 ¥ 10-34 J s 1 Ê h ˆ = = 1.93 ¥ 10–10 m Dp ÁË 4 p ˜¯ {(9.1 ¥ 10-31 kg )(3 ¥ 105 m s -1 )}(4)(3.14)

91. It is Y 2 and not Y 92. Y 2 cannot be imaginary. Y 2 is positive for a real wave function and Y *Y (where Y * is the complex conjugate of Y) is positive. 93. The quantum number s is not the outcome of the solution of Schrödinger equation. 94. The value of n = 0 is not allowed. 95. The signs of lobes of an orbital indicates the sign of wave function. –

is

(1s)

110. Dv =

2(2s)2(2p)6(3s)2(3p)6.

h /4 p (6.626 ¥ 10-34 J s) / (4 ¥ 3.14159) = = 5.79 ¥ 105 m s–1 mD x (9.1 ¥ 10-31 kg) (10-10 m)

111. The number of radial nodes = n – l – 1 = 4 – 0 – 1 = 3 112. We have E = eV and E2 = p2/2m Hence

p= l=

2meV = [(2) (9.1 ¥ 10–31 kg) (1.6 ¥ 10–19 C) (100 V)]1/2 = 5.396 ¥ 10–24 kg m s–1 h 6.626 ¥ 10-34 J s = = 1.228 ¥ 10–10 m = 122.8 pm -24 -1 p 5.396 ¥ 10 kg m s

113. The R-part of the wave function depends on the quantum number n and l. 114. At the nodal point, R = 0. This will be possible if 3 – 2 (r/a0) + 2r2/9a 02 = 0. This will give two values of r. Hence, there are 2 nodes. 115. The orbital is 3s, since it has two nodes. 116. Writing the quadratic expression as 2r¢2 – 18r¢ + 27 = 0 Its roots are

r¢ =

18 ± 182 - 8 ¥ 27 18 ± 108 18 ± 10.39 = = = 1.90, 7.10 4 4 4 R with respect to r will be equal to zero.

Ê dR d ÈÏ 2r 2r 2 ˆ ¸ - r / 3a0 ˘ ÈÊ 2 4r ˆ Ê 2r 2r 2 ˆ = + + 2 ˜ ˝e ÍÌconst ¥ Á 3 ˙ = ÍÁ - + 2 ˜ + Á 3 a0 9a02 ˜¯ dr dr Î Ó a0 9a0 ¯ ˛ Ë ˚ ÎË a0 9a0 ¯ Ë

Ê 1 ˆ ˘ – r/3a0 ÁË - 3a ˜¯ ˙ e 0 ˚

3.34 Complete Chemistry—JEE Main

2 4r 1 2r 2r 2 + 2 + 2 =0 a0 9a0 a0 3a0 27a03

Hence

-

or

2r 2 10r +3=0 2 27a0 9a0

2

or

Ê rˆ Ê rˆ 2 Á ˜ - 30 Á ˜ + 81 = 0 Ë a0 ¯ Ë a0 ¯

Solving for r/a0, we get 30 ± 15.87 30 ± 900 - 8 ¥ 81 30 ± 252 r = = = = 3.53, 11.46 4 4 4 a0

118. At node, R = 0. There will be one node. 119. The orbital is 3p. 120.

r Ê r ˆ 2= 0. This gives r = 6a0. Á 3a0 ˜¯ a0 Ë

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The number of d electrons retained in Fe2+ (atomic number 26) ion is (a) 3 (b) 4 (c) 5 2. Which of the following groupings represents a collection of isoelectronic species? (Atomic numbers of Cs in 55 and that of Br is 35.) (b) Na+, Ca2+, Mg2+ (c) N3–, F–, Na+ (a) Ca2+ Cs+, Br–

(d) 6

[2003]

(d) Be, Al3+, Cl–

[2003]

3. The orbital angular momentum for an electron revolving in an orbital is given by l (l + 1)(h/ 2p ). The angular momentum for an s-electron will be given by (b) (1/2) (h/2p) (c) zero (d) (h/2p) [2003] (a) 2 (h/ 2p) 4. The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 m/s is approximately (b) 10–33 m (c) 10–31 m (d) 10–16 m [2003] (a) 10–25 m 5. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in atom of hydrogen? (a) 2 Æ 5 (b) 3 Æ 2 (c) 5 Æ 2 (d) 4 Æ 1 [2003] 6. Which of the following sets of quantum number is correct for an electron in 4f orbital? (a) n = 4, l = 3, m = +4, s = +1/2 (b) n = 4, l = 4, m = –4, s = –1/2 (c) n = 4, l = 3, m = +1, s = +1/2 (d) n = 3, l = 2, m = –2, s = +1/2 [2004] 7. Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively (a) 12 and 4 (b) 12 and 5 (c) 16 and 4 (d) 16 and 5 [2004] 1, would be (Rydberg constant = 1.097 ¥ 107 m–1) (a) 91 nm (b) 192 nm (c) 406 nm (d) 9.2 ¥ 10–8 nm [2004] 9. Which one of the following sets of ions represents the collection of isoelectronic species? (a) K+, Ca2+, Sc3+, Cl– (b) Na+, Ca2+, Sc3+, F– (c) K+, Cl2–, Mg2+, Sc3+ (d) Na+, Mg2+, Al3+, Cl– (Atomic numbers: F = 9, Cl = 17, Na =11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21) [2004] 10. In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the (i) n = 1, l = 0, m = 0 (v) n = 3, l = 2, m = 0

(ii) n = 2, l = 0, m = 0

(iii) n = 2, l = 1, m = 1

(iv) n = 3, l = 2, m = 1

Atomic Structure

(a) (ii) and (iv) (b) (iv) and (v) (c) (i) and (ii) 11. Of the following sets which one does not contain isoelectronic species? (b) CN–, N2, C22– (c) SO32–, CO32–, NO–3 (a) PO43–, SO42–, ClO4–

(d) (ii) and (iii)

3.35

[2005]

(d) BO33–, CO32–, NO–3 [2005]

12. Which of the following statements in relation to hydrogen atom is correct? (a) 3s and 3p orbitals are of lower energy than 3d orbital (b) 3s, 3p and 3d orbitals all have the same energy (c) 3s orbital is lower in energy than 3p orbital (d) 3p orbital is lower in energy than 3d orbitals [2005] 13. According to Bohr’s theory, the angular momentum of an electron in 5th orbit is (a) 2.5 h/p (b) 25 h/p (c) 2.5 h/p (d) 10 h/p [2006] 14. Uncertainty in the position of an electrons (mass = 9.1 ¥ 10–31 kg) moving with a velocity 300 m s–1, accurate upto 0.001% will be (b) 19.2 ¥ 10–2 m (c) 5.76 ¥ 10–2 m (d) 1.92 ¥ 10–2 m [2006] (a) 3.84 ¥ 10–2 m 15. Which one of the following sets of ions, represents a collection of isoelectronic species? (a) Li+, Na+, Mg2+, Ca2+ (b) K+, Cl–, Ca2+, Sc3+ (c) Ba2+, Sr2+, K+, Ca2+ (d) N3–, O2–, F–, S2– [2006] 16. Which of the following sets of quantum numbers represents the highest energy of electron in an atom? (a) n = 3, l = 1, m =1, s = +1/2 (b) n = 3, l = 2, m =1, s = +1/2 (c) n = 4, l = 0, m =1, s = +1/2 (d) n = 3, l = 1, m =1, s = +1/2 [2007] 6 J mol–1. The energy required to excite the electron in the 17. The ionization enthalpy of hydrogen atom is 1.312 ¥ 10 atom from n = 1 to n = 2 is (b) 8.51 ¥ 10–2 J mol–1 (c) 6.56 ¥ 105 J mol–1 (d) 7.56 ¥ 105 J mol–1 (a) 9.84 ¥ 105 J mol–1 [2008] 18. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty to which the position of the electron can be located is (h = 6.6 ¥ 10–34 kg m2 s–1, mass of an electron, me = 9.1 ¥ 10–31 kg) (b) 3.84 ¥ 10–3 m (c) 1.52 ¥ 10–3 m (d) 5.10 ¥ 10–3 m [2009] (a) 1.92 ¥ 10–3 m 19. Calculate the wavelength associated with a proton moving at 1.0 ¥ 103 m s–1. (Mass of a proton =1.67 ¥ 10–27 kg and h = 6.63 ¥ 10–34 J s). (a) 2.5 nm (b) 14.0 nm (c) 0.032 nm (d) 0.40 nm [2009] + –18 –1 2+ 20. Ionization energy of He is 1.96 ¥ 10 J atom n = 1) of Li is –17 –1 –16 –1 –17 –1 (b) 4.41 ¥ 10 J atom (c) –4.41 ¥ 10 J atom (d) –2.2 ¥ 10–15 J atom–1 (a) 8.82 ¥ 10 J atom [2010] 21. A gas absorbs a photon of 355 nm and emits two wavelengths. If one of the emissions is at 680 nm, the other is at (a) 518 nm (b) 1035 nm (c) 325 nm (d) 743 nm [2011 (cancelled)] 22. The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following? (a) n = 3 to n = 1 (b) n = 2 to n = 1 (c) n = 3 to n = 2 (d) n = 4 to n = 3 [2011] n and l (i) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3, l = 1 can be placed in order of increasing energy is (a) (iii) < (iv) < (ii) < (i) (b) (iv) < (ii) < (iii) < (i) (c) (ii) < (iv) < (i) < (iii) (d) (i) < (iii) < (ii) < (iv) [2012] –18 2 2 24. Energy of and electron is given by E = –(2.178 ¥ 10 J) (Z /n ). Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be: (h = 6.62 ¥ 10–34 J s and c = 3.0 ¥ 108 m s–1) (b) 2.816 ¥ 10–7 m (c) 6.500 ¥ 10–7 m (d) 8.500 ¥ 10–7 m [2013] (a) 1.214 ¥ 10–7 m

3.36 Complete Chemistry—JEE Main

25. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is (a) 5, 0, 1, +1/2 (b) 5, 0, 0, +1/2 (c) 5, 1, 0, +1/2 (d) 5, 1, 1, +1/2 state of Li2+ is (a) –27.2 eV

(b) 30.6 eV

(c) –30.6 eV

[2014]

(d) 27.2 eV [2014, online]

27. If l0 and l1 are the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is 2hc Ê l0 - l ˆ 2h Ê 1 1 ˆ 2h 2hc (b) (c) (d) (a) ( l0 - l ) ( l0 - l ) Á ˜ m Ë ll0 ¯ m ÁË l0 l ˜¯ m m [2014, online] Ê 1 ˆ 1 28. Based on the equation DE = –(2.0 ¥ 10–18 J) Á 2 - 2 ˜ Ë n2 n1 ¯ the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level n = 2 will be (Given: h = 6.625 ¥ 10–34 J s, c = 3 ¥ 108 m s–1) (a) 1.325 ¥ 10–7 m (b) 1.325 ¥ 10–10 m (c) 2.650 ¥ 10–7 m (d) 5.300 ¥ 10–10 m [2014, online] 29. The de-Brogile wavelength of a particle of mass 6.63 g moving with a velocity of 100 m s–1 is (b) 10–35 m (c) 10–31 m (d) 10–25 m [2014, online] (a) 10–33 m 30. Excited hydrogen atom emits light in the ultraviolet region at 2.47 ¥ 1015 Hz. With this frequency, the energy of a single photon is (Given : h = 6.63 ¥ 10–34 J s) (b) 2.680 ¥ 10–19 J (c) 1.640 ¥ 10–18 J (d) 6.111 ¥ 10–17 J (a) 8.041 ¥ 10–40 J [2014, online] 31. Ionization energy of gaseous Na atoms is 499.5 kJ mol–1. The lowest possible frequency of light that ionizes a sodium atom is (Given : h = 6.626 ¥ 10–34 J s, NA = 6.022 ¥ 1023 mol–1) (b) 4.76 ¥ 1014 s–1 (c) 3.15 ¥ 1015 s–1 (d) 1.24 ¥ 1015 s–1 (a) 7.50 ¥ 104 s–1 [2014, online] 32. If the principal quantum n (a) ns Æ np Æ (n – 1)d Æ (n – 2)f (b) ns Æ (n – 2)f Æ (n – 1)d Æ np (c) ns Æ (n – 1)d Æ (n – 2)f Æ np (d) ns Æ (n – 2)f Æ np Æ (n – 1)d [2015, online] 33. At temperature T, the average kinetic energy of any particle is (3/2) kT. The de Broglie wavelength follows the order: (a) Thermal proton > Visible photon > Thermal electron (b) Thermal proton > Thermal electron > Visible photon (c) Visible photon > Thermal electron > Thermal neutron (d) Visible photon > Thermal neutron > Thermal electron [2015, online] 34. Which of the following is the energy of a possible excited state of hydrogen? (a) +13.6 eV (b) –6.8 eV (c) –3.4 eV (d) +6.8 eV [2015] 35. The total number of orbitals associated with the principal quantum number 5 is (a) 20 (b) 25 (c) 10 (d) 5 [2016, online]

ANSWERS 1. (d) 7. (b)

2. (c) 8. (a)

3. (c) 9. (a)

4. (b) 10. (b)

5. (a) 11. (c)

6. (c) 12. (b)

Atomic Structure

13. 19. 25. 31.

(a) (d) (b) (d)

14. 20. 26. 32.

(d) (c) (c) (b)

15. 21. 27. 33.

(b) (d) (c) (c)

16. 22. 28. 34.

(b) (b) (a) (c)

17. 23. 29. 35.

(a) (b) (a) (b)

3.37

18. (a) 24. (a) 30. (c)

HINTS AND SOLUTIONS 2

(2s)2 (2p)6 (3s)2 (3p)6 (3d)6 (4s)2. With the removal of 4s electrons, Fe2+ is formed. Hence, 3d orbitals retain 6 electrons. 2. All the three species in choice c contain 10 electrons. h (6.63 ¥ 10-34 J s) = = 1.1 ¥ 10-33 m 4. l = mu (60 ¥ 10-3 kg)(10 m s -1 ) 6. For a 4f orbital, we have Principal quantum number, n = 4; Azimuthal quantum number, l = 3 Magnetic quantum number, m = any one value from +3, +2, +1, 0. –1, –2, –3 For an electron in this orbital, spin quantum number, s = either +1/2 or –1/2 (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)5 (4s)1 24Cr is l = 1 corresponds to p orbital. The number of electrons in p orbitals is 6 + 6 = 12 l = 2 corresponds to d orbital. The number of electrons in d orbitals is 5  = R Ê 1 - 1 ˆ = R ÊÁ 1 - 1 ˆ˜ = R ; 8. We have DE •Á 2 •Ë 2 • Ë n1 n22 ˜¯ •2 ¯ 1

l=

1 • R

= 0.91 ¥ 10-7 m = 91nm

9. The isoelectronic species have the same number of electrons. Each of the species K+, Ca2+, Sc3+ and Cl– has the same number (=18) of electrons. n + l possess the same electronic energy. 11. The number of electrons of each species in the choice a is 50. The number of electrons of each species in the choice b is 14. The number of electrons of each species in the choice c are: SO32–(42), CO32–(32) and NO3– (42). The number of electrons of each species in the choice d is 32. 12. For a single electron species, the energy of electron depends only on the principal quantum number. Hence, 3s, 3p and 3d orbitals will have the same energy. 13. In Bohr’s theory, the angular momentum of electron is given by L = n(h/2p), where n is the quantum number. For 5th orbit, n = 5. 14. p = mv = (9.1 ¥ 10–31 kg) (300 m s–1) = 27.3 ¥ 10–29 kg m s–1 Dp = (0.001%)p = (10–5) (27.3 ¥ 10–29 kg m s–1) = 27.3 ¥ 10–34 kg m s–1 Dx =

Ê 6.626 ¥ 10-34 J s ˆ 1 Ê hˆ 1 ÁË ˜¯ = ˜¯ = 0.0192 m Dp 4p 4 ¥ 3.14 (27.3 ¥ 10-34 kg m s -1 ) ÁË

15. 19K+, 17Cl–, 20Ca2+ and 21Sc3+ have the same number of 18 electrons. 16. Larger the value of n + l, higher the energy of atomic orbital. 17. The energy difference between the two orbits of hydrogen atom is given by Ê 1 1ˆ DE = N A R• hc Á 2 - 2 ˜ Ë n1 n2 ¯ For ionization, n1 = 1 and n2 = •. Hence, For n =1 Æ n = 2, the energy difference is

NAR•hc = 1.312 ¥ 106 J mol–1

3.38 Complete Chemistry—JEE Main

3 3 ( N A R• hc) = ¥ 1.312 ¥ 106 J mol-1 = 9.84 ¥ 105 J mol-1 4 4 18. According to the uncertainty principle Dp Dx ≥ (h/4p) i.e (m Du) Dx ≥ (h/4p) DE =

Hence, Dx ≥

19. We have

1 (6.6 ¥ 10-34 kg m 2 s -1 ) h = = 1.92 ¥ 10-3 m -31 -1 4p m( Du ) 4(22 / 7)(9.1 ¥ 10 kg){(600 m s )(0.005 / 100)} (6.63 ¥ 10-34 J s) h h = = = 4.0 ¥ 10-10 m = 0.40 nm p mu (1.67 ¥ 10-27 kg)(1.0 ¥ 103 m s -1 )

l=

20. (IE)He+ = Z12 R• hc = 19.6 ¥ 10-18 J R• hc =

This gives

19.6 ¥ 10-18 J 19.6 ¥ 10-18 J = 4 Z12

(9)(19.6 ¥ 10-18 J/ 4) = -4.41 ¥ 10-17 J n12 (12 ) (355 nm)(680 nm) 1 1 1 = + or l = = 743 nm 21. The conversion of energy required that (680 - 355) nm 355 nm 680 nm l ( E1 ) Li2 + = -

Z 22 R• hc

=-

22. The expression of emission of special transitions is For the given transition in He+, we have

Ê 1 1ˆ DE = Z 2 R• hc Á 2 - 2 ˜ Ë n2 n1 ¯

Ê 1 1ˆ 3 DE = (2) 2 R• hc Á 2 - 2 ˜ = R• hc Ë2 4 ¯ 4

In hydrogen atom, this transition will correspond to the transition n2 = 1 ¨ n1 = 2, since Ê1 1ˆ 3 DE = (1)2 R• hc Á 2 - 2 ˜ = R• hc Ë1 2 ¯ 4 23. Energy of an electron in an atom increases with increase in the value of n + l. For the same value of n + l, the energy is larger for the larger value of n. Hence, we have Choice (i); n+l=4+1=5 Choice (iii); n+l=3+2=5 Choice (ii); n+l=4+0=4 Choice (iv); n+l=3+1=4 Hence, the correct order of increasing energy is (iv) < (ii) < (iii) < (i) 1 1 24. DE = E2 - E1 = (-2.178 ¥ 10-18 J) ÊÁ - ˆ˜ = 1.6335 ¥ 10-18 J Ë 4 1¯ l=

hc (6.62 ¥ 10-34 J)(3 ¥ 108 m s -1 ) = = 1.214 ¥ 10-7 m -18 DE (1.6335 ¥ 10 J) 1

l = 0, m = 0 and ms = +1/2 26. The energy of electron in Bohr orbits is

E=-

Z2 R• n2

(where R• is Rydberg constant)

E = –R• = –13.6 eV 2

3 9 9 R = - 2 R• = - 2 (13.6 eV) 2 • n n n When n = 2, E = –30.6 eV(given choice). For Li2+ (Z = 3), we have

27. E0 = hc/l0

and

E=-

E = hc/l

1 Ê 1 1 ˆ hc(l0 - l ) ; mu 2 = E - E0 = hc Á - ˜ = Ë l l0 ¯ ll0 2

1/ 2

È 2hc (l0 - l ) ˘ u=Í ll0 ˙˚ Î m

. Thus it has n = 5,

Atomic Structure

3.39

Ê 1 -18 Ê 1 1ˆ 1ˆ -18 -18 28. DE = -(2.0 ¥ 10-18 J) Á 2 - 2 ˜ = -(2.0 ¥ 10 J) ÁË - ˜¯ = (2.0 ¥ 10 J)(3 / 4) = 1.5 ¥ 10 J 4 1 Ë n2 n1 ¯ l= 29. l =

hc (6.625 ¥ 10-34 J s)(3 ¥ 108 m s -1 ) = = 1.325 ¥ 10 –7 m DE (1.5 ¥ 10-18 J)

h (6.626 ¥ 10-34 J s) = = 10-33 m mu (6.63 ¥ 10-3 kg)(100 m s -1 )

30. E = hn = (6.63 ¥ 10-34 J s)(2.46 ¥ 1015 s -1 ) = 1.64 ¥ 10-18 J 31. Ionization energy per gaseous atom is Ê 495.5 ¥ 103 J mol-1 ˆ E (8.228 ¥ 10-19 J) -19 = ¥ = = E=Á 8 . 228 10 J; n = 1.24 ¥ 1015 s -1 h (6.626 ¥ 10-34 J s) Ë 6.022 ¥ 1023 mol-1 ˜¯ 32. The guiding principle is Energy increases with increase in the value of n + l. For the same value of n + l, smaller the value of n, lesser the energy For the given orbitals, we have ns np (n – 1)d (n – 2)f lÆ 0 1 2 3 n+l n n+1 n+1 n+1 ns Æ (n – 2)f Æ (n – 1)d Æ np 33. Lesser the mass of particle, larger its de Broglie wavelength (l = h/mu). Hence, the order of wavelength is Visible photon > Thermal electron > Thermal neutron 34. The energy of orbits in hydrogen atom is given by the expression 13.6 eV En = n2 For n = 2, E2 = –3.4 eV 35. The number of orbitals in the principal quantum number n is n2. Hence, answer is 25.

4 Chemical Bonding and Molecular Structure SECTION 1

Bond Formation and VSEPR Theory

The formation of a chemical bond between two atoms implies that the system consisting of these two atoms at stable internuclear distance is energetically more stable than the two isolated atoms. A general study on the reactivity of different elements revealed that noble gases have little tendency to combine with other elements. This leads to the ns)2 np)6

Kössel and Lewis Theory of Chemical Combination According to this theory, atoms can combine either by transfer of outer-shell electrons, known as valence electrons, eight electrons) in their respective valence shells. formation of ionic bond between the two involved atoms.

Representation of a Bond by Lewis Structure In Lewis structure, a bond between the two atoms is shown by Lewis electron-dot symbols in which valence electrons are shown by dots around the letter symbol of the atom. The dots are placed as follows. Place a single dot on the four sides of the letter symbol followed by the second dot till all the valence-electrons have been accounted for. Illustrations: Lithium 1

Nitrogen 2

3

Li

N

Beryllium 2

Oxygen 2

4

Be

O

Boron 2

Fluorine 2

Carbon

B

1

5

2

F

2

Neon 2

6

C Ne

Formation of Covalent Bond(s) A covalent bond involves mutual sharing of valence electrons between two atoms. The sharing of two, four and six electrons leads to the formation of a single, double and triple bond, respectively. A covalent bond is formed if the atoms have lesser number of valence electrons as compared to the nearby noble formation of covalent bond is

4.2 Complete Chemistry—JEE Main

Electronegative element + Electronegative element æÆ Covalent bond Exception to the octect rule is the hydrogen atom which can accommodate only two electrons which corresponds to 2) atom.

Illustrations Formation of Single Bond(s) Cl2

Cl

+

Cl

Cl Cl

or

Cl

Cl2

2H

+

O

H O H

or

H

O

H

NH3

3H

+

N

H N H

or

H

N

H

Cl

H H

4H

CH4

C

+

H

H H C H

or

H

C

H

H H Formation of Double Bond(s) O2

O

+ O

CO2

O

+

C 2H 4

O

+

C

O

O

O

4 H + 2 C

H C H

C

O

C H H

or O

O

or O

C

O

or H

C

C

H

H

H

Formation of Triple Bond(s) N2 C 2H 2

N

N

+

N N H C

2 H + 2 C

Exceptions to the Octect Rule

or C H

N or

N H

C

C

H

The octect rule is generally obeyed by the elements of second and third periods

with the following exceptions.

The Incomplete Octect BeCl2

Cl

Be

+

+

Cl

Cl Be Cl 4 electrons

AlCl3

3 Cl

+

Al

Cl Cl Al Cl 6 electrons

Odd-Electron Molecules All atoms of a compound containing odd number of electrons will not satisfy octect rule as even number of electrons are required for pairing of electrons.

Chemical Bonding and Molecular Structure

NO

N

O

+

N

N O

O

+

4.3

7 electrons 2 O

NO2

+

N

O

+

N

+

O

O N O – + O N O 7 electrons

The Expanded Octect Elements of third period and beyond can accommodate more than 8 electrons due to the availability of vacant d orbitals. Cl PCl5

6

5 Cl

6 F

5 Cl

P

+

6 F

+

+

Cl

P

Cl

Cl

F

F F

F

+

F

Coordinate Covalent (or Dative) Bond

Cl

P

F

If a pair of electrons shared between two atoms comes exclusively from ). Once a coordinate bond is formed, it behaves like

a covalent bond. H H3N—BF3

H

N

F +

H

B F

F

H

H

F

N

B

H

F

F

Writing a Lewis Structure The structure of a molecule or ion may be written by following the steps listed below. 1. Calculate the total numbers of valence electrons of the atoms in the molecule. For an anion, add the number of negative charges and for a cation, subtract the number of positive charges. 2. Write the skeleton structure of the molecule or ion connecting every bonded pair of atoms by a single bond, i.e. a pair of electron dots.

If there are fewer than eight electrons on the central atom, move one or two pairs of electrons from a surrounding

Illustration Lewis structure of COCl2. Step 1 Valence electrons are 4 + 6 + 2 ¥ 7 = 24

4.4 Complete Chemistry—JEE Main

Carbon being the most electropositive atoms occupies the central position to which other atoms are bounded. O Cl C Cl Step 3 Assign 8 electrons each to surrounding atoms O Cl C Cl Step 2

Step 4 There were 24 valence electrons and all of them have been distributed. However, the central C atom has only 6 electrons. In order that this atom also has 8 electrons, move one pair of electrons from O to the bond connecting C atom, thus forming a double bond. O

O

O Cl C Cl

Cl C Cl

or

Cl

C

Cl

Formal Charge and Lewis Structure The formal charge on an atom is the difference between the valence electrons in an isolated atom and the number of electrons assigned to that atoms in a Lewis structure. The equation for computing formal charge is 1 2 bonding) electrons in a Lewis structure

The sum of the formal charges of atoms in a Lewis structure is equal to the charge on the molecular species.

Illustration

COCl2 molecule

Lewis structure

Cl C Cl O

Atom

Valence electrons in a free atom

Electrons in Lewis structure Nonbonding

bonding

Cl

7

6

2

1 + 2 ¥ 2) =

O

6

4

4

1 + 2 ¥ 4) =

C

4

8

Formal Charge

1 + 2 ¥ 8) =

Utility of Formal Charge Computing formal charge of atoms in a molecule or ion helps deciding a plausible Lewis structure of the species. The guiding principles are as follows. Amongst the several Lewis structures, the species having the lowest magnitude of formal charge is the preferred structure. Amongst Lewis structures having similar distribution of formal charges, the one having negative formal charges on the more electronegative atoms is the preferred structure.

Polar Covalent Bond Each atom in a molecule has its own ability to attract the bonded pair of electrons. This ability is known as electronegativity. 2, O2, F2, Cl2, etc.) is shared equally by both of electrons is closer to the atom having larger electronegativity. Consequently, this atom acquires a partial negative charge while the other atom acquires equal partial positive charge. Because of the charge separation, the covalent bond between these two atoms is said to be a polar covalent bond. The polarization of bonded pair of electrons between two atoms is expressed in terms of physical m m dq r where d q is the partial charge separation between two atoms and r is the distance between the two atoms.

Dipole Moment

Chemical Bonding and Molecular Structure

4.5

Representation of Dipole Moment Dipole moment is a vector quantity, i.e. it has magnitude as well as direction. In chemistry, dipole moment is indicated by the crossed arrow as shown in the following.† positive end negative end that is, it is directed from positive end to the negative end. dq

Unit of Dipole Moment –18

–18

–18

r) = C m

esu cm. This value of dipole moment is known as 1 debye

Ï Ê 1.6 ¥ 10-19 C ˆ ¸ ˝ ) Ì(1esu ) Á Ë 4.8 ¥ 10-10 esu ˜¯ ˛ Ó

–2

m) = 3.33 ¥

Cm

Dipole Moment of a Polyatomic Molecule Each bond in a molecule has a dipole moment, known as bond moment. The dipole moment of a molecule is obtained by the vector addition of these bond moments. Illustration The bond moment of O H2O molecule will be

H bond is 1.52 D. The bond angle of H2 H

mH2O = 2mOH

O

Nonpolar Polyatomic Molecule

H

The dipole moment of a nonpolar polyatomic molecule is zero inspite of the fact that the bond moments of the molecule is not zero. This is due to the fact that the individual bond moments in the molecule is symmetrically placed so that their vector additions is zero. Illustration F Cl F

Be

F

B

F

O

C Cl

F

C

O

Cl Cl

Comparison of Dipole Moments of NH3 and NF3 Both NH3 and NF3 have pyramdial shapes with lone pair of electrons on nitrogen atom.

N

N H

F

F F H In NH3, orbital dipole acts in the same direction as the sum of bond vectors of the three N H bond bonds. F bond bonds. In NF3, orbital dipole acts in the opposite direction to the sum of bond vectors of the three N These facts make the dipole moment of NH3 m = 1.57 D) larger than that of NF3 m H

Per Cent Ionic Character of a Polar Band Per cent ionic character =

The per cent ionic character of a polar band A

m AB ¥ 100 mionic

where mionic = e rAB † In actual practise, the dipole moment is represented as a simple arrow pointing from negative end to the positive end, that is negative end positive end In chemistry, the crossed arrow is used which indicates the direction of the shift in electronegativity in the molecule.

4.6 Complete Chemistry—JEE Main

Illustration will be

The bond moment of O

H bond is 1.52 D. If bond length O

Per cent ionic character =

(1.52 D) (3.33 ¥ 10-30 C m / 1 D) mOH ¥ 100 ¥ 100 = (1.6 ¥ 10-19 C) (95 ¥ 10-12 m) e rOH

Concept of Resonance distribution of electrons over a given skeleton of atoms in a molecule. None of the individual structures adequately explains the characteristics of the molecule. However, these can be explained if the actual structure of the molecule is considered as the superposition of individual structures. This phenomenon is known as resonance and the individual ) inserted between the resonating structures.

Illustrations

O

O

O

O

O

O

written as O bond lengths equal in size. 2–

O C

O

O

O

O

O

written as

C O

C

2–

O

O

2–

C O

2–

O

O

Formation of an Ionic Bond leads to the formation of positive and negative ions, respectively. The electrostatic attraction between the positive and negative ions results in the formation of an ionic bond between the involved ions. Illustrations +

Na 1

F 2

Na+ +

Cl 2

+ Mg 5

2

5

+

–

Cl 2

Mg2+ + 2 F

F 2

6

5

2

– 6

Energies Involved in the Formation of One Molecule of Sodium Chloride

The formation of Na+Cl–

– Ei = 8.24 ¥ J Æ Na+ – Æ Cl– Eea = –5.78 ¥ J + – Æ Na+Cl– PE where PE is the potential energy in the formation of ionic bond. This is evaluated by the expression QQ PE = 1 2 (4pe 0 )r

¥

where Q1 = –Q2 we have PE =

C and r = rNa+ + rCl–

(1.60 ¥ 10-19 C)(-1.60 ¥ 10-19 C) = - 8.34 ¥ 10-19 J (4)(3.14)(8.854 ¥ 10-12 C2 N -1 m -2 )(276 ¥ 10-12 m)

Q1 and Q2 as point charges,

Chemical Bonding and Molecular Structure

Æ Na+Cl– DE = Ei + DEea

we have

¥

4.7

DE =? J = –5.88 ¥

Comment The negative value of DE indicates that the formation of an isolated ionic bond Na+Cl–

Essential Requirement for the formation of an Ionic Bond Ei Ei Hence Electropositive element + Electronegative element

For DE

Ionic Bond

Formation of 1 mol of Solid Ionic Compound from Constituent Elements Taking an example of sodium chloride, we have the following steps in the formation of solid compound. –1 Æ DH1 1 –1 Æ DH2 2 Cl2 – –1 Æ Na+ DH3 – Æ Cl– –1 DH4 + – + – Æ Na Cl DH5 = –788 kJ mol–1 1 2

Cl2

Æ Na+

DH = –313 kJ mol–1

DH is negative, the formation of solid NaCl is energetically favourable. From the values of DH’s listed above, it Æ Na+

–

the ionic solid. compound into gaseous constituent ions. Larger the value of lattice energy, more stable the ionic compound. listed above for the formation of solid ionic compound. In this cycle, DH DH1, DH2, DH3 and DH4 from the value of DH gives the value of DH5. The lattice energy is negative of the value of DH5.

Fajan Rules

An ionic compound has partial covalent character and vice versa. The partial covalency in an ionic compound my be explained qualitatively with the help of Fajan rules described in the following. High Charge and Small Size of the Cation cationic electronic charge to penetrate partially into the anionic electronic cloud resulting into the partial covalent bond character to the ionic bond. High Charge and Large Size of the Anion The electronic cloud of such an anion is most easily polarized by For two cations of the same size and charge, the cations of electronic n – 1)dx ns n – 1)s2 n – 1)p6 ns nucleus by the electronic cloud of transition metal ions as compared to that in the alkali and alkaline metal ions.

Hg2+ ion has larger polarizing effect than Ca2+ ion. Lithium salts have more covalent character than the alkali salts. I– ion is more easily polarized than Cl– ion by Ag+ ion.

Illustrations

The Valence Shell Electron Pair Repulsion (VSEPR) Theory

4.8 Complete Chemistry—JEE Main

The number of electron pairs in the valence shell of the central atom of a molecule decides the shape of the A multiple bond is treated as if it is a single electron pair. The repulsive interaction of electron pairs decrease in the order

Table 1 Molecule

Shapes of some molecules on the basis of VSEPR model Basic shape †

Number of valence electrons around the central atom

electron pairs

bounding pairs

BeCl2

4

2

2

Linear

BCl3

6

3

3

Triangular planar

CH4

8

4

4

NH3

8

4

3

1

H 2O

8

4

2

2

PF5

5

5

4

5

4

1

5

3

2

12

6

6

12

6

5

CIF3

6

IF5

lone pairs

Tetrahedron

Trigonal bypyramid

Octahedron

1

A few examples of molecules containing lone pair electrons along with their geometry are described in the following. N O

O

H

H

O H

H

H

† The actual shapes of molecules containing lone pairs is a little distorted from the basic shape. This is due to the fact that lone pair-bonded pair repulsion is larger than bonded pair-bounded repulsion.

Chemical Bonding and Molecular Structure

F

F

F

Cl

F F

F

F

Note: Lone pair equatorial position as it involves only two lp-bp repulsion. It is not occupied in axial position as it will involve three lp-bp repulsions

Note: Lone pair occupy equatorial positions so as to have lesser lp-bp repulsions

Note: F

F

F

Xe F

F F

F

O

F

F

I F

F

F

F

Xe F

F

I

F

I

Cl

I

F

F F

F

O

Xe

Xe

F

F

O

F

F

MULTIPLE CHOICE QUESTIONS ON SECTION 1 Identify the correct choice in the following questions. 1. Which of the following statements regarding Hg2+ 2+ ion has more polarizing effect on an anion as compared to Ca2+ ion 2+ ion has less polarizing effect on an anion as compared to Ca2+ ion 2+ ion and Ca2+ have equal polarizing effect on an anion 2+ ion and Ca2+have no polarizing effect on an anion 2. The increasing order of covalency in silver halides is

2+

3. The increasing order of melting point of carbonates of alkaline-earth metals is 3 3

3

3

3

3

3 3

3

3

3

3

4. Which of the following resonating structures of N2O is feasible in its linear structure? N N O N N O N N O + – 2– + + 5. Which of the following statements regarding BrF3 molecule is correct?

Br

4.9

N N O – +

4.10 Complete Chemistry—JEE Main

6. Which of the following pairs of species are not isostructural? + + – + , IF 5 4, PF 4 4 and NH 4 2, I 3 5 7. Which of the following order regarding bond angle O N O in the given species is correct? + – – + + – 2 2 2 NO 2 2 2 2 2 2 2 8. Which of the following facts regarding Fajan’s rules is not correct?

– 3

+ 2

– 2

is not correct?

I 3 is correct? F bond has the same length

Cl

11. The structure of ICl4 is Cl

Cl

Cl

Cl

Cl I

I

Cl

Cl Cl 12. Which of the following statements is correct? 2 2 has nonlinear structures 2 2 has linear structures 2 and CO2 has linear structures 2 and CO2 has non-linear structures 13. The shape of XeF4 is Cl

Cl

Cl

Cl

I

I

Cl

Cl

Cl

14. Which one of the following compounds of xenon has trigonal bipryramid structure? 3 2F 2 4 15. The number of pairs of electrons around I in IF3, IF5 and IF7 ,respectively, are

Cl

3F 2

16. Which of the following species has pyramidal shape? 2– 3

3

– 3

3

17. Which of the following species has linear structure? –

+ 2

2

2

18. Which of the following statements regarding bond angles in the given molecules is correct? 3 3 and OF2 2 3 3 and OF2 2 3 3 and OF2 2 3 3 and OF2 2 3

3

3

3

3

3

3

3

3

3

3

3

Chemical Bonding and Molecular Structure

s – bond and 1 p

4.11

s – bond, 1 p – bond and 1 coordinate bond

ANSWERS

HINTS AND SOLUTIONS 1. Hg2+ ion belongs to transition metal and Ca2+ ion to alkaline-earth metal. Thus Hg2+ ion has more polarizing effect on an anion. 2. As the size of halide is increased, it is more polarizable by Ag+ and hence larger is the covalency of silver halide. 2+) of Group 2 of periodic table, the polarizing power on CO2– is decreased. The 3 MCO3 becomes more ionic, hence their melting point increases.

F 5. There are 28 valence electrons. These are distributed as

F

Br F

F

Hence, its structure is pentagonal bipyramid with two lone pairs occupying equatorial positions. Br Due to lone pair-bonding pair, the angle F 6. XeF4 has 6 pairs of electrons around Xe while PF+4 has 4 pairs of electrons around P. + O

N

–

O

O

NO2+

N O NO2

NO–2 The molecule NO2 bonding electrons move away so as to reduce the repulsion between them.

F 3

is

Cl F

F

O

N O NO–2

Br F

F

4.12 Complete Chemistry—JEE Main

11. There are six pairs of electron around I. These are oriented to make an octahedron. The two lane pairs occupy axial positions. F F Xe Lewis structure 13. Valence electrons of XeF4 = 8 + 4 ¥ 7 = 36. F F There are 6 pairs of electrons around. These are directed to form octahedron. The four F atoms occupy apical positions, therefore, the shape of XeF4 is square planar. 14.

XeO2F2

XeO3 8 + 3 ¥ 6 = 26

XeOF4

8 + 2 ¥ 6 + 2 ¥ 7 = 34 O

O O

Xe

8 + 6 + 4 ¥ 7 = 42

F

O

F

Xe

O F

15.

IF3 Valence electrons = 4 ¥ 7 =28 F

F Xe

F

F Trigonal bipyramid

IF5 Valence electrons = 6 ¥ 7 = 42 F

F

F

F

F

Valence electrons = 24

F

CO32– Valence electrons = 24

O

F

F

PCl3 Valence electrons = 26 Cl

Trigonal planar

O

F

F I

F

F

3

IF7 Valence electrons = 8 ¥ 7 = 56

I

I

O

O

O

F Trigonal bipyramid

Pyramidal

O

Xe

O O

O

F F

O O

F Xe

F

OF Xe

O

F Xe

F

O

8+3¥6+2¥

F

F Xe

O

XeO3F2

P

Cl

F F

NO–3 Valence electrons = 24 Trigonal planar

Cl

O 3-pairs of electrons

P triagonal planar

Cl

Cl Cl Pyrimidal

NCO–

17. N

C

NO–2 O

2-pairs of electrons on C, inear structure

O

N

OF2

2

O

3-pairs of electrons on N, not linear

F

O

F

4-pairs of electrons on O

Chemical Bonding and Molecular Structure

4.13

18. Bond angle in NF3 is smaller than in NH3. Also bond angle in OF2 is smaller than in OH2. This is primarily due to ionicity of N F and O away from N and O nuclei. This causes expansion of lone pair causing bond angles F N F and F O F smaller than those in H N H and H O 3 3 2 2

– C

SECTION 2

+ O.

VB and MO Theories

Valence Bond Theory The quantitative description of chemical bond is provided by the quantum mechanical theories. Two theories, namely, The essential guidelines of VB method are as follows. A molecule is considered to be a collection of atoms with electrons occupying their respective atomic orbitals. The formation of molecule is analysed in terms of interactions amongst electrons-electrons, electrons-nuclei and nuclei-nuclei. For a molecule to be stable, the electrostatic attractions must predominate over the electrostatic repulsion. The difference in these two is released in the form of heat. Thus, a molecule is energetically more stable than the individual atoms.

Potential Energy Diagram The potential energy diagram display the variation of potential energy during the formation of chemical bond between two atoms when these are brought close to each other from a larger distance. HB) is as follows. The description of potential energy variation of H2 A At larger distance, DVinteraction A and HB exists. As the atoms HA and HB are brought near to each other, there occurs a net decrease in potential energy as the B and electron B – nucleus A) are greater than the B and nucleus A and nucleus B) The decrease in potential energy is continued till the atoms HA and HB are brought up to stable internuclear distance. As the atoms are brought more near to each other, an increase in potential energy is observed as the repulsive interactions become greater than the attractive interactions. These interactions are shown in Fig. 1 which explain the formation of HA–HB bond at the stable internuclear distance.

Essential Criteria for the Formation of Covalent Bond If the formation of H2 2 A

molecule = 74 pm B) overlap each other in the bonded region.

Fig. 1

4.14 Complete Chemistry—JEE Main

Electron associated with atom HA can go to the atom HB and vice versa through the overlap region. Also in the overlap two electrons must have opposite spins.

internuclear distance. 2. Each of the two orbitals must possess one unpaired electron with opposite spin.

Fig. 2

Stablility of the Molecule The intervening electronic charge between the two nulcei has an affect of decreasing nuclear repulsion and maximises electron-nuclei attractions. This lead to the stable H2 molecule. Sigma an Pi Bonds In a molecule, two types of overlapping of orbitals having directional characteristics may be distinguished. End to End Overlap—Sigma Bond (s Bond) In this overlapping, the electronic charge is concentrated between Sideways Overlap—Pi Bond (p Bond)

In this overlapping, the electronic charge is concentrated above the

Fig. 3

Fig. 4

Illustrations 2 Oxygen molecule 2p atomic orbital, each containing one electron. Thus it can form two bonds

Fig. 5

2

x)

2

s and p

y)

1

z)

1.

There are two

Chemical Bonding and Molecular Structure 2 Nitrogen molecule three 2p orbitals, each containing one electron. Thus it can form three bonds

2

x)

1

y)

1

one s and two p

z)

1

4.15

. There are

Fig. 6

The concept of Hydridization According to the valence band theory. 2 , 2s2) should form no chemical bond as it does not contain any unpaired electron. 4 2 2 1 , 2s , 2p ) should form a single bond as it contains only one unpaired electron. 5 2 2 1 1 , 2s , 2px , 2py ) should form two bond as it contains two unpaired electrons. 6 Experimentally it if found that Be is divalent, B is trivalent and C is tetravalent, To explain this, the concept of hydridization is introduce. In this concept we have Two or more atomic orbitals of the same atom mix each other to provide a new set of identical number of degenerate orbitals. These orbitals, known as hybrid orbitals, are completely identical in size, shape and orientations.

Kind of Hybridization sp Hybridization

One s orbital and one p oribtal of the same valence shell of an atom are mixed to provide

indicates the sign of wave function of the orbital.

Fig. 7 2

2

sp Hybridization In sp valence shell of an atom combine to give three degenerate equivalent sp2 hybrid orbitals. These are oriented at an

Fig. 8

4.16 Complete Chemistry—JEE Main

sp3 Hybridization In sp3 valence shell of an atom combine to give four degenerate equivalent sp3 hybrid orbitals. These four orbitals are directed ¢

Fig. 9

Hybrid Orbitals involving d orbitals sp2d or dsp2 Hybridization The resultant four In dsp2, d orbital belongs to penultimate shell while in sp2d, it belong to the valence shell. sp3d or dsp3 Hybridization

sp3d2 or d2sp3 Hybridization The resultant six hybrid orbitals are directed to the corners of regular octahedron

sp3d3 or d3sp3 Hybridization The resultant seven orbitals are directed to the corners of a regular pentagonal xy,

Fig. 10

Fig. 11

dxz and dyz.

Fig. 12

Fig. 13

Comment on Hybridization The concept of hybridizaton follows from molecular geometry and not vice versa since the choice of appropriate is known.

Chemical Bonding and Molecular Structure

4.17

A Few Illustrative Examples Involving Hybridization Methane It involves sp3

Ethene

It involves sp2

Fig. 15 Fig. 14

Ethyne

It involves sp3 hybrid orbitals of nitrogen. ¢

Ammonia

Fig. 16 Fig. 17

¢

Water It involves sp3

Fig. 18

Note:

4.18 Complete Chemistry—JEE Main

[Ni(CN)4]2– It involves dsp2 hybrid orbitals of Ni2+

2–

4s 28Ni

4p

2+

dsp2 hybridization

Fig. 19

PCl5 It involves sp3 3s

3p

3d

15P

sp3d hybridization

Fig. 20

SF6 It involves sp3d2

21) 3s

3p

3d

16

sp3d2 hybridization

Fig. 21

Molecular Orbital Theory

Molecular orbital theory provides the explanation for the formation of bond in a molecule on the lines very similar to those of atomic orbitals. The essential guidelines of this theory are as follows. Like atomic orbitals in an atom, there exists molecular orbitals in a molecule. The only difference is that an atomic around more than one nucleus and thus belongs to the molecule as a whole). Electrons in a molecule occupy molecular orbitals in accordance with aufbau principle, Pauli’s exclusion principle and Hund’s rule.

the electron.

Molecular Orbitals Built as a Linear Combination of Atomic Orbitals Consider the formation and breaking of hydrogen molecule shown in Fig. 22. a b) transform into molecular orbitals Y1 as the two atoms are brought near to each other and vice versa.

Fig. 22

Chemical Bonding and Molecular Structure

4.19

Thus, molecular orbitals may be constructed as the linear combination of valence atomic orbitals. a b), two combinations can be taken as shown in the following. Y1 a b Y2 a b Thus, the linear combination of two atomic orbitals leads to two molecular orbitals. In general, the number of molecular orbitals formed is always equal to the number of atomic orbitals being combined.

Bonding Molecular Orbital Consider the molecular orbitals Y1 a b of electron in an orbital is described by the square wave function, we have 2 2 Y12 a)} b)} a b)} 2 2 )} )} give the probability of electron associated with atoms a and b, respectively. The term a b )} enhances the probability in the bonded region of the molecule, causing a decrease in the nucleusa b nucleus repulsion relative to increase in electron-nuclei attractions. This make the orbital more stable relative to those of atomic orbitals. Hence, this orbital is known as bonding molecular orbital. For the molecular orbital Y2

Antibonding Molecular Orbital Y 22

a)}

2

b)}

2

a

a

b),

the probability distribution is

b)}

a b)} diminishes the probability in the bonded region of the molecule causing increase in nuclear-nuclear repulsion relative to electron-nuclei attraction. This make the orbital less stable relative to those of atomic orbitals. Hence, this orbital is known as antibonding molecular orbital.

Shapes of Molecular Orbitals Y1 and Y2 The shapes of molecular orbitals of Y1 and Y2 are as follows. Bonding Molecular Orbital Y1 The positive combination of atomic orbitals in bonding molecular orbital is said to

Fig. 23

AntiBonding Orbital Y2 The negative combination of atomic orbitals in antibonding molecular orbital is said to be

Fig. 24

Energies of Molecular Orbitals Y1 and Y2 The energies of bonding orbital Y1 and antibonding orbital Y2 relative a b) are shown in Fig. 25. The decrease in energy for a bonding molecular orbital and increase in energy for an antibonding orbital depend on extent of overlapping between the involved atomic orbitals. The larger the overlapping, the greater are the stability and instability of bonding and antibonding orbitals, respectively.

4.20 Complete Chemistry—JEE Main

Fig. 25

Guidelines for Constructing Molecular Orbitals difference of the two atomic orbitals. Example The extent of mixing decreases with the decrease in the overlap of atomic orbitals placed at the stable internuclear distance. If the overlap is zero, there is no mixing of atomic orbitals. Example z Only the valence atomic orbital are considered for the construction of molecular orbitals. The inner core atomic orbitals are strongly bounded to their respective nuclei and thus are considered as nonbonding orbitals. Example In the compounds of carbon, we have

Fig. 4.26

1s2 2s2 2p1x 2p1y 2p z considered as nonbonding orbital

participate in the formation of molecular orbital

Sigma and Pi Molecular Orbitals Sigma Molecular Orbital

In this orbital, electron density is concentrated symmetrically around the line joining s.

Examples

Fig. 27

Chemical Bonding and Molecular Structure

Pi Molecular Orbital

4.21

In this orbital, electron density is concentrated above and below the line joining the two p.

Examples

Fig. 28

Designation of Molecular Orbital The designation of a molecular orbital starts by starting its s or p nature followed by the atomic orbitals into which it separates at larger distance. The antibonding orbital is designated by placing an asterisk on the symbol s or p. Molecular Orbitals of Homonuclear Diatomic Molecules of Second Period considered are x

y

z

x

y

z

1. Combination Involving 2s(A) and 2s(B)

Fig. 29a

2. Combination Involving 2pz(A) and 2pz(B)

Fig. 29b

The valence atomic orbitals to be

4.22 Complete Chemistry—JEE Main

3. Combination Involving 2px(A) and 2px(B)

Fig. 29c

4. Combination Involving 2py(A) and 2py(B) The molecular orbitals formed are similar to p2px and p*2px in the direction of y-axis.

Relative Energies of Molecular OrbitalsCorrelation Diagram diagram expected for the orbitals of atoms of second period. In this diagram, E s E p2p) since the end-to-end overlap of 2p orbitals is expected to be larger than side-ways overlap. found to be applicable only for O2 and F2 molecules. The relative energies of molecular orbitals is s s s2pz p2px = p2py p*2px = p*2py s*2pz For the molecules Li2 to N2, the energies difference

Es E p2p) The relative energies of molecular orbitals is s s p2px = p2py s2pz p*2px = p*2py

Fig. 30

Correlation diagram for O2 and F2

s*2pz

Electronic Structure and Molecular Characteristics The following guidelines are helpful in describing molecular characteristics. 1. The valence electrons are distributed among the molecular orbitals in accordance with aufbau principle, Pauli’s exclusion principle, and Hund’s rule. Aufbau principle: Pauli exclusion principle: No more than two electrons can occupy a single molecular orbital. The two electrons have opposite spins. Hund’s rule: The degenerate orbitals are singly occupied by electrons with parallel spins followed by double occupancy.

Chemical Bonding and Molecular Structure

Fig. 31

4.23

Correlation diagram for Li2 to N2

bond order one half of the net excess of bonding electrons, i.e. Number of (bonding - antibonding) electrons Bond order = 2 3. The strength of a bond depends on the bond order of the molecule. The larger the bond order, the stronger the bond and larger the dissociation energy of the molecule. 4. Addition of an electron in the bonding orbital or removal of an electron form the antibonding orbital increases bond order and hence increases stability of a molecule. bond order and hence decreases stability of a molecule. 6. Paramagnetism in a substance is due to the presence of unpaired electrons in its molecules. Explanation Paramagnetic Substance

A substance which is attracted towards a magnet is known as paramagnetic substance.

Paramagnetism

The phenomenon of attracting a substance towards a magnet is known molecules of a substance.

Diamagnetic Substance

A substance containing no unpaired electrons is known as diamagentic substance. The pairing of electrons annule the tiny magnets internally.

7. Bond length is inversely related to the bond order.

4.24 Complete Chemistry—JEE Main

Molecular Characteristics Species H2

Valence electrons 2

Bond Order † s1s)2 2

H+2

1

s1s)1 2

=1 =

1 2

He2

4

s1s)2 s*1s)2

2–2 2

He+2

3

s1s)2 s*1s)1

2–1 1 = 2 2

Li2

2

s2s)2 2

=1

Characteristics Diamagnetic as there is no unpaired electron 2, bond length increases, bond dissociation energy decreases Paramagnetic as there is one unpaired electron. Does not exist as bond order is zero

Expected to exist, paramagnetic Diamagnetic

Be2

4

s2s)2 s*2s)2

2–2 2

Does not exist

B2

6

s2s)2 s*2s)2 p2px)1 p*2py)1

4–2 =1 2

Paramagnetic

C2

8

s2s)2 s*2s)2 p2px)2 p2py)2

6–2 =2 2

Diamagnetic

N2

s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2

8–2 =3 2

N2+

s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)1

1 7–2 =2 2 2

s and 2p bonds) Maximum bond dissociation energy Minimum bond length, Diamagnetic 2 , bond dissociation energy decreases, bond length increases Paramagnetic

1 s2s) 2 s*2s) 2 p2p x ) 2 p2p y ) 2 s2p z ) 2 8 – 3 =2 p*2px)1 2 2

— same —

12

s2s) 2 s*2s) 2 s2p z ) 2 p2p x ) 2 p2p y ) 2 8 – 4 = 2 p*2px)1 p*2py)1 2

Paramagnetic as it contains two unpaired electrons

11

s2s) 2 s*2s) 2 s2p z ) 2 p2p x ) 2 p2p y ) 2 8 – 3 = 2.5 p*2px)1 2

2, bond order is increased. Thus bond dissociation energy increases and bond length decreases Paramagnetic

N2–

11

O2 O2+

Note: These effects are just opposite to those observed in going from N2 to N2+. O2–

13

1 s2s) 2 s*2s) 2 s2p z ) 2 p2p x ) 2 p2p y ) 2 8 – 5 =1 p*2px)2 p*2py)1 2 2

2, bond order is decreased. Thus, bond dissociation energy decreases and bond length increases

Note: The species O2– is expected to be less stable than O+2 due to larger electron-electron repulsion. F2

14

s2s) 2 s*2s) 2 s2p z ) 2 p2p x ) 2 p2p y ) 2 8 – 6 = 1 p*2px)2 p*2py)2 2

Diamagnetic

Ne2

16

s2s) 2 s*2s) 2 s2p z ) 2 p2p x ) 2 p2p y ) 2 8 – 8 p*2px)2 p*2py)2 s*2pz)2 2

Does not exist

†

Chemical Bonding and Molecular Structure

Heteronuclear Diatomic Molecules Figure 32 displays the correlation diagram for heteronuclear diatomic molecules. The sequence of energy levels is very electronegative than the atom A. Examples NO s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 p*2px)1 NO+ s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 lesser bond length.

Fig. 32

MULTIPLE CHOICE QUESTION ON SECTION 2 Identify the correct choice in the following questions. 1. The hybridization of chlorine orbitals in the compound ClF3 is 3

2

3

3d 2

2. The hybridization of iodine orbitals in the compound IF7 is 3 3d 2 2d 3 3. The hybridization of nitrogen orbitals in the compound NO3– anion is 2

3

3d 3 2

d

3

4. Which of the following ions does not involve sp hybridization? – 2

– 2 3

6. The hybridization of P in

– 4

3– 4

is

PO43–

2

3

2

d

2

3

2

d

is

7. The hybridization of Xe in XeO3 is 2

3

8. The hybridization of Xe in XeOF4 is 3 3 2 d not correct?

2

2

2 3

d

4

4.25

4.26 Complete Chemistry—JEE Main 2– 2 ion,

respectively are

11. The bond order of CN– is 12. Which of the following pairs of species has identical value of bond order? +, N –, O + + 2 2 2 13. Which one of the following is expected to have maximum bond length? +

2

and O+2

2+

–

14. Which of the following is expected to be diamagnetic? + 2

2

2

+ 2

– 2

+ 2

– 2

+ 2

16. Which one of the following species will have maximum bond dissociation energy? + 2

2

– 2

_ 2 2

17. Which of the following statements regarding carbon monoxide is correct?

ANSWERS

HINTS AND SOLUTIONS 1. sp3d hybridization is involved. 3s

3p

3d

sp3d

2. sp3d3 hybridization is involved. 5s 5p

5d

sp3d3

3. NO3– has trigonal planar shape. It involves sp2 hybridization of N. 4. ICl–2 involves sp3 5s 5p 5d

–

ion is

sp3d hybridization

Two single electrons form s bond with two chlorine atoms. There are three lone pairs of electrons.

Chemical Bonding and Molecular Structure 3

3s

4.27

is 3p

sp2 hybridization Form s–bonds with oxygen

3d

Form double bonds with oxygen atoms

–

to form 3– ion is 3p

3s

sp3 hybridization Form s–bonds with four oxygen atoms

5s

Form double bonds with oxygen atoms

5d

5p

sp3 hybridization The unpaired electrons form s bonds with three oxygen atoms

5s

3d

These form 3 p–bonds with oxygen

5d

5p

sp3d2 hybridization The unpaired electrons form s–bonds with four F atoms and one oxygen atom

Form p–double bonds with oxygen

s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)2 p*2py)2 Bonding electrons = 8 Antibonding electrons = 6 11. Valence electrons in CN– s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 (Bonding - antibonding) electrons 8 - 2 Bond order = = =3 2 2 12. Valence electrons of N2 s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 2– 2

Valence electrons of NO+

s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 +

Valence electrons of CN– Valence electrons of N–2 Valence electrons of O2+

+

s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)1 s2s) s*2s) p2px) p2py) s2pz) p*2px)

+

NO NO2+

4.28 Complete Chemistry—JEE Main

NO– –

will having maximum bond length.

+ – + 15. Bond orders are O–2 2 2 2 2 2 2 2 s 2s) s*2s) p*2px) p2py) 16. C2 + C2 s2s)2 s*2s)2 p2px)2 p2py)2 s2py)1 C–2 2– C2 Larger the bond order, larger the dissociation energy. – + 17. The CO molecule is represented as C ¨ O

Dipole moment is low due to pull of bonding electrons towards more electronegative oxygen atom.

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE CHAPTER General Characteristics 1. Which of the following molecules does not follow the octet rule? 2

3

2

2. In which one of the following compounds does hydrogen bonding occur? 3. Which of the following compounds would show evidence of the strongest hydrogen bonding? 4. A covalent bond is most likely to be formed between two elements which

5. A covalent bond is formed between

6. An electrovalent bond is formed between

7. Which of the following orbitals of a diatomic molecule AB oriented along x-axis will not have positive overlap? x

x

x

z

8. Which of the following occurs when two hydrogen atoms bond with each other?

3

2

2O

2

is about 11. Which of the following unit conversion of dipole moment is correct? Cm D

Chemical Bonding and Molecular Structure

4.29

12. Which of the following unit conversion is correct? ¥ –12 ¥ –14 C m –16 ¥ ¥ –18 C m 13. o-Nitrophenol is more volatile than p-nitrophenol. It is due to o o-nitrophenol p p-nitrophenol 14. The maximum number of hydrogen bonds that can be formed by water molecules is 15. Which of the following interactions is not a part of van der Waals interactions?

16. Which one of the followings contributes maximum to the van der Waals interactions?

VSEPR and Hybridization

5

which contains 5 bonding pairs and one lone pair has a shape of

21. Hybridization involves

4

molecule involves 2

3 4

3d

]2–

is

24. The geometry arrangement of F atoms around P in PF5 is 25. The geometrical arrangement of NH3

3) 6]

26. The sp3d hybridization of central atom of a molecule would lead to

27. The sp3d2 hybridization of central atom of a molecule would lead to

28. The bond angle in PH3 is expected to be

3+

is

orbitals

4.30 Complete Chemistry—JEE Main + 2, 2,

NO3– and NH 4+ respectively are

sp3

3,

2– 4

sp2

2,

sp, sp3 2– 4

4

31. Which of the following species involves 2

3,

sp2

sp2, sp

2Cl2

hybridization?

2

2

2

32. The species ClO2– involves 2

3

33. Which of the following species involves 3

sp3

2

hybridization

hybridization? – 3

3

3

34. The species ClO3– involves 2

3

3

hybridization

35. Which of the following statements is correct? 3

involves sp3 hybridization and NF3 involves sp2 hybridization

3

involves sp2 hybridization and NF3 involves sp3 hybridization 3

and NF3 involve sp3 hybridization

3

and NF3 involve sp2 hybridization

36. The hybridization in OF2 is 2 2

3

2

is 145 pm and its bond moment is 5.8 ¥

C m. Its per cent ionic character

is about 38. Which of the following species involves the smallest bond angle? 3

2

2

2

2

4

2

2

Molecular Orbitals not true? between different atoms are considered

41. Ethyne molecule contains bonds and 1 bonds and 3

bonds and 2 42. Which of the following molecules is paramagnetic? 2

2

bond bonds

2

2

43. In which of the following species, the bond length is expected to be the greatest? 2– 2

44. The bond order of O2 molecule is 45. The bond order of N2 molecule is

– 2

2

+ 2

Chemical Bonding and Molecular Structure

4.31

46. The number of molecular orbitals obtained by mixing two atomic orbitals from each of the two atoms is 47. Which of the following order of energies of molecular orbitals of N2 is correct? E *2px) = E *2py) E 2px) = E 2py E E *2px) = E *2py) E 2px) = E 2py E E *2px) = E *2py) E 2px) = E 2py E E *2px) = E *2py) E 2px) = E 2py E 48. Which of the following order of energies of molecular orbitals of O2 is correct? E *2px) = E *2py) E 2px) = E 2py E E *2px) = E *2py) E 2px) = E 2py E E *2px) = E *2py) E 2px) = E 2py E E *2px) = E *2py) E 2px) = E 2py E not correct from the viewpoint of molecular orbital? 2 is not a stable molecule + 2 is not stable but He 2 is expected to exist 2 is maximum amongst the homonuclear diatomic molecules belonging to the second period 2 molecule is E *2px) = E *2py E *2p) E E E 2px) = E 2py E – 2

2

+ 2

– 2

2

+ 2

– 2

2

+ 2

– 2

2

+ 2

2

+ 2

– 2

2

+ 2

51. Which of the following orders regarding the bond length is correct? – 2

2

+ 2

– 2

2

+ 2

– 2

52. Which of the following facts regarding change in bond length is correct? + + 2 to N2 , decreases in going from O2 to O2 + + 2 to N2 , increases in going from O2 to O2 + + 2 to N2 and O2 to O2 + + 2 to N2 and O2 to O2 53. Which of the following orders is correct for the bond dissociation energy of O2, O2+, O2– and O22–? + 2

2

– 2

2– 2

2

+ 2

2

– 2

2– 2

2

+ +

2

– 2

2– 2

2

– 2

2– 2

54. Which of the following species has the shortest bond length? + 2

2

– 2

2– 2

– 2

2– 2

55. Which of the following species has the largest bond length? + 2

ANSWERS

2

4.32 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS 1. 2. 3. 4.

Boron in BCl3 does not satisfy octect rule. There are 6 electrons around B. Hydrogen attached to the electronegative nitrogen shows hydrogen bonding. Larger the number of —OH groups, the more extensive the hydrogen bondings. Atoms having similar electronegativity form covalent bond.

6. An electrovalent bond is formed between electropositive and electronegative atoms z

8. Potential energy is lowered.

=

(1.03 ¥ 3.3356 ¥ 10-30 C m) ¥ 100 = 16.9% (1.602 ¥ 10-19 kg)(127 ¥ 10-12 m)

11. 1 D = 3.3356 ¥ 12. 4.8 ¥

C m. ¥

C. Hence

Ê 1.602 ¥ 10-19 C ˆ -2 (10 m) = 3.3356 ¥ –12 C m 1 esu cm = Á Ë 4.8 ¥ 10-10 esu ˜¯ 13. o–Nitrophenol involves intramolecular whereas p-nitrophenol involves intramolecular hydrogen bondings. 14. Water molecule is involved in four hydrogen bondings. 15. Ion-dipole interaction is not counted in van der Waals forces. 16. Dispersion forces have maximum contribution. 17. Four lone pairs of electrons lead to square planar arrangement. 18. Three lone pairs of electrons lead to triangular arrangement.

21. Hybridization involves mixing atomic orbitals of the same atom. Cl

Cl

Cl

Cl

¥ 7) electrons. These are distributed as

4

3d

3d 23. Ni

2+

4s

4p

ion

in complex dsp2 hybridization

24. PF5

¥ 5 ¥ 7) valence electrons. These are distributed as

F

F P

F

F

F

Chemical Bonding and Molecular Structure

25. 26. 27. 28.

4.33

There are 6 ligands attached to Cu. There arrangement will be octahedron. sp3d hybridization leads to trigonal bipramide geometry. sp3d2 hybridization leads to octahedral geometry. PH3 ¥ 1) valence electrons. There are distributed as H P H. There are 4 pairs of electrons around H

P. + 2

has 16 valence electrons. There arrangement will be O N O . Two pairs of electrons around N implies – ¥ 3 ¥ 6 + 1) valence electrons. These are arranged linear geometry involving sp hybridization. NO 3 as O N O . There are three pairs of electrons around N. Its geometry will be trigonal planar and will result from 2– 4

sp2

O hybridization. NH +4 has sp3 hybridization. 2– 4

2Cl2

have 4 pairs of electrons around the central atoms which lead to tetrahedral geometry.

4 2

3

hybridization is

involved. 32. ClO–2 has four pairs of electrons around Cl which leads to tetrahedral geometry, hence sp3 hybridization is involved. 33. Only PCl3 has 26 valence electrons and P has four pairs of electrons around it which leads to tetrahedral geometry to PCl3. Hence, P involves sp3 hybridization. 34. The species ClO3– has 26 valence electrons with four pairs of electrons around Cl leading to tetrahedral geometry, hence sp3 hybridization 35. BF3 has 24 valence electrons with three pairs of electrons around B leading to trigonal planar geometry, hence sp2 hybridization. NF3 has 26 valence electrons with four pairs of electrons around N leading to tetrahedral geometry, hence sp3 hybridization. 3 36. OF2 hybridization. 5.8 ¥ 10 -30 C m ¥ 100 = 25% 37. Per cent ionic character = (1.602 ¥ 10 -19 C)(145 ¥ 10 -12 m) 38. H2O will have minimum bond angle due to repulsion between lone pair-bonding pair electrons. It will be smaller than NH3 as the latter has only one lone pair while H2O has two lone pairs. 2 2 involves sp

42. 43. 44. 45. 46.

s-bonds and two p-bonds. Only O2 has two unpaired electrons and hence paramagnetic. The bond order of O2– 2 is the minimum and thus it has largest bond distance. The bond order of O2 The bond order of N2 These will be as many molecular orbitals as the number of atomic orbitals being mixed.

4.34 Complete Chemistry—JEE Main

54. N2 has the shortest bond length as its bond order is largest. 55. N2– 2

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The pair of species having identical shapes for molecules is BF3, PCl3 5, IF5 4 4 2, CO2 2. Which of the following pair of molecules will have permanent dipole moments for both members? 2 and CO2 4 and NO2 2 and CO2 2 and O3 3. Which one of the following compounds has the smallest bond angle in its molecule? 3 2 2O 2 , BF is 2 3 3 4 2

4

3

3

2

4

[2003] [2003]

3

[2004] 2 3 3 4 + The bond order in NO is 2.5 while that in NO is 3. Which of the following statements is true for these two species? 2

5.

3

[2003]

3

4

3

+ + is

+

equal to that in NO

[2004]

3BO3) are respectively 2 and sp2 2 and sp3 3 and sp2 sp3 and sp3 7. Which one of the following has the regular tetrahedral structure? 2– 4 4 4 4] 3

+ 2

3

2

– 2

+ 2

hybridization

[2004] [2004]

sp3d2 hybridization [2004] 2

[2005]

[2005] 11. Based on lattice energy and other consideration which one of the following alkali metal chlorides is expected to have the highest melting point? [2005] , CF and XeF are 4 4 4

[2005] 13. The number and type of bonds between two carbon atoms in calcium carbide are [2005][2011][2014] [2005] 2

2– 2

2

+ 2

[2006]

Chemical Bonding and Molecular Structure

4.35

not equal? – 4

[2006] 4 4 4 17. In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed? + + + Æ NO+ [2007] 2 Æ C2 2 Æ O2 2 Æ N2 18. Which of the following hydrogen bonds is the strongest? H… H… H…O O H… F [2007] 2– 2

+ 2

O

2

[2007]

the increasing order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+? 2+

2+

+

+

2+

2+

2+

2+

+

2+

2+

2+

2+ 2+ + Ca2+ [2007] 21. Which one of the following pairs of species have the same bond order? + – – and CN+ and NO+ and CN+ O2– and CN– [2008] –1 –1 22. The bond dissociation energy of B—F in BF3 is 646 kJ mol whereas that of C—F in CF4 is 515 kJ mol . The

p – pp interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4. p – pp interaction between B and F in BF3 than that between C and F in CF4. s bond between B and F in BF3 as compared to that between C and F in CF4. 23. Using MO theory predict which of the following species has the shortest bond length? 2– 2+ O2+ 2 2 2 24. Among the following maximum covalent character is shown by the compound 2

[2009] [2009]

2

AlCl3 25. Which of the following has maximum number of lone pairs associated with Xe?

[20011 (Cancelled)]

2

3

4

XeF2

6

[2011]

26. The structure of IF7 is [20011 (Cancelled)] NO3–,

27. The hybridization of orbital of N atom in 2, sp3 2, sp, sp3 28. The molecule having smallest bond angle is 3 2– 3

NO2+,

and

NH+4,

respectively 2, sp3 3, sp2

3

and NO–3

3

2

[20011 (Cancelled)]

3

4

2

5

[2012]

3 3– 6

and Br5

2

[2013]

6

[2013]

2

not likely to exist? +, 2

He22– – 2, Li 2 – 2

He22–

Li+2 2 33. Which one of the following properties is not shown by NO? 2

+ 2

and

–, 2 Li+2 – 2

2+, 2

2

– 2,

He2 – 2

+ 2

It combines with oxygen to form nitrogen dioxide

Li–2

He2+

[2013] 2

+ 2

[2013]

[2014]

4.36 Complete Chemistry—JEE Main

+ O22– [2014, online] 2 – The correct order of bond dissociation energy among N2, O2, O 2 is shown in which of the following arrangements? – – – – O2 2 2 2 2 2 2 2 2 2 2 2 – 2

2

35.

[2014, online] not correctly stated? 7

5

ICl3

3

[2014, online]

3H 4 3

2

2

sp2 and sp3 [2014, online]

38. Which of the following molecules is paramagnetic? O3

2

[2014, online]

Assertion 2, electron density is increased between the nuclei Reason The bonding MO is yA + yB, which shows the destructive interference of combining electron waves

[2015, online] 4

e

¥

esu)

[2015, online]

42. Choose the incorrect formula out of the four compounds for an element X below. 2Cl3 2O 3 2 4) 3 43. The group of molecules having identical shape is 5, IF5, XeO2F2 3, PCl3, XeO3 + 4, XeF4, CCl4 3, XeOF2, XeF 3

4

[2015, online]

2 2

[2016, online] 6 2 6? [2016, online]

2O

[2016, online]

+ 2

[2016]

2

3

4

3

46. The species in which the N atom is in a state of sp hybridization is – 2

ANSWERS

– 3

2

Chemical Bonding and Molecular Structure

4.37

HINTS AND SOLUTIONS 1. 2. 3. 4.

Both XeF2 and CO2 have linear structures. Both NO2 and O3 have bent structure. Their dipole moment will not be zero. ¢ The bond angles of NH3 2, H2O and H2 ¢ Bond angle is H2 3

¢. Bond angle in BF3 is

4

5. The bond order and bond length have inverse dependence, i.e. the larger the bond order, shorter the bond length. + 2

6. Boric acid is a planar molecule as boron involves sp hybridization and oxygen involves sp3 hybridization. ¥ 7) valence electrons. These are distributed as shown in Fig. 1a. There are six pairs of 7. XeF4 electrons around Xe. These are distributed octahedrally. Thus, the shape of XeF4 is shown in Fig. 1b. F

F

F

F xe

Xe F

F

F

F

Fig 1a

Fig 1b

¥ 7) electrons. These are distributed as shown in Fig. 2a.

4

F

F

F

F

F

F

F F

Fig 2a

Fig 2b

BF–4 ¥ 7 + 1) electrons. These are distributed as shown in Fig. 3a. There are four pairs of electrons around B. They assume tetrahedral geometry. The shape of BF–4 F

F

F B

B

F

F

F

Fig 3a 4]

2–

F F

Fig 3b

, we have 3d 28Ni

4s

4p

atom

Ni2+ ion Ni2+ ion is complex dsp2 hybridization 2– 4] is shown in Fig. 4. Thus BF–4 has the regular tetrahedral structure.

CN

NC Ni

CN

NC Fig. 4

2

4.38 Complete Chemistry—JEE Main 3d 2

are 4, 6 and 12 such angles in dsp

2,

dsp3

and

sp3d2

hybridization. There

hybridizations, respectively.

s1s)1 He+2 s1s)2 s*1s)1 H2+ – 2 1 H2 s1s) s*1s) H2 s1s)2 A diamagnetic substance does not possess unpaired electron. Dihydrogen does not possess unpaired electron; hence it is diamagnetic. 11. Melting point of LiCl is less than that of NaCl and later it generally decreases on descending the group. The low melting point of LiCl is due to its covalent character. 4 = 6 + 4 ¥ 7 = 34. These are distributed as follows. F

F 4

F

will be trigonal bipyramide with

F

Number of Valence electrons in CF4 = 4 + 4 ¥ 7 = 32. These are distributed as follows. F

F C

F

There are four paired of electrons around C. The shape of CF4 is tetrahedral. There is no lone pair of electrons on the C atom.

F

Obviously the choice b is correct. The molecule of XeF4 will have different geometry with two lone pairs of ¥ 7) valence electrons and these are distributed as follows. F

F The shape of XeF4 will be square planar.

Xe F

F

13. The carbide in calcium carbide is [C

C]2–, which involves one sigma and two pi bonds. 3PO2) is O

There are two hydrogen atoms attached to P atom.

H

P

H

OH O 2;

s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)1 p*2py)1

O22–;

s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)2 p*2py)2

B 2; N2+;

s2s)2 s*2s)2 p2px)1 p2py)1 s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)1 No. of valence electrons

Lone pairs around the central atom

BF–4

32

4

Tetrahedral

4

34

5

Trigonal bipyramide

32

4

Tetrahdral

36

6

Octahedral geometry: square planar)

4

XeF4

Arrangement of lone pairs

Chemical Bonding and Molecular Structure 4,

4.39

all bonds will not be identical as its structure is F F F F

17. The order of increasing energy of molecular orbitals is s2s) s*2s) p2px) p2py) s2pz) p*2px p*2py) s*2pz) The valence electrons in C2, NO, O2 and N2 s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 p*2px)1 In converting NO to NO+, the electron is removed from antibonding molecular orbital and the resultant ion possesses no unpaired of electrons. 18. Larger the electronegativity of atom, stronger the hydrogen bond. s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)2 p*2py)2 O22–; O2+ and NO s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)1 O 2; s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)1 p*2py)1 2– Hence, O2 is diamagnetic as it does not contain unpaired electron. of the cation. – NO+ 4+5–1=8 O –2 6 + 6 + 1 = 13 CN+ + and CN– have the same number of valence electrons and thus are expected to have the same The species NO

p –pp interaction between B and F due to back bonding of lone pair of electrons on F to an empty p orbital on B. There is no such interaction in CF4. F B

F

F

s2s)2 s*2s)2 s2p)2 p2px)2 p2py)2; BO = 3 O22+ 2 2 2 2 2 1 + KK s2s) s*2s) s2p) p2px) p2py) p*2px) ; BO = 2.5 O2 KK s2s)2 s*2s)2 s2p)2 p2px)2 p2py)2 p*2px)2 p*2py)1; BO = 1.5 O2– KK s2s)2 s*2s)2 s2p)2 p2px)2 p2py)2 p*2px)2 p*2py)2 O22– The bond length is inversely proportional to the bond order. Hence O22+ is expected to have shortest bond length. hence larger covalent character to the bond between cation and anion. 25. We have 1 ¥ 6 = 26 Lone pair of electrons around Xe = XeO3 2 1 ¥ 7 = 36 Lone pair of electrons around Xe = XeF4 2 1 XeF6 ¥ Lone pair of electrons around Xe = 2 1 XeF2 ¥ 7 = 22 Lone pair of electrons around Xe = 2

¥ 8) =1 ¥ 8) =3 ¥ 8) =1 ¥ 8) =3

4.40 Complete Chemistry—JEE Main

¥ 7) valence electrons in IF7. There with be 7 pairs of electrons around iodine. These will be oriented in the directions of pentagonal bipryamid as shown in the following structure. F F F F I F F F 27. The ion NO–3 ¥ 6 + 1) valence electrons. These will be distributed as – O N O There are 3 electron-pairs is around N. The shape of NO3– will be trigonal planar. Hence sp2 hybridization of N orbitals is involved. O The ion NO+2 + O N O

¥ 6 – 1) valence electrons. These will be distributed as There are 2 electron pairs around N. The shape of NO+2 will be linear. Hence, sp hybridization of N orbitals is involved.

It can be shown that the NH+4 involves sp3 ¥ 1 – 1). These are distributed as + H These are four electron-pairs around N. The shape of NH4+ will be tetrahedral. Hence, sp3 hybridization H N H of N orbitals is involved. H 28. The bond angle Cl A increase in the size of the central atom A. Thus, out of NCl3, PCl3, AsCl3 3, the latter is expected to have the smallest bond angle. 5 and BrF5), the number of paired of electrons around the central atoms of the two compounds in each pair are identical. Hence, the compounds in each pair have identical structures. In PF5 and BrF5, the number F F F

F Five paired of electrons around P, dsp3 hybridization of P orbitals

2

s2s)2 s2s)2 s2s)2 s3s)2

s*2s)2 s*2s)2 s*2s)2 s*3s)2

p2px)2 p2px)2 s2pz)2 p3px)2

F Br

F

C2 N2 O2

F

P F

F F 3 2

d

hybridization of Br orbitals

p2py)2 p2py)2 s*2pz)2 p2px)2 p2py)2 p*2px)1 p*2py)1 p3py)2 s*3pz)2 p*3px)1 p*3py)1

The molecules C2 and N2 do not possess unpaired electrons. Hence, these are expected to exhibit diamagnetic behaviour. 31. The species H22+ 2 + – 2, Li 2, Li 2 and their bond orders are 2 + 1 Li 2 s2s) Li–2 s2s)2 s*2s)1 Li2 s2s) The species Li–2 will have larger electronic repulsion as compared to Li+2 and thus will be less stable. + – Thus, the stability order is Li2 2 2.

Chemical Bonding and Molecular Structure

4.41

s2s)2 s*2s)2 p2px)2 p2py)2 s2p)2 p*2px)1 Molecule is paramagnetic as it contains one unpaired electron. Bond order =

(Bonding - antibonding) electrons 8 - 3 1 = =2 2 2 2

NO is a neutral oxide. It combines with O2 to give NO2. N2 N22+ O2–

s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2 s2s)2 s*2s)2 p2px)2 p2py)2 s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)2, p*2py)1

O22–

s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)2 p*2py)2

s2s)2 s*2s)2 p2px)2 p2py)2 s2pz)2; BO = 3 N2 2 2 2 2 2 1 1 KK s2s) s*2s) s2pz) p2px) p2py) p*2px) p*2py) ; BO = 2 O2 KK s2s)2 s*2s)2 s2pz)2 p2px)2 p2py)2 p*2px)2 p*2py)1 O2– The larger the bond order, larger the dissociation energy. F F F 36. IF7; Valence electrons = 7 + 7 ¥ 7 = 56 I F F F F F

BrF5; Valence electrons = 7 + 5 ¥ 7 = 42

F

F

F F

Br F

BrF3; Valence electrons = 7 + 3 ¥ 7 = 28

F

F F

Five pairs of electrons around Br

Br

F

F Br

F

F F Br

F

F Cl

Cl I

ICl3 Cl 1

37. The allene is H2C hybridized.

I Cl

2

C

Cl Planar dimer Cl

1

CH2. Carbon labelled as 1 are sp2 hybridized and carbon labelled as 2 is sp

38. The molecule NO contains odd number of valence electrons. It is expected to be paramagnetic.

4

F

¥ 7). These are distributed as follows. F Xe

F

F

O

There are 6 pairs of electrons around Xe. Their orientations will be square bipyramidal. The geometry of XeOF4

4.42 Complete Chemistry—JEE Main

is square pyramid as the lone pair of electrons occupy apical position F F F

Xe

F

O

p = qr, we have p 0.38 ¥ 10-18 esu cm = 2.35 ¥ –11 esu q= = r 1.617 ¥ 10-18 cm 2.35 ¥ 10-11 esu Fractional charge on each atom = 4.802 ¥ 10-10 esu 42. The incorrect formula is X2Cl3. 43. Only the species given in the choice d have the same number of valence electrons of 28.

F

Cl

F F

O

Xe

F

F

F

Xe

F

+

F

lone pair of electrons occupy the equatorial positions giving T-shape to the molecules. 44. Na+ and F–

2

K+, Cl– and Ca2+

2

6.

2

2

6

2

6.

45. CH4 pair-bonding pair repulsion, the angle H — X — H is decreased from the expected bond angle of tetrahedral ¢. 46.

Species

Number of valence electrons

NO–2

5 + 2 ¥ 6 + 1 = 18

NO–3

5 + 3 ¥ 6 + 1 = 24

Distribution of valence electrons

O

N

O

O

N

O

Number of lone pairs in central atom

Hybridization

-

3

sp2

-

3

sp2

+

2

sp

O NO2

5 + 2 ¥ 6 = 17

NO+2

5 + 2 ¥ 6 – 1 = 16

O

N

O

O

N

O

Thus, N is NO+2 has two lone pairs of electrons around it and is thus sp hybridized.

5 Solutions

SECTION 1

Composition of a Solution

Solutions are homogeneous mixtures of two or more than two constituents. The constituent present in largest amount is called the solvent while those present in lesser amounts are known as solutes†. Composition of a solution In order to deal with a solution, it is necessary to know its composition. This may be expressed in the following ways. (i) Mass percentage The mass fraction of B in solution is Mass of B in solution wB = (1) Total mass of solution Mass percentage of B = wB × 100 (ii) Volume percentage The volume fraction of B in a solution is Volume of liquid B fB = (2) Volume of solution Volume percentage of B = fB ¥ 100 Note: Total volume of the solution may or may not be equal to the sum of volumes of its constituents depending upon the fact whether the solution is ideal or not. (iii) Mass by Volume Percentage For this, we have Mass of constituent B Mass by volume percentage of B = × 100 (3) Volume of solution (iv) Parts per million For this, we have Number of parts of B ¥ 106 (4) Parts per million of B = Total number of parts of all constituennts It is abbreviated as ppm. Parts per million may refer to mass to mass, volume to volume and mass to volume way of expressing composition. (v) Amount (or Mole) Fraction nB n Amount of constituent B in the solution = = B (5) xB = Total amount of consstituents in the solution  B nB ntotal

† Throughout this unit, the solvent is represented by the subscript 1 and solute by the subscripts 2,3,....., and so on.

5.2

Complete Chemistry—JEE Main

The sum of amount (or mole) fractions of constituents in a solution is 1. Ê n ˆ  n  B xB =  B Á B ˜ = B B = 1 Ë Â B nB ¯  B nB Amount (or mole fraction) is independent of temperature. (vi) Molarity M=

(6)

n Amount of solute in solution = 2 3 Volume of solution expressed inn dm (i.e. L) V

(V expressed in dm3, i.e. L)

(7)

The unit of molarity is mol dm–3 ( i.e. mol L–1). It is often represented by the symbol M. The molarity of a solution depends on temperature owing to the variation of volume with temperature. (vii) Molality n Amount of solute = 2 (m1 expressed in kg) m= (8) Mass of solvent expressed in kg m1 The molality carries the unit of mol kg–1 . It is independent of temperature.

Illustration Calculate the mole fraction of solute, molality and molarity of a solution containing 24 mass per cent of acetic acid. The density of solution is 1.05 g cm–3. In 100 g of solution, we have Mass of acetic acid, m2 = 24 g; Mass of water, m1 = 76 g = 0.76 kg Amount of acetic acid, n2 =

m2 m 24 g 76 g = = 0.40 mol ; Amount of water, n1 = 1 = = 4.22 mol -1 M 2 60 g mol M1 18 g mol-1

Mole fraction of acetic acid is

x2 =

n2 0.40 mol = = 0.087 n1 + n2 4.22 mol + 0.40 mol

Molality of solution is

m=

n2 0.40 mol = = 5.26 mol kg -1 m1 0.076 kg

To calculate molarity, we need to know volume of the solution, which is Mass of solution 100 g = = 95.24 cm3 V= Density of solution 1.05 g cm -3 n 0.40 mol = 4.20 mol dm -3 (i.e. 4.20 mol L–1 or 4.20 M) Molarity of solution is M = 2 = 3 V 0.09524 dm Interconversion Expressions Let we represent Amount of solute = n2 Molar mass of solute = M2 Mass of solute, m2 = n2M2 Amount of solvent = n1 Molar mass of solvent = M1 Mass of solvent, m1 = n1M1 m1 + m2 n1M1 + n2 M 2 = r r

Density of solution = r

Volume of solution, V =

n2 V n2 Molality of solution, m = m1

; (V expressed in dm3)

(9)

;(m1 expressed in kg)

(10)

Molarity of solution, M =

n2 n1 + n2 Molarity into Molality and Vice Versa We have n2 n2 n2 / n1 M1 m mr = = = = M= V (n1 M1 + n2 M 2 ) / r [1 + (n2 / n1 M1 ) M 2 ] / r (1 + mM 2 ) / r 1 + mM 2 Mole fraction of solute, x2 =

(11)

(12)

Solutions

Rearranging Eq. (12), we get

m=

M r - MM 2

Molarity into Mole Fraction and Vice Versa We have (n2 / n1 ) r n2 n2 = = M= V (n1M1 + n2 M 2 ) / r M1 + (n2 / n1 ) M 2 MM1 n2 This gives = n1 r - MM 2 n2 n /n MM1 / ( r - MM 2 ) MM1 = 2 1 = = Now x2 = n1 + n2 1 + n2 / n1 1 + MM1 / ( r - MM 2 ) M ( M1 - M 2 ) + r x2 r x2 r x2 r = = Rearranging Eq. (14), we get M = M1 - x2 ( M1 - M 2 ) (1 - x2 ) M1 + x2 M 2 x1M1 + x2 M 2

5.3

(13)

(14) (15)

Molality into Mole Fraction and vice versa We have (n2 / n1 ) (n2 / m1 ) M1 n2 n /(m1 / M1 ) = = 2 = x2 = n1 + n2 1 + (n2 / n1 ) 1 + n2 /(m1 / M1 ) 1 + (n2 / m1 ) M1 which in view of Eq. (10) becomes x2 = Rearranging Eq. (16), we get

m=

mM1 1 + mM1

(16)

x2 x = 2 M1 (1 - x2 ) x1M1

(17)

MULTIPLE CHOICE QUESTIONS ON SECTION 1 1. An antifreeze solution is prepared by mixing 217.0 g of ethylene glycol ( C2H6O2) with 200 g water. The molality of solution is (a) 15.5 mol kg–1 (b) 17.5 mol kg–1 (c) 18.5 mol kg–1 (d) 20.1 mol kg–1 2. If the density of a solution containing 217.0 g ethylene glycol ( C2H6O2) in 200 g water is 1.10 g cm–3, the molarity of the solution is (a) 8.8 mol dm–3 (b) 10.1 mol dm–3 (c) 9.23 mol dm–3 (d) 12.12 mol dm–3 3. A solution of glucose (C6H12O6) is 18% by mass. The mole fraction of glucose will be (a) 0.18 (b) 0.25 (c) 0.0214 (d) 0.015 4. The mass of an equimolar mixture of Na2CO3 and NaHCO3 is 1.0 g. The volume of 0.1 M solution of HCl required to react completely with this mixture is (a) 52.1 mL (b) 104.2 mL (c) 125.2 mL (d) 156.3 ml 5. A solution contains 441.0 g of H2SO4 in 1 L of solution. If the density of solution is 1.25 g mL–1, the molarity and molality of the solution, respectively, are (a) 4.5 mol L–1 and 5.56 mol kg–1 (b) 4.5 mol L–1 and 4.5 mol kg–1 (c) 5.56 mol L–1 and 4.5 mol kg–1 (d) 5.56 mol L–1 and 5.56 mol kg–1 6. A solution of HCl is prepared by mixing 60 mL of 0.12 M HCl and 40 mL of 0.25 M HCl solution. The molarity of HCl in the resultant solution is (a) 0.185 M (b) 0.150 M (c) 0.20 M (d) 0.172 M 7. The mole fraction of glucose in water is 0.2. The molality of the solution will be (a) 10.2 mol kg–1 (b) 12.4 mol kg–1 (c) 14.2 mol kg–1 (d) 16.6 mol kg–1 –1 8. The density of 40% by mass of acetic acid is 1.25 g mL . The molarity of the solution is (a) 5.2 M (b) 6.8 M (c) 7.5 M (d) 8.33 M

5.4

Complete Chemistry—JEE Main

9. Mass of Mohr’s salt (molar mass 392 g mol–1) required to prepare 250 mL of 0.15 M solution is (a) 9.8 g (b) 12.7 g (c) 14.7 g (d) 16.8 g 10. The mole fraction of glucose in 0.15 M solution (density = 1.10 g mL–1) is (b) 2.0 ¥ 10–3 (c) 2.82 ¥ 10–3 (d) 2.40 ¥ 10–2 (a) 2.40 ¥ 10–3 11. The mole fraction of methanol (CH3OH) in aqueous solution is 0.10. If density of the solution is 0.97 g mL–1, the molarity of the solution will be (a) 4.0 M (b) 4.5 M (c) 5.0 M (d) 5.5 M –3 12. The density of 1.50 M solution of acetic acid (CH3COOH) is 1.05 g cm . The molality of the solution will be (b) 1.36 mol kg–1 (c) 1.46 mol kg–1 (d) 1.56 mol kg–1 (a) 1.126 mol kg–1 –1 13. The molality of an aqueous solution of methanol (CH3OH) is 1.44 mol kg . The mole fraction of methanol in the solution will be (a) 0.0253 (b) 0.0213 (c) 0.0273 (d) 0.0198 –1 14. The density of 1.40 molal solution of acetic acid (CH3COOH) is 1.084 g mL . The molarity of solution will be (a) 1.30 M (b) 1.40 M (c) 1.50 M (d) 1.60 M 15. The molality m of a solution is related to its molarity M by the expression M M M /r r - MM 2 (a) m = (b) m = (c) m = (d) m = r - MM 2 r + MM 2 r + MM 2 M where r is the density of solution and M2 is the molar mass of the solute. 16. The molarity M of a solution is related to the mole fraction x2 of solute by the expression x2 r x2 r (a) M = (b) M = M1 + x2 ( M1 - M 2 ) M1 - x2 ( M1 - M 2 ) x2 r x2 r (d) M = (c) M = M1 + x2 ( M1 + M 2 ) M1 - x2 ( M1 + M 2 ) where r is the density of solution and M1 and M2 are the molar masses of solvent and solute, respectively. 17. The molality (m) of a solution in terms of mole fractions of solvent (x1) and solute (x2) is given by the expression x2 x1 x1M1 x2 M1 (b) m = (c) m = (d) m = (a) m = x2 M1 x2 x1 x1 M1 where M1 is the molar mass of the solvent.

ANSWERS 1. (b) 7. (c) 13. (a)

2. (c) 8. (d) 14. (b)

3. (c) 9. (c) 15. (a)

4. (d) 10. (b) 16. (b)

HINTS AND SOLUTIONS 1. Molar mass of C2H6O2 = 62 g mol–1 m 217.0 g = = 3.5 mol Amount of C2H6O2, n = M 62 g mol-1 3.5 mol n = = 17.5 mol kg -1 Molality of solution, m = m1 0.200 kg 2. Total mass of solution, m = 417.0 g m 417.0 g = = 379.1 cm3 Volume of solution, V = r 1.10 g cm -3

5. (a) 11. (c) 17. (d)

6. (d) 12. (d)

Solutions

m 217.0 g = = 3.5 mol M 62 g mol-1 n 3.5 mol = 9.23 mol dm -3 Molarity of solution, M = = V 379.1 ¥ 10-3 dm3

Amount of C2H602, n =

M2 =180 g mol–1 3. Molar mass of C6H12O6, Mass of glucose in the 100 g solution, m2 = 18 g m2 18 g = = 0.1 mol Amount of glucose in 100 g solution, n2 = M1 180 g mol-1 Mass of water in 100 g solution, m1 = 82 g m1 82 g = = 4.56 mol Amount of water in 100 g solution, n 1= M1 18 g mol-1 n2 0.1 mol = = 0.0214 Mole fraction of glucose, x2 = n1 + n2 (4.56 + 0.1) mol 4. Let n be the amount of each of Na2CO3 and NaHCO3 in the given 1.0 g of mixture. We will have n M Na 2CO3 + n M NaHCO3 = 1.0 g n(108 g mol–1 + 84 g mol–1) = 1.0 g 1.0 g 1.0 = mol n= -1 192 192 g mol

i.e, or, From the reaction

Na2CO3

+

(1.0/192) mol

NaHCO3 (1.0/192) mol

2HCl

Æ

2NaCl + CO2 + H2O

(2.0/192) mol

+

HCl

Æ NaCl + CO2 + H2O

(1.0/192) mol

2.0 1.0 3.0 mol + mol = mol 192 192 192 3.0 molˆ˜ /(0.1 mol L-1 ) = 0.1563 L = 156.3 m L Volume of 0.1 M HCl required, V = n / M = ÊÁ ¯ Ë 192 Amount of HCl required =

5. Mass of 1 L solution, m = rV = (1.25 g mL–1) (1000 mL) = 1250 g Mass of water in solution, m1 = m – m2 = (1250 – 441) g = 809 g = 0. 809 kg m2 441 g = = 4.5 mol Amount of H2SO4 in solution, n2 = M 2 98 g mol-1 n2 4.5 mol = = 4.5 mol L-1 Molarity of solution, M = 1L V Molality of solution, m =

n2 4.5 mol = = 5.56 mol kg -1 m1 0.809 kg

6. The molarity of resultant solution will be M1V1 + M 2V2 (0.12 M )(60 m L ) + (0.25 M )(40 m L ) = = 0.172 M (100 m L ) V1 + V2 7. Mole fraction of glucose = 0.2 implies that Amount of glucose, n2 = 0.2 mol; Amount of water, n1 = 0.8 mol M=

Mass of water, m1= n1M1= (0.8 mol) (18 g mol–1) =14.4 g = 14.4 ¥ 10–3 kg n2 0.2 mol = = 14.18 mol kg -1 Molality of solution, m = -3 m1 14.4 ¥ 10 kg

5.5

5.6

Complete Chemistry—JEE Main

8. 40 mass percent of acetic acid implies that Mass of water, m1 = 60 g Mass of acetic acid, m2 = 40 g; m2 40 g = = (2/3) mol Amount of acetic acid, n2 = M 2 60 g mol-1 m1 + m2 100 g = = 80 m L Volume of solution, V = r 1.25 g m L-1 Molarity of solution, M =

n2 (2 / 3) mol = = 8.33 M V 80 ¥ 10-3 L

M = n2/V = (m2/M2)/V, we get m2 = MVM2 = (0.15 mol L–1) (0.250 L) (392 g mol–1) =14.7 g

10. Let we have Amount of glucose, n2 = 0.15 mol; Volume of solution, V = 1000 mL Mass of glucose, m2 = n2M2 = (0.15 mol) (180 g mol–1) = 27.0 g Mass of solution, m = Vr = (1000 mL) (1.10 g mL–1) = 1100 g Mass of water, m1 = m1 – m2 = 1100 g – 27 g = 1073 g m 1073 g = 59.61 mol Amount of water, n1 = 1 = M1 18 g mol-1 n2 0.15 mol = = 0.002 Mole fraction of glucose, x2 = n1 + n2 (59.61 + 0.15) mol 11. For x2 = 0.10, let we have n2 = 0.1 mol and n1 = 0.9 mol The mass of solution will be m = n1M1 + n2M2 = (0.9 mol) (18 g mol–1) + (0.1 mol) (32 mol–1) =19.4 g The volume of solution will be m 19.4 g = = 20.0 ml = 20 ¥ 10 -3 L V= r 0.97 g m L-1 The molarity of solution will be n 0.1 mol M= 2 = = 5.0 mol L-1 V 20 ¥ 10-3 L 12. Let there be 1 L of solution. we have n2 = 1.5 mol Mass of solution, Mass of acetic acid,

m = Vr=

(103

and

V=1L

3

cm ) (1.05 g cm–3) = 1050 g

m2 = n2M2 = (1.5 mol) (60 g mol–1) = 90.0 g

m1 = m – m2 = 1050 g – 90 g = 960 g = 0.960 kg n 1.5 mol = 1.56 mol kg -1 Molality of solution, m= 2 = m1 0.960 kg 13. Let the mass of solvent be 1 kg. We will have n2 =1.44 mol and m1 = 1000 g m1 1000 g = = 55.56 mol The amount of water will be n1 = M1 18 g mol-1 n2 1.44 mol = = 0.0253 The mole fraction of the solute will be x2 = n1 + n2 (55.56 + 1.44) mol 14. Let the mass of solvent be 1 kg. We will have m1 = 1000 g and n2 = 1.40 mol Mass of acetic acid, m2 = n2M2 = (1.40 mol) (60 g mol–1) = 8.40 g Mass of water,

Mass of solution, Volume of solution, Molarity of solution

m = m1 + m2 = 1000 g + 84 g = 1084 g m 1084 g = = 1000 m L = 1L V= r 1.084 g m L-1 n2 1.40 mol = = 1.40 M M= 1L V

Solutions

SECTION 2

5.7

Liquid Solutions

When a solute (solid, gas or liquid) is dissolved in a liquid solvent, a solution is formed.

Solid in Liquid In general, it is found that like dissolves like. For example, an ionic compound (such as sodium compound) or a polar compound (such as hydrogen chloride) is soluble in polar solvent water whereas these are insoluble in nonpolar solvent benzene. Similarly, naphthalene is soluble in benzene while it is insoluble in water. The solubility of a solid in a solvent depends on temperature. For an endothermic dissolution process (DH = +ve), solubility increases with increase in temperature and for an exothermic process (DH = –ve), solubility decreases with increase in temperatures. A solution becomes saturated when no more solute dissolves in the solution. At a saturation stage, a dynamic equilibrium exists between undissolved solute and solute in solution.

Gas in Liquid Gases are soluble in liquids. For a given pair of gas and liquid, solubility depends on temperature and pressure. It is found that the solubility of a gas in a liquid increases with ease of liquefaction of the gas. For example, in any solvent, H2 and He are much less soluble than CO2, NH3 and SO2. Solubility is very much enhanced if there occurs a chemical interaction between the gas and the liquid.

Effect of Pressure—Henry’s Law The solubility of a gas in a liquid increases with the increase in external pressure, provided the gas does not react chemically with the liquid. According to Henry’s law, the mass of dissolved gas in a given volume of solvent is proportional to the partial pressure of the gas with which it is in equilibrium (Fig. 1), i.e. m2 μ p or m2 = kp2 where k is constant of proportionality.

(18)

Alternative Form of Henry’s Law If m2 is the mass of the gas dissolved in the mass m1 of the solvent at gaseous pressure p2, we have m2 = kp2 Dividing throughout by m1, we get m2 Ê k ˆ p2 = m1 ÁË m1 ˜¯ Dividing the masses with the respective molar masses, we get m2 / M 2 Ê kM1 ˆ n2 or = k ¢p2 p2 = m1 / M1 ÁË m1M 2 ˜¯ n1

Fig. 1

(19)

Since the solution of a gas in a liquid is a dilute solution, we can approximate mole fraction x2 as shown in the following. n2 n  2 (as n2 0 Examples are: (i) Carbon tetrachloride and heptane (ii) Ethyl ether and acetone (iii) Heptane and ethyl alcohol (iv) Benzene and acetone

Fig. 5

Solutions Exhibiting Negative Deviation from Raoult’s Law If the molecular interactions A· · ·B are stronger than those of A· · ·A and B· · ·B, the solution exhibits negative deviation from Raoult’s law. In this case, the vapour pressure curves of the constituents A, B and the solution lie below those of ideal lines (Fig. 6). The negative deviation arises from the fact that the tendency of molecules of each kind to escape from the solution into the vapour is weaker than that prevailing in and ideal solution. For this type of solution DmixH < 0. Examples are: (i) Pyridine and formic (or acetic or propanic) acid (ii) Mixture of halomethane (e.g. chloroform) with an oxygen or nitrogen compound (e.g. a ketone, ether, ester or amine) (iii) Aqueous solutions of HCl, HNO3 and perchloro acids Common feature of Nonideal Solutions Before describing the common feature of nonideal solutions displayed in Fig. 5 and Fig. 6, it is worth correlating Henry’s law and Raoult’s law. Fig. 6

Solutions

Correlation Between Henry’s Law and Raoult’s Law p2 = kH x2

5.11

Henry’s law is (20)

This law is applicable to solute in dilute solution. In a case, where Henry’s law is applicable over the entire range of concentration starting from x2 = 0 (pure solvent) to x =1 (pure liquid solute), we will have p2 p*2 = kH = x2 1 where p* is the vapour pressure of pure liquid solute (x2 = 1). With this, Eq. (20) becomes p2 = p*2 x2 (21) which is Raoult’s law. The Raoult’s law is a special case of Henry’s law; all systems which obey Raoult’s law must satisfy Henry’s law, but the reverse is not true unless Henry’s law is applicable over the entire range of concentration. In general, if Henry’s law is applicable to solute in the limited range of dilute solution (known as ideally dilute solution), Raoult’s law is applicable to solvent in the same range of composition. For nonideal solutions shown in Figs (5) and (6), the associated Raoult’s law line and Henry’ law line are also shown. It may be seen that Raoult’s law line meets the partial vapours pressure curve tangentially in the range xAÆ1 and Henry’s law line meets the curve in the region xAÆ0 . The same is true for the component B. In general, we have Ideal solution kH,A = pA* and kH,B = p*B (23 a) Nonideal solutions Positive deviation kH,A > pA* and kH,B > p*B (23 b) (23 c) Negative deviation k < p* and k < p*B H,A

A

H,B

Formation of Azeotropic Solution If the deviations from Raoult law are large, a maximum/minimum in the vapour pressure versus composition curve is observed. Because of increase/decrease in vapour pressure, the boiling point versus composition curve of the solution exhibits a minimum/maximum at which composition of liquid and vapour phases are identical (Figs. 7 and 8). The composition corresponding to this solution is known as azeotropic composition and the corresponding temperature is known as azeotropic temperature.

Fig. 7

Fig. 8

MULTIPLE CHOICE QUESTIONS ON SECTION 2 Identify the correct choice in the following questions. 1. Henry’s law for the solubility of gas in water is not applicable to the gas (a) H2 (b) O2 (c) N2

(d) NH3

5.12

Complete Chemistry—JEE Main

2. The Henry’s law constant does not depend upon (a) nature of solute (b) nature of solvent (c) temperature of the system (d) external pressure of the gas on the system 3. Show that Henry’s law (p = kH x2) for the dissolution of a gas in a solvent when expressed in molality (m) for the solution is approximately given as (b) p = (kH /M1)m (c) p = (kH M2)m (d) p = (kH /M2)m (a) p = (kH M1)m where M1, and M2 are the molar masses of solvent and gas, respectively. 4. Water saturated with air (20% O2 and 80% N2) at 1 atm and 298 K contains 8.9 ¥ 10–3 g L–1 dissolved oxygen. Its solubility under 25 atm pressure of O2 will be (a) 8.9 ¥ 10–3 g L–1 (b) 1.1125 g L–1 (c) 1.5642 g L–1 (d) 1.834 ¥ 10–2 g L–1 5. The solubility of a gas in a liquid for the given external pressure (a) increases linearly with increase in temperature (b) decreases linearly with increase in temperature (c) increases exponentially with increase in temperature (d) decreases exponentially with increase in temperature 6. The solubility of N2 in blood at 37 ºC and at a partial pressure of 0.80 atm is 5.6 × 10–4 mol L–1. The Henry’s law constant for this system is (b) 1.62×103 atm L mol–1 (a) 1.43 × 103 atm L mol–1 (c) 5.6 × 10–4 atm L mol–1 (d) 2.86×102 atm L mol–1 7. The solubility (x2) of gases in a given liquid under the constant external partial pressures is (a) directly proportional to Henry’s law constant (b) inversely proportional to Henry’s law constant (c) directly proportional to inverse of Henry’s law constant (d) does not depend on Henry’s law constant of the gas 8. Henry’s law constants at 25 ºC for N2 and O2 in water are 4.34 × 105 Torr/(g N2/100 g H2O) and 1.93 ¥105 Torr/ (g O2/100 g H2O), respectively. If the pN 2 = 608 Torr and pO2 = 152 Torr above the solution, the ratio of mass of O2 to N2 in water (a) is larger than the corresponding ratio in air (b) is smaller than the corresponding ratio in air (c) is equal to the corresponding ratio in air (d) cannot be compared with the corresponding ratio in air 9. For an ideal binary liquid solution, which of the following facts is not correct? (a) DmixH = 0 (b) DmixV = 0 (c) DmixG < 0 (d) DmixS < 0 10. For an ideal solution of liquids A and B, which of the facts is correct? (a) kH, A > pA* and kH, B > p*B (b) kH, A < pA* and kH, B < p*B (d) kH, A > pA* and kH, B < p*B (c) kH, A = pA* and kH, B = p*B where kH¢s stand for Henry’s law constants and p*¢s are the vapour pressures of pure components. 11. For a binary ideal liquid solution of A (having vapour pressure pA* ) and B (having vapour pressure p*B ), xA and yA are the mole fractions of A in liquid and vapour phases, respectively. Which of the following facts holds good? (b) yA varies linearly with 1/xA (a) yA varies linearly with xA (c) 1/yA varies linearly with xA (d) 1/yA varies linearly with 1/xA 12. Liquids A and B have vapour pressures pA* and p*B , respectively, with pA* > p*B . These liquids from an ideal solution. If xA and yA are mole fractions of A in liquid and vapour phases then the slope of (b) the linear plot of yA versus xA is negative (a) the linear plot of yA versus xA is positive (d) the linear plot of 1/yA versus 1/xA is negative (c) the linear plot of 1/yA versus 1/xA is positive 13. Liquids A and B have vapour pressures pA* and p*B , respectively, with pA* > p*B . The plot of 1/ptotal versus yA (mole fraction of A in vapour phase) (a) is linear with positive slope (b) is linear with negative slope (c) increases nonlinearly (c) decreases nonlinearly

Solutions

5.13

14. Methanol (p* = 90 mmHg at 300 K) and ethanol (p* = 51 mmHg at 300 K) form very nearly an ideal solution. The total vapour pressure of a solution obtained by mixing 23 g ethanol with 32 g methanol would be (a) 17 mmHg (b) 60 mmHg (c) 77 mmHg (d) 82 mmHg 15. An ideal solution of benzene (p* = 75 mmHg) and toluene (p* = 22 mmHg) has a vapour pressure of 50 mmHg. The mole fraction of benzene in liquid and vapour phases, respectively are (a) 0.53, 0.795 (b) 0.795, 0.53 (c) 0.47, 0.205 (d) 0.205, 0.47 16. An ideal liquid solution of 1 mol of A and 3 mol of B has the vapour pressure of 550 Torr. If one more mole of B is added, its vapours pressure becomes 560 mmHg. The vapour pressures of pure A and pure B, respectively are (a) 600 Torr and 400 Torr (b) 400 Torr and 600 Torr (c) 650 Torr and 350 Torr (d) 350 Torr and 650 Torr 17. An ideal solution of liquid A ( pA* = 100 mmHg) and liquid B ( p*B = 30 mmHg) has 0.6 mole fraction of A in the vapour phase. The mole fraction of A in the liquid phase will be (a) 0.21 (b) 0.25 (c) 0.31 (d) 0.41 * * p p 18. An ideal solution of liquid A ( A = 100 mmHg) and liquid B ( B = 30 mmHg) has 0.4 mole fraction of A in the vapour phase. The total vapour pressure of the solution is (a) 35.27 mmHg (b) 41.67 mmHg (c) 52.0 mmHg (d) 58.2 mmHg 19. A binary liquid solution exhibits positive deviation from ideal behaviour. Which of the following facts hold good? (a) DmixH > 0 and DmixG < 0 (b) DmixH < 0 and DmixG < 0 (d) DmixH < 0 and DmixG > 0 (c) DmixH > 0 and DmixG > 0 20. A binary liquid solution exhibits negative deviation from ideal behaviour. Which of the following facts hold good? (b) DmixH > 0 and DmixS < 0 (a) DmixH > 0 and DmixS > 0 (d) DmixH < 0 and DmixS < 0 (c) DmixH < 0 and DmixS > 0 21. Which of the facts regarding dilute solution of a nonideal binary liquid solution is correct? (a) Solute follows Raoult’s law and solvent follows Henry’s law (b) Solute follows Henry’s law and solvent follows Raoult’s law. (c) Both solute and solvent follows Raoult’s law (d) Both solute and solvent follows Henry’s law 22. Which of the following facts is correct for a binary solution of liquid A (vapour pressure pA* ) and liquid B (vapour pressure p*B ) exhibiting positive deviation from ideality? (b) kH,A < pA* and kH,B < p*B (a) kH, A > pA* and kH,B < p*B (c) kH, A > pA* and kH,B > p*B (d) kH,A < pA* and kH,B > p*B 23. For a binary liquid solution of A and B, the mole fraction of A in liquid and vapour phases are 0.70 and 0.35 at 600 Torr external pressure. The vapour pressure of A and B, respectively, are (a) 1300 Torr and 300 Torr (b) 450 Torr and 950 Torr (b) 950 Torr and 450 Torr (d) 300 Torr and 1300 Torr 24. The mole fraction of liquid A in a binary liquid solution of A (vapour pressure pA* = 300 Torr) and B (vapour pressure, p*B = 800 Torr) is 0.6. The external pressure on this is slowly reduced from 760 Torr. The pressure at (a) 400 Torr and 0.24

(b) 500 Torr and 0.50

(c) 500 Torr and 0.36

(d) 600 Torr and 0.25

25. The mole fraction of liquid A in a binary liquid solution of A ( pA* = 300 Torr) and B ( p*B = 800 Torr) is 0.6. The pressure at which last droplet of liquid is vapouried and the mole fraction of A in this last droplet, respectively, are (a) 400 Torr and 0.80 (b) 400 Torr and 0.20 (c) 500 Torr and 0.75 (d) 500 Torr and 0.45 26. A solution of liquids A and B exhibits positive deviation from Raoult’s law provided (a) Molecular interactions A· · ·A, B· · ·B and A· · ·B are identical (b) Molecular interactions A· · ·B are larger than those between A· · ·A and B· · ·B (c) Molecular interactions A· · ·B are weaker than those between A· · ·A and B· · ·B (d) Molecular interactions A· · ·A are larger than those between A· · ·A and weaker than those between B· · ·B

5.14

Complete Chemistry—JEE Main

ANSWERS 1. 7. 13. 19. 25.

(d) (b) (b) (b) (b)

2. 8. 14. 20. 26.

(d) (a) (c) (a) (c)

3. 9. 15. 21.

(a) (d) (a) (b)

4. 10. 16. 22.

(b) (c) (b) (c)

5. 11. 17. 23.

(d) (d) (c) (d)

6. 12. 18. 24.

HINTS AND SOLUTIONS 1. NH3 interacts with water. 2. The Henry’s law constant does not depend upon the external pressure of the gas on the system. 3. The expression of mole fraction in terms of molality of the solution is mM1 x2 = 1 + mM1 For a dilute solution mM1 1 for an electrolyte, the molar mass of an electrolyte is reduced by a factor i of the molar mass of the corresponding nonelectrolyte. For example, for a dilute solution of sodium chloride, we will have

5.22

Complete Chemistry—JEE Main

M=

( M 2 )0 (23 + 35.5) g mol-1 = = 29.25 g mol-1 2 2

Molecules of ethanoic acid and benzoic acid dimerise in benzene due to hydrogen bondings. O ... H O  C CH3 2 CH3COOH    CH3 - C O H ... O ( M 2 )0 60 g mol-1 = = 120 g mol-1 i 1/2 Number Average Molar Mass If a solution contains more than one solute species, the molar mass determined by substituting the value of total molality /concentration in the appropriate expression of colligative property gives the The van’t Hoff factor in this case is 1/2. Hence

Mn =

M2 =

N1M1 + N 2 M 2 +  Âi Ni M i = N1 + N 2 +  Âi N i

where Ns represent the number of species and Ms represent the respective molar masses.

MULTIPLE CHOICE QUESTIONS ON SECTION 3 1. Relative lowering of vapour pressure of solvent when a nonvolatile solute is added is (a) equal to the mole fraction of solvent in solution (b) equal to the mole fraction of solute in solution (c) independent of mole fraction of solvent (d) dependent on the nature of solute 2. A solution of 30 g of an unknown compound in 116 g of solvent (molar mass = 58 g mol–1) has vapour pressure of 0.475 atm. If the vapour pressure of pure solvent is 0.5 atm, the molar mass of unknown compound is (a) 100 g mol–1 (b) 200 g mol–1 (c) 250 g mol–1 (d) 300 g mol–1 3. A solution contains 20 g glucose (C6H12O6) and 34.2 g of sucrose (C12H22O11) in 108 g water. If the vapour pressure of pure water is 35 mmHg, the decrease in vapour pressure of the solvent will be (a) 0.85 mmHg (b) 0.98 mmHg (c) 1.19 mmHg (d) 2.25 mmHg –1) is 400 K. If its D H = 40.0 kJ mol–1, the value of its 4. The boiling point of a solvent (molar mass = 50 g mol vap boiling point elevation constant is (a) 0.85 K kg mol–1 (b) 1.66 K kg mol–1 (c) 1.25 K kg mol–1 (d) 1.82 K kg mol–1 5. A solution of 0.5 g of a solute (molar mass = 150 g mol–1) in 50 g of a solvent yields a boiling point elevation of 0.40 K. Another solution of 0.60 g of an unknown solute in the same mass of solvent exhibits a boiling point elevation of 0.8 K. The molar mass of unknown solute is (a) 60 g mol–1 (b) 90 g mol–1 (c) 120 g mol–1 (d) 180 g mol–1 6. Given is the mass fraction, w2, of solute in a solution. If r is the density of solution, the expression of osmotic pressure is w r RT w RT w RT (a) P = w2rRT (b) P = 2 (c) P = 2 (d) P = 2 M2 M 2r r –3 7. The density of 9.0% by mass solution of glucose (C6H12O6) is 1.05 g cm at 300 K. The osmotic pressure of the solution is (a) 9.22 atm (b) 10.40 atm (c) 11.2 atm (d) 12.92 atm 8. A 0.120 molal solution of CsCl (with ionizes in the solution as CsCl  Cs + + Cl- ) freezes at –0.4 °C. The van’t Hoff factor and degree of dissociation of CsCl in the solution, respectively, are (Given Kf(water) = 1.86 K kg mol–1) (c) 1.92, 0.92 (d) 1.2, 0.2 (a) 1.79, 0.79 (b) 1.5, 0.5

Solutions

5.23

9. The osmotic pressure of a solution containing 0.10 g of haemoglobin in 10.0 cm3 of solution is 30.0 Torr at 300 K. The molar mass of haemoglobin is (a) 4500 g mol–1 (b) 5452 g mol–1 (c) 6232 g mol–1 (d) 6932 g mol–1 10. A 0.101 M solution of Ba3(PO4)2 is isotonic with 0.004 M glucose solution. The degree of dissociation and van’t Hoff factor of Ba3(PO4)2, respectively, are (c) 4, 0.75 (d) 4, 0.85 (a) 4, 0.5 (b) 4, 0.65 11. If potassium ferrocyanide is 85% ionized in a solution, its van’t Hoff factor will be (c) 3.9 (d) 4.4 (a) 2.4 (b) 3.4 12. Which of the following pairs of solutions will be isotonic to each other? (a) 0.01 M Glucose + 0.01 M Sodium chloride (b) 0.01 M Potassium ferrocyanide + 0.02 M Calcium nitrate (c) 0.01 M Potassium ferricyanide + 0.015 M Potassium nitrate (d) 0.01 M Magnesium chloride + 0.01 M Mercurous nitrate 13. The van’t Hoff factor of solutes A, B and C in aqueous solutions are 0.8, 1.6 and 1.2, respectively. The freezing point of equimolar solutions follow the order (a) A > B > C (b) A > C > B (c) B > A > C (d) B > C > A 14. Arrange the aqueous solutions of A(0.01 M sodium chloride), B(0.01 M glucose) and C(0.01 M acetic acid) in increasing order of boiling points. (a) A > B > C (b) A > C > B (c) B > A > C (d) B > C > A 15. Which of the following solute will have minimum value of van’t Hoff factor? (a) 0.1 molal solution of acetic acid in water (b) 0.1 molal solution of benzoic acid in benzene (c) 0.1 molal solution of glucose in water (d) 0.1 molal solution of sodium chloride in water

ANSWERS 1. (b) 7. (d) 13. (b)

2. (d) 8. (a) 14. (b)

3. (c) 9. (c) 15. (b)

4. (b) 10. (c)

5. (b) 11. (d)

6. (b) 12. (d)

HINTS AND SOLUTIONS 1. Relative lowering of vapour pressure is equal to the mole fraction of solute in the solution. m2 30 g = = 300 g mol–1 2. M2 = * -1 (m1 / M1 )(- Dp1 / p1 ) (116 g /58 g mol )(0.025 atm / 0.5 atm) 3. Amount of glucose, n2 = Amount of water, n1 =

m2 20 g 1 = = mol ; M 2 180 g mol-1 9

Amount of sucrose, n3 =

m3 34.2 g 1 = = mol -1 M 3 342 g mol 10

m1 108 g = = 6 mol M1 18 g mol-1

Mole fraction of solutes in solution, 0.211 mol n2 + n3 [(1 / 9) + (1 / 10)] mol = 0.034 = = x= 6.211 mol n1 + n2 + n3 6 mol + [(1 / 9) + (1 / 10) mol] Vapour pressure of solution, p1 = x2p*1 = (0.034) (35 mmHg) = 1.19 mmHg

5.24

Complete Chemistry—JEE Main

4. Kb =

M1 RTb*2 (0.05 kg)(8.314 J K -1 mol -1 )(400 K ) 2 = = 1.66 K kg mol–1 D vap H (40 ¥ 103 J mol -1 )

5. The boiling point elevation constant of solvent is DTb DTb 0.40 K = = m (m2 / M 2 ) / m1 (0.5 g/150 g mol-1 ) /(0.05 kg)

Kb =

Molar mass of unknown solute is 1 ˘ È 0.6 g ˘ Ê K ˆ Ê m ˆ È (0.40 K )(0.05 kg) M2 = Á b ˜ Á 2 ˜ = Í = 90 g mol–1 Ë DTb ¯ Ë m1 ¯ Î (0.5 g /150 g mol-1 ) 0.8 K ˙˚ ÍÎ 0.05 kg ˙˚ (m2 / M 2 ) RT Ê m2 ˆ RT r w2 RT r Ê n2 ˆ = = 6. P = cRT = ÁË ˜¯ RT = V (m1 + m2 ) / r ÁË m1 + m2 ˜¯ M 2 M2 7. P =

w2 RT r (0.09)(0.082 L atm K -1 mol)(300 K )(1.05 ¥ 103 g L-1 ) = = 12.92 atm M2 (180 g mol-1 )

8. The depression in freezing point assuming solute as nonelectrolyte is (–DTf )0 = Kf m = (1.86 K kg–1 mol–1)(0.12 mol kg–1) = 0.223 K i=

The van’t Hoff factor is

a=

The degree of dissociation is

- DTf 0.4 K = = 1.79 ( - DTf )0 0.223 K

i - 1 1.79 - 1 = = 0.79 n -1 2 -1

˘ È (0.082 L atm K -1 mol -1 )(300 K ) ˘ m RT ˆ È 0.10 g –1 –1 9. M2 = ÊÁ 2 ˆ˜ ÊÁ = ˙ = 8.2 ¥ 760 g mol = 6232 g mol Ë V ¯ Ë P ˜¯ ÍÎ10.0 ¥ 10-3 L ˙˚ ÍÎ (30 / 760) atm ˚ 10. i =

P (0.004 M ) RT = = 4; (P )0 (0.001 M ) RT

a=

i -1 4 -1 3 = = = 0.75 n -1 5 -1 4

i -1 fi i = 4.4 5 -1 12. The total concentration of species, assuming complete ionization wherever needed, are as follows. 11. a = 0.85;

0.85 =

0.01 M Glucose NaCl Æ Na + + Cl0.01 M

0.01 M ˘ not isotonic 0.02 M ˙˚

0.01 M

K 4 [Fe(CN)6 ] Æ 4K + + [Fe(CN)6 ]40.04 M

0.01 M

Ca(NO3 )2 Æ Ca 2+ + 2 NO30.02 M

0.04 M

K 3 [Fe(CN)6 ] Æ 3K + + [Fe(CN)6 ]30.03 M

0.01 M

KNO3 Æ K + + NO30.015 M

0.015 M

MgCl2 Æ Mg 2+ + 2Cl0.01 M

0.02 M

Hg 2 (NO3 )2 Æ Hg 2+ 2 + 2 NO3 0.01 M

0.02 M

0.05 M ˘ ˙ not isotonic ˙ 0.06 M ˙˚ 0.04 M ˘ ˙ not isotonic ˙ 0.03 M ˙˚ 0.003 M ˘ ˙ isotonic ˙ 0.03 M ˙˚

Solutions

5.25

13. (–DTf ) = i (–DTf )0. For equimolar solutions, large the van’t Hoff factor, larger the decrease in freezing point and hence minimum will be the freezing point. 14. DTb = Kb m, larger the molality (or concentration for dilute solution), larger the value of DTb. Hence, larger will be the boiling point. Though, all the three solutions are equimolar, but the number of species available (or the total concentration of species) will follow the order A > C > B. The same order holds good for boiling points. 15. Benzoic acid in benzene dimerises and hence its van’t Hoff factor is 1/2 which will be minimum amongs other values.

MULTIPLE CHOICE QUESTIONS FOR THE ENTIRE CHAPTER Composition of a Solution 1. Which of the following is true? (a) Molarity of a solution is independent of temperature. (b) Molality of a solution is independent of temperature. (c) Mole fraction of a solute in a solution is dependent on temperature. (d) The unit of molality is mol dm–3. 2. The molality of a solute in the solution is the (a) mass of solute per dm3 of solution (b) amount of solute per dm3 of solution (c) amount of solute per kg of solution (d) amount of solute per kg of solvent 3. The unit of molality is (c) mol kg–1 (d) mol–1 (a) mol (b) mol dm–3 4. The unit of molarity is (c) mol dm–3 (d) mol kg–1 (a) mol (b) mol–1 5. The expression relating molarity of a solution with its molality is mM 2 1 + mM 2 mr 1 + mr (c) M = (d) M = (a) M = (b) M = 1 + mr 1 + mM 2 mM 2 mr where the various symbols have their usual meanings. 6. The expression relating mole fraction of solute in a solution with its molality is 1 - mM1 1 + mM1 mM1 mM1 (a) x2 = (b) x2 = (d) x2 = (c) x2 = mM1 1 - mM1 mr 1 + mM1 where the various symbols have their usual meanings. 7. The expression relating mole fraction of solute in a solution with its molarity is MM1 MM1 MM1 MM1 (b) x2 = (c) x2 = (d) (a) x2 = M ( M1 - M 2 ) - r M ( M1 + M 2 ) + r M ( M1 - M 2 ) + r M ( M1 + M 2 ) - r 8. 9. 10. 11.

where the various symbols have their usual meanings. The density of a 10.0% by mass KC1 solution in water is 1.06 g cm–3. Its molarity is (d) 1.489 mol kg–1 (a) 1.489 M (b) 1.420 M (c) 1.420 mol kg–1 What mass of ethanol be added to 1.0 kg water to have the mole fraction of ethanol equal to 0.20? (a) 63.89 g (b) 6.39 g (c) 638.89 g (d) 683.89 g The volume of 96% H2SO4 (density 1.83 g mL–1) required to prepare 2.0 L of 3.0 M H2SO4 solution is (a) 33.47 mL (b) 3.347 mL (c) 334.7 mL (d) 343.7 mL –3 The density of 95.2 mass % H2SO4 is 1.53 g cm . The molarity of this solution is (b) 13.6 mol dm–3 (c) 14.8 mol dm–3 (d) 16.2 mol dm–3 (a) 12.5 mol dm–3

5.26

Complete Chemistry—JEE Main

12. The mole fraction of a solute in carbon tetrachloride is 0.235. The molality of the solution is about (b) 1.5 mol kg–1 (c) 1.0 mol kg–1 (d) 0.5 mol kg–1 (a) 2.0 mol kg–1

Ideal and Nonideal Solutions 13. Which of the following is true? (a) The ideal behaviour of a liquid solution is due to the fact that the different molecules present in it do not interact with one another. (b) Henry’s law deals with the variation of solubility of gas with temperature. (c) The constituents of an ideal solution follow Raoult’s law under all conditions. (d) The addition of a nonvolatile solute to a volatile solvent decreases the boiling point of the latter. 14. For an ideal binary liquid solution with pA > pB, which of the following is true? (a) (xA)liquid = (xA)vapour (b) (xA)liquid > (xA)vapour (c) (xA)liquid < (xA)vapour (d) (xA)liquid and (xA)vapour do not bear any relationship with each other 15. Which of the following behaviours is not true for an ideal binary liquid solution? (a) Plot of pA versus xA (mole fraction of A in liquid phase) is linear (b) Plot of pB versus xB is linear (c) Plot of ptotal versus xA (or xB) is linear (d) Plot of ptotal versus xA (or xB) is nonlinear 16. For a dilute solution, Raoult’s law states that (a) the lowering of vapour pressure is equal to the mole fraction of solute (b) the relative lowering of vapour pressure is equal to the mole fraction of solute (c) the relative lowering of vapour pressure is proportional to the amount of solute in solution (d) the vapour pressure of the solution is equal to the mole fraction of solvent 17. Raoult’s law is obeyed by a binary liquid solution when (a) the forces of attractions between like molecules are greater than those between unlike molecules (b) the forces of attractions between like molecules are smaller than those between unlike molecules (c) the forces of attractions between like molecules are more or less identical with those between unlike molecules (d) the volume occupied by unlike molecules are different 18. An azeotropic solution of two liquids has a boiling point lower than either of the boiling points of the two liquids when it (a) shows negative deviation from Raoult’s law (b) shows positive deviation from Raoult’s law (c) shows no deviation from Raoult’s law (d) is saturated 19. An azeotropic solution of two liquids has a boiling point higher than either of the boiling points of the two liquids when it (a) shows negative deviation from Raoult’s law (b) shows positive deviation from Raoult’s law (c) shows no deviation from Raoult’s law (d) is saturated 20. Which of the following behaviours is true for an ideal binary liquid solution? (a) Plot of ptotal versus yA (mole fraction of A in vapour phase) is linear (b) Plot of ptotal versus yB is linear (c) Plot of l/ptotal versus yA (or yB) is linear (d) Plot of l/ptotal versus yA (or yB) is nonlinear 21. A binary solution of ethanol and n-heptane is an example of (a) an ideal solution (b) a nonideal solution with positive deviations from Raoult’s law (c) a nonideal solution with negative deviations from Raoult’s law. (d) a solution exhibiting positive deviations at low concentrations and negative deviations at higher concentration

Solutions

5.27

22. An azeotropic solution of two liquids has boiling point higher than either of them when it (a) shows negative deviations from Raoult’s law (b) shows positive deviations from Raoult’s law (c) shows no deviations from Raoult’s law (d) follows Henry’s law 23. Which of the following solutions is expected to show positive deviations from Raoult’s law? (a) Pyridine-formic acid (b) Ethanol-hexane (c) Acetone-chloroform (d) Hydrochloric acid-water 24. Which of the following solutions is expected to show negative deviations from Raoult’s law? (a) Ethanol-hexane (b) Ethanol-water (c) Acetone-chloroform (d) Ethyl ether-acetone 25. The dissolution of sulphuric acid in water will exhibit (a) negative deviations from ideal behaviour (b) positive deviations from ideal behaviour (c) ideal behaviour (d) negative or positive deviation depending upon the mass of H2SO4 dissolved. 26. The formation of an ideal solution involves (a) increase in enthalpy (b) increase in entropy (c) increase in free energy (d) increase in energy

Vapour Pressure 27. The vapour-pressure lowering of a solvent is proportional to (a) the mole fraction of the solute (b) the mole fraction of the solvent (c) the molality of the solvent (d) the normality of the solution 28. The vapour pressure of a solvent in a solution is proportional to (a) the mole fraction of the solute (b) the mole fraction of the solvent (c) the molality of the solvent (d) the normality of the solute 29. The vapour pressure of pure liquid is 70 Torr at 27 °C. The vapour pressure of a solution of this liquid and another liquid (mole fraction 0.2) is 84 Torr at 27 °C. The vapour pressure of pure liquid B at 27 °C is (a) 140 Torr (b) 280 Torr (c) 160 Torr (d) 200 Torr 30. The vapour pressure of pure benzene and toluene are 160 Torr and 60 Torr, respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene would be (a) 0.50 (b) 0.84 (c) 0.73 (d) 0.27 31. The mass of sucrose to be added to 300 g of water to lower its vapour pressure by 1.0 mmHg at 25°C is (Given: *(water) = 23.8 mmHg) (a) 249.9 g (b) 329.4 g (c) 215.2 g (d) 342.2 g 32. A solution contains 1 mol of pentane (p* = 450 mmHg) and 4 mol of hexane (p* = 150 mmHg), the mole fraction of pentane in vapour phase will be (a) 0.454 (b) 0.429 (c) 0.641 (d) 0.75 Depression of Freezing Point (a) ice (b) solid solution of sugar and ice (c) sugar (d) a compound formed from sugar and water 34. In the phenomenon of osmosis through the semipermeable membrane (a) solvent molecules pass from solution to solvent (b) solvent molecules pass from solvent to solution (c) solute molecules pass from solution to solvent (d) solute molecules pass from solvent to solution 35. The freezing point of a 0.05 molal solution of a nonelectrolyte in water Kf = 1.86 K kg mol–1 is (a) –1.86 °C (b) –0.93 °C (c) –0.093 °C (d) 0.93 °C 36. Which of the following exhibits the greatest freezing-point lowering? (b) 0.1 mol kg–1 CaCl2 (c) 0.1 mol kg–1 HC2H3O2 (d) 0.1 mol kg–1 NaC2H3O2 (a) 0.1 mol kg–1 NaCl

5.28

Complete Chemistry—JEE Main

37. The freezing point depression constant is given as (b) Kf = M1R/Tf*2 DfusHm (a) Kf = M1RTf*2/DfusHm *2 (c) Kf = MTf /RDfusHm (d) Kf = DfusHm/MRTf*2 where the various symbols have their usual meanings. 38. The unit of freezing point depression constant is (b) K kg–1 mol–1 (c) K kg mol–1 (d) K kg–1 (a) K mol–1 39. Arrange the following (0.1 molal) solutions in the increasing order of freezing point. Soln of Na2SO4; Soln of glucose Soln of Ca3(PO4)2; (I)

(II)

(III)

(a) I > II > III (b) I > III > II (c) III > II > I (d) III > I > II Elevation of Boiling Point 40. The boiling point elevation constant is given as (a) Kb = M1R T*b 2/DvapHm (b) Kb = M1R/Tb*2 DvapHm (c) Kb = M1Tb*2/R D vav Hm (d) Kb = DvapHm/M1RTb*2 where the various symbols have their usual meanings. 41. The boiling point of water at 735 Torr is 99.07 °C. The mass of NaCl added in 100 g water (Kb = 0.51 K kg mol–1) to make its boiling point 100 °C is (a) 10.68 g (b) 5.34 g (c) 2.67 g (d) 26.7 g 42. The elevation of boiling point of water produced by dissolving 1.17 g sodium chloride in 100 g water (Kb = 0.512 K kg mol–1) is (a) 0.103 K (b) 0.205 K (c) 0.309 K (d) 0.410 K 43. The normal boiling point of pure ethyl acetate is 77.06 °C. A solution of 50.0 g of naphthalene (C10H8) dissolved in 150 g of ethyl acetate boils at 84.27 °C. The boiling point elevation constant of ethyl acetate is (a) 2.77 K kg mol–1 (b) 1.88 K kg mol–1 (c) 2.12 K kg mol–1 (d) 1.54 K kg mol–1 44. The boiling point of a solution at a height of 10 km above sea level is (a) found to be greater than that at sea level (b) found to be lesser than that at sea level (c) found to be the same as that at sea level (d) dependent upon the air dissolved in solution 45. A solution of sucrose in 100 g of water boils at 100.25 °C. If this solution is boiled till its temperature is 101 °C, the mass of water evaporated from the solution is (a) 25 g (b) 50 g (c) 75 g (d) 85 g 46. Molal elevation boiling point constant Kb depends (a) only on the nature of solvent (b) only on the nature of solute (c) on the nature of both solute and solvent (d) remains constant at a place at different heights from the sea level.

Osmotic Pressure 47. The osmotic pressure of a solution (density = 1.02 g cm–3) containing 50 g glucose (C6H12O6) in 1 kg of water at 300 K is (a) 67.39 kPa (b) 673.43 kPa (c) 6.74 kPa (d) 673.85 Pa 48. The osmotic pressure of 0.1 M sodium chloride solution at 27 °C is (a) 4.0 atm (b) 2.46 atm (c) 4.92 atm (d) 1.23 atm 49. Which of the following pairs of solutions are expected to be isotonic? (a) 0.1 M urea and 0.1 M NaCl (b) 0.1 M urea and 0.1 M MgCl2 (c) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4 (d) 0.1 M NaCl and 0.1 M glucose 50. 250 mL of a solution containing 10.0 g of sodium chloride and glucose produces 15 atm of osmotic pressure at 27 °C. The mass percentage of sodium chloride in the mixture is (a) 25% (b) 33.8% (c) 52.4% (d) 66.2%

Solutions

51. Isotonic solutions have the same (a) freezing point (b) boiling point (c) osmotic pressure 52. Through the semipermeable membrate, there can occur

5.29

(d) vapour pressure

Van’t Hoff Factor 53. The van’t Hoff factor of the compound K3Fe(CN)6 in dilute solution is (a) 1 (b) 2 (c) 3 (d) 4 54. The expression relating degree of dissociation of the weak electrolyte AxBy with its van’t Hoff factor is i -1 i -1 x + y -1 x + y -1 (a) a = (b) a = (c) a = (d) a = x + y -1 x + y -1 i -1 i +1 55. Assuming 100% ionization, the increasing order of the freezing point of the solution will be (a) 0.10 mol kg–1 Ba3(PO4)2 < 0.10 mol kg–1 Na2SO4 < 0.10 mol kg–1 KCl (b) 0.10 mol kg–1 KCl < 0.10 mol kg–1 Na2SO4 < 0.10 mol kg–1 Ba3(PO4)2 (c) 0.10 mol kg–1 Na2SO4 < 0.10 mol kg1 Ba3(PO4)2 < 0.10 mol kg1 KCl (d) 0.10 mol kg–1 KCl < 0.10 mol kg–1 Ba3(PO4)2 < 0.10 mol kg–1 Na2SO4 56. When mercuric iodide is added to an aqueous solution of potassium iodide, the (a) freezing point is lowered (b) freezing point is elevated (c) freezing point does not change (d) effect on freezing point cannot be predicted 57. The freezing point of equimolal solution will be lowest for (a) anilinium chloride (b) calcium nitrate (c) glucose (d) sodium phosphate 58. The van’t Hoff factor of benzoic acid in benzene is found to be (a) 1 (b) 2 (c) 0.5 (d) 1.5 59. Which of the following electrolytes would have the same van’t Hoff factor as that of potassium ferricyanide? (c) A12(SO4)3 (d) A1(NO3)3 (a) NaCl (b) Na2SO4 60. The van’t Hoff factor for 0.1 molal Ba(NO3)2 solution is 2.74. Its degree of dissociation is (a) 0.74 (b) 0.87 (c) 0.91 (d) 1.0 61. The van’t Hoff factor of acetic acid in water will be (a) equal to one (b) equal to two (c) in between one and two (d) more than two 62. The molar mass of calcium nitrate as determined from the colligative properties is found to be 65.6 g mol–1. The degree of dissociation of the salt is (a) 0.25 (b) 0.50 (c) 0.75 (d) 0.85 63. 5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is (a) 120 g mol–1 (b) 125 g mol–1 (c) 130 g mol–1 (d) 134 g mol–1 64. The molar mass of benzoic acid in benzene as determined by colligative properties is found to be (b) 183 g mol–1 (c) 244 g mol–1 (d) 305 g mol–1 (a) 122 g mol–1 Determination of Molar Mass 65. The expression to compute molar mass of a solute from the relative lowering of vapour pressure of a solvent is (b) M2 = m2/{(m1/M1) (–Dp1/p*1)} (a) M2 = m2/{(m1/M1) (–Dp1)} (d) M2 = (m1/M1) (–Dp1/p*1)/m2 (c) M2 = m2(–Dp1/p*1)/(m1/M1) where the various symbols have their usual meanings. 66. The expression to compute molar mass of a solute from the elevation of boiling point of a solvent is DTb m2 K b m2 K b m1 DTb m1 (b) M2 = (c) M2 = (d) M2 = (a) M2 = K b m1 DTb m1 K b m2 DTb m2 where the various symbols have their usual meanings.

5.30

Complete Chemistry—JEE Main

67. The expression to compute molar mass of a solute from the depression in freezing point of a solvent is K f m1 (-DTf ) m2 K f m2 (-DTf ) m1 (a) M2 = (-DT ) m (b) M2 = (c) M2 = (d) M 2 = K f m1 (- DTf ) m1 K f m2 f 2 where the various symbols have their usual meanings. 68. The molar mass of a solute from the osmotic pressure which it produces when a mass m of solute is dissolved in volume V is given by the expression m RT m m PR Ê mˆ P (a) M = ÊÁ ˆ˜ (d) M = ÊÁ ˆ˜ (c) M = ÊÁ ˆ˜ P RT (b) M = ÁË ˜¯ ËV ¯ P ËV ¯ ËV ¯ T V RT 69. The vapour pressure of a solution having 2.0 g of a solute X (molar mass 32 g mol–1) in 100 g of CS2 (vapour pressure 854 Torr) is 848.9 Torr. The molecular formula of the solute is (a) X (b) X2 (c) X4 (d) X8 70. Pure Benzene freezes at 5.4 °C. A solution of 0.223 g of phenyl acetic acid (C6H5CH2COOH) in 4.4 g of benzene (Kf = 5.12 K kg mol–1) freezes at 4.47 °C. From this observation, one can conclude that (a) phenyl acetic acid exists as such in benzene (b) phenyl acetic acid undergoes partial ionization in benzene (c) phenyl acetic acid undergoes complete ionization in benzene (d) phenyl acetic acid dimerizes in benzene 71. The vapour pressure of a solution of 5 g of a nonelectrolyte in 100 g water at a particular temperature is 2950 Pa and that of pure water at the same temperature is 3000 Pa. The molar mass of the solute is (a) 54 g mol–1 (b) 119 g mol–1 (c) 179 g mol–1 (d) 229 g mol–1 72. The vapour pressure of a pure liquid A is 10.0 Torr. at 27 °C. One gram of B is dissolved in 20 g of A, the vapour pressure is lowered to 9.0 Torr. If the molar mass of A is 200 g mol –1, the molar mass of B is (b) 85 g mol–1 (c) 100 g mol–1 (d) 115 g mol–1 (a) 75 g mol–1 73. A solution contains 10 g of a solute and 116 g of acetone at 30 °C. Its vapour pressure is 290 Torr. If the vapour pressure of pure acetone is 300 Torr at 30 °C, the molar mass of solute is (a) 75 g mol–1 (b) 100 g mol–1 (c) 125 g mol–1 (d) 150 g mol–1 74. The molar mass of acetic acid dissolved in benzene determined from the depression in freezing point is found to be (a) 30 g mol–1 (b) 60 g mol–1 (c) 120 g mol–1 (d) 240 g mol–1 75. The freezing point of a solution containing 36 g of a compound (empirical formula : CH2O) in 1.20 kg of water is found to be – 0.93 °C. The molecular formula of the solute is (a) CH2O (b) C2H4O2 (c) C3H6O3 (d) C4H8O4 76. The molar mass of sodium chloride determined by osmotic pressure would be (b) more than 58.5 g mol–1 (c) less than 58.5 g mol–1 (d) 117.0 g mol–1 (a) 58.5 g mol–1 77. A solution containing 8.6 g L–1 of urea (molar mass 60 g mol–1) is isotonic with a 5% solution of unknown solute. The molar mass of the solute will be (a) 348.9 g mol–1 (b) 174.5 g mol–1 (c) 87.3 g mol–1 (d) 34.89 g mol–1 78. Which of the following physical properties is used to determine the molar mass of a polymer solution? (a) Relative lowering of vapour pressure (b) Elevation of boiling point (c) Depression of freezing point (d) Osmotic pressure 79. The molar mass of sodium chloride determined by osmotic pressure measurement will come out to be (b) 117.0 g mol–1 (a) 58.5 g mol–1 (c) 29.25 g mol–1 (d) Anywhere in between 29.25 and 58.5 g mol–1

Solutions

5.31

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79.

(b) (b) (c) (a) (a) (a) (a) (a) (c) (a) (c) (d) (d) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(d) (b) (b) (c) (b) (b) (c) (b) (b) (b) (c) (a) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(c) (c) (d) (b) (a) (a) (c) (c) (c) (d) (d) (d) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76.

(c) (c) (b) (a) (b) (b) (a) (a) (b) (c) (c) (d) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77.

(b) (c) (c) (b) (a) (c) (b) (b) (d) (d) (b) (a) (a)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78.

(c) (a) (b) (c) (c) (b) (b) (c) (a) (b) (c) (c) (d)

HINTS AND SOLUTIONS 8. M = 9. 0.2 =

(10 g/74.5 g mol-1 ) = 1.42 ¥ 10–3 mol cm–3 = 1.42 mol dm–3 (100 g / 1.06 g cm -3 ) m 0.2 ¥ 1000 1 (m / 46 g mol-1 ) = ¥ = 13.89 i.e. m = 13.89 ¥ 46 g = 638.89 g ; -1 -1 46 g 18 0.8 (m / 46 g mol ) + (1000 g / 18 g mol )

10. Volume of 96% H2SO4 = (VM) Mm (100/98)/r ˆ 1 Ê 100 ˆ Ê ˜¯ Á = (2.0 L)(3.0 mol L–1)(98 g mol–1) ÁË -1 ˜ = 334.7 mL 96 Ë 1.83 g mL ¯ 11. Let there be 100 g of solution. Its volume will be

V=

This solution contains 95.2 g of H2SO4. Its amount is n =

Hence, the molarity of solution is

M=

100 g 100 ˆ 3 0.10 m = = ÊÁ dm3 ˜ cm = -3 Ë r 1.53 g cm 1.53 ¯ 1.53 m 95.2 g = M m 98 g mol-1 n (95.2 / 98) = = 14.86 mol dm -3 V (0.1 / 1.53) dm3

12. Let Amount of solute = 0.235 mol; Amount of solvent = (1 – 0.235) mol = 0.765 mol Mass of solvent, m = nMm = (0.765 mol) (154 g mol–1) = 117.81 g n 0.235 mol =  2.0 mol kg -1 Molality of solution = m 117.81 ¥ 10-3 kg 25. The forces of attraction between solute and solvent are larger than those existing in solute-solute and solventsolvent. The solution thus exhibits negative deviation from ideal behaviour. 29. p = xA pA* + xB pB* fi 84 Torr = (0.8) (70 Torr) + (0.2) (p*B) fi p*B = 140 Torr 30. pb = xb pb = (0.5) (160 Torr) = 80 Torr; ptotal = xb p*b + xt p*t = (0.5) (160 Torr) + (0.5 Torr) (60 Torr) = 110 Torr pb 80 Torr = = 0.73 yb = ptotal 110 Torr

5.32

Complete Chemistry—JEE Main

31. Since p = x1 p1, we get p 23.8 - 1.0 x1 = * = = 0.958. 23.8 p1

Now

x1 =

n1 (300 / 18) mol = n1 + n2 (300 / 18) mol + (m/342 g mol-1 )

Equating these two, we get {(300 / 18)(1 - 0.958)}(342 g ) (300 / 18) = 249.89 g = 0.958 or m = 0.958 (300 / 18) + (m / 342 g) 32. In vapour phase, we will have ppentane = {1/(1 + 4)} (450 mmHg) = 90 mmHg; ppentane = {4/(1 + 4)} (150 mmHg) = 120 mmHg ppentane 90 = = 0.429 ypentane = ppentane + phexane 90 + 120 39. Larger the sum of stoichiometric number of ions of the molecule, larger ions it produces in solution, and thus larger decrease in freezing point. Consequently, the freezing point is the lowest. DT (100 - 99.07) mol kg -1 41. m = b = Kb 0.51 Assuming NaCl to be completely dissociated, we will have Molality of NaCl =

1 Ê 100 - 99.07 mol kg -1 ˆ˜ Á ¯ 2Ë 0.51

1 1 0.93 ˆ Mass of NaCl in 100 g water = ÊÁ ˆ˜ ÊÁ (58.5) g = 5.33 g Ë 10 ¯ 2 Ë 0.51˜¯ 42. Molality of sodium chloride =

(1.17 / 58.5) mol = 0.2 mol kg -1 0.1 kg

Molality of ions in solution = 2 ¥ 0.2 mol kg–1 DTb = Kbm = (0.512 K kg mol–1) (2 ¥ 0.2 mol kg–1) = 0.205 K 43. Molality of solution is m2 / M 2 (50 g ) /(128 g mol-1 ) = = 2.604 mol kg -1 m= 0.150 kg m1 DTb (84.27 - 77.06) K = = 2.77 K kg mol -1 -1 m 2.604 mol kg 44. At 10 km above sea level, the atmospheric pressure will be less than that existing at sea level. Hence, the boiling point of the solution will be less. 45. We have from DTb = m Kb, Kb =

( DTb ) 2 m2 n2 /(mwater ) 2 (mwater )1 = = = ( DTb )1 m1 n2 (mwater )1 (mwater ) 2 Hence

(mwater)2 = (mwater)1 {DTb)1/(DTb)2} = (100 g) (0.25K/1K) = 25 g. Mass of water evaporated = 100 g – 25 g = 75 g

(50 / 180) mol = 0.27 mol L-1 (1.05 kg/1.02 kg L-1 ) Osmotic pressure, P = cRT = (0.27 mol L–1) (8.314 kPa L–1 K–1 mol–1) (300 K) = 673.43 kPa 48. Osmotic pressure, P = cRT= (2 ¥ 0.1 mol L–1) (0.082 atm L K–1 mol–1) (300 K) = 4.92 atm 50. Let m be the mass of NaCl in the mixture. mNaCl m 10.0 g - m = ; Amount of NaCl, nNaCl = Amount of glucose, nglucose = -1 M NaCl 58.5 g mol 180 g mol-1 Total amount of species in the mixture

47. Concentration of glucose =

Solutions

2m 10.0 g - m ˆ 1 ntotal = {nNa + + nCl- } + nglucose = ÊÁ + ˜¯ Ë 58.5 180 g mol -1 Total concentration of species in the solution, c =

{(2m / 58.5) + (10.0 g - m) / 180}/ g mol-1 0.250 L

Now P = cRT, we have {(2m / 58.5) + (10.0 g - m) / 180}/ g mol-1 15 atm = 0.250 L (0.082 L atm K -1 mol -1 )(300 K) Ê 2m + 10.0 g - m ˆ g = 15 ¥ 0.250 = 0.1524 Ë 58.5 180 ¯ (0.082)(300)

or

(180 ¥ 0.1524 - 10.0)(58.5) g = 3.38 g (360 - 58.5) 3.38 ¥ 100 = 33.8 % Mass percentage of NaCl = 10 y+ x54. A x B y  xA + yB m=

or

ay

ax

1-a

Total amount of species = 1 + (x + y – 1)a Van’t Hoff factor, i = 55. Assuming complete dissociation Hence

1 + ( x + y - 1)a 1

or a =

i -1 x + y -1

(–DTf )Ba3(PO4)2 > (–DTf)Na2SO4 > (–DTf ) KCl

(Tf)Ba3(PO4)2 < (Tf)Na2SO4 < (Tf)KCl

58. Benzoic acid dimerizes in benzene. Its van’t Hoff factor will be 1/2. 60. a =

i -1 2.74 - 1 = = 0.87 ( x + y) - 1 3 -1

62. We have Ca ( NO3 ) 2  Ca 2+ + 2 NO3c(1 – a)

ca

c(2a)

Total concentration = c(l – a) + ca + c(2a) = c(l + 2a) Theoretical molecular mass (1 + 2a ) 164 1 + 2a = fi fi a = 0.75 = Observed molecular mass 1 65.6 1 63. Mole fraction of water in solution is (100 / 18) mol x= (5 g / M ) + (100 / 18) mol Hence,

p = x1 p*1 2980 Pa =

(100 / 18) mol (3000 Pa ). (5 g / M ) + (100 / 18) mol

This gives

M = 134.1 g mol-1

64. Benzoic acid dimerizes in benzene. So, its molar mass is twice as that of monomer. (100 g / 76 g mol-1 ) 69. Mole fraction of solvent = (100 g/76 g mol-1 ) + (2 g / M ) Hence,

(100 / 76 mol-1 ) (854 Torr ) 848.9 Torr = (100 / 76 mol-1 ) + (2 g / M )

Solving for M, we get

M= 253 g mol–1

5.33

5.34

Complete Chemistry—JEE Main

Number of S atoms in a molecule of solute = 70. m =

( - DTf ) (5.4 - 4.47) = mol kg -1 Kf 5.12

Solving for M, we get

253 amu =8 32 amu 0.223 g/M 0.933 = mol kg -1 (4.4 / 1000) kg 5.12

i.e.

M = 279 g mol–1

279 =2 136 (m / M 2 ) Dp (3000 - 2950) Torr 1 71. x2 = – = = . Now x2  2 . * 3000 Torr 60 m1 / M1 p1 1 (5 g / M ) = . This gives M = 54 g mol -1 Hence, 60 (100 g / 18 g mol -1 ) Number of molecules of phenyl acetic acid =

Dp 1 Torr 1 (m / M 2 ) Now x2  2 = = * (m1 / M1 ) p1 10 Torr 10 (1g / M 2 ) 1 = . This gives M 2 = 100 g mol -1 Hence, 10 (20 g/200 g mol -1 )

72. x2 = -

73. x2 = -

Dp 10 Torr 1 = = . * Torr 300 30 p 1

Now x2 

(m2 / M 2 ) . (m1 / M1 )

(10 g / M 2 ) 1 = . This gives M 2 = 150 g mol-1 30 (116 g / 58 g mol-1 ) 74. Acetic acid dimerizes in benzene. Its molar mass would be 2 ¥ 60 g mol–1, i.e. 120 g mol–1. Ê 36 g/M ˆ 75. –DTf = Kf m fi 0.93 K = (1.86 K kg mol–1) Á Ë 1.20 kg ˜¯ 1.86 ¥ 36 g mol–1 = 60 g mol–1. This gives M= 0.93 ¥ 1.20 Molar mass 60 g mol-1 n= = =2 Molar Empirical formula mass 30 g mol-1 Hence, molecular formula is C2H4O2 77. Isotonic solution has the same osmotic pressure and hence the same concentration in the solution. Hence,

(8.6 g/60 g mol -1 ) (5 g/M ) 5 ¥ 60 = . Hence, M = g mol -1 = 348.8 g mol -1 1L 0.1 L 8.6 ¥ 0.1 79. The dissolution of sodium chloride produces equal amount of Na+ and Cl– ions. Since the molar mass is the number average, the molar mass will be 29.25 g mol–1 n1M1 + n2 M 2 M1 + M 2 (23 + 35.5) g mol-1 = = = 29.25 g mol-1 Mn = (n1 + n2 ) 2 2

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Which one of the following aqueous solutions will exhibit highest boiling point? (a) 0 01 M Na2SO4 (b) 0.01 M KNO3 (c) 0.015 M urea

(d) 0.015 M glucose [2004] 20 2. 6.02 ¥ 10 molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (a) 0.001 M (b) 0.01 M (c) 0.02 M (d) 0.1 M [2004] 23 –1 (Avogadro constant, NA = 6.02 ¥ 10 mol )

Solutions

5.35

3. Which of the following liquid pairs shows a positive deviation from Raoult’s law? (a) water-hydrochloric acid (b) benzene-methanol (c) water-nitric acid (d) acetone-chloroform [2004] 4. Which one of the following statements is false? (a) Raoult’s law states that the vapour pressure of a component over a solution is proportional to its mole fraction. (b) The osmotic pressure (P) of a solution is given by the equation P = MRT, where M is the molarity of the solution. (c) The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl2 > KCl > CH3COOH > sucrose. (d) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression. [2004] 5. If a is the degree of dissociation of Na2SO4, the van’t Hoff factor used tor the calculation of molecular mass is (a) 1 + 2a (b) 1 – 2a (c) 1 + a (d) 1 – a [2005] 6. Benzene and toluene form nearly ideal solutions. At 20 °C, the vapour pressure of benzene is 75 Torr and that of toluene is 22 Torr. The partial vapour pressure of benzene at 20 °C for a solution containing 78 g of benzene and 46 g toluene in Torr is (a) 37.5 (b) 53.5 (c) 50 (d) 25 [2005] 7. Equimolar solutions in the same solvent have (a) same boiling point and same freezing point (b) different boiling point and different freezing point (c) same boiling point but different freezing point (d) same freezing point but different boiling point [2005] 8. Among the following mixtures, dipole-dipole as the major interaction, is present in (a) benzene and carbon tetrachloride (b) benzene and ethanol (c) acetonitrile and acetone (d) KCl and water [2006] 9. 18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100 °C is (a) 752.40 Torr (b) 759.00 Torr (c) 7.60 Torr (d) 76.00 Torr [2006] –1) in the same 10. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm–3, molar mass of the substance will be (a) 90.0 g mol–1 (b) 115.0 g mol–1 (c) 105.0 g mol–1 (d) 210.0 g mol–1 [2007] 11. In a saturated solution of the sparingly soluble electrolyte AgIO3 (relative molecular mass = 283) the equilibrium which sets in is AgIO3 (s)  Ag + (aq ) + IO3- (aq ) If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 ¥ 10–8 M, what is the mass of AgIO3 contained in 100 mL of its saturated solution? (b) 2.83 ¥ 10–3 g (c) 1.0 ¥ 10–7 g (d) 1.0 ¥ 10–4 g [2007] (a) 28.3 ¥ 10–2 g 12. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mmHg at 300 K. The vapour pressure of propyl alcohol is 200 mmHg. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mmHg) at the same temperature will be (a) 350 (b) 300 (c) 700 (d) 360 [2007] 13. At 80 °C, the vapour pressure of pure liquid A is 520 mmHg and that of pure liquid B is 1000 mmHg. If a solution of A and B boils at 80 °C and 1 atm pressure, the amount per cent of A in the mixture is (a) 50 mol per cent (b) 52 mol per cent (c) 34 mol per cent (d) 48 mol per cent [2008]

5.36

Complete Chemistry—JEE Main

14. The vapour pressure of water at 20 °C is 17.5 mmHg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20 °C, the vapour of the resulting solution will be (a) 17.325 mmHg (b) 17.675 mmHg (c) 15.750 mmHg (d) 16.500 mmHg [2008] 15. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. The vapour pressures (in mmHg) of X and Y in their pure states, respectively, will be (a) 400 and 600 (b) 500 and 600 (c) 200 and 300 (d) 300 and 400 [2009] 16. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution? (a) The solution is nonideal showing negative deviation from Raoult’s law (b) n-Heptane shows positive deviation while enthanol shows negative deviation from Raoult’s law (c) The solution formed is an ideal solution (d) The solution is nonideal showing positive deviation from Raoult’s law. [2009] 17. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the decrease in freezing point of water (DTf) when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, (Kf = 1.86 K kg mol–1) is (a) 0.0186 K (b) 0.0372 K (c) 0.0558 K (d) 0.0744 K [2010] 18. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa, respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane (molar mass = 100 g mol–1) and 35.0 g of octane (molar mass of octane = 114 g mol–1) will be (a) 144.5 kPa (b) 72.0 kPa (c) 36.1 kPa (d) 96.2 kPa [2010] 19. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at –6 °C will be (Kf for water = 1.86 K kg mol–1 and molar mass of ethylene glycol = 62 g mol–1) (a) 304.60 g (b) 804.32 g (c) 204.30 g (d) 400.00 g [2011 (Cancelled)] 20. The degree of dissociation (a) of a weak electrolyte AxBy is related to van’t Hoff factor (i) by the expression i -1 i -1 x + y +1 x + y -1 (a) a = (b) a = (d) (c) a = x + y -1 x - y +1 i -1 i -1 [2011 (Cancelled)] 21. A 5.2 molal aqueous solution of methyl alcohol, CH3OH is supplied. What is the mole fraction of methyl alcohol in the solution? (a) 0.050 (b) 1.100 (c) 0.190 (d) 0.086 [2011 (Cancelled)] –1 22. A 5 % solution of cane sugar (molar mass: 342 g mol ) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is (a) 136.2 (b) 171.2 (c) 68.4 (d) 34.2 [2011] 23. The molality of a urea solution in which 0.0100 g of urea, [(NH2)2CO], is added to 0.3000 dm3 of water at STP is (b) 5.55 ¥ 10–4 mol kg–1 (c) 33.3 mol kg–1 (d) 3.33 ¥ 10–2 mol kg–1 (a) 0.555 mol kg–1 [2011] 24. Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of hte solution lowered to – 2.8 °C? (a) 72 g (b) 93 g (c) 39 g (d) 27 g [2012]

Solutions

5.37

25. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2 (aq), 0.250 M KBr (aq) and 0.125 M Na3PO4(aq) at 50 °C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? (a) 0.500 M C2H5OH(aq) has the highest osmotic pressure. (b) They all have the same osmotic pressure (c) 0.100 M Mg3(PO4)2 (aq) has the highest osmotic pressure [2014, online] (d) 0.125 M Na3 PO4(aq) has the highest osmotic pressure 26. Dissolving 120 g of a compound of molar mass 60 g mol–1 in 1000 g water gave a solution of density 1.12 g mL–1. The molarity of the solution is (a) 1.00 M (b) 2.00 M (c) 2.50 M (d) 4.00 M [2014] 27. The observed osmotic pressure for a 0.10 M solution of Fe(NH4)2(SO4) at 25 °C is 10.8 atm. The expected and experimental (observed) values of van’t Hoff factor (i) will be respectively (R = 0.082 L atm K–1 mol–1) (a) 5 and 4.42 (b) 4 and 4.00 (c) 5 and 3.42 (d) 3 and 5.42 [2014, online] 28. For an ideal solution of two components A and B, which of the following is true? (a) Dmix H < 0 (b) Dmix H > 0 (c) A· · ·B interactions is stronger than A· · ·A and B· · · B interactions. (d) A· · ·A, A· · ·B and B· · ·B interactions are identical 29. A solution at 20 °C is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressures of pure benzene and toluene are 74.7 Torr and 22.3 Torr, respectively, then the total vapour pressure of the solution and benzene mole fraction in equilibrium with it will be respectively, (a) 35.0 Torr and 0.480 (b) 38.0 Torr and 0.589 (c) 30.5 Torr and 0.389 (d) 35.8 Torr and 0.280 [2015, online] 30. Determination of molar mass of acetic acid in benzene using freezing point depression is affected by (a) dissociation (b) association (c) partial ionization (d) complex formation [2015, online] 31. The vapour pressure of acetone at 20 ºC is 185 Torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 ºC, its vapour pressure was 193 Torr. The molar mass of the substance is (a) 32 g mol–1 (b) 64 g mol–1 (c) 128 g mol–1 (d) 488 g mol–1 [2015] –1 32. The solubility of N2 in water at 300 K and 500 Torr partial pressure is 0.01 g L . The solubility (in g L–1) at 750 Torr partial pressure is (a) 0.0075 (b) 0.005 (c) 0.02 (d) 0.015 [2016, online] 33. An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is (a) 0.50 (b) 0.33 (c) 0.67 (d) 0.80 [2016, online] 34. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in Torr) for the aqueous solution is (a) 76.0 (b) 752.4 (c) 759.0 (d) 7.6 [2016]

ANSWERS 1. 7. 13. 18. 24. 30.

(a) (a) (a) (b) (b) (all)

2. 8. 14. 19. 25. 31.

(b) (c) (a) (b) (b) (b)

3. 9. 15. 20. 26. 32.

(b) (a) (a) (b) (b) (d)

4. 10. 16. 21. 27. 33.

(d) (d) (d) (b) (a) (a)

5. 11. 16. 22. 28. 34.

(a) (b) (c) (c) (d) (b)

6. 12. 17. 23. 29.

(c) (a) (c) (b) (b)

5.38

Complete Chemistry—JEE Main

HINTS AND SOLUTIONS 1. Larger the number of species in solution, larger the boiling point. 0.01 M Na2SO4 solution has 0.03 M of species (= 0.02 M Na+ + 0.01 M SO42– ). N 6.02 ¥ 1020 = = 10-3 mol 2. Amount of urea, n = 23 -1 N A 6.02 ¥ 10 mol n 10-3 mol = = 10-2 mol L-1 V 0.1 L 3. The molecular interactions between benzene and methanol are weaker than those existing between benzene and benzene, and methanol and methanol. Consequently, the solution will exhibit positive deviations from Raoult’s law. 4. The freezing point depression is given by –DTf = Kf m. Different solvents have different values of Kf and hence show different freezing point depression. 5. Na2SO4 ionizes as Na2SO4 Æ 2Na+ + SO 2– 4 Concentration of urea solution, c =

n(1 – a)

n(2a)

na

Total amount of species in solution will be n¢ = n(1– a) + 2na + na = n(1+ 2a) n¢ The van’t Hoff factor is i = = 1 + 2a n m1 78 g n1 = = = 1 mol 6. Amount of benzene, M1 78 g mol-1 m2 46 g = = 0.5 mol Amount of toluene, n2 = M 2 92 g mol-1 n1 1 x1 = = Amount fraction of benzene, n1 + n2 1.5 Ê 1ˆ Partial pressure of benzene, p1 = x1 p1* = Á ˜ (75 Torr ) = 50 Torr Ë 1.5 ¯ 7. The solutions will have same boiling point and freezing point provided the solutes dissolved are nonelectrolytes. 8. Both acetonitrile and acetone are polar and have permanent dipole moment. 18 g 178.2 g = 0.1 mol; Amount of water = = 9.9 mol 9. Amount of glucose = -1 180 g mol 18 g mol-1 Mole fraction of water =

9.9 g = 0.99 0.1 + 9.9

Vapour pressure of water in solution at 100 °C is

p = xp* = (0.99) (760 Torr) = 752.40 Torr

10. Isotonic solutions have the same osmotic pressure. Hence, from P = cRT, we get c1 = c2 fi n1 = n2 i.e. 11. For

5.25 g 1.5 g 5.25 ¥ 60 = fi M= g mol-1 -1 M 1.5 60 g mol

AgIO3 (s)  Ag + (aq ) + IO3- (aq ),

we have

Ksp = [Ag+] [IO–3 ] = s2

Hence s =

i.e.

M = 210 g mol–1

Ksp = 1.0 ¥ 10 -8 M 2 = 1.0 ¥ 10 -4 M

Mass of AgIO3 in 100 mL solution is m = (1.0 ¥ 10–4 mol L–1) (283 g mol–1) (0.1 L) = 2.83 ¥ 103 g 12. Using the expression p = xA pA + xB pB; (A is ethyl alcohol and B is propyl alcohol)

Solutions

5.39

290 mmHg = 0.6 pA + 0.4 ¥ 200 mmHg

we get

(290 - 0.4 ¥ 200) mmHg = 350 mmHg 0.6 13. Let xA be the amount fraction of A in the liquid solution. We will have pA =

This give

ptotal = xA p*A + xB p*B = xA p*A + (1 - xA ) p*B = xA ( p*A - p*B ) + p*B ptotal - p*B 760 mmHg - 1000 mmHg 240 1 = = = 520 mmHg - 1000 mmHg 4880 2 p*A - p*B nA 1 mol ¥ 100 = ¥ 100 = 50 Hence, the amount per cent of A is 2 mol nA + nB xA =

This gives

n2 =

14. Amount of glucose,

m2 18 g = = 0.1 mol M 2 180 g mol -1

m1 178.2 = = 9.9 mol M1 18 g mol-1 n1 9.9 g x1 = = = 0.99 n1 + n2 (9.9 + 0.1) mol

n1 =

Amount of water, Mole fraction of water,

p = x1 p *1 = (0.99)(17.5 mmHg) = 17.325 mmHg. p = xX p*X + xY p*Y

Vapour pressure of solution, 15. (a) For an ideal solution

Ê 1 ˆ p* + Ê 3 ˆ p* fi p* + 3 p* = 2200 mmHg 550 mmHg = ÁË ˜ Y ˜ X ÁË X Y 1 + 3¯ 1 + 3¯ Ê 1 ˆ p* + Ê 4 ˆ p* fi p* + 4 p* = 2800 mmHg 560 mmHg = ÁË ˜ Y ˜ X ÁË X Y 1 + 4¯ 1 + 4¯ Solving for p X* and p*Y, we get

p*X = 400 mmHg and p*Y = 600 mmHg

16. The interaction between unlike molecules is weaker than those involved between like molecules. This results into positive deviations from Raoult’s law 17. Na 2SO 4 (aq) Æ 2 Na + (aq) + SO24- (aq);

van't Hoff factor = 3

– DTf = i Kf m = (3) (1.86 K kg mol–1) (0.01 mol kg–1) = 0.0558 K

Hence,

18. Amount of heptane, n1 = Amount of octane, n2 =

m1 25.0 g = = 0.25 mol M1 100 g mol-1 m2 35.0 g = = 0.307 mol M 2 114 g mol-1

Mole fraction of heptane, x1 =

n1 0.25 = = 0.45 n1 + n2 0.25 + 0.307

Mole fraction of octane, x2 = 1 – xx = 0.55 According to Raoult’s law p = x1 p1* + x2 p*2 = (0.45) (105 kPa) + (0.55) (45 kPa) = 72.0 kPa 19. Since

- DTf = K f m = K f m2 =

m2 / M 2 , m1

we have

(- DTf )m1M 2 (6 K )(4 kg)(62 g mol-1 ) = = 800 g Kf (1.86 K kg mol-1 )

5.40

Complete Chemistry—JEE Main

20. We have

A x B y  xA y + + yB x -

m(1 – a)

m(xa)

m(ya)

Total molality of the solution = m(1 – a) + m(xa) + m(ya) = m{(x + y – l) a + 1} i=

The van’t Hoff factor is or

i = (x + y – l)a + 1

DTb K m{( x + y - 1)a + 1} = b Kb m ( DTb )0 i.e.

a=

i +1 x + y -1

21. In 5.2 molal aqueous solution, we have n2 = 5.2 mol and n1 = (1000 g/18 g mol–1) = 55.56 mol The mole fraction of methyl alcohol in the solution is x2 =

n2 5.2 mol 5.2 = = = 0.086 n1 + n2 (55.56 + 5.2) mol 60.76

22. Isotonic solutions have identical osmotic pressure and hence identical concentrations, i.e. the same amounts in the 5g 1g 342 ˆ = ; This gives M = ÊÁ g mol -1 = 68.4 g mol -1 -1 Ë 5 ˜¯ M 342 g mol 23. Assuming density of water equal to 1 kg dm–3, we have m=

n2 m2 / M 2 (0.0100 g/60 g mol-1 ) = = = 5.55 ¥ 10-4 mol kg -1 m1 m1 (0.3000 kg )

24. Since – DTf = Kf m, we get. - DTf 2.8 K m= = = 1.505 mol Kf 1.86 K kg mol-1 Since the molality, m = n/m1 , we get n = mm1 = (1.505 mol kg–1) (1 kg) = 1.505 mol Finally, the mass of ethylene glycol, C2H6O2 (the molar mass = 62 g mol–1) required will be m = n Mm = (1.505 mol)(62 g mol–1) = 93.3 g 25. All the four solutions have the same amount of the species present in the solution. 0.500 M C2H5OH 0.300 M Mg2+ + 0.200 PO3– 0.100 M Mg3(PO4)2 4 0.250 M KBr 0.250 M K+ + 0.250 M Br– 0.125 M Na3PO4 0.375 M Na+ + 0.125 M PO3– 4 Since all of them have the concentration of total species, their osmotic pressure will have the same value. 26. Total mass of the solution = 120g + 1000g = 1120g 1120 g m Volume of the solution, V= = = 1000 mL = 1.0 L r 1.12 g mL-1 m 120 g = = 2.0 mol M 60 g mol -1

Amount of compound,

n=

Molarity of solution,

M=

n 2.0 mol = = 2.0 mol L-1 V 1L

Solutions

5.41

27. Fe(NH4)2(SO4)2 ionizes as Fe(NH4)2 (SO4)2 Æ Fe2+ + 2NH +4 + 2SO2– 4 The number of species produced is 5 times than that of Fe(NH4)2 (SO4)2. Therefore, the expected van’t Hoff factor is 5. The expected osmotic pressure assuming no ionization is p = cRT = (0.10 mol L–1) (0.082 L atm K–1 mol–1) (298.15 K) = 2.45 atm The observed van’t Hoff factor is 10.8 atm = 4.4 i= 2.45 atm 28. For an ideal solution, interactions among A· · ·A, B· · ·B and A· · ·B are identical. nb 1.5 = = 0.3 29. Mole fraction of benzene, xb = nb + nt 1.5 + 3.5 Mole fraction of toluene, xt = 1 – xb = 0.7 The vapour pressure of the solution is p = xb p*b + xt p*t = (0.3 ¥ 74.7 + 0.7 ¥ 22.3) Torr = 38.02 Torr Mole fraction of benzene in vapour phase is yb =

pb (0.3 ¥ 4.7) Torr = = 0.589 p (38.02 Torr )

30. All are correct. 31. According to Raoult’s law, p1= x1 p*1 = (1 – x2)p*1 i.e.

x2 = 1 –

p1 * 1

p

= 1−

183 Torr

=

2

185 Torr 185 n2 (1.2 g/M ) = x2 = n1 + n2 (100 g / 58 g mol−1 ) + (1.2 g/M )

Also,

(1.2 g/M ) 2 = −1 (100 g / 58 g mol ) + (1.2 g/M ) 185

Hence, This gives

1.2 g ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 100 g ⎞ ⎟ ⎟⎜ ⎟=⎜ ⎜1 − M ⎝ 185 ⎠ ⎝ 185 ⎠ ⎝ 58 g mol−1 ⎠ Approixmate value Writing x2

or

n2/n1,we get

−1 ⎛ 183 ⎞ ⎛ 185 ⎞ ⎛ 58 g mol ⎞ M = (1.2 g) ⎜ ⎟ = 63.68 g mol–1 ⎜ ⎟⎜ ⎟ ⎝ 185 ⎠ ⎝ 2 ⎠ ⎝ 100 g ⎠

2 1.2 g/M . = 185 (100 g / 58 g mol−1 )

58 185 ⎞ ⎛ −1 × M = ⎜1.2 × ⎟ g mol = 64.38 g mol–1 100 2 ⎠ ⎝ 32. According to Henry’s law, s = kH p. Hence Êp ˆ Ê 750 Torr ˆ s2 p = 2 i.e. s2 = Á 2 ˜ s1 = Á (0.01 g L-1 ) = 0.015 g L–1 ˜ s1 p1 Ë 500 Torr ¯ Ë p1 ¯ This gives

33. We have MX2

m(1 – a)

2+ –     M + 2X

ma

2ma

Total molality in the solution = m(1 + 2a) Hence This gives

m(1 + 2a ) =2 m a = 0.50

5.42

Complete Chemistry—JEE Main

35. Amount of glucose, n2 =

m2 18 g = = 0.1 mol M 2 180 g mol-1

m1 178.2 g = = 9.9 mol M1 18 g mol-1 n2 0.1 = Amount fraction of glucose, x2 = = 0.01 n1 + n2 0.1 + 9.9 The relative lowering of vapour pressure of water is Dp - *1 = x2 p1 Amount of water, n1 =

p1 = x2 p*1 = (0.01) (760 Torr) = 7.6 Torr p1 = p*1

p1 = (760 – 7.6) Torr = 752.4 Torr

Note: In the above calculations, temperature of aqueous solution is assumed to be 100 °C.

6 Chemical Thermodynamics

Section 1

Basic Definitions and First Law of Thermodynamics

The subject of thermodynamics deals basically with the intersection of one body with another in terms of the quantities of heat and work.

1.

Neither energy nor matter can be transferred to or from it. 2. the system. 3. The boundary may be imaginary or real; it may be rigid or nonrigid; it may be a conductor of heat (diathermic 4. distinguished based on the size of system.

5.

A process is the path along which a change of state of a system takes place. The following ways of carrying a process may be considered.

state.

6.2 Complete Chemistry—JEE Main

6.

If the change in property of a system does not depend upon the path adopted in going form one state to another, the property is said to be state function. Energy of a system can be changed in

The internal energy (symbol: U the form of work (symbol: w dU = dq + dw

q

DU = q + w

DU, whereas heat and work are simple written as q and w. This follows from the fact that U is a state function whereas q and w are path dependant. DU = 0,

For an isolated system, both q and w i.e. U

constant. system against an external pressure is known as mechanical work. Consider a gaseous system as shown in Fig. 1 If the piston is pushed through a distance dl Work = Force ¥ Distance = (Pressure ¥ ie. dw = (pext A l pext (Adl pextdV where dV recommendations, the expression of w is w = –pextdV For expansion of the gas, dV w work is done by the system. For compression of the gas, dV and dw dU = dq + dw dU = dq – pextdV dU = dqV

or

DU = qV

that is, but qV is a state function. For pext = p dU = dq – pdV Adding and subtracting Vdp on the right side, we get

q is not state function

Chemical Thermodynamics 6.3

dU = dq – pdV + (Vdp – Vdp = dq – (pdV + Vdp + Vdp = dq – d(pV + Vdp or dU + d(pV = dq + Vdp or d(U + pV = dq + Vdp or where H = U + pV and is known as enthalpy function. dH = dqp

dH = dq + Vdp

If heat is added at constant p DH = qp

or

q is not a state function but

qp is a state function. T is the increase in temperature by adding heat dq, then the heat capacity is dq C= dT For gaseous system, two types of heat capacities may be considered. These are CV = (∂q/∂T V Heat capacity at constant pressure: Cp = (∂q/∂T p dU = dqV = CV dT dH = dqp = Cp dT

DH DU H = U + pV Hence, For an ideal gas Hence,

Hence, D(pV  0

DH = DU + D(pV

DH  DU D(pV  D(nRT nR(DT DH  DU + nR DT

Cp Hence

CV H = U + pV dH = dU + d(pV

For one mole of an ideal gas,

H = dU + nR dT Cp dT = CV dT + nRdT or Cp,m – CV,m = R

Cp – CV = nR

compression of a system is dw = –pext dV Let an ideal gas undergo expansion from V1 to V2 (or compression from V2 to V1 For expansion, the external pressure must satisfy the expression. pext £ p2. Let pext = p2 † w = –pext (DV p2(V2 – V1 V 2 > V 1, w For compression, the external pressure must satisfy the expression pext ≥ p1. Let pext = p1 † w = –pext DV = –p1(V1 – V2 V1 p1

6.4 Complete Chemistry—JEE Main

or compression is carried out in more than one step, the pext has to be adjusted in each step. Let pext pressure in each step.

compression. This trend is continued as the number of steps is increased.

For expansion pext = p – dp The expression of work is – dp V  – p dV dw = – pext dV = – (p + (The product dpdV For an ideal gas, Hence

p = nRT/V. dw = –nRT

dV V Vf

w = - nRT

Ú

Vi

For expansion Vf (=V2 system For compression, Vf (=V1

Vf dV = - nRT ln Vi V

Vi(=V1 w Vi (=V2 w

For compression

pext = p + dp

Chemical Thermodynamics

under the isotherm.

compression. For adiabatic expansion or compressure, heat is neither enters q dU = dw

or

nCV dT = –pextdV

w = –pext (V2 – V1

or,

nCV(T2 – T1

For expansion V2 > V1, w < 0, hence T1 > T2 For compression, V2 < V1, w > 0 and hence, T2 > T1 from the expression w = DU = nCV, m (T2 – T1 T2 pext = p ± dp. Thus or Replacing

dU = dw

nCV, mdT = –pext dV

or

nCV, mdT = – (p ± dp V  –pdV p = nRT/V,

we get

nRT ˆ nCV, mdT = - ÊÁ dV Ë V ˜¯ T2

CV , m Ú

T1

or

or

ÊT ˆ ln Á 2 ˜ Ë T1 ¯

V

2 dT dV R Ú = V T V

CV , m/ R

R

1

ÊV ˆ = ln Á 1 ˜ Ë V2 ¯

or

T

dT dV = –R T V

T2 V = - R ln 2 T1 V1 CV , m / R

Ê T2 ˆ ÁË T ˜¯ 1

or

=

V1 V2

CV, m / R

V = constant

CV ,m 1 1 = = = C -C ( C / C ) 1 g -1 p ,m V,m p, m V , m T1/(g Ê RT ˆ TÁ Ë p ˜¯

The relation between p and V is

pext

CV , m ln

or

T2CV, m / RV2 = T1CV, m / RV1 CV , m

CV, m

or

V = constant

or

TV g – 1

g -1

= constant

Ê pV ˆ (V g -1 ) = constant ˜ ÁË R ¯

T g p 1– g = constant

or or

pV g

or

T g p 1– g

6.6 Complete Chemistry—JEE Main

or w and Dq = 0

dw = –pextdV = 0 DV = 0, DH = 0

For such an expansion,

Ek = Hence,

3 Ê ∂Ek ˆ CV = ÁË ˜¯ = nR ∂T V 2

CV, m =

3 R; 2

C p, m =

and

CV , m =

3 5 R + R = R; 2 2

3 nRT 2

3 R 2

g =

C p, m CV , m

=

5 = 1.667 3

motions. RT towards molar energy of the molecule. N–5 and 3N

N is the RT towards molar energy of the molecule. The following is the summary of molar energy contributions by each motion towards the total molar energy of the molecules. It also includes, their heat capacities and the ratio of heat capacities. Translational

Rotational

Vibrational

Total

CV, m

Cp,m

g =Cp, m/CV, m

3 RT 2

1 2 ¥ RT 2

RT

7 RT 2

7 R 2

9 R 7

9 = 1.286 7

linear

3 RT 2

1 2 ¥ RT 2

4RT

13 RT 2

13 R 2

15 R 2

15 = 1.154 13

nonlinear

3 RT 2

1 3 ¥ RT 2

3RT

6RT

6R

7R

7 = 1.167 6

Molecules Diatomic Triatomic

CV, m.

MULTIPLE CHOICE QUESTIONS ON SECTION 1

Chemical Thermodynamics 6.7

5. The work done on a gaseous system is 12 J and also 20 J of heat is added to it. The change in internal energy of the gaseous system is 6. The heat capacity of 12.0 g of helium is about –1

–1

–1

–1

7. 10 J of heat is supplied to 36.0 g water at 373.15 K and 1 atm pressure. The heat capacity of water system will be –1

–1

–1 3 to

20 dm3 against an external pressure

9. The difference between Cp and CV of 0.2 mole of an ideal gas is about –1

–1

–1

–1 3

to 5 dm3 at 300 K. The work

11. A gas undergoes an isothermal expansion against a constant external pressure in stepwise manner such that the be largest when the expansion is carried out in 12. A gas undergoes an isothermal compression against a constant external pressure in stepwise manner such that the compression is carried out in 13. 28.0 g of nitrogen undergoes expansion from 1 atm, 10 dm3 to 0.5 atm, 20 dm3 against a constant external pressure

V1 to V2 against a constant external pressure. It is DU = 0 and w

DU = 0 and w

DU = 0 and w DU > 0 and w > 0 V1 to V2 against a constant external pressure. If T1

T2 > T1

T2 < T1

V1/T1 = V2/T2

V1 to V2, which of the following facts holds good: T2 < T2¢ T2 = T2¢ T2 > T2¢ where g is the ratio Cp, m/CV, m of the gas.

V1 T1 = V1 T2

to T2 T2 = T1 16. If T2 and T2¢

g

p 1, V 1, T 1 g

g

p1 V1 = p2 V2 p 1V 1 = p 2V 2 where g is the ratio Cp,m/Cv,m of the gas.

p1/T1 = p2/T2

g

p 2, V 2, T 2 V1/T1 = V2/T2 V1 to V2. If 298 K and 278

R

R

R V1 to V2

then in this process DU = 0 and w

DU = 0 and w

DU = 0 and w

R V2 to V1, DU > 0 and w > 0

6.8 Complete Chemistry—JEE Main

R R 21. The ratio g = Cp, m/CV, m for a diatomic gaseous molecules is

R

R

g = Cp, m/CV, m for diatomic gaseous molecules will be 23. The molecule of a gas has three independent translational motions along with each of the three Cartesian axis. The contribution of each translation motion of a molecule towards total energy is kT kT kT kT

25. Which of the following statements is

qp and qV are state functions

ANSWERS

HINTS AND SOLUTIONS 1. In an isolated system, neither heat nor matter can transferred to or from the system.

5. DU = q + w = 20 J + 12 J = 32 J 6. Amount of helium, n =

8. 9. 10.

Helium is monatomic gas. Its CV, m

R

9 R = ÊÁ molˆ˜ (8.314 J K -1 mol -1 ) = 37.41 J K -1 ¯ Ë2 At 1 atm pressure, 373.15 K is the normal boiling point of water. Heat supplied at this temperature is utilized in q q C= = =• Dt 0 3 – 10 dm3 ¥ w = –pext (V2 – V1 –1 mol–1 –1 Cp – CV = nR Ê 5 dm3 ˆ V2 –1 w = –nRT ln mol–1 ÁË 20 dm3 ˜¯ = 6916.6 J = 6.92 kJ V1 Heat capacity,

7.

m 12 g = = 3 mol . M 4 g mol-1

CV = nCV, m

Chemical Thermodynamics 6.9

11. 12. 13. 14.

Larger the number of steps, larger the magnitude of work done by the gas. Lesser the number of steps, larger the work done on the gas. For expression, pext £ p . The magnitude of work will be maximum when pext = p . DU compared to the magnitude of work done by the system during expansion, w = wexp + wcomp > 0 T 2 < T 1. T2 < T2¢.

g

pV = constant holds good. Ê 3ˆ R ÁË ˜¯ 2 19. DU = 0 as U a state function. w

18. w = nCV, m (T2 – T1

R

R. 20. For a monatomic gas, CV, m R and Cp, m R. Thus g =Cp, m/CV, m = 9/7 =1.286 21. For a diatomic gas, CV, m R and Cp, m R. . Thus g =Cp, m/CV, m = 7/5 =1.4 22. In this case, CV, m kT. V1 24. w = wexp + wcomp = –pext (V2 –V1 nRT ln V2 3 –10 dm3 –1 mol–1 = –101.325 J + 1729. 0 J = 715.74 J

Section 2

Thermochemistry

chemical reaction, and the surroundings. x, pronounced as x= The unit of x is mol.

amount of a reactant consumed or a product produced stoichioometric number of the reactant or product

Let the decomposition of N2 amount of N2 5 equal to n0. To start with, extent of reaction x = 0. The enthalpy of the system to start with is Hinitial = n0 Hm( N2 5 When the reaction has proceeded to the extent x, then 2 N 2 O5 (g)  4 N 2 O5 (g) + O 2 (g) n0 - 2x

4x

x

xHm( H = (n0 – 2x Hm( N2 5 2 The enthalpy change is DH = H – Hinitial = x[–2Hm( N2

xHm( 5

Hm

2

2

Hm

2

5

is studied with the initial

Complete Chemistry—JEE Main

D rH =

DH = –2Hm( N2 x

D rH =

Unit of DH J kJ = or Unit of x mol mol

5

Hm

Hm

2

2

In general, enthalpy of reaction may be expressed as D rH =

 B n B H m (B) -  B n B H m (B)

For exothermic reactions, heat is released and thus DrH requires that the enthalpy of reactants is larger than that of products. For endothermic reactions, heat is absorbed and DrH than that of reactants. follows that For example 2 may be formed in two ways æÆ 2 1 2

æÆ

2 1 2

Add 2

D r H = –393.5 kJ mol–1

2

2

Dr H = –110.3 kJ mol–1

æÆ æÆ

Dr H = –283.2 kJ mol–1

2

D r H = –393.5 kJ mol–1

2

The chemical equations may be treated as algebraic expressions which can be added or subtracted to yield the required chemical equation. The corresponding enthalpy changes are also manipulated in the

required to completely separate one mole of solid ionic compound into gaseous ions. For example – æÆ Na+ æÆ æÆ Dissociation of

1 2

1 2

Cl2

+

;

Dr H2 Dr H3

æÆ Cl–

Dr H4

–

æÆ

Dr H5

Cl2

æÆ

Dr H6

– –

–

æÆ

Cl2

Formation of Cl– Condensation of Na+

Dr H1 Na+

Add 1 2

According to Hess’s Law DrH6 = DrH1 + DrH2 + DrH3 + DrH4 + DrH5 Except DrH5, all other DrHs can be determined experimentally. Hence –DrH5 = DrH1 + DrH2 + DrH3 + DrH4 – DrH6 The term –DrH5 represents lattice energy.

Chemical Thermodynamics

enthalpy of reaction is known as standard enthalpy of reaction, if the external pressure on the chemical system is 1 bar (1 atm = 1.01325 bar = 1.03125 ¥ 105 DrH° = ÂPnP H°m, P – ÂRnR H°m, R H°m = U°m + pVm,

DrH° = ÂPnP(U°m + pVm P – ÂRnR(U°m + pVm

we get

= (ÂPnPU°m, P – ÂRnRU°m, R + (ÂPnP(pVm P – ÂRnR(pVm = DrU° + ÂPnP(pVm P – ÂRnR(pVm pVm  0 pVm  RT

R

R

R

for one mole of solid or liquid for one mole of gaseous species DrH°  DrU°

DrH° = DrU° + [ ÂPnP(RT – ÂRnR(RT = DrU° + (RT

 Pn P –  Rn R

= DrU° + (Dng RT where Dng is the change in stoichiometric number of gaseous species in going from reactants to products. 2

2

Æ 2NH3

DrH° = –92.38 kJ mol–1. Calculate the energy of reaction at

298 K. Dng DrU = DrH – (Dng RT = –92.38 kJ mol–1

–1

mol–1

= –92.38 kJ mol–1 + 4955 J mol–1 = –92.38 kJ mol–1 + 4.96 kJ mol–1 = –87.42 kJ mol–1

formation of substance can be built.

For examples, D fH 2

D fH

DfH°(Br2

D fH

D fH

The standard enthalpy of formation (symbol: Df H the requisite amounts of elements in their stable state of aggregation. The formation of one mole of a substance implies that the compound appears as product with stoichiometric number equal to one. For examples 1 H (g) 2 2

+ 12 Br2 (l) æÆ HBr (g)

Df H° = –36.40 kJ mol–1

C(graphite) + 2H 2 (g) æÆ CH 4 (g) 3 2

2

æÆ

3

Df H° = –74.80 kJ mol–1 Df H° = –393.72 kJ mol–1 The standard enthalpy of a reaction may be computed by using the

Complete Chemistry—JEE Main

DrH° = Â B n B D f H ∞(B) - Â B n B D f H ∞(B) Æ

2

DrH° = DfH

D fH

2

D fH

D fH –1

= 65.3 kJ mol–1

D cH

CH4

æÆ DcH° = DfH 2

2 2

2

D f H°(H2

DfH°(CH4

D fH

2

–1

= –694.46 kJ mol–1

Measurement of Enthalpy of Combustion a substance. A known mass of the substance in a closed steel The calorimeter is surrounded by a known mass of water. The sample is ignited electricity to bring about the combustion

qcomb = –(mwater cwater + mbomb c bomb DT where c of substance is equal to energy of combustion. qcomb qcomb = D cU = nsubs msubs / M subs The enthalpy of combustion will be DcH = DcU + (Dng RT of neutralization is the enthalpy change when one mole of H+ –

H+

–

æÆ H2

acid or base is made to ionize completely in the solution. – æÆ H3 + 2 – æÆ H + + CN– 3 DrH° = DneutH° + DionzH

DneutH° = –55.84 kJ mol–1

DionzH° = 43.71 kJ mol–1 –1 =12.13

kJ mol–1

Chemical Thermodynamics

2 2

æÆ HCl(10H2 æÆ HCl(200H2

DsolH° = –69.49 kJ mol–1 DsolH° = –74.20 kJ mol–1

æÆ

DsolH° = –75.15 kJ mol–1

The integral enthalpy of dilution is the change in enthalpy when a solution containing 1 mole of solute is diluted from one concentration to another. –1 = –4.71 mol–1 DdilH HCl(10H2 2 æÆ HCl(200H2 of a solution of known concentration so that there occurs no appreciable change in concentration of the solution. of a solution of known concentration so that there occurs no appreciable change in concentration of the solution. It is the enthalpy change when one mole of one allotropic form changes to another. æÆ DtrsH° = 1.90 kJ mol–1 It is the enthalpy change (symbo: DatH atoms in the ground state in the gaseous phase. æÆ DatH° = 625 kJ mol–1 For example H2 The bond enthalpy (symbol: e dissociate the said bond present in different gaseous molecules into free atoms in the gaseous state. For example æÆ DrH°1 = 501.87 kJ mol–1 H2 æÆ DrH°2 = 423.38 kJ mol–1 Hence

1 2

e

(DrH°1 + DrH°2 = 12

kJ mol–1 = 462.62 kJ mol–1

DrH°1 and DrH°2 of all bond dissociation enthalpies. The data on the bond enthapies may be employed to estimate the approximate formation of a compound. Enthalpy of formation of ethane from the following data: = 348 kJ mol–1, e = 412 kJ mol–1 e = 436 kJ mol–1 and e æÆ C2H6 D rH 2 We may proceed as follows: DfH° ææÆ 2

H ¨æÆ H

= – 80 kJ mol–1

eH–H – 3eC–C – 6eC–H

2H 6

Make bond –eC–C – 6eC–H H

C· · · C H

According to Hess’s law DfH° = 2DsubH

= 716 kJ mol–1

Æææ

ææÆ

ææÆ

carbon DsubH°

Break bond 3eH – H

DsubH

H H –1

Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS ON SECTION 2

DcH°(H2

–1

DfH°(CH4 kJ mol–1 DcH°(C2H4

D cH

;

;

DcH°(CH4

–1

will be kJ mol–1

DrH° of the reaction kJ mol–1 DfH(Mg3N2 D rH kJ mol–1

C 2H 4 kJ mol–1

–1;

N2

–1;

5

kJ mol–1 DcH°(H2

–1

æÆ C2H6

2

kJ mol–1

DfH°(NH3 H3 –1 kJ mol

–1

Df H°(H2

kJ mol–1

DcH°(C2H6

–1;

The Dr H° for the reaction kJ mol–1

–1

kJ mol–1

–1

æÆ Mg3N2 2 kJ mol–1

Df H

3

H 2O

æÆ

–1;

kJ mol–1 Df H°(N2

–1

5

3

mol–1

kJ mol–1

kJ mol–1

–1

heat capacity and density of both the solution are 4.18 J K–1 g–1 and 1 g mL–1

Df H

2

DrU°

–1

and 2

Df H 1 2

2

–1

3

æÆ

3

mol–1 kJ mol–1 kJ mol–1 kJ mol–1 7. For the reaction. 2A2 5B2 æÆ 2A2B5 DrH° < DrU° DrH° = DrU° DrH° = –3DrU° DrH° > DrU° –1 8. A 0.1215 g sample of solid magnesium (molar mass: 24.30 g mol = 1.7448 kJ °C–1 combustion of magnesium at 298 K is about. kJ mol–1 kJ mol–1 kJ mol–1 kJ mol–1 –1 –1 e = 374 kJ mol ; eC = C = 632 kJ mol ; e = 411 kJ mol–1 If the enthalpy formation of benzene from gaseous atoms is 5536 kJ mol–1, the resonance energy benzene is kJ mol–1 kJ mol–1 kJ mol–1 kJ mol–1 –1 –1 –1 e = 388 kJ mol ; e = 436 kJ mol ; DfH (NH3 The bond enthalpy of N∫∫N is about kJ mol–1 kJ mol–1 kJ mol–1 kJ mol–1 H2 4 5H2O æÆ H2 4◊5H2 DrH° = –58.0 kJ mol–1 –1 g–1 is obtained when 1 mol of H 2 4 is added to 5 mol of H2

Chemical Thermodynamics

12. If DcH° of DfH°

–1

, DfH°

kJ mol–1

–1

2

kJ mol–1

and Df H°(H2

kJ mol–1 4

–1

kJ mol–1 –1 and –51.3 kJ mol–1

4

kJ mol–1 kJ mol–1 –1 14. If Df H°(C2H6 = –84 kJ mol , DsubH° e = 414 kJ mol–1, then the bond enthalpy e kJ mol–1 kJ mol–1 +

kJ mol–1 kJ mol–1 –1, D (H = 435 kJ mol–1 and the bond enthalpy at 2 will be about kJ mol–1 kJ mol–1 DfH°m

–

– 3

–

ANSWERS

HINTS AND SOLUTIONS 1 2

æÆ H2 DcH° = –286.1 kJ mol–1 æÆ DcH° = –394.9 kJ mol–1 2 2 DcH° = –882.0 kJ mol–1 4 2 æÆ 2 2 The required chemical equation for which Df H 2

Hence, 2H 4

DfH°(CH4 æÆ 2 7 2

2H 6

2

2

1 2

2

2

æÆ

= –85.1 kJ mol–1 DcH° = –1410.9 kJ mol–1

2 2

DcH° = –1559.8 kJ mol–1

2

DcH° = –285.8 kJ mol–1

The required chemical equation for which DrH° is required is Hence,

D rH

–1

æÆMg3N2

2 1 2

3 2

N2

H2

;

æÆ NH3 3

2 1 2

1 2 3 2

N2 2

5 2

1 2

2 2

æÆ N2

The required reaction Hence, D rH

= –136.9 kJ mol–1 DfH° = –463 kJ mol–1

æÆ Mg3N2 2 –371 kJ mol–1 DfH° = –286 kJ mol–1 DfH° = –174 kJ mol–1

3

DfH° = –14.5 kJ mol–1

5

N2

2

–1 =

æÆ

H2

C2H 4

DfH° = –46 kJ mol–1

DrH° = (– 463 + 2 ¥ æÆ H2 2

Hence,

æÆ CH4

–1

æÆ H2

2

2

5

2

æÆ

3 –1 =

–76.5 kJ mol–1

æÆ C2H6

Complete Chemistry—JEE Main –1

n1 = VM Amount of HCl, n2 = VM

–1

+

–

, heat liberated in this process is –1

q = nDneutH Increase in temperature of the solution will be DT =

q 7320 J = = 5.0 ∞C mc (350 g)(4.18 J g -1 ∞C -1 )

Hence, temperature of solution is 25.0 °C + 5.0 °C = 30.0 °C æÆ

2 3 2

æÆ

2

DfH° = –296.8 kJ mol–1

;

2

DfH° = –395.7 kJ mol–1

3 1 2

2

Hence,

æÆ

2

D rH

–1

3

= –98.9 kJ mol–1

Dng = –1/2. Hence ¥ 10–3 kJ K–1mol–1 DrU° = DrH° – (Dng RT = –98.9 kJ mol–1 –1 = –97.66 kJ mol–1 Dng = –3. From the expression DrH° = DrU° + (Dng RT, it follows that DrH° < DrU° qwater = mwatercwaterDT

–1

°C–1

qbomb = (mbombcbomb DT = CbombDT = (1744.8 J °C–1 qtoal = qwater + qbomb q 3000 J = 600 ¥ 103 J mol–1 DcU = total = -1 m / M (0.1215 g / 24.30 g mol ) 9. Enthalpy of formation of benzene from the gaseous atoms using bond enthalpies data is –1 = –5484 kJ mol–1 DH = –3e –3eC == C – 6e –1 Resonance energy = DH DH = –52 kJ mol–1 10.

1 2

3 2

D fH ææÆ

3

Æææ

eN ∫∫ N

H2 ææÆ

ææÆ

1 2

3 2

N2 e

–3e

¨æÆ H

H N H

According to Hess’s law

Df H =

1 2

eN ∫∫ N = 2DfH – 3e

or

eN ∫∫ N +

+ 6e

3 2

e

– 3e

= (–2 ¥ 46.11 – 3 ¥ 436 + 6 ¥

–1 =

927.8 kJ mol–1

mcp DT = – D f H° where

m = m H2

+ mH2 = (nM

DT =

where 12. C6H6

4

15 2

2

æÆ

Df H°(C6H6

4

+ (nM

–1

water

-Df H 58 ¥ 103 J = 73.74 ∞C = mc p (188 g)(4.184 J º C-1 g -1 ) 2

DcH°(C6H6 = 6 Df H Hence

H2

2 2,

Df H°(H2

= (–3260 + 6 ¥ 393 + 3 ¥

Df H(C6H6 –1

= – 47 kJ mol–1

–1

Chemical Thermodynamics

13. H+ NH4 NH4

æÆ H2 æÆ NH4+

–

DrH°1 = –57.3 kJ mol–1 DrH2° = –51.3 kJ mol–1 DrH° =

+ H+ 2 – æÆ NH4+ DrH° = DrH2° – DrH°1 = 6.0 kJ mol–1 3H2 ææÆ

ææÆ

2DsubH

DfH° ææÆ

2H 6

Æææ

14.

–3eC–C –6eC–H

3DatH°(H2 H ¨æÆ H

C

H C

H

H According to Hess’s law DfH° = 2DsubH

DatH°(H2

eC–C = 2DsubH

DatH°(H2

H

eC–C – 6eC–H eC–H – DfH° = [ 2 ¥ 720 + 3 ¥ 435 – 6 ¥

–1

= 345 kJ mol–1 Df H°(H+

=0

Section 3

Criteria of Spontaneity

Many such examples can be cited. All these processes are said to be spontaneous. In these processes, the spontaneous

First law of thermodynamics holds good for both spontaneous and nonspontaneous processes.

1 2

2

æÆ

D f H° = 6.3 kJ mol–1

Thus DrH° criterion of spontaneity is obtained from the phenomenon of mixing of two gases as shown in Fig. 8.

Before mixing

After mixing

Complete Chemistry—JEE Main

The mixing of two gases shown in Fig. 8 is accompanied with increase in disorderlines of the two gases. This . For an . DSsystem + DSsurroundings When this isolated system attains an equilibrium state where the net interactions between system and surroundings is zero, there will occur in . Mathematically, the criteria of spontaneity and equilibrium of an isolated system may be stated as For spontaneity DS > 0 For equilibrium DS = 0

‘‘

’’

of entropy is dS =

dqrev T T.

where dq

The characteristics of entropy function may be understood from the example V1 to V2 dU = dq + dw U = f(T

T, dU = 0. dq q

or

= – dw = pdV = nRT = nRT ln

DS =

V2 V1

dV V

qrev V = nR ln 2 T V1

V1 and V2 are state functions, DS

S is a state function. S

states that. .

–1.

æÆ heat

where T *m and T *B

Tm* æÆ liquid (Tm* æÆ liquid (Tb* æÆ gas (Tb* æÆ gas (T heat

heat

heat

heat

Chemical Thermodynamics

D fus H ∞ T*

DSsÆl =

and

DSlÆ v =

D vap H ∞ Tb*

m

The molar enthalpy of fusion of solid water is 6008.2 J mol–1 D fus Sm∞ =

D fus H m∞ 6008.2 J mol-1 = = 22.0 J K -1 mol-1 * 273.15 K Tm –1

D vap Sm∞ =

D vap H m∞ Tb*

=

40585 J mol-1 = 108.8 J K -1 mol-1 373.15 K

DfusH°m and D H°m at temperatures T*m and T*b The change in entropy of the surrounding will be same but of opposite sign and thus DsysS°m + DsurrS°m = 0. mations take place at temperatures other than T*m and T*b in nature. In this case, DsysS°m + DsurrS°m > 0. Entropy change in a chemical reaction is obtained by using the expression DrS° = CH4

 Bn B Sm∞ , B -  Bn B Sm∞ ,B 2

DrS° = Sm

æÆ

2

2

Sm° (H2

2

Sm(CH4

= (213.74 + 2 ¥ 69.91 – 186.26 – 2 ¥

Sm –1

2

mol–1

= –242.70 J K–1 mol–1 equation. DtotalS = DsysS + DsurrS > 0

At constant p, heat is exchanged between system and surroundings at the temperature of transformation. Hence, we can write q q D H DsurrS = p ,surr = - p ,sys = - sys T T T Hence,

DtotalS = DsurrS + DsysS = -

or

TDtotalS = – (DsysH – TDsysS

At constant T or

TDtotalS = –Dsys(H – TS

where G as DH and DS

D sys H T

+ DsysS

DsysG H and S are state functions. Dropping the subscript ‘sys’

DG = – TDtotalS

Complete Chemistry—JEE Main

Hence, For spontaneous process, For equilibrium process, For nonspontaneous process,

DtotalS > 0 DtotalS = 0 DtotalS < 0

and thus and thus and thus

DG < 0. DG = 0. DG > 0.

Four cases may be distinguished D rH and DrS In this case, DG°(= DH° – TDS is always spontaneous. Example is 2H2 2 æÆ 2H2 DrH° = –196.1 kJ mol–1, DrS° = 125.8 J K–1 mol–1, 2 –1 At 298 K, DrG° = –233.6 kJ mol DrH° and DrS° In this case, DG nonspontaneous. Example is æÆ DrH° = 66.4 kJ mol–1, DrS° = –121.8 J K–1 mol–1, N2 2 2 –1 At 298 K, DrG° = 102.7 kJ mol , D rH and DrS T DrG° = DrH° – TDrS DrH° > TDrS° i.e. T < DrH°/ DrS° DrH° < TDrS° i.e. T > DrH°/ DrS° = zero if DrH° = TDrS° i.e. T = DrH°/ DrS° For example, DrH° = 57.2 kJ mol–1, DrS° = 175.8 J K–1 mol–1 2N2 4 æÆ 2 D r H ∞ 57.12 ¥ 103 J mol-1 = = 325.4 K Dr S∞ 175.8 J k -1mol-1 Thus if T > 325.4 K; spontaneous reaction T < 325.4 K; nonspontaneous reaction T = 325.4 K; equilibrium reaction and DrS D rH T DrG = DrH – TDrS ÁDH Á< ÁTD S Á i.e. T > ÁDrH Á/ ÁDrS Á ÁDH Á> ÁT D S Á i.e. T < ÁDrH Á/ ÁDrS Á = zero if ÁDH Á= ÁTD S Á i.e. T = ÁDrH Á/ ÁDrS Á For example, 2H2 æÆ 2NH3 DrH° = –92.2 kJ mol–1, DrS° = –198.8 J K–1 mol–1 2 3 -1 D r H ∞ 92.2 ¥ 10 J mol = = 463.8 K Dr S∞ 198.8 J k -1mol-1 Thus if T > 463.8 K; nonspontaneous reaction T < 463.8 K spontaneous reaction T = 463.8 K; equilibrium reaction DrH° and DrS°are assumed to be independent of temperature.

DG G = H – TS = (U + pV

TS

dG = dU + pdV + VdP – TdS – SdT = (dq + dw w wmech = – pdV wnonmech q = TdS dG = dwnonmech + Vdp – SdT At constant T and p

pdV + Vdp – TdS – SdT

Chemical Thermodynamics

dGT, p = dwnonmech

DGT, p = wnonmech

or

Ecell cell, then the electrical work produced is welec = –nFEcell where n spontaneous cell reaction and F is Faraday constant (=96487 C mol–1 DGT,p = dwnonmech DGT,p = –nFEcell For a spontaneous reaction, DGT,p < 0, hence Ecell > 0 For a Daniell cell (E Cu2+

Cathode

æÆ Zn2+ – æÆ 2+

–

æÆ Zn2+

–1 The decrease in free energy is DG –1 Thus, 212300 J mol of electrical work is produced per unit extent of cell reaction.

–1

The standard free energy of formation of elements in their stable form aggregations at 1 bar pressure is taken to be zero. of the substance is formed from the stable form of elements at their stable states of aggregation. 1 æÆ DG° = –95.3 kJ mol–1 The reaction to be referred is 12 H2 2 Cl2 DG° of this reaction is said to be the standard free energy of formation (symbol: DfG

The free energy change in a chemical reaction may be computed by the expression DrG° =

 B n B D f G°B -  B n B D f G°B

For example, CH4 DrG° = DfG

2

æÆ 2

2

2

DfG°(H2

DfG°(CH4 ¥

= –817.7 kJ mol–1 Ê pˆ DfG = DfG° + RT ln Á ˜ Ë p∞ ¯

For a gaseous phase For a substance in solution For a condensed phase where p

c DfG = DfG° + RT ln ÊÁ ˆ˜ Ë c∞ ¯ DfG = DfG°

c° (= 1 mol dm–3

D fG –1

2

6.22 Complete Chemistry—JEE Main

Free energy of a reaction may be expressed in terms of standard free energy Expression of free energy change of the reaction DrG = 2DfG(NH3

DfG(N2

N2(g, pN2

2(g,

pH2 æÆ 2NH3(g, pNH3

DfG(H2

Expressing DfG in terms of DfG°,we get Ê pNH3 ˆ ˘ È È Ê pN 2 ˆ ˘ È Ê pH 2 ˆ ˘ DrG = 2 Í D f G∞( NH3 , g) + RT ln Á ˙ - Í D f G∞( N 2 , g) + RT ln Á ˙ - 3 Í D f G∞(H 2 , g) + RT ln Á ˙ ˜ ˜ Ë p∞ ¯ ˚ Ë p∞ ˜¯ ˚ Ë p∞ ¯ ˚ Î Î Î ˘ È ( pNH3 / p∞) 2 = [2D f G∞( NH3 , g) - D f G∞( N 2 , g) - 3D f G∞(H 2 , g)] + RT ln Í 3˙ ÍÎ ( pN 2 / p∞)( pH 2 / p∞) ˙˚ = DrG° + RT ln Q°p where Q°p is known as standard reaction quotient. D rG becomes 0 = DrG° + RT ln K°p or where K°p (= Q°p

DrG° = – RT ln K°p = –2.303 RT log K°p DfG°(N2

DrG° = 2DfG°(NH3 Hence

log K°p = –

DfG°(NH3 DfG°(H2 –1 = –32.90 kJ mol–1

–1

at 298 K. Hence,

(-32.90 ¥ 103 J mol-1 ) = 5.77 (2.303)(8.314 J K -1 mol-1 ) (298 K) 5.

K°p

MULTIPLE CHOICE QUESTIONS ON SECTION 3 1. In the mixing of two ideal gases at the same temperature, the entropy of mixing is not zero because

V1 to V2 is V1 V2 V1 V2 DS = nR ln DS = nR ln DS = nRT ln DS = nRT ln V2 V1 V2 V1 3. The transformation H2 æÆ H2 DsysS°= –20.5 J K–1. Which of the following facts will hold good for the change in entropy of surroundings (DsurS DsurrS° > ÁDsysS° Á DsurrS° < ÁDsysS° Á DsurrS° = 0 DsurS° = ÁDsysS°Á 4. A body of Cu at 100 °C is in contact with another body of Cu at 0 °C. There occurs transfer of heat from a hot body to a cold body till both acquires the same temperature. If DhS° and DcS° are the changes in entropy of hot DhS° + DcS

DhS° + DcS

DhS° + DcS

DhS° = DcS°

Chemical Thermodynamics 6.23

6. For a reaction A2 æÆ spontaneous T T 7. For a reaction A2 reaction will be spontaneous T T  H2 8. The reaction H2 will be about –1 mol–1 –1

DrH° = 50.0 kJ mol–1 and DrS° = 125 J K–1 mol–1. The reaction will be T > 400 K 2

p = 1 atm DrH° = –100 kJ mol–1 and DrS° = –250.0 J K–1 mol–1. The

æÆ 2AB2,

T > 400 K T > 500 K DrH° = 40.6 kJ mol–1 –1

mol–1 and

S transformation is in equilibrium is

mol–1 æÆ

–1

mol–1 Df H –1 mol–1

S

Á Zn2+ –1

–1

–1,

–1

–1

11. The standard free energy change of reaction 2Fe2+ æÆ 2Fe3+ 2 The standard emf of the cell producing this reaction will be about 2+

13. The standard free energy change of the reaction 4NH3 –1, D G° DfG°(NH3 f

2 3

–1

9.65

2

15. If DfH°

S°

at 300 K. The standard

2

–1.

–1

2

–1,

–1.

æÆ 2 amd DfG(H2

2

æÆ N2

–

æÆ A2+

–1

æÆ

mol–1

Á

Mg2+ –1

–1

DrS°

3 –1

æÆ

–1

æÆ æÆ 2 mol–1 and S°

1 2

2 2 –1

T > 500K T

–

+

DrG – DrG° of the reaction –1

+

N2

– 3

æÆ

2 –1

–1

2 –1

mol–1,

6.24 Complete Chemistry—JEE Main

ANSWERS

HINTS AND SOLUTIONS 1. In the mixing of two gases, disorderliness or entrorpy of both the gases increases. nRT 2. dU = 0 at constant temperature. Hence dq = -dw = + pdV = dV V Thus

dS =

dq dV = nR T V

or

DS = nR ln

V2 V1

æÆ H2 2 DsurrS° > 0. From this, it follows that DsurrS° > ÁDsysS° Á. DtotalS° = DhS° + DcS° > 0.

DtotalS° = DsysS° +

DG < 0 and thus w 6. For spontaneous process DrG = DrH – TDrS < 0. –1 mol–1 Hence T > DrH/DrS = (50 ¥ 103 J mol–1 DrH – (–TDrS DrH > (–TDrS or DrG° = DrH° – TDrS –1 T < (–DrH D rS ¥ 103 J mol–1 mol–1 D H ∞ 40.6 ¥ 103 J mol-1 DrG° = DrH° – TDrS DrS° = r = = 108.8 J K -1 mol-1 373 K T 9. DrH° = DfH D fH = 0.33 kJ mol–1 – 0 = 0.33 kJ mol–1 –1 mol–1 = 0.8 J K–1 mol–1 DrS° = S S T=

D r H ∞ 0.33 ¥ 103 J mol -1 = = 412.5 K Dr S ∞ 0.8 J K -1 mol -1

10. DrG° = -nFE

–1

–1 =

– 50.18 kJ mol–1

n = 2. Hence,

E° = -

D r G∞ 114.0 ¥ 103 J mol-1 = = 0.59 V nF (2)(96500 C mol-1 )

RT 12. DG° = -nFE° and DG° = -RT ln K°. Hence nFE° = RT ln Kº or E° = ln K°. nF -1 -1 (8.314 J K mol )(300 K ) [2.303 log 109.65 ] E° = (2)(96500 C mol -1 ) 13. DrG° = 4DfG° DfG°(H2 DfG°(NH3 DfG° 2 –1

= –1226.2 kJ mol–1 14. Larger the change in gaseous molecules, larger the entropy change. æÆ –1 = – 1.89 kJ mol–1 DrH° = DfH° DfH° –1 mol–1 = 3.36 J K–1 mol–1 DrS° = S° S° DrH° < 0 and DrS° > 0, the reaction is spontaneous at all temperatures as DrG°

Chemical Thermodynamics

17. H+ DrG = DrG° + RT ln Hence

2 ( pNO / p∞) 2

( pN 2 / p∞)( pO2 / p∞) 2

DrG – DrG° = RT ln

(0.2) 2 = –RT (0.2)(0.2)2 –1 mol–1

= 4015.2 J mol–1

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE CHAPTER Identify the correct choice in the following questions.

q 4. Which of the following is qp

2 2

w qV

q /T q/T

qw q + dw

at room temperature and pressure 2

H + pV U + pV 9. The relation DsubH°= DfusH°+ D H° is true at all

U – pV

H – pV

10. Which of the following expressions is DS = q /T U = d q + dw DU = Dq + Dw DH = qp 11. If one gram of kerosene liberates 46.0 kJ of heat when it is burned, to what temperature can 0.25 g of kerosene raise the temperature of 75 cm3

6.26 Complete Chemistry—JEE Main

14. A gas expands from 1.5 L to 6.5 L against a constant pressure of 0.5 atm and during this process the gas also absorbs 100 J of heat. The change in the internal energy of the gas is 15. If a gas expands adiabatically from 1.0 L to 12.0 L against a constant pressure of 0.75 atm. The DU of the gas is 16. Molar heat capacity of water in equilibrium with ice at constant pressure is –1 mol–1 –1 mol–1 The change in internal energy of the gas is

æÆ 3 æÆ

2

2

4 –

2

æÆ æÆ H2

2

+

H+

2

+ 2H2

pg V pV pV g 20. An ideal gas undergoes expansion as shown in the following.

p2 > p3

p2 < p3

pCp, m V CV, m = constant

p2 = p3

p2 = p3

DH DU 21. For which of the following reactions, will DH be equal to DU 1 H2 2 2 2 2 2 N2 4 2 3 1 –1 DH = 30.66 kJ mol 2 2 2 DH = DU DH > DU DH < DU 23. CH4 DH1 2 CH4 DH2 DH1 = DH2 DH1 > DH2 24. The enthalpy change for the reaction DU for the reaction will be –1

DH1 < DH2 3 –1

2

2

DH =

–1

26. An ideal gas is heated from temperature T1 to T2 DH = DU DH = DU + pDV

1 DU 2

DH1 = 2DH2 at 1000 K is 176 kJ mol–1 –1

–1

DU at 300 K was found

25. The reaction of cyanamide, NH2 DH at 300 K for the combustion reaction to be – 743 kJ mol–1 3 N2 NH2 2 2 2 2 would be –1

2

–1

DH = DU + V DP

–1

DH = DU + nR (T2 – T1

Chemical Thermodynamics 6.27

27. The difference between D r H and D rU of the reaction 2C6H6 –1

28. Identify the reaction in which D r H < D rU. æÆ 2 4 2 3 æÆ H2 2 2 29. Identify the reaction in which D r H > D rU. æÆ 6H12 6 2 2 æÆ 3 2 2

2

2

2

2

D cA D f U°(NH3

D cS –1

H2

2

2

2

2

2H2

2

2

2

2

4

2

2

4

2

2

2

2

–1

1 H2 2 2 of enthalpy of combustion at 25 °

H2

2

4

2

æÆ 2H2 æÆ

2

2

2

–1

2

4

–1

æÆ

2

–1

2

2

æÆ

2 3

–1

1 2 1 2

2

–1

D cH

D cU 31. If D f H°(NH3

2

æÆ

2

–1

2

2

2

2

2

2

2 2

35. The standard enthalpy of combustion at 25 °C of H2 and – 3920 kJ mol–1 –1

–1

–1

36. The molar enthalpies of combustion of C2H2

–1 –1,

2 2H 2

–1

–1

–1

37. The enthalpy change of the reaction H+ of H+ 2

– –1

–1

–1 –1.

H2

If the enthalpies of formation –

–1

–1

–1

38. The enthalpy change for the reaction H2 2 æÆ H2 2 –1 , the enthalpy of formation of H2 formation of H2 2 –1

–1.

–1

3

2

æÆ H2

–1

2

2

4

–1

DionizH

, the enthalpy of formation of H2

–1

. If the enthalpy 4

–1

and DionizH°(CH3

Ka(CH3 Ka Ka Ka(CH3 Ka Ka 41. Enthalpy of neutralization of a strong acid and a strong base is –1

–1 –1

–1

–1 –1

If the enthalpy of

–1 –1

. Predict which one of the

Ka(CH3 Ka(CH3 –1

–1

6.28 Complete Chemistry—JEE Main

42. In an endothermic reaction

DH DU 43. In an exothermic reaction

DH DU 2

2

45. The standard enthalpies of formation at 300 K for CCl4 2 DU° at 300 K for the reaction CCl4 –93 kJ mol–1 –1

– 3 –

2 –1

–1

–

+

2 –1

DH1 = –41.84 kJ mol–1 DH2 = – 57.32 kJ mol–1 2– H+ 3

2– 3

H2

2

H2 – 3 –1

–1

–1

–1 –1 and – 395.3 kJ mol–1

–1

–1

1 Df H 2 Df H

Df H Df H

–1

2 2

1 Df H 2

49. The enthalpies of formation of Al2 3 enthalpy change of the reaction Fe2

2 2

3

–1

Df H

Df H

Df H

Dc H°(C, –1 and – 834

3

æÆ Al2

kJ mol–1

2

The –1 –1,

2 2

–1

graph

–1

2

2

æÆ

2

–1

51. The bond dissociation enthalpies of H2

1 Df H 2 Dc H

3

–1

–1

–1

–1

–1

–1

2 –1

–1

–1 –1

–1

53. The bond enthalpies of H H and Cl the bond enthalpy of HCl would be –1

–1

–1

–1

DfH

Cl are 430 and 242 kJ mol–1, –1

–1

–1

–1

,

Chemical Thermodynamics 6.29

54. The enthalpy of combustion of H2 are 433 kJ mol–1 and 492 kJ mol–1

–1

2

–1

and bond enthalpies of H H is

–1

–1

e (N∫ The enthalpy change of the reaction N2 –1

e æÆ 2NH3

–1, 2

–1

–1

Df H

–1

and e

–1.

–1

and D f H

–1

–1

–1

–1

–1,

–1

–1

then

–1

Answer Q.57 to Q.60 based on the following informations. Enthalpies of formation of ethane, ethyene and benzene from –1. the gaseous atoms are – 2839.2, –2275.2 and –5536 kJ mol–1 –1

–1

–1

–1

–1

–1

–1

58. The bond enthalpy of C=C is –1

59. Enthalpy of formation of kekule structure of benzene from gaseous atoms is –1

–1

–1

–1

60. Resonance energy of benzene is –1

–1

–1

–1

61. The unit of entropy is –1

–1

62. For a chemical reaction at equilibrium at temperature T T > DH/DS T < DH/DS 2

–1

T = DH/DS

T = DS/DH

2

–1 –1 –1

mol–1 mol–1

–1 –1

mol–1 mol–1

DS DS

DH DH

DS DS

DH DH

DS DS

DH DH

DS DS

DH DH

DSsurr

DSsys

2 2

2 2

2NH3

DStotal

2

DStotal

2H2

2 2

2

Complete Chemistry—JEE Main

DSsys = DSsurr DSsys = DSsurr

DSsys = – DSsurr DSsys = DSsurr = 0

DSsys = DSsur

DSsys = – DSsur

DSsys

DSsurr

G = H – TS

G = H + TS

G = U – pV – TS

DSsys

DSsurr

G = U + pV + TS

75. The expression connecting DG° and K°eq of a reaction at equilibrium is o o o o DG° = – RT ln K eq DG° = RT log K eq DG° = – RT log K eq DG° = RT ln K eq DU°300K = – 10.0 kJ mol–1 and DS°300K = – 45 J K–1 mol–1. The reaction

DS DS DS 78. Which of the following statements regarding the entropy is 2

2

3

DG° = – nFE

DG° = – RT ln K°p  DG° of the reaction A RT

RT

E° =

RT log K°p nF

DS 83. The term R ln Kp° is equal to DG 4NH3

B is RT

RT log 4 DS1 DS2

DS1 > DS2 DS1 may be greater or smaller than DS2

DG

DU

DG

205.0

–1

Dp = 0

DG/T

2

– 46.2 192.5 – 16.6

DG = DG° + RT ln Qp sure

Predict which of the following is true. DS1 = DS2 DS1 < DS2

–1

2

H2 H2

2

H2

Df H°/kJ mol–1 S°/J K–1 mol–1 Df G°/kJ mol–1

T of the gas

DG°/T

2

90.4 210.6

– 241.8 188.7 – 228.6

–1

–1

86. The standard entropy change of the reaction is –1 –1 mol–1 mol–1

–1

–1

–1

–1

–1

mol–1

–1

mol–1

Chemical Thermodynamics

87. The standard free energy change of the reaction is –1

–1

–1

–1

–1

–1

–1

–1

89. If the standard free energy change for a reaction is 1.546 kJ mol–1 constant for the reaction is 90. Which of the following species is 2

2 –1 –1

–1

mol–1

–1

mol–1

–1

2

mol–1

2 2

2

at constant temperature and pressure. The entropy change of the

system 94. The combustion reaction occurring in an automobile is reaction is accompanied with DH DS DG DH DS DG Æ H2 95. For the transformation H2

96. For a spontaneous endothermic reaction DG DH

2C8H18

2

DH DH

æÆ

2

2

DS DG DS DG DStotal (= DSsys + DSsurr

DG –1

DS > DH/T and 40 J K–1 mol–1, the boiling point of the

liquid will be  by the expression DH° – TDS° = + RT ln K°p DH° = – RT ln K°p

ANSWERS

DH° and DS° at temperature T DH° – TDS° = – RT ln K°p DH° + TDS° = – RT ln K°p

6.32 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS

5. Isolated system cannot exchange heat and matter. 6. Closed system can exchange heat and not matter.

Dq and Dw

q and w but simply as q and w. 1 11. Heat liberated = 4 Hence –1

11500 cal 4184 .

g–1 t

i.e.

t – 25 °C =

11500 °C = 36.65 °C 4184 . ¥ 75

t = 36.65°C + 25°C = 61.65 °C Ê 8.314 J ˆ ÁË 0.082 atm L ˜¯ = – 243.3 J

12. w = – pext DV 13. 0.082 atm L ∫ 8.314 J.

1 atm L =

8.314 J = 101.3 J 0.082

14. DU = q + w

Ê 8.314 J ˆ ÁË 0.082 atm L ˜¯ = – 153.3 J

15. DU = w = – pext DV

Ê 8.314 J ˆ ÁË 0.082 atm L ˜¯ = – 835.9 J

17.

20. 21. 22.

Cp = (dq/dT p temperature. Hence Cp = dq /zero = •. DU = q + w = 100 J – 25 J = 75 J 3 requires energy. pV g = constant where g = Cp, m /CV, m In adiabatic expansion, temperature of the gas decreases where in isothermal expansion, temperature remains p μ T, p2 > p3. DH = DU only when Dvg = 0. DH = DU + (Dng RT suggests that DH > DU. Dng

Chemical Thermodynamics 6.33

CH4 DHa DHb DH2

2

2H2 CH4 Hence

DH1 =DHa + DHb + DH2

DHa and DHb

DH1 > DH2.

24. Dng = 1. Hence DU = DH – (Dng RT = 176 kJ mol–1 –1

10–3 kJ K–1 mol–1

= 167.7 kJ mol–1

1 25. DH = DU + (Dng RT = – 743 kJ mol–1 + ÊÁ ˆ˜ (8.314 ¥ 10–3 kJ K–1 mol–1 Ë 2¯ 26. H = U + pV. Hence DH = DU + D(pV DU + nRDT –1 mol–1 27. D rH – DrU = (Dng RT

–1

–1

Dc U° of the reaction. 31. The reaction is 1 N2 2 Hence

3 H2 2

Æ NH3

Dng = – 1

Df U° = DfH° – (Dng RT

¥ 10–3

–1

= – 47.71 kJ mol–1

33. In enthalpy of combustion, the stochiometric number of the species being combusted is equal to one. The products 2

2

35. C6H10

C6H12

2

6H12

2

17 2

6H10

2 2

2

1 H2 2 2 DH = DH + DH

2

5 2

C 2H 2

2 2

2 2

– DH

2

–1

DH1 = – 1300 kJ mol–1

2

DH2 = – 394 kJ mol–1

2

1 H2 2 2 Hence, for the equation

DH3 = – 286 kJ mol–1

H2

2

C 2H 2

DH = 2DH2 + DH3 – DH1 = [– 2 – 37. DH = Df H (H2 Df H(H+ Df H – 57.3 kJ mol–1 = – 285.84 kJ mol–1 – Df H

Df H°

–

= – 121 kJ mol–1

–1

–1

–

= – 228.54 kJ mol–1

= 226 kJ mol–1

6.34 Complete Chemistry—JEE Main 1 2

38. The reaction to be considered is H2 H2

æÆ H2

2

H2

DH = – 187.4 kJ mol–1

2

DH

2

+

1 2

3

= – 285.7 kJ mol–1 DH = – 298.2 kJ mol–1

2

æÆ

2 1 2

H2

–1

æÆ

2

DH = – 98.7 kJ mol–1

3

æÆ H2

2

DH = – 287.3 kJ mol–1

æÆ H2

2

æÆ H2 DH = – 98.3 kJ mol–1

2

æÆ H2

2

2

The formation of H2

DH = – 130.2 kJ mol–1

4 4

2

2

æÆ H2

4 2

DH

–1

= – 814.4 kJ

mol–1

DionzH, weaker the acid. 42. In an endothermic reaction, heat is added to the system, Hproducts > Hreactants and thus DH Hproducts < Hreactants and DH (or DU DH = Df H

Df H

2

Df H(CCl4

Df H(H2 –1

= –175 kJ mol–1 DU = DH – (Dng RT = – 175 kJ mol–1 10–3 kJ K–1 mol–1 = – 175 kJ mol–1 – 7.48 kJ mol–1 = – 182.5 kJ mol–1 DH = DH1 – DH2 2

2

2

2

–1

DHa = – 393.5 kJ mol–1 DHb = – 395.3 kJ mol–1

Hence DH = DHb – DHa 2

–1

D cH

2

1 2 2 Hence, for the equation

D cH

2

1 2

2

D cH

DH = DcH 49. D r H = Df H(Al2

D f H(Fe2

3

2

2

2

= 1.80 kJ mol–1

–1

3

= –836 kJ mol–1

æÆ

2

DH = – 393.5 kJ mol–1

2

æÆ

DH = – 110.5 kJ mol–1

æÆ H2

DH = – 241.8 kJ mol–1

4

Chemical Thermodynamics 2

DH 51. H2

æÆ

2

–1

2

= 41.2 kJ mol–1

2

1 [e e 2 = – 92 kJ mol–1 DH for the equation

Df H

DH = Df H(H,

1 [(435 + 243 – 2 2

e

D fH

DHf

–1

1 H2 2

DH1 =

1 2

430 kJ mol–1

1 Cl2 2

DH2 =

1 2

242 kJ mol–1

1 H2 2

1 Cl2 2

HCl DH3 = – 91 kJ mol–1

DH = DH1 + DH2 – DH3 54. For the equation 1 H2 2 2

–1

–1

= 427 kJ mol–1

DH = – 249 kJ mol–1

H2 1 e 2

e

e

–1

1 È ˘ –1 ÍÎ433 + 2 (492) ˙˚ kJ mol – 2e 1 2

e

–1

1 È ˘ –1 –1 ÍÎ+ 249 + 433 + 2 (492) ˙˚ kJ mol = 464 kJ mol

55.

Hence, DH = e (N ∫∫

e

e

¥ 436 – 6 ¥

= – 93 kJ mol–1 1 F2 2

Æ

Æ 1 F2 2

D f H° = 92.0 kJ mol–1 D f H° = 536 kJ mol–1

Æ

DH

–1

DH for the reaction Æ This reaction can be obtained by the following manipulations

–1

= 424 kJ mol–1

6.36 Complete Chemistry—JEE Main

57. 58. 59. 63.

–1 Hence e = 521.5 kJ mol–1 e C–C = Enthalpy to break C2H6 into gaseous atoms –6 ¥ e C–H –1 = (2839.2 – 6 ¥ = 373.98 kJ mol–1 e C=C = Enthalpy to break C2H4 into gaseous atoms –4 ¥ e C–H –1 = (2275.2 – 4 ¥ = 631.72 kJ mol–1 e( e C–C – 3e C=C – 6e C–H –1 = –53.68 kJ mol–1 Dng

64. DS =

–1

40.63 ¥ 103 J mol-1 DH = = 108.9 J K–1 mol–1 373 K T

65. For a spontaneous process, DG DG

DH

DG = DH – TDS DS

66. DG DH DS 69. Increase in entropy will occur if Dng reactants. 70. Decrease in disorderliness would result into the decrease in entropy. heat absorbed by the system is equal to the heat lost by the surroundings, DSsys and DSsurr but carry opposite sign. DSsys DSsurr DStotal = DSsys + DSsurr 74. Enthalpy of an ideal gas depends only on the temperature. 10–3 kJ K–1 mol–1 76. DH° = DU° + (Dng RT = – 10.0 kJ mol–1 = – 12.49 kJ mol–1 DG° = DH° – TDS° = – 12.49 kJ mol–1

10–3

–1

mol–1

–1

DS in entropy during decrease in temperature.

80. K°eq =

1 . Hence, 4

1 DG° = – RT ln ÊÁ ˆ˜ = RT ln 4 Ë 4¯

82. Adiabatically implies q DfH°(H2 84. DH° = 4Df H = – 904.4 kJ mol–1

w = 0. Hence DU = q + w = 0 DfH°(NH3

85. DU° = DH° – (Dng RT = – 904.4 kJ mol–1 86. DS° = 4S S°(H2 S°(NH3 = [4

–1

10–3 kJ mol–1 S

2 –1

mol–1

= 179.6 J K–1 mol–1 87. DG° = DH° – TDS° = [– 904.4 – 298 (179.6 10–3 88. DG° = 4 Df G DfG°(H2 DfG°(NH3 –1 = 4 D G f D fG

1 (347.3 kJ mol–1 4

–1

–1

= – 957.9 kJ mol–1

–1

Chemical Thermodynamics 6.37

89. DG° = – RT ln K° 1.546 103 J mol–1 = – (8.314 J K–1 mol–1

K°

1546 . ¥ 10 i.e. K 2.303 ¥ 8.314 ¥ 773 3

i.e.

log K° =–

DS = DH/T = (37300 J mol–1 H2 The reaction H2 Æ H2 For a spontaneous reaction, DG < 0. Hence DH 25 ¥ 103 J mol-1 97. Tb = = 625 K = DS 40 J K -1 mol-1

–1

91. 92. 95. 96.

98. For a reaction at equilibrium

mol–1 DStotal

DH – TDS < 0 or DH < TD S.

DG° = – RT ln K°p where DG° = DH° – TDS°.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is DG = RT ln K°c DG = RT ln K°c DG° = RT ln K°c DG° = RT ln K°c 2. The enthalpy change for a reaction does not depend upon

T and p in which on p – V work in being done, the change in G SV,E = 0, dGT,p

S SV,E < 0, dGT, p

SV,E > 0, dGT,p

SV,E = 0, dGT,p = 0

kJ mol–1 H2C==CH2 + H2 at 298 will be

æÆH3

3

–1

–1

–1

–1

5. The internal energy change when a system goes from state A to B is 40 kJ mol–1. If the system goes from A to B –1

6. An ideal gas expands from 1 ¥ 10 1 ¥ 105 N m–2. The work done is

–3

–1

m to 1.0 ¥ 10 3

–2

–1

3

m at 300 K against a constant pressure of

7. The enthalpies of combustion of carbon and carbon monoxide are –393.5 and –283 kJ mol–1 enthalpy of formation of carbon monoxide is –1

–1

8. For a spontaneous reaction the DG, equilibrium constant (K

–1

E°cell

–1

6.38 Complete Chemistry—JEE Main

9. Consider the reaction N2 + 3H2 æÆ 2NH3 carried out at constant temperature and pressure. If DH and DU are the enthalpy and internal energy changes for the reaction, which of the following expression is true. DH < DU DH > DU DH DH = DU D fH 10. If the bond dissociation energies of XY, X2 and Y2 for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2, will be –1

–1

–1

11. The standard enthalpy of formation (DfH

–1 –1

4

. The additional

2 2 and enthalpy of sublimation of carbon

Ti, is the initial temperature and Tf Tf =Ti Tf > (Tf Tf >Ti = (Tf Tf 13. (DH – DU

Tf =Ti R = 8.314 J K–1 mol–1

–1

–1

–1

–1

14. The enthalpy changes for the following processes are listed below: –1 Æ Cl2 Æ I2

Æ

–1

–1

Æ I2

I2

–1 2

–1

2

–1

–1

–1

DU –1 and

R = 8.3 J mol

–1 K–1 –1

–1

–1

–1

16. Identify the correct statement regarding a spontaneous process:

3

kJ mol–1 and 160.2 J K–1 mol–1

1 Cl 2 2

DdissH°

æÆ

DcgH°

æææÆ

æææÆ Cl– 2

–1

DdissH°(Cl2

DhydH°

æææÆ Cl– –

–1 –1;

DcgH

DH° and DS° are + 179.1 DH° and DS° do not change with

2

–1 –1;

and DhydH°(Cl–

–1 –1;

Chemical Thermodynamics 6.39 2, 1 2

X2 +

3 2

Y2 and XY3 are 60, 40 and 50 J–1 K–1 mol–1

Y2 æÆ XY3, DH = –30 kJ mol–1. This reaction will be at equilibrium at temperature equal to

æÆ H+

H2

+

H2

1 2

æÆ H2

2

DH = 57.32 kJ mol–1

–

DH = –286.20 kJ mol–1

;

DfG°(H+ –

ions at 25 °C is

–1

–1

–1

–1

21. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is 3 2

CH3

2

æÆ

H2

2

3

–394.4 kJ mol–1 the fuel cell will be

2

2

–1

22. The standard enthalpy of formation of NH3 is – 46.0 kJ mol–1. If the enthalpy of formation of H2 from its atom is –436 kJ mol–1 and that of N2 is –712 kJ mol–1 3 is –1

–1

2

2 Al2 3

3

æÆ

4 3

3

–1

–1

at 500 °C is as follows. DrG = –966 kJ mol–1

2;

T, DH and DS temperature at equilibrium, the reaction would be spontaneous when Te > T T > Te T = Te dm3

3

Te is the Te is 5 times T

at 27 °C

–1

–1

–1

y change (DH

2H 5

1366.5 kJ mol–1 –1

2

–1

–1

æÆ 2

DH ∞ - T DS ∞ RT

2

–1

27. The incorrect expression among the following is DGsystem = -T DS total K∞ =

2

–1

w

= –nRT ln

Vf Vi

K° = exp(–DG°/RT q and w for the process will be

–1

mol–1

(R = 83.14 J K q = +208 J, w

q = –208 J, w

q = –208 J, w

q = +208 J, w = +208 J

Complete Chemistry—JEE Main

29. For complete combustion of ethanol, C2H5 æÆ the heat produced as measured 2 2 2 in bomb calorimeter, is 1364.47 kJ mol–1 at 25°C. Assuming ideality the enthlapy of combustion, DcH, for the reaction will be (R = 8.314 J K–1 mol–1 –1

–1

30. The entropy (S CH4

mol–1 –1 mol–1

–1

–1

2

–1

mol–1 –1 mol–1

2

H2

The entropy change (DS –1 mol–1

CH4 –1 mol–1

æÆ

2

2

2 –1

31. The standard enthalpy of formation (DfH°298K 2

–1

mol–1

4 is –74.9 kJ mol

–1

mol–1

–1

molecule

2

32. Consider the reaction If N2 5 for N2 5 is 54 kJ mol–1

2

2 2

æÆ 2N2

DrH = –111 kJ mol–1 D rH

5

5

–1

–1

–1

–1

¥ 1023 mol–1 ¥ nm 4 ¥ 10 nm

¥ 10 ¥ 104 34. The following reaction is performed at 298 K. 3

2

DH of sublimation

103

2

Kºp = 1.6 ¥ 1012

2

R –1

¥ 1012 +R

–1

¥

1012

12

–1

–

ln (1.6 10 ) R (298 K )

¥ 86600 J mol–1 – R

¥ 1012 Kp

1 Æ 2 2 Identify the correct statement: T

1 2

Æ

T T

2

Æ

M Æ MO

is spontaneous.

Æ H S –1 –1 – 29.8 kJ mol and – 0.100 kJ K mol at 298 K. The equilibrium constant for the reaction at 298 K is: –1

–10

C Æ CO 20 1200 T/K 10

Chemical Thermodynamics

the correct statement about the reaction among the following. H S H S 38. If 100 mol of H2 2 1 bar pressure is: (R = 8.3 J K–1 mol–1 2H2 2H2 2 2

H H

S S 2

39. The heats of combustion of carbon and carbon monoxide are – 393.5 and – 283.5 kJ mol–1 of formation (in kJ mol–1

ANSWERS

HINTS AND SOLUTIONS 1. The correct expression is DG° = –RT ln K°c 2. The enthalpy change of a reaction does not depend upon the nature of intermediate path. or dsysS – dsysH/T > 0 sysS + dsurrS > 0 H = U + pV TdS – d(U +pV TdS – dU – pdV –Vdp > 0 p (i.e. dp TdS – dU – pdV > 0 At constant U and V, this reduces to dSU,V > 0. G = H – TS TdS – d(G +TS TdS – dG – TdS – SdT > 0 > 0 or dG < 0 At constant T T, p T, p 4. H2C==CH2

DH = 4e

2

+ eC==C + e

H3

or

– dG – SdT > 0.

3

–e

¥

– 6e 5

6. The work done is w = –pDV = – (10 N æÆ 2 æÆ 2 2

8. For a spontaneous reaction, DG

m–2 2 2

æÆ

–1

= 125 kJ mol–1

U=0 –2 m3 – 10–3 m3 ¥ 102 N m = –900 J DH = –393.5 kJ mol–1 DH = –283 kJ mol–1 DH

–1

= –110.5 kJ mol–1 E°cell producing the

6.42 Complete Chemistry—JEE Main

DH = DU + (Dng RT 1 2

DfH° =

1 2

e X2 +

1 2

eX2 = – 4(DfH

Hence

1 2

X2

eY2 – eXY =

1 2

Dng = –2. Hence, æÆ

Y2

e X 2+

1 2

( 12 e X 2 ) – eX = – 2

– 4(–200 kJ mol–1 æÆ CH4 2

Hence, in addition to DfH°(CH4 of carbon.

DH < DU.

1 eX 4 2

–1

DfH° = –74.84 kJ mol–1 æÆ 4 2 and enthalpy of sublimation

Tf 1 2

–1

1I 2 2

1I 2 2

1 2

Æææ

1 2

Cl2 Æææææææ

1 2



mol–1 1 2

Æææ

14.

2

Cl Hence,

–1

DfH° ææÆ ææææææÆ

For this reaction, Dng = 1/2. Hence DrH° – DrU° = (Dng RT

> (Tf

–211.3 kJ

ææÆ

◊◊◊ Cl

Df H° = ÎÈ 12 (62.76) + 12 (151.0) + 12 (242.3) - 211.3˘˚ kJ mol-1 = 16.73 kJ moll-1 2

Æ H2

For this process, Dng = +1. Hence DrU = DrH – (Dng RT = (41 ¥ 103 mol–1

–1

mol–1

= 41 ¥ 103 J mol–1 – 3095.9 J mol–1 = 37904 J mol–1 = 37.904 kJ mol–1 16. For a spontaneous process DStotal (= DSsystem + DSsurroundings Hence, DSsystem

DSsurroundings = 0.

DH ∞ 179.1 ¥ 103 mol-1 = = 1118 K Teq = DS ∞ 160.2 J K -1mol-1 At T > Teq, DG 18. According to Hess’s law DH =

DS = S(XY3

1 2

1 2

-1 DdissH° + DegH° + DhydH° = ÎÈ 12 (240) - 349 - 381˘˚ kJ mol = –610 kJ mol–1

S(X2

3 2

S(Y2

(50 - 12 ¥ 60 - 32 ¥ 40) J K -1 mol -1 = – 40 J K–1 mol–1

DG = 0 = DH – T DS = –30 ¥ 103 J mol–1 –T (–40 J K–1 mol–1 30 ¥ 103 T= K = 750 K 40

Chemical Thermodynamics 6.43

20. For the reaction, H2 æÆ H+ DrH = Df H (H+ 57.32 kJ

mol–1

= 0 + Df H

–

Df H

Df H (H2

–

–

– Df H 21. For the reaction CH3 Dr H° = Df H

–1 –1

3 2

2

2

æÆ Df H°(H2

2

2

Df H°(CH3

= (–394.4 – 2 ¥

–1

= –702.6 kJ mol–1 h=

D r H ∞ -702.6 kJ mol -1 = = 0.9678 Dc H ∞ -726 kJ mol -1

H2 æÆ NH3 æÆ H2 æÆ N2 We require to calculate æÆ 3 1 2

N2

DH° = –46.0 kJ mol–1 DH° = –436 kJ mol–1 DH° = –712 kJ mol–1

3 2

DH°1 3 2

This is obtained by the manipulation Hence,

1 2

-1 -1 DH°1 = ( + 46.0 + 32 ¥ 436 + 12 ¥ 712) kJ mol = 1056 kJ mol

e

=

1 3

(1056 kJ mol–1

–1

23. The basic reactions area: 2 (2Al3+ ) 3

+ 4e- æÆ 34 Al

and

2 (3O 2- ) 3

æÆ O2 + 4e2 Al2 O3 æÆ 43 3

Al + O 2

Thus, using the expression DG= –nFE, we get E=–

DG 966 ¥ 103 J mol-1 == -2.50 V nF (4)(96500 C mol-1 )

Thus, DG = DH – TDG Reactions at equilibrium: DG = DH – TeDS = 0 For a reaction to be spontaneous: DG = DH – TDS < 0 DG T > Te as both DH and DS V2 100 ˆ -1 –1 = (2 mol)(8.314 J K -1 mol -1 ) ÊÁ 2.303 ¥ log ˜ = (2) (8.314) (2.303) J K = 38.3 J K Ë 10 ¯ V1 DrH = DrU + (Dng RT DrU = DrH – (Dng RT Dng = –1. Hence ¥ 10–3 kJ K–1 mol–1 DrU = –1366.5 kJ mol–1 –1  1364.0 kJ mol–1 27. The correct thermodynamic expressions are

25. DS = nR ln

DGsys = –TDStotal; is incorrect as

w

= –nRT ln

ln K° = -

DH ∞ - T DS ∞ RT

Vf ; Vi

DG° = DH°–TDS° = –RT ln K°

6.44 Complete Chemistry—JEE Main

28. For an isothermal expansion of an ideal gas, DU absorbed, q = 208 J and w = – 208 J. Dng = –1. Hence DH° = DU° + (Dng RT = (–1364. 47 kJ mol–1

q = –w

¥ 10–3 kJ K–1mol–1

= –1366.95 kJ mol–1 4

æÆ

2

DS° = S

2

2

S°(H2

2

S°(CH4

= (213.6 + 2 ¥ 69.9 –186.2 – 2 ¥

S

2

–1

mol–1

= –242.8 J K–1 mol–1 æÆ CH4

2

DfH°298K= –74.9 kJ mol–1

CH4 æÆ for which we must require the enthalpy change for the reactions æÆ DsubH H2 æÆ DatH°(H2 2

2N2

æÆ 2N2 æÆ 2N2 5 2

5

DrH = –111 kJ mol–1 DrH = –2 ¥ 54 kJ mol–1

5

Add 2

æÆ 2N2

2

Æ Æ

CH4 C 2H 6 From DH1, we get From DH2, we get

E=

e C -- H NA

DrH = – (111 + 2 ¥

5

–1

= –219 kJ mol–1

DH1 = 4e DH2 = e + 6e –1 –1 eC–H = 90 kJ mol –1 = 80 kJ mol–1 eC–H = DH2 – 6eC–H = (620 – 6 ¥

=

(80 kJ mol-1 ) (6.02 ¥ 1023 mol-1 )

= 1.33 ¥ 10–22 kJ = 1.33 ¥ 10–14 J

hc (6.626 ¥ 10-34 J s) (3 ¥ 108 m s -1 ) = 1.495 ¥ 10–6 m = E (1.33 ¥ 10-19 J )

l=

= 1.495 ¥ 103 nm Dr Also, DrGº = 2DfG Hence, D fG

D fG

2

[DrGº + 2 Df DG° = – RT ln Kp 2

1 2

1 2

2

=

Æ

M (s) + 12 O 2 (g) Æ MO(s); C(s) + MO(s) Æ CO(g) + M (s)

1 2

RT ln Kºp = –R

12

12

–1

[–R DG°1 DG∞2 = postive DG∞3 = DG∞1 - DG∞2

Chemical Thermodynamics

DG°2 < DG°1 at T > 1200 K, DG°3 and hence the reaction will be spontaneous. H°– T S° = (– 29.8 kJ mol–1 G° = – RT ln K°p,

G

H

S

T < 1200 K, DG°2 > DG°1, and thus DG°3 –1

mol–1

we get K°p = 1 G H–T S

G

G

H > T S. At high temperature,

H Kp Æ reactants) Qp < Kp

Æ

Qp = Kp

Illustration The reaction sets in is

2

N2

2

2

Qc =

3

[ NH3 ]2 [ N 2 ][H 2 ]3

=

3

Kc

¥

2

(mol/L)–2

(8.0 mol/ 20 L) 2 (1.5 mol/ 20 L)(2.0 mol/ 20 L)3

¥

3

(mol/L)–2

Since Qc > Kc

Le Chatelier’s Principle

If a system at equilibrium is subjected to a change, the system adjusts to a new equilibrium stage in such a way so as to oppose or reduce the said change.

Chemical and Ionic Equilibria 7.7

is added 2

2

D rH

3

–1

(i) Effect of Changing Pressure Note: 2

(ii) Effect of Changing Temperature

Since DrH

2

3

æÆ æÆ

Note:

∞ d ln K eq dT

=

Dr H ∞

(13)

RT 2

∞ = - D r H ∞ + constant ln K eq RT

(14) K°eq

T

Fig. 1

DrH°/R

7.8 Complete Chemistry—JEE Main

D rH

T

D rH

T

(iii) Effect of Adding One of the Components

2

3

(iv) Effect of Adding Inert Gas Inert Gas Added at Constant Volume

Ê nB ˆ Ê ntotal RT ˆ nB RT pB = xB ptotal = Á Á ˜= Ë ntotal ˜¯ Ë V ¯ V V xB is decreased while ptotal nB Inert Gas Added at Constant Pressure

Ê nB ˆ p = (decreases) (constant) = decrease pB = xB ptotal = Á Ë ntotal ˜¯ total Dn

Qp

Kp Dn

Kp

Qp

Qp becomes

Kp

Qp becomes

Kp Dn

Note:

2

2

MULTIPLE CHOICE QUESTIONS ON UNIT 1

2

2

2

2

æÆ 2SO3

Kp

¥

bar–1

Kc

will be ¥ ¥

¥ ¥

13

dm3 mol–1 dm3 mol–1 2

2

2

11

dm3 mol–1 dm3 mol–1 2

Chemical and Ionic Equilibria 7.9

K p= K c (a) 2SO3 (c) N2

2SO2

2

2

2

2

3

2

(a) is endothermic

(b) is exothermic

T T

Kºp T T T T 2

–1

Ls

2

–1

–2

Kc

2

L2 s–1

–1

L s–1

L–1

–2

–1

Kc

2

–1

Kc

2

2

CaO(s) + CO2 Kp 3 dissociated is

3(s)

3

2

2

V and T

Kp ptotal/Kp (a) 2

(b) 4

(i) BrF3

(c) 6

2

¥ ¥ ¥

Kp Kp Kp

BrF5 ClF3 BrF3

2 2

The Kp

3

¥

¥

38

(d) 8

35

atm–1 12 atm–1 atm–1 5

38

¥

¥ 16 2O(l)

38

Kc ¥ 2N2

–3

2

L2 s–1

mol L–1

¥

–3

mol L–1

¥

–3

mol L–1

æÆ 2 2

¥

–3

2

mol L–1

7.10 Complete Chemistry—JEE Main 3(s)

CaCO3(s)

in the reaction

CaO(s) + CO2 2

(ii) CO2 2O(l) CO (aq) 2 3 + O(l) 2 Kc (a) Kc1Kc2Kc3/Kc4

3(aq)

(b)

+ CO2 2 K /K2 Kc1Kc2 c3 c4

[F2] = 4 ¥ ¥

Br2

–4

Kp

2

¥

–4

2

2

(ii) N2

(a)

¥

2

bar4 ¥

2 2

Ba2+(aq) + 2F– +(aq) + F– + (aq) 2

Kp =

p

pH 2 pI2 ( pHI )

2

=

–8

Ks Ka Ba2+

(c) K p 2 K 1p/32 / K 3p1/ 2 ¥ ¥

–6

3

–4

HINTS AND SOLUTIONS 2

2]

= 4 ¥ ¥

–8

–5

bar and

bar

2

ANSWERS

2

2

Kp1 Kp2 Kp3

(b) K p1 K 1p/22 / K 3p 3/ 2

Kc –3

–14

3

K p 3 / K 3p1/ 2

2 K K /K2 (d) Kc1 c2 c3 c4

Qc

¥ will be 5

2

2(s)

+ CO32– (c) Kc1Kc2/Kc3Kc4

–4

Kp K 1p/22

+ (aq) 4

2

2 3

2O(l)

Kc

2

Qc > Kc Qc > Kc Qc < Kc Qc < Kc 5

Kc1 Kc2 Kc3 Kc4

2CO3 +(aq) + CO2– 3 –

2

(a) (b) (c) (d)

–

+ 4

2O(l)

3

p

(0.08 atm)(0.08 atm) = 4.0 (0.04 atm) 2

2

=p2

(d) K p 2 K 1p/32 / K 1p/12

Chemical and Ionic Equilibria 7.11 2

Dn

2SO3

2 Dn (RT)

Kp = Kc Kc = Kp (RT)–Dn ¥ 11 3 ¥ dm mol–1

Since

¥

bar–1

–2

bar dm3 mol–1

ˆ Ê 40 g ÁË 100 g ¥ 18 g˜¯ = 7.2 g;

–1

7.2 g = 0.4 mol 18 g mol-1

H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g)

(1- 0.4) mol (1- 0.4) mol

Since Dn

0.4 mol

0.4 mol

Kc = Kn Kc =

(0.4 mol)(0.4 mol) = 0.44 (0.6 mol)(0.6 mol) Dn Kp = Kc

ln K°eq = -

K°eq is

Dr H ∞ + constant T

or

D rH

K°eq = Ae - D r H ∞ / T

K°eq

D rH

K°eq =

with increase in 1/T

positive + constant T

K°eq will increase linearly

2ICl(g)  I2 (s) + Cl2 (g)

0.75 M -x

x

x

[I2 ][Cl2 ] x2 = = 16 [ICl2 ] (0.75 - x)2 x = 4 or 5x 0.75 M - x 0.6 M ¥ 100 = 80% 0.75 M

K c=

Keq = k /kb

kb =

I 2 (g)  2I(g) 1.0 mol

Kc =

0.5 mol

x

kf 1.54 mol-1 L s -1 = = 3.08 mol-2 L2 s -1 K eq 0.50 mol L-1

[I]2 (0.5 mol/ V )2 = = 0.01 mol L-1 [I 2 ] (1.0 mol/ V )

V = 25 L pCO2

Kp 2

is

n=

pV (1.16 atm)(10 L) 1 = = mol RT (0.082 L atm K -1 mol-1 )(1160 K ) 8.2

Ê 1 molˆ m = nM = ÁË ˜¯ 8.2

3

3

dissociated =

12.2 g 20 g

–1

7.12 Complete Chemistry—JEE Main 2

p –p

p +p

2p

p fi p 2 ( pI ) (2 p) (2 ¥ 0.05 atm) 2 Kp = ( p ) = p - p = (0.07 - 0.05) atm = 0.5 atm I2 0

p +p

2

p p(1 – 1/3)

Ê Ë

ptotal = p 1 +

p/3

pA pB ( p / 3)( p / 3) p = = 6 pA B p(1 - 1 / 3)

Kp =

Kp =

p/3

K p (iii ) K p (i ) K 2p ( ii )

=

1ˆ 3¯

=

The ratio

4 3

p

ptotal/Kp =

(7.8 ¥ 1027 atm -1 )(8.6 ¥ 1035 atm -1 ) (7.8 ¥ 1012 atm -1 ) 2

(4 / 3) p =8 ( p / 6)

¥

38

Ê 1 atm ˆ = 0.041 atm K p = pwater = 31.2 Torr = (31.2 Torr) Á Ë 760 Torr ˜¯ Dn

2O(l)

2

Kc =Kp (RT)–1 =

(0.041 atm) (0.082 L atm K -1 mol-1 )(300 K )

¥

–3

mol L–1

2

2 3

Kc = K2c1Kc2Kc3/K2c4 Qc =

[ClF]2 (3.65 M) 2 = = 666.1 [Cl2 ][F2 ] (0.2 M)(0.1 M) 2 5] =

[BrF5]

Since Qc > Kc

[Br2 ][F2 ]5 (4 ¥ 10-5 bar)(4.0 ¥ 10-4 bar)5 4.096 ¥ 10-25 bar 6 = 4.0 ¥ 10-8 bar 2 = = Kp 1.024 ¥ 10-14 bar 4 (1.024 ¥ 10-14 bar 4 ) ¥

–4

bar 1 2

3 2 Kp =

BaF2(s)

K p 2 K 1p/32 K 3p1/ 2 Ba2+(aq) + 2F– +

BaF2

(aq) + 2F

+ (aq)

–

Ba2+

Ks K2a

Chemical and Ionic Equilibria 7.13

K=

Ks K a2

=

1.0 ¥ 10-6 M 3 = 2.37 M (6.5 ¥ 10-4 M) 2

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE UNIT

Equilibrium Constants Kp and Kc

3(s)

CaO(s) + CO2

2

3 3 3(s)

CaO(s) + CO2

2 to

Kc not Keq direction Keq direction Keq Keq direction N 2O 4

a

2NO2 Kp is

(a) a =

Kp p

(b) a =

4 + Kp p

Ê Kp p ˆ (c) a = Á ˜ Ë 4 + K p p¯

Kp 4 + Kp

Calculations of Equilibrium Constant 2

2

2

2

2

will be 2.27

2

1/ 2

2O 4

Ê Kp ˆ (d) a = Á ˜ Ë 4 + Kp ¯

1/ 2

7.14 Complete Chemistry—JEE Main 1 2

2

1 2

2

will be

2

2

7.4 Kc

SO2

3

Kc

SO3

2

2

2

2

2

–1

–1

–1

are

Kc –1

–1

2

2

Kp –1

–1

–1

2 2

4(s)

2

¥ 3

CaCO3(s)

mol–1 L s–1 ¥

2

4

2

C2

4

6

DG

the Kp

2

–1

(c)

atm3

the Kp

CaO(s) + CO2

C2 6 pº = 1 atm)

–2

¥ 2)

–2 –5

the Kp

2

2

CaCO3

2

–2

Kp

2

–1

–1

4

–1

–1

mol–1 L s–1

(b) 4 ¥

¥

4

1

2

¥

–2

Chemical and Ionic Equilibria 7.15

(c) 2 2

12 2 2

2

2

2

and CO2

2

¥

¥

–5

–4

mol–1 dm3 s–1

mol–1 dm3 s–1 K°p

(a) 8

PCl3 K°p (c) 32

5

(b) 16 2

–1

DH

2

–1

2

(d) 64

–1

–1

–1

2O 4 N2O4 decomposed to NO2

Relation Between Kp and Kc 2

(a) Kp > Kc 2

(a) Kp > Kc

(b) Kp < Kc C2 2

4

(a) Kp (RT)D Kc Kp

(c) Kp = Kc

(d) cannot be predicted

(c) Kp = Kc

(d) Kp =

(c) Kp = Kc R

(d) Kc = R Kp

6

(b) Kp < Kc 4NO2 2O 5 2 (b) Kp < Kc Kp and Kc = Kc (b) Kp = Kc (RT)D N 2O 4 2 NO2 ¥ –6 2SO2 2SO3 2

(a) Kp > Kc

(c) Kp = Kc

Kp and Kc (d) Kp = K2c Kp and Kc

2

–3

Kc

Kp –1 –1

Kc

be ¥

2

¥

mol–1 cm3

2

¥

–1

¥

–3

Kp < Kc 2

(c) 2SO3

2

2

2SO2

2

2

3

2

CO2

2

Le-Chatelier Principle 2

2

DH

–1

not

2

–1

7.16 Complete Chemistry—JEE Main 2

2

DH

3

–1

not

3 3

2O(s)

2

2O(s)

N2

2

2 –1

DH

3

2 3

increase at

1 2

SO2

O2

SO3

3

3 3

3 1 2

SO2

O2

SO3

3 3 3

3

2

CaO(s) + CO2

CaCO3

3

N2 2 is added to the system 3 is added to the system 2

2

3

2

is expected to

Chemical and Ionic Equilibria 7.17

N2

2

3

N2

2

3

3

2 2

(c) N2

3

N2

2

3

2

(c) O2

in the system at

3 2

DH

2

increased when

2

2

3

2SO2

DH

2SO3

3

2

3 is

(a) increased

(b) decreased 2

Miscellaneous Problems N 2O 4 D H D G

2 NO2

–1 –1 2O 4

–1

–1

–1

–1

–1

mol–1

–1

–1

mol–1

–1

–1

–1

–1

mol–1

–1

mol–1

K°p

¥

K°c ¥

–2

pº = 1 atm) –3

p is

K°p (a) K°p =

2a Ê p ˆ 1 - a ÁË p∞ ˜¯

4a Ê p ˆ Á ˜ 1 + a 2 Ë p∞ ¯ 2

(b) K°p =

4a Ê p ˆ Á ˜ 1 - a 2 Ë p∞ ¯ 2

(c) K°p =

(d) K°p =

2a Ê p ˆ 1 + a ÁË p∞ ˜¯

7.18 Complete Chemistry—JEE Main

Ê 2a ˆ (a) K°p = Á Ë 1 - a ˜¯

K°p Ê pˆ ÁË p∞ ˜¯

Ê 4a ˆ Ê p ˆ (b) K°p = Á Á ˜ Ë 1 - a 2 ˜¯ Ë p∞ ¯ 2

Ê 2a ˆ (c) K°p = Á Ë 1 + a ˜¯

p is Ê pˆ ÁË p∞ ˜¯

Ê 4a 2 ˆ Ê p ˆ (d) K°p = Á Á ˜ Ë 1 - a ˜¯ Ë p∞ ¯

2O 4

(a) remains constant

2O 4

(b) increases

(c) decreases

(d) cannot be predicted K°p

2O 4

(a) remains constant

(b) increases

(c) decreases

(d) cannot be predicted 2O 4 (d) cannot be predicted

2O 4

(a) remains constant

(b) increases 2O 4

(c) decreases

(b) increases

(c) decreases

2

2O 4

(a) remains constant

2O 4 2O 4 (a) remains constant

2O4 at 1 atm (d) cannot be predicted

2 2O 4

(b) increases

(c) decreases

K °p

2

(d) cannot be predicted 1 N 2O 4 2

2O 4

will be

2O 4 –1 –1

–1

–1

–1 –1

–1

–1 2O 4

N 2O 4

2

ANSWERS

2

–1

–1

Chemical and Ionic Equilibria 7.19

HINTS AND SOLUTIONS

Keq = p(CO2 2 NO2 N 2O 4 1–a

ntotal = 1 + a

2a

1- a p p N 2O 4= 1+ a

2a p pNO2 = 1+ a

and

Ê Kp p ˆ a= Á ˜ Ë 4 + K p p¯

Kp =

2 pNO 2

pN 2 O 4

2a ˆ 2 Ê 1 + a 1 ˆ 4a2 p˜ Á p = = ÊÁ ˜ Ë1+ a ¯ Ë1- a p¯ 1- a2

1/ 2

Knew = 1/Kold Knew = (Kold)2

2

SO3

2

2

1 mol – x

1 mol – x

1 mol + x

1 mol + x

(1 mol + x) (1 mol - x)2

Ê 1 mol + x ˆ ÁË 1 mol - x ˜¯

2

Kc = or

(1 mol + x) = 4 (1 mol – x) x

2

1 mol – x

or

2

x

2

2 mol – x

2x

Since D

Kc = Kp = Kn Kn = x

1 mol – x

(2 x)2 (1 mol - x) (2 mol - x) x x

(2 x)2 (1 mol - x) (2 mol - x) x

2x

2x 2L Kc =

that is

–1

x

2 [C]2 (0.70 mol 2 L ) = [A] [B]2 (0.65 mol 2 L ) (0.8 mol 2 L )2

–1

L

2

p

p

ptotal p

2p

p ptotal

p) + 2p

p

7.20 Complete Chemistry—JEE Main

Kp =

(2 ¥ 0.139 atm) 2 ( pNOCI )2 = ( pNO )2 ( pCl2 ) (0.373 - 2 ¥ 0.139)2 (0.310 - 0.139) atm3

2

–1

2

x

x

Since D

Kp = Kn = x

x 0.2 mol - x 0.0127 ¥ 0.082 ¥ 363 ˆ p = ÊÁ ˜¯ atm Ë 1

nH 2

x

4(s)

2

nH 2S

3

x

2

p

2p

Kp = p2

3

–5

pCO2

Ê 2.9 ¥ 10-5 atm ˆ p = Á ˜¯ 4 Ë

atm3 = (2p)2 p

1/ 3

ptotal = 2p + p = 3p = 3

Kp = pCO 2 DG° = – RT ln Kºp a –x

–1

mol–1

–1

a – 2x

kf 1.62 ¥ 10-2 mol-1 L s -1 = kb 4.0 ¥ 104 mol-1 L s -1 Keq = p pB p pB Keq =

2 1 [C] 1

1

2 1

=

p pB

[B]2 [C]22 = 2[B]1

2

=

[C]1

2

= 8[B]1

2

=

[C]1

2

= 2 [C1

=

[B]1

2

[C]31 = [B]2 [C]32 [C]31 = [B]22 [C]32

2

2

2

2 1/3

8

=

[C]1 2

8

2

x

x

x

x

2

x (0.4 mol - x) 2 Keq = k /kb ¥ Kp Ea(b)

x -4 mol–1 dm3 s–1

-5

mol–1 dm3 s–1

-1 2

Since Dn

2 2

Kp = Kn 2O 4

2NO2 ¥

p Kp = Kc (RT)D

¥

D

ntotal Kp = Kc

2

Chemical and Ionic Equilibria 7.21

D

Kp = Kc/RT Kp < Kc . Kp = Kc (RT)3

D Kp = Kc

RT Kp > Kc

(RT)D

D –3

mol

–1

–1

mol–1

=–1 –1

Kc = Kp RT

–1

RT –1

D

mol–1

=+1

L–1

Kp = Kc (RT)D

–1

2

mol–1 L

DH D DV

2O(s)

2

D r H° = 2D H°(NO2) – D H°(N2O4) = (2 ¥ D r G° = 2D G°(NO2) – D G°(N2O4) = (2 ¥ –1 D r S° = (D r H°– Dr G°)/T K°p = exp(– D G°/RT) = ¥ 3 J mol–1 –3

2O 4

n(1–a)

–1

–1

–1 –1

–1

mol–1

–1

mol–1

mol–1

Dn = + 1

Kc° = K°p (c°RT/p°)–Dn ¥

–1

3

–1

mol–1

–1

–3

2NO2 n(1 + a)

n(2a

Ê n (1 - a ) ˆ p N 2O 4 = Á p Ë n (1 + a ) ˜¯ total ( pNO /p∞)2 K°p =

Ê n (2a ) ˆ pNO2 = Á p Ë n (1 + a ) ˜¯ total 2

4a 2 ptotal ˆ Ê 1 + a p∞ ˆ Ê 2a ptotal / p∞˜ Á = Á = ( pN 2O4 /p∞) ¯ Ë 1 - a ptotal ˜¯ Ë1 + a 1 - a 2 p∞ 2

Ê 4a 2 ˆ K°p = Á Ë 1 - a 2 ˜¯

peq = (1 + a) pinitial Ê 4a 2 ˆ Ê pinitial ˆ K°p = Á Á ˜ Ë 1 - a ˜¯ Ë p° ¯

fi

4a 2 1-a

4a 2 Ê pinitial ˆ Ê (1 + a ) pinitial ˆ = ÁË ˜¯ p∞ 1 - a ÁË p∞ ˜¯ a

a

ptotal = (1 + a) pinitial K°p K°p

2O 4

7.22 Complete Chemistry—JEE Main

K°p (new) = 1/ K ºp (old) = 1/ 0.129 ptotal = (1 + a) pinitial pinitial E

= D r H + Ea(b)

–1

–1 –1

2O 4

2NO2

1 mol + x

x

1 mol – 2x

p N 2O 4 =

2 pNO 2

Kp =

Ê 1 mol - 2 x ˆ pNO2 = Á (1 atm) Ë 2 mol - x ˜¯

1 mol + x 2 mol - x

(1 mol - 2x)2 (2 mol - x) (1 mol + x)

=

pN 2 O 4

x

x

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN ¥ (a) 3 ¥

¥

–2

–2

N 2O 4 mol L–1

2NO2 ¥

3

2SO2

2

DH

2SO3

2

2O 4

Kc ¥

2

¥

–1

–3

[2003]

–1

[2003] 4(s) + 5O2

(a) K c =

[P4 O10 ]

(b) K c =

5

[P4 ][O 2 ]

2

[P4 O10 ] 5[P4 ][O 2 ]

(c) 2

Kc ¥

T is 4 ¥

2

–4

1 2 O2 –4

[2004]

Keq

(c) exothermic (d) endothermic

[2004]

[2004]

RT

¥

2

1 [O 2 ]5

Kp/Kc

COCl2

1 2 N2

(d) K c =

(c) Kc = [O2]5

(b) RT

(a) 1/RT

P 4O

[2005]

Chemical and Ionic Equilibria 7.23 2

that R (a) Kp = Kc

–1

Kp and Kc

2

–1

Kc

Kp Kc (c) Kp (d) Kp is less than Kc

[2005] 4 3

2

4 2

2

PCl3

PCl5

2

2

[2005]

x ˆ P (d) ÊÁ Ë x - 1˜¯

[2006]

2

P 3 will be x ˆ P (a) ÊÁ Ë 1 - x ˜¯

x ˆ P (b) ÊÁ Ë x + 1˜¯

5

2x ˆ P (c) ÊÁ Ë 1 - x ˜¯

K°c the reaction ¥ –2

2SO2

2SO3

2

1 2 O2

SO2

3

will be

is x

K°c

¥ –3 2Y and Z

Kp1 and Kp2

¥ ¥

–2

–2

[2006]

[2008] CO2

2 4

2

4

2

CO2

2

(a) K21 = K3K32

K1 K2 K3

2

2

(b) K3 = K1 K 2

(c) K1 = K2K3

(d) K3 = K1K2

2

[2008]

2

K is

Kc

Kc

2

2

(1/2)N2 ¥ 1 2 + 2 O2

x (a) 1

2

[2011, Cancelled] T is 4 ¥ –4 2

(c) 4 ¥

–4

[2012]

Kp = Kc(RT)x

SO3 (b) –1

(c) –1/2

(d) 1/2 4(s)

2

Kp

¥ ¥

–5 –2

[2014]

3

2

atm3 ¥

¥

–2

2

and

–2

¥

–2

atm [2014, online]

2

[2014, online]

7.24 Complete Chemistry—JEE Main

[2014, online] 2O 4

2O 4

2

2NO2 2O 4

2O 4

[2015, online]

[2015, online] –1 –1

R

mol–1 Q > Kc Q < Kc

Q > Kc Q < Kc

[2015]

–1

[2016] T

Kp [2016 online]

ANSWERS

HINTS AND SOLUTIONS Kc =

[ NO 2 ]2 (1.2 ¥ 10-2 M) 2 = = 3.0 ¥ 10-3 M [ N 2 O 4 ] (4.8 ¥ 10-2 M)

4(s)

+ 5O2 2

2

2

dT DH/R Dn

=

DH ∞ RT 2

Kp = Kc (RT)Dn = Kc/RT or Kp /Kc = 1/RT

COCl2 Dn Kc = 4 ¥ –4 (1/2)N2

∞ d ln K eq

Kc = 1/[O2]5 as only

P 4O

K¢c =

2

which on integeration gives

Dn

Kc

∞ =ln K eq

K°eq and 1/T

DH Kp and Kc is Kp = Kc(RT)

1

where Dn

=

1 4 ¥ 10-4

= 50

DH ∞ + constant RT

Chemical and Ionic Equilibria 7.25

(RT) Dn Kp = Kc

Since

–1

mol–1

–1

–1

4

3

Kp > KC 2

t teq

p

p

peq

p) + p

p 2

Kp PCl3 + Cl2

5

n(1 – x)

nx

nx

ntotal = n(1 + x)

x Ï nx ¸ P pPCl = Ì ˝P = 3 Ó n(1 + x ) ˛ 1+ x 2SO3 2 2 K c∞ =

3

1 1 = = 416 2 ∞ ¥ ( . 4 9 10-2 ) 2 ( Kc )

2Y p (1 – a) p (2a)

peq1 = p (1 – a) + p (2a) = p (1 + a)

pY2 ˆ

Ê { pX (2a )}2 4 pXa 2 K p1 = Á = = pX (1 - a ) 1- a Ë pX ˜¯ P + Q

Z pZ(1 – a)

p Za

p Za

peq2 = pZ(1 – a) + pZa + pZa = pZ(1 + a)

Ê pP pQ ˆ ( pZa )( pZa ) pZa 2 = = K p2 = Á ; Now Ë pZ ˜¯ 1- a pZ (1 - a ) pX 1 = ; Also Kp1/Kp2 pZ 36

This leads to

DG(iii) = DG(i) + DG(ii) K3 = K1 K2

K p1 K p2 peq1 peq2

=

4 pXa 2 /(1 - a )

pZa /(1 - a ) pX 1 = = pZ 36 2

–RT ln K3° = –RT ln K1° – RT ln K2°

or

2

p)

2p

p) + 2p

p

p pCO2 Kp = N2

p pCO = 2 ¥ 2 pCO

pCO2

=

2

(0.6 atm) = 1.8 atm 0.2 atm

[ NO]2 [ N 2 ][O 2 ] [ N ]1 / 2 [O 2 ]1 / 2 = 2 [ NO]

K c1 =

2

(1/2)N2 Kc 2 = x

K c2

2

1 K c1

=

1 4 ¥ 10-4

x = S Pn P - S Rn R 1 1 x = 1 - ÊÁ1 + ˆ˜ = Ë 2¯ 2

=

1 = 50.0 2 ¥ 10-2

=

4 pX pZ

SO3

1 2 O2

7.26 Complete Chemistry—JEE Main 4(s)

2

3

2

2p

Kp = (p

2p 3

)2 (pCO2) = (2p)2 (p) = 4p3 1/ 3

1/ 3

Ê 2.9 ¥ 10-5 atm3 ˆ Ê Kp ˆ =Á p= Á ˜ ˜¯ = 0.01935 atm Ë 4 ¯ 4 Ë ptotal = p + pCO = 2p + p = 3p = 3 ¥ 3

2

Æ Æ t

n

teq

n –n / 2

kn =

2

2

2

2

(nHI )

2

N 2O 4

=

–2

atm

Kp = Kc = Kn

2

n /4

(nH )(nI )

Dn

2

¥

n /4

(n0 / 4)(n0 / 4) (n0 / 2) 2

2NO2

n (1 – a)

n (2a)

n = n (1 – a) + n (2a) = n (1 + a mass of 1 mol of N 2 O 4 (92 g mol-1 ) = V (1.2 mol) RT /p (92 g mol-1 ) p (92 g mol-1 ) (101.325 kPa ) = = (1.2 mol) (RT ) (1.2 mol) (8.314 kPa dm3 K -1 mol-1 ) (300 K )

r=

DrGº = –RT ln Kºc ΔrGº (2492.2 J mol−1 ) =− = −1 ln Kºc = – RT (8.314 J K −1mol−1 )(300 K ) [B][C] (2 mol)(1 mol / 2) 4 Qc = [A]2 (1 mol / 2)2

Now

–3

Kºc = e–1

–1

Since Qc > Kºc Keq x

x

x

x

[C][D] (1 M + x)2 = [A][B] (1 M - x)2 1M + x 1M - x 11x x

Keq =

or

eq

x

x)

x p +pY

Since p = pY

p = pY

Kp = p pY = (5 bar)2 = 25 bar2

Chemical and Ionic Equilibria 7.27

Ionic Equilibrium

UNIT 2

SECTION 1

Concepts of Acids and Bases

Arrhenius Concept + – + +

–

Brönsted-Lowry Concept + + +

accepts proton 2O

acid1

Æ

Cl–

base2

3O

+

1

2

loses proton

2O

æÆ

Cl–

acid1

3 2

as compared to Cl

O+

3O

+

1

3O

2

+

–

– –

+

3O

2

–

7.28 Complete Chemistry—JEE Main

Common Conjugate Acid-Base Pairs

Table 1 Acid Formula

Name Perchloric acid

4

Formula

Conjugate base Name

ClO–4

Perchlorate ion

– 4

2SO4

æÆ

–

Nitric acid 3O

Br– Cl– NO–3 2O – SO42 – 2PO 4 – NO 2 – 3COO

3 + – 4

Phosphoric acid

3PO4 2

Nitrite ion

– 3

Bicarbonate ion

–

2S + 4

3

CN– – 2

3

2O

Cyanide ion

–

2O

æÆ Cl–

3O

+

NH +4

2O

3

base1

acid2

acid1

–

stron

2

General Criterion on the Extent of a Reaction Levelling Effect 3

3O

O+ 4

3O

2O

æÆ

3O

2O

æÆ

3O

+

2O

æÆ

3O

+

+ ClO–4 + Cl– + Br–

+

2O

2

3O

+

–

3PO4

2O

3O

+

+ H 2 PO 4-

2

2O

3O

+

+ NO -2

+

æÆ

3 2CO3

Cabonic acid

Bromide ion Chloride ion Nitrate ion

Chemical and Ionic Equilibria 7.29 – – –

Æ Æ

2O

NH -2

2O

–

2

–

3

–

CN–

2O

–

2O

– –

2

3O

Phenomenon of Hydrolysis

3O

(i) Complete Hydrolysis

+

–

+

–

2

(a)

PH +4

2O

acid1

base2

æÆ

3O

+

3

acid2

PH +4

base1

3O

–

2O

base1

æÆ

acid2

+

3

2

acid1

+

2

–

base2

–

2

2 3O

(ii) No hydrolysis 2

Na+

(a)

2O

acid1

–

3O

base2

Na+ (b) Cl–

3

¨æ

base1

+

acid1

base2

–

2 –

2

NH +4

2O

acid1

base2

NH +4 – 3COO base1

3O

+

3O

base1

acid2

3O 2O

–

Lux-Flood Concept

CaO

+ SiO2 Æ CaSiO3

+

and

3O

+

3O

+ 3O

4

–

) and ) exhibits hydrolysis to a limited

–

2 –

3

acid1 –

+

+

4

acid2

3COO

–

2 –

Cl– (iii) Hydrolysis to a Limited Extent

(a)

–

acid2

¨æ

acid2

+

+

base1 3O

2O

O+

base2 3

2

7.30 Complete Chemistry—JEE Main

Lewis Concept

3

3

base

Æ

3

ÆBF3

acid

MULTIPLE CHOICE QUESTIONS ON SECTION 1

(a)

HSO -4

(a) ClO–

HSO 4- + CN -  SO 42- + HCN (b) HSO -4 SO 24 – ClO - ClO - and ClO 3 2 4 (b) ClO -2

–

SO 24

(d) CN–

(c) ClO3

(d) ClO -4

(a) NH +4 > HSO -4 > H3O +

(b) NH +4 > H3O + > HSO 4-

(c) H3O + > HSO -4 > NH +4

(d) H3O + > NH +4 > HSO -4

(a) NO -2 > CH3COO - > HCO3-

(b) NO -2 > HCO3- > CH3COO -

(c) HCO3- > NO 2- > CH3COO -

(d) HCO3 > CH3COO > NO 2

(a) NH -2 > OH - > CN (c) CN - > NH 2- > OH -

(b) NH -2 > CN - > OH (d) CN - > OH - > NH 2 2SO4

(a) NH -2

(c) NO -2

–

3O

3PO4

4

3

+

2O

+ 3O + 3O + 3O + 3O

– 3PO4

2 3PO4 (a) HN3

+ CN–

3PO4

2O 2O 2O

2O

+ CN– H 2 PO 43PO4

2 3PO4

2 (b) N 2

–

2

(c) N3-

2

(d) N3–

Chemical and Ionic Equilibria 7.31

not (a) BF3 (a) BF3

3 2+

4

3

–

(d) FeCl3

3

5

2O

5

(c) BF3

(b) CN–

2O

3

(d) BF3

3

2

(d) NO -2

–

ANSWERS

HINTS AND SOLUTIONS HSO -4 –

to

3O

NH +4 2

HCO3- > CH3COO - > NO 23O

+

2CO3

3 2

NH -2 > OH - > CN -

HSO -4

+

3

+ –

æÆ

2O

3O

2O

+

3O

+

–

2 3PO4

2 3

3 3 –

2+

and SnCl4

N3-

3PO4

as compared

7.32 Complete Chemistry—JEE Main

SECTION 2

The pH Scale and pH of Acid and Base Solutions

The pH Scale 3O

3O

Note:

+

+

–14

] by mol dm

3O –3

+

] /mol dm–3}

Equilibrium Constant of Water 2O

2

Kc =

+

–

(1)

[H3O + ][OH - ]

(2)

[H 2 O]2

3O

2O] =

3O

+

¥

– –1

(1000 g / 18 g mol-1 ) n m/M = = (1 L) V V

–1

Kc(water) is Kc =

(10-7 M )(10-7 M ) (55.56 M )2

Ionization Constant of Water + 2O

Ki =

+

2O]

Kw is

M2

[H ][OH ] [ H 2 O] ¥

(3) –16

=

3O

2

Kw

¥

+

–

]

(4) ¥

[H3O + ] [OH - ] M M

Ê OH - ˆ Ê H O+ ˆ ÊK ˆ log Á w2 ˜ = log Á 3 2 ˜ + log Á ËM ¯ Ë M ˜¯ Ë M ¯ pKºw

+

2

Kw = Ki

Kw

=

3O

-

(10-7 M )2 (55.56 M )

Ionic Product of Water

–18

– +

Ki = Kc

¥

¥

–14

2

–3

)

Chemical and Ionic Equilibria 7.33

Ê [OH - ] ˆ - log Á Ë M ˜¯

ÊK ˆ pKºw = - log Á w2 ˜ ËM ¯

(6)

Nature of Aqueous Solution + – 3O 3O

+

–

K w) 3O

+

–

3O

+

–

Kw [H3O + ] = K w

1 2 pK°w

or

[OH - ] = K w

1 2 pK°w

[H3O + ] > K w

1 2 pK°w

or

[OH - ] < K w

1 2 pK°w

[H3O + ] < K w

1 2 pK°w

or

[OH - ] > K w

1 2 pK°w

3O

+

+ 3O + 3O

– –

Variation of the pH of Water with Temperature

Table 1

qC Kw

25 ºC ¥

–15

¥

2

–15

2

¥

–14

¥

2

¥

–14

pKºw (pKºw/2)

pH of Solution of a Monoprotic strong Acid + 3O £ +] +] +] total acid water +]

–]

water +]

total

=c

+] 2 total

–c

–6

–] total

water

= c + Kw

– Kw

+ 3O

c

water

+]

3O

total

+

–14

2

7.34 Complete Chemistry—JEE Main

+]

total

=

c + c2 + 4K w 2

(8)

–6

Illustration

+] =

10-6 M + (10-6 M ) 2 + 4 (10-14 M 2 ) = 10-6 M + 3.22 ¥ 10-6 M 2 2 +

–6

–6

Note:

pH of a Solution of a Monoprotic Weak Acid +

–

a

c

+

c (1– a)

Ka =

–

ca

ca

(c a )(c a ) ca 2 = c(1 - a ) 1- a

ca2 + Kaa – Ka

or a

[H + ][A - ] [HA]

Ka =

3O

+]

a = (–Ka +

K a2 + 4 K a c) ) / 2c

(11)

(= ca a

Ka Ka =

ca 2  ca 2 1- a

a = Ka / c

or

(12)

cÆ

Ostwald Dilution Law –5

Ka

Illustration Ka = c + ]=ca

a=

Ê 1.8 ¥ 10-5 M ˆ ÁË 0.1 M ˜¯

a

1/ 2

–2 –2

–3

+

pH of a Solution of a Strong or Weak Base –

– –]/mol

–]

–]

total

base –] 2 total

–]

–]

water –] base

–2

dm–3

–2

–] –]

base

3O

+]

water

–]

base

+ Kw

–]

total

total – Kw

2 10-7 M + (10-7 M )2 + 4(10-14 M 2 ) [OH - ]base + [OH - ]base + 4K w = 2 2 10-7 M + 2.24 ¥ 10-7 M = 2

total=

Chemical and Ionic Equilibria 7.35

¥

– –3

¥

Kb

4

NH 4 OH  NH +4 c (1- a )

–5

+ OH -

ca

a=

= –]

Kb =

ca

(c a )(c a ) c(1 - a )

Kb =

and

ca2 + Kba – Kb

or

[ NH +4 ][OH - ] [ NH 4 OH] a=

- K b + K b2 + 4 K b c 2c

(-1.8 ¥ 10-5 M ) + [(1.8 ¥ 10-5 M )2 + 4(1.8 ¥ 10-5 M )(10-3 M )]1 / 2 2(10-3 M ) -1.8 ¥ 10-5 M + 2.69 ¥ 10-4 M 2(10-3 M ) ¥

= ca

¥

–3

–3

pH of a Solution of Diprotic Acid –

2O

2

–

3O

2–

2O

2

[HA - ][H3O + ] Ka1 = [H 2 A]

+

+ 3O

Ka2 =

(13)

[A 2- ][H3O + ]

(14)

[HA - ]

Ka1>>Ka2 3O

Computation of [H3O+] + 3O ]total

3O 3

O +]

+

3

2

O +]

+

+ 3O

–

+]

water

+ 3O

Since Ka1>>Ka2 3O

– Kb then pKºa < pKºb Ka < Kb then pKºa > pKºb Ka = Kb then pKºa = pKºb

Illustration Kb 4

4

¥

–5

Kh =

Kw (1.0 ¥ 10-14 M 2 ) = = 1.389 K a K b (4.0 ¥ 10-10 M )(1.8 ¥ 10-5 M )

Ka

¥

7.42 Complete Chemistry—JEE Main

a=

Kh 1 + Kh

=

1.389 1 + 1.389

¥

pKºa

¥

Kºb

–5

1 1 [pKºw + pKºa – pKºb] = 2 2

MULTIPLE CHOICE QUESTIONS ON SECTION 3

¥ ¥

(a) a = K w K a / c where Ka

(b) a = K w / K a c

(c) a = K w c / K a

(a) Na2CO3

¥

¥

(d) a = K a c / K w

–5

¥

5

¥ ¥

–5

–4

–4

¥

– 3

–11

4

= Ka

¥

–4

Ka 3 Ka(C6 Ka Ka Ka

¥

2CO3

Kb ¥

¥

Ka1 be

–5

¥

–8

2CO3

4CN

3

Kb

–5

¥

¥

–2

¥

–3

4

–4

3

1 2 (pK°a1 + pK°a2

K°a1 + K°a2 ¥

¥

3 2CO3

¥

–8

2+

Ka

4

1 2 (pK°a1 – K°a2)

K°a1 – K°a2

2CO3

4

3

4

¥

¥

–8

3 4

4Cl CO 2 3

2CO3 3

Chemical and Ionic Equilibria 7.43

ANSWERS

HINTS AND SOLUTIONS ¥

Kb = Kw /Ka –

ca

-

[HAc][OH ] (ca )(ca ) ca 2 = = c(1 - a ) 1 - a [Ac- ] Kb = ca2 or a = K b / c

a

c–x

a = K w / Ka c 2O

Kb = Kw /Ka where Ka

–

x

x

x2 x2 [HF][OH - ] =  Kb = c-x c [ F- ] –]

x

K b c = ( K w / K a )c

=x=

¥

– 3

–

1 2 [pK°w + pK°a

– 3

2O

Kh 1 + Kh

¥

2

=

1 2 [14

c

Kh =

¥

–11

(1.0 ¥ 10-7 M ) = 5.56 ¥ 10-3 (1.8 ¥ 10-5 M )

5.56 ¥ 10-3 = 5.52 ¥ 10-3 -3 1 + 5.56 ¥ 10

+

– 3

– 3

– 3

2O

2O

3

–4

1/2

c

Kw (1.0 ¥ 10-14 M 2 ) = K a K b (1.8 ¥ 10-5 M 2 ) 2 a=

–14

K°a

1 2 [pK°w + pK°a 2– 3

¥

c)

–6

Ka

Kh =

¥

–5

–

ca

Kb =

¥

2

2O

c(1 – a)

–

–14

O+

+

2CO3

CO32– –

K a2 =

Kb =

[H3O + ][CO32- ] [HCO3- ]

[H 2 CO3 ][OH - ]

K [OH - ][H + ] = w Kb = + [HCO3 ][H ]/[H 2 CO3 ] K a1

[HCO3- ]

1 2 [14

¥

–6

7.44 Complete Chemistry—JEE Main

Ka2 + 3O ] =

3O

K a2 [HCO3- ] [CO32- ]

+ 2

] = K a1 K a2

=

K a2 {[H 2 CO3- ][OH - ]/ K b } [CO32- ]

+] 2

[CO32- ]

–3

+]

–3

2

{K b (A - )}c

=

–14

2

–

Kb

or

–5

–

Kw / Kb

a=

[CO23–

1 2 (pK°a1+ pK°a2)

= Ka1Ka2

–

Kb Ka

[H 2 CO3 ] K w ( K w / K a1 ) [H3O + ]

2CO3]

+

that is Finally

[CO32- ]

[H 2 CO3 ]

Since both Ka2 and Kb 3O

K a2

=

+]2/c

–5

Kh

Kh =

1 + Kh

a 1- a

a ˆ 2 Ê 0.5 ˆ 2 = 1.0 =Á Kh = ÊÁ Ë 1 - 0.5 ˜¯ Ë 1 - a ˜¯ 2+

+

2O

Kh = +]

+

[Cu(OH)+ ][H + ] [H + ]2 [H + ]2 =  [Cu 2+ ] [Cu 2+ ]0 - [H + ] [Cu 2+ ]0

= (Kh

2+]

¥

)1/2 ¥

–8

1/2

¥

–5

–5 + 4

4

NaNO3

2– 3

2CO3

SECTION 4

Buffer Solutions

Common-Ion Effect c1

c 2O –

co – x’

3O

–

x’ 2O

3O

c –x +

+

x

(44)

x’ +

–

c1 + x

-

[H3O ][Ac ] (c1 + x)( x) = [HAc] c0 - x 2 + x (c + K ) – K c x 1 a a o Kt =

(45)

x

3O

c

c1

+

Chemical and Ionic Equilibria 7.45

Ka = 3O

(c1 )( x) c0

+

Êc ˆ x = Ka Á 0 ˜ Ë c1 ¯

or

3O

+]

Ê [acid] ˆ log Á Ë [salt] ˜¯

Ka

= Ka

Ê [salt] ˆ K°a + log Á Ë [acid] ˜¯

4

4

Ê [salt] ˆ pK°b + log Á Ë [base] ˜¯ Ka

Illustration +

(48) ¥

–5

– -

[H ][Ac ] [H + ][Ac- ] [H + ]2 =  [HAc] [HAc]0 - [H + ] [HAc]0 +

Ka =

[acid] [salt]

¥

+

–5

¥

1/2

Ê [salt] ˆ pK°a + log Á Ë [acid] ˜¯

¥

or

[H + ] = K a [HAc]0

–3

¥

+

Ê 0.05 ˆ ˜ ÁË 0.10 ¯

–5

Acidic Buffer Solution æÆ

–

(acid)

–

2

(added)

(salt)

æÆ

–

+

(salt)

(added)

(acid)

–

–

+

–

–

Illustration

3

3

Ê 0.05 M - 0.01 M ˆ log Á Ë 0.10 M + 0.01 M ˜¯

Ê [salt] ˆ K°a + log Á Ë [acid] ˜¯

Basic Buffer Solution æÆ

+

4

+ 4

2

(base)

(salt) + 4

(salt)

Buffer capacity

+

æÆ

4

(base)

–3

7.46 Complete Chemistry—JEE Main

[salt] +

∂cbase/∂

∂cacid/∂ ∂(salt)/∂ a = [salt] + [acid]

b = [salt]

and

Ê [salt] ˆ 1 b ˆ ln ÊÁ = pK°a + K°a + log Á ˜ ˜ Ë [acid] ¯ 2.303 Ë a - b ¯

Then

1 a ∂(pH) = 2.303 b(a - b) ∂b

b( a - b) ˆ ∂b = 2.303 ÊÁ ˜ Ë a ¯ ∂(pH)

or

i.e.

∂[salt] Ê [salt][acid] ˆ = 2.303 Á Ë [salt] + [acid] ˜¯ ∂(pH)

∂b/∂

b/a – b) = 1 or b = a

Buffer Range pKa – 1 to pKa + 1 (or pKb – 1 to pKb

MULTIPLE CHOICE QUESTIONS ON SECTION 4 Ka

– 2PO4

¥

–8

2PO4 and Na2

¥

Kb

–4

Ka

2– 4

4

3) 2

¥

3) 2

2

–5

–1

3PO4

2

4

is

(a) 1 ¥

¥

¥

¥

–11

–12

a and b b(a – b)/a 4

25° C is

a(a – b)/b Kb

ab/(a + b ¥

–5

4

a + b)/ab

Chemical and Ionic Equilibria 7.47

K°a

ANSWERS

HINTS AND SOLUTIONS ¥ ¥ ¥

K°a K°b K°a

–8

K°a – 1 to pK°a Kb – 1 to 14 – pKb

–4 –5

K°a Ê [acid] ˆ log Á Ë [salt] ˜¯

[acid] =2 [salt] –1

∂b b( a - b) ˘ = 2.303 ÈÍ ∂(pH) Î a ˙˚ where a = [acid] + [salt]

b

and

∂b È (0.05 M)(0.15 M - 0.05 M) ˘ = 2.303 Í ˙ ∂(pH) (0.15 M) Î ˚ Ka + log

[salt ] Ê 0.05 M ˆ = pKa + log Á Ë 0.10 M ˜¯ [acid]

Ka

¥

pKa

pKa

–12

2.303

pK°b + log

[base][acid] ba ˆ = 2.303 ÊÁ ˜ Ë [base] + [acid] b + a¯ [salt ] [base]

¥

–5

Ê 0.2 M ˆ ) + log Á Ë 0.1 M ˜¯

¥ n1 = VM1 n2 = VM2 n3 =

m 0.04 g = = 10-3 mol -1 M 40 g mol

–3

¥

–1 –3

–3 –1

mol –3

mol

7.48 Complete Chemistry—JEE Main

–3

= 2¥

–3

–3

1 2 [ pK°w + pK°a 1 =2

mol

1

¥

c/ mol dm–3)] = 2

SECTION 5

–3)]

Solubility Product

Expression of Solubility Product

+(aq)

CaF2(s) Ca3(PO4)2

+ Cl–(aq) + 2F–(aq) 3Ca2+(aq) + 2PO3– 4 (aq)

Ca2+(aq)

Ks

+] [Cl–] Ks Ks(CaF2) = [Ca2+] [F–]2 Ks(Ca3(PO4)2) = [Ca2+]3 [PO43–]2 xB y

General Expression of Solubility Product Ks

xB y)

y+]x

=

=x

y+

+ yBx–

[Bx–]y

s y+

x

xB y

+ yBx–

(xs)

Ks =

(xs)x (ys)y

(ys)

= xx yy sx+y

s = (Ks/xxyy)1/(x+y)

or

Illustrations ¥

3– 4

s Ca3(PO4)2

3Ca2+ + 2 [PO43–] (3s)

s

¥

Ks =

[Ca2+]3[

(2s)

(3s)3

(2s)2 ¥

2CrO4 2CrO4(s)

+(aq)

+

(2s)

Ks

¥

s PO42–]2 =

¥

s5 –12

3

s

(s

] [ CrO42–] = (2s)2 (s) = 4s3 ¥

–12

3

/4)1/3

¥

–32

5

2CrO4

CrO42– (aq)

+ 2

s = (Ks / 4)1/3

5

¥

–4

Chemical and Ionic Equilibria 7.49

Ionic Product

Illustrations –4

Ks

¥

2

–12

3) 2 3

2

+

–

2+

2

–] 2

Ks

–4

–5

2

–5 –14

3

2

–4

3) 2 –] ions need to initiate precipitation is

–]

Ê Ks (Mg(OH) 2 ) ˆ = Á ˜¯ Ë [Mg 2+ ] ¥

1/ 2

2

Ê 8.9 ¥ 10-12 M 3 ˆ =Á Ë (10-4 M ) ˜¯

1/ 2

= 2.988 ¥ 10-4 M

–4

Preferential precipitation of salts

1. Salts having the same stoichiometry –

Illustration Ks

¥

2

–

¥

Ks –

–13

– 2 and K

–

+

¥

s

–

+]

Ê Ks (AgCl) ˆ 1.7 ¥ 10-10 M 2 = = 1.7 ¥ 10-9 M = Á Ë [Cl- ] ˜¯ (0.1 M )

For Br–

+]

Ê Ks (AgBr) ˆ 5.0 ¥ 10-13 M 2 = = 5.0 ¥ 10-12 M = Á Ë [Br - ] ˜¯ (0.1 M )

+

+

Ks

+

Cl–

–

Ks

Ks

–

Quantitative Predictions For

2

–

–

Ê Ks (AgI) ˆ 8.5 ¥ 10-19 M 2 = = 8.5 ¥ 10-18 M ]= Á Ë [I - ] ˜¯ (0.1 M ) +

–

– +

–

¥

–12

–

ions

7.50 Complete Chemistry—JEE Main

Ê Ks (AgI) ˆ (8.5 ¥ 10-19 M 2 ) -7 = Á [Ag + ] ˜ = (5.0 ¥ 10-12 M ) = 1.7 ¥ 10 M Ë Br - ¯

–]

–

Br–

¥

+

–

ions

–

Ê Ks (AgBr) ˆ (5.0 ¥ 10-13 M 2 ) = = 2.94 ¥ 10-4 M [Br–] = Á Ë [Ag + ]Cl ˜¯ (1.7 ¥ 10-9 M )

2. Salts having different stoichiometry

Illustration Ks

¥

2

and Ks

¥

2CrO4

+

2CrO4 –12 3

+

[Ag + ] =

–

Ks (AgCl)

=

-

[Cl ]

-10

ions is

1.7 ¥ 10 M = 1.7 ¥ 10-9 M (0.1 M ) 2

+

2– 4

Ê K (Ag 2 CrO 4 ) ˆ [Ag + ]= Á s Ë [CrO 42- ] ˜¯

1/ 2

Ê 1.9 ¥ 10-12 M3 ˆ =Á ˜¯ 0.1 M Ë

ions is

1/ 2

= 4.36 ¥ 10-6 M

+ –6

–

[Cl- ]=

Ks (AgCl) [Ag + ]CrO2 -

=

4

1.7 ¥ 10-10 M 2 = 3.90 ¥ 10-5 M 4.36 ¥ 10-6 M –

2CrO4 2– 4

¥

+

The CrO42–

+ +

–

2CrO4

Qualitative Analysis of Cations +

Group I Group II

2+ 2 and

3+

Group III Group IV Group V Group VI

3+

Pb2+ 2+

3+

2+

2+

2+

2

2+

3+

3+

and Cr3+ 2+

2+

2+

2+

and Zn2+ and Sr2+

+

MULTIPLE CHOICE QUESTIONS ON SECTION 5

Na2SO4 ¥

¥

–1

4

¥

–8

¥

–8

–4

¥

–8

and Sn2+ are

Chemical and Ionic Equilibria 7.51

¥

3(PO4)2

¥

–14

¥

5

Ka1

+

¥

2

¥

2+

and Fe2+

2

¥

Ks –15

¥

5

2

2–

–21

–32

mol L–1

2

3

Ka2

–22

2

5

¥

2

Ks(FeS) = 4 ¥

¥

3(Ks

–32

2

–14

and Ka1 Ka2 2S in

2

2(Ks

3

(c) SrCO3

2

¥

¥

–11

¥

5

¥

2

Ks

¥

2(Ks

–32

–15

2

–

¥

–

2– 3

¥

¥

Ks 3

3

s =(Ks 2CO3

3

2)

=

3

+

3– 4

2 –12

¥

Ks

2

–

Ks 2CO3 (Ks

Ks

3) 2

2

3)

4

2)

¥

(Ks

–22

4)

1/5

Ks (c) Ca3(PO4)2

2

¥ oxalate is ¥

¥

Ks(PbF2 ¥ ¥

¥

3 –8

3

and Ks

¥ ¥

–3 –3

¥

3

¥

2

3

–3

–11

3 2

¥ ¥

–3

¥

3

¥ ¥

–3 –3

2

–3 –3

3 3+

–

3+

3+

+

–

3+ +

2

(aq)

Ba2+

Kc

+ +

– –

] ] 2

not 2+ 2–

and S are also considered

2

7.52 Complete Chemistry—JEE Main

ANSWERS 1. (b) 7. (b) 13. (d)

2. (c) 8. (b) 14. (a)

3. (d) 9. (c)

4. (a) 10. (d)

5. (b) 11. (a)

6. (a) 12. (b)

HINTS AND SOLUTIONS Ê 9.32 ¥ 10-4 g ˆ Ê 1000 mL L-1 ˆ s= Á = 4 ¥ 10–5 mol L–1 Ë 233 g mol-1 ˜¯ ÁË 100 mL ˜¯

1. Solubility (in mol/L) is

For the equilibrium BaSO 4 (s)  Ba 2+ (aq ) + SO 24- (aq ), Ks (BaSO 4 ) = [Ba 2+ ][SO 24- ] = s2 = (4 ¥ 10–5 mol L–1)2 Solubility of BaSO4 in 0.1 M Na2SO4 is 2. Ca 3 (PO 4 ) 2  3 Ca

2+

(3 s )

+

[Ba 2+ ] =

2 PO34 (2s)

Ks (BaSO 4 ) [SO 24- ]

=

(4.0 ¥ 10-5 mol L-1 )2 = 1.6 ¥ 10-8 M -1 (0.1 mol L )

K s = [Ca 2+ ]3 [PO34- ]2 = (3s )3 (2s )2 = 108 s 5 = (108)(1.6 ¥ 10–7 M)5 = 1.13 ¥ 10–32 M5 3. H 2S  H + + HS-

;

Ka1 = [H+][HS–]/[H2S]

HS-  H + + S2-

;

Ka2 = [H+][S2–]/[HS–]

H 2S  2H + + S2-

;

Ka = [H+]2[S2–]/[H2S]

Ka = Ka1 Ka2 = (10–7 M) (10–14 M) = 10–21 M2

Obviously,

[S2–] =

K a [H 2S] [H + ]2

=

(10-21 M 2 )(0.1 M ) = 2.5 ¥ 10-21 M (0.2 M ) 2

Number of S2– ions = [S2–]NA = (2.5 ¥ 10–21 mol L–1) (6.022 ¥ 1023 mol–1) = 1506 L–1 4. H 2S  2H + + S2- ; Since

Ka1 Ka2 = 1.0 ¥ 10–21 M2

Ka1 Ka2 = [H+]2 [S2–]/[H2S], [S2–] =

K a1 K a 2 [H 2S]

[H + ]2 Ionic products of given ions are

=

we have

(1.0 ¥ 10-21 M 2 )(0.1 M ) = 1.0 ¥ 10–20 M (0.1 M) 2

For ZnS

Qc = [Zn2+] [S2–] = (0.1 M) (10–20 M) = 10–21 M2

For FeS

Qc = [Fe2+] [S2–] = (0.1 M) (10–20 M) = 10–21 M2

Since only for Qc(ZnS) > Ks(ZnS), only ZnS is precipitated. 5. Since Ks(SrCO3) > Ks(AgCl) and Ks(SrF2) > Ks(Fe(OH)2), we compute solubility of SrCO3 and SrF2 only. SrCO3  Sr 2+ + CO32- ; s

s

SrF2  Sr 2+ + 2F- ; s

K s = [Sr 2+ ][CO32- ] = s 2 Ks = [Sr 2+ ][F- ]2 = 4s 3

2s

s = Ks = (7 ¥ 10-10 M 2 )1/ 2 = 2.65 ¥ 10-5 M s = ( K s / 4)1 / 3 = (8.0 ¥ 10-11 M 3 / 4)1/ 2 = 2.71 ¥ 10-4 M

Thus, SrF2 has largest solubility. 2+

– 2

6. Ks (Mg(OH)2) = [Mg ] [OH ] . Hence

Ê Ks ˆ [OH - ] = Á Ë [Mg 2+ ]˜¯

1/ 2

Ê 9 ¥ 10-15 M 3 ˆ =Á Ë 0.1 M ˜¯

1/ 2

= 3.0 ¥ 10-7 M

Chemical and Ionic Equilibria 7.53

pOH = –log(3.0 ¥ 10–7) = 6.52

pH = 14 – 6.52 = 7.48

7. M (OH) 2  M 2+ + 2OH s

2s

Ks = [M2+] [OH–]2 = (s) (2s)2 = 4s3; s = (Ks /4)1/3 = (3.2 ¥ 10–17 M3/4)1/3 = (8 ¥ 10–18 M3)1/3 = 2 ¥ 10–6 M [OH–] = 2s = 4 ¥ 10–6 M; pOH = –log(4 ¥ 10–6) = 5.4 pH = 14 – 5.4 = 8.6 + 8. The concentration of Ag ions needed to precipitate given ions are as follows. AgCl

:

[Ag + ] =

Ks (AgCl)

AgI

:

[Ag + ] =

Ks (AgI)

Ag2CO3 :

[Cl- ] [I- ]

=

=

1.8 ¥ 10-10 M 2 = 1.8 ¥ 10-9 M 0.1 M

8.0 ¥ 10-17 M 2 = 8.0 ¥ 10-16 M 0.1 M

Ê K (Ag 2 CO3 ) ˆ [Ag ] = Á s Ë [CO 2- ] ˜¯

1/ 2

+

3

Ag3AsO4 :

Ê K (Ag3 AsO 4 ) ˆ [Ag ] = Á s Ë [AsO3- ] ˜¯

Ê 8.0 ¥ 10-12 M3 ˆ =Á ˜¯ 0.1 M Ë

1/ 3

+

4

1/ 2

Ê 1.0 ¥ 10-22 M 4 ˆ =Á ˜¯ 0.1 M Ë

= 8.94 ¥ 10-6 M 1/ 3

= 1.0 ¥ 10-7 M

Ag+ 9. The exponent 1/5 implies that ions are obtained from the ionization of salt, the salt is Ca3(PO4)2. 10. Ks = [Ag+]2 [Ox2–] = (0.1 M)2 (1.1 ¥ 10–9 M) = 1.1 ¥ 10–11 M3 11. Let x and y be the respective solubilities of PbF2 and MgF2. Then

([Pb2+] = x, [Mg2+] = y and [F–] = 2(x + y)

For PbF2 :

Ks = [Pb2+][F–]2 = 4x(x + y)2 = 7.2 ¥ 10–8 M

For MgF2 :

Ks = [Mg2+][F–] = 4y(x + y)2 = 14.4 ¥ 10–8 M

Solving for x and y, we get x = 1.26 ¥ 10–3 M and y = 2.52 ¥ 10–3 M (Note: 12. Since the solution is electrically neutral, sum of total positive charge will be equal to total negative charge. Each Al3+ carries 3+ charge, hence 3 [Al3+] + [H+] = [OH–]. 13. Let s be the solubility of BaF2 in a buffer solution of pH = 2 (i.e.

[H+] = 10–2 M). Then

BaF2 (s) + 2H + (aq) Æ Ba 2+ (aq) + 2HF(aq) 2s

s 2+

Kc =

2

2

[Ba ][HF] ( s )(2 s ) = ; + 2 [H ] (10-2 M ) 2

4s3 = (10–2 M)2 Kc = (10–2 M)2 (2.56 M)

1/ 3

È (10-4 M 2 )(2.56 M ) ˘ -6 -2 1/ 3 s= Í ˙ = (64 ¥ 10 M ) = 4 ¥ 10 M 4 Î ˚ 14. The solubility of PbS is increased if the hydrolysis of Pb2+ and S2– ions are also considered. The solubility of Zn(OH)2 is pH dependent due to the reaction between OH– from Zn(OH)2 and H+ ions (i.e. pH) present in the solution.

SECTION 6

Acid-Base Indicators

Indicators, in general, are either organic weak acids or weak bases with a characteristic of having different colours in the ionized and unionized form. For example,

7.54 Complete Chemistry—JEE Main

Phenolphthalein

Methyl orange

Indicator Constant

Representing a weak-acid indicator as HIn, we write

HIn  H + + In K HIn =

or

[H + ][In - ] HIn

pH = pK°HIn + log

or

[H + ] = K HIn

[HIn ] [In - ]

[In - ] [HIn]

(50)

The constant KHIn is known as indicator constant.

Indicator Range we have The solution gets characteristic colour of In– if the ratio [In–]/[HIn] ≥ 10 The solution gets characteristic colour of HIn if the ratio [HIn] / [In–] ≥ 10. With these parameters in Eq. (50), we have pH = pK°HIn + 1 and For colour of HIn pH = pK°HIn – 1 For colour of In– Diagrammatically, we have

1 [In -1 ] ¥ 100 = ¥ 100 = 9.1 that is only 9.1% of indicator is present in the ionized At pH = pK°HIn – 1, (1 + 10) [In ] [HIn ] form. [In -1 ] 10 ¥ 100 = ¥ 100 = 91.0 that is 91.0% of indicator is present in the ionized form. At pH = pK ∞HIn + 1, 10 +1 [In ] + [HIn ]

Chemical and Ionic Equilibria 7.55

Selection of Indicators

In the titration of an acid solution (say, 0.1 M HCl) versus a base solution (say, 0.1 M NaOH) or vice versa, the pH of the solution near the equivalence point changes steeply. The centre of this steep change lies at pH = 7 only when a strong acid is titrated wita a 14.0 strong base. If the titration involves a weak acid or a weak 12.0 base, the pH is either greater than or lesser than 7 due to the hydrolysis of the salt formed. 10.0

1. Titration of a Strong Acid Versus Strong Base Figure 1 displays the typical titration curve depecting pH of the solution versus volume of base added. The steep change of pH at the equivalence point depends upon the concentrations of acid and base being titrated. A typical range lies from pH = 3 to pH = 11. Therefore, any indicator whose pH range is within this interval can be used. The commonly used indicators are: 1. phenolphthalein (pH range : 8.3 to 11, colour change is from colourless to pink). 2. methyl orange (pH range : 3.1 to 4.4, colour change is from red to yellow). 3. methyl red (pH range : 4.2 to 6.3, colour change is from red to yellow).

phenolphthalein

pH

8.0 6.0

bromothymol blue

4.0

methyl orange

2.0 0.0

0

10

20

30 40 Vh/cm3

50

60

Fig. 1 14.0 12.0 10.0 8.0

pH

The selection of acid-base indicator to locate end point of a titration is decided by the following factors: 1. The steep portion of the titration curve at the equivalence point must encompass an interval of pH values at least as large as the pH transition range of an indicator. 2. The pH transition range of the indicator must coincide with the steep portion of the titration curve. The following is the description of selection of indicators in different types of titrations.

6.0

methyl red

4.0 2.0 0.0

0

5

10

15

20 25 Va/cm3

30

35

40

Fig. 2

2. Titration of a Weak Base Versus a Strong Acid Figure 2 displays the typical titration curve depicting pH of the solution versus volume of acid added. A typical range of pH at the equivalence point is from pH = 6.5 to pH = 4 The commonly used indicators are methyl red and methyl orange.

14.0 12.0 10.0 phenolphthalein 8.0

pH

3. Titration of a Weak Acid Versus a Strong Base Figure 3 displays the typical titration curve depicting pH of the solution versus volume of base added. A typical range of pH at the equivalence point is from pH = 6 to pH = 10.5. The commonly used indicator is phenolphthalein. 4. Titration of a Weak Acid Versus a Weak Base In this titration, the steep rise in pH near the equivalence point does not occur. Also, the rise of pH does not encompass an interval equal to the pH transition range of any of the indicators. Hence, no suitable indicator can be chosen to locate the end point of the titration.

bromothymol blue

6.0 4.0 2.0 0.0

0

5

10

15

20 Vb/cm3

Fig. 3

25

30

35

40

7.56 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS ON SECTION 6 ∞ - 1, an indicator is present in 1. At pH = pK HIn (a) 80% ionized form (b) 85% ionized form (c) 9.1% ionized form (d) 50% ionized form ∞ + 1, an indicator is present in 2. At pH = pK HIn (a) 80% ionized form (b) 91% ionized form (c) 25% ionized form (d) 50% ionized form 3. When 50% of an indicator is present in ionized form, the pH of the solution will be equal to ∞ - pK In ∞(b) pK°Hin – 1 (c) pK°Hin (d) pK HIn (a) pK°Hin 4. Which of the following indicators (buffer range provided in brackets) cannot be used to detect end point of titration between strong acid and strong base? (a) Trinitrobezene (12.0 to 14.0) (b) Phenol red (6.0 to 7.6) (c) p-Nitrophenol (5.0 to 7.0) (d) Phenolphthalein (8.3 to 10.0) 5. An indicator is 20% ionized in a solution of pH = 2.4. The percentage ionization in a solution of pH = 3.3 will be about (a) 33.3% (b) 45.0% (c) 50.0% (d) 66.67%

points are 8.34 and 3.97, respectively. The steep portion at these two

pH

6. The titration of a weak base versus weak acid cannot be carried out by using acid-base indicator because (a) there is no variation in pH of the solution (b) the variation in pH is gradual with no steep change in pH near the equivalence point. (c) the change in pH near the equivalence point does not encompasses an interval equal to the pH transition range of the indicator (d) no indicator can be found which changes colour at pH of equivalence point. 7. 50 mL of 0.1 M solution of a weak acid is titrated against 0.1 M NaOH solution. After the addition of 10 mL of NaOH, the pH of the solution is found to be 4.14. After the addition of 40 mL of NaOH solution, the pH of the solution will be (a) 4.74 (b) 5.10 (c) 5.34 (d) 6.20 8. Which of the following indicators can be used to detect end point in the titration of a weak acid versus strong base? 14.0 (a) Methyl red (4.2 to 6.3) (b) Methyl violet (0.1 to 3.2) 12.0 (c) Methyl orange (3.1 to 4.4) 10.0 (d) Thymolphthalein (9.3 to 10.5) 9. The pH titration curve of 0.1 M Na2CO3 solution versus 0.1 M HCl is 8.0 6.0 4.0

(a) Methyl violet (0.1 to 3.2) and phenolphthalein (8.3 to 10.0) 2.0 (b) Phenolphthalein (8.3 to 10.0) and methyl orange (3.1 to 4.4) 0.0 (c) Trinitrobenzene (12.0 to 14.0) and phenol red (6.8 to 8.4) 0 10 20 30 40 Va/cm3 (d) Methyl orange and methyl violet 10. The colour of phenolphthalein in acid and base ranges of the indicator respectively, are (a) pink, colourless (b) pink, red (c) colourless, pink (d) red, pink 11. The colour of methyl orange in acid and base ranges of the indicator, respectively, are (a) yellow, red (b) red, yellow (c) yellow, blue (d) blue, red

50

60

Chemical and Ionic Equilibria 7.57

ANSWERS 1. (c) 7. (c)

2. (b) 8. (d)

3. (a) 9. (b)

4. (a) 10. (c)

5. (d) 11. (b)

6. (c)

HINTS AND SOLUTIONS 1. Since pH = pK°HIn + log ([In–]/[HIn]), we will have 2. Here

log([In–]/[HIn]) = –1

or

[In–]/[HIn] = 1/10.

Percentage ionization = [1/(1 + 10)] ¥ 100 = 9.1%

log([In–]/[HIn]) = 1

or

[In–]/[HIn] = 10/1.

Percentage ionization = [10/(10 + 1)] ¥ 100

3. At 50% ionization,

[In–

]/[HIn] = 1.

Hence,

91%

pH = pK°HIn

4. Indicator range of trinitrobenzene is beyond the steep portion of pH near the equivalence point. 5. We have 0.2 ˆ or pK°HIn = 3.0 2.4 = pK°HIn + log ÊÁ Ë 0.8 ˜¯ 3.3 = 3.0 + log ([In–/HIn])

i.e.

log ([In–]/[HIn]) = 0.3

This gives

[In–]/[HIn] = 2.

The percentage ionization = [2/(2 + 1)] ¥ 100 = 66.67% 6. During the titration, a buffer solution of weak acid and salt of its cojugate base is formed. Its pH is given by pH = pK °a 4.14 = pK °a + log[10/(50 – 10)]. This gives pK°a = 4.14 – log(1/4) = 4.14 + 0.60 = 4.74 After the addition of 40 mL of NaOH solution, we will have pH = 4.74 + log[40/(50 – 40)] = 4.74 + 0.60 = 5.34 8. In the titration of a weak acid (say, acetic acid) versus strong base (say, sodium hydroxide), the pH at the equivalence point is more that 7 due to hydrolysis of congugate base of the weak acid. The steep rise in pH around equivalence point encompasses the range 7 to 10. Thus, thymolphthalein can be used as the indicator to locate end point.

10. Phenolphthalein is colourless in acid range and pink in base range of the indicator. 11. Methyl orange is red in acid range and yellow in base range of the indicator

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE UNIT pH of a Solution (a) pH = – ln ([H+]/mol dm–3) (b) pH = – log ([H+]/mol dm–3) (d) pH = – ln (mol/dm–3/[H+]) (c) pH = – log [mol dm–3/[H+]) 2. Which of the following expressions is true? (a) [H+] + [OH–] of water is a constant and is independent of temperature (b) [H+] [OH–] of water is a constant and is independent of temperature (c) pH + pOH = 14 and is independent of temperature. (d) Irrespective of whatever may be the temperature, pH + pOH = pK°w

7.58 Complete Chemistry—JEE Main

3. What will be the pH of water at 50 °C? Given: pK°w = 13.26 at 50°C. (a) 6.0 (b) 7.0 (c) 6.63 (d) 13.26 4. What will be the change in the pH of water if 10–3 mol of NaOH is added to 1.0 dm3 of water at 25°C? (a) Increased by 2 (b) Increased by 4 (c) Decreased by 2 (d) Decreased by 4 + ions present in 1 mL of a solution having pH = 13 is 5. The number of H (a) 1013 (b) 6.022 ¥ 1013 (c) 6.022 ¥ 107 (d) 6.022 ¥ 1010 6. Give pK°w(H2O) = 13.54 at 40 °C. Predict the nature of solution having pH = 7 at 40 °C. (a) acidic (b) alkaline (c) neutral (d) cannot be predicted 7. Which of the following expressions is not true?

8. 9. 10. 11.

12.

13.

14.

15. 16. 17. 18.

(a) For a neutral solution

[H+] = [OH–] =

(b) For an acidic solution

[H+] >

Kw

Kw and

[OH–]
K w (d) For a neutral solution [H+] = [OH–] = 10–7 M at all temperatures –5 10 M HCl solution at 25 °C is diluted 1000 times. The pH of the solution will (a) be equal to 8 (b) lie between 7 and 8 (c) lie between 6 and 7 (d) remain unchanged 10–5 M NaOH solution at 25 °C is diluted 100 times, the pH of the solution will (a) be equal to 7 (b) lie between 7 and 8 (c) lie between 6 and 7 (d) remain unchanged At 90 °C, pure water has [H3O+] = 10–6 mol dm–3. The value of pKw at 90 °C is (a) 6 (b) 12 (c) – 6 (d) – 12 Which of the following statements is correct? (a) pK°w increases with increase in temperature (b) pK°w decreases with increase in temperature (c) pK°w = 14 and is independent of temperature (d) pK°w increases at low temperatures but decreases at high temperatures For pure water, (a) pH increases with increase in temperature (b) pH decreases with increase in temperature (c) pH = 7 and is independent of temperature (d) pH increases at low temperatures but decreases at high temperatures For pure water, the product [H+] [OH–] (a) increases with increase in temperature (b) decreases with increase in temperature (c) is constant and is independent of temperature (d) increases in the low temperature range but decreases in the high temperature range The pH of a solution having [H+] = 10–8 M (a) lies between 6 and 7 (b) lies between 7 and 8 (c) is 8 (d) is 7.5 The pH of a neutral water is 6.5. The temperature of water (a) is 25 °C (b) is more than 25 °C (c) is less than 25 °C (d) cannot be predicted 3 –4 One dm solution of 10 M HCl solution is diluted 1000 times. Its pH will be (a) 7 (b) Less than 7 (c) More than 7 (d) eqaul to 8 100 mL of 0.1 M HCl solution is mixed with 50 mL of 0.05 M NaOH. The resultant solution will have pH (a) 1.5 (b) 1.8 (c) 2.12 (d) 3.42 Which of the following has the minimum [H+]? (c) 0.1 M NH3 (d) 0.1 M C2H5OH (a) 0.1 M NaOH (b) 0.1 M NaC2H3O2

Chemical and Ionic Equilibria 7.59

19. The pH of HCl solution is 1.0. If 900 mL of water is added to 100 mL of this solution, the pH of the resultant solution will be (a) 0.1 (b) 2.0 (c) 4.0 (d) 7.0 20. Which of the following solutions will have pH close to 1.0? (a) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH (b) 55 mL of 0.1 M HCl + 45 mL of 0.1 M NaOH (c) 10 mL of 0.1 M HCl + 90 mL of 0.1 M NaOH (d) 75 mL of 0.2 M HCl + 25 mL of 0.1 M NaOH 21. The number of H+ ions in 1 cm3 of a solution of pOH =13 is (b) 6.022 ¥ 1019 (c) 6.022 ¥ 1010 (d) 6.022 ¥ 1013 (a) 6.022 ¥ 104 22. The pH of a solution is 3. If its pH is to be raised to 6, then the [H+] of the original solution has to be (a) doubled (b) halved (c) increased 1000 times (d) decreased 1000 time

Weak Acids 23. The pH of a solution of 10–1 M acetic acid (Ka = 1.8 ¥ 10–5 M) is (a) 2.87 (b) 1.87 (c) 3.87 24. The expression of Ostwald dilution law is

(d) 1.34

(b) a = Ka c (c) a = K a c (d) a = c K a (a) a = K a c where the various symbols have their usual meanings. 25. For a fairly concentrated solution of a weak electrolyte AxBy, the degree of dissociation is given as (a) a =

K eq c ( x + y )

(c) a =

K eq c xy

(

(b) a = K eq c x + y -1 x x y y (d) a =

)

1 ( x+ y)

K eq xy c

26. The pH of 0.1 M NH4OH (Kb = 1.8 ¥ M) at 25 °C is (a) 2.87 (b) 1.8 (c) 11.13 (d) 12.2 27. Which of the following shows the highest percentage dissociation? (b) 0.1 M HCN (a) 1.0 M HCN (Kºdiss = 4.0 ¥ 10–10) –4 (d) 0.1 M HNO2 (c) 1.0 M HNO2 (Kºdiss = 4.5 ¥ 10 ) 28. The pH of a 0.02 M Ca(OH)2 at 25°C solution is (a) 12.0 (b) 12.3 (c) 12.6 (d) 13.6 29. A monoprotic acid is 0.001% ionized in 0.1 M of its solution. The ionization constant of the acid is (b) 10–8 M (c) 10–11 M (d) 10–13 M (a) 10–5 M 30. Which one of the following will have the largest value of pH? (a) 1.0 M acetic acid (b) 0.1 M acetic acid (c) 0.01 M acetic acid (d) 0.001 M acetic acid 3 –5 31. The pH of 1 dm solution of 0.5 M acetic acid (K°a = 1.8 ¥ 10 ) is 2.52. If this solution is diluted 4 times, its pH will change by (a) + 0.30 (b) + 0.60 (c) + 0.90 (d) – 1.2 –3 –5 32. The degree of dissociation of 10 M ammonium hydroxide (K°b = 1.8 ¥ 10 ) solution is about (a) 0.25 (b) 0.20 (c) 0.13 (d) 0.09 10–5

Hydrolysis of Salts 33. A solution of 0.1 M CuSO4 solution is (a) acidic in nature (b) alkaline in nature (c) neutral in nature (d) acidic at low temperature, neutral at room temperature and alkaline at high temperature

7.60 Complete Chemistry—JEE Main

34. A solution of NaCN is (a) acidic in nature (b) alkaline in nature (c) neutral in nature (d) acidic at low temperature, neutral at room temperature and alkaline at high temperature 35. A solution of NaCl is (a) acidic in nature (b) alkaline in nature (c) neutral in nature (d) acidic at low temperature, neutral at room temperature and alkaline at high temperature 36. A solution of ammonium chloride is (a) acidic in nature (b) alkaline in nature (c) neutral in nature (d) acidic at low temperature, neutral at room temperature and alkaline at high temperature 37. A solution of ammonium acetate (Ka(acetic acid) = Kb(ammonium hydroxide)) is (a) acidic in nature (b) alkaline in nature (c) neutral in nature (d) acidic at low temperature, neutral at room temperature and alkaline at high temperature 38. A solution of ammonium cyanide (Ka(HCN) < Kb(NH4OH)) is (a) acidic in nature (b) alkaline in nature (c) neutral in nature (d) acidic at low temperature, neutral at room temperature and alkaline at high temperature 39. The expression to compute pH of NH4Cl solution is (a) pH = 12 pK°w – 12 pK°b(NH4OH) – 12 log ([salt]/M) (b) pH =

1 2

pK°w +

1 2

pK°b(NH4OH) –

1 2

log ([salt]/M)

(c) pH =

1 2

pK°w +

1 2

pK°b(NH4OH) +

1 2

log ([salt]/M)

(d) pH =

1 2

pK°b(NH4OH) –

1 2

1 2

log ([salt]/M)

pK°w +

40. The expression to compute pH of sodium acetate solution is (a) pH =

1 2

pK°w –

1 2

pK°a(CH3COOH) –

1 2

log ([salt]/M)

(b) pH =

1 2

pK°w +

1 2

pK°a(CH3COOH) –

1 2

log ([salt]/M)

(c) pH =

1 2

pK°w +

1 2

pK°a(CH3COOH) +

1 2

log ([salt]/M)

(d) pH =

1 2

pK°a (CH3COOH) –

1 2

log ([salt]/M)

1 2

pK°w +

41. Mixing 0.1 mol of NaOH, 0.1 mol of HC2H3O2, and one litre of water, yields a solution which (a) is acidic (b) is neutral (c) is basic (d) may be acidic, neutral or basic depending upon the temperature 42. Which of the following solutions has the maximum pH value? (a) 0.2 M HNO3 (b) 0.2 M HCl (c) 0.2 M HC2H3O2 (d) 0.2 M NaC2H3O2

Chemical and Ionic Equilibria 7.61

43. Which of the following is acidic? (c) NH4Cl solution (d) CH3COONH4 solution (a) KCN solution (b) NaHCO3 solution 44. Which of the following is alkaline? (a) KCl solution (b) CH3COONH4 solution (d) KCN solution (c) FeCl3 solution 45. The pH of 0.1 M solution of the following salts increases in the order (b) HCl < NH4Cl < NaCl < NaCN (a) NaCl < NH4Cl < NaCN < HCl (d) HCl < NaCl < NaCN < NH4Cl (c) NaCN < NH4Cl < NaCl < HCl 46. Which of the following salt solutions will have lowest pH value? (b) CaCl2 (c) Ca(OH)2 (d) CH3COONa (a) CaCO3 47. The expression to be used to calculate the pH of ammonium acetate solution is (b) pH = 12 (pK°w + pK°a + pK°b) (a) pH = 12 (pK°w + pK°a – pK°b) (c) pH =

1 2

(pK°w – pK°a + pK°b)

(d) pH =

1 2

(pK°w – pK°a – pK°b)

where the various symbols have their usual units. 48. The expression to be used to calculate the pH of sodium bicarbonate solution is (a) pH = pK°a1 / pK°a2 (b) pH = pK°a1 + pK°a2 (c) pH =

1 2

(pK°a1 + pK°a2 )

(d) pH =

1 2

(pK°a1 – pK°a2)

where K°a1 and K°a2 stand for the standard ionization constants of H2CO3– and HCO3–, respectively.

Solubility Product 49. (a) Ksp = [Ca2+] [PO3– 4 ]

ÈCa 2+ ˘˚ ÈÎPO34- ˘˚ (b) Ksp = Î [ Ca 3 (PO 4 )2 ]

2 (c) Ksp = [Ca2+]3 [PO3– 4 ]

3 (d) Ksp = [Ca2+]2 [PO3– 4 ]

50. The solubility of mercurous chloride in water will be given as (a) Ksp(Hg2Cl2)

(b)

Ksp(Hg 2 Cl2 ) 4

(c) [Ksp(Hg2Cl2)/4]1/3

(d) [Ksp(Hg2Cl2)]1/3

51. If s0, s1, s2 and s3 are the solubilities of AgCl in water, 0.01 M CaCl2, 0.01 M NaCl and 0.05 M AgNO3 solutions, respectively, then (b) s0 > s2 > s1 > s3 (c) s0 > s2 > s3 > s1 (d) s0 > s1 = s2 > s3 (a) s0 > s1 > s2 > s3 52. Given: Ksp(AgCl) = 1.7 ¥ 10–10 M2, Ksp(Ag2CrO4) = 1.9 ¥ 10–12 M3 AgNO3 is added to a solution containing 0.1 M each of NaCl and Na2CrO4? (a) chloride ion (b) chromate ion (c) Both Cl– and CrO2– precipitate together (d) cannot be predicted 4 53. The solubility of magnesium hydroxide in pure water is 8.1 ¥ 10–5 mol per 500 mL of water. The solubility product of magnesium hydroxide is (b) 5.51 ¥ 10–13 M3 (c) 8.84 ¥ 10–12 M3 (d) 1.77 ¥ 10–11 M3 (a) 6.72 ¥ 10–9 M2 –17 2 54. Given: Ksp(AgI) = 8.5 ¥ 10 M . The solubility of AgI in 0.1 M KI solution is (b) 8.5 ¥ 10–16 M (c) 8.5 ¥ 10–17 M (d) 8.5 ¥ 10–18 M (a) 8.5 ¥ 10–15 M 55. The concentration of S2– in a 0.1 M H2S (Ka1Ka2 = 1.1 ¥ 10–21 M2) which is also 0.1 M in H+ is (a) 1.1 ¥ 10–19 M (b) 1.1 ¥ 10–20 M (c) 1.1 ¥ 10–21 M (d) 1.1 ¥ 10–22 M 56. The pH of Mg(OH)2 solution is 10.45 at 25 °C. The solubility product of magnesium hydroxide will be (b) 1.12 ¥ 10–11 M3 (c) 3.36 ¥ 10–11 M3 (d) 5.60 ¥ 10–12 M3 (a) 2.24 ¥ 10–11 M3 –23 5 57. The solubility product of A2X3 is 1.08 ¥ 10 M . Its solubility is (b) 1.0 ¥ 10–4 M (c) 1.0 ¥ 10–5 M (d) 1.0 ¥ 10–6 M (a) 1.0 ¥ 10–3 M

7.62 Complete Chemistry—JEE Main

58. The molarity of Mg 2+ ions in a saturated solution of Mg 3(PO 4) 2 whose solubility product is 1.08 ¥ 10–13 M5 is (b) 2.0 ¥ 10–3 M (c) 3.0 ¥ 10–3 M (d) 4.0 ¥ 10–3 M (a) 1.0 ¥ 10–3 M 2+, Zn2+, Mn2+ and Co2+ is saturated with 59. –14 2– H2S (which provides 1.0 ¥ 10 M of S in the solution) is (b) Zn2+ (c) Mn2+ (d) Co2+ (a) Pb2+ –27 2 –21 2 –13 (Given: Ksp(PbS) = 2.3 ¥ 10 M , Ksp(ZnS) = 1.0 ¥ 10 M , Ksp(MnS) = 7.9 ¥ 10 M2 and Ksp(CoS) = 7 ¥ 10–23 M2.) –3 60. A solution is saturated with respect to SrCO3 and SrF2. The [CO2– 3 ] was found to be 1.2 ¥ 10 M. The concentration – in the solution would be of F (b) 3.7 ¥ 10–2 M (c) 5.8 ¥ 10–7 M (d) 2.6 ¥ 10–2 M (a) 1.3 ¥ 10–3 M (Given: Ksp(SrCO3) = 7.0 ¥ 10–10 M2 and Ksp(SrF2) = 7.9 ¥ 10–10 M3.) 61. The maximum pH of a solution which have 0.10 M of Mg2+ from which Mg(OH)2 is not precipitated is (a) 4.96 (b) 6.96 (c) 7.04 (d) 9.04 –11 3 (Given: Ksp(Mg(OH)2) = 1.2 ¥ 10 M .) 62. Which of the following is most soluble? (a) CuS (K°sp = 8 ¥ 10–37) (c) Bi2S3 (K°sp = 1 ¥ 10–70)

(b) Ag2S (K°sp = 6 ¥ 10–51) (d) MnS (K°sp = 7 ¥ 10–16)

Buffer Solution 63. The pH of a solution obtained by mixing 20 mL of 0.02 M sodium acetate and 10 mL of 0.01 M acetic acid (Ka = 1.8 ¥ 10–5 M) is (a) 4.75 (b) 5.05 (c) 4.45 (d) 5.34 64. Which of the following constitutes a buffer solution when dissolved in 500 mL water? (a) 0.05 mol NaOH and 0.05 mol HCl (b) 0.05 mol NaCl and 0.05 mol HCl (c) 0.05 mol propanoic acid and 0.05 mol sodium propanoate (d) 0.05 mol acetic acid and 0.05 mol sodium hydroxide 65. Which of the following can be used as a buffer solution? (a) Sodium chloride solution (b) Ammonium chloride solution (c) Sodium acetate solution (d) Ammonium acetate solution 66. The solution containing weak acid and salt of its conjugate base has maximum buffer capacity when (a) [salt] = [acid] (b) [salt] > [acid] (c) [salt] < [acid] (d) [salt] + [acid] is minimum 67. Adding sodium hydroxide to a solution of acetic acid (b) increases [H3O+] (c) increases [HC2H3O2] (d) decreases the pH (a) increases [C2H3O–2] 68. In a buffer solution, pH = pKa when the solution contains (a) [salt] = [acid] (b) [salt] > [acid] (c) [salt] < [acid] (d) [salt] + [acid] has a maximum value 69. Which of the following buffer solutions containing the indicated ratio of [salt]/[acid] has a maximum buffer capacity? (a) 1/2 (b) 1 (c) 2/1 (d) 4/1 70. A buffer solution is prepared by using a weak acid having pKa = 4. The approximate range it can be used effectively is (a) 2 to 4 (b) 3 to 4 (c) 3 to 5 (d) 4 to 5 71. A buffer solution of a weak acid and salt of its conjugate base is diluted to twice its volume. Its pH (a) increases (b) decreases (c) remains the same (d) does change but cannot be predicted

Chemical and Ionic Equilibria 7.63

Indicator 72. An acid-base indicator has Kºa = 3.0 ¥ 10–5. The acid form of the indicator is red and the basic form is blue. The change in [H+] required to change the indicator from 75% red to 75% blue is (b) 9 ¥ 10–5 M (c) 1 ¥ 10–5 M (d) 4 ¥ 10–5 M (a) 8 ¥ 10–5 M 73. Indicator range of phenolphthalein is (a) 4.2 to 6.3 (b) 5.0 to 7.0 (c) 8.3 to 10.0 (d) 12.0 to 14.0 74. Indicator range of methyl red is (a) 4.2 to 6.3 (b) 5.0 to 7.0 (c) 8.3 to 10.0 (d) 12.0 to 14.0 75. The indicator thymol blue has pH = 2.0 when half of the indicator is present in the ionized form. The pK°Indicator is (a) 1.0 (b) 2.0 (c) 3.0 (d) 4.0 76. The indicator constant of an indicator is 3.0. The percentage of the indicator in the unionized form in a solution of pH = 2.6 is (a) 40% (b) 55% (c) 62% (d) 71% 77. The indicator constant of an indicator is 3.0. The percentage of the indicator in the unionized form in a solution of 3.4 is (a) 28.6% (b) 35% (c) 42% (d) 51% 78. The indicator constant of an indicator is 10–4. If pH of solution in which this indicator is present is 3.5, which of the following facts will be correct? (a) Ionized indicator is present in the larger amount (b) Unionized indicator is present in the larger amount (c) Both ionized and unionized indicator are present in equal amounts (d) Percentage of indicator present in ionized form cannot be predicted

Acids and Bases 79. The conjugate acid of NH2– is (b) NH2OH (c) NH+4 (d) N2H4 (a) NH3 80. The compound that is not a Lewis acid is (b) AlCl3 (c) BeCl3 (d) SnCl4 (a) BF3 81. Which of the following acids is expected to be strongest? (b) H3PO4 (c) H3AsO3 (d) H3AsO4 (a) H3PO3 82. In the given anions, the weakest Brönsted base is (b) ClO2– (c) ClO–3 (d) ClO– (a) ClO–4 83. Which of the following acts as a Brönsted acid as well as a Brönsted base? (b) OH – (c) NH3 (d) HCO3– (a) Na2CO3 84. In the reaction HCN + H2O H3O+ + CN–, the conjugate acid-base pair is (a) HCN, H2O (b) H3O+, CN – (c) HCN, H3O+ (d) HCN, CN– 85. Which of the following statements is not correct? (a) A substance which supplies OH– in aqueous medium is a base. (b) A substance which can accept a proton in aqueous medium is a base. (c) A substance which can donate a pair of electrons is a base. (d) A substance which can accept a pair of electrons is a base. 86. Which of the following species is not a Lewis acid? (a) BF3 (b) Cu2+ (c) NH3 (d) AlCl3 87. In aqueous solution, which of the following orders regarding acid strength is correct? (b) HCN < CH3COOH < HClO4 < HCl (a) HCN < CH3COOH < HCl < HClO4 (d) HClO4 < CH3COOH < HCl < HCN (c) HCN < HClO4 < HCl < CH3COOH

7.64 Complete Chemistry—JEE Main

88. In aqueous solution, which of the following orders regarding base strength is correct? (b) CH3COO– > HS– > NH3 > CO2– (a) CH3COO– < HS– < NH3 < CO32– 3 (c) CH3COO– > HS– < NH3 < CO32– (d) CH3COO– > HS– > CO32– > NH3 89. The decreasing base strength of OH–, NH2–, HC∫∫C– and CH3CH2– is (b) HC∫∫C– > CH3CH2– > NH2– > OH– (a) CH3CH2– > NH2– > HC∫∫C– > OH– – – – – (c) OH > NH2 > HC∫∫C > CH3CH2 (d) NH2– > HC∫∫ C– > OH– > CH3CH2– 90. The correct order of increasing base strength of Cl–, ClO4– and CH3COO– is (b) CH3COO– < Cl– < ClO4– (a) ClO4– < CH3COO– < Cl– – – – (c) ClO4 < Cl < CH3COO (d) Cl– < ClO4– < CH3COO– 91. The strongest base amongst OH–, F–, NH2– and CH3– is (a) NH2– (b) CH3– (c) F– (d) OH–

Problems for Revision 92. 93. 94. 95. 96. 97. 98. 99.

100. 101.

102. 103. 104.

Based on acetic acid (K°a = 1.8 ¥ 10–5) solution, answer the following questions. The degree of dissociation of 0.05 M acetic acid solution is (a) 0.0019 (b) 0.019 (c) 0.19 (d) 0.45 The concentration of H+ in 0.05 M acetic acid solution is (b) 9.5 ¥ 10–3 M (c) 4.5 ¥ 10–4 M (d) 9.5 ¥ 10–4 M (a) 4.5 ¥ 10–3 M The pH of 0.05 M acetic acid solution is (a) 3.02 (b) 3.52 (c) 4.02 (d) 4.52 If 0.05 M acetic acid solution is diluted 10 times, its degree of dissociation changes to (a) 0.02 (b) 0.04 (c) 0.06 (d) 0.08 If 0.05 M acetic acid solution is diluted 10 times, its pH becomes (a) 3.53 (b) 2.22 (c) 3.22 (d) 4.22 It is desired to change [H+] of 0.05 M acetic acid solution to its half value, the dilution factor of the solution is (a) 4.0 (b) 5 (c) 6.0 (d) 6.5 It is desired to change pH of 0.05 M acetic acid solution to 1.5 times pH value, the solution should be diluted to (a) 540.5 (b) 870.5 (c) 1033.3 (d) 1150.0 If to 100 mL of 0.1 M acetic acid solution, 0.82 g of sodium acetate is added, the pH of the solution (assume no dilution factor) becomes (a) 3.75 (b) 4.75 (c) 5.25 (d) 5.75 If the solution obtained in Q.99 is diluted 10 times, it pH becomes (a) 4.75 (b) 3.75 (c) 5.75 (d) 6.25 In going from solution of Q.99 to the solution of Q.100, its buffer capacity (a) increases (b) decreases (c) remains constant (d) may increase or decrease depending upon temperature The buffer range of the acidic buffer obtained from acetic acid and sodium acetate is (a) 3.0 to 5.0 (b) 3.2 to 5.2 (c) 3.75 to 5.75 (d) 4.2 to 6.2 If to a solution of Q. 99, 0.002 mol of HCl is added, its pH changes to (a) 2.0 (b) 3.0 (c) 3.5 (d) 4.6 If to a 100 mL of solution of Q. 100, 0.002 mol of HCl is added, its pH changes to (a) 1.5 M (b) 2.0 (c) 3.0 (d) 4.0

Chemical and Ionic Equilibria 7.65

105. The pH of a solution of a salt increases with increase in the concentration of the salt. The salt may be (a) ammonium chloride (b) sodium acetate (c) ammonium acetate (d) sodium bicarbonate 106. The pH of a solution of a salt decreases with increase in the concentration of the salt. The salt may be (a) ammonium chloride (b) sodium acetate (c) ammonium acetate (d) sodium bicarbonate 107. The pH of a solution of a salt remains more or less constant with change in the concentration of the salt. The salt may be (a) ammonium chloride (b) sodium acetate (c) ammonium acetate (d) sodium sulphate 108. The pH of sodium bicarbonate solution depends on (a) the ionization constant of H2CO3 (b) the ionization constant of HCO 3– (c) the ionization constants of both H2CO3 and HCO3– (d) the concentration of sodium bicarbonate 109. The pH of a solution of salt formed from a weak acid and a weak base decreases with (b) increase in pKb of base (a) increase in pKa of acid (c) increase in concentration of the salt (d) decrease in concentration of the salt. 110. The solubility of Zn(OH)2 in water with increase in pH of the solution (from acidic to neutral to alkaline conditions) shows the pattern (a) initially decreases followed by an increase (b) initially increases followed by a decrease (c) remains constant 111. In the titration of H2 (a) Ka1/Ka2 £ 1

(b) Ka1/Ka2 £ 10

Vbase are observed provided (c) Ka1/Ka2 £ 100 (d) Ka1/Ka2 ≥ 1000

112. The solubility of sparingly soluble salt AxBy in water is given by the expression 1/( x + y ) (a) s = Ksp

x + y) (b) s = xx yy K 1/( sp

(c) s = (Ksp/xxyy)1/(x + y)

(d) s = (Ksp/xyyx)1/(x + y)

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79. 85.

(b) (d) (b) (b) (b) (a) (c) (c) (c) (b) (d) (a) (c) (a) (d)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 86.

(d) (c) (c) (d) (c) (c) (b) (d) (c) (b) (d) (a) (a) (c) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75. 81. 87.

(c) (b) (b) (b) (d) (a) (a) (b) (b) (c) (d) (b) (b) (b) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76. 82. 88.

(b) (b) (b) (d) (c) (b) (c) (b) (a) (c) (c) (c) (d) (a) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77. 83. 89.

(c) (b) (c) (a) (c) (c) (c) (a) (d) (a) (d) (c) (a) (d) (a)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 90.

(b) (b) (a) (a) (d) (a) (d) (c) (b) (b) (a) (a) (b) (d) (c)

7.66 Complete Chemistry—JEE Main

91. 97. 103. 109.

(b) (b) (d) (b)

92. 98. 104. 110.

(b) (c) (b) (a)

93. 99. 105. 111.

(d) (b) (b) (d)

94. 100. 106. 112.

(a) (a) (a) (c)

95. (c) 101. (b) 107. (c)

96. (a) 102. (c) 108. (c)

HINTS AND SOLUTIONS 3. pK°w = pH + pOH For water pH = pOH. Hence, pH = 12 pK°w = 6.63 4. pH changes from 7 to 11. Hence, change is 4. 5. pH equal to 13 implies that [H+] = 10–13 mol L–1 Amount of H+ in 1 mL solution = 10–16 mol Number of H+ in 1 mL solution = (10–16 mol) (6.022 ¥ 1023 mol–1) = 6.022 ¥ 107 6. 7. 8. 9. 10.

For neutral water pH = 12 pKw = 6.77. Since given pH (= 7) is greater than 6.77, it is an alkaline solution. The statement (d) is applicable only at 25°C and is not independent of temperature. The solution will remain acidic as it represents 10–8 M HCl. Hence, its pH will be less than 7 at 25 °C. Solution will become 10–7 M in NaOH. Since the solution remains alkaline, its pH will be more than 7. For pure water, [H3O+] = [OH–] Kw= [H3O+] [OH–] = 10–12 (mol dm–3)2;

pK°w = – log K°w = 12

11. Kw increases with increase in temperature. pK°w (= – log K°w) will decrease with increase in Kw. 12. For a pure water, [H+] increases with increase in temperature due to more ionization of water. pH (= – log [H+]) will decrease. 15. pH equal to 6.5 means [H+] > 10–7 M, i.e. there occurs more ionization of water as compared to that occurring at 25 °C. More ionization means temperature is greater than 25 °C 19. [H+] in pH = 1.0 is 10–1 M. Solution becomes diluted ten times. Hence, its [H+] becomes 10–2 M. Its pH would become 2. 20. The pH of choice a will be 7 as the acid completely neutralizes the base. The choice b will contain 10 mL of excess 0.1 M HCl. The total volume of solution becomes 100 mL. Hence, its effective molarity will be 0.01 M and thus its pH = 2. The choice c will contain 80 mL of excess 0.1 M NaOH. Hence, its pH > 7. The choice d will contain 50 mL of excess 0.2 M HCl. Because of dilution to 100 mL, its effective molarity will be 0.1 M and hence pH = 1. 21. For pOH = 13, [OH–] = 10–13 M and [H+] = 10–1 M Ê 10-1 mol ˆ Number of H+ in 1 cm3 solution = Á 3 3 ˜ (6.022 ¥ 1023 mol–1) = 6.022 ¥ 109 cm–3 Ë 10 cm ¯ 22. For pH = 3, [H+] = 10–3 M. For pH = 6, [H+] = 10–6 M. Hence, [H+] has to be decreased 1000 times. CH3COO– + H+

23. CH3COOH

Ka =

[CH3COO - ] [H + ] [H + ]2  [CH3COOH] [CH3COOH]0

[H+] = (Ka[CH3COOH]0]1/2 = (1.8 ¥ 10–5 ¥ 0.1)1/2 M = 1.34 ¥ 10–3 M pH = – log (1.34 ¥ 10–3) = 2.87 25. Ax By

c(1 – a)

x Ay+ + yBx– c(xa)

Keq=

c(ya)

(c x a ) x (c y a ) y  c x+y–1 x xyy ax+y c (1 - a )

or

Ê

K eq

ˆ

a = Á x + y -1 x y ˜ Ëc x y ¯

1/( x + y )

Chemical and Ionic Equilibria 7.67

26. NH4OH

or

NH+4 + OH– [NH +4 ] [OH - ] [OH - ]2 Keq =  [NH 4 OH] [NH 4 OH]0

[OH–] =

(since [NH+4] = [OH–])

K eq [NH 4 OH ]0 = 1.8 ¥ 10-5 ¥ 0.1 M 2 = 1.34 ¥ 10–3 M

pOH = – log {[OH–]/M} = 2.87;

pH = 14 – pOH = 11.13

27. Ostwald dilution law; more dilute the solution, more the dissociation. 28. Assuming complete dissociation pOH = 1.40; Thus pH = 14 – pOH = 12.6 [OH–] = 2 ¥ 0.02 M; H+ + A–

29. HA c(1 – a)

ca

Ka =

[H + ] [A - ] (ca ) 2 =  ca 2 c (1 - a ) [HA]

ca

Ka = (0.1 M) (10–5)2 = 10–11 M

Hence,

30. The less concentrated solution will produce lesser H+ and hence larger pH. 31. Solution concentration becomes 0.125 M a=

K ∞a 1.8 ¥ 10-5 M = = 0.012 ; c 0.125 M

pH= –log (0.0015) = 2.82;

[H+] = ca = (0.125 M) (0.012) = 0.0015 M

D(pH) = 2.82 – 2.52 = 0.30

1.8 ¥ 10-5 M Ka = 0.134 = 10 -3 M c Solution is acidic due to the hydrolysis reaction Cu2+ + H2O Cu(OH)+ + H+ Solution is alkaline due to the hydrolysis reaction CN– + H2O HCN + OH– + – Solution is neutral because no hydrolysis of Na and Cl take place. NH4OH + H+ Solution is acidic due to the hydrolysis reaction NH4+ + H2O + A solution of ammonium acetate is neutral because both NH4 and Ac– ions are hydrolysed to equal extent. Solution is alkaline as CN– is hydrolysed more than NH4+ ions. NH+4 + H2O NH4OH + H+

32. a 33. 34. 35. 36. 37. 38. 39.

2

+ + K w [ NH 4 OH ] ÈÎH ˘˚ ÈÎH ˘˚ =  K h= Kb ÈÎ NH 4+ ˘˚ ÈÎ NH 4+ ˘˚ 0

[H+]2 =

or

Kw [NH4]0 Kb

2 log ([H+]/M) = log K°w – log K°b + log ([NH4+]/M); or – log ([H+]/M)= – pH = 40. Ac– + H2O

1 2

pK°w –

1 2

pK°b –

1 2

1 2

log K°w +

1 2

log K°b –

1 2

log ([NH4+]/M)

log ([salt]/M)

HAc + OH– 2

K w [ HAc] ÈÎHO ˘˚ ÈÎOH ˘˚ =  Kh = ; Ka ÈÎAc- ˘˚ ÈÎAc- ˘˚ 0

which on taking log and multiplying by – 1 gives pH =

K [OH ] = w [Ac–]0 ; Kb – 2

1 2

pK°w +

1 2

pK°a +

1 2

2

Kw Ê Kw ˆ – ÁË + 2 ˜¯ = K [Ac ]0 [H ] a

log ([salt]/M)

42. Sodium acetate solution will be alkaline and thus will have maximum value of pH. 45. HCl is the strongest acid. Its pH will be minimum. NaCl does not hydrolyse in solution. Its pH = 7 NH +4 in NH4Cl on hydrolysis produces H+. Hence, its pH will be slightly less than 7. CN– in NaCN on hydrolysis produces OH–. Hence, its pH will be slightly greater than 7.

7.68 Complete Chemistry—JEE Main

46. In CaCO3, both ions show hydrolysis. In CaCl2, only Ca2+ is hydrolysed. Both Ca(OH)2 and CH3COONa will be alkaline in nature. Hence, the lowest pH will be that of CaCl2 solution. NH4OH + HAc 47. NH +4 + Ac– + H2O Kh =

Kh =

Also

[NH 4 OH ] [NH +4 ][OH - ]

pH = 48. HCO–3 + H2O

[HAc] -

+

[Ac ][H3O ]

Kw Ka K b

◊ [H3O + ][OH - ] =

[H3O + ] = Ka

or

Kw Ka K b

H3O+ + CO2– 3;

Kh =

Ka2 =

Kh/Ka2 = Kw/Ka1Ka2. Hence,

50. Hg2Cl2

– Hg2+ 2 + 2 Cl

x

51. s0 =

Ksp ;

H 3O + =

or

[H 2 CO3 ][OH - ] [HCO3- ]

=

K w Ka Kb

Kw K a1

[H3O + ][CO32- ] [HCO3- ]

Kh Kw [H CO ] [OH - ] Kw = = 2 -3  + Ka 2 K a1 K a 2 [CO3 ] [H3O ] [H3O + ]2

Also

Kw Kb Ka

1 (pK°w + pK°a – pK° ) b 2

H2CO3 + OH– ;

HCO–3 + H2O

[Ac- ]2

[ NH +4 ][Ac- ]

[HAc] = [Ac- ]

Hence

[HAc]2

[ NH 4 OH][HAc]

[H3O+]2 = Ka1 Ka2

– 2 2 Ksp = [Hg2+ 2 ] [Cl ] = (x) (2x) ;

(since [H2CO3] ; [CO32–])

pH =

or

1 (pK a∞1 + pK a∞2 ) 2

13 x = ( Ksp / 4)

2x

s1 = Ksp/0.02 M;

s2 = Ksp/0.01 M;

s3 = Ksp/0.05 M. Obviously s0 > s2 > s1 > s3.

-10 52. [Ag+] needed to start precipitation of Cl– = 1.7 ¥ 10 M 0.1

[Ag+] needed to start precipitation of CrO2– 4 =

1.9 ¥ 10-12 M 0.1 –

to precipitate.

53. Solubility per litre of water = 16.2 ¥ 10–5 mol L–1 Ksp = [Mg2+] [OH–]2 = (16.2 ¥ 10–5 M) (2 ¥ 16.2 ¥ 10–5 M)2 = 1.77 ¥ 10–11 M3 54. [Ag+] =

55. [S2–] =

Ksp [I- ]

=

8.5 ¥ 10-17 M 2 = 8.5 ¥ 10-16 M 0.1 M

K a1 K a2 [H + ]2

56. pH = 10.45;

[H 2S] =

(1.1 ¥ 10-21 M3 )(0.1 M ) = 1.1 ¥ 10–20 M [0.1 M ]2

pOH = 14 – 10.45 = 3.55;

[OH–]/M = antilog (– 3.55) = 2.82 ¥ 10–4

Ê 2.82 ¥ 10-4 ˆ Ksp = [Mg2+] [OH–]2 = Á M˜ (2.82 ¥ 10-4 M ) 2 = 1.12 ¥ 10–11 M 2 Ë ¯

Chemical and Ionic Equilibria 7.69

57. A2X3

2A3+ + 3X2– (2x)

Ksp = [A3+]2 [X2–]3 = (2x)2 (3x)3 = 108x5

(3x)

1/ 5 Ê 1.08 ¥ 10-23 5 ˆ Ê Ksp ˆ x= Á M ˜ =Á ˜ Ë 108 ¯ 108 Ë ¯

58. Mg3(PO4)2 3Mg2+ + 2 PO3– 4 ; If x is the solubility of Mg3(PO4)2, then (3x) (2x) = 1.08 ¥ 10 3

2

–13

1/ 5

= 1.0 ¥ 10-5 M 2 Ksp = [Mg2+]3 [PO3– 4 ]

Ê 1.08 ¥ 10-13 5 ˆ x= Á M ˜ 108 Ë ¯

5

M;

1/ 5

= 0.001 M

[Mg2+] = 3x = 3 ¥ 10–3 M 59. Lesser the value of Ksp, earlier the salt precipitates. Hence, Pb2+ 60. [Sr2+] =

Ksp (SrCO3 ) [CO32- ]

=

7.0 ¥ 10-10 M 2 = 5.8 ¥ 10–7 M 1.2 ¥ 10-3 M 1/ 2

[F–]

È Ksp (SrF2 ) ˘ = Í ˙ 2+ Î [Sr ] ˚

Ê 7.9 ¥ 10-10 M 3 ˆ =Á Ë 5.8 ¥ 10-7 M ˜¯

1/ 2

= 3.7 ¥ 10–2 M

61. For Mg(OH)2 not to be precipitated 1/ 2

È Ksp (Mg (OH)2 ) ˘ [OH–] < Í ˙ [Mg 2+ ] Î ˚ pOH > 4.96;

Ê 1.2 ¥ 10-11 M 3 ˆ [OH–] < Á ˜¯ 0.10 M Ë

;

1/ 2

= 1.095 ¥ 10–5 M

pH < 14 – 4.96 = 9.04

[salt] (0.02 ¥ 2 3) M = – log (1.8 ¥10–5) + log = 4.74 + 0.60 = 5.34 (0.01 ¥ 1 3) M [acid] [salt ] For an acidic buffer solution pH = pK∞a + log [acid] Obviously, pH = pK°a if [salt] = [acid]. The buffer capacity is maximum when the ratio of concentrations of salt to acid is one. It decreases on either sides of this ratio. The approximate buffer range is in the range pK°a – 1 to pK°a + 1. The pH of a buffer solution is not changed on dilution as the ratio of concentrations of salt to acid remains unchanged. For 75% red, we have [salt] 25 = – log (3 ¥ 10–5) + log ÊÁ ˆ˜ = 4.52 – 0.48 = 4.04 pH = pK°a + log [acid] Ë 75 ¯ [H+] = 10–pH = 10–4.04 = 9.12 ¥ 10–5 M

63. pH = pK°a + log 68. 69. 70. 71. 72.

For 75% blue, we have 75 pH = – log (3 ¥ 10–5) + log ÊÁ ˆ˜ = 4.52 + 0.48 = 5.00; Ë 25 ¯ The change in [H+] is D [H+]= 9.12 ¥ 10–5 M – 1.0 ¥ 10–5 M = 8.12 ¥ 10–5 M 75. pH = pK°HIn + log 76. 2.6 = 3.0 + log

[ In - ] [ HIn ]

[ In - ] [ HIn ]

.

,

since [In–] = [HIn], Hence

[ In - ] [ HIn ]

pK°HIn = pH

= antilog (–0.4)

Per cent of HIn = (1/1.40) (100) = 71.4%

[H+] = 10–5 M

or

[ In - ] [ HIn ]

= 0.40

or

[ In - ] + [HIn ] [ HIn ]

= 1.40

7.70 Complete Chemistry—JEE Main

77. 3.4 = 3.0 + log

[ In - ] [ HIn ]

.

Hence,

[ In - ] [ HIn ]

= antilog (0.4)

or

[ In - ] [ HIn ]

= 2.5

or

[ In - ] + [HIn ] [ HIn ]

= 3.5

Per cent of HIn = (1/3.5) (100) = 28.6% H+

78. HIn + In–; Since pH < pK°HIn, the solution contains larger H+ than that obtained from ionization of HIn alone. Hence, the above equilibrium is shift to left, thus more of unionized indicator is present. 81. Phosphorus is more electronegative than arsenic. 82. Strongest acid amongst HClO, HClO2, HClO3 and HClO4 is HClO4. Hence, the conjugate base ClO–4 will be the weakest. 86. :NH3 contains a lone pair of electrons. It can be donated and hence it acts as a Lewis base. 87. HCN is the weakest acid. The next stronger acid is CH3COOH. Out of HCl and HClO4, the former is a weaker acid. 88. Weaker the acid, stronger the conjugate base. The order of acid strength is CH3COOH > H2S > NH +4 > HCO3–. Hence, the conjugate base follows the reverse order. 89. The weaker the acid, stronger the conjugate base. The order of acid strength is CH3CH3 < NH3 < HC ∫ CH < H2O. Hence, the conjugate base follows the order CH3CH2– > NH2– > HC∫C– > OH–. 90. The correct order of acid strength is CH3COOH < HCl < HClO4. Hence, the conjugate base follows the order CH3COO– > Cl– > ClO 4–. 91. The weakest acid amongst H2O, HF, NH3 and CH4 is CH4. Hence, the strongest base will be CH 3–. 92. CH3COOH c(1 – a)

CH3COO– + H+ ca

ca

[CH3COO - ] [H + ] (ca ) (ca )  ca 2; K a= = c(1 - a ) [CH3COOH]

a=

Ka / c =

1.8 ¥ 10-5 / 0.05 = 0.019

93. [H+] = ca = (0.05 M) (0.019) = 9.5 ¥ 10–4 M 94. pH = – log {[H+]/M} = – log (9.5 ¥ 10–4) 95. c = 0.005 M;

a=

3.02

K a / c = ( 1.8 ¥ 10-5 / 0.005 ) = 0.06

96. pH = – log {(0.06) (0.005)} = 3.53 97. [H+] = 0.00095 M/2 = 0.000425 M CH3COOH K a=

CH3COO– + H+ [CH3COO - ] [H + ] [H + ]2  [CH3COOH] [CH3COOH ]0

[CH3COOH]0= [H+]2/Ka = (0.000425 M)2/(1.8 ¥ 10–5 M) = 0.01 M Dilution factor= 98. pH = 1.5 ¥ 3.02 = 4.53

0.05 M =5 0.01 M

[H+]= (10–pH) M = 2.95 ¥ 10–5 M; c = [H+]2/Ka = (2.95 ¥ 10–5 M)2/(1.8 ¥ 10–5 M) = 4.84 ¥ 10–5 M 0.05 M Dilution factor = = 1033.3 4.84 ¥ 10-5 M 99. M(CH3COONa) = 82 g mol–1 n(CH3COONa)= m/M = 0.82 g / 82 g mol–1 = 0.01 mol c(CH3COONa)= n/V = 0.01 mol/0.1 L = 0.1 M pH = pKa + log [salt]/[acid] = –log (1.8 ¥ 10–5) + log (0.1/0.1) = 4.75 100. pH of buffer solution does not change on dilution. 101. Buffer capacity decreases on diluting the solution.

Chemical and Ionic Equilibria 7.71

102. Buffer range, pH = pKa ± 1. Thus, buffer range is from 3.75 to 5.75. 103. On adding 0.002 mol HCl, 0.002 mol of sodium acetate is changed into acetic acid. n(CH3COONa) = (0.1 M) (0.1 dm3) – 0.002 mol = 0.008 mol n(CH3COOH) = (0.1 M) (0.1 dm3) + 0.002 mol = 0.012 mol n (salt) 0.008 ˆ pH = pK°a + log = 4.75 + log ÊÁ ˜ = 4.75 – 0.176 = 4.57 Ë n (acid) 0.012 ¯ 104. In the solution of Q.100, [CH3COOH] = [CH3COONa] = 0.01 M n(CH3COOH) = n(CH3COONa) = (0.01 M) (0.1 dm3) = 0.001 mol On adding 0.002 mol HCl, the entire CH3COONa is replaced by CH3COOH. Thus, solution contains 0.002 mol acetic acid and 0.001 mol HCl. The solution concentrations are [CH3COOH] = 0.02 M and [HCl] = 0.01 M

105.

106.

107.

108.

109.

110. 112.

The major source of H+ in the solution will be HCl. Hence, pH = – log (0.01) = 2 For a salt formed from strong base and weak acid, pH of the solution is given by 1 pH = [pK°w + pK°a + log (c/c°)] (where c° = 1 M) 2 For a salt formed from weak base and strong acid, pH of the solution is given by 1 pH = [pK °w – pK °b – log (c/c°)] (where c° = 1 M) 2 For a salt formed from weak acid and weak base, pH of the solution is given by 1 pH = (pK °w + pK °a – pK °b) 2 For a salt such as NaHCO3, pH of the solution is given by 1 pH = (pK°a1 + pK °a2) 2 where Ka1 and Ka2 are the ionization constants of H2CO3 and HCO3–, respectively. For a salt formed from a weak acid and a weak base, pH of the solution is given by 1 (pK °w + pK °a – pK °b) pH = 2 The solubility of an amphoteric salt is larger in acidic and alkaline media as compared to the neutral medium. Ka1/Ka2 ≥ 103 y+ x– A xB y xA + yB . If s is the solubility of the salt, then in solution, we will have xAy+ + yB x–; Ksp = [Ay+ ]x [Bx– ] y = (xs)x (ys)y; This gives s = (Ksp /xx yy )1/(x + y). A xB y (xs)

(ys)

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

(b) H2O (c) H2S (d) NH3 (a) PH3 2. When rain is accompanied by a thunderstorm the collected rain water will have a pH value (a) which depends on the amount of dust in air (b) slightly lower than that of rain water without thunderstorm (c) slightly higher than that when the thunderstorm is not there

[2003]

[2003]

7.72 Complete Chemistry—JEE Main

3. The solubility of a sparingly soluble salt AB2 in water is 1.0 ¥ 10–5 M. Its solubility product will be (b) 4 ¥ 10–15 M3 (c) 4 ¥ 10–10 M3 (d) 1 ¥ 10–15 M3 [2003] (a) 1.0 ¥ 10–10 M3 4. The conjugate base of H2PO4– is (b) P2O5 (c) H3PO4 (d) HPO2– [2004] (a) PO3– 4 4 –1 5. The molar solubility (in mol L ) of a sparingly soluble salt MX4 is s. The corresponding solubility product is Ksp. The solubility s is given in terms of Ksp by the relation (a) s = (Ksp/128)1/4 (b) s = (128 Ksp)1/4 (c) s = (256 Ksp)1/5 (d) s = (Ksp/256)1/5 [2004] 6. Hydrogen-ion concentration in mol/L in a solution of pH = 5.4 will be (a) 3.68 ¥ 10–6 (b) 3.98 ¥ 10–6 (c) 3.98 ¥ 108 (d) 3.88 ¥ 106 [2005] –12 7. The standard solubility product of a salt having general formula MX2 in water 4 ¥ 10 . The concentration of M2+ ions in the aqueous solution of the salt is (b) 4.0 ¥ 10–6 M (c) 2.0 ¥ 10–6 M (d) 1.0 ¥ 10–4 M [2005] (a) 1.6 ¥ 10–4 M – 8. What is the conjugate base of OH ? (b) H2O (c) O (d) O2– [2005] (a) O2 9. The pK°a of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is (a) 4.5 (b) 2.5 (c) 9.5 (d) 7.0 [2007] 10. In a saturated solution of the sparingly soluble electrolyte AgIO3 (relative molecular mass = 283) the equilibrium which sets in is AgIO3 (s)  Ag + (aq ) + IO3- (aq )

12.

13.

14.

15.

If the standard solubility product constant K°sp of AgIO3 at a given temperature is 1.0 ¥ 10–8, what is the mass of AgIO3 contained in 100 mL of its saturated solution? (a) 28.3 ¥ 10–2 g (b) 2.83 ¥ 10–3 g (c) 1.0 ¥ 10–7 g (d) 1.0 ¥ 10–4 g [2007] –5 –10 respectively, 2A are 1.0 ¥ 10 and 5.0 ¥ 10 overall standard dissociation constant of acid will be (a) 5.0 ¥ 10–5 (b) 5.0 ¥ 1015 (c) 5.0 ¥ 10–15 (d) 0.2 ¥ 1015 [2007] The pK°a of a weak acid, HA, is 4.80. The pK°b of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be (a) 9.22 (b) 9.58 (c) 4.79 (d) 7.01 [2008] Four species are listed below. (i) HCO–3, (ii) H3O+, (iii) HSO–4 , (iv) HSO3F Which one of the following is the correct sequence of their acid strength? (a) (iii) < (i) < (iv) < (ii) (b) (iv) < (ii) < (iii) < (i) (c) (ii) < (iii) < (i) < (iv) (d) (i) < (iii) < (ii) < (iv) [2008] –4 2+ Solid Ba(NO3)2 is gradually dissolved in a 1.0 ¥ 10 M Na2CO3. solution. At what concentration of Ba will a precipitate begin to form? (Given: Ksp = 5.1 ¥ 10–9 M2 for BaCO3.) (a) 8.1 ¥ 10–8 M (b) 8.1 ¥ 10–7 M (c) 4.1 ¥ 10–5 M (d) 5.1 ¥ 10–5 M [2009] –7 In aqueous solution the ionization constants for carbonic acid are K °1 = 4.2 ¥ 10 and K °2 = 4.8 ¥ 10–11. Select the correct statement for a saturated 0.34 M solution of the carbonic acid. (a) The concentration of H+ is double that of CO2– 3 (b) The concentration of CO23– is 0.034 M

(c) The concentration of CO23– is greater than that of HCO–3 (d) The concentrations of H+ and HCO–3 are approximately equal. 16. Solubility product of silver bromide is 5.0 ¥ 10–13 M2. The quantity of potassium bromide (molar mass taken as 120 g mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is (a) 5.0 ¥ 10–8 g (b) 1.2 ¥ 10–10 g (c) 1.2 ¥ 10–9 g (d) 6.2 ¥ 10–5 g [2010]

Chemical and Ionic Equilibria 7.73

17. At 25 °C, the solubility product of Mg(OH)2 is 1.0 ¥ 10–11 M3. At which pH will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions? (a) 8 (b) 9 (c) 10 (d) 11 [2010] 18. The correct order of increasing basicity of the given conjugate bases (R ∫ CH3) is (a) RCOO– < HC ∫ C– < NH2– < R– (b) RCOO– < HC ∫ C– < R– < NH2– (c) R– < HC ∫ C– < RCOO– < NH2– (d) RCOO– < NH 2– < HC ∫ C– < R– [2010] – 19. Three reactions involving H2PO 4 are given below? (i) H3PO4 + H2O Æ H3O+ + H2PO–4 + (ii) H2PO–4 + H2O Æ HPO2– 4 + H 3O

(iii) H2PO4– + OH– Æ H3PO4 + O2– In which of the above reactions does H2PO4– act as an acid? (a) (i) only (b) (ii) only (c) (i) and (ii) only 20. The K°sp for Cr(OH)3 is 1.6 ¥ 10 (a) ( 2 1.6 ¥ 10-30 )M 21. An acid HA ionizes as HA (a) 1 ¥ 10–10 M

–30

(d) (iii) only

[2010]

. The molar solubility of this compound in water is

(b) ( 4 1.6 ¥ 10-30 )M

(c) ( 4 1.6 ¥ 10-30 / 27)M

(d) (1.6 ¥ 10–30/27)M

H+ + A–. The pH of 1.0 M solution is 5. Its dissociation constant would be (c) 5 ¥ 10–8 M

(b) 5 M

(d) 1 ¥ 10–7 M

22. The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant Ka of this acid is (b) 1 ¥ 10–3 M (c) 1 ¥ 10–5 M (d) 1 ¥ 10–7 M [2013] (a) 3 ¥ 10–1 M 23. How many liters of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? (a) 0.1 L (b) 0.9 L (c) 2.0 L (d) 9.0 L [2013] 24. Among the following oxoacids, the correct order of acid strength is (b) HOCl > HClO2 > HClO3 > HClO4 (a) HClO2 > HClO4 > HClO3 > HOCl (c) HClO4 > HOCl > HClO2 > HClO3 (d) HClO4 > HClO3 > HClO2 > HClO

[2014]

25. Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate (K°a = 1.0 ¥ 10–5) will be (a) 5.0 (b) 6.0 (c) 8.0 (d) 9.0 [2014, online] 26. Zirconium phosphate [Zr3(PO4)4] dissociates into three zirconium cations of charge +4 and four phosphate anions of charge –3. If molar solubility of zirconium phosphate is denoted by s and the solubility product by Ksp then which of the following relationship between s and Ksp is correct? (a) s = Ksp/(6912)1/7

(b) s = (Ksp/144)1/7

(c) s = [Ksp/6912]1/7

(d) s = (Ksp/6912)7 [2014, online]

ANSWERS 1. 7. 13. 19. 25.

(b) (d) (d) (b) (d)

2. 8. 14. 20. 26.

(b) (d) (d) (c) (c)

3. 9. 15. 21.

(b) (c) (d) (a)

4. 10. 16. 22.

(d) (b) (c) (c)

5. 11. 17. 23.

(d) (c) (c) (d)

6. 12. 18. 24.

(b) (d) (a) (d)

7.74 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS 3. Ksp = [A2+][B–]2 = (1.0 ¥ 10–5 M) (2 ¥ 1.0 ¥ 10–5 M)2 = 4 ¥ 10–15 M3 4. We have to consider the reaction H2PO–4 HPO2–4 + H+ acid

5. We have

MX4

M4+

base

4X–

+

s

4s

Ksp = [M4+] [X–]4 = (s) (4s)4 = 256 s5; Hence s = (Ksp/256)1/5 6. We have pH = –log{[H+]/M}. Hence [H+] = 10–pH M = 10–5.4 M = 3.98 ¥ 10–6 M 7. For the MX2 salt, we have MX2 M2+ + 2X– Ksp = (s) (2s)2 = 4s3 s

Hence,

2s

1/ 3 Ê 4 ¥ 10-12 ˆ Ê Ksp ˆ M˜ = s= Á ÁË Ë 4 ˜¯ 4 ¯

1/ 3

= 10-4 M

OH– Æ O2– + H+

8. The reaction to be considered is

acid

9. From Henderson’s equation, we get

base

pH = pKa + log

[salt ] [acid]

50% of acid ionization implies that [salt] = [acid] Hence, pH = pKa, and pOH = pKw – pH = 14 – 4.5 = 9.5 10. For AgIO3(s) Ag+(aq) + IO–3 (aq), we have Ksp = [Ag+] [IO–3 ] = s2 Hence

s=

Ksp = 1.0 ¥ 10-8 M 2 = 1.0 ¥ 10-4 M

Mass of AgIO3 in 100 mL solution is

m = (1.0 ¥ 10–4 mol L–1) (283 g mol–1) (1.0 L) = 2.83 ¥ 10–3 g

11. The overall standard dissociation constant of the acid is K°a = K°a1 K°a2 = (1.0 ¥ 10–5) (5.0 ¥ 10–10) = 5.0 ¥ 10–15 12. For the salt formed from a weak acid and a weak base, the pH of the solution is given by 1 pH = (pK°w + pK°a – pK°b) 2 1 Hence, at 25 ° C, we will have pH = (14 + 4.80 – 4.78) = 7.01 2 13. HCO–3 is the weakest acid. This is followed by HSO–4. The strongest acid in aqueous solution is H3O+. The choice d indicates that the given acids follow the correct order. 14. We have BaCO3(s) Ba2+(aq) + CO32– (aq) ;Ksp = [Ba2+] [CO2– 3 ] 2+ The minimum [Ba ] required to start precipitation of BaCO3 is Ksp 5.1 ¥ 10-9 M 2 = = 5.1 ¥ 10-5 M [Ba2+] = [CO32- ] 1.0 ¥ 10-4 M 15. Carbonic acid ionizes as H+ + HCO3– K°1 = 4.2 ¥ 10–7 (i) H2CO3 + – 2– (ii) HCO3 H + CO3 K°2 = 4.8 ¥ 10–11 Since K°2 CH∫CH > NH3 > CH4. Their conjugate bases will follow the reverse order, i.e. RCOO– < HC ∫ C– < NH2– < CH–3. 19. The reaction (ii) provides H3O+. Hence, in this reaction H2PO4– acts as an acid. 20. We have Cr(OH)3 æÆ Cr3+ + 3OH– s 3s 3+ – 3 Hence, Ksp = [Cr ] [OH ] = (s) (3s)3 = 27s4 s=

Thus 21. HA

H+ + A–

For

pH = 5,

Ka = [H+]

Ksp / 27 = 1.6 ¥ 10-30 M 4 / 27 4

4

[H + ][A - ] [H + ]2  [HA] [HA]0

=

10–5

M

Hence,

(10-5 M ) 2 = 10-10 M Ka = 1.0 M

22. For a weak acid, we have HQ Also [H+]

H + + Q –;

[Q–] and [HQ] = [HQ]0. Thus

Ka =

[H + ][Q - ] [HQ]

Ka =

[H + ]2 (1 ¥ 10-3 M )2 = = 1 ¥ 10-5 M [HQ]0 0.1 M

23. pH = 1 fi [H+] = 10–1 M pH = 2 fi [H+] = 10–2 M To change the solution having 10–1 M H+ ions to 10–2 M H+, the solution will have to be diluted 10 times. Thus 9 L of water will have to be added to 1 L of the original solution. 24. The larger the number of oxygen atoms attached to chlorine, greater the electron pull towards oxygen, hence, more easy to remove hydrogen from the acid. The given acids are O3Cl – OH Cl – OH; OCl – OH; O2Cl – OH; (HOCl)

(HClO2)

(HClO3)

HClO4

25. The acetate ion undergoes hydrolysis as CH3COOH + OH– CH3COO– + H2O Kh =

or

[CH3COO - ]

∫

Kw 1.0 ¥ 10-14 M 2 [H + ][OH - ] = = 1.0 ¥ 10-9 M = 1.0 ¥ 10-5 M [CH3COO - ][H + ]/[CH3COOH] K a

[CH3COOH] = [OH–].

Also Hence,

[CH3COOH ][OH - ]

[OH - ]2 or [OH - ] = (K h [CH3COOH]0 )1/2 [CH3COOH ]0 – [OH ] = [(1.0 ¥ 10–9 M) (0.1M)]1/2 = 1.0 ¥ 10–5 M [H+] = Kw / [OH–] = 1.0 ¥ 10–14 M2/(1.0 ¥ 10–5 M) = 1.0 ¥ 10–9 M pH = – log ([H+]/M) = 9 Kh =

3Zr4+ + 4 PO3– 4 (3s) (4s) Ksp = [Zr4+]3 [PO43–]4 = (3s)3 (4s)4 = 6912 s7

26. Zr3(PO4)4

Hence, s = (Ksp/6912)1/7

8 Redox Reactions and Electrochemistry UNIT 1

REDOX REACTIONS AND ELECTROLYSIS

SECTION 1

Redox Reactions

For example

æÆ Zn Oxidised

Reduced

Concept of Oxidation Number

4

2 4

8.2 Complete Chemistry—JEE Main

2

4

8

4 4

3 2

2O 2

2O 2

2 2

compound OF2.

8

x

2O 3 2

x x

4

2

x

x

2

x x

2 4O 6

x

2

x x

Explaining Redox Reaction in Terms of Concept of Oxidation Number

Zn oxidizing agent decrease reducing agent

increase

Balancing Redox Reactions Oxidation-State Change Method

2 3

2O

3 2

Redox Reactions and Electrochemistry 8.3

Step Step Step 3

3

2O

3

2O

Step

3 2

2

3 2

2O

3

2O

3 2 3

2O

3 2

2O

Ion-Electron or Half-Equation Method

Step Step

reduction

ææææ Æ oxidation Æ 2O ææææ 3

3 2O

Step

2 2

3

2O

2

Step

3e

3 2

Step

2O 2

3 2

Step

2O

2 3

2e 2O] × 2 ]¥3 2O

2

2 4O 6

Comment on Fractional Oxidation Number 2 4O 6

Equivalent Mass of a substance†

O

O

O

O

expressed as

Fig. 1

†

( 1n substance) that is

Molar mass of substance 1 Molar mass of ÊÁË substanceˆ˜¯ = n n

is 2.5. The

.

8.4 Complete Chemistry—JEE Main

Molar mass of a substance n eq mol-1 n g mol-1 eq mol-1

Unit of molar mass eq mol-1

Illustrations

2O 7

2

in its oxidation reaction in acidic medium. Æ

2O 7 2 4

2

294 g mol-1 6 eq mol-1

Molar mass of K 2 Cr2 O7

2O 7

6 eq mol-1

in its oxidation reaction.

In acidic medium

Æ

4

2

Molar mass of KMnO 4 5 eq mol-1

4

In weakly acidic, alkaline or neutral medium Æ 4

2O

2

Molar mass of KMnO 4 3 eq mol-1

4

In alkaline medium

Æ

4

4

Molar mass of KMnO 4 1 eq mol-1

4

MULTIPLE CHOICE QUESTIONS ON SECTION 1

3

3

3

Æ

2O 3

3

2

Æ

4

x

5

4

2

2

Æ

3

2

4

is Æ

2

2O 7

4 2

2

2O 4

2

2

Æ

2O 4

2

3

2

y

x and y

2O

2O

are

8

2O 3

4O 6

2O 8

8

2O 8

2O 3

4O 6

8

2O 3

4O 8

4O 6

8

2O 8

4O 6

2O 3

2O

2

Redox Reactions and Electrochemistry 8.5 2 2O 3 ◊

2 2O 3

2

2

Æ

2 4O 6

2 2O 5

O O

O

O 3 3O

6]

6]

6]

2

2 4

4 2

2

6

6

2

2

2

6

2

4

4

6

4 4

2

4

4

2

2 6

4 4

2 2

6] 6]

ANSWERS

HINTS AND SOLUTIONS 3

x

x 2

Æ

4

4

2O

2 8

O 2O 8

O

O O

O

O

O O

2O 3

4O 6

and

8.6 Complete Chemistry—JEE Main

2O 3

Æ

2 ¥ Molar mass of Na 2S2 O3 ◊ 5H 2 O

2 2O 3◊

4O 6

2 eq mol-1

2

8. The superoxide is O 2 O 2 2O 5

O

is O

2O2 is

O

x 2

O x

4

2

4

2

4

2

6

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

dichromate solution is [2005] 2 5

2 2

4 2

Æ 4

3

Æ

2 4

2

2

2

4 2 3

Æ l2 4

Æ

4 2O

2

[2006]

4

[2011] 2

[2011] 2O 2 2O 7

4 2

[2014]

3

3

2

Æ

4

statements

2

3

2

3

2O 7

[2014] 2O 4

Æ

2

[2014] 4

6]

2O 3

2

2

[2014]

Redox Reactions and Electrochemistry 8.7 2]

3 4

is [2005]

D rG Æ Æ

2 2

D rG D rG

[2011]

ANSWERS

HINTS AND SOLUTIONS Æ

2O 7 2 4

2

2

4

2

2

Æ

2O 7

2

2O 7 ∫

2

n 2

n m

2O 7

n Mm

æÆ 2 O H 2O 2

4

2O 2

4

Æ

2

2O 2

2 2O 7

3 2O 2

Æ

2O

4

2O 2 Æ

2O

2

2O 2 Æ

2O 7

2O

5

5 2.

Æ

5

2

3

2

2

2

æÆ

n/

Ê 0.45 molˆ (294 g mol-1 ) = 22.05 g ˜¯ ÁË 6

2 2O

∫

VM

2O 7 2

3

2O 7

4

4

2

2

4

and

8.8 Complete Chemistry—JEE Main 2O 7

Æ

2O 7

4

6]

2O

2O 3

x D rG D rG

2

2

x

x

SECTION 2

Electrolytic Cell

Fig. 2

Electrolytic cell

Table 1 Cathode

Anode

Anions Reduction

Working of an Electrolytic cell

Oxidation

Redox Reactions and Electrochemistry 8.9

Electrode processes Cathodic Process cathodic processes take place in the order decreasing reduction potential. Anodic Process

anodic processes take place in the order of decreasing oxidation potential (i.e. increasing reduction potential).

Illustration

Electrolysis of acidic solution of CuSO4

Cathodic Process Æ Æ

E E

2

Anodic Process Æ 4

Eoxid

; Æ

2O 8

Eoxid

;

Æ O2

Eoxid

;

Comment 2

A Few Examples of Electrolysis

Table 2 Solution

2

4

Electrodes

Electrolysis of Some of the Aqueous Solutions of Salts

Cathodic reaction 2

Æ

2

2

Æ

2

4

2 4

Æ Æ Æ

Anodic reaction Æ

2

2O

Æ O2 2

2O

Æ O2 2 Æ 2

Æ

Faraday’s Laws of Electrolysis The mass of a chemical substance involved at an electrode reaction is directly proportional to the amount of current passed through the cell. 2. The masses of different substances produced by a given amount of current are proportional to the equivalent masses of the substances.†

†

8.10 Complete Chemistry—JEE Main

mB

It M B F ne / nB

mB and nB reaction at the electrode.

MB

t and n e

I

Value of Electronic Charge Æ

¥

23

mol

†

ions

2

1 Cl Æ Cl2 (g) + e2 -

e=

(96 487 C mol-1 ) = 1.602 ¥ 10-19 C ( 6.022 ¥ 1023 mol-1 )

Illustrations 4

F

m

. ne

Æ Thus nB It M (2.5 A)(50 ¥ 60 s) (63.5 g mol-1 ) = F ne 2 (96500 C mol-1 )

ne

F

. nB

Q

It

M

Ê 96500 C mol-1 ˆ F m ÊÁ ˆ˜ n e = (12.5 g) Á Ë It ¯ 19300 C ˜¯ Ë M

F

.

nB -1 Ê It ˆ Ê M ˆ (0.01 A)(3 ¥ 60 ¥ 60 s) Ê 192 g mol ˆ ne ÁË ˜¯ ÁË ˜¯ = F m (96500 C mol-1 ) ÁË 0.0715 g ˜¯

MULTIPLE CHOICE QUESTIONS ON SECTION 2 2

F.

† F

Redox Reactions and Electrochemistry 8.11

mol

4

F

F

F

M 2

F

F

3

4

M and M

Æ 1 Æ 2 Æ

2

E°oxd

;

2

E°oxd

;

E°oxd > Æ Æ Au; Æ

Au

E E E > Au

> Au 2

3

solution is

ANSWERS

HINTS AND SOLUTIONS Æ n

2

2

m 1 È It M ˘ It 1 (1 A) (30 ¥ 60 s ) 1 = = = Í ˙ M M Î F n e ˚ F n e (96500 C mol-1 ) 2

9 mol 965

8.12 Complete Chemistry—JEE Main

V It M F ne

m

t

9 nVm = ÊÁ molˆ˜ (22.414 mol-1 ) = 0.21 L ¯ Ë 965 Ê 96500 C mol-1 ˆ Ê ˆ F Ên ˆ 2 m ÊÁ ˆ˜ Á e ˜ = (0.635 g) Á Á ˜ -1 ˜ Ë I ¯ Ë M¯ 10 A Ë ¯ Ë 63.5 g mol ¯ n=

the m

m

to produce

F.

mFn e (1 g)(96500 C mol-1 )(2) = = 64.33 g mol-1 It (10 A)(300 s)

It M Fne

ItM (10 A)(160.83 s)(135 g mol-1 ) = =3 Fm (96500 C mol-1 )(0.75 g)

ne It F M ne

m

mAg mCu

It =

ion r

It M Fne M

Q

m 20.0 g = = 0.5 mol M 40 g mol-1 ¥ F,

Ê M Ag ˆ Ê n Cu ˆ Á n ˜ ÁË M ˜¯ ; Ë Ag ¯ Cu

n e Fm n e F ( rV ) = M M

m

Ê 108 g mol-1 ˆ Ê ˆ 2 (0.2 g) Á ÁË ˜ -1 ˜ 1 ¯ Ë 63.5 g mol ¯

(1)(96500 C mol-1 )(10.5 g cm -3 )(100 ¥ 10-2 cm3 ) (108 g mol-1 ) 2

Æ

2

2

2

2 mol H 2 O(l) 4 ¥ 2 mol gases = mol H 2 O 3 3 mol gases

2

Æ

2

Æ

2

2

H–

2F n

Ê ˆ 2 mol OH 2 mol OH ¥Q =Á [(0.257 A)(25 ¥ 60 s)] -1 ˜ 2F Ë 2 ¥ 96500 C mol ¯

¥

mol

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE UNIT Electrolyte Cells and Electrolysis I

Qt

I

Q/t

I

Qt

I

t/Q

Redox Reactions and Electrochemistry 8.13 4 2O

4 2O

O2

2

4

2

2 2

2 2O

O2

2 4 2O

4 2O

O2

2

4

2

2

not

not

not

and anode are 2

2

2

2

2

2

4

2

at anode

2

at anode

2

3 3 Æ 2O Æ

2

2

3

2

Æ

2 2

Æ

4

2O 8

4

at anode

Faraday Laws of Electrolysis m

QM F | ne |

m

ItF | n e | M

m

It | n e | MF

m

ItMF/ | e|

8.14 Complete Chemistry—JEE Main

F

NA

F¥

F 4

NA

NA

F

solution is 3

3 2.

2

2

2

2

4

4

3

4

Redox Reactions 2O 2

2 2

2

2O

3 3

2O 7

2

2

4

ANSWERS

3 2

3

3

3

2 4

4

3 2

4

2

2O

2

2

2

2O

4

2

2F 2 3

2O 7

Redox Reactions and Electrochemistry 8.15

HINTS AND SOLUTIONS 4 2O. 2

4

2 2

4 2O

Æ Æ

4

. 4

Æ

2

.

2O 8

m F | n e | (V r ) F | n e | (1 cm3 ) (8.94 g cm -3 ) (96500 C mol-1 ) 2 = M M 63.5 g mol-1 m Q 9.65 ¥ 3 ¥ 60 ¥ 60 = = mol M F | ne | 96500 ¥ 2 (0.5 dm3 ) (2 mol dm -3 ) - 0.54 mol 0.5 dm3

Q

t

(80 ¥ 0.005 ¥ 10.5 g) (96500 C mol-1 ) (1) (3 A) (108 g mol-1 )

m F | n e | (V r ) F | n e | = IM IM 1Ê 2Á Ë

1 mol ˆ 96500 C ˜¯

1 2

2

1 2

2O

2

1 2

O2

2.

2

Æ Æ

¥

Q n 2

It Q F

¥5¥ 0.75 ¥ 5 ¥ 60 mol 96500

Æ

2 2

2

V

1 Ê 0.75 ¥ 5 ¥ 60 molˆ˜ Á ¯ 4Ë 96500 1.08 g = 0.01 mol 108 g mol-1 Ê 63.5 g mol- 1 ˆ ÁË ˜¯ 2

n

8.16 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 3

[2003] not 4 2

4

4

¥

[2003]

are [2005] Æ

cathodic reaction is Al ¥

¥

4

¥

¥

7

7

[2005] [2014]

2O 2 2O 2 2O 2

Æ2

2O

2O 2

Æ 2O

Æ O2

2O 2

2 Æ O2 2

2O

[2014]

F [2014] NA NA

NA

NA

NA

[2014]

[2016, online] 4

[2015]

ANSWERS

Redox Reactions and Electrochemistry 8.17

HINTS AND SOLUTIONS

m Q

Q F M/ n e

mF n e (5.12 ¥ 103 g)(96500 C mol-1 )(3) 5.12 ¥ 9.65 = = ¥ 107 C = 5.49 ¥ 107 C M 9 (27 g mol-1 ) 2O 2

It (10.0 A)(2 ¥ 60 ¥ 60 s) = F (96500 C mol-1 ) 0.746 mol 3 0.250 mol n

m 6.35 g = = 0.10 mol M 63.5 g mol -1 NA

Anode

NA

CH 2 COO CH 2 ææ Æ ΩΩ (g ) + 2 CO 2 (g ) + 2e Ω CH2 CH2COO 2

2 2

Æ

4

2

2

8.18 Complete Chemistry—JEE Main

UNIT 2

Electrolytic Conduction

Ohm’s Law I ER I

E and R

Fig. 1

l

Resistivity of the Solution t

A

l R μ ÊÁ ˆ˜ Ë A¯

or r

l R = r ÊÁ ˆ˜ Ë A¯

unit of A ˆ R ÊÁ Ë unit of l ˜¯

r W

Wm

R. G

Conductance and Conductivity G

R k

k

r k is W m . The unit W k

r

=

k

.

Ê l ˆ = GÊ l ˆ ÁË ˜¯ Á ˜ R Ë A¯ A

lA

Kcell k

G Kcell

Equivalent Conductivity

V †

†

Redox Reactions and Electrochemistry 8.19

A

G

l

Ê Veq / eq -1 ˆ Aˆ Ê k Á ˜ =k Á ˜ Ël¯ l2 ¯ Ë

L

†

L

Gl

2

L

kV k

3

V

2

c

V L

Al.

Gl 2 = k (Veq / eq -1 )

or

Gl2

V

c

k c

of the substance, when placed between two parallel electrodes which are at a unit distance apart and large enough to contain between them the whole volume. Molar Conductivity Vm d Lm k Vm k c 2 mol . The name molar conductivity follows c Lm is from the fact that it is conductivity divided by molar concentration.

Relationship Between Lm and Leq

n z

n ˙z

c

n z

L

k / cm Lm k k = = = -1 -1 ceq (n + z+ eq mol )cm n + z+ eq mol n + z+ eq mol-1

cm

Variation of Conductivity of a Solution

Variation of Molar Conductivity of a Solution Vm

c

on dilution. Variation for Strong Electrolyte Lm b

L•m - b c L•m Kohlrausch equation.

c= .

Variation for Weak Electrolyte Fig. 2

† 2

mol .

zB

zB is

8.20 Complete Chemistry—JEE Main

Lc

a

L•

Lc

L•

Kohlrausch’s Law of Independent Migration of Ions aÆ

cÆ

L•m = n + l+• + n - l-•

l+• and l-• are the

n and n

Application of Kohlrausch’s Law

Lm

L•m

Illustration L•m

L•m

L•m

l•m

l•m

l•m

l•m

L•m

2

L•m l•m

mol ; L•m

2 2

Thus Limiting Molar Conductivities of Ions

n + l+• L •m

l•m

t- =

and

L•m

mol 2

mol

l• and l•.

Molar conductivities of H+ and Cl– Ions

t

t

L•m

t

2

l•m

t

L•m

¥

2

l•m

t L•m

¥

2

xlm ( 1x A x+ )

and

l•m

x

mol 2

mol

lm• ( 13 Al3+ )

2

mol

lm• (B y - ) = ylm ( 1y B y - )

then

3lm• ( 13 Al3+ ) = 3 ¥ 63

l•m

(

)

(

x+ • • 1 yl m 1x A and l m B y 1 A x + and 1 B y x y

)

mol

2 mol mol l •m ( 1x A x+ ), l •m ( 1y B y - ) and so on†

For Example

†

mol

n - l -• L •m

t and t

Illustration

l•m

L•m

L•m

t•

l•m

2

mol

2

mol

2

mol

Redox Reactions and Electrochemistry 8.21

Illustration L•m

Molar and Equivalent Conductivities of Al2 (SO4)3

2

lm•

4 3

lm•

4

2 ÈÎ3lm• ( 13 Al3+ )˘˚ + 3 ÈÎ2lm• ( 12 SO 42- )˘˚ 6 ÈÎlm• ( 13 Al3+ ) + lm• ( 12 SO 42- )˘˚

mol

2

mol

1 1 ∫ L•m ÊÁ Al2 (SO 4 )3 ˆ˜ = Lm (Al2 (SO 4 )3 ) ¯ 6 Ë6

2

mol

2

L•

2

4 3

2

mol

Comments on Some Values of Molar Ionic Conductivities 1. Abnormal High Molar Ionic Conductivities of H+ and OH– : lm•

2

• mol and lm

2

Fig. 3

2. The Molar Conductivities of Alkali-Metal Ions

lm•

2

mol

Conductometric Titrations

Illustration Na + OH -

ææææ Æ -H O 2

mol

¥

2

mol

8.22 Complete Chemistry—JEE Main

k

k

and

Fig. 4

Fig. 5 Conductometric titration curve of a weak acid (e.g., ) versus a strong base (e.g., )

Fig. 6 Conductometric titration curve of a mixture of a strong acid (e.g., ) and a weak acid (e.g. ) with a strong base (e.g. )

ion.

MULTIPLE CHOICE QUESTIONS ON UNIT 2 W the solution is W cm

W cm ¥

2

¥

W cm mol 2 mol

mol cm2 mol ¥

2

¥

2

mol

¥ W 2

mol

W mol

3 2

2

mol

2

mol

Redox Reactions and Electrochemistry 8.23 •

lm

2 2

mol and lm• ( 14 Fe(CN)64- ) = 111.0 2

mol

lm• ( 12 Zn 2+ ) = 52.8 2

2

2

mol

L•m

mol 2

mol

L•m

mol

2

2

mol

6 2

mol 2

6] 2

3

4

mol

L•m

mol

6 2 mol

mol

¥ 2

a

2

mol ¥

¥

¥

¥ ¥

2 2

2

¥

mol

¥

3

3

mol ¥

2 3

¥

3

2

4

m

. is

mol

4 2

Lc and

2

mol

2

mol

2

mol

L•

mol

c

i Ka =

cLc

Ka =

L• - Lc

cL2c L• - Lc

Ka =

cL2c L• (L• - Lc )

Ka =

cL2•

Lc (L• - Lc )

represented L

Lc bc b is a constant

l• > l• > l•

L = Lc - b c

L

l• > l• > l•

l• > l• > l•

Lc

L = Lc + b c

bc

l• > l• > l•

solution is 2

2

mol

mol

2

2

mol not

not 4

2

mol

8.24 Complete Chemistry—JEE Main

ANSWERS

HINTS AND SOLUTIONS k Kcell R W W cm 2. Lm k c ¥ W cm ¥ mol cm3 1Ê lˆ Ê 1 ˆ Ê 1.0 cm ˆ ÁË ˜¯ = ÁË 3. k ¥ 3 ˜ R A 7 ¥ 10 W ¯ ÁË 3.0 cm 2 ˜¯ k 4.76 ¥ 10-5 S cm -1 3 cm mol = L ¥ c 0.1 mol dm -3 K cell 0.4 cm -1 = 4. k ¥ R 200 W k 2 ¥ 10 -3 S cm -1 c ¥ mol cm ¥ = Lm 20.0 S cm 2 mol -1 5. L•m

4

6. Lm

l•m

6

lm Zn

6] 2

3

or lm

1 2

6

7. c

l•m ( 14 Fe (CN)64- )

2

¥

¥

3 ¥ 2lm ( 12 Zn ) + 2+

lm

6

2+ 1 ÎÈ Lm ( Zn 3 [Fe(CN)6 ]2 ) - 6lm ( 2 Zn ) ˘˚

2

1 2

k 4.0 ¥ 10 S cm -1 = L 200 S cm 2 mol -1 2

L•m

Ksp 4

c

] [F ]2 L•m 4

¥

Ka

a

mol

[922.8 - 6 ¥ 52.8] Scm 2 mol-1

¥

2

mol

¥

¥

mol dm

2c

¥

c3 L•m

a

¥

3

L•m

3 2

Lc

L•

ca

c( Lc / L• )2 [H + ][A - ] (ca )(ca ) c a 2 = = = c(1 - a ) 1 - a 1 - ( Lc / L• ) [HA]

2

mol

2

mol

2lm (Fe(CN)36- )

¥

mol cm

a c

2

2 mol

mol

is

-5

c

3 cm

¥

k 5.56 ¥ 10-8 S cm -1 5.56 ¥ 10-8 S cm -1 = = c 55.56 mol dm -3 55.56 ¥ 10-3 mol cm m -3 Lwater 1.0 ¥ 10-6 S cm 2 mol-1 = a • (349.8 + 197.6) S cm 2 mol-1 lH• + lOH -

L

mol

c L2e L• ( L• - Lc )

L = Lc - b c . .

mol

2

mol

Redox Reactions and Electrochemistry 8.25

0.00632 S cm -1 0.05 mol dm -3

k c

L

mol dm3

4

3

mol

2

mol

2

conductance.

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE UNIT Ohm’s Law W

W

W cm

W cm2 mol

W

W cm

W cm2

W cm2 mol W cm

¥

W

lA

Al

2

cm2 W

l2 A

lA

¥

W

W

W W

Equivalent and Molar Conductivites W cm

W k

m

k L n z

m

n

Lm

W cm2

m k c is the molar concentration.

m

n z L

m

m

z

W cm2 mol k

n L z

m

m

k

z L n

8.26 Complete Chemistry—JEE Main

W

4

¥

W cm L•

L•

¥ L•

2

¥

W cm2 W cm2

4

W cm2 Lc

W cm

W cm2

¥ L•

W cm2

W cm 2

4

W cm2

W cm2 mol

4

l• 4 L• ( 13 Al3+ )

W cm2 mol and l• W cm2 mol 2• W cm2 mol and l ( 12 SO 4 ) W cm2 mol W cm2 mol W cm2 mol W cm2 mol W 2

mol and

cm2

2

mol

2

mol

L•

2

2

mol 2

4

¥

2

¥

2

is

¥

2

mol

mol

L•m

mol 4

mol

4 3

W cm2 mol . The molar

L•m

L•m

2

¥ l•

2

¥

2 2

mol

¥ and l•

2 2

mol ¥

¥

¥

¥

i li Âi li

ti

ti

ci li Âi ci li

ti

ci ki Âi ci ki

ti

Lm and c is linear Lm c is linear

Lm and c Lm and c L L

2

lm

4 3

2

4 3

lm

lm lm

L L

4 4

2

L•m L•m L•m L•m

2 2 2 2

L•m L•m L•m L•m

2 2 2 2

L•m L•m L•m L•m

Âi li li

2 2 2 2

4 4 4 4

L•m L•m L•m L•m

2 2 2 2

4 4 4 4

2

4 3

2

4 3

lm lm

lm lm

4 4

Redox Reactions and Electrochemistry 8.27

ANSWERS

HINTS AND SOLUTIONS 3. Kcell

kR k ÊÁ Ë

G

¥

Aˆ ˜ l¯

W cm

W Ê 1.50 cm 2 ˆ ÁË 0.5 cm ˜¯

¥

¥

2

K cell 0.25 cm -1 = R 25.0 W 2.54 È -3 ˘ Í (159 / 2) ¥ 1000 eq cm ˙ Î ˚

1 1 = G 2.91 ¥ 102 S

R

k

k

L c

L•

2

W cm2 L•

4

L•

3+ l• ( 13 Al )

l• l•

4

l•

2

L•

W cm2

¥ 63 W cm2 mol

2l• ( 12 SO 4 )

¥

l•

4 3

W cm2

L•m

L•m L(HC)

a

L•(HC)

3.83 383 Ka

a

ca

ca

W cm2 mol

W cm2 mol

dm3

L•m

2

W cm2 mol 2

mol

mol

¥ [H + ][C- ] (ca )(ca ) = c(1 - a ) [HC]

kc

c

k 3.06 ¥ 10-6 S cm -1 = L 1.53 S cm 2 mol -1 4

¥

K cell 2.0 cm -1 = W cm R 200 W k 0.01 W -1 cm -1 = L W c 0.1 mol dm -3

L

Ksp

W cm2 mol

¥

4

L•m =

W cm2 mol

W cm2 mol l•

k

a

4

W cm

L 11.5 11.5 = = • ( 73 . 4 197 . 6 ) 271 .0 + L

a

c

2

¥

4

¥ ¥

ca 2

2

¥

cm mol dm

2

¥ ¥

mol2 dm

k 5.54 ¥ 10 -8 S cm -1 = cL•m (H 2 O) (55.56 mol dm -3 ) {(349.8 + 197.8) S cm 2 mol -1} 5.54 ¥ 10-8 S cm -1 (55.56 ¥ 10 -3 mol cm -3 ) (547.6 S cm 2 mol -1 )

¥

mol dm

8.28 Complete Chemistry—JEE Main

ti

c l Conductivity of ith ion = i i Total conductivity of the solution S i ci li

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAINS L

Lμ

μ 2

2

2

mol

2

mol

mol 2

mol

mol [2004] 2

3

mol 2

2

mol

μ

L Lμ Lμ

2

mol

2

mol

mol [2005]

[2005] mol

Lμ

and

2

L μ 2O

Lμ

Lμ

[2006] [2006] W. The

W ¥ ¥

¥ ¥

2

mol 2 mol

2 2

mol mol

[2006]

2

L•CH3COONa

2

L•

2

L•

L•C

L• L•

2

W W 2 mol

¥

mol

3

[2007]

2

¥

2

2

mol

mol [2011]

Lc = L• + B c

Lc and L• Lc L•

Lc

Bc

W 2

mol is ¥ 2

¥

Lc and

c L• Bc

L•

Lc = L• - B c [2014] . The resistance

W ¥

¥

3

[2014]

Redox Reactions and Electrochemistry 8.29

ANSWERS

HINTS AND SOLUTIONS Lμ

Lμ

L•m

Lμ

L•m

Lμ L•m

L•

L• L•

2

L•m

L•

2

mol 2

L• L•

mol 2

mol

mol

L• ion.

W kR

R Kcell

L•

k W

129 m -1 520 W

K R

k

and

and L L•

3

k c

(129 / 520) S m -1 0.02 ¥ 103 mol m -3 L•

3

¥

2

mol

L•

L• r l

R Kcell

r Kcell

R r

Rk

k

K cell 65 m -1 = 0.25 S m -1 R 260 W

L

k 0.25 S m -1 0.25 S m -1 = = c 0.4 mol dm -3 0.4 mol (10-1 m) -3

W

¥

2

mol

Lc and L• Lc = L• - B c increases on dilution. Kcell

k

Rk

W

k 70 m -1 1 = = S m -1; R 280 W 4

Lm

k (1 / 40) S m -1 (1 / 4) S m -1 = = c (0.5 mol dm -3 ) (0.5 ¥ 103 mol m -3 )

¥

2

mol

8.30 Complete Chemistry—JEE Main

UNIT 3

Zn is oxidised to Zn

GALVANIC CELLS

ion at one electrode. ion is

Fig. 1 Daniell cell

Principle of Working of a Galvanic Cell

ion. æÆ Zn

Resume of working of a Galvanic Cell two half-cells salt bridge 4

negatively charged positively charged

3.

Redox Reactions and Electrochemistry 8.31

to anode.

Cell Diagram

eliminated.

c

c2

Cell Reaction

Illustration c

c2 + 2e Æ Zn2+ + 2e Æ Add

c2

Æ Zn

c Æ

Zn

+

+ 2e + 2e Æ

Zn Add

2+

+

+ Zn2+

Æ

+

Reduction Potentials and Cell Potential Oxidation and reduction Potentials oxidation potential reduction potential

.

Correlation Between Oxidation and Reduction Potentials An atom having larger oxidation potential produces an ion with the lesser reduction potential and vice versa.

8.32 Complete Chemistry—JEE Main

Cell Potential Ecell † Ecell Ecell Ecell

Ecell = E

or E

E

E c Ecell = E

u

c

EZn

u

E

Spontaneity of Cell Reaction

E E > EL Ecell E the cell reaction will be spontaneous in nature.

if the cell diagram has a positive cell potential, E

E E E EA+ | A so that Eºcell is positive. In the given reaction, the reducing tendency of ions is B+ > A+ and the oxidation tendency of atoms is A > B. Hence, A (which acts as reducing agent) reduces B+ (which acts as oxidising agent). Thus Larger the reduction potential of a species, better the oxidising agent, and the reduced species is poor reducing agent. In the given reaction B+ is better oxidising agent than A+ and A is better reducing agent than B Illustration Predicting displacement ability of Atoms Given : and

Eº(Al3+ |Al) = –1.66 V; Eº(Fe2+ |Fe) = –0.44 V; Eº(Zn2+ |Zn) = –0.76 V

Eº(Cu2+|Cu) = 0.34 V Eº(Mg2+ |Mg) = –2.37 V

8.36 Complete Chemistry—JEE Main

The decreasing order of reduction potentials is

Eº(Cu2+ |Cu) > Eº(Fe2+ |Fe) >Eº(Zn2+ | Zn) > Eº(Al3+ | Al) > Eº(Mg2+ |Mg)

Thus, the decreasing order of reducing ability and hence the decreasing order of oxidising agent is Cu2+ > Fe2+ >Zn2+ >Al3+ >Mg2+ The increasing order of oxidising ability and hence the increasing order of reducing agent is Cu < Fe < Zn E(A+ |A), then the reaction B+ + A Æ B + A+ will be spontaneous and hence. A can displace the metal B from its salt. Thus Mg can displace Cu, Fe, Zn and Al Al can displace Cu, Fe, and Zn Zn can displace Cu and Fe Fe can displace Cu

Electrode Potentials and Electrolysis In general, the reduction process at cathode occurs in the order of decreasing reduction potential, i.e. the species with higher reduction potential is reduced in preference to the other species. For example, In the electrolysis of sodium chloride, we have Na+ (aq) + e–Æ Na(s); Eº = – 2.71 1 Eº= – 0.83 V H2O(l) + e–Æ H2(g) + OH–(aq); 2 Since EºH2 | OH– |Pt > EºNa+ |Na, it is water which is reduced in preference to Na+. On the other hand, the oxidation process at anode occurs in the order of increasing reduction potential (i.e. decreasing oxidation potential) i.e. the species with lower reduction potential (i.e. higher oxidation potential) is oxidised in preference to the other species. For example, In the electrolysis of sodium bromide, we have 1 – – 2 Br2(l) + e Æ 2Br (aq);

O2(g) + 4H+ + 4e– Æ 2H2O(l);

Eº = 1.06 V Eº = 1.23 V

Since EºBr– | Br2 < EºO2 | H+ | pt, . Br– ion is oxidised in preference to H2O. Note: If the two potentials are close to each other, it is not possible to predict with certainty the species that will undergo oxidation or reduction. Sometimes, a higher potential is required for the desired change due to imperfection at the electrode reaction.

Extraction of an Element from its oxide

FeO + C Æ Fe + CO we write the following two reactions to obtain this reaction. (i) 2Fe + O2 Æ 2FeO DG1 (ii) 2C + O2 Æ 2CO DG2 Subtracting Eq. (i) from Eq. (ii), we get 2C + 2FeO Æ 2CO + 2Fe DG3 = DG2 – DG1 For this reaction to be feasible, we must have DG1 > DG2. We can

773 0

DfG/4.184 kJ (mol O2)–1

Figure 2 displays the variation of DG of oxidation reactions per mole of oxygen consumed for a few elements with increasing temperature. Ellingham diagram. This diagram assumes that DrH and DrS of the oxidation reactions are independent of temperature. Figure 2 may be used to predict the temperature conditions at which an element is obtainable by the reduction of its oxide. Taking an example of the reaction

2HgO 1273

T/K 1773

2Cu2O

2273

2ZnO

2 Cr O 3 2 3

2FeO

CO2

–100

2 Al O 2 3 3 –200

2 TiO

2C

aO

2CO

gO

2M –300

Fig. 2

Ellingham diagram

Ellingham diagram. The requirement is that the plot of DG versus T for the oxidation of Fe must lie above that of C. The temperature (Teq) at which these two plots intersect each other represent the reaction at equilibrium, since DG of the reaction is zero. At T > Teq, DG1 > DG2 indicating that the reduction of FeO with C is feasible whereas if T < Teq, DG1 < DG2, the reaction is not feasible.

Redox Reactions and Electrochemistry 8.37

Some Commercial Cells

Examples of Primary cells Leclanche Dry Cell 2 4

2

4

2

2

Reactions Æ

2+ 2

+ 2e– –

2 –

2

2

Æ Æ

2 2+

2

– Æ 3 + Cl – 3 + 2Cl Æ –

4 2+

–

3

(aq) –

3

2 3)2Cl2

Comments 4

Zinc-Mercury Cell

–

Reactions –

Æ

Æ Æ

2

–

2

–

–

Comment Alkaline Cell

2

Reactions –

Æ

2

–

2 2

2

+ 2e– Æ Æ 2

– 2

Silver Oxide Cell

2

Reactions – 2

Æ –

2 2

2

Æ Æ

2

+ 2e– – 2

8.38 Complete Chemistry—JEE Main

Examples of Secondary Cells Lead Storage Cell

Reactions 2– 4 2

Æ

– 4 + 2e 2– + 2e– Æ 4 +

+ 2

4 2– (aq) 4

Comment

2

2

Æ

4

2

4

2

4

Nickel-Cadmium Cell 2 2

2 –

2

Reactions –

2

Æ –

2

2

2

+ 2e–

Æ

2

2

Æ

2

2

– 2(s)

2

Comment Hydrogen-Oxygen Fuel Cell

Reactions –

2 2

2

2

–

2

Æ Æ –

2 2

Comment

Æ

–

O2

Rust Fe2O3

water H+

2

2 2

Air

cathode

2

Phenomenon of Corrosion

anode e–

O2 + 4H+ + 4e–

Corrosion of Iron

Fe

2H2O 2+

+

Iron

2

Reactions Æ

Add Side reaction

Fe2+ + 2e–

2+(aq)

+ 2e–] ¥ 2

O 2 (g) + 4H + (aq ) + 4e- Æ 2H 2 O(l) 2Fe(s) + O 2 (g) + 4H + (aq ) Æ 2Fe2+ (aq ) + 2H 2 O(l)

2+ 2+

2

2

Æ

2

Cell potential E°cell = E°R – E° Factors Helping Corrosion Some of the factors are :

3

+(aq)

Fig. 3

Redox Reactions and Electrochemistry 8.39

+

2+

2

3

Prevention from Corrosion

2+(aq)

+(aq)

2

MULTIPLE CHOICE QUESTIONS ON UNIT 3

– 4

+ +

2

–

+

2e–

Æ Æ

2+

E E2

2

4

(a) Ecell = E – E2

(b) Ecell = E2 – E

2+

(c) Ecell

2+

c

–

E – 3

–

2

– 3

2+

E

–

–

–

E

1 HgO(s) + 12 H 2 O(l) + e 2

2 2+

–

2

E

+

E

E

E2

2+ 2

n 2+

2 2+

Ecell

3+ – 4

E

E – 2E2

E

2Cl2

Æ 12 Hg (l) + OH - (aq )

E +

(aq) + e– Æ

1 2

2

(a) E° + (RT/F + E

– 2

¥ Eº (Al3+

E

2+

(b) E° – (RT/F – E

K° +

–

¥

(b) E° + (2RT/F

¥ E

E

2+

K°

K°

– 2

–20

2+

3+

K°

¥

–22

2+

E

3+

–23

8.40 Complete Chemistry—JEE Main n+

+ ne– Æ

If n > m (a) E°3 =

m+

E° E °3

nE1∞ + mE ∞2 n+m

(b) E°3 =

+ me– Æ

E °2 + (n –m)e– Æ m+ mE1∞ + nE ∞2 (c) E°3 = n+m

nE1∞ - mE ∞2 n-m

–

2

K (=E°2

K E°

–

Æ

1 2

2

E°3 = –

K E°I–

mE1∞ - nE ∞2 n-m

E°Cl–

(=E°),

(=E°3) is

(a) E° > E°2 > E°3

(b) E° > E°2 > E°3

(c) E°3 > E°2 > E°

2+

E E

3+

2+

3+

2+

2

E°3 > E° > E°2

2–

2+

2

¥

4 2+

2 2+



¥

20

4

Eºcell

K°c

2 2+

Æ

2+

¥

–20

ANSWERS

HINTS AND SOLUTIONS 2+ Æ Æ 2 Ecell = ER – E = E – E2 – 4

+

–

2+

2 +

+ 2e–

+

– 4

2+

+

–

Æ

3+

2+

2+ 2

2+

2

RT Ê [Mn 2+ ] ˆ E = E° – ln 5F ÁË [MnO 4- ][H + ]5 ˜¯ +

E

Redox Reactions and Electrochemistry 8.41

+(aq)

+ 2e– Æ

2

R

+

(H 2 (g )) L Æ 2H (aq) + 2e(H 2 (g)) L Æ (H 2 (g)) R RT Ê ( pH 2 ) R ˆ 0.059 V 0.5 ˆ 0.059 V log ÊÁ ˜ Ecell = - 2 F ln Á ( p ) ˜ = Ë 2 0.25 ¯ = – Ë H2 L ¯ 2

– 3 –

– 3

–

Æ

Æ

1 2

2 –

2

–

2

Æ

–

–

E E

–

2

1 2

–

2

E

-5FE ∞ = -4 F (0.54 V) - 1F (0.45 V). 1 Hg 2 Cl2 (s) + e2

E = E° –

E -

Æ Hg (l) + Cl (aq )

RT ln ([Cl- ] / c∞) F

1 2

1 2

–

2

Æ

–

1 2

E = E° –

+

Ê [H + ][OH - ]/ M 2 ˆ RT RT = E° – Á ˜ + F Ë ¯ [H ]/ M F RT RT Ê + ln K w∞ ˆ˜ + = Á E∞ ¯ Ë F F

+

E = E° –

+

E° – (RT/F – 2

+ e– Æ

K° =

3+ 2+

+ e– Æ + 2e– Æ

Cr 3+ + 3e- Æ Cr DG 3 = –3FEº

–

K

1 2

2

E ∞ = - 0.8 V ∞ = - 1.18 V Ecell RT

K° K

DG° F DG2 = – (2)F

Eº

+

E

DG3∞ = - F ( - 0.42 - 2 ¥ 0.90) V 3+

2+

(aq) + e– Æ

nEcell ∞ (1)(-1.18 V) = = -20 ( RT /F ) (0.059 V)

2+

–

–

K°

Ag Æ Ag + + eAg (CN) -2 Æ Ag + + 2CN DG° = –nFEcell ° = –RT

RT F

3+

E

¥

–20

8.42 Complete Chemistry—JEE Main n+

+ ne– Æ

DGº = –nFEº

m+ +

me– Æ

DGº2= –mFEº2

n+ +

(n – m)e– Æ

DGº3= –F(nEº – mEº2)

m+

DG°3 = –(n – m) FE°3 –(n – m) FE°3 = –F(nE° – mE°2 E = E° –

E°3 = (nE° – mE°2)/(n – m)

RT ln{( pH 2 /bar )1/ 2 ([OH - ]/ M )} F

+

1/ 2 + 2 1/ 2 RT ÏÔ ([H ][OH ]/ M )( pH 2 /bar ) ¸Ô Ê E ∞ - RT ln K ∞ ˆ - RT ln ÏÔ ( pH 2 /bar ) ¸Ô Ì ˝ E = E° – w˜ ln Ì + ˝ = ÁË ¯ F F F ÔÓ [H ]/M Ô˛ [H + ]/ M ÔÓ Ô˛

+

+

+ e– Æ 12

2

2

1/ 2 RT ÏÔ (pH 2 /bar ) ¸Ô E=– ln Ì ˝ F Ô [H + ]/ M Ô Ó ˛

E° =

RT F

K° =

(2)

2.303 RT F

E°X–

= E°

K° +

+

K°

RT ∞ (AgX)} ln {Ksp F E°X–

3+

E

3+

2+

E

3+

= E°

2+

2+

3+

–

RT Ê [Fe2+ ] ˆ ln nF ÁË [Fe3+ ] ˜¯

4 2+

K°c =

Æ E°cell

2+

nE ∞ (2)( -0.59 V) = = -20. RT/F (0.059 V)

K°c

¥

–20

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE UNIT Characteristics of Galvanic Cells not

Redox Reactions and Electrochemistry 8.43

(b) (c) not

2

2(s)

4

4(s)

2

4

2

4 2

4

Oxidation and Reduction Abilities 2+

2+ 2+

E

E

2+

2+ 2+

E

2+

+

E

+

2+

2 2+

2

Æ

2+

2+

2

2+

2+

2+

2+(aq);

+

E 2

2+

8.44 Complete Chemistry—JEE Main +

E

2+

E

+

+

2

2+

2+

2

not (a) If E (b) If E

n+

n+

n+

+

2 2 2 2

2+

E

2+

2+

E

E

2X– + Y2 –+Y 2 –+X 2 (a) E° – (c) E° –

> E°Y – E°Y– 2 2

> E°X – > E°X– 2 2

2+

2+

E

2X–

2

> E° – > E° – 2 2

(b) E° – E° –

2 2

+

2+

E°Y– > E°Y – 2

2

E E

2+

E E

+

3+

2

2+

E

3+

E

E E

2

2+

2+

2

3+

2+

3+ 3+

2+

2+(aq)

2+

2+

2+

2+ 2+

2+

2

2+

2+

2

E

2

E° – E° –

2+

2+ +

2

3+

+

E E

E°X– E°X–

2

+ +

+

2Y– + X2

+

E

+

2+

E

2+

2+

2+

Redox Reactions and Electrochemistry 8.45 2+

E

2+

E

2+

+

2

2+

2+

2 +

E

3) 2

2+

E

3

3) 2

2+

E

2+

2+

E

4+

2+

2+ 2

E

2+

E

4+

4+

2+

2+

4+

2+

+

2 2+ 2 2+

Cell Representation, Reactions and Potentials 2+

Ce4+(aq) + e– Æ Ce3+(aq);

E

+ e– Æ

E

3+(aq) 3+

E E

2+(aq);

2+

3+

4+

2+

E(Ce4+, Ce3+

4+

–(aq)

2

DG

Æ

2(l)

+ 2Cl–(aq);

is

2

+

–

– 4

E

4+

+

E

3+

2

2+

+

+

E

3+

+

+

8.46 Complete Chemistry—JEE Main 2+

E

–

(a) E (a) p

2

(c) p

2

2+

E

+

E°(Cl–

2

+

E

–

E

2

+

p

2

+

p

2

E°(I– 2)

2 + + 4

E

2

E E°

2

2

2

2

E°

3+

1 2

E°

2+

Æ

2

+(aq)

2+

+ Cl–

3

2

2

3

2 2+

2+

3+

+

+

3+ 2+

(a) E = E° + (RT/nF (c) E = E

3+]) 2+

3+

+ e– æÆ 2+ is (b) E = E° – (RT/nF E=E

2+

3+]) 2+

3+])

–

Applications of Galvanic cells E

+

E

–

¥

¥ +(aq)

E

¥

¥

¥

¥

2+ +

E

¥ E°cell K° ° /RT eq = nFEcell ° /RT K°eq = nFEcell

¥ K° eq

2+

K°eq = – nFEcell ° /RT K°eq = – nFEcell ° /RT

8

Redox Reactions and Electrochemistry 8.47 2+

2+

E°cell

2+

Kºeq

2+

K°eq

K°eq

K°eq

2+

Ecell 8

K°eq

2+ 4

20 2+

æÆ 2

2+

2

+ 2e– – æÆ

2+ – 2 2+ 3+ 3+

æÆ

2

2+

E E

2+

+ e– – æÆ

Ecell

3+

2

–

2–

2+ 2+

Miscellaneous Problems – 4 – 4

E

2+

+

2+

+

E

2+

– 4

2+

Æ

2+ 3+

3+

2

2+

– 4

2+

2+

3+

3

+

3+

– 4

– 4

2+]

2+

8.48 Complete Chemistry—JEE Main 3+

3+

by +

3

by

– 4

2+

2+

3+

3

+

2+ –3

ANSWERS

HINTS AND SOLUTIONS

9.65 ¥ 60 ¥ 60 96500 60 ¥ 60 2 ¥ 10000

2+

65.4 ¥ 60 ¥ 60 g 2 ¥ 10000 E

2+

2+

E 2+

Cell

2+

Cell reaction

2+ +

+ 2+

x

Æ

2+

x

63.5 ¥ 60 ¥ 60 g 2 ¥ 10000

2+(aq)

x

2+

2+

Æ

2+

Redox Reactions and Electrochemistry 8.49

E E° =

RT 2F

[Fe2 + ] [Cd 2 + ]

x 2 ¥ 0.04 ˆ Ê ÁË 0.1 M - x ˜¯ = 0.059 x= 2+]

x ˆ Ê ÁË 0.1 M - x ˜¯

0.059 V 2 x 0.1 M - x

22.7 ¥ 0.1 M 23.7

eq + 2 2+

2+ –

2Y–

+ Y2

– –

+

E°Y–

3+

E

E°X –

2

2+

2+ 2+

E

Y+ + e– = Y;

E

X + + e –;

E

Y

Y + + e –;

E

+

e –;

> E°Y– 2

> E°X – 2

2

2+

2

> E° –

2

+

E

E

X

+

2+

3+

2

> E° –

2

> E°Y– 2

2

+

+

E

2+

+

X + + e– = X; + e–

E

2+

> E X–

2

E° – 2X–

2+

2

+ X2

E° –

3+

2+

3+

+ Y2

+ X2

+

+

E

Reducing tendency

˘ ˙ ˙ ˙ ˙ ˚

X + > Y + > Z+ Oxidizing ability

˘ ˙ ˙ ˙ ˙ ˚

Oxidation tendency Z>Y>X reducing ability Z>Y>X

X + > Y + > Z+

2+

2+ 2+

2+

E

2+

2+

+ 2+

2+

+

2 4+

2+

4+

2+

2+

2+

Æ

8.50 Complete Chemistry—JEE Main

+ Ce4+ Æ 40 mL 2+] = (50 + 10) mL 2+

E

3+

= E°

2+

3+

E E

3+

= E°

2+

3+

3+

–

RT F

[Fe2+ ] [Fe3+ ]

–

RT F

[Ce3+ ] [Ce4+ ]

ECe4+, Ce3+ = E°Ce4+, Ce3+

E

3+

2+

+ ECe4+, Ce3+

= E°

3+

E=E

3+

=

2+

3+]

3+

10 mL (50 + 10) mL

+ E°Ce4+, Ce3+ –

2+

3+

1 E° 2

=

= ECe4+, Ce3+

2+

2+

+ Ce3+

[Fe2+ ] [Fe3+ ]

RT F

–

2+

3+

3+

2+

Ce4+

+ E°Ce4+, Ce3+ ] =

2+

[Fe2+ ] [Ce3+ ] [Fe3+ ] [Ce4+ ]

RT F

1 2

4+

=

50 mL (50 + 60) mL

ECe4+, Ce3+

= E°Ce4+, Ce3+

3+]

D G° = – nFEcell °;

4+]

+ R

+ e– Æ

Cell reaction Cell potential

+ R

Æ

1 2

¥

Ecell °

Left half-cell

2

+

[H + ]L

RT F

Ê 0.01 ˆ ÁË -6 ˜¯ 10

+

[H ]R – 4

+

–

Æ

2+

2+

2

4

E = E° –

RT F

[Mn ] [H 2 O]

E = E° –

RT F

[Mn 2 + ] [H 2 O]4

[MnO -4 ] [H + ]8

[MnO -4 ]

10 mL (50 + 60) mL

[Ce3+ ] [Ce4+ ]

RT F

Ecell ° = – D G°/nF;

Right half-cell

Ecell = –

–

=

-

RT 5F

1 [H + ]8

3

1 2

2

Æ

+

+ e–

Redox Reactions and Electrochemistry 8.51

[Mn 2 + ] [H 2 O]4

RT F

E2 = E° –

[MnO 4- ]

+ e– Æ

+

3+

+ 3e Æ –

3+

+ 2e Æ –

– 2F (E° 3+

1 ([H + ] / 100)8

+

DG° = – F

E

DG° = – 3F

+eÆ +

+

) E° 3+

+

2+

DG°2 = – F E°

2+

DG°3 = – F E°

+ e–

3+

F

DG° = – 2F E° +

- 8 (0.059 V) ¥2 5

=

DG° = – 2F (E°

;

) = – 3F

+

8

E

2+ 2+

RT 5F

([H + ]/ 100)8 RT =– + 8 F [H ]

RT F

E2 – E =

-

+

+

DG°3 = DG° – DG°2 – F E° E° E

+

2

+

= – 2 F E°

+

=2¥

2+

+ F E°

2+

+

p1H/ 22 RT = ln + F [H ] p

+

2

E

+

2

1 RT ln E = E° – 2 F [Cu 2+ ]1 E2 = E° –

1 1 RT RT ln ln = E∞ 2+ 2 F [Cu ]2 2F 0.1[Cu 2+ ]1

E2 – E = -

RT 1 1 RT RT ln + ln = 2+ 2+ 2F 2F 0.1[Cu ]1 2 F [Cu ]1 + +

+

e–

1 2 E=0

2;

p1H/ 22 RT E= ln + F [H ]

12 Ê pH 2 ˆ [H + ] = = 10-7 ÁË atm ˜¯ mol dm -3

E= 3+

2

( pH 2 /atm)1/ 2 RT RT 1 =ln ln + F F [H ]/M (10-7 )

+ 3e– + 2e–

2+ 3+

p

+ e–

2+

E° E°2

DG° = – 3FE° DG°2 = – 2FE° 2

E°3

DG°3 = – FE° 3

0.059 V RT = 2 2F

8.52 Complete Chemistry—JEE Main

DG°3 = DG° – DG°2

FE°3 = – 3FE° + 2FE°2

E°3 = 3E° – 2E°2 +

2

+ 2e– Æ + e– Æ 2+ + e– Æ E

2

DG° = – 2F DG°2 = – F DG°3 = DG° – DG°2 = – F

2+ +

+

F

+

+ 2e– æÆ

2+

E = E° –

2+ 2

[Hg 2+ 2 ]

RT 2F

[Hg 2+ ]2 2Cl2

(0.1) [Hg 2+ 2 ]

RT 2F RT E2 – E = – 2F E2 = E° –

[Hg 2+ ]2 RT 2F 2Cl2

RT 2F

E = E° –

+ 2e– æÆ

–

–] 2 –

RT 2F

E2 = E° – E2 – E = –

–] 2}

RT 2F

RT F +

Æ + + e–

–

RHC LHS

Æ Æ E°cell = E°

–

+

–

– E°

–

+

nFE°cell = – RT K ° (1) (- 0.7 V) nEcell ∞ K° = = (0.059 V) ( RT / F ) 2+

RHC LHC Cell reaction Cell potential

+

+ e– Æ Æ 2+ + 2e– +Æ 2+ E°cell = E° + – E° +

E°cell = E°

+

– E°

2+ 2+

K°

¥

Redox Reactions and Electrochemistry 8.53

nFE°cell = – RT

K°eq

nEcell ° 2.303 (RT /F )

K °eq = ¥

K °eq

K°eq =

(2) (0.19 V) (0.059 V)

E°cell 2+

2+

E°cell K°eq =

2 ¥ (-0.6264) n F Ecell ∞ n Ecell ∞ = = 0.059 2.303 RT (2.303 RT F ) Ê 2 ¥ 0.48 ˆ ˜ ÁË 0.059 ¯

o Ê n Ecell ˆ ÁË 0.059 V ˜¯

Keq

DG° = – nFE°cell 3+

E = E° –

+

2+

2+

[Fe2+ ] [Fe3+ ]

RT F

3+

E°cell = E°

(aq) + e– æÆ

– 4

– E°

2+ 3+

+

2+

– 4

2+

E°cell Ecell = E°cell –

ˆ [Mn 2+ ] [Fe3+ ]5 RT Ê ln Á + 8 2+ 5 ˜ 5 F Ë [MnO 4 ] [H ] [Fe ] ¯ 2+

3+

– 4

+

Ecell Ecell RT DE = F RT DE = F DE = Ecell

0.059 V ˆ RT Ê 1 ˆ ln Á 8 ˜ = + ÊÁ ˜ (8 ¯ Ë Ë 5 ¯ 5F 2 RT È (0.5) (2.0)5 ˘ ln Í ˙ 5F Î (1.0) (2.0)8 (0.5)5 ˚

Ê 0.059 V ˆ log ˜¯ ÁË 5 0.059 ¥ 0.301 V 5

Ê 0.059 V ˆ ˜ ÁË 5 ¯ 0.059 V ˆ DE = - ÊÁ ˜ Ë 5 ¯

–3

È ˘ 1 Í 3 4˙ Î ( 2.0) (0.5) ˚

8.54 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

[2003]

[2003] Æ

2+

2+

E°cell [2003]

¥

¥

¥

–2

[2003]

[2004] E

E°

3+

E°

2+

2+

3+(aq)

,R

F

2+

¥ +(aq) Æ

2+(aq)

is [2004]

)

¥

¥

Æ

2+

¥

2

2

30

[2004]

4

E E E E E°

3+

[2004]

2+

[2004] –

Æ

–

Æ

E

+

+ e–

E

RT/F

K°

[2006] 2+

2+

E°cell 2+

2+

2+

2+])

is ¥

E

3+

E 3+

4

[2007]

2+ 2+

[2008]

Redox Reactions and Electrochemistry 8.55

3+

E° 3+ (aq) + e– Æ

E°

2+

2+

[2009] 2

2 4 Al2O3 Æ Al + O 2 ; 3 3

3

D r G = 966 kJ mol-1 [2010]

E°

2+

[2010] (a) p (c) p

+

2

p p

+

2

+

2

+

2

[2011, (Cancelled)] 2+

2+

2+

Æ X2+

2+

[2012] E°

E°

3+

– 4

E°

2+

2–

E°Cl2

3+

2

–

[2013] (a) Cl

–

3+

+ 2e– Æ 3+ + e– Æ 2+ Æ

2+

2+

Ecell

– 4

E E

2+; 3+

[2014] E°

+

[2014] + e– Æ 2+(aq) + 3e– Æ Al(s) – – (aq) 2(aq) + 2e Æ 3+(aq)

E E E

Al3+(aq)

2+

–

–

2+

–

2+

2+

–

[2014, online] E 2+

2+

E

E

[2015, online] –

2

(a) Cl

– –

– 4

–

2

–

2

–

RT/F –

–

2+

(b)

Cl– –

–

[2015, online]

8.56 Complete Chemistry—JEE Main 4

[2016, online]

[2016, online] [2016]

ANSWERS

HINTS AND SOLUTIONS

Ecell = Ecell ∞ -

0.059 V ˆ RT [ Zn 2+ ] Ê 1 ˆ ln = (1.10 V) - ÊÁ ˜¯ log ÁË ˜¯ = 1.07V 2+ Ë 2 F [Cu ] 2 0.1

DG° = –RT K∞ =

K° = –nFE°cell

nFEcell ∞ nEcell ∞ 2(0.295 V) = = (2.303) RT (2.303 RT ) /F (0.059 V) –(aq)

2 2

Æ –

2 2

2

–

2

Æ

–

Æ

¥

K

(aq)

2

2 3(aq)

E°cell = E°

3+

Æ

– E°

2+

2+

2+

2+

DG = –nFE°cell = – RT K°eq 1 n Ecell (0.591 V) = 10 ∞ = K°eq = 2.303( RT /F ) (0.059 V) +

(aq) Æ

2+ 2+

Ecell = E°cell –

2+

2

RT ([ Zn ] / c∞)( pH 2 / p∞)] ln nF ([H + ]/ c∞)2

Keq

3+

2+

Redox Reactions and Electrochemistry 8.57 2

3+

+

4

+ e– + e–

2+

3+

Ecell

3+

E° E°

2+

3+

3+

+ e– + e–

2+ 2+

2+

E E

2+

3–

Æ

E

+

+ I–

E°cell

is

∞ -0.952 V ∞ = Ecell log Ksp = = -16.13 RT /F 0.052 V 2+

2+ 2+(aq)

2+

Keq =

[ Zn 2+ (aq)] [Cu 2+ (aq )]

E°cell DG° = –nFE°cell = –RT

K°eq

K °eq =

∞ nEcell (2.303RT /F )

RT/F K °eq =

(2)(1.10 V) = 37.3 (0.059 V)

K °eq

Ecell = ER – E + 2e– Æ 1 0.059 V ˆ RT ln = - 0.42 V + ÊÁ ER = E°R – ˜ log (0.01) = - 0.42 V - 0.06 V = - 0.48 V 2+ Ë 2 F [Fe ]/ c∞ 2 ¯ 2+

Right half-cell

+ 3e– Æ

3+

Left half-cell

E = E° –

1 0.059 V ˆ RT ln = - 0.72 + ÊÁ ˜ log (0.1) = - 0.72 V - 0.02 V = - 0.74 V Ë 3F [Cr 3+ ]/ c∞ 3 ¯

Ecell + 3e– Æ 2+ + 2e– Æ

DG°(i) = – 3F DG°(ii) = – 2F

3+

3+

+ e– Æ

2+

E ∞Fe3+ , Fe2 + 1Pt 2 (2Al3+ ) + 3

DG° = DG°(i) – DG°(ii) F DG∞ ( -0.770 V) F == = 0.770 V nF (1) F

4e- Æ 34 Al and

2 3

(3O 2- ) Æ O 2 + 4e2 Al2 O3 3

DG = –nFEcell Ecell = –

DG 966 ¥ 103 J mol-1 == -2.50 V nF (4)(96500 C mol-1 )

Æ 34 Al + O 2 .

F

F

8.58 Complete Chemistry—JEE Main

E 2+

E

2+

E

+(aq)

+ e– Æ E= -

E

2+

E

2+

2 1/ 2 RT ( pH 2 /atm) ln F ([H + ]/M )

E

p 2+

+ 2

Æ X2+

2+

2+

2+

2+

2+

2+

– 4 2+

2+

+ 2e– Æ

2+

3+

+eÆ

2+

Æ

3+

2+

E° 2+]

¥2

E°2

3+

E°cell = E – E2

E n+

+ ne– Æ

n+

–

2+

– 4

E = E∞ -

–

+

–

2+

Æ

2+

2

RT Ê [Mn 2+ ] ˆ ln Á 5F Ë [MnO -4 ] [H + ]8 ˜¯ +

E = E∞ -

RT RT Ê [Mn 2+ ] ˆ ln (1024 ) ln 5F 5F ÁË [MnO -4 ] ˜¯

2+

E

– 4

Ê 0.059 V ˆ ÁË ˜ 5 ¯ 2

2

–3

9 Chemical Kinetics

The subject of chemical kinetics deals with the rates of chemical reactions including their dependence on the concentration of reacting species and the experimental conditions.

General Appearance of Changes in the Concentration of Species in a Reaction In general, the concentration of a reactant decreases while that of a product increases with the progress of a reaction. For example, for the reaction 2N2O5(g) Æ 4NO2(g) + O2(g) the changes in the concentrations of N2O5, NO2 and O2 are shown in Fig. 1

Average Rates of Changes in Concentration of Species in a Reaction The average rate of change in the concentration of the species B of a reaction over a time interval Dt D[B]/Dt. The changes D[B]/Dt of various species in a balanced chemical equation can be correlated with each other based on the principle described in the following. The magnitude of average rates of change in the concentration of species in a balance chemical equation become equal to each other when divided by the corresponding stoichiometric numbers. For example, for the chemical equation 2N2O5(g) Æ 4NO2(g) + O2(g),

Fig. 1

1 D[ N 2 O5 ] 1 D[ NO 2 ] D[O 2 ] = = (1) 2 4 Dt Dt Dt Note: D[N2O5]/Dt has a negative value as the concentration of N2O5 decreases with time while D[NO2]/Dt and

we have

-

D[O2]/Dt have positive values. The negative sign before D[N2O5]/Dt makes this term also positive.

Instantaneous Changes in Concentration of Species in a Reaction In chemical kinetics, the instantaneous changes in concentration of species of a reaction play important role in determining the rate of reaction. d[B] D[B]ˆ = lim ÊÁ ˜ Ë D t Æ 0 dt Dt ¯ t

(2)

9.2 Complete Chemistry—JEE Main

Concentration of reactant

The instantaneous rate (or simply the rate) of decrease (or increase) in the concentration of a reactant (or a product) may be determined by determining the slope of concentration curve versus time of the reactant (or product) at the given time (Figs 2 and 3)

[A]¢1 [A]¢2

[A]

t1¢

t

t2¢ Time

Fig. 2

Fig. 3

The instantaneous changes in concentration at a given time, like average, are related to each other through their stoichiometric number. For example, for the reaction 2N2O5(g) Æ 4NO2 (g) + O2(g), we have 1 d[ N 2 O5 ] 1 d[ NO 2 ] d[O 2 ] = = 2 4 dt dt dt Illustration The rate of formation of N2 in the reaction 4NH3 + 3O2 Æ 2N2 + 6H2O the rate of formation of H2O and the rates of consumptions of NH3 and O2. The identity of changes in concentrations of the given reaction is -

Now

(3) is 0.68 mol L–1 s–1. Determine

1 d[ NH3 ] 1 d[O 2 ] 1 d[ N 2 ] 1 d[H 2 O] == = 4 dt 3 dt 2 dt 6 dt d[H 2 O] Ê 1 d[ N 2 ] ˆ = 6Á = (3) (0.68 mol L–1 s–1) = 2.04 mol L–1 s–1 Ë 2 dt ˜¯ dt –

d[ NH3 ] Ê 1 d[ N 2 ] ˆ = 4Á = (2) (0.68 mol L–1 s–1) = 1.36 mol L–1 s–1 Ë 2 dt ˜¯ dt

–

d[O 2 ] Ê 1 d[ N 2 ] ˆ = 3Á = (1.5) (0.68 mol L–1 s–1) = 1.02 mol L–1 s–1 Ë 2 dt ˜¯ dt

Rate of a reaction The rate of a reaction is the rate of decrease (or increase) in the concentration of a reactant (or a product) divided by the corresponding stoichiometric number in the balanced chemical equation of the reaction. For example, for the reaction 4NH3(g) + 3O2(g) Æ 2N2(g) + 6H2O(g) r= -

1 d[ NH3 ] 1 d[O 2 ] 1 d[ N 2 ] 1 d[H 2 O] == = 4 dt 3 dt 2 dt 6 dt

Notes: There is a negative sign before the rate of change in concentration of a reactant. This results in a positive value of rate of reaction since d[B]/dt of a reactant is itself negative It is mandatory to state the chemical equation to which the rate of reaction is referred to. For example, for the decomposition of N2O5

Chemical Kinetics 9.3

r= -

2N2O5 Æ 4NO2 + O2; N2O5 Æ 2NO2 +

1 d[ N 2 O5 ] 1 d[ NO 2 ] d[O 2 ] = = 2 4 dt dt dt

1 d[ N 2 O5 ] 1 d[ NO 2 ] 1 d[O 2 ] O2; r ¢ = – = = 2 dt 2 dt (1/ 2) dt

r¢ = 2r.

Obviously,

IUPAC Recommendations about Rate of Reaction If a reaction involves change in volume, the concentrations of its reactant or product changes not only due to the reaction in progress but also due to the volume change. To avoid such a The rate expression for the reaction is 2N2O(g) Æ 4NO2(g) + O2(g)

Illustration

∑

1 dn( N 2 O5 ) 1 dn( NO 2 ) dn(O 2 ) = = dt dt dt 2 4 Rate of Reaction in terms of Extent of Reaction Let the decomposition of N2O5 is initiated with the amount n0 of N2O5. If x is the extent of reaction at time t, we will have 2N2O5(g) Æ 4NO2(g) + O2 x=-

t=0

n0

0

0

t

n0 – 2x

4x

x

∑

Hence

1 dn( N 2 O5 ) 1 d(n0 - 2x ) dx == 2 2 dt dt dt 1 dn( NO 2 ) 1 d(4x ) dx = = = 4 4 dt dt dt

x=-

=

dn(O 2 ) dx = dt dt

(4)

Thus, the rate of reaction is equal to the change in extent of reaction with time.

Rate of Reaction at Constant Volume Condition

Dividing Eq. (4) by constant volume, we have

∑

x 1 d(nN 2O5 /V ) 1 d{n0 - 2x / V } d (x / V ) dx r= === = 2 dt 2 dt dt dt V

=

1 d(nNO2 /V ) 1 d(4x / V } dx == dt dt dt 4 4

d(x / V } dx = dt dt dt where x is the extent of reaction divided by constant volume. =

d(nO2 /V )

=

(5)

Differential Rate Law Differential rate law (or simply rate law) expresses the dependence of rate of reaction on the concentration of reacting species involved in a chemical reaction. A early generalization in this regard is due to Gulber and Waage. This generalization, known as law of mass action, is stated as follows. The rate of reaction is proportional to the product of effective concentration of the reacting species, each raised to a power which is equal to the corresponding stoichiometric number of the substance appearing in the chemical equation. The law of mass action is strictly applicable to elementary reactions†. † An elementary reaction involves one step for the conversion of reactants into products. The reactants collide at one place and are simultaneously converted into products. The number of molecules appearing in the reactant side is known as molecularity of elementary reaction. Elementary reactions with molecularity greater than three are very rare.

9.4 Complete Chemistry—JEE Main

For example,

A Æ products

–

d[A] = k[A] dt

2A + B Æ products

–

1 d[A] d[B] == k[A]2[B] 2 dt dt

For a general chemical reaction

Order of Reaction

n 1A 1 + n 2A 2 Æ n 3A 3 + n 4A 4

the experimental data on the

r = k[A1]a[A2]b where the dimensionless exponents a, b,. . .may or may not be equal to the corresponding stoichiometric numbers n1, n2,.... These exponents may have positive or negative integral values, fractional values or zero values. The exponent a is known as partial order of the reaction with respect to A1 and so on. The overall order of the reaction is n=a+b+... (6) The constant k is known as rate constant. Its unit is unit of rate of reaction mol L-1 s -1 ... = = (mol L–1)1 – (a + b + ) s–1 (7) unit of k = a +b + -1 a + b + (unit of concentration) (mol L ) It may be emphasized that the rate equation with its rate constant and order of various reacting species is an

Integrated Rate Laws The integrated rate law expresses the concentration of reacting species with time. This information is obtained by integrating the corresponding differential rate law.

[A]t

Ú

[A]0

[A]0

Slope = –k

[A]

For a zero-order reaction A Æ product d[A] =k[A]0 = k (8) the rate of reaction is r = – dt The unit of k is that of rate reaction r, that is mol L–1 s–1. If [A]0 and [A]t are the concentrations at t = 0 and t = t, respectively, we have

Zero-order Reaction

t

[A]0/k

d[A] = - k Ú dt 0

t This gives [A]t = [A]0 – kt (9) Fig 4 The variation of [A] with time is linear with slope equal to –k (Fig. 4). Half-Life: The half-life period of a reaction is the time required for the concentration of reactant to decrease to half of its original concentration, i.e. at t = t1/2, [A]t = [A]0 /2. From Eq. (9), we have

[A]0/2 = [A]0 – k t1/2

i.e. t1/2 = [A]0/2k

(10)

that is, half-life period is directly proportional to the initial concentration of the reactant. Time of completion of the reaction Here[A]t = 0. Hence tcompletion = [A]0/k Example

Enzyme-catalysed recations

substrate æenzyme æææ Æ product r = k[enzyme][subtrate]0 to the substrate.

Chemical Kinetics 9.5

First-Order Reaction A Æ product r=-

the rate of reaction is

d[A] = k[A] dt

(12)

The unit of k is (unit of r/unit of [A]), i.e. s–1. If [A]0 and [A]t are the concentrations at t = 0 and t = t, respectively, then

Ú

[ A ]0

This gives

t

d[A] = - k Ú dt [A] 0

Ê [A]t ˆ ln Á = - kt Ë [A]0 ˜¯

(13)

or ln ( [A]t/cº ) = ln ( [A]0/cº ) = –kt where cº = 1 M. The variation of ln ( [A]t /cº )1/2 or ( log ( [A]t /cº )) with time is linear with slope equal to –k (or –k/2.303), (Fig. 5). Half-Life At t = t1/2, [A]t = [A]0/2. Hence

log {[A]/mol dm–3}

[A]t

Slope – k /2.303

Ê [A]0 / 2 ˆ ln Á = –k t1/2 Ë [A]0 ˜¯ or

t1/2 =

t/s

ln 2 2.303 log 2 0.693 = = k k k

Fig. 5

(14)

t

t

t

1/ 2 1/ 2 1/ 2 100% æææ Æ50% æææ Æ 25% æææ Æ12.5%

3t1/2 = 3 ¥ 2.0 min = 6.0 min

Concentration

Thus, half-life period is independent of the initial concentration of the reactant (Fig. 6) and has a constant value dependent only on the rate constant. Note etc.) to occur is independent of the initial concentration of the reactant. Illustration to reduce the initial concentration to 12.5% is

Examples N 2 O5 (g ) Æ 2 NO 2 (g) + 12 O 2 (g) H 2 O 2 (aq ) Æ H 2 O(l) + 12 O 2 (g)

0

1 2 3 Number of half-life

4

Fig. 6

All radioactive decays.

equation in terms of physical quantities which are directly proportional to the concentrations. For a reaction involving gaseous species, concentration terms may be replaced by partial pressures. Let the reaction AÆ products be started with the pressure p0 of A. If Dng is the change in stoichiometric number of gaseous species, then A Æ products p0 – p

with This gives

(Dng +1)p

ptotal = (p0 – p) + (Dng + 1)p = p0 + (Dng)p. p = (ptotal – p0)/ Dng,

9.6 Complete Chemistry—JEE Main

p0 – p = p0 -

and

ptotal - p0 ( Dn g + 1) p0 - ptotal = Dn g Dn g

The integrated rate law is Ê p - pˆ ln Á 0 = –kt Ë p0 ˜¯

i.e.

Ê ( Dn g + 1) p0 - ptotal ˆ ln Á ˜ = –kt Dn g p0 Ë ¯

If p• p• = (Dng + 1)p0 with this, the rate law becomes Ê p - ptotal ˆ ln Á • = –kt Ë p• - p0 ˜¯

Illustration

For the reaction (CH3)3COOC(CH3)3 (g) Æ 2 CH3COCH3(g) + C2H6(g), we have di-tertiarybutyl peroxide

Dng = 2 + 1 – 1 = 2;

Ê 3 p0 - ptotal ˆ p• = (Dng + 1)p0 = 3p0. The rate law becomes ln Á ˜¯ = –kt Ë 2 p0 The acid hydrolysis of an ester

H+

Æ RCOOH + R¢OH RCOOR¢ + H2O æææ may be studied by titrating the product RCOOH against an alkali solution. It Vt and V• are the volume of alkali solution required for the neutralization of RCOOH at time t and [Ester]t = V• – Vt [Ester]0 = V• Hence, the rate law becomes Ê [Ester ]t ˆ Ê V - Vt ˆ ln Á ∫ ln Á • = - kt ˜ Ë [Ester ]• ¯ Ë V• ˜¯ order kinetics, is C12H2O11(aq) + H2O(s) æÆ C6H12O6 + C6H12O6 sucrose (dextrorotatory)

glucose fructose (leavorotatory)

The reaction can be followed by measuring the rotation of plane polarized light. If qt and q• are the respective angles of rotation at time t and at the end of reaction, then and [sucrose] μ q•– qt [sucrose]0 μ q•– q0 Hence, the rate law is Ê q - q0 ˆ Ê [sucrose]0 ˆ = - kt ∫ ln Á • ln Á ˜ Ë [sucrose]t ¯ Ë q • - q t ˜¯

Second-Order Reactions For a second-order reactions A + B æÆ products, the rate law is d[A] d[B] == k[A][B] dt dt Let a0 and b0 be the initial concentrations of A and B, respectively, at let x be the extent of reaction divided by constant volume, at time t, we will have A + B æÆ products b0 t = 0 a0 t a0 – x b0 – x Since,

-

d(a0 - x) dx d[A] =we write the rate law as = , dt dt dt dx = k (a0 - x)(b0 - x) dt

Chemical Kinetics 9.7

The integrated expression is Ê b - xˆ Ê b0 ˆ Ê b0 - a0 ˆ log Á 0 = log Á ˜ + Á ˜ kt ˜ Ë a0 - x ¯ Ë a0 ¯ Ë 2.303 ¯

Special Case

If [A]0 = [B]0 then the rate expression becomes

d[A] = k [A]2 dt 1 1 = kt The integrated expression is [A]t [A]0 -

Half- Life Period

Equation (16) gives 1 1 = kt1/2 [A]0 / 2 [A]0

(15) (16)

i.e.

t1/2 =

1 k[A]0

(17)

that is, half-life period is inversely proportional to the initial concentration of the reactant. Note: Equations (15) to (17) are applicable for a reaction A Æ products provided the reaction is second order with respect to A. Examples of Second-Order Reactions 2CH3CHO Æ CH4 + 2CO CH3COOC2H5 + OH– Æ CH3COO– + C2H5OH S2O82– + 2I– Æ 2SO2– 4 + I2

Pseudo First-Order Reaction

Consider the hydrolysis of an ester. Its reaction is +

CH3COOC2H5 + H2O æH ææ Æ CH3COOH + C2H5OH This reaction is of third order. Its rate law is d(ester ) = k’ [ester][H2O][H+] dt The concentration of H+ remains constant as it acts as a catalyst (it is consumed in the reaction in one of its elementary step and is regenerated in another step). The concentration of H2O remains constant as it is present in large amount. These two concentration terms can be merged into rate constant to give d(ester ) =k ¢ [ester]; where k ¢ = k[H2O][H+] dt Note The rate of reaction increases with increase in the concentration of H+ as the rate constant is dependent on this term.

Fractional-order Reactions

For a reaction following the differential rate equation

d[A] = k[A]n +1 / 2 dt

; (where n is integer)

the integer rate law is ¸ 1 Ï 1 1 Ì n -1 / 2 n -1 / 2 ˝ = kt n - 1 / 2 Ó ([A]0 - x) [A]0 ˛ with

1 2n -1 / 2 - 1 t1/2 = k (n - 1 / 2) [A]0n -1 / 2

As mentioned above, the rate constant and order of a reaction are determined experimentally based on the kinetic data observed during the progress of the reactions.

9.8 Complete Chemistry—JEE Main

Methods to Determine Order and Rate Constant Integration Method In this method, the data are substituted in the integrated rate equations for different-order reactions. The equation which gives almost a constant value of rate constant decides the order of the reaction. Graphical Method In this method, the data are plotted according to the different integrated rate equations to yield a straight line. The rate constant of the reaction is obtained from the slope of the resultant straight-line plot. This method can be employed only when the rate law involves only one concentration term. For t0.5 = 0.693/k and it is independent of the initial concentration of the reactant. For the second order t0.5 μ1/[A]0 and so on.

Half-Life Method

Van’t Hoff’s Differential Method In this method, the order of a reaction with respect to each of its reactants can be determined. A number of kinetic experiments are carried out by varying the initial concentration of only one species keeping all others constant. From the variation in the rate of reaction, the order of the reaction with respect to the species whose concentration is varied is determined. The experiment is repeated this way with the variation of concentration of other species one by one. Ostwald Isolation Method In this method, the concentrations of all species except one are taken in large amounts. During the course of the reaction, the variation in these excess concentration terms is negligible and thus remain virtually constant. These constant terms can be merged with the rate constant and, thus, the rate of reaction depends only on one term which is not taken in the excess amount. In this manner the experiment is repeated taking one term each time not in excess and all other terms in excess amounts and the rate of reaction with respect to the species not taken in large amounts is determined as in the case of the van’t Hoff differential method. The following data were obtained at 303 K for the reaction

Illustration

Experiment Number 1

[ A ]0 mol dm 0.1

Rate of reaction

[ B]0 -3

mol dm 0.1

2A + BÆC + D.

-3

mol dm -3 min -1 6.03 ¥ 10–3

2

0.3

0.2

7.2 ¥ 10–2

3

0.3

0.4

2.88 ¥ 10–1

4

0.4

0.1

2.4 ¥ 10–2

Derive the expression of rate law. What is the value of rate constant?

rate in experiment 3 is four times than that in experiment 2. This means, doubling the concentration of B (and keeping concentration of A constant) makes rate of reaction four times indicating that the reaction is second order with respect B. Similarly, comparing experiments 1 and 4 (where the concentration of B remains constant whereas that of A in of reaction also becomes four times. This indicates that the order of the reaction with respect to A is one. Hence, the differential rate law is given by rate = k[A][B]2

Temperature Dependence of Rate constant The rate constant, hence the rate of reaction, increases with increase in temperature. It is found that this increase is about 2 to 3 times with an increase of 10 ºC of the reaction. The Arrhenius equation relating the rate constant, k with kelvin temperature, T, is (18) k = Ae–Ea/RT where A is known as pre-exponential factor. Its unit is the same as that of rate constant k, i.e. (mol–1 dm3)n –1 s–1 where n is the order of reaction, R is a gas constant (= 8.314 J K–1 mol–1) and Ea is known as energy of activation. Writing Eq. (18) in the logarithm form, we have

Chemical Kinetics 9.9

or

k E 1 A ln ÊÁ ˆ˜ = ln ÊÁ ˆ˜ - a Ë kº¯ Ë kº¯ R T k Ea 1 A log ÊÁ ˆ˜ = log ÊÁ ˆ˜ Ë kº¯ Ë k º ¯ 2.303R T

(19) (20)

where kº represents the unit of rate constant k, so that the quantities k/kº and A/kº become unitless. According to Eq. (20), log (k/kº) varies linearly with l/T with slope equal to –Ea/(2.303 R) and intercept equal to log (A/kº). This is shown in Fig. 7. For the two sets of data, we have Ea Ê 1 1 ˆ Êk ˆ log Á 2 ˜ = Ë k1 ¯ 2.303R ÁË T2 T1 ˜¯

(21) Fig. 7

The conversion of reactants into products requires reacting molecules to come close and collide together. During the collision, the molecular rearrangement takes place involving breaking of some bonds and making others. This rearrangement takes place only when the colliding molecules have energy equal to or greater than some minimum energy. If the energy of colliding molecules is less than this energy, no rearrangement takes place and the reactant molecules do not pass over to the products. The difference between the minimum energy activation (Fig. 8). It can be shown on the basis of kinetic theory of gases that the number of molecules having energy Ea or more than this is proportional to the Boltzmann factor exp(–Ea/RT) and thus the rate constant is directly proportional to this factor. On increasing temperature, this factor increases (Fig. 9) so is the rate constant.

Fig. 8

Fig. 9

¥ 10–2 s–1 at 300 K. Calculate the rate constant at

Illustration 350 K if Ea = 50.0 kJ mol–1 Ea 1 A Êk ˆ log Á 1 ˜ = log ÊÁ ˆ˜ Ë kº¯ Ë k º ¯ 2.303R T1

and

Ea 1 Ê Aˆ Êk ˆ log Á 2 ˜ = log ÁË ˜¯ Ë kº¯ kº 2.303R T2

From these, we get (50.0 ¥ 103 J mol-1 ) È 350 K - 300 K ˘ Ea Ê 1 1 ˆ Ea (T2 - T1 ) Êk ˆ log Á 2 ˜ = - ˜= = ˙ Á -1 -1 Í Ë k1 ¯ (2.303)(8.314 J K mol ) Î (350 K ) (300 K) ˚ 2.303R Ë T2 T1 ¯ 2.303R T2 T1 = 1.244 log (k2/kº) = log (k1/kº) + 1.244 = log (4.00 ¥10–2) + 1.244 = –1.398 + 1.244 = – 0.154 k2 = 0.70 s–1

9.10 Complete Chemistry—JEE Main

The collision theory of gaseous molecules provides the rationalization of rate law based on the collisions between the reacting molecules. A reaction involves the rearrangement of bonds present in the reacting molecules to those present in the products. This is possible provided The reacting molecules collide with each other to provide the close proximity for the bond rearrangement. Reacting molecules must have some minimum energy to initiate the process of bond rearrangement. Rate of reaction = (Number of collisions per litre per second) (Fraction of molecules having minimum energy) (22) = ZAB e–Ea/RT where

2 (8kBT / pm )1 / 2 N A* N B* . ZAB = p s AB

Here

2 is collision diameter (=(sA+ sB)/2, where sA and sB are radii of A and B, respectively). s AB

kB is the Boltzmann constant, m is reduced mass (=mAmB /(mA+ mB)). NA* and NB* are the respective numbers of molecules of A and B per unit volume Equation (22) predicts rate law which is much larger than the observed rate law for most of reactions. This fact is taken into account by introducing steric factor which requires that the reacting molecule not only colloide but also have 2AB(g) Æ A2(g) + B2(g)

Fig. 10

With steric factor p r = p ZABe–Ea/RT

(22)

The steric factor is usually less than 1 and consequently predicts the reduced rate of reaction.

Reaction Mechanism The pathway by which the reaction occurs is called its mechanism. The determination of rate law of reaction helps proposing its mechanism. The overall balanced equation for a reaction represents the net chemical change that occurs as the reaction proceeds to completion. The overall equation may proceed involving one elementary reaction or more than one elementary reactions. In the latter case, the net reaction is obtained by the algebraic sum of elementary reactions.

Chemical Kinetics 9.11

Illustration

2NO + 2H2 Æ 2H2O + N2

The reaction

proceeds through the following steps.

2NO Æ N2O2 N 2O 2 Æ N 2O + H 2O N 2O + H 2 Æ N 2+ H 2O Rate-Determining Step If a reaction involves more than one elementary step, the slowest-step is the rate-determining step. The products obtained in the slowest step quickly combine with other substances to give products of the overall reaction.

Illustration k

Over-all reaction

H 2 (g ) + I 2 (g ) æ æ Æ 2HI(g)

Mechanism

eq 1   I 2 (g )    2 I(g )

K

(fast equilibrium)

Keq 2

  I(g) + H 2 (g )    H 2 I(g ) k¢

H 2 I(g) + I(g) ææÆ 2HI(g)

(fast equilibrium) (slow)

From the slow step, we write r =k¢ [H2I] [I] From the second fast-equilibrium reaction, we have Keq2 =

[H 2 I(g)] [I(g)][H 2 (g)]

(1)

fi

[H2I(g)] = Keq2 [I(g)][H2(g)]

Substituting [H2I(g)] in Eq. (1), we get r = k’Keq2 [H2(g)] [I(g)]2

Keq1 =

[I(g)]2 [I2 (g)]

fi

(2)

[I(g)]2 =Keq1[I2(g)]

Substituting [I(g)]2 in Eq. (2), We get r = k¢Keq1Keq2[H2(g)][I2(g)] = k[H2(g)] [I2(g)] The main characteristics of a catalyst are as follows; A catalyst is a substance that increases the rate of a reaction. In a reaction, catalyst goes through a cycle in which it is used up and regenerated so that it is used over and over again. A catalyst, though involved in a reaction, does not appear in the over-all chemical equation. Its presence is shown by writing it over the arrow connecting reactants and products. A catalyst operates by providing an alternate path that has a lower energy of activation for the reaction (Fig. 11). A catalyst lowers the energies of activation of both the forward and reverse reactions without affecting the enthalpy of reaction. This leads to increase in the rate of both forward and backward reactions and thereby helps attaining the equilibrium position of the reaction more rapidly. From Fig. 11, it follows that (24) DrH = Ea(f) – Ea(b) where Ea(f) and Ea(b) are the activation energies of the forward and backward reactions, respectively. For exothermic reactions Ea(f) < Ea(b) For enothermic reactions Ea(f) > Ea(b) These characteristics are shown in Fig. 12

9.12 Complete Chemistry—JEE Main

A catalyst does not change the equilibrium constant of a reaction. Thus, the relative concentrations of reactants and products at equilibrium are not affected by the use of a catalyst. A catalyst is said to be inhibitor if it slows down the rate of reaction.

Fig. 11

Fig. 12

MULTIPLE CHOICE QUESTIONS FOR REVISION 1. Ammonia reacts with oxygen giving nitrogen and water. If the rate of formation of N2 is 0.70 mol L–1 s–1, the rate at which O2 is consumed is (b) 0.70 mol L–1 s–1 (c) 2.10 mol L–1 s–1 (d) 0.35 mol L–1 s–1 (a) 1.05 mol L–1 s–1 2. The rate constant of a reaction is 1.5 mol–3/2 L3/2 s–1, the order of the reaction is (a) 1 (b) 1.5 (c) 2.5 (d) 3.0 1/2. The order of the reaction is 3. The half-life of a reaction A Æ product is found to be inversely proportional to [A]0 (a) 1 (b) 1.5 (c) 2.5 (d) 3.0 4. The reaction (CH3)2O(g) Æ CH4(g) + H2 constant volume with p0 as the initial pressure of (CH3)2O. If p is the pressure at time t, then a linear plot will be observed between (a) ln(p/p0) versus t (b) ln{(p – p0)/ p0}versus t (d) ln{(3p – p0)/ (2p0)}versus t (c) ln{(2p – p0)/ p0}versus t

Chemical Kinetics 9.13

5. As per IUPAC (International Union of Pure and Applied Chemistry) the rate of reaction is expressed as (a) d[reactant]/dt (b) d[product]/dt (c) dx/dt (d) (dx/dt)/V where x is the extent of reaction. (b) mol L–1 s–1 (c) mol L–1 (d) mol–1 L s–1 (a) s–1 7. The half-life for the thermal decomposition of acetone is 80 s and is independent of initial concentration of (a) 186.1 s (b) 206.1 s (c) 150.1 s (d) 226.1 s 8. A living plant disintegrates at the rate of 15 disintegrates per minute per gram of the carbon. After 1.94 ¥ 103 y. the = –0.097) (a) 5270 y (b) 5730 y (c) 6026 y (d) 6750 y 9. The rate of reaction becomes double when its temperature is raised from 300 K to 330 K. The activation energy (b) 23.96 kJ mol–1 (c) 28.96 kJ mol–1 (d) 33.96 kJ mol–1 (a) 18.96 kJ mol–1 10. The reaction NH+4 + OCN– OC(NH2)2, proceeds through the following mechanism. Keq

  NH +4 + OCN -    NH 4 OCN k

NH +4 OCN ææ Æ OC(NH 2 ) 2

(slow)

The rate constant of the reaction is (a) k (b) kKeq 11. The kinetic data for the reaction

(fast)

OH -

(c) k/Keq

(d) Keq/k

OCl– + I– æææÆ OI– + Cl–

[OCl - ]

[I- ]

[OH - ]

mol dm -3

mol dm -3

mol dm -3

10-4 ¥ d[IO - ]/ dt mol dm -3 s -1

0.0017 0.0034 0.0017 0.0017

0.0017 0.0017 0.0034 0.0017

1.0 1.0 1.0 0.5

1.75 3.50 3.50 3.50

The rate low for the formation of OI– is (a) r = k [OCl–][I–] (c) r = k [I–]2[OCl–]

(b) r = k [OCl–]2 (d) r = k [OCl–][I–]/[OH–]

(b) kT + 10K/kT (c) kT/kT + 20K (d) kT + 20K /kT (a) kT2/kT1 13. Which of the following statements is not correct? (a) Larger the activation energy, lesser the value of rate constant of a reaction (b) Larger the temperature, larger the value of rate constant of a reaction (c) Larger the activation energy, larger the effect of a given temperature increase on the rate constant (d) At the lower temperature, increase in temperature causes lesser change in the value of rate constant than at higher temperature For a reaction A(g) Æ products, Initial pressure of A, pA/Torr Half-life period, t1/2/s

the following data are available. 429 273 150 350 550 1001

9.14 Complete Chemistry—JEE Main

14. The order of the reaction is (a) 0 (b) 1

(c) 2

(d) 3

(c) 6.66 ¥ 10–6 Torr–1 s–1

(d) 3.33 ¥ 10–6 Torr–1 s–1

15. The rate constant of the reaction is (a) 6.66 ¥ 10–2 Torr–1 s–1

(b) 6.66 ¥ 10–4 Torr–1 s–1

16. The rate of reaction at initial pressure of 500 Torr of A is (a) 1.665 Torr s–1

(b) 2.665 Torr s–1

(c) 5.656 Torr s–1

(d) 9.25 Torr s–1

(a) 5 min

(b) 10 min

(c) 20 min

(d) 12 min

18. The half-life of a reaction A Æ product order of the reaction is (a) 0

is increased two times when the concentration of A is double. The

(b) 1

(c) 2

(d) 3

19. For a reaction A + B 2 C + D, if the activation energy of the forward and backward reactions are 25 kJ mol–1 and 35 kJ mol–1 respectively. The enthalpy change of the reaction is (a) 10 kJ mol–1

(b) –10 kJ mol–1

(c) 60 kJ mol–1

(d) –60 kJ mol–1

20. A catalyst lowers the activation energy from 20 kJ mol–1 to 15 kJ mol–1 of the forward direction of the reaction A B. Which of the following statement regarding the reaction is correct? (a) Activation energy of the backward direction is increased by 5 kJ mol–1. (b) Activation energy of the backward direction is decreased by 5 kJ mol–1. (c) Enthalpy of reaction is increased by 5 kJ mol–1. (d) Equilibrium constant of the reaction is decreased. 21. The reaction

k

NO 2 + CO ææ Æ NO + CO 2

follows the mechanism

k

1 NO 2 + NO 2 ææ Æ NO3 + NO

(slow)

k

2 NO3 + CO ææ Æ NO 2 + CO 2 The rate low of the reaction will be (a) –d[NO2]/dt = k [NO2] [CO] (c) –d[NO2]/dt = k1[NO2]2

22. The reaction

k

2NO + O 2 ææ Æ 2NO 2

(fast) (b) –d[NO2]/dt = k [NO3] [CO] (d) –d[NO2]/dt = k[NO2]2 [CO]

follows the mechanism

Keq

  NO + NO    N 2 O2 k2

  N 2O2 + O2    2NO 2 The rate law of the reaction will be (a) r = k [NO]2[O2] with k = k2Keq (c) r = k [NO2]2[O2] with k = Keq/k2

(in rapid equilibrium) (slow) (b) r = k [NO2]2[O2] with k = k2/Keq (d) r = k [NO]2[O2] with k = 2k2Keq

ANSWERS 1. 7. 13. 19.

(a) (a) (d) (b)

2. 8. 14. 20.

(c) (c) (c) (b)

3. 9. 15. 21.

(b) (a) (c) (c)

4. 10. 16. 22.

(d) (b) (a) (a)

5. (c) 11. (d) 17. (b)

6. (b) 12. (b) 18. (a)

Chemical Kinetics 9.15

HINTS AND SOLUTIONS 1. The reaction is 4NH3 + 3O2 Æ 2N2 + 6H2O d[O 2 ] 3 d[ N 2 ] Ê 3 ˆ 1 d[ N 2 ] 1 d[O 2 ] =Hence, = = Á ˜ (0.7 mol L-1 s -1 ) = 1.05 mol L-1 s -1 . Ë 2¯ 2 dt 3 dt 2 dt dt –3 1–n –1 2. The unit of rate constant is (mol dm ) s , where n is the order of reaction. Hence, 1– n = –3/2. This gives n =1 + 3/2 = 5/2 i.e 2.5. 3. The half-life is t1/2 μ 1/ [A]n–1 0 , where n is the order of reaction. Hence, n –1 = 1/2 i.e. n = 3/2. 4. For a gaseous reaction at constant volume the rate constant is ln {(Dng + 1) p0 – p}/ (Dng p0)} versus t is a straight Dng = 2. Hence the plot of ln{3(p0 – p)/2p0} versus t will a straight line 6. The rate law of any reaction is expressed as –d[reactant]/dt or d[product]/dt divided by corresponding stoichiometric number. Its unit will be mol L–1 s–1. k = 0.693/t1/2 = 0.693/ (80 s). For the reaction to be 80%, [A]t/[A]0 = 0.2. Hence t= 8. k = -

2.303 log{[A]t /[A]0 } (2.303)(0.301 - 1) (2.303)(log 2 ¥ 10-1 ) == 186.1 s = (0.693 / 80 s) k (0.693 / 80 s)

2.303 log ( [A]t /[A]0 ) 2.303 log (12 / 15) == (2.303)(0.097) / (1.94 ¥ 103 y) = 1.15 ¥ 10-4 y -1 t 1.94 ¥ 103 y t1/ 2 =

9. We have

Ea =

0.693 0.693 = = 5026 y k 1.15 ¥ 10-4 y -1

{log(k2 / k1 )}(2.303R)(T1T2 ) (log 2)(2.303)(8.314 J K -1 mol-1 )(300 K )(330 K ) = T2 - T1 (30 K) =

(0.30)(2.303)(8.314)(300)(330) J mol-1 = 18956 J mol–1 = 18.96 kJ mol–1 (30)

10. From slow step, we write r = k[NH4OCN] where [NH4OCN] from the fast equilibrium step is [NH4OCN] = Keq[NH+4][CN–]. Hence, the rate law is r = (kKeq)[NH+4][CN–] The rate constant is kKeq. 11. On doubling the [OCl–] keeping the [I–] and [OH–] constant, the rate of formation of [OI–] is doubled, hence the order of reaction with respect to [OCl–] is one. On doubling the [I–] keeping the [OCl–] and [OH–] constant, the rate of formation of [OI–] is doubled, hence the order of reaction with respect to I– is one. On changing the [OH–] to a half-value, the rate of formation of [OI–] is doubled. Hence, the order of reaction with respect to [OH–] is –1. Hence the rate law is r = k[OCl–] [I–][OH–]–1 = k[OCl–] [I–]/[OH–] by 10 K. 13. We have

Ê kT ˆ E Ê 1 1 ˆ E T -T E DT ln Á 2 ˜ = - a Á - ˜ = a 2 1 = a R Ë T2 T1 ¯ R T2T1 R (T1 + DT )T1 Ë kT1 ¯

9.16 Complete Chemistry—JEE Main

Since the lower temperature occurs in the denomination, the value kT2/kT1 will be larger on increasing the temperature by DT. t1/2 = constant 14. For zero-order reaction t1/2 μ [A]0 For second-order reaction t1/2 μ 1/[A]0 For third-order reaction t1/2 μ 1/[A]20 For the given data, t1/2[A]0 (429 Torr) (350 s) = 150 150 Torr s (273 Torr) (550 s) = 150 150 Torr s (150 Torr) (1001 s) = 150 150 Torr s Thus, the reaction is second order. 1 1 = = 6.66 ¥ 10–6 Torr–1 s–1 15. For the second-order reaction, k = t1 / 2 ( pA )0 (429 s)(350 Torr ) 16. The rate of reaction at 500 Torr is r = k (pA)20 = (6.66 ¥ 10–6 Torr–1 s–1) (500 Torr)2 = 1.665 Torr s–1 t

t

t

1/ 2 1/ 2 1/ 2 [A]0 æææ Æ [A]0 / 2 æææ Æ [A]0 / 4 æææ Æ [A]0 / 8

50% completed

75% completed

87.5% completed

Hence, 3t1/2 =15 min or t1/2 = 5 min. For 75% completion, t = 2t1/2 = 10 min 18. For a zero-order reaction, the half-life is directly proportional to the initial concentration of A. Hence, the reaction is of zero order. 19. DH = Ea(f) – Ea(b) = (25 – 35) kJ mol–1 = –10 kJ mol–1. 20. Energy of backward reaction is also decreased by 5 kJ mol–1 without affecting the value of DH and Keq of the reaction. 2 2]/dt = k [NO2] . Note that the rate law is not divided by 2 as one molecule of NO2 is recovered in the second step. 1 d[ NO 2 ] k [ N O ][O ] = 2 2 2 22. Since the second step is slow, the rate law is 2 2 dt [ N 2O2 ] From the fast equilibrium, we have Keq = . This gives [N2O2] = Keq [NO]2 [ NO]2 1 d[ NO 2 ] Hence, = k2 Keq [NO]2 [O2] 2 dt

MULTIPLE CHOICE QUESTIONS ON THE ENTIRE CHAPTER General Characteristics 1. The rate constant does not depend upon (a) temperature (b) concentration of reactants and products (c) activation energy (d) catalyst 2. For the reaction 2A Æ 3B, the rate of reaction may be represented as (a) r = – d[A]/dt = d[B]/dt (b) r = – d[A]/dt = (1/3) d[B]/dt (c) r = – (1/2) d[A]/dt = (1/3) d[B]/dt (d) r = (1/2) d[A]/dt = – (1/3) d[B]/dt 3. The order of a reaction (a) is never equal to zero or fraction (b) is a theoretical parameter and can be predicted from the expression of the over-all chemical equation (c) is always equal to the total stoichiometric number of the chemical equation (d) is always determined experimentally

Chemical Kinetics 9.17

4. The units of rate of reaction and its rate constant (a) are identical (b) are quite independent of each other (c) depend on the experimental conditions of the reaction 5. In a reaction 2X + Y Æ X2Y, the reactant X disappears at (a) half the rate as that of disappearance of Y (b) the same rate as that of disappearance of Y (d) twice the rate as that of appearance of X2Y (c) the same rate as that of appearance of X2Y 6. The units of rate constant and rate of reaction are identical for (c) second-order reaction 7. The order of an elementary reaction (a) is equal to its molecularity (c) depends upon temperature 8. The radioactive decay follows

(d) reversible reaction (b) cannot be predicted (d) depends on the experimental condition

(c) second-order kinetics (d) fractional-order kinetics 9. During the course of a chemical reaction, the rate of a reaction (a) remains constant throughout (b) increases as the reaction proceeds 2A + B Æ D + E involves the following mechanism (fast) AÆB BÆC (slow) A+CÆD+E The rate expression would be (b) k[B] (c) k[A] (d) k[A] [B] (a) k[A]2[B] 11. The hydrolysis of an ester was carried out separately with 0.05 M HCl and 0.05 M H2SO4. Which of the following will be true? (b) kHCl < kH2SO4 (c) kHCl = kH2SO4 (d) kHCl = (1/2) kH2SO4 (a) kHCl > kH2SO4 12. The hydrolysis of an ester was carried out separately with 0.1 N HCl and 0.1 N H2SO4. Which of the following will be true?

10. The reaction

(a) kHCl > kH2SO4

(b) kHCl < kH2SO4 (c) kHCl = kH2SO4 (d) kHCl = (1/2)kH2SO4s kf f¢ B is started with A such that initially one can write A ækæ 13. The equilibrium reaction A ÆB kb In such a case, which of the following will be true? (a) kf = k ¢f (b) kf > k¢f (c) kf < k¢f (d) kf may be greater or smaller than k¢f depending upon the concentration of A k 14. The equilibrium reaction A f B is started with B such that initially one can write kb In such a case, which of the following will be true? (a) kb = k ¢b (b) kb > k¢b

k¢

b B ææ ÆA

9.18 Complete Chemistry—JEE Main

(c) kb < k¢b 15.

16.

17.

18.

(d) kb may be greater or smaller than kb depending upon the concentration of B For a reaction N2 + 3H2 Æ 2NH3, the rate of formation of ammonia was found to be 2.0 ¥ 10–4 mol dm–3 s–1. The rate of consumption of H2 will be (a) 1.0 ¥ 10–4 mol dm–3 s–1 (b) 2.0 ¥ 10–4 mol dm–3 s–1 (c) 3.0 ¥ 10–4 mol dm–3 s–1 (d) 4.0 ¥ 10–4 mol dm–3 s–1 For a chemical reaction X Æ Y, the rate of reaction increases by a factor of 1.837 when the concentration of X is increased by 1.5 times. The order of the reaction with respect to X is (a) 1 (b) 1.5 (c) 2 (d) 2.5 In a reaction 2A + B æÆ A2B, the reactant A will disappear at (a) half the rate at which B disappears (b) the same rate as that of B (d) twice the rate at which B decreases (c) the same rate as that of appearance of A2B The terms rate of reaction and rate of appearance (or disappearance) of reactant (or product) (a) represent one and the same physical quantity (b) differ by a constant factor (c) are positive parameters and have same value

balanced chemical equation 19. For the reaction 2A + 3B æÆ 4C, (a) r = - 2 (c) r = -

the rate of reaction may be represented as

d [A] d [B] d [ C] =-3 =4 dt dt dt

(b) r = - 6

1 d [A] 1 d [B] 1 d [C] = = 2 dt 3 dt 4 dt

(d) r = -

d [A] d [B] d [ C] =-4 =3 dt dt dt

1 d [A] 1 d [B] 1 d [C] == 2 dt 3 dt 4 dt

20. The unit of rate of reaction is (b) mol s–1 (c) mol L–1 s–1 (d) mol–1 L s–1 (a) s–1 21. The half-life of a reaction A æÆ B varies as the inverse of concentration of A. The order of the reaction would be 22. If kf and kb are the rate constants of forward and backward reactions in an equilibrium reaction, the equilibrium constant of the reaction is given by (b) Keq = kb/kf (c) Keq = kf kb (d) Keq = 1/kf kb (a) Keq = kf /kb 23. The reaction A + 2B + C Æ D occurs by the following mechanism A+B

k1 k2

E

(rapid equilibrium)

E+C

k3

F

(slow)

F+B

k4

D

(very fast)

The rate law for this reaction is (a) r = k [C] (b) r = k [A] [B]2 [C] 24. The rate equation for 2N2O5 Æ 4NO2 + O2 0.1 M N2O5 will be (a) 6.3 ¥ 10–6 mol dm–3 s–1

is

25. For the reaction

A

(a) d[B]/dt = – k2 [B] (c) d[B]/dt = k1[A] – k–1[B]

(d) r = k [A] [B] [C]

r = (6.3 ¥ 10 s ) [N2O5]. The initial rate of decomposition of –4 –1

(b) 6.3 ¥ 10–5 mol dm–3 s–1

(c) 6.2 ¥ 10–4 mol dm–3 s–1 k1 k –1

(c) r = k [D]

(d) 6.3 ¥ 10–3 mol dm–3 s–1 B and A

k2

C,

the rate of change of concentration of B is given as d[B] (b) = k [C] dt (d) d[B]/dt = k1[A] + k–1[B] – k2[B]

Chemical Kinetics 9.19 k

k

1 2 26. For the reactions A ææ Æ B and A ææ Æ C, (b) r = k2/k1 (a) r = k1/k2

27. A reaction A Æ B (a) 0.693/ k

the ratio of [B]/[C] is given by (c) r = k1k2

(d) r = k1/k 22

is of the order 2.5 with respect to A. The half-life of the reaction will be given (b) 1/([A]0 k) 1.5 (d) [1/(1.5 k [A]1.5 0 )] (2 –1)

(c) k/[A] 0

28. For the reaction 5Br– +BrO3– + 6H+ Æ 3Br2(aq) + 3H2O(l), if the rate of disappearance of Br– is 2.5 ¥ 10–2 mol L–1 s–1, the rate of appearance of Br2 will be (b) 1.5 ¥ 10–2 mol L–1 s–1 (a) 2.5 ¥ 10–2 mol L–1 s–1 (c) 1.25 ¥ 10–2 mol L–1 s–1

(d) 3.0 mol L–1 s–1

Zero-order and Fractional-order Reactions 29. A substance (initial concentration a) reacts according to zero order kinetics. The time it takes for the completion of the reaction is (a) a/k (b) a/2k (c) k/a (d) 2k/a 30. The half-life of a zero order reaction A æÆ B is (a) directly proportional to the concentration of A (b) independent of the concentration of A (c) inversely proportional to the concentration of A (d) determined by the concentration of B 31. A zero order reaction A æÆ B is half completed in time equal to (d) 2k/[A]0 0/2k 32. In a kinetic study, the plot of [reactant] versus time is a straight line with a negative slope. The reaction follows (c) second-order kinetics

(d) fractional-order kinetics 1 2 -1 = . The reaction follows k (1 / 2) [A]01 / 2 1/ 2

33. For a reaction half-life period is given by t1 / 2

(c) fractional order kinetic with order = 3/2 (d) fractional order kinetics with order = 1/2 k 34. A reaction A ææ Æ product follows half-order kinetics with respect to A. A linear plot is observed between 1/2 versus t with slope equal to –k (b) [A]1/2 versus t with slope equal to –k/2 (a) [A] (c) [A]1/2 versus t with slope equal to –2k (d) [A]3/2 versus t with slope equal to –k k 35. A reaction A ææ Æ product follows half-order kinetics with respect to A. Its half-life period will be (a)

[A]01 / 2 ( 2 - 1) k

(b)

2[A]01 / 2 ( 2 - 1) k

(c)

2[A]01 / 2 ( 2 - 1) k

(d)

2 2[A]01 / 2 ( 2 - 1) k

First-order Reactions 36. The reaction 2N2O5 (g) Æ 4 NO2(g) + O2(g) provides a linear plot when ln pN2O5 is plotted against t with a negative slope. The decomposition of N2O5 follows ÆB

is given as

(a) t1/2 = 0.693 k (b) t1/2 = 0.693 ln k (c) t1/2 = 0.693/k 38. The unit of rate constant of the decomposition reaction 2N2O5 Æ 2NO2 + O2 to N2O5 is (b) mol–1 dm3 s–1 (c) mol dm–3 s–1 (a) s–1 (a) 0.223 min–1

(b) 0.0223 min–1

(c) 2.23 min–1

(d) t1/2 = log 2/k

(d) mol–2 dm6 s–1 (d) 22.3 min–1

9.20 Complete Chemistry—JEE Main

40. The half-life of a radioactive isotope is 3 h. What mass out of 100 g is left after 15 h? (a) 12.5 g (b) 6.25 g (c) 3.125 g (d) 1.562 g (a) 8

(b) 9

(c) 10

(d) 12

required to reduce initial concentration by a factor of 1/16 will be (a) 10 min (b) 20 min (c) 30 min (d) 40 min –5 s–1. The rate of the reaction after 3.85 h k = 5.0 ¥ 10 when the reaction is started with 1.0 M of A will be (a) 4.0 ¥ 10–5 mol dm–3 s–1

(b) 3.0 ¥ 10–5 mol dm–3 s–1

(c) 2.5 ¥ 10–5 mol dm–3 s–1

(d) 2.0 ¥ 10–5 mol dm–3 s–1

44. The gaseous reaction A(g) Æ mmHg, the total pressure after 10 min is found to be 180 mmHg. The rate constant of the reaction is (a) 1.15 ¥ 10–3 s–1

(b) 2.30 ¥ 10–3 s–1

(c) 3.45 ¥ 10–3 s–1 2O5, it is found that

2N2O5(g) Æ 4NO2(g) + O2(g) N2O5(g) Æ 2NO2(g) +

1 O2(g) 2

Which of the following is true? (a) k = k¢ (b) k < 2k¢ 2N2O5(g) Æ 4NO2(g) + O2(g) N2O5(g) Æ 2NO2(g) + 1 O2(g) 2 Which of the following expressions is true? (a) k = k¢ (b) k > k¢

–

d[ N 2 O5 ] = k [N2O5] dt

–

d[ N 2 O5 ] k¢ [N2O5] dt

pA = 90

(d) 4.60 ¥ 10–3 s–1

(c) k = 3k¢ (d) k = 2k¢ O , the following information is available. 2 5 rate = k[N2O5] rate = k¢ [N2O5] (c) k = 3k¢

(d) k¢ = 2k

k

ææ Æ 4NO2 + O2, the half-life period is given by (a) 2.303/k (b) 0.693/k (c) 2.303/2k (d) 0.693/2k 48. The reaction n1A + n2B æÆ reaction is started with [A]0 and [B]0, the integrated rate expression of this reaction would be 2O 5

(a) ln

[A]0 = k1t [A]0 - x

(b) ln

[A]0 = k1t [A]0 - n1 x

(c) ln

[A]0 = n1k1t [A]0 - n1 x

(d) ln

[A]0 = n1k1t [A]0 - n1 x

where x is the extent of reaction divided by constant volume. æÆ products (a) the degree of dissociation of reactant is equal to 1 – exp (– kt). (b) the pre-exponential factor in the Arrhenius equation has the dimension of time. (c) the time taken for the completion of 75% reaction is thrice of half-life time (d) a plot of reciprocal of concentration of the reactant versus time is a straight line. 128 mmol L–1 is (a) (128 mmol L–1)/24 (a) [A] versus t

(b) (128 mmol L–1)/25

(c) (128 mmol L–1)/26

(d) (128 mmol L–1)/27

(b) log {[A]/M} versus t

æÆ products? (c) 1/[A] versus t (d) [A]2 versus t

Chemical Kinetics 9.21

52. For a reaction A Æ B, it is found that [A]0 t1/ 2 is constant. The order of the reaction with respect to A is (a) 0 (b) 1 (c) 2 (d) 3 53. The gaseous phase reaction A(g) Æ pressure p0 of A, then the pressure pt of the system at time t is given by the expression (a) ln

3 p0 - pt = – kt 2 p0

(b) ln

3 p0 - pt = – kt 2 p0

(c)

pt - p0 = – kt p0

(d)

pt - p0 = – kt p0

54. The rate constant of acid hydrolysis of an ester with pH = 3 is 1.1 ¥ 10–4 s–1. Its rate constant at pH = 2 will be (b) 11 ¥ 10–4 s–1 (c) 1.1 ¥ 10–2 s–1 (d) 1.1 ¥ 10–1 s–1 (a) 1.1 ¥ 10–4 s–1 55. The rate law for the reaction 2NO(g) + H2 (g) Æ N2O(g) + H2O(g) is given by dp(N2O)/dt = k (pNO)2 (pH2). If (pNO)0 is very much larger than (pH2)0, then the reaction follows

Second-order Reactions 56. The half-life of a second-order reaction A Æ B is given as (a) t1/2 = 0.693/k (b) t1/2 = k/[A0] (c) t1/2 = [A]0/k (d) t1/2 = 1/k[A]0 57. How will the rate of reaction 2NO(g) + O2(g) Æ 2NO2(g) get affected if the volume of the reacting system is 2. (a) Diminishes to one-fourth of its initial value (b) Diminishes to one-eighth of its initial value (c) Increases four times (d) Increases eight times 58. The half-life period and the initial concentration for a reaction are as follows. 420 294 735 t1/2/s a/ mmHg 350 500 200 The order of the reaction is (a) zero (b) one (c) two (d) three 59. For a reaction A Æ products follows a linear plot of 1/[A]t verus t. The reaction follows 60. For a reaction 2NO(g) + H2(g) Æ N2O(g) + H2O(g) the rate law is dp(N2O)/dt = k (pNO)2 (pH2). The halflife of the reaction when (pNO)0 = 10 mmHg and (pH2)0 = 1200 mmHg is found to be 830 s. The half-life when (pNO)0 = 20 mmHg and (pH2)0 = 1200 mmHg will be (a) 830 s (b) 415 s (c) 1245 s (d) 208 s 61. For a second order reaction A Æ product, the plot of (b) 1/[A]t versus time is linear with slope = – k (a) [A]t versus time is linear with slope = k 2 versus time is linear with slope = k (d) 1/[A]t versus time is nonlinear (c) 1/[A]t

Temperature Dependence 62. The rate constant of a reaction follows (a) exponential increase with increase in temperature (b) exponential decrease with increase in temperature (c) linear increase with increase in temperature (d) linear decrease with increase in temperature 63. The plot of log (k /k°) versus 1/T is linear with a slope of (b) Ea/R (c) – Ea/2.303 R (d) Ea/2.303 R (a) – Ea/R (a) (b) (c) (d)

the difference in energies of reactants and products the sum of energies of reactants and products the difference in energy of intermediate complex with the average energy of reactants and products the difference in energy of intermediate complex and the average energy of reactants

9.22 Complete Chemistry—JEE Main

65. The logarithm of rate constant of a reaction (a) increases linearly with increase in inverse of temperature (b) decreases linearly with increase in inverse of temperature (c) increases linearly with increase in temperature (d) decreases linearly with increase in temperature 66. The activation energy for a reaction which doubles the rate when the temperature is raised from 300 K to 310 K is (b) 53.6 kJ mol–1 (c) 56.6 kJ mol–1 (d) 59.6 kJ mol–1 (a) 50.6 kJ mol–1 67. For the decomposition of N2O5(g), it is given that 2N2O5(g) Æ 4NO2 (g) + O2(g) N2O5(g) Æ 2NO2(g) +

68.

69.

70.

71.

72.

1 2

O2(g)

activation energy Ea activation energy E a¢

then (b) Ea > E a¢ (c) Ea < E a¢ (d) Ea = 2E a¢ (a) Ea = E a¢ By increasing the temperature by 10 °C, the rate of forward reaction at equilibrium is increased by a factor of 2. The rate of backward reaction by this increase in temperature (a) remains unaffected (b) increases by a factor greater than two (c) decreases by a factor lesser than two (d) is also increased by a factor of two For an exothermic reaction A æÆ B, the activation energy is 65 kJ mol–1 and enthalpy of reaction is 42 kJ mol–1. The activation energy for the reaction B æÆ A would be (b) 107 kJ mol–1 (c) 65 kJ mol–1 (d) 42 kJ mol–1 (a) 23 kJ mol–1 The rate constant, the activation energy and the Arrhenius parameters of a chemical reaction at 25 °C are 2.0 ¥ 10–5 s–1, 100 kJ mol–1 and 6.0 ¥ 1014 s–1, respectively. The value of rate constant at very high temperature approaches ¥ 1014 s–1 (d) 12 ¥ 10–9 s–1 (a) 2.0 ¥ 10–5 s–1 The rate constant of a reaction (activation energy E1) is twice as the rate constant of a second reaction (activation energy E2). Which of the following conditions will hold good if the Arrhenius parameters A1 =A2? (b) E1 < E2 (c) E1 > E2 (d) E1 = (1/2) E2 (a) E1 = E2 –1 The activation energy of a reaction is 65.8 kJ mol , on changing the temperature from 2 °C to 27 °C, its rate constant changes by

73. The order of a reaction with respect to OH– is –1. The OH– species (a) act as a catalyst (b) act as an inhibitor (c) will always appeared in the chemical equation (d) help in increasing the rate of reaction 74. The use of a catalyst helps in (a) increasing the rate of forward reaction only (b) increasing the rate of backward reaction only (c) increasing the rates of both forward and backward reactions (d) increasing the relative amounts of products 75. A catalyst lowers the activation energy of the forward reaction by 10 kJ mol–1. What effect it has on the activation energy of the backward reaction? (a) Increase by 10 kJ mol–1 (b) Decrease by 10 kJ mol–1 (c) Remains unaffected (d) Cannot be predicted 76. A catalyst is a substance which (a) increases the equilibrium concentrations of the products (b) decreases the energy of activation (c) does not alter reaction mechanism (d) increases the frequency of collision of reacting species

Chemical Kinetics 9.23

77. In the Haber’s process of the synthesis of ammonia, the use of catalyst helps (a) increasing rate constant without changing the equilibrium amount of NH3 (b) increasing rate constant with increasing the equilibrium amount of NH3 (c) decreasing rate constant with increasing the equilibrium amount of NH3 (d) decreasing rate constant with decreasing the equilibrium amount of NH3

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

(b) (a) (a) (d) (c) (c) (c) (c) (a) (a) (b) (a) (b)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(c) (b) (a) (c) (a) (a) (a) (a) (c) (d) (a) (d) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(d) (c) (c) (c) (d) (c) (b) (a) (b) (b) (c) (b) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76.

(d) (b) (b) (a) (b) (b) (c) (d) (d) (c) (d) (c) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77.

(d) (b) (d) (d) (a) (b) (c) (d) (a) (c) (b) (b) (a)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72.

(a) (a) (d) (b) (a) (b) (b) (c) (b) (b) (b) (d)

HINTS AND SOLUTIONS 5. rate = –

d[X 2 Y] 1 d[X] = dt 2 dt

or

-

Ê d[X 2 Y] ˆ d[X] = 2 ÁË ˜ dt ¯ dt

10. Slow step is the rate determining step. 11. For hydrolysis of an ester

kapp = ktrue [H+]

Since [H+] in HCl is smaller than in H2SO4, it follows that (kapp)HCl < (kapp)H2SO4 12. In 0.1 N HCl and 0.1 N H2SO4 acids, [H+] in the former is larger since HCl ionizes in a single step whereas H2SO4 ionizes in two steps. 13. kf does not depend upon the concentration of A. 15. We will have It is given that 16. (1.5)n = 1.837;

–

d[ N 2 ] 1 d[H 2 ] 1 d[ NH3 ] =– = dt 3 dt 2 dt d[ NH3 ] = 2 ¥ 10–4 mol dm–3 s–1. Hence, dt log 1.837 n= = 1.5 log 1.5

23. From the slow step, we write

d [H 2 ] 3 d [ NH3 ] = = 3 ¥ 10-4 mol dm -3 s -1 dt dt 2

rate = k3[E][C] Keq =

Hence,

-

k1 [E] = k2 [A][B]

fi

[E] = Keq [A][B]

rate = k3 Keq [A][B][C] = k[A][B][C]

24. Initial rate = (6.3 ¥ 10–4 s–1) [N2O5]0 = (6.3 ¥ 10–4 s–1) (0.1 M) = 6.3 ¥ 10–5 M s–1

9.24 Complete Chemistry—JEE Main

26.

d [B] = k1[A] dt

27. -

and

d [ C] = k2[A]. dt

d[A] = k[A]2.5. dt

For t0.5,

-

Hence,

[A] = [A]0 /2.

d[A] = k dt. [A]2.5

Thus

1 1 = (1.5)kt 1.5 [A] [A] 1.5 0

Hence

1 1 = 1.5kt0.5 1.5 ([A]0 /2) [A]1.5 0 28. We have -

k d [B]/ dt = 1 k2 d[C]/dt

Hence,

fi

t0.5 =

1 (21.5 – 1) 1.5 1.5 k [A] 0

1 d[Br2 ] 1 d[Br - ] = 3 dt 5 dt 3 d[Br - ] d[Br2 ] 3 = = - (– 2.5 ¥ 10–2 mol L–1 s–1) = 1.5 ¥ 10–2 mol L–1 s–1 5 dt dt 5

Hence,

d [A] 29. – =k dt

[A]

i.e.

–

Ú

d [A] = k

Ú dt

[A]0 – [A] = kt

i.e.

0

[ A ]0

t = a/k d [A ] = k . Thus dt

At tcompletion, [A] = 0. Hence, 32. For a zero-order reaction 33. We have

[A ]t

d [A ] = k [ A ]3 2 dt

0

34. -

d [A ] = k [ A ]1 2 . This gives dt

35.

[ A ]0 2 - [ A ]1 2 = 0

ln 39. log or

k t1 2 2

k=–

and

t1/2 =

1 1 2 -1 {([A]0 / 2) -1 2 - [A ]0-1 2 } = k (1 2) k (1 2) [A ]10 2 k 2

[A ]1t 2 - [A ]10 2 = - t

2 [ A ]10 2 k ˆ - 1˜ = - t1 2 or t1/2 = or [ A ] 2 k 2 ¯ i.e. ln [A]t = ln[A]0 – kt 0 – ln [A]t = kt 12Ê 0 Á Ë

[A]t = – kt. At t0.5, [A]0

[A]t k t =– [A]0 2.303

d [A ]

Ú 3 2 = - kt [A ] [A ]

or

1 ÈÎ[A]t-1 2 - [A]0-1 2 ˘˚ = kt (1 2)

Thus

[A]t – [A]0 = –kt

i.e.

1

[A]t = [A]0/2. log 0.8

=–

Hence,

t0.5 =

(

)

2 -1

ln 2 0.693 = k k

k (10 min) 2.303

2.303 log (0.8) = 0.0223 min–1 10 min [A]0 5

0.1 = – kt99.9 100 1 = – kt0.5. 2

2

Hence

=

100 g = 3.125 g 32

t0.99 log (10-3 ) -3 = = = 10 t0.5 log (0.5) -0.3

Chemical Kinetics 9.25

43. log ([A]t /M) = log ([A]0/M) –

kt kt (5.0 ¥ 10-5 s -1 ) (3.85 ¥ 60 ¥ 60 s) =– =– = – 0.300 2.303 2.303 2.303

[A]t = 0.500 mol dm–3 Rate = k[A]t = (5 ¥ 10–5 s–1) (0.5 mol dm–3) = 2.5 ¥ 10–5 mol dm–3 s–1 44. A(g) Æ 2B(g) + C(g) p0 – p

p

2p

Total pressure = (p0 – p) + (2p) + p = p0 + 2p i.e.

180 mmHg = 90 mmHg + 2p p -p k Now log 0 t =– 2.303 p0 log

p = 45 mmHg

or

45 k =– (10 ¥ 60 s). Thus 90 2.303

k=

0.301 ¥ 2.303 = 1.155 ¥ 10–3 s–1 10 ¥ 60 s

45. For the reaction 2N2O5(g) Æ 4NO2(g) + O2(g), the rate expression is d [N 2 O5 ] 1 d[ N 2 O5 ] – = k¢¢ [N2O5] i.e. – = 2k¢¢ [N2O5] = k [N2O5] dt dt 2 d[ N 2 O5 ] 1 O2(g), the rate expression is – = k¢ [N2O5] and for the reaction N2O5(g) Æ 2NO2(g) + dt 2 Since –d[N2O5]/dt does not depend upon the way the chemical reaction is formulated, we will have k = k¢ 2N2O5(g) Æ 4NO2(g) + O2(g);

46. For

N2O5(g) Æ 2NO2(g) +

For

1 2

1 d[N 2 O5 ] = k[ N 2 O5 ] 2 dt d[N 2 O5 ] = k¢ [ N 2 O5 ] dt -

O2(g);

Since –d[N2O5]/dt does not depend upon the way the chemical reaction is formulated, we will have k¢ = 2k. 47. For

2N2O5 Æ 4NO2 + O2, we have

1 d[N 2 O5 ] = k[ N 2 O5 ] 2 dt On integrating, we get -

ln

[N 2 O5 ]0 = 2kt . [N 2 O5 ]t

-

1 d[A] = k [A] n1 d t

-

or

For a half-life,

d[N 2 O5 ] = 2k dt [N 2 O5 ]

[N2O5]t = (1/2) [N2O5]0.

Hence

t1/ 2 =

ln 2 0.693 = 2k 2k

48. We have

i.e.

ln

[A] = – n1 kt [A]0

ln

[A] = – kt [A]0

[A]

i.e.

Ú

[A]0

or

or

d[A] = – n1 k [A]

ln

t

Ú

dt

0

[A]0 = n1 kt [A]0 - n1 x

[A] = e– kt [A]0

i.e

[A]0 (1 - a ) = e- kt [A]0

i.e

a = 1 – e– kt

50. 30 min is equal to six times the half-life of 5 min. Hence, the concentration will be reduced by 26 times. 0/M}

– kt

9.26 Complete Chemistry—JEE Main

52. t1/2 μ 1/[A] 0n–1 where n is the order of the reaction. If n = 3, then t1/2 μ 1/[A] 02 or (t1/2) ([A]0)2 = constant 53. A(g) 2B(g) + C(g) p0 – p

2p

p

pt = (p0 – p) + 2p + p = p0 + 2p ln

[A] = – kt [A]0

fi

ln

p0 - p = – kt p0

fi

ln

p0 - ( pt - p0 )/ 2 = – kt p0 +].

kpH = 2 = (kpH =3 )

fi

ln

3 p0 - pt = – kt 2 p0

Hence

[H + ]pH = 2 1.1 ¥ 10-4 s -1 -2 ÈÎ10 ˘˚ = 11 ¥ 10–4 s = [H + ]pH =3 ÈÎ10-3 ˘˚

55. When (pNO)0 is very much larger than (pH2)0 56. For a second order reaction 1 1 – = kt At t0.5, [A]t = [A]0/2. [A]0 [A]t

2.

Hence,

t0.5 =

1 k [A]0

57. On increasing the volume to a twice value, the concentrations of each species is reduced by a factor of 2. Hence rate2 1 [ NO] ˆ 2 Ê [O 2 ]ˆ and rate2 = k ÊÁ i.e. = rate1= k [NO]2 [O2] ˜ Á ˜ Ë 2 ¯ Ë 2 ¯ 8 rate1 58. For a second-order kinetics, [A]0 t0.5 is constant. Hence, a t0.5/mmHg s = [420 ¥ 350 = 294 ¥ 500 = 200 ¥ 735] 1 1 59. For a second-order kinetics = kt [A ]t [A ]0 60. Since pH 2 >> ( pNO )0 the reaction follows second-order kinetics with respect to NO. Thus, if the pressure of NO

(

)

is doubled, it half-life becomes half. 62. According to the Arrhenius equation k = A exp(– Ea/RT) ,that is, k increases exponentially with increase in temperature. 63. Arrhenius equation in the logarithm form is E Ea 1 A Êkˆ Ê Aˆ Êkˆ or log Á ˜ = log ÊÁ ˆ˜ – ln Á ˜ = ln Á ˜ – a ¯ Ë kº¯ Ë kº¯ Ë ¯ Ë k º RT 2.303 R T kº 2.303RT1T2 k 2.303 ¥ 8.314 ¥ 300 ¥ 310 log 2 = ÊÁ log 2ˆ˜ J mol-1 = 53600 J mol–1 = 53.6 kJ mol–1 ¯ Ë DT 10 k1 68. At equilibrium, Rate of forward reaction = Rate of backward reaction

66. Ea =

Obviously, Ea (b) = Ea(f) + DH = (65 + 42) kJ mol–1 k = Ae–Ea/RT. When T Æ •, k Æ A. E Ê 1 1ˆ 65.8 ¥ 103 ¥ 25 65.8 ¥ 103 J mol-1 Ê 1 1 ˆ = = 2.40 72. ln (k2 /k1) = – a Á - ˜ = ˜ Á R Ë T2 T1 ¯ 8.314 ¥ 300 ¥ 275 (8.314 J K -1 mol-1 ) Ë 300 K 275 K ¯ 70. We have

k2/k1 = exp(2.40) = 11.0

Chemical Kinetics 9.27

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The rate law for a reaction between the substances A and B is given by rate = k[A]n [B]m On doubling the concentration of A and reducing the concentration of B to half, the ratio of the new rate to the earlier rate of the reaction will be as (b) 1/2(m + n) (c) (m + n) (d) (n – m) [2003] (a) 2(n – m) 2. For the reaction system 2NO(g) + O2(g) Æ 2NO2(g) volume is suddenly reduced to half its volume by 2 and second order with respect to NO, the rate of reaction will (a) increase to four times of its initial value (b) diminish to one-fourth of its initial value (c) diminish to one-eighth of its initial value (d) increase to eight times of its initial value [2003] 3. In respect of the equation k = A exp(–Ea/RT) in chemical kinetics, which one of the following statements is correct? (a) R is Rydberg’s constant (b) k is equilibrium constant [2003] (c) A is adsorption factor (d) Ea is the energy of activation 4. The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of it remaining undecayed after 18 hours would be (a) 16.0 g (b) 4.0 g (c) 8.0 g (d) 12.0 g [2003]

6.

7.

8.

9.

10.

11.

taken for the concentration to change from 0.1 M to 0.025 M is (a) 30 minutes (b) 15 minutes (c) 7.5 minutes (d) 60 minutes [2004] The rate equation for the reaction 2A + B Æ k[A][B]. The correct statement in relation to this reaction is that the (a) unit of k must be s–1 (b) t1/2 is a constant (c) rate of formation of C is twice the rate of disappearance of A (d) value of k is independent of the initial concentrations of A and B. [2004] The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is (a) 1.042 g (b) 2.084 g (c) 3.125 g (d) 4.167 g [2004] Consider an endothermic reaction X ÆY with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general (a) Eb = Ef Eb and Ef (c) Eb < Ef (d) Eb > Ef [2005] A reaction involving two different reactants can never be (a) second order reaction (b) bimolecular reaction [2005] t1/4 can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its value. If the rate constant k, the t1/4 can be written as (a) 0.69/k (b) 0.75/k (c) 0.10/k (d) 0.29/k [2005] A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will (a) double (b) remain unchanged (c) triple (d) increase by a factor of 4 [2006]

9.28 Complete Chemistry—JEE Main

k = A e–E/RT In this equation, E represents (a) the fraction of molecules with energy greater than the activation energy of the reaction (b) the energy above which all the colliding molecules will react (c) the energy below which colliding molecules will not react (d) the total energy of the molecules at a temperature, T [2006] 13. The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr. NOBr2(g) NO(g) +Br2 (g) 2NOBr(g) NOBr2(g) + NO (g) If the second step is the rate determining step, the order of the reaction with respect to NO(g) is (a) 2 (b) 1 (c) 0 (d) 3 [2006] 2AB are 180 kJ mol–1 and 14. The energies of activation for forward and reverse reactions for A2 + B2 200 kJ mol–1, respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol–1. The enthalpy change of the reaction (A2 + B2 Æ 2AB) in the presence of catalyst will be (b) 120 kJ mol–1 (c) 280 kJ mol–1 (d) 20 kJ mol–1 [2007] (a) 300 kJ mol–1 15. Consider the reaction, 2A +B Æ Products When concentrations of B alone was doubled, the half-life did not charge. When the concentrations of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is (b) no unit (c) mol L–1 s–1 (d) s–1 [2007] (a) L mol–1 s–1 16. For a reaction 1 A Æ 2B, 2

the rate of disappearance of A is related to the rate of appearance of B by the expression

(a) –d[A] / dt = 4d[B] / dt (c) –d[A] / dt = (1/4)d[B] / dt

(b) –d[A]/ dt = (1/2)d[B] / dt (d) –d[A]/ dt = d[B] / dt

[2008]

(a) 46.06 minutes (b) 460.6 minutes (c) 230.6 minutes (d) 23.03 minutes [2009] + – 18. Consider the reaction Cl2(aq) + H2S(aq) Æ S(s) + 2H (aq) + 2Cl (aq) The rate equation for the reaction is rate = k[Cl2][H2S] Which of the following mechanisms is/are consistent with this rate equation? (A) Cl2 + H2S Æ H+ + Cl– + Cl+ + HS– (slow) + + + – (fast) Cl + HS Æ H + Cl + S (B) H2S H+ + HS– (fast equilibrium) (slow) Cl2 + HS– Æ 2Cl– + H+ + S (a) A only (b) B only (c) Both (A) and (B) (d) Neither (A) nor (B) [2010] 19. The time for half period of a certain reaction A Æ Products is 1 h when the initial concentration of the reactant A is 2.0 mol L–1. How much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction? (a) 1 h (b) 4 h (c) 0.5 h (d) 0.25 h [2010] 20. The rate of a chemical reaction doubles for every 10 °C rise of temperature. If the temperature is raised by 50 °C, the rate of reaction increases by about (a) 64 times (b) 10 times (c) 24 times (d) 32 times [2011, (Cancelled)] k

1 A ææ Æ B;

k

2 A ææ Æ C;

Activation energy Ea1 Activation energy Ea2

Chemical Kinetics 9.29

If Ea/2 = 2 Ea1, then k1 and k2 are related as (a) k1= 2k2 eEa2/RT

(b) k1= k2 eEa1/RT

(c) k2= k1 eEa2/RT

(d) k1= Ak2 eEa1/RT

[2011]

Æ products, the concentration of A changes from 0.1 M to 0.025 M in 40 min. The rate of reaction when the concentration of A is 0.01 M is (a) 1.73 ¥ 10–5 mol dm–3 min–1 (b) 3.47 ¥ 10–4 mol dm–3 min–1 (c) 3.47 ¥ 10–5 mol dm–3 min–1 (d) 1.73 ¥ 10–4 mol dm–3 min–1 [2012] 23. The rate of a reaction doubles when its temperatures changes from 300 K to 310 K. The activation energy of such a reaction will be (R = 8.314 J K–1 mol–1 and log 2 = 0.301) (b) 48.6 kJ mol–1 (c) 58.5 kJ mol–1 (d) 60.5 kJ mol–1 [2013] (a) 53.6 kJ mol–1 24. For the non-stoichiometric reaction experiments, all at 298 K.

2A + B Æ C + D,

the following kinetic data were obtained in three separate

Initial concentration of A

Initial concentration of B

Initial rate of formation of C in mol L–1 s–1

0.1 M

0.1 M

1.2 ¥ 10–3

0.1 M

0.2 M

1.2 ¥ 10–3

0.2 M

0.1 M

2.4 ¥ 10–3

The rate law for the formation of C is (a) d[C]/ dt = k[A] (b) d[C]/ dt = k[A] [B]

(d) d[C]/ dt = k[A] [B]2 [2014] 2SO3, the rate of change of [O2] was 25. In the reaction of formation of SO3 by contact process 2SO2 + O2 –4 –1 –1 measured as d[O2]/dt = –2.5 ¥ 10 mol L s . The rate of change of [SO2] in mol L–1 s–1 will be (b) –2.50 ¥ 10–4 (c) –3.75 ¥ 10–4 (d) –5.00 ¥ 10–4 (a) –1.25 ¥ 10–4 [2014, online] 26. For the reaction, 2N2O5Æ 4NO2 + O2, the rate equation can be expressed in term of d[ N 2 O5 ] = k [ N 2 O5 ] dt The constant k and k¢ are related as (a) k = k¢ (b) 2k = k¢ -

27.

28.

29.

30.

(c) d[C]/ dt = k[A]2 [B]

and

d[ NO 2 ] = k ¢[ N 2 O5 ] dt

(c) k = 2k¢

(d) k = 4k¢

[2014, online] at 100 °C and 1.3¥ 150 °C. What is The constant (k) for a particular reaction is 1.3¥ 10 the energy of activation (Ea) in kJ mol–1 for the reaction? (R = gas constant = 8.314 J K–1 mol–1). [2014, online] For the reaction 3A + 2B Æ C + D, the differential rate law can be written as 1 d[A] d[C] d[A] d[C] (a) (b) = = k[A]n [B]m = = k[A]n [B]m dt dt dt 3 dt 1 d[A] d[C] 1 d[A] d[C] (c) + (d) [2014, online] == k[A]n [B]m = = k[A]n [B]m dt dt 3 dt 3 dt The reaction 2N2O5(g) Æ 4NO2(g) + O2 N2O5 was found to increase from 50 mmHg to 87.5 mmHg in 30 min. The pressure exerted by the gases after 60 min will be (Assume temperature remains constant) (a) 106.25 mmHg (b) 116.25 mmHg (c) 125 mmHg (d) 150 mmHg [2015, online] For the equilibrium A(g) B(g), DH is – 40 kJ/mol. If the ratio of activation energies of the forward (Ef) and reverse (Eb) is 2/3, then (a) Ef = 60 kJ/mol ; Eb = 100 kJ/mol (b) Ef = 30 kJ/mol ; Eb = 70 kJ/mol (d) Ef Eb = 30 kJ/mol [2015, online] (c) Ef = 80 kJ/mol ; Eb = 120 kJ/mol –4

M–1 s–1

10–3

M–1 s–1 at

9.30 Complete Chemistry—JEE Main

31. A + 2B Æ C, the rate equation for the reaction is given as Rate = k [A] [B] If the concentration of A is kept the same but that of B is doubled, what will happen to the rate itself? (a) halved (b) the same (c) doubled (d) quadrupled [2015, online] (a) low probability of simultaneous collision of all the reacting species (b) increase in entropy and activation energy as more molecules are involved. (c) shifting of equilibrium towards reactants due to elastic collisions (d) loss of active species on collisions 33. Match the catalysts to the correct processes. Catalyst Process (A) TiCl3 (i) Wacker process (B) PdCl2

(ii) Ziegles-Natta polymerization

(C) CuCl2

(iii) Contact process

[2015]

(D) V2O5 (iv) Deacon’s process (a) (A) – (iii), (B) – (ii), (C) – (iv), (D) – (i) (b) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii) (c) (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i) (d) (A) – (iii), (B) – (i), (C) – (ii), (D) – (iv) [2015] 34. The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown kI = 5.2 ¥ 109 L mol–1 s–1 (i) O3 (g) + Cl∑ (g ) Æ O 2 (g ) + ClO∑ (g ) ∑ ∑ ∑ (ii) ClO (g ) + O (g ) Æ O 2 (g ) + Cl (g) kII = 2.6 ¥ 1010 L mol–1 s–1 [2016, online] The closest rate constant for the overall reaction O3(g) + O (g) Æ 2O2(g) is (d) 2.6 ¥ 1010 L mol–1 s–1 (a) 1.4 ¥ 1020 L mol–1 s–1 (b) 3.1 ¥ 1010 L mol–1 s–1 (c) 5.2 ¥ 109 L mol–1 s–1 35. The rate law for the reaction below is given by the expression k[A][B] A+B Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be (a) 3 k (b) 9 k (c) k/3 (d) k [2016, online] 36. Decomposition of H2O2 2O2 decreases from 0.5 M to 0.125 M in one such decomposition. When the concentration of H2O2 reached 0.05 M, the rate of formation of O2 will be (b) 2.66 L min–1 at STP (a) 6.93 ¥ 10–4 mol L–1 min–1 –2 mol L–1 min–1 (d) 6.93 ¥ 10–2 mol L–1 min–1 [2016] (c) 1.34 ¥ 10

ANSWERS 1. 7. 13. 19. 25. 31.

(a) (c) (a) (d) (d) (c)

2. 8. 14. 20. 26. 32.

(d) (c) (d) (d) (b) (a)

3. 9. 15. 21. 27. 33.

(d) (c) (a) (d) (b) (b)

4. 10. 16. 22. 28. 34.

(b) (d) (c) (b) (d) (c)

5. 11. 17. 23. 29. 35.

(a) (d) (a) (a) (a) (a)

HINTS AND SOLUTIONS m

Ê [B] ˆ 1. r1 = k [A]n [B]m and r2 = k [2A]n ÁË ˜¯ = 2(n – m) k [A]n [B]m. 2 2. r1 = k [O2] [NO];

r2 = k (2[O2])(2 [NO])2 = 8k [O2] [NO]2.

Hence,

Hence,

r2 =8 r1

r2 = 2(n – m) r1

6. 12. 18. 24. 30. 36.

(d) (c) (a) (a) (c) (a)

Chemical Kinetics 9.31

4. 18 hours implies six half-lives. Hence, the mass remaining undecayed will be m = 256/26 = 4.0 g 5. The decrease of concentration from 0.8 M to 0.4 M implies that t1/2 = 15 min. The change in concentration from 0.1 M to 0.025 M will require two half-lives. Hence, t = 2t1/2 = 30 min. 6. For the reaction 2A + B Æ C, the rate law is r = k [A] [B] The unit of k will be mol–1 L s–1. The half-life period, t1/2 is not constant. The rate of formation of C will be half of the rate of disappearance of A. The rate constant will be independent of initial concentrations of A and B. 7. Twenty four hours is equivalent to six half-lives. Mass of remaining radioisotope = (1/26) (200 g) = 3.125 g. 8. Endothermic reaction implies the absorption of heat when reactants are converted into products. This implies that the enthalpy of products (EP) is more that of reactants (ER) i.e. EP > ER. and Eb = Etransition state– EP , it follows that Eb < Ef Since Ef = Etransition state– ER 9. Unimolecular reaction involves one molecule of reactants. [A]t 1 3 0.29 = - kt. Thus t1 / 4 = - ln  10. We have ln [A]0 k 4 k 2 2 11. r1 = k[CO] and r2 = k(2[CO]) , Hence, r2 / r1 = 4 12. The choice (c) is correct. 13. The steps are [ NOBr2 ]  Keq = NO(g) + Br2(g)    NOBr2(g); [ NO][Br2 ] k Æ 2NOBr(g); NOBr2 + NO(g) ææ (slow step) The rate expression is rate = k[NOBr2][NO] = k{Keq [NO] [Br2]}[NO] = k Keq [NO]2 [Br2] Hence, the order of the reaction with respect to NO is 2. 14. The enthalpy of reaction is not changed in the presence of a catalyst. Hence DrH = Ea(b) – Ea(f) = 200 kJ mol–1 – 180 kJ mol–1 = 20 kJ mol–1

the concentration of A. Thus, the reaction is second order. Hence, the unit of rate constant is unit of rate of reaction mol L-1 s -1 = = mol–1 L s–1 unit of rate constant = (unit of concentration)2 (mol L-1 s -1 ) 16. For the given chemical equation, we have 1 d[A] 1 d[B] 1 d[A] 1 d[B] d[A] 1 d[B] i.e. Hence, = = . = VA dt VB dt (1 / 2) dt 2 dt dt 4 dt 0.693 0.693 = = 0.1 min -1 17. k = 6.93 min t1 / 2 t =-

[A]t 2.303 2.303 1 ˆ 2 ¥ 2.303 = == 46.06 min log log ÊÁ Ë 100 ˜¯ 0.11 min -1 k [A]0 0.1 min -1

18. For the mechanism A, the rate equation is r = k[Cl2] [H2S] For the mechanism B, the rate equation is r = k[Cl2] [HS–] From the fast equilibrium, we have K eq [H 2S] [H + ][HS- ] fi [HS- ] = Keq = [H 2S] [H + ] Thus r = k Keq [Cl2] [H2S]/[H+] 19. For zero-order reaction Its half-life time as

[A]0 – [A]t = kt

[A]0 – [A]0 /2 = kt1/2

For given concentration, we have [A]0 – 0.25 M = kt2

i.e. and

t1/2= [A]0 /2k [A]0 – 0.5 M = kt1

9.32 Complete Chemistry—JEE Main

0.25 M = k(t2 – t1) 0.25 M 0.25 M 0.25 M = = = 0.25 h or Dt = t2 – t1 = [A]0 / 2t1 / 2 2.0 M/(2 ¥ 1 h ) k Since initial concentration is decreased four times, the half-life period will also decrease four times, i.e. from 1 h to 0.25 h. 20. We will have r1 ææ Æ 2r1 ææ Æ 4r1 ææ Æ 8r1 ææ Æ 16r1 ææ Æ 32r1 These give

(t )

1

In short, 21. We have

(t + 10 ∞ C)

r=

25

(t + 30 ∞ C)

1

(t + 40 ∞ C)

1

(t + 50 ∞ C)

1

1

r1 = 32 r1

k1 = A1e- Ea1 / RT

and

k2 = A2 e- Ea 2 / RT = A2 e-2 Ea1 / RT

k1 A1 e- Ea1 / RT A = 1 e Ea1 / RT = k2 A2 e-2 Ea1 / RT A2

Hence 22. For Hence,

(t + 20 ∞ C)

1

or

ÊAˆ k1 = Á 1 ˜ k2 e Ea1 / RT = Ak2 e Ea1 / RT Ë A2 ¯

t

t

1/ 2 1/ 2 Æ 0.05 M of A æææ Æ 0.025 M of A M of A æææ 2t1/2 = 40 min or t1/2= 20 min. The rate constant of the reaction would be 0.693 0.693 = = 0.03465 min -1 k= 20 min t1 / 2

The rate of rejection when [A] = 0.01 M would be r = k [A] = (0.03465 min–1) (0.01 mol dm–3) = 3.465 ¥ 10–4 mol dm–3 min–1 23. We have fi

k = A e- Ea / RT ln (k2/k1) =

Ea R

fi

ln (k / k ∞) = ln ( A / k ∞) - Ea / RT

Ê 1 1 ˆ Ea Ê T2 - T1 ˆ ÁË T - T ˜¯ = R ÁË T T ˜¯ 1 2 1 2

fi

Ea =

RT1T2 È k ˘ 2.303 log 2 ˙ Í T2T1 Î k1 ˚

Substituting the given values we get È (8.314 J K -1 mol-1 )(300 K )(310 K ) ˘ Ea = Í ˙ (2.303 log 2) = 53.599 J mol–1 = 53.6 kJ mol–1 (310 K - 300 K ) Î ˚ 24. On changing the concentration of B to twice without changing the concentration of A, there is no change in the rate of reaction. Hence, the order of the reaction with respect to B is zero. On changing the concentration of A to twice without changing the concentration of B, the rate of reaction becomes twice. Hence, the order of the reaction with respect to A is one. 25. For the reaction 2SO2 + O2 Æ 2SO3, the rate of reaction is -

d[O 2 ] 1 d[SO 2 ] =– = 2.5¥ 10–4 mol L–1 s–1. 2 dt dt

26. For the reaction

2N2O5 Æ 4NO2 + O2,

Hence,

-

d[SO 2 ] = 5.0 ¥ 10–4 mol L–1 s–1 dt

the rate of expression is

1 d[ N 2 O5 ] 1 d[ NO 2 ] = = k ¢¢[ N 2 O5 ] 2 dt 4 dt d[ N 2 O5 ] 1 d[ N 2 O5 ] k It is given that = k[ N 2 O5 ]; Hence, = [ N 2 O5 ] dt dt 2 2 -

(1)

d[ NO 2 ] 1 d[ NO 2 ] k ¢ = [ N 2 O5 ] Hence, = k ¢[ N 2 O5 ]; 4 dt 4 dt k k¢ i.e. 2k =k ¢ Equation (1) requires that = 2 4 Ea Ê 1 1 ˆ Ê k2 ˆ 27. We have ln Á ˜ = - Á - ˜ Ë k1 ¯ R Ë T2 T1 ¯ Hence,

RT2 T1{2.303 log(k2 / k1 )} (8.314 J K -1 mol-1 )(423 K)(373 K)(2.303) log (1.3 ¥ 10-3 / 1.3 ¥ 10-4 ) Ea = = (423 K - 373 K) (T2 - T1 )

Chemical Kinetics 9.33

=

8.314 ¥ 423 ¥ 373 ¥ 2.303 J mol-1 = 60420 J mol-1 = 60.42 kJ mol–1 50

the corresponding stoichiometric number appeared in the balanced chemical equation. For a reactant, it carries negative sign as its concentration decreases with time. 29. We have 2N2O5(g) æÆ 4NO2(g) + O2(g) 4p p p0 – 2p Total pressure, ptotal = (p0 – 2p) + 4p + p = p0 + 3p After 30 min, ptotal = 87.5 mmHg. Hence p - p0 (87.5 - 50) mmHg = 12.5 mmHg. p = total = 3 3 Partial pressure of N2O5 at 30 min will be pN2O5 = p0 – 2p = (50 – 2 ¥ 12.5) mmHg = 25.0 mmHg. Since the initial pressure of 50 mmHg of N2O5 is reduced to 25 mmHg, the half-life of the reaction will be 30 min. After 60 min (which is equal to two half-lives), the partial pressure of N2O5 will be pN2O5 = 12.5 mmHg. For this value, the value of p will be p0 - pN 2O5 (50 - 12.5) mmHg = 18.75 mmHg. = p= 2 2 Finally, the pressure of the gas will be p = p0 + 3p (50 + 3 ¥ 18.75) mmHg = 106.25 mmHg 30. Only for the choice (c), Ef/Eb = 2/3 31. The rate of reaction will also be doubled. 32. There is low probability of simultaneous collision of all the reacting species. Thus, the reactions with order more than three are not known. 33. TiCl3 is used in Ziegler-Natta polymerization (A-ii) V2O5 is used in Contact process (D-iii) PdCl2 is used in Wacker process (B-i) CuCl2 is used in Deacon’s process (C-iv) 34. Slow step is the rate-determining step. Hence, the rate constant will be 5.2 ¥ 109 L mol–1 s–1 35. The rate constant remains constant irrespective of the concentrations of A and B. 36. The decomposition reaction is

t

Hence,

1 H 2O2 Æ H 2O + O2 2 t

1/ 2 1/ 2 0.5 M ææ Æ 0.25 M ææ Æ 0.125 M 2t1/2 = 50 min, i.e. t1/2 = 25 min

The rate constant of the reaction is 0.693 0.693 k= 2.772 ¥ 10–2 min–1 = 25 min t1/ 2 1 d[O 2 ] = k [H2O2] The rate expression is (1 / 2) dt When [H2O2] = 0.05 M, the rate of formation of O2 will be d[O 2 ] 1 = (2.772 ¥ 10–2 min–1) (0.05 mol L–1) = 6.93 ¥ 10–4 mol L–1 min–1 dt 2

10 Surface Chemistry The term ‘adsorption’ implies the presence of excess concentration of any particular component (known as adsorbate) at the surface of liquid or solid phase (known as adsorbent) as compared to that present in the bulk of the material. This is due to the presence of residual forces at the surface of the body. The process of adsorption is an exothermic process and is associated with a decrease in entropy of the system, such that |DH | < T DS as DG of the process is negative. On the basis of the forces of attraction between adsorbent and adsorbate, two types of adsorption, namely, physisorption The extent of adsorption of gases increases with increase in the pressure of the gases and it decreases with increase in temperature of the gas. The variation in the mass (x) of gas adsorbed by a given mass (m) of an adsorbent (say, charcoal) with change in the pressure of the gas at a constant temperature may be expressed by the following equations (known as adsorption isotherms). (k is constant and n Freundlich equation (x/m) = kp n; k p); (k and k2 are constants) Langmuir equation (x/m) = k k2 p Table 1

2. 3. 4. 5. 6. 7.

Characteristics of Physisorption and Chemisorption

Physisorption The forces of attractions are of van der Waals type (weak forces) Predominates at low temperature All gases show this adsorption at low temperatures Heat of adsorption is low, about 40 kJ mol Reversible in nature Low activation energy ( Adsorption is multilayer

5 kJ)

Chemisorption The forces of attraction are of a chemical nature (strong forces) Predominates at high temperature Heat of adsorption is large ( Usually irreversible Large activation energy

80 to 420 kJ mol )

Adsorption is monolayer

According to Freundich equation, log (x/m) varies linearly with log p. The actual plots show a slight curvature, especially at low temperatures. Langmuir equation predicts the linear variation between p/(x/m) and p. At low pressure, the Langmuir equation is reduced to the form x/m = kp and at high pressures, it is reduced to x/m = k; hence at intermidiate pressures, an equation of the type x/m = kp n (Freundlich equation) may be applicable for the adsorption. The above expressions are also applicable for the adsorption of acetic acid or oxalic acid from its solution by activated charcoal. Colloids or sols are the substances whose sizes lie in between the solutes present in a true solution (e.g. salt, sugar,

but not through perchment paper or animal membrane.

10.2 Complete Chemistry—JEE Main

Table 2

Types of colloidal systems

Dispersed phase Solid

Dispersed medium Solid

Colloidal system Solid sol

Solid

Liquid

Sol

Solid Liquid Liquid

Gas Solid Liquid

Solid aerosol Gel Emulsion

Liquid Gas Gas

Gas Solid Liquid

Liquid aerosol Solid foam gases Foam or Froth

Examples Coloured gems and glasses, some alloys, minerals Starch or proteins in water, paints, gold sol Smoke, dust, storm Jellies, Cheese, butter, boot polish medicines Mist, fog, cloud, insecticide sprays Styrene foam, rubber, occluded Whipped cream, lemonade froth, soap suds

former is less stable and gets coagulated by adding electrolytes, heating or agitation. They carry charges. On the other hand, lyophilic sols are quite stable and are not easily coagulated. They do not carry charges. nature of the particle. A multimolecular colloid consists of an aggregate of small particles held together by van der Waals forces. A macromolecular particle is itself a large molecule (e.g. starch, cellulose and proteins). An associated colloidal system behaves as normal solution at low concentration but becomes colloidal at higher concentrations. Examples include soap and synthetic detergents. The long chain RCOO– of these molecules associate at higher concentrations and form micelles. A mixture of colloidal particles and true electrolytes may be separated by using dialysis or electrodialysis method. The mixture is taken in a bag made from parchment paper or a cellophane membrane. This bag is dipped in warm water. Through the parchment paper or cellophane membrane only true particles are able to pass through. The colloidal particles remain within the bag. The colloidal particles in a solution is a two-phase system. These exhibit Brownian movement (zig-zag motion in all possible directions), Tyndall effect (scattering of light), electrophoresis (preferential movement in the presence of electrical potential), osmotic pressure, diffusion and sedimentation. The colloidal particles in solution acquire charges due to preferential adsorption of ions. For example, ferric hydroxide sol is positively charged due to adsorption of Fe – ions and becomes positively charged in a solution containing silver nitrate due to the adsorption of Ag ions. The colloidal particles get coagulated with the addition of electrolytes due to the neutralization of charges. According to Hardy-Schulze rule, larger the charge on the ion, larger its coagulation ability. Certain lyophilic sols can protect the lyophobic sols from coagulation by electrolytes. Zsigmondy introduced the term

solution. Soap (or detergent) molecules involve a long chain fatty acid terminating in a carboxylate anion (or sulphonate acid points outwardly. The hydrocarbon chain is easily miscible with the grease on the clothes and encapsulates it along with the dirt to form a micelle. These micelles are removed by rinsing with water. Emulsions are sols of liquid in liquid. Two types of emulsions may be distinguished, namely, oil-in-water and waterin-oil. To make emulsions stable, emulsifying agent such as soaps and detergents are added. Any substance which can decrease the surface tension of water to a large extent is known as surfactant. Examples are soap and detergents. Such substances have larger concentrations at the surface of water as compared to the bulk of the solution.

Surface Chemistry 10.3

MULTIPLE CHOICE QUESTIONS Adsorption and Catalysis Physical adsorption (a) involves the weak attractive interactions between the adsorbent and adsorbate (b) involves the chemical interactions between the adsorbent and adsorbate (c) is irreversible in nature (d) increases with increase in temperature 2. Chemisorption (a) involves the weak attractive interactions between the adsorbent and adsorbate (b) is irreversible in nature (c) decreases with increase in temperature (d) involves multilayer adsorption 3. Which of the following is not correct? (a) The extent of adsorption depends on the nature of the adsorbent and adsorbate (b) The extent of adsorption depends on the pressure of the gas (c) The extent of adsorption depends on the temperature (d) The extent of adsorption has no upper limit 4. Which of the following statements regarding adsorption is not true? (a) The phenomenon of adsorption implies the presence of excess concentration of adsorbate at the surface of adsorbent (b) The phenomenon of adsorption is due to the presence of residual forces at the surface of the body (c) During adsorption, there occurs a decrease in free energy of the system (d) During adsorption, there occurs an increase in entropy of the system 5. Which of the following statements regarding adsorption is not correct? (a) The process of adsorption is an exothermic process (b) The substances being adsorbed is known as adsorbate (c) The substance on which adsorption occurs is known as adsorbent (d) The activation energy in chemisorption is smaller as compared to that of physisorption 6. Which of the following statements is not correct? (a) Physical adsorption is monolayer (b) Physical adsorption is reversible in nature (c) Physical adsorption involves low activation energy (d) The extent of physical adsorption decreases with increase in temperature 7. Which of the following statements is not correct? (a) Chemisorption is monolayer (b) Chemisorption is irreversible in nature (c) Chemisorption involves larger activation energy (d) The extent of chemisorption initially decreases with increase in temperature x =kp m

n

the value of n is (a) always greater than one (b) always smaller than one (c) always equal to one (d) greater than one at low temperature and is smaller than one at high temperature

10.4 Complete Chemistry—JEE Main

9. Which of the following graphs would yield a straight line plot? (a) x/m versus p (b) log x/m versus p (c) log x/m versus log p

(d) x/m versus log p

(a) The extent of physical adsorption increases linearly with increase in pressure in the low pressure region (b) The extent of physical adsorption attains a limiting value at the high pressure region (d) Physical adsorption involves the reversible process adsorption desorption

GS

where G, S and GS represent, respectively, the unabsorbed gaseous molecules, adsorption sites and adsorbed gaseous molecules. (a) (b) (c) (d)

Catalyst is not involved in the reaction The concentration of a catalyst remains constant throughout the progress of chemical reaction The mechanism of catalytic reactions may vary from reaction to reaction NO acts as a homogeneous catalyst in the oxidation of SO2 into SO3

(b) Catalyst operates by providing alternate path for the reaction that involves a lower energy of activation (c) Catalyst lowers the energy of activation of the forward reaction without affecting the energy of activation of the backward reaction (d) Catalyst does not affect the overall enthalpy change of the reaction (a) The catalyst changes not only the rate of forward reaction but also that of the backward reaction (b) The catalyst changes the value of equilibrium constant of the reaction (c) The mechanism of a catalytic reaction depends on the type of the catalyst, i.e. whether it is homogeneous or heterogeneous (d) Enzymes are essentially proteins which are responsible for the catalysing reactions occurring in living matter

(a) (b) (c) (d)

Most heterogeneous catalytic reactions involve the solid surface of the catalyst Heterogeneous catalysts primarily function by lowering the activation energy of the reaction A solid catalyst present in the powder form is more effective as it has larger surface area The catalyst may be deactivated by heating it to a high temperature in vacuum

(a) plot of p/(x/m p is linear x/m) versus p Colloidal Solution

(b) plot of p/(x/m) versus p is linear x/m p is linear

Which of the following statements is not true for a lyophobic sol?

20. Which of the following statements is not true for a lyophilic sol?

Surface Chemistry 10.5

(a) A colloidal solution is a heterogeneous two-phase system (b) Liquid-liquid colloidal system is known as emulsion (c) Silver sol in water is an example of lyophilic sol (d) Metal hydroxides in water are examples of lyophobic sol 22. The diameters of colloidal particles may range from 23. Which of the following colloidal system represents a gel? (a) Solid in liquid (b) Solid in gas (c) Liquid in solid (d) Liquid in gas 24. Which of the following colloidal system represents a sol? (a) Solid in liquid (b) Solid in gas (c) Liquid in solid (d) Liquid in gas 25. Which of the following represents a multimolecular colloidal particles? (a) Sol of sulphur (b) Starch (c) Soaps (d) Proteins 26. Which of the following represents a macromolecular colloidal particles? (a) Sol of gold (b) Cellulose (c) Soaps (d) Synthetic detergents 27. Which of the following represents an associated colloids? (a) Sol of gold (b) Starch (c) Proteins (d) Soaps 28. Which of the following does not represent macromolecular colloidal particles? (a) Nylon (b) Plastics (c) Rubber (d) Soaps 29. Which of the following statements is not correct? (a) Peptization is the process by which certain substances are converted into the colloidal state when shaken in water containing a minute amount of an electrolyte (b) Metal sols of gold, silver, platinum, etc. can be prepared by Bredig’s arc method (d) Dialysis is a process with the help of which impurities (made up of ions and molecules) present in a sol can be conveniently removed 30. Which of the following ions is most effective in the coagulation of a ferric hydroxide sol? (a) Cl– (b) Br– (c) NO–3 (d) SO2– 4 (a) dialysis (c) mechanical dispersion 32. Which of the following sols is negatively charged? (a) Arsenius sulphide (c) Ferric hydroxide 33. Which of the following sols is positively charged? (a) Silver iodide in potassium iodide solution (c) Gold

(b) peptization (d) oxidation (b) Aluminium hydroxide (d) Silver iodide in AgNO3 solution (b) Ferric hydroxide (d) Silver

(a) Na

(b) Mg

(c) Ca

(d) Al

(a) Na

(b) Mg

(c) Ca

(d) Al

(a) Cl– (b) Br– (c) SO2– (d) [Fe(CN)6]3– 4 37. The addition of soap in water (a) increases its surface tension (b) decreases its surface tension (c) increases its surface tension at low concentration and decreases at high concentration. (d) decreases its surface tension at low concentration and increases at high concentration.

10.6 Complete Chemistry—JEE Main

38. The presence of electric charge on colloidal particles can be illustrated by the technique of

40. Which of the following solutions changes the colour from red to blue of a colloidal gold solution?

(a) lyophilic colloid (b) lyophobic colloid (c) gel (d) emulsion 42. At isoelectric point, (a) a colloidal particle moves towards cathode during electrophoresis (b) a colloidal particle moves towards anode during electrophoresis (c) a colloidal particle does not move either towards cathode or towards anode during electrophoresis (d) pH of medium becomes 7. 43. The Brownian motion is due to the (c) convection current 44. Tyndall effect is due to

(d) impact of solvent molecules on the colloidal particles

(c) refraction of light by colloidal particles (d) absorption of light by colloidal particles 45. An As2S3 sol carries a negative charge. The maximum precipitating power for this sol is shown by (a) K2SO4 (b) CaCl2 (c) Na3PO4 (d) AlCl3 46. Milk is (a) fat dispersed in water (b) water dispersed in fat (c) fat and water dispersed in an oil (d) a homogeneous solution of fat and water 47. Which of the following statements is correct?

48. Peptization process involves

ANSWERS

25. (a)

26. (b)

27. (d)

28. (d)

29. (c)

30. (d)

43. (d)

44. (a)

45. (d)

46. (a)

47. (b)

48. (c)

Surface Chemistry 10.7

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN (a) (b) (c) (d)

Both enthalpy and entropy of adsorption are negative Adsorption on solids is reversible Adsorption increases with increase in temperature Adsorption is spontaneous

[2003]

respectively. Which of the following statements is not correct? (a) Mixing of the sols has no effect (b) Coagulation in both the sols can be brought about by electrophoresis (d) Sodium sulphate solution causes coagulation in both the sols. [2005] 3. The volume of a colloidal particle, Vc, as compared to the volume of a solute particle in a true solution, Vs, could be –3 3 23 (b) Vc/Vs (c) Vc/Vs Vc/Vs [2005] (a) Vc/Vs (a) (b) (c) (d)

the mass of gas striking a given area of surface is independent of the pressure of the gas the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered the adsorption at a single site on the surface may involve multiple molecules at the same time the mass of gas striking a given area of surface is proportional to the pressure of the gas [2006]

of their protective powers is (a) B < D < A < C (b) D < A < C < B (c) C < B < D < A 6. Which of the following statements is incorrect regarding physisorptions? (a) Under high pressure, it results into multimolecular layer on adsorbent surface (b) Enthalpy of adsorption (DadsH) is low and positive

(d) A < C < B < D [2008]

[2009] 7. According to Freundlich adsorption isotherm, which of the following is correct? (a) (x/m) μ p0 (b) (x/m) μ p (c) (x/m) μ p n (d) All the above are correct for different ranges of pressure [2012] 8. The coagulating powers of electrolytes having ions Na , Al and Ba for arsentic sulphide sol increases in the order: (b) Na < Ba < Al (a) Al < Ba < Na (c) Ba < Na < Al (d) Al < Na < Ba [2013] 9. The following statements relate to the adsorption of gases on a solid surface. incorrect statement among them: (a) Enthalpy of adsorption is negative (b) Entropy of adsorption is negative (c) On adsorption, the residual forces on the surface is increased (d) On adsorption decrease in surface energy appears as heat [2015, online]

10.8 Complete Chemistry—JEE Main

molar concentration? (a) CH3 – (CH2) N (CH3) Br–

(b) CH3 – (CH2) – OSO–3 Na

(c) CH3 – (CH2)8 – COO– Na

(d) CH3 – (CH2) N (CH3) Br–

[2015, online]

charcoal is [2015]

(a) Adsorption is monolayer. (b) Adsorption increases with increase in temperature. . (d) Energy of activation is low.

[2016, online]

and stir well.

and stir well.

(a) Gelatin

[2016, online]

(b) Starch

(c) Oleate

(d) Gum Arabic [2016, online] x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct? (k and n are constants) n n appears as the slope n) appears as the intercept (d) Both k n) appear in the slope term [2016]

ANSWERS

HINTS AND SOLUTIONS 3 3. The probable answer could be Vc/Vs as the size of the colloidal particle will be greater than that of a particle in a true solution. 4. Langmuir adsorption involves monolayer formation 5. Larger the gold number, lesser the protective power. 6. Adsorption is an exothermic process and thus DadsH has a negative value.

Surface Chemistry 10.9

7. Freundlich adsorption isotherm is generally represented as (x/m) = kp n with n imited range of pressure. At low pressure, x/m is generally vary linearly with pressure. Hence, (x/m) μ p. x/m) does not At high pressure, (x/m) attains a constant value. Hence, (x/m) μ p0 increase as fast as pressure. Hence (x/m) μ p n with n of pressure. 8. The larger the charge, larger the coagulating power. Thus the order is Na < Ba < Al . 9. On adsorption, the residual forces on the surface is decreased. character increases. n = V(N – N2) = (0.05 L)[(0.06–0.042) mol L

–4

mol

Mass of acetic acid adsorbed is m = nM

–4

mol) (60 g mol

–3

chloride solution. A better protective colloid requires minimum gold number. Hence, gelatin is the better protective colloid. x = kp n m Ê xˆ Taking logarithm, we get log Á ˜ = log k log p Ë m¯ n n.

11 Chemical Families–Periodic Properties

Dobereiner’s Triads (1829)

Newland Law of Octaves (1864)

Mendeleev Law

the properties of elements are periodic function of their atomic masses

Modern Periodic Table the properties of elements are periodic function of their atomic numbers

Periods

Table 1

11.2 Complete Chemistry—JEE Main

Chemical Families–Periodic Properties 11.3

ns

Table 2

np

n

Number of elements in the different periods of modern periodic table

Period Number*

Orbitals being

Number of elements

Groups

Nobel Gases n

n

n n

Group n

n

n

n

ns

n

n

n

n n n n representative elements

transition elements

n

Z Z

11.4 Complete Chemistry—JEE Main

n n– ns inner transition elements

Periodicity in Properties

Van der Waals radius Metallic radius Covalent bond radius

ns

Chemical Families–Periodic Properties 11.5

|

–

Ed (AB) - [ Ed (AA) + Ed (BB) / 2

| Ed

H=

=

Oxidation States

IE + EA 2

11.6 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS

Chemical Families–Periodic Properties 11.7

11.8 Complete Chemistry—JEE Main

Questions 41 to 44 are based on the following passage. The long form of the periodic table is outlined below.

General Characteristics

Chemical Families–Periodic Properties 11.9

n

n

11.10 Complete Chemistry—JEE Main

Chemical Families–Periodic Properties 11.11

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

– –

– –

– –

–

–

11.12 Complete Chemistry—JEE Main

ANSWERS

Chemical Families–Periodic Properties 11.13

HINTS AND SOLUTIONS

11.14 Complete Chemistry—JEE Main

– – –

n

Chemical Families–Periodic Properties 11.15

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN [2003] [2003]

[2003] H3 [2003] –

[2004]

[2004] – –

–

–

DH DH

11.16 Complete Chemistry—JEE Main

– –

[2004]

[2004] [2004] –

[2004] [2006] correct

[2006]

[2007]

[2009]

[2009] –

–

– –

–

[2010]

[2011 Cancelled]

[2011]

Chemical Families–Periodic Properties 11.17

–

–

–

–

[2012]

[2013] [2013]

Æ 19 Æ Æ Æ

[2014, online]

[2014, online] -

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

[2014, online] -

[2014, online]

[2015, online] –

[2015] [2016]

[2016, online]

ANSWERS

11.18 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS ¢

–

+

–

.

Chemical Families–Periodic Properties 11.19

-

-

-

-

–

n n

n

12

General Principles and Processes of Isolation of Metals

Only a few metals such as gold, silver, copper and platinum occur in nature in the free or native state. Most of the metals occur in the form of their compounds such as oxides, carbonates, sulphides and silicates. Such naturally occurring sources are called minerals ore. Thus, an ore is a mineral but a mineral may not be an ore. For example, iron pyrites, FeS2, is a mineral but not an ore. In this unit, we shall describe some commercially important ores of Fe, Cu, Al and Zn and the principles involved in the extraction of the metals from their important ores.

Iron Iron occurs in the free state as meterorites which also contain 20 to 30% nickel. In the combined state, iron occurs in the following minerals. Magnetite, Haematite, Limonite, Fe2O3 3Fe2O3 ◊ 3H2O Fe3O4 Spathic iron ore, Iron pyrites, Copper pyrites, FeS2 CuFeS2 FeCO3 Iron is extracted by the carbon reduction method.

Extraction of Iron Iron is extracted from its principal ore, haematite. After the preliminary washing, concentration and roasting, the ore is smelted in the presence of coke and limestone in a blast furnace (Fig. 1). Roasted ore (8 parts) with desulphurized coke (4 parts) and limestone pieces (1 part) is fed into the blast furnace from the top. P is a temperature gradient as we move from the bottom (temperature about 2000 K) to the top (temperature about 500 K) of the blast furnace. The blast furnace may be broadly divided into three main parts as described in the following. 1. Zone of Fusion The lower portion where coke burns and produce carbon dioxide and a lot of heating is known as zone of fusion: DH = - 406 kJ mol -1 C + O2 Æ CO2 Here the temperature is about 1775 K. A little above this, where temperature is about 1475 K - 1575K, iron coming from above melts. 2. Zone of Heat Absorption The middle portion (temperature 1075 K - 1275 K), CO2 rising up is reduced to CO with the absorption of heat: CO2 + C Æ 2CO DH = 163 kJ mol -1 combines with silica (present as impurity-gangue) to form calcium silicate (fusible slag): CaCO3 Æ CaO + CO2 CaO + SiO2 Æ CaSiO3

12.2 Complete Chemistry—JEE Main

Fig. 1

Blast furnace

3. Zone of Reduction The upper portion (675 K-975 K) where iron oxide is reduced to spongy iron by carbon monoxide rising up the furnace: Fe2O3 + 3CO Æ 2Fe + 3CO2 The reduction is believed to take place in stages: 3Fe2O3 + CO Æ 2 Fe3O4 + CO2 Fe3O4 + CO Æ 3FeO + CO2 FeO + CO Æ Fe + CO2 molten iron from oxidation. These two can be removed from different holes (Fig. 12.1). Waste gases escaping at the top consists of about 30% CO, 10% CO2 and the rest nitrogen. Iron obtained from the blast furnace is known as pig iron. Pig iron contains about 2-5% carbon as well as other impurities (usually Si, Mn, S and P). Pig iron is converted into various articles. Wrought iron, which is the purest form of iron, can be obtained by heating cast iron in a reverberatory furnace lined with iron oxide. Wrought iron contains about 0.2% carbon.

Zinc The chief ore of zinc are (i) zinc blende (ZnS), (ii) calamine (ZnCO3), zincite (ZnO) and franklinits (ZnO ◊ Fe2O3). The extraction of ore is carried out by the carbon reduction of ZnO which is obtained by roasting the ore and concentrating 4 which is obtained by roasting the ore at moderate temperature and then dissolving it in sulphuric acid. Copper Copper occurs in the native state as well as in the compound form. The natural ores of copper are Malachite, Cu(OH)2 ◊ CuCO3 Copper glance, Cu2S Copper pyrites, CuFeS2 Azurite, Cu(OH)2 ◊ 2CuCO3 Cuprite or ruby copper, Cu2O Copper may be extracted by self-reduction method.

General Principles and Processes of Isolation of Metals 12.3

Extraction of Copper The ores of Copper are: copper pyrites (CuFeS2), malachite (CuCO3 2), cuprite (Cu2O) and copper glance (Cu2S). Copper is mainly extracted from copper pyrites. After the concentration of its ore reactions occurring are (i) 2CuFeS2 + O2 Æ Cu2S + 2FeS + SO2 (major reaction) (ii) 2Cu2S + 3O2 Æ 2Cu2O + 2SO2 (iii) 2FeS + 3O2 Æ 2FeO + 2SO2 (minor reactions) The ore is then mixed with a little coke and sand and smelted in a water-jacketed blast furnace. The minor reactions that occurred during roasting continue here. Ferrous oxide combines with sand to form a fusible slag. Cuprous oxide for oxygen than copper. (iv) FeO + SiO2 Æ FeSiO3 (v) Cu2O + FeS Æ Cu2S + FeO Molten mass collected from the bottom of furnace contains largely cuprous sulphide and a little ferrous sulphide. This molten mass is known as matte. A blast of sand and air is blown in the converter through tuyeres which are situated a little above the bottom. This causes removal of S and As as oxides and ferrous oxide as slag (reaction iv). At the same time Cu2S is oxidized mostly into Cu2O (reaction ii) and partly into CuO and CuSO4. All these react with Cu2S giving copper. The reactions are (vi) 2Cu2S + 3O2 Æ 2Cu2O + 2SO2≠ 2Cu2S + 5O2 Æ 2CuSO4 + 2CuO 2Cu2O + Cu2S Æ 6 Cu + SO2≠ CuSO4 + Cu2S Æ 3Cu + 2SO2≠ Cu2S + 2CuO Æ 4Cu + SO2≠

Fig. 2

Bessemer converter

Aluminium Aluminium does not occur free in nature. In the combined state, it occurs in the following forms. Oxides: Corundum, Al2O3 2O3 ◊ H2O and bauxite, Al2O3 ◊ 2H2O. Fluorides: Cryolite, Na3AlF6 Silicates: Feldspar, KAlSi3O8, mica (KAlSi3O10(OH)2) and kaolinite (Al(OH)4, Si2O5) Basic Sulphates: Alunite or alumstone, K2SO4 ◊ Al2(SO4)3 ◊ 4Al(OH)3 Basic Phosphates: Turquoise, AlPO4 ◊ Al(OH)3 ◊ H2O. Aluminates: Aluminates of Mg, Fe and Mn. Aluminum is the third most abumdant element of earth’s crust. Extraction of aluminium Aluminium is isolated from the electrolysis of bauxite, Al2O3 ◊ 2H2 upon the impurity present in the ore. If the bauxite contains iron oxide as the impurity, one can use Baeyer’s or Hall’s process as described below.

Baeyer’s Process Finally ground ore is roasted to convert ferrous oxide to ferric oxide and then digested with concentrated caustic soda solution at 423 K. Al2O3 dissolves while Fe2O3 and from the solution Al(OH)3 is precipitated by adding a weak acid. The ignition of Al(OH)3 gives Al2O3. Al2O3 + 2OH - + 3H2O Æ 2Al(OH)–4 aluminate ion dissolves

Al(OH)–4

+ H Æ Al(OH)3 + H2O +

precipitates heat

2Al(OH)3 ææÆ Al2O3 + 3H2O

12.4 Complete Chemistry—JEE Main

Hall’s process In this process the ore is fused with sodium carbonate when soluble metaaluminate (NaAlO2) is produced. This is extracted with water leaving behind iron oxide. Carbon dioxide at 323-333 K is passed through water extract to get Al(OH)3 which on heating gives Al2O3. fused

Al2O3 + Na2CO3 ææÆ 2NaAlO2 + CO2 extracted with water

2NaAlO2 + 3H2O + CO2 Æ 2Al(OH)3 + Na2CO3 heat

2Al(OH)3 ææÆ Al2O3 + 3H2O. If the impurity is silica, the Serpek’s process is used to purify bauxite.

Serpek’s Process The powdered ore is mixed with coke and heated to 2075 K in a current of nitrogen. Silica present is reduced to silicon which volatilizes off and alumina gives aluminium nitride. The hydrolysis of the latter gives Al(OH)3, heating of which gives Al2O3. SiO2 + 2C Æ Si ≠ + 2CO2≠ Al2O3 + 3C + N2 Æ 2AlN + 3CO AlN + 3H2O Æ Al(OH)3 + NH3 2Al(OH)3 ææÆ Al2O3 + 3H2O heat

After obtaining pure Al2O3, it is dissolved in fused cryolite, Na3AlF6 2, and is electrolysed in an iron tank lined with blocks of carbon which serve as the cathode. The anode consists of a number of graphite rods suspended vertically inside the tank (Fig. 3). Aluminium gets settled at the bottom of the tank and can be removed. The reactions occurring at the electrodes are Cathode Al3+ + 3e– Æ Al – Anode 2O2– 2 Æ O2 + 4e C + O2 Æ CO2 Anode is replaced periodically because of its consumption.

Fig. 3

Electrolytic cell for the production of aluminium

MULTIPLE CHOICE QUESTIONS Iron 1. In the metallurgy of iron, when limestone is added to the blast furnace, the calcium ions ends up in (a) slag (b) gangue (c) metallic Ca (d) calcium carbonate 2. Which of the following minerals does not contain iron? (a) Magnetite (b) Magnesite (c) Haematite (d) Limonite

General Principles and Processes of Isolation of Metals 12.5

3. The principal ore of iron is (a) haematite (b) iron pyrites (c) copper pyrites (d) spathic iron 4. The principal reducing agent in the metallurgy of iron is (a) carbon (b) carbon monoxide (c) carbnon dioxide (d) aluminium 5. The principal reaction in the zone of fusion of blast furnace employed in the metallurgy of iron is (a) C + O2 Æ CO2 (b) 2C + O2 Æ 2CO (c) CO2 + C Æ CO (d) Fe2O3 + 3CO Æ 2Fe + 3CO2 6. The principal reaction(s) in the zone of heat absorption of blast furnace employed in the metallurgy of iron is/are (a) C + O2 Æ CO2 (b) 2C + O2 Æ CO – CO2

7.

8.

9. 10.

SiO2

(c) 2CO + O2 Æ 2CO2 (d) CaCO3 ææÆ CaO ææÆ CaSiO3 Iron is mainly extracted by (a) carbon reduction method (b) self-reduction method (c) the method of electrolysis (d) leaching with aqueous solution of NaCN followed by reduction Which of the following sequences of carbon content is correct? (a) steel < cast iron < wrought iron (b) steel < pig iron < wrought iron (c) steel < wrought iron < cast iron (d) wrought iron < steel < cast iron In the metallurgy of iron, the material obtained from the bottom of blast furnace is (a) slag (b) pig iron (c) cast iron (d) wrought iron The chemical processes in the production of steel from haematite ore involve (a) reduction (b) oxidation (c) reduction followed by oxidation (d) oxidation followed by reduction

Aluminium 11. In the electrolysis of alumina, cryolite is added to (a) lower the melting point of alumina (b) decrease the electrical conductivity (c) minimise the anode effect (d) remove impurities from alumina 2) which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na3AlF6) is (a) that of a catalyst (b) to make the fused mixture very conducting (c) to lower the melting temperature of the mixture (d) to decrease the rate of oxidation of carbon at the anode. 13. In the electrolytic extraction of aluminium, the solvent is (a) molten Al2O3 (b) water (c) Fe2O3 + Al (d) molten cryolite 14. Mineral that does not contain Al is 15. The main ore of aluminium is (a) bauxite (b) alumina (c) potash alum (d) cryolite 16. Which of the following statements regarding the metallurgy of aluminium by electrolytic method is not correct? (a) Electrolyte is Al2O3 dissolved in Na3AlF6 containing a little of CaF2 (b) Anode consists of a number of graphite rods which are periodically replaced

17. In the commercial electrochemical process for aluminium extraction the electrolyte used is (a) Al(OH)3 in NaOH solution (b) an aqueous solution of Al2(SO4)3 (c) a molten mixture of AlO(OH) and Al(OH)3 (d) a molten mixture of Al2O3 and Na3AlF6

12.6 Complete Chemistry—JEE Main

18. Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out (a) in the presence of NaCl (c) in the presence of cryolite which forms a melt with lower melting temperature (d) in the presence of cryolite which forms a melt with higher melting point

Copper and Silver 19. Copper is mainly extracted from (a) cuprite (b) azurite (c) copper pyrites 20. The composition of cuprite is (a) Cu2S (b) CuFeS2 (c) Cu(OH)2 · CuCO3 21. Copper is mainly extracted by (a) carbon reduction method (b) self-reduction method (c) the method of electrolysis (d) leaching with aqueous solution of NaCN followed by reduction

(d) malachite (d) Cu2O

(a) silver (b) lead (c) copper (d) iron 23. In the metallurgy of copper, the material obtained from the bottom of blast furnace is (a) slag (b) copper (c) matte (d) cuprous oxide (a) Zn, Cu, Ag, Au

(b) Zn, Ag, Au

(c) Cu, Ag, Au

(d) Au

Additional Problems 25. Of the following, the metals that cannot be obtained by electrolysis of the aqueous solution of their salts are (a) Ag (b) Mg (c) Cu (d) Au 26. The principal reducing agent in the metallurgy of tin is (a) carbon (b) carbon monoxide (c) carbon dioxide (d) aluminium 27. Calamine is (a) ZnS (b) ZnCO3 (c) ZnO (d) ZnSO4 28. Zinc is obtained from ZnO by (a) carbon reduction method (b) reduction by H2 (c) boiling giving Zn and O2 (d) treating of KCN followed by electrolysis 29. The impure zinc contains impurities such as cadmium, arsenic, iron and lead. From this, zinc may be recovered by (a) distillation at 500 K (b) distillation at 1170 K – 1270 K followed by passing a column at 1070 K (c) distillation at 3100 K (d) solvent extraction method

ANSWERS 1. 7. 13. 19. 25.

(a) (a) (d) (c) (b)

2. 8. 14. 20. 26.

(b) (d) (b) (d) (a)

3. 9. 15. 21. 27.

(a) (b) (a) (b) (b)

4. 10. 16. 22. 28.

(b) (d) (c) (a) (a)

5. 11. 17. 23. 29.

(a) (a) (d) (c) (b)

6. 12. 18. 24.

(d) (c) (c) (d)

General Principles and Processes of Isolation of Metals 12.7

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The substance not likely to contain CaCO3 is (a) dolomite (c) calcined gypsum

(b) a marble statue (d) sea shells

[2003]

(a) magnetite (b) cassiterite (c) galena (d) malachite [2004] 3. Aluminium oxide may be electrolysed at 1000 °C to furnish aluminium metal (atomic mass : 27 amu). –1 , the electricity required to prepare The cathodic reaction is Al3+ + 3e– 5.12 kg of aluminium is (b) 5.49 × 101 C (c) 5.49 × 107 C (d) 1.83 × 107 C [2005] (a) 5.49 × 104 C 4. Which of the following factors is of no for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? (a) CO2 is more volatile than CS2 (b) Metal sulphides are thermodynamically more stable than CS2. (c) CO2 is thermodynamically more stable than CS2 (d) Metal sulphides are less stable than the corresponding oxides. [2008] K 1700 K ææÆ TiI4 (g) æææÆ Ti(s) + 2I2(g) Ti(s) + 2I2(g) æ523

[2012] 6. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is (a) Cr (b) Ag (c) Ca (d) Cu [2014] 7. The form of iron obtained from blast furnace is: (a) Steel (b) Cast Iron (c) Pig Iron (d) Wrought Iron [2014, online] 8. In the isolation of metals, calcination process usually results in: (a) metal carbonate (b) metal oxide (c) metal sulphide (d) metal hydroxide [2015, online] 9. Calamine is an ore of: (a) Aluminium (b) Copper (c) Iron (d) Zinc [2015, online] 10. In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is false? (a) CO and CO2 are produced in this process (b) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity (c) Al3+ is reduced at the cathode to from Al [2015] (d) Na3AlF6 serves as the electrolyte 11. Extraction of copper by smelting uses silica as an additive to remove (b) FeS (c) FeO (d) Cu2S [2016, online] (a) Cu2O (a) Siderite

(b) Galena

(c) Malachite

(d) Magnetite

ANSWERS 1. (c) 7. (c)

2. (c) 8. (a)

3. (c) 9. (d)

4. (d) 10. (d)

5. (d) 11. (c)

6. (c) 12. (b)

[2016]

12.8 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS 3. Using the expression m = (Q/F) (M/|ne|), we get m F |ν e | (512 . × 103 g) ( 96500 C mol −1 ) ( 3) 5.12 × 9.65 = = ¥ 107 C= 5.49 × 107 C Q= 9 M ( 27 g mol −1 )

reverberatory furnace and blowing a current of air over the molten surface. The impurity is oxidized which is swept away by the air current. Impurities are oxidized and escape as gases. To reduce any formed oxide of the metal, a pole of green wood is added which liberates hydrocarbon gases at high temperature (mainly CH4) helping the reduction of oxide. metal forms volatile metallic iodide and impurities do not react with iodine. The metallic iodide is heated to a high choice d). 6. The alkaline-earth metal Ca cannot be obtained by the electrolysis of an aqueous solution of its salt. In the electrolysis, water is electrolysed. 7. Iron obtained from the blast furnace is known as pig iron. 9. Calamine is the ore of Zn 10. Na3AlF6 also decreases melting point and brings conductivity. The electrolyte is molten Al2O3. The reactions occurring at electrodes are : Cathode Anode

Al3+(melt) + 3e– Æ Al(l) C(s) + O2–(melt) Æ CO(g) + 2e– C(s) + 2O2–(melt) Æ CO2 (g) + 4e–

11. Silica is added to remove FeO as FeSiO3. 12. Siderite Malachite

FeCO3

Galena

PbS

Cu(OH)2 ◊ CuCO3

Magnetite

Fe3O4

13 Hydrogen

Hydrogen 1

+ +

–

–

Similarities to Alkali Metals Table 1 Element 1

H

2

1

Na

2

2

6

1

K

2

2

6

2

6

1

2

2

6

2

6

10

2

Cs

2

6

2

6

10

2 2

6 6

1 10

2

6

+ +

H æÆ H+ + e– æÆ + + e– 2O, 2

2

HCl æÆ H+ + Cl– 2H+ + 2e– æÆ H2

Similarities to Halogens

H2

NaCl æÆ Na+ + Cl– + + e– æÆ

1

) is the

13.2

Table 2 Element H

1 2, 7 2, 8, 7

Br

2, 8, 18, 7 –

H + e æÆ H –

–

+

– –

H–, Na+H–, Ca2+

–

æÆ

–

+

)2

Cl–, Na+Cl– , Ca2+

–

)2

2

and H2

, ZrH

2) n

2) n

)n

Ionization energy/kJ mol-1

2500 2000 F 1500 1000

Cl H

500 0

Na

Li 1

5

10 Atomic number

K 15

20

Fig. 1 2 2

, SiCl NCl and PCl

, SiH and PH are similar to

13.3 + Cl2 , - HCl æææææ ææææ Æ CH 4 ¨ æ + H , - HCl

Methane

+ Cl , - HCl

2 æææææ Æ

CH3Cl Methyl chloride

CH 2 Cl2 Methylene chloride + Cl2

Ø- HCl

+ Cl , - HCl

2 ¨ææææ æ CHCl3

CCl4 Carbon tetrachloride

Chloroform

Occurrence 2

Isotopes of Hydrogen

H 2, D 2

Table 3

2

D2

H2

2

T/K

T/K D

–1

D e

–1

–1

r 2

Nuclear Isomers

Ortho

ortho

ortho

MULTIPLE CHOICE QUESTIONS

13.4

2

+ H2

2

2

OH

and H2

2O

> H 2S > H 2 2O < H 2S > H 2

2

2

2O

> H2S < H2Se 2O < H2S < H2Se

K°w

2O

2 2

2

2

2–

2

–

2O

2

– 2

2

13.5 2O

–

+ 6H+ + H2O2 – + 2H+ + H2O2 2 + 2H2

2+

+ 2H+ + H2O2 + H 2O 2 + 2OH– 6

2+

+ 8H2

6

with 2O 2 2+ + 2OH– + H2O2 + 2H2O + O2 + H 2O 2 + H 2O + O 2 O 2 2 + H 2O 2 2O2 + H2O O 2 2 2O

2

+ 2H2

—O—

ortho orthoortho ortho 2O 2 2O 2 2O 2 2O 2 2O 2

2O 2

2 2O 2

2

+O

2O 2

+ 2H2

2H2O + 2O2

2O 2 2O 2

2

+O

2O2 + H2O

2O 2

13.6 2 2O

2

ANSWERS

HINTS AND SOLUTIONS 2

2S

2

is

smaller than H2 K°w = – log K°w

K°w

2O

2H2O Æ 2O 2 2O 2

is

K°w

+

–

+ O2

2O 2

2O 2 2H2O2

2

2

–1

H 2O 2

2

Æ H 2O + O 2 ¥

–1

2O 2 –1

2

–1

)

–1 2O 2 2O 2

2O 2

2 2O 2

– 2 2O 2

H2O2 Æ H2O + O and O + O Æ 2O2 O Æ O2

Æ

¥

2

13.7

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 2

2

2

2 2Cl2

[2008]

2

[2012] 2O 2

+ – 2O2 + 2H + 2e Æ 2H2 – – 2O2 + 2e Æ 2OH

– 2e– Æ O2 + 2H+ – – 2O2 + 2OH – 2e Æ O2 + 2H2O 2O 2

[2014] 2O 2

MnO -4

Cr2 O72-

SO32-

[2014 online]

[2015 online] wrong [2015 online]

ANSWERS

HINTS AND SOLUTIONS

2O 2

With MnO -4

2O 2

2MnO 4- + 6H + + 5H 2 O 2 Æ 2Mn 2+ +8H 2 O + 5O2 2H 2 O 2 oxidizes SO23 to SO 4 , I to I 2 and

SO32- + H 2 O 2 Æ SO 24- + H 2 O 2 I - + 2H + + H 2 O 2 Æ 2 H 2 O + I 2 Cr2 O72- + 2H + 4H 2 O 2 Æ 2CrO5 + 5H 2 O (perchromate)

Cr2 O72- to CrO3

14 s-Block Elements (Alkali and Alkaline Earth Metals) The Group 1 Elements – Alkali Metals Group 1 of the periodic table is composed of the alkali metals lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs) and francium (Fr). Their physical properties are recorded in Table 1. Table 1

General Physical Properties of Alkali Metals

Property

Li [He](2s)

Na 1

[Ne](3s)

K 1

[Ar](4s)

Rb 1

[Kr](5s)

Cs 1

[Xe](6s)

Fr 1

[Rn](7s)1

Atomic number

3

11

19

37

55

87

Relative atomic mass

6.941

22.990

39.098

85.468

132.905

223

Abundance in earth’s crust (ppm)

18

22700

18400

78

2.6

—

Atomic radius, r/pm

152

186

227

248

265

—

ionic radius, r/pm

76

102

138

152

167

—

453.5

370.8

336.2

312.0

301.5

—

1620.0

1154.4

1038.5

961.0

978.0

—

0.54

0.97

0.86

1.53

1.90

—

520

496

419

403

376

375

1.0

0.9

0.8

0.8

0.7

–3.03

–2.71

–2.93

–2.93

–2.92

T/K T/K Density at 293 K, r

cm

–3

I/kJ mol–1 Standard electrode potential, E°/V at 298 K for M+(aq) + e– Æ M(s)

—

Description of Physical Properties number. The descriptions of these properties are as follows. 1. All are metals with one electron in their outermost orbitals and thus they form unipositive M+ ions(where M is an alkali metal). 2. At normal temperature, all the metals adopt a body-centred cubic type of lattice with a coordination number of 8. 3. The elements are soft and low melting This is due to the contribution of only one electron per atom towards 4. Their atomic and ionic radii

5. Their melting and boiling points

14.2 Complete Chemistry—JEE Main

6. Their densities than water. Their ionization energies

Fig. 1 Metallic and Ionic Radii of Alkali Metals 550

Fig. 2 550

Ionization Energies

500

IE/kJ mo–1

IE/kJ mo–1

500

450

450

350

Ionization Energies

450

450

Li

Na

K

Rb

Cs

Fig. 3 Densities of Alkali Metals

350

Li

Na

Fig. 4

reducing agents. 9. The elements impart characteristic colours

10. are called photoelectrons. 11. The electronegativity values ionic bonds.

K

Rb

Cs

s-Block Elements (Alkali and Alkaline Earth Metals) 14.3

Table 2

Flame Colours of Alkali Metals

Element

Colour

Wavelength l/nm

Li

crimson

670.8

Na

yellow

589.2

K

lilac

766.5

Rb

red-violet

780.0

Cs

blue

455.5

12. The standard electrode potentials, E°(M+ Li+ > Na+ > K+ > Rb+ > Cs+. 13. Alkali metals dissolve in liquid ammonia

 (2e–   e2– 2 ). The deep blue solution is moderately stable at temperatures where ammonia is still liquid. In the presence of Na + NH3(1) Æ NaNH2 + 1 H2 2 14. The conducting ability of alkali ions follows the order Li + < Na+ < K+ < Rb+ < Cs+

Isolation of Sodium and Potassium Sodium occurs in rock salt (NaCl), chile salpeter (NaNO3 ◊ 2 ◊ 6H2O), kainite (KSO4 ◊ 4 ◊ 2 ◊ 6H2O), salt peter (KNO3) and potassium feldspar (K2O ◊ Al2O3 ◊ 6SiO2). Seawater is an important source of sodium chloride and potassium chloride. Sodium and potassium are obtained by the electrolytic reduction method. Sodium is produced by the electrolysis of 2

+

(1)

Na+

Chemical Properties of Alkali Metals them from air. 2. Burning in Air or Oxygen

Æ 2Li2O 2 Æ Na2O2 Æ MO2 2 2

(K,Rb,Cs)

14.4 Complete Chemistry—JEE Main

Explanation

The stability of an ionic compound depends on the relative sizes of its ions. An ionic compound

3. Reaction with Water 2M

+

2H2O Æ 2MOH + H2

(Li, Na, K, Rb, Cs)

4. Reaction with Nitrogen 6Li + N2 Æ 2Li3N 2 Li3N ⎯heat ⎯→ 6Li + N2 Li3N + 3H2O Æ 3LiOH + NH3

Compounds of Alkali Metals 1. Oxides

2–

Na2O2 + 2H2O Æ 2NaOH + H2O2 2]

–

2O2 and O2 are released. 1 KO2 + 2H2O Æ KOH + H2O2 + O2 2

2. Hydroxides increases from LiOH to CsOH. Explanation

is due to the reaction  C2H5OH + OH–   OC2H–5 + H2O

caustic soda.

3. Reaction with Halogens 2M

+

X2

Æ 2MX

s-Block Elements (Alkali and Alkaline Earth Metals) 14.5

MF > MCl > MBr > MI

lithium chloride is due to its covalent character.

Melting points

LICl Sodium halides

Fig. 5

NaCl KCl Alkali chloride

RbCl

CsCl

Fig. 6

The covalent character of LiX (where X = CI, Br, I) increases as the size of halide increases. This is due to the fact that the smaller sized Li+ increases as LiCl < LiBr < LI. Df H For Chlorides, bromides and iodides, Df H For the same alkali metal, Df H low solubility due to smaller enthalpy of hydration Cs+ ion.

4. Carbonates and Bicarbonates 4HCO3.

The solubilities of both

Manufacturing of Sodium Carbonate This compound is manufactured by Solvay process from limestone (CaCO3 CaCO3 Æ CaO + CO2 NH3 + H2O Æ NH+4 + OH– NaCl +

NH+4

+ OH + CO2 Æ NH4Cl + NaHCO3 –

(i) (ii) (iii)

14.6 Complete Chemistry—JEE Main

to obtain sodium carbonate 2NaHCO3 æheat ææ Æ Na2CO3 + H2O + CO2

(iv)

from reaction (i) with ammonium chloride obtained in reaction (iii). CaO + 2NH4Cl Æ 2NH3 + H2O + CaCl2 The by-product in the above process is calcium chloride. Sodium carbonate crystallises as Na2CO3 ◊ 10H2O. It is commonly known as washing soda K Na2CO3 ◊ 10H2O æ375 ææÆ Na2CO3 ◊ H2O + 9H2O K Na2CO3 ◊ H2O æ>373 æææ Æ Na2CO3 + H2O

(white)

Na2CO3 is also known as soda ash.

Fig. 7 Carbonates at 298 K

5. Nitrates

500 °C 2NaNO3     2NaNO2 + O2

800 °C   4NaNO3   2Na2O + 5O2 + 2N2 Nitrites produce brown fumes of NO2 when treated with dilute acids. 2NaNO2 + 2HCl Æ 2NaCl + H2O + NO2 + NO

2NO + O2 Æ 2NO2 2CO3

solution.

Na2CO3 + NO2 + NO Æ 2NaNO2 + CO2 KNO3 + Zn Æ KNO2 + ZnO 2KOH + 4NO Æ 2KNO2 + N2O + H2O 4KOH + 6NO Æ 4KNO2 + N2 + 2H2O.

6. Hydrides

3

4LiH + AlCl3 Æ Li[AlH4] + 3LiCl

7. Reaction with Compounds Containing Acidic Hydrogen 2M + C2H5OH Æ 2C2H5OM + H2 2M + 2HC ∫∫ CH Æ 2MC ∫∫ CH + H2.

Anomalous Behaviour of Litliium and Its Diagonal Relationship with Magnesium 2

diagonal relationship).

s-Block Elements (Alkali and Alkaline Earth Metals) 14.7

4Li + O2 Æ 2Li2O 2

Æ

6Li + N2 Æ 2Li3N 2

Æ

3N 2

Li2CO3 Æ Li2O + CO2 3

Æ

2LiOH Æ Li2O + H2O

2

2

Æ

2O

4LiNO3 Æ 2Li2O + 4NO2 + O2 3) 2

Æ

2

+ O2

2NaNO3 Æ 2NaNO2 + O2 2

are soluble in alcohol. The other alkali metal chlorides remain insoluble.

MULTIPLE CHOICE QUESTIONS 1. The number of alkali metals known so far is (a) 4 (b) 5

(c) 6

(d) 7

(b) Francium is a radioactive element (d) Sodium is used in the photoelectric cells (a) Alkali metals tarnish in air (b) They are kept under kerosene

(a) LiCl < NaCl < KCl < RbCl (c) LiCl < NaCl > KCl > RbCl

(b) LiCl > NaCl > KCl > RbCl (d) LiCl > NaCl < KCl < RbCl

(b) Alkali metals form covalent hydrides with H2 when heated

14.8 Complete Chemistry—JEE Main

(c) Alkali metals in liquid ammonia imparts deep blue solution

2

(a) RbO

are soluble in alcohol whereas chlorides of other alkali metals are insoluble

(b) Rb2O

(a) Na+ ions (c) free protons 13. Ionic radii of alkali metal ions in water are in the order (a) Li+ > Na+ > K+ > Rb+ (c) Li+ > Na+ > K+ < Rb+

(c) Rb2O2

(d) RbO2

(b) conduction electrons (d) a body-centred cubic lattice (b) Li+ < Na+ < K+ < Rb+ (d) Li+ > Na+ < K+ < Rb+

(a) Li+

(b) Na+

(c) K+

(d) Rb+

(a) Li

(b) Na

(c) K

(d) Rb

(a) Li

(b) Na

(c) K

(d) Rb

(a) Li (b) Na (c) K 18. The number of water molecules of crystallization in sodium carbonate is (a) 5 (b) 10 (c) 7 19. Chile salt petre is (a) KNO3 (b) NaNO3 (c) NH4NO3

(d) Rb (d) 2 (d) LiNO3

(a) LiCN

(b) NaCN

(c) KCN

(d) RbCN

(a) LiNO3

(b) NaNO3

(c) KNO3

(d) RbNO3

(c) Sodium thiosulphate (d) Sodium chloride 23. In the titration of sodium thiosulphate and iodine, the equivalent mass of the former is equal to (a) molar mass (b) molar mass/2 (c) molar mass/3 (d) molar mass/4

(b) Francium is an radioactive element (c) Atomic number of francium is 86

s-Block Elements (Alkali and Alkaline Earth Metals) 14.9

(c) The abundance of potassium in the earth’s crust is more than that of sodium

(a) Alkali metals cannot be obtained by the chemical reduction of their compounds (b) Alkali metals are usually obtained by the electrolytic reduction

(a) Sodium and potassium are soft and silvery-white metals

(d) Sodium and potassium are kept under kerosene to avoid the contact with air and moisture O2 2Na2O ædry æÆ 2Na2O2

(a) 4Na + O 2 (dry)

(b) 4K + O2

O2 2K2O ædry æÆ 2K2O2

(c) 2M + 2H2O (d) 2M + 2NH3

2MOH + H2; where M is Na or K 2MNH2 + H2; where M is Na or K 2

· 6H2O

(b) K2SO4 4 (d) K2O · Al2O3 · 6SiO2

(c) KNO3

2

· 6H2O

2 · 6H2O (c) K2SO4 2 · 6H2O 32. Solvay process is used for the manufacture of (a) Na2CO3 · 10H2O (b) K2CO3

(b) K2SO4 (d) K2SO4 (c) NaOH

(d) Na2O2

(a) Na2CO3 · 10H2O (b) NaHCO3 34. Castner-Kellner cell is used for the production of (a) NaOH (b) Na2CO3

(c) K2CO3

(d) NaOH

(c) NaHCO3

(d) NaCl

(a) (b) (c) (d)

4

· 6H2O

4

2

· 6H2O

The main raw material required in Solvay process are sodium chloride and ammonium chloride NaHCO3 is less soluble in water as compared to NH4Cl The byproduct in Solvay process is calcium chloride Na2CO3 · 10H2

(c) Sodium bicarbonate is used to cure acidity in the stomach

(a) Li (a) Na2CO3 and NaHCO3 (c) Na2CO3and NaOH

(b) Na

(c) Rb (b) Na2CO3 and NaOH (c) NaHCO3 and NaCI

(d) Cs

14.10 Complete Chemistry—JEE Main

39. Sodium is manufactued (a) in Nelson cell (c) by Solvay process 40. The Glauber’s salt is 4 . 7H2O

(b) by Downs process (d) by Hall’s process (b) ZnSO4 . 7H2O

(c) FeSO4 . 7Hs2O

(c) NaSO4 . 10H2O

(a) CaCO3 æheat ææ Æ CaO + CO2 (b) NaCl + NH3 + H2O + CO2 (c) CaO + 2NH4Cl

NH4Cl + NaHCO3

2HN3 + H2O + CaCl2

(d) Na2CO3 + CO2 + H2O æheat ææ Æ 2NaHCO3

(a) Li2CO3

(b) Na2CO3

(c) K2CO3

(d) CsCO3

(a) Li2CO3

(b) Na2CO3

(c) K2CO3

(d) CsCO3

(b) KO2

(c) K2O2

(d) KO

(a) K2O 46. The compound KO2 is a

47. The useful product and by-product obtained in the Solvay's process are (a) quick lime and CO2 (b) NaHCO3 and NH4Cl (c) Na2CO3 and CaCl2 (d) NaHCO3 and CO2 48. Brine is (a) concentrated solution of NaCl (b) concentrated solution of KCl (c) concentrated solution of Na2CO3 (d) concentrated solution of NaHCO3 49. Metallic character of alkali metals (a) increases with increase in atomic number (b) decreases with increase in atomic number (c) increases followed by a decrease with increase in atomic number (d) decreases followed by an increase with increase in atomic number (a) 1

(b) 2

(c) 13

(a) Li2O2 (b) Na2O2 (c) K2O2 52. In the reaction Na2S + O2 + H2O Æ A + NaOH the product A is (a) Na2SO3 (b) Na2S2O3 (c) Na2S4O6 53. The correct order of density of alkali metal follows the order (a) Li < Na < K (b) Li < K < Na (c) Na < Li < K 54. The molar ionic conductivity of alkali metal ions follows the order (a) Li+ > Na+ > K+ (b) Li+ > K+ > Na+ (c) K+ > Na+ > Li+

(d) 16 (d) Rb2O2 (d) Na2SO4 (d) Na < K < Li (d) K+ > Li+ > Na+

s-Block Elements (Alkali and Alkaline Earth Metals) 14.11

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49.

(c) (b) (a) (b) (c) (d) (a) (a) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(d) (a) (a) (c) (a) (a) (c) (a) (a)

3. 9. 15. 21. 27. 33. 39. 45. 51.

(c) (d) (a) (c) (d) (b) (b) (b) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52.

(b) (b) (a) (c) (c) (a) (d) (b) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53.

(c) (d) (d) (a) (b) (a) (d) (c) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54.

(c) (b) (b) (c) (a) (a) (d) (a) (c)

HINTS AND SOLUTIONS

2). 7. Alkali metals form ionic hydrides. +

> Na+ > K+ > Rb+

14. 23. 24. 25. 26. 27.

Li+ S4O62– + 2I–. Equivalent mass = Molar mass The reaction is 2S2O32– + I2 Atomic number of francium is 87. The abundance of sodium is more than that of potassium. Chile saltpeter is NaNO3. Downs cell is used for the production of sodium.

29. 35. 36. 41.

2). KO2. K + O2 The main raw materials needed are limestone (CaCO3) and sodium chloride. NaHCO3 is less soluble as compared to Na2CO3. The reverse reaction is actually involved.

2O 2)

2

43. Li2CO3 44. Same as Q. 43. 45. Li forms Li2O, Na forms Na2O2

2).

– 2

47. The useful product is Na2CO3 and by-product is CaCl2 48. Brine is concentrated solution of NaCl. 50. The atomic numbers of alkali metal are

3,11,19,37,55,87,

whereas potassium produces

14.12 Complete Chemistry—JEE Main

51. Li2O2 52. The reaction is

2Na2S + 2O2 + H2O Æ Na2S2O3 +2NaOH –3

–3

–3

). 54. Because of hydration of ions in solution, the molar conductivity follows the order K+ > Na+ > Li+

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 2O 2 2O 2. 3+

(c) Na2O2

to

CrO 24

[2014, online]

(a) LiO2, Na2O2 and K2O (c) Li2O, Na2O2 and KO2

(b) Li2O2, Na2O2 and KO2 (d) Li2O, Na2O and KO2

[2016]

ANSWERS 1. (d)

2. (c)

HINTS AND SOLUTIONS O -2 . Na2O2

3+

in acidic medium to CrO 24 and is a

derivative of H2O2. Æ 2Li2O 2 Æ Na2O2 2 Æ KO2 2

The Group 2 Element – Alkaline Earth Metals strontium (Sr), barium (Ba) and radium (Ra). Their physical properties are recorded in Table 3. Table 3 Property

Be

Atomic number Relative atomic mass

Abundance in earth’s crust (ppm) Atomic radius, r/pm Ionic radius for 6-coordinate, r/pm T/K T/K

Density at 293 K, r/g cm–3 I/kJ mol–1 I II Standard electrode potential

E°/V at 298 K for M2+(aq) +2e– Æ 2M(s) *

General Physical Properties of Alkaline Earth Metals

For 4-coordinate.

[He](2s) 4 9.012 2 112 27* 1580

Ca 2

1.85

[Ne](3s) 12 24.305 27640 160 72 922 1378 1.74

899 1757

–1.70

–2773

2

[Ar](4s) 20 40.08 46668 197 100

Sr 2

Ba 2

Ra

135 1000 (2123) 3.62

[Rn](7s)2 88 226.025 10–6 — 148 (973) (1973) 5.5

1.55

[Kr](5s) 38 87.67 384 215 118 1041 1654 2.63

738 1450

590 1145

549 1064

503 965

509 975

–2.37

–2.87

–2.89

–2.90

–2.92

1112 1767

[Xe](6s) 56

2

137.33

390 222

s-Block Elements (Alkali and Alkaline Earth Metals) 14.13

Description of Physical Properties 1. All are metals with two electrons in their outermost orbitals and thus they form bivalent M2+ ions. 2. The elements of Group 2 are soft to one electron contributed by the elements of Group 1. 3. Their melting and boiling points 4. The atomic and ionic radii 5. Because of the smaller atomic radii, the elements are more dense

second ionization energies

Fig. 9

I/kJ mol–1

Fig. 8

Mg

Fig. 10 Densities of Alkaline Earth Metals

Fig. 11

Atomic and Ionic Radii of Alkaline

14.14 Complete Chemistry—JEE Main

7. Calcium, strontium and barium impart characteristic colours

Table 4

Flame Colours of Ca, Sr and Ba

Element

Colour

Ca

Brick red

Sr

Crimson red

Ba

Green

8. The standard electrode potential, E°(M2+ 9. Alkaline-earth metals dissolve in liquid ammonia which slowly decompose to amides. M + 6NH3

⎯evaporation ⎯⎯⎯→

M(NH3)6 —Æ M(NH2)2 + 4NH3 + H2

Concentrated solutions of metals in ammonia are bronze coloured, due to the formation of metal clusters. 10. The crystalline compounds of Group 2 contain more water of crystallization 2 ◊ 6H2O, CaCl2 ◊ 6H2O and BaCl2 ◊ 2H2O all have water of crystallization. The number of molecules of water of crystallization decreases as the ions become

Chemical Properties

3. Burning in Air or Oxygen 2M + O2 Æ 2 MO

Æ

M + O2 Æ MO2

3N 2

4. Reaction with Water M + 2H2O Æ M(OH)2 + H2. 5. Reaction with Nitrogen 3M + N2 Æ M3N2. The formation of nitride ion, N3–, from N2

2

M2+ and N3–.

is a very stable molecule. 3N 2 +

ion.

M3N2 + 6H2O Æ 3M(OH)2 + 2NH3.

Isolation of Magnesium and Calcium

7H2O), carnallite (KCI ◊

2 ◊ 6H2

3 2 SiO4

◊ CaCO3

3 2Si4O10(OH)2

4 ◊ Si O (OH) 3 2 5 4]

s-Block Elements (Alkali and Alkaline Earth Metals) 14.15

Calcium is present in limestone, marble or chalk (CaCO3 4 ◊ 2H2 2), ◊ CaCO ) F]. Sea shells, corals and pearls are essentially CaCO . 3 3 5 4 3 3 Their fused anhydrous chlorides are electrolysed to obtain them in the metallic form.

Compounds of Alkaline Earth Metals 1. Oxides All the elements in Group 2 burn in O2 of the carbonates, MCO3. electropositive character of the metals.

BaO2 and SrO2 2 and CaO2

2

with H2O2

2. Hydroxides 2

Be(OH)2 2

is a weak base.

Ca(OH)2 and Sr(OH)2 Ba(OH)2 Ca(OH)2 is known as slaked lime. Its solution in water Ca(OH)2 is known as lime water while that of Ba(OH)2 is known as baryta solution → CaCO3 + H2O Ca(OH)2 + CO2 ⎯⎯ lime water turns milky

Fig. 12

Excess CO

2 ⎯⎯⎯⎯⎯ →

Ca(HCO3)2 milkiness disappears

14.16 Complete Chemistry—JEE Main

Explanation 1. Lattice Energy its tendency to split into ions to pass over to the solution, i.e. less will be its solubility. 2. Hydration Energy Hence, 2+

to Ba2+ ions. This factor is predominant as compared to the decrease in the hydration

3. Carbonates and Bicarbonates Alkaline earth metals form solid carbonates. Bicarbonates are known only in solution because of the less basic character of alkaline earth metals. follows. BeCO3 < 100 °C

CaCO3 900 °C

3

540 °C

Explanation

As the size of M

Explanation Ba2+

In case of carbonates, the size of CO32–

SrCO3 1290 °C

BaCO3 1360 °C

2+

2– 3

2+

decreases

2+

to

to Ba2+

4. Sulphates BeSO4

4

are soluble. CaSO4

SrSO4, BaSO4 and RaSO4 are virtually insoluble. Explanation In case of sulphates, the size of SO42– 2+

2+

to Ba2+ do

2– 4

decreases

to Ba2+ MSO4 ⎯heat ⎯→ MO + SO3

BeSO4 500 °C Explanation

4

895 °C

CaSO4 1149 °C

SrSO4 1374 °C

As the size of M2+

5. Nitrates

2O 4

N O

to 50 °C °C 2 4→ Be(NO ) .2N O ⎯warm Be(NO3)2 ⎯125 BeCl2 ⎯⎯⎯ ⎯⎯⎯⎯ ⎯⎯ → [Be4O(NO3)6] 3 2 2 4 in vacuum →

s-Block Elements (Alkali and Alkaline Earth Metals) 14.17 Basic beryllium nitrate

Basic beryllium acetate, [Be4O(CH3COO)6], has a similar structure.

6. Halides 2

at the appropriate temperature.

700 °C   BeO + C + Cl2   BeCl2 + CO It cannot be prepared in aqueous solution due to the formation of hydrated ion [Be(H2O)4]2+. The halides obtained has the molecular formula [Be(H2O)4]Cl2, [Be(H2O)4]F2

[Be(H2O)4]Cl2

⎯heat ⎯→ Be(OH)2 + 2HCl + 2H2O

Beryllium halides are covalent and fume in air due to hydrolysis: BeCl2 + 2H2O Æ Be(OH)2 + 2HCl In vapour phase it is present as BeCl2 and (BeCl2)2 and in solid phase it is polymerized. Cl—Be —

CI—Be—CI monomer

Cl

Be–Cl — Cl

dimer

2

◊ 6H2

2

◊ 6H2O. Anhydrous CaCl2

2

7. Complexes

2– 4

, [Be(H2O)4]2+, [Be(C2O4)2]2–, [Be4O(Ac)6], etc.

14.18 Complete Chemistry—JEE Main 3

hybridization of beryllium orbitals. These are the orbitals (2s and three 2p) which are available in the valence shell of

ethylenediamine tetraacetate (see unit 12).

Anomalous Behaviour of Beryllium

few points of differences are as follows. 1. Beryllium forms bonds with appreciable covalent character. + 3O . This is due to the formation of a bond.

4. Be(OH)2

Diagonal Relationship of Beryllium

1. Both beryllium and aluminium become passive by concentrated nitric acid due to the formation of an inert layer 2. Both beryllium and aluminium react with alkali to form beryllate, [Be(OH)4]2– and aluminate, [Al(OH)6]3–. 3. Both Be(OH)2 and Al(OH)3 are amphoteric. 4. Both beryllium and aluminium form polymeric halides.

5. Both the metals form nitrides which liberate ammonia when treated with water. Be3N2 + 6H2O Æ 3Be(OH)2 + 2NH3 AlN + 3H2O Æ Al(OH)3 + NH3. Be2C + 4H2O Æ 2Be(OH)2 + CH4 A14C3 + 12H2O Æ 4Al(OH)3 + 3CH4. 2+

3+

7. Both Be and Al

2– 4] ,

[Be(C2O4)2]2–, [AlF6]3– and [Al(C2O4)3]3–.

s-Block Elements (Alkali and Alkaline Earth Metals) 14.19

MULTIPLE CHOICE QUESTIONS 1. The number of alkaline earth metals is (a) 4 (b) 5

(c) 6

(d) 7

(a) Alkaline earth metals are harder and denser than alkali metals (c) Alkaline earth metals are more reactive than alkali metals

(b) Beryllium differs considerably from the rest of the alkaline earth metals (c) Anhydrous salts of beryllium are covalent whereas those of Ca, Ba and Sr are ionic

(a) Be(OH)2 (a) BeSO4

2

4

(c) Ca(OH)2

(d) Ba(OH)2

(c) CaSO4

(d) BaSO4

(a) Beryllium does not react with water (c) Calcium reacts even with cold water

(c) The sulphates of alkaline earth metal are stable to heat

(c) The halides of Ca, Sr and Ba are essentially ionic

(c) Anhydrous CaCl2

(b) Be and Al react with alkali to form beryllate, Be(OH)42–, and aluminate Al(OH)63–, respectively

14.20 Complete Chemistry—JEE Main

(c) Beryllium and aluminium nitrides liberate ammonia when treated with water (d) Beryllium and aluminium carbides liberate acetylene when treated with water 13. If one mole of Ba(OH)2 is added to 10 litres of water at 25 °C, the pH of the resultant solution would be (a) 13.3 (b) 13.0 (c) 0.7 (d) 1.7 14. Soap scum is often composed of (a) sodium carbonate (b) calcium carbonate (c) sodium sterate (d) calcium sterate (a) Be(OH)2

2

(b) CaF2

3

(c) Ca(OH)2

(d) Ba(OH)2

(c) CaCl2

18. Hardness in water is due to 2+ (b) HCO3– (a) Ca2+

2

(c) Na+

(d) K+

(a) Be2+

2+

(c) Ca2+

(d) Sr2+

(a) Be2+

2+

(c) Ca2+

(d) Sr2+

2+

from Be2+ to Ba2+ to Ba2+.

(c) The salts of alkaline earth metals are less hydrated than those of alkali metals. (d) Carbonates of alkaline earth metals are insoluble in water. 2 · 2H2 2·

2 · 4H2

6H2

2 · 6H2

4 · 7H2

3)2 · 6H2

(c) BaCO3 (a) BaO (b) Ba(OH)2 26. The number of water of crystallization in barium chloride is (a) 2 (b) 4 (c) 5 (a) Be2

(b) B2

2 · 8H2O

(c) C2

3

4) 2

(d) Ba(NO3)2 (d) 6 (d) N2

(a) The alkaline earth metals are harder and denser than alkali metals (c) Alkaline earth metals are too reactive to occur freely in nature

(a) Gypsum

(b) Dolomite

(c) Talc

(d) Fluorspar

s-Block Elements (Alkali and Alkaline Earth Metals) 14.21

2Æ

2Æ

(d) Cr2O3 + 3Ca Æ 2Cr + 3CaO

ææ Æ 2CaO + C (c) 2Ca + CO2 æheat

(a) Ammonia soda process (c) Dow process 36. The composition of Sorel’s cement is

(b) Downs cell (d) Castner-Kellner cell

2 3

2

nH2O nH2O

2 3

· 2H2

2

· 4H2

2

· 6H2

3– 4 4

4

4

4

· 6H2

40. Quick lime is (b) Ca(OH)2 (a) CaCO3 41. Limestone is (b) Ca(OH)2 (a) CaCO3 42. Slacked lime is (b) Ca(OH)2 (a) CaCO3 43. Milk of lime is (a) solid Ca(OH)2 (c) suspension of CaCO3 in water (a) Limestone (a) CaSO4 · (a) CaSO4

1 2

H 2O

2

2

· 7H2O

ions is due to the formation of 3 4) 2

(c) CaO

(d) CaSO4

(c) CaO

(d) CaSO4

(c) CaO

(d) CaSO4

(b) suspension of Ca(OH)2 in water (d) suspension of CaO in water

(b) Marble

(c) Chalk

(d) Talc

(b) CaSO4 · H2O

(c) CaSO4 · 2H2O

(d) CaSO4

(c) CaSO4 · H2O

(d) CaSO4 · 2H2O

(b) CaSO4 ·

47. The composition of dolomite is (b) CaCO3 3

1 2

H 2O

3

· CaCO3

3 · CaCO3 · 6H2O

14.22 Complete Chemistry—JEE Main

48. The composition of asbestos is 2O5(OH)2

2Si2O5

· 4H2

2O5)2(OH)4

2Si2O5(OH)4

51. Solubility of alkaline earth metal sulphates decreases down the Group 2 because

(c) enthalpy of hydration of bivalent metal ions decrease rapidly (d) the compounds become more and more ionic 3

4

2

3 4)2 is treated with H2SO4, superphosphate of lime is obtained. The composition of this fertiliser is (b) Ca(H2 4)2 (a) Ca3 4)2 + CaSO4 (c) Ca(H2 4)2 · H2O + 2CaSO4 · 2H2O (d) Ca(H2 4)2 + CaSO4

(a) CaO

(b) CaCO3

(a) Dolomite

(b) Gypsum

(c) CaC2O4

2+ 2+ 2+

(d) Ca(OH)2

(c) Epsomite + ions because + are insoluble. + ions +

2+

(d) Talc

2+ +

ions

57. The covalent halides are formed by 2+

(b) Na+ ions

(c) Be2+

(a) Ca > Ba > K

(b) K > Ca > Ba

(c) Ca > K> Ba

(d) K > Ba > Ca

(a) 11

(b) 20

(c) 31

(d) 40

(c) Ca(OH)2

(d) Ba(OH)2

(a) Al

3+

ions

(a) Be(OH)2

2

3+

ions

2.6H2O, the residue contains 2

64. Thermal stability of carbonates of alkaline-earth metals follows the order (a) CaCO3 3 > BaCO3 (b)CaCO3 > BaCO3 3 (c) BaCO3 > CaCO3 (d) BaCO 3 3

3

> CaCO3

(b) Ca(OH)2 (c) Ba(OH)2 (a) Be(OH)2 66. Identify the correct order of thermal stability of hydrides of Ca, Sr and Ba. (b) CaH2 > BaH2 > SrH2 (c) BaH2 > CaH2 > SrH2 (a) CaH2 > SrH2 >BaH2

(d) NaOH (d) BaH2 > SrH2 > CaH2

s-Block Elements (Alkali and Alkaline Earth Metals) 14.23

67. The formula of basic beryllium acetate is (b) Be4O(CH3COO)4 (a) Be2O(CH3COO)2

(c) Be4O(CH3COO)6

(d) Be(CH3COO)2

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67.

(c) (b) (a) (a) (b) (d) (c) (b) (d) (a) (d) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62.

(c) (d) (d) (d) (a) (c) (c) (d) (a) (b) (a)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

(a) (b) (a) (d) (a) (c) (b) (a) (c) (a) (d)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64.

(d) (b) (c) (c) (d) (a) (c) (d) (d) (b) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65.

(d) (a) (d) (c) (b) (c) (a) (c) (c) (b) (a)

HINTS AND SOLUTIONS 2. Alkaline earth metals are less reactive than alkali metals.

10. Beryllium chloride vapour contains BeCl2 but the solid is a polymer with a chain structure. 12. Both beryllium and aluminium carbides form methane on hydrolysis: 2Be (OH)2 + CH4 Be2C + 4H2O Al4C3 + 12H2O 4Al(OH)3 + 3CH4 13. Ba(OH)2 completely ionizes in water. Hence, [OH – 15. Because of the small size of Be2+ ions, the Be2+ and OH – 19. Because of the small size of Be2+

27. Be2 2 3O 4. 3Si4O10

(OH)2.

is zero).

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66.

(a) (d) (a) (b) (c) (b) (b) (d) (c) (b) (a)

14.24 Complete Chemistry—JEE Main

ææ Æ 4CaO + CaC2. 32. 5Ca + 2CO2 æheat 2+

57. Beryllium halide is covalent, rest of alkaline-earth metals form ionic halides.

19K

(3s)2 (3p)6 Æ (3s)2 (3p)5

20Ca

(3s)2 (3p)6 (4s)1Æ (3s)2 (3p)6

56Ba

(5s)2 (5p)6 (6s)1Æ (5s)2 (5p)6

stable (3p)6 60. Atomic numbers of elements of Group 2 are 4,12,20,38.....

2 due to the smaller size of Be. In fact, Be(OH)2

67. Basic beryllium acetate is Be4O(CH3COO)6. its structure is

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

(d) inter-ionic attraction (a) NaNO3

(b) KClO3

(c) CaCO3

[2005] (d) NH4NO3

(a) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 2 (b) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 2 2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 2 < Sr(OH)2 < Ca(OH)2 < Ba(OH)2 2

(a) CaSO4

[2012]

[2015 online]

(b) BeCl2

(c) SrCl2

(d) CaCl2

[2015 online]

(b) BeSO4

(c) BaSO4

(d) SrSO4

[2015]

s-Block Elements (Alkali and Alkaline Earth Metals) 14.25

6. The correct order of the solubility of alkaline-earth metal sulphates in water is [2016 online] (a) Quick lime

(b) Milk of lime

(c) Slaked lime

(d) Lime stone [2016 online]

ANSWERS 1. (c) 7. (a)

2. (c)

3. (c)

4. (b)

5. (b)

6. (a)

HINTS AND SOLUTIONS 2. The thermal decomposition of CaCO3 CaCO3 ææ Æ CaO basic oxide

+

CO 2 acidic oxide

4. BeCl2 is essential covalent. This is due to the small size of Be. It can easily polarize Cl– covalent bond. 2+ 5. 2+ to Ba2+ 4 to BaSO4. 2.

The paste of lime

15 Study of the p-Block Elements (Groups 13, 14 and 15) The Group 13 Elements – Boron Family Group 13 of the periodic table is composed of elements boron (B), aluminium (Al), gallium (Ga), indium (In) and thallium (Tl). Their physical properties are recorded in Table 1. Table 1 Property

Atomic and Physical Properties of Group 13 Elements

B

Al 2

Atomic number Relative atomic mass Atomic radius, r/pm

Ionic radius, r(M3+)/pm r(M+)/pm Melting point, T/K Boiling point, T/K

Density at 293 K, r/g cm–3 Ionization energy, l/kJ mol–1 I II III Electronegativity

Enthalpy of fusion/kJ mol–1 Enthalpy of vaporization/kJ mol–1

1

Ga 2

In

1

Tl 10

2

1

[He](2s) (2p) [Ne](3s) (3p) [ A r ] [Kr](4d) (5s) (5p) [Xe](4f)14(5d)10(6s)2(6p)1 10 2 1 (3d) (4s) (4p) 5 13 31 49 81 10.81 26.982 69.72 114.82 204.383 170 (80-90) 143 (125) 167 (27) 53.5 62.0 80.0 88.5 120 140 150 2453 933.4 302.8 249.6 576.5 3923 2740 2676 2353 1730 2.35 2.70 5.90 7.31 11.85 800.5 2426.5 3658.7 2.0 23.60 504.5

577.4 1816.1 2744.1 1.5 10.50 290.8

578.4 1978.8 2962.3 1.6 5.59 270.3

558.2 1820.2 2704.0 1.7 3.26 231.8

2877.4 1.8 4.31 166.1

(–0.87)*

–1.66 + 0.55

–0.56 (–0.79)†

–0.34 –0.18

+1.26 –0.34

589.1 1970.5

Standard electrode

potential, E°/V at 298 K for M3+(aq) + 3e– Æ M(s) M+(aq) + e– Æ (M(s) * †

For H3BO3 + 3H+ + 3e– Æ B + 3H2O. In acidic medium

Description of Physical Properties ns)2(np)1, where n varies from 2 to 6. They are expected to form compounds with +3 oxidation state.

15.2 Complete Chemistry—JEE Main

3. The metallic radii of atoms do not increase regularly on descending the group (Fig. 1).

A Few Comments (i) Boron is not a metal, the reported radius is half the closest approach in its structure. (ii) Gallium has an unusual structure, the reported radius is half the closet approach. (iii) Ga, In and Tl follow immediately after the ten transition elements. Thus, the outer shell is preceded by d10 near the nucleus as compared to their expected normal locations. Thus, their atomic radii are smaller than the expected values. This contraction in size is sometimes called the d-block contraction. (iv) The atomic radius of Tl is a little larger than In. This is due to the intervention of 4f electrons, which shield the nuclear charge more poorly. The small value of atomic radii is a result of lanthanide contraction. 4. The ionic radii of M3+ increase down the group. The reported value of B3+ is an estimated value as B3+ is not found to exist. 5. On descending the group, +1 oxidation state becomes more stable than +3 state due to the inert pair effect. Explanation of Inert Pair Effect In inert pair effect, s electrons do not take part in bonding. The reason behind this is the energy factor. The energy required to unpair them exceeds the energy evolved when they form bonds. Thus, they remain intact. 6. The melting points of the Group 13 elements do not show a regular trend (Fig. 2).

Fig. 1 Atomic and Ionic Radii

Fig. 2 Melting Points of the Group 13 Elements

The very high melting point of boron is due to its unusual crystal structure. The small size and high ionization energy makes boron a nonmetallic element, thus metallic bondings do not exist. The structure of boron is icosahedral (20-faced) with boron atoms at all 12 corners (Fig. 2). Gallium again has an unusual structure. Each metal atom has one close neighbour at a distance of 243 pm and six more distant neighbours at distances between 270 pm and 279 pm. In this structure, more or less diatomic molecules exist which accounts for the low melting point (about 30°C). 7. The variation of boiling point corresponds to the expected pattern as no unusual structures exist in liquid phase (Fig. 3). 8. The densities increase on descending the group (Fig. 4). 9. The ionization energies do not follow the expected trend of decreasing values on descending the group (Fig. 5). The sum of the three ionization energies for each element is very high. This explains their tendency to form covalent compounds. Boron has no tendency to form ions, but the other elements have this tendence in solution. 10. The standard electrode potential, E°(M3+| M), becomes more negative in going from B to Al indicating the increase in metallic nature from B to Al. After Al, the values become less and less negative and for Tl, it becomes positive indicating the reduction M3+ + 3e– Æ M becomes more easy. This explains the +3 oxidation state becomes less stable in aqueous solution on descending the group.

Study of the p-Block Elements (Groups 13, 14 and 15)

Fig. 3

Icosahedral Structure of Boron

Fig. 5

Densities of the Group 13 Elements

Fig. 4

15.3

Melting Points of the Group 14 Elements

Fig. 6 Ionization Energies of the Group 13 Elements

On the other hand, the more negative value of E°(M +|M) for Tl than In indicates the more stability of Tl+ than In+.

Isolation of Boron The chief minerals of boron are borax (Na2[B4O5(OH)4] ◊ 8H2O), i.e. Na2B4O7 ◊ 10H2O, colemanite (Ca2[B3O4(OH)3]2 ◊ 2H2O) i.e. Ca2B6O11 ◊ 5H2O and kernite (Na2[B4O5(OH)4] ◊ 2H2O) i.e. Na2B4O7 ◊ 4H2O. Boron is isolated by converting its mineral into boron trioxide followed by is reduction with magnesium. Na2[B4O5(OH4)] ◊ 8H2O + 2HCl Æ 2NaCl + 5H2O + 4H3BO3 borax

orthoboric acid red heat

2H3BO3 ææææ Æ B2O3 + 3H2O B2O3 + Mg Æ 2B + 3MgO Crystalline boron is obtained by the reduction of boron trichloride with zinc or dihydrogen at high temperature. 1200 K

2BCl3 + 3Zn æææÆ 3ZnCl2 + 2B 1200 K

2BCl3 + 3H2 æææÆ 2B + 6HCl

Chemical Properties of Group 13 Elements 1. Reaction with Oxygen All elements burn in oxygen at high temperatures forming M2O3. The element Tl also forms some T12O.

15.4 Complete Chemistry—JEE Main

The reaction of aluminium with oxygen is strongly exothermic. 3 Df H = – 1670 kJ mol–1 2Al(s) + O2(g) Æ Al2O3(s) 2 This reaction is known as the thermite reaction. metallurgical extraction of other metals from their oxides. 3Mn3O4 + 8Al Æ 9Mn + 4Al2O3 Cr2O3 + 2Al Æ Al2O3 + 2Cr. 2. Reaction with Water Aluminium, in principle, can react with water. But the reaction does not continue due to the formation of thin protective layer of Al2O3. Reactions with Acids and Alkalis Aluminium reacts with dilute mineral acids. 2Al + 6HCl Æ 2A13+ + 6C1– + 3H2. With concentrated nitric acid, the metal becomes passive due to the formation of protective layer of oxide. Aluminium also dissolves in aqeous NaOH indicating that the element is amphoteric. 2Al + 2NaOH + 6H2O Æ 2NaAlO2 ◊ 2H2O + 3H2 Sodium aluminate

Compounds of Group 13 Elements 1. Oxides All the elements of Group 13 form oxides (M2O3). Their acidic character reduces on descending the group. B2O3 is acidic Al2O3 and Ga2O3 are amphoteric ln2O3 and T12O3 are basic The oxides M2O3 are known as sesquioxide (sesqui means one and a half) as they can be represented as MO3/2. 2. Hydroxides All the elements of Group 13 form hydroxides, M(OH)3. Their acidic character decreases on descending the group. Boric acid, B(OH)3 or H3BO3 is obtained by the action of hydrochloric acid or sulphuric acid on borax. On cooling Na2[B4O5(OH)4] ◊ 8H2O + 2HC1 Æ 2NaCl + 4H3BO3 + 5H2O Na2[B4O5(OH)4] ◊ 8H2O + H2SO4 Æ Na2SO4 + 4H3BO3 + 5H2O It may also be obtained by the hydrolysis of most boron compounds like nitrides, sulphides, etc. 2BN + 6H2O Æ 2H3BO3 + 2NH3 B2S3 + 6H2O Æ 2H3BO3 + 3H2S Boric acid is a white crystalline substance, soft and soapy to touch. It is moderately soluble in cold water. On heating it decomposes to form metaboric acid at 375 K, tetraboric acid at 435 K and boron trioxide at red heat. 375 K H3BO3 ⎯−⎯⎯ H O→ HBO2 2

Metaboric acid

4HBO2

K ⎯−435 ⎯⎯ H O→ H2B4O7 2

Tetraboric acid

Red heat H2B4O7 ⎯⎯⎯⎯ −H O → 2

2B2O3 Boric oxide

Boric acid is a very weak monobasic acid. It does not liberate hydrogen ion but accepts a hydroxyl ion, i.e. it behaves as Lewis acid.   H3O+ + [B(OH)4]– B(OH)3 + 2H2O  (or H3BO3)

Metaborate ion

pKa = 9.25

Study of the p-Block Elements (Groups 13, 14 and 15)

15.5

The structures of B(OH)3 and [B(OH)4]– are as follows.

The boric acid cannot be titrated satisfactorily with NaOH as sharp end point is not obtained. However, in the presence of a cis-diol (glycerol, mannitol or sugars), boric acid acts as a strong acid and can be titrated with NaOH in the presence of phenolphthalein indicator. With a cis-diol, the product [B(OH)4]– in the above reaction forms a strong complex, causing the reaction to move in the forward direction.

Boric acid contains triangular planar BO3– 3 units. In the solid the B(OH)3 units are hydrogen bonded gether into two dimensional sheets with almost hexagonal symmetry (Fig. 7).

Fig. 7

Hydrogen Bonded Structure of Boric Acid

3. Halides All the elements of Group 13 form trihalides. The boron halides are covalent. BF3 is gaseous, BC13 liquid and BI3 is solid. They are all hydrolyzed by water. BF3 is hydrolyzed incompletely while BCl3 and BI3 are completely hydrolyzed. Hydrolysis of BF3: In this, the HF produced reacts with the H3BO3

15.6 Complete Chemistry—JEE Main

4BF3 + 12H2O Æ 4H3BO3 + 12HF 12HF+ 3H3BO3 Æ 3H+ + 3[BF4]– + 9H2Q 4BF3 + 3H2O Æ H3BO3 + 3H+ + 3[BF4]–

Hydrolysis of BCl3 (and BI3): BCl3 + 3H2O Æ H3BO3 + 3HCl BF3 is useful organic catalyst for Friedel-Crafts reactions. It is also used extensively as a polymerization catalyst. It can be obtained from B2O3 or borax. B2O3 + 6HF + 3H2SO4 Æ 2BF3 + 3H2SO4 ◊ H2O 6H 2 O Na2B4O7 + 12HF ⎯−⎯⎯ → [Na2O(BF3)4]

2 SO 4 ⎯2H ⎯⎯ → 4BF3 + 2NaHSO4 + H2O

attaining an octet of electrons. The structure of dimer (AlCl3)2 is

This dimeric structure is retained in non-polar solvents such as benzene. In water, the high enthalpy of hydration is ◊ 6H2O]3+ and 3X– ions. Anhydrous aluminium chloride is used as a catalyst in the Friedel-Crafts reaction. It is also used in petroleum cracking and in the manufacture of dyes, drugs and perfumes. 3

by p bonding. The other elements in the group have larger atoms and cannot get effective p overlap, so they polymerize

3

The sp2 hybridization leads to the triangular planar geometry. The observed bond length of 130 pm is shorter than the sum of the covalent radii (B = 80 pm and F = 72 pm). This is explained by the formation of pp–pp bond in which the lone pair on F is extended to the empty p orbital of B. This is known as back bonding. Due to the resonating structures

all the three B—F bond distances are identical.

Acid Character of Boron Trihalides Due to back bonding, the electron density on boron is increased. The tendency to form pp–pp bond is maximum in BF3 and falls rapidly on passing to BCl3 to BBr3 due to the increase in size of halogen 2p orbital. Thus, tendency to accept electron pair is minimum for BF3 and BC13 and is maximum for BBr3. Hence, the increasing order of acid strength is BF3 < BCl3 < BBr3

Study of the p-Block Elements (Groups 13, 14 and 15)

Borax

15.7

Borax is usually written as Na2B4O7 ◊ 10H2O. Its structure is OH B– O

O O

HO — B

B — OH O

O B–

OH It contains two tetrahedral units and two triangular units. Its actual formula is Na2[B4O5(OH)4] ◊ 8H2O Borax is used as a primary standard for titrating acids. Na2B4O7 ◊ 10H2O + 2HCl Æ 2NaCl + 4H3BO3 + 5H2O Since the products contains boric acid, methyl orange (pH range 3.1-4.4) is used as the indicator. One mole of borax reacts with two moles of acid. The actual reactions are 2B(OH)3 + 2[B(OH)4]– [B4O5(OH)4]2– + 5H2O

2[B(OH)4]– + 2H3O+ Æ 2B(OH)3 + 4H2O Borax is also used for making buffer solution as its aqueous solution contains equal amounts of weak acid and its salts. Borax Bead Test In the borax bead test, B2O3 loop of platinum wire. The mixture fuses to give a glass-like metaborate bead. CoO + B2O3 Æ Co(BO2)2 Cobtalt metaborate (blue colour)

Metaborate beads of many transition metals have characteristic colours and thus borax bead test is used to identify the metals.

4. Hydrides Elements of group 13 do not react directly with hydrogen. Their hydrides are prepared by indirect methods. Boron forms a large number of hydrides, called boranes, of the general formula BnHn+4 (known as nido-boranes) and BnHn+6 (known as arachno-boranes). The most important of these boranes is B2H6, called diborane. It is prepared by the following methods: 3Mg + 2B æDæ Æ Mg3B2 + H3PO4 ⎯⎯ → Mixture of boranes ⎯Heat ⎯→ B2H6 Magnesium boride

Diborane

→ B2H6 + 6HC1 2BCl3 + 6H2 ⎯⎯⎯⎯⎯ Silent electric discharge

4BC13 + 3LiAlH4 ⎯⎯ → 2B2H6 + 3LiCl + 3A1C13 2NaBH4 + I2 ⎯⎯ → B2H6 + H2 + 2NaI (in polyether solvent) 250 atm

B2O3 + 3H2 + 2Al ææææ Æ B2H6 + Al2O3 150∞C

Structure of Diborane In diborane there are 12 valency electrons, three from each B atom and six from the H atoms. The structure of diborane is as follows.

15.8 Complete Chemistry—JEE Main

Hb Ht

tH

B

Terminal B—H is 2c – 2e bond with bond length 119 pm

B

tH

Bridged B—H is 3c – 2e bond with bond length 133 pm

Ht Hp

Hts are terminal hydrogen atoms, they are in the same plane including B atoms. The other H atoms are in the perpendicular plane, they lie centrally between two B atoms. The terminal B—H bonds are the normal two-centre twoelectron bonds. The bridging H—B—H bonds are the three-centre two-electron bonds. The higher boranes have an open cage structure. B2H6 + 2(CH3)3N: Æ 2[(CH3)3N Æ BH3] Their reactions with ammonia depend on the experimental conditions. low temperature

B2H6 + NH3

B2H6 ◊ 2NH3

excess

higher temperature

(BN)x

Boron nitride

If the ratio of B2H6 and NH3 is 1:2, the product obtained at higher temperature is borazine (B3N3H6). The structure of boron nitride is very similar to graphite (as the sum of valence electrons of B and N is equal to that of two C atoms). Borazine is known as inorganic benzene. H N B

B N B

N B N

B N B

N B

B N

N

N B

B

H

H

N B

B N H

H N B

H

H

H

H

–

+

N B

B

–

N+

N B

+ –

H H

H

Borazine Boron nitride Diborane is a powerful electrophilic reducing agent for certain functional groups. For example B H

2 6 RCN æææ Æ RCH2NH2

B H

2 6 Æ RNH2 RNO2 æææ

B H

2 6 Æ RCH2OH RCHO æææ The other elements of the group 13 form only a few stable hydrides. Thus, (AlH3)n exists as a white polymer containing aluminium atoms joined together by hydrogen bridges, Al—H—Al. Aluminium hydride can be prepared from LiH and AlCl3 in ether solution. 3nLiH + nAlCl3 Æ (AlH3) + 3nLiCl

However, if excess of LiH is present, the product obtained is lithium aluminium hydride, LiAlH4. Boron also forms the compounds called borohydrides containing the tetrahedral BH 4– unit. Borohydrides are obtained by the action of diborane on alkali metal hydrides (in ether). 2MH + B2H6 Æ 2M+[BH4]– (where M = Li or Na)

Study of the p-Block Elements (Groups 13, 14 and 15)

15.9

Sodium borohydride may be prepared as follows: 4NaH + B(OCH3)3 Æ 3NaOCH3 + Na[BH4] Methyl borate

Alkali metal borohydrides are valuable reducing agents in inorganic chemistry. Aluminium and gallium also form compounds analogous to borohydrides, e.g. Li[AlH4], Li[GaH4] LiAlH4 is used as a reducing agent in organic chemistry for selectively reducing the functional groups.

Diagonal Relationship of Boron with Silicon Boron shows anomalous behaviour in its groups because of its small size and nonavailability of d orbitals. It resembles silicon, the second member of the next higher group. 1. Both boron and silicon are non-metals, and exist in allotropic forms. They have high melting points and are semiconductors. The other members of Group 13 are metals. respectively. The hydrides are readily hydrolyzed. The lower hydrides can be obtained by the reduction of chlorides with LiAlH4. 4BCl3 + 3LiAlH4 Æ 3AlCl3 + 3LiCl + 2B2H6 SiCl4 + LiAlH4 Æ AlCl3 + LiCl + SiH4 In contrast, aluminium hydride is a polymeric solid. 3. Both boron and silicon form halides which are readily hydrolyzed. BCl3 + 3H2O Æ B(OH)3 + 3HCl Boric acid

SiCl4 + 4H2O Æ Si(OH)4 + 4HCl Silicic acid

The aluminium halides are only partly hydrolyzed in water. 4. B2O3 is an acidic oxide, like SiO2. B2O3 + 6NaOH Æ 2Na3BO3 + 3H2O SiO2 + 2NaOH Æ Na2SiO3 + H2O. Al2O3 is an amphoteric oxide. 5. Both boron and silicon react with several metals to form metal borides and silicides. 3Mg + 2B Æ Mg3B2 Magnesium boride

2Mg + Si Æ Mg2Si Magnesium silicide

Borides and silicides get decomposed by dilute acids to give volatile hydrides (boranes, silanes). 6. Both boron and silicon, as well as their oxides, react with alkalis to form borates and silicates containing BO4 and SiO4 tetrahedral units, respectively.

MULTIPLE CHOICE QUESTIONS 1. Which of the following elements does not belong to Group 13 of the periodic table? (a) Boron (b) Aluminium (c) Gallium (d) Tin 2. The number of elements in Group 13 is (a) 4 (b) 5 (c) 6 (d) 7 3. The atomic number of indium which belongs to 5th period is (a) 47 (b) 48 (c) 49 (d) 50

15.10 Complete Chemistry—JEE Main

4. Which of the following statements is not true? (a) Boron is a nonmetal while all other elements of Group 13 are metals (b) Boron trichloride acts as a Lewis acid (c) Boron does not form the hydrated B3+ ion (d) The stability of the +1 oxidation state decreases down the group whereas that of +3 increases 5. Which of the following facts is not correct? (a) Elements of Group 13 do not react directly with hydrogen (b) The boron hydrides are known as boranes (c) The structure of diborane is H3B – BH3 (d) Aluminium hydride exists as white polymer 6. The geometry of BH 4– unit is (a) square planar (b) tetrahedral (c) octahedral (d) trigonal pyramidal 7. Which of the following statements is not correct? (a) LiAlH4 and LiGaH4 are reducing agents in inorganic chemistry (b) BF3 is a useful catalyst in Friedel-Crafts reactions (c) LiAlH4 is used as a reducing agent in organic chemistry for selectively reducing the functional groups 8. Which of the following oxides is an acidic oxide? (a) B2O3 (b) Al2O3 (c) Ga2O3 9. Which of the following oxides is an amphoteric oxide? (a) B2O3 (b) Al2O3 (c) In2O3 10. Which of the following oxides is a basic oxide? (a) B2O3 (b) Al2O3 (c) Ga2O3 11. Which of the following equations is not correctly formulated? 2NaCl + 4H3BO3 + 5H2O (a) Na2[B4O5 (OH)4] · 8H2O + 2HCl

(d) In2O3 (d) Tl2O3 (d) In2O3

(borax)

(b) 2BN + 6H2O

2H3BO3 + 2NH3

375 K

(c) H3BO3 æææÆ HBO2; - H2O

(metaboric acid)

435 K

4HBO2 æææÆ H2B4O7; - H2O

(tetraboric acid)

red heat

H2B4O7 æææÆ 2B2O3 - H 2O

(Boric oxide)

(d) H3BO3 is a weak monobasic acid as it liberates hydrogen ions as H + + H2BO3– H3BO3 12. The diagonal relationship of boron is with (a) carbon (b) silicon (c) magnesium (d) phosphorus 13. Which of the following facts regarding boron and silicon is not true? (a) Boron and silicon are semiconductors (b) Boron and silicon form halides which are not hydrolysed (c) Boron and silicon react with magnesium to form magnesium boride and magnesium silicide which are decomposed by acids to give volatile borane and silane, respectively (d) Both boron and silicon react with alkalis to form borates and silicates containing BO4 and SiO4 tetrahedral units, respectively (a) B (b) Al (c) Ga (d) In 15. Which of the following ions is hydrolysed to the maximum extent? (a) Al3+ (b) Ga3+ (c) Tl+ (d) Tl3+ 16. The corrosion of aluminium is prevented by the formation of impervious surface coating of (a) AlN (b) Al(OH)3 (c) Al2(CO3)3 (d) Al2O3 17. The precipitate of Al(OH)3 dissolves in NaOH solution. It is due to the formation of (a) Al(H2O)4 (OH)+2 (b) Al(H2O)3 (OH)3 (c) Al(H2O)2 (OH)–4 (d) Al(OH)63–

Study of the p-Block Elements (Groups 13, 14 and 15)

15.11

18. Which of the following does not represent an alum? (a) KAl(SO4)2 · 12H2O (b) NH4 Al(SO4)2 · 12H2O (c) NH4 Cr(SO4)2 · 12H2O (d) Al2(SO4)3 · 18H2O 19. Which of the following ions shows the largest tendency to form complexes with Al3+? (a) H2O (b) F – (c) Cl– (d) SO42– 20. The treatment of Mg3B2 with acids produces (a) B2O3 (b) MgB2O4 (c) B4H10 (d) MgB4O7 21. B3+ cannot exist in aqueous solution because of its (a) strong reducing ability (b) strong oxidizing ability (c) small size and large charge (d)large size and small charge 22. The following solid exists in the form of ions (a) BI3 (b) BCl3 (c) BF3 (d) CaB2O4 – 23. The geometry of BO3 ion is (a) tetrahedral (b) octahedral (c) square planar (d) triangular planar 24. The reaction in which metal oxide is reduced by Al is known as (a) Silberschmidt reaction (b) Bayer reaction (c) Goldschmidt reaction (d) Rosenmud reduction 25. Which of the following elements has the largest melting point? (a) B (b) Al (c) Ga (d) In 26. Which of the following elements has the minimum melting point? (a) B (b) Al (c) Ga (d) In 27. Which of the following statements regarding boron is not correct? (a) Boron compounds act as Lewis acids (b) Boron is a good conductor of electricity (c) Colemanite is one of the chief minerals of boron (d) Boron exists in two isotopic forms 10B and 11B 28. Which of the following statements regarding boron is not correct? (a) Boron has high melting and boiling points (b) The hydrides of boron are known as boranes (c) The structure of B2H6 is H3B — BH3 (d) In Borax, two boron atoms are in triangular geometry and two boron atoms are in tetrahedral geometry 29. Which of the following is boric acid? (a) HBO2 (b) H3BO3 (c) H2B4O7 (d) B2O3 30. Which of the following is metaboric acid? (a) HBO2 (b) H3BO3 (c) H2B4O7 (d) B2O3 31. Which of the following is tetraboric acid? (a) HBO2 (b) H3BO3 (c) H2B4O7 (d) B2O3 32. Which of the following statements is not correct? (a) Boric acid is a very weak monobasic acid (b) Boric acid contains planar BO3 units which are bonded together through hydrogen bonds forming a layer structure (c) Boric acid is used as a food preservative (d) Boric acid is a tribasic acid 33. Which of the following statements is not correct? (a) The aqueous solution of boric acid serves as mild antiseptic (b) Boron-10 has a high ability to absorb neutrons (d) Metaboric acid is formed by the dehydration of tetraboric acid (b) NH2OH (c) BCl3 (d) H3N Æ BCl3 (a) NH3 35. Which of the following is the correct formula of borax? (a) Na2[B4O5(OH)4] · 2H2O (b) Na2[B4O5(OH)4] · 4H2O (c) Na2[B4O5(OH)4] · 6H2O (d) Na2[B4O5(OH)4] · 8H2O

15.12 Complete Chemistry—JEE Main

36. Orthoboric acid is (a) monobasic (b) diabasic (c) tribasic (d) tetrabasic 37. Which of the following minerals does not contain aluminium? (a) Cryolite (b) Feldspar (c) Beryl (d) Olivine 38. The composition of beryl is (a) Be3Cr2Si6O18 (b) Be3Al2Si6O18 (c) KAlSi3O11(OH)2 (d) KAlSi3O6 39. Which of the following formulae represents mica? (a) Be3Cr2Si6O18 (b) Be3Al2Si6O18 (c) KAlSi3O11(OH)2 (d) KAlSi3O6 40. Which of the following statements regarding aluminium is not correct? (a) Aluminium is a light metal with considerable strength (b) Aluminium is a good conductor of heat and electricity (c) The corrosion of the aluminium metal is prevented by the formation of a coat of aluminium oxide on its layer (d) Aluminium dissolves in hydrochloric acid but not in concentrated sodium hydroxide 41. Chemically, potash alum is (a) K2SO4 · Al2(SO4)3 · 6H2O (b) K2SO4 · Al2(SO4)3 · 12H2O (c) K2SO4 · Al2(SO4)3 · 18H2O (d)K2SO4 · Al2(SO4)3 · 24H2O 42. Which of the following metals has the largest abundance in the earth’s crust? (a) Aluminium (b) Calcium (c) Magnesium (d) Sodium 43. The chief ore for the extraction of aluminium is (a) bauxite (b) cryolite (c) beryl (d) mica 44. Aluminium chloride at higher temperatures exists as (a) trigonal planar AlCl3 (b) dimer Al2Cl6 (c) Al3+(3Cl–) (d) (AlCl2)+ Cl– 45. In the gaseous phase, aluminium chloride at low temperatures (420 – 480 K) exists as Cl Cl (a) trigonal planar AlCl3 Cl

(b) dimer Al2Cl3, i.e. Cl Al

(c) dimer Al2Cl3, i.e. Cl

Cl Al

Cl

Al ... Al

Cl Cl

(d) Al

3+

Cl Cl

–

(3Cl )

Cl

46. Which of the following statements regarding the structure aluminium chloride is correct? (a) All the bond angles Cl—Al—Cl and Al—Cl—Al in Al2Cl3 are identical (b) All the bond lengths Cl —Al in Al2Cl3 are identical (c) All the bond lengths Cl—Al as well as all the bond angles Cl—Al—Cl and Al—Cl—Al are different (d) The bond lengths of terminal Al —Cl and bridged Al—Cl bonds are different and also outer bond angle Cl—Al—Cl and bridged bond angles Cl—Al—Cl and Al—Cl—Al have different values. 47. Which of the following alloys does not contain aluminium? (a) Duralumin (b) Magnalium (c) Alnico (d) Constantan 48. Which of the following oxides has the largest percentage in a usual sample of Portland cement? (a) MgO (b) CaO (c) SiO2 (d) Al2O3 49. Which of the following oxides has the lowest percentage in a usual sample of Portland cement? (a) SO2 (b) MgO (c) SiO2 (d) Al2O3 50. The most common ore of aluminium is (a) bauxite (b) alumina (c) potash alum (d) cryolite 51. In the aluminothermite process, aluminium acts as 52. Hydrogen gas will not reduce heated (a) copper(II) oxide (b) iron(III) oxide 53. Which of the following is an amphoteric oxide? (a) B2O3 (b) Al2O3

(c) tin(IV) oxide

(d) aluminium(III) oxide

(c) Ga2O3

(d) PbO

Study of the p-Block Elements (Groups 13, 14 and 15)

15.13 2)

55. 56.

57.

58.

59.

60.

61.

62.

is added to (a) make solution conducting (b) make solution anhydrous (c) lower the temperature of the melt (d) prevent hydrolysis of aluminium salt Heating of Al2(SO4)3 gives (a) Al2O3 and O2 (b) Al2O3 and SO2 (c) Al2O3 and SO3 (d) Al2O3 S and O2 Chrome alum is (a) K2SO4 · Cr2 (SO4)3 · 6H2O (b) K2SO4 · Cr2 (SO4)3 · 12H2O (c) K2SO4 · Cr2 (SO4)3 · 18H2O (d) K2SO4 · Cr2 (SO4)3 · 24H2O Heating of diborane with excess of ammonia to a high temperature yields (a) borane (b) diammoniate of borane (c) borazine (d) borone nitride Which of the following statements is incorrect? (a) Boron does not form ionic compounds with the formation of B3+ ions (b) Borane exhibits inert pair effect (c) Aluminium carbide on reacting with water liberates methane (d) Borane nitride on hydrolysis liberates NH3 Which of the following statements is not correct? (a) Boron BF3 involves sp2 hybridization (b) BF3 acts as a Lews acid (c) BF3 is a volatile liquid (d) forms an adduct with NH3 Which of the following statements is correct? (a) Borax involves two boron atoms in sp2 hybridization and two boron atoms in sp hybridization (b) Borax involves two boron atoms in sp2 hybridization and two boron atoms in sp3 hybridization (c) Borax involves all the four atoms in sp2 hybridization (d) Borax involves all the four atoms in sp3 hybridization Borax has the molecular formula (b) Na2[B4O5(OH)4].6H2O (a) Na2[B4O3(OH)4].6H2O (d) Na2[B4O6(OH)2].8H2O (c) Na2[B4O5(OH)4].8H2O In the reaction H3BO3 æ375 ææ Æ A æ435 ææ Æ B æred æææ ÆC K K heat

The compounds A,B and C, respectively, are (b) B2O3, HBO2, H2B4O7 (a) B2O3, H2B4O7, HBO2 (c) H2B4O7, HBO2, B2O3 (d) HBO2, H2B4O7, B2O3 63. In borax, the number of —OH group attached to boron atoms is (a) 2 (b) 3 (c) 4 64. The number of water of crystallization in borax is (a) 6 (b) 8 (c) 10

(d) 6 (d) 12

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61.

(d) (d) (b) (b) (a) (c) (d) (a) (d) (c) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62.

(b) (a) (a) (c) (c) (d) (b) (a) (a) (d) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

(c) (b) (a) (c) (b) (d) (c) (c) (c) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64.

(d) (d) (d) (d) (c) (c) (d) (d) (d) (b) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(c) (d) (c) (d) (b) (d) (d) (d) (b) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(b) (b) (d) (c) (a) (a) (a) (b) (c) (b)

15.14 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS 3. The atomic number of indium is 2 + 8 + 8 + 18 + 13 = 49. 4. The stability of the +1 oxidation state increases down the group. This is because of the participation of only p electron. The two s-electrons (s2) do not take part in the chemical reaction (inert pair effect). 5. The structure of diborane is

The two BH2 groups lie in the same plane while the two bridging hydrogen atoms lie in a plane perpendicular to this plane. dimers. 11. H3BO3 is a weak monobasic acid. It does not liberate H + but accepts OH–, i.e. it is Lewis acid. 13. The halides of B and Al are hydrolysed. B(OH)3 + 3HCl and SiCl4 + 4H2O Si(OH)4 + 4HCl BCl3 + 3H2O Boric acid

Silicic acid

20. The treatment of borides with acids produces boranes. 27. Boron has a very low electrical conductivity. 28. The structure of B2H6 is H H

H B

B

H H H The two BH2 groups lie in the same plane while the two bridging hydrogen atoms lie in a plane perpendicular to this plane.

29. Boric acid is H3BO3. It is a monobasic acid. It does not liberate hydrogen ion but accepts a hydroxyl ion, i.e., it behaves as a Lewis acid. 375 K

30. Metaboric acid is formed by heating of boric acid. H3BO3 æææÆ HBO2 - H2O 37. Olivine is (Mg, Fe)2 SiO4. 40. Aluminium dissolves both in HCl and conc. NaOH 2Na[Al(OH)4] + 3H2O 2Al + 6HCl Al2Cl3 + 3H2 and 2Al + 2NaOH + 6H2O 43. Bauxite is Al2O3 ·2H2O. 46. The structure of Al2Cl6 is Cl Cl Cl 101° 118° Al 79°

Cl

Cl

Al 221 pm

206 pm

Cl

47. Constantan is an alloy of Cu (60%) and Ni (40%). 55. The reaction is Al2(SO4)3 Æ Al2O3 + 3SO3 56. The chrome alum involves 24 water of crystallization. excess NH

3 Æ (BN)x 57. B2H6 æææææ high temp

58. Inert pair effect operates on the elements at the end of the group.

Study of the p-Block Elements (Groups 13, 14 and 15)

15.15

59. BF3 is a gaseous substance. 60. The structure of borax is OH B

–

O

O HO

–

B

O

B

OH

O

O B– OH

61. See Q.60 62. The reactions are 63. See Q.60

- H2O ææ Æ H2B4O7 ææææ ææ Æ HBO2 æ435 Æ B 2O 3 H3BO3 æ375 K K red heat

64. The formula of borax is

Na2[B4O5(OH)4] .8H2O

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Aluminium chloride exists as dimer, Al2O6 in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives (b) [Al(H2O)6]3+ + 3Cl– (c) [Al(OH)6]3– + 3HCl (d) Al2O3 + 6HCl [2004] (a) Al3+ + 3Cl– 2. The structure of diborane is (a) two 2c – 2e bonds and two 3c – 3e bonds (b) four 2c – 2e bonds and four 3c – 2e bonds (c) four 2c – 2e bonds and two 3c – 2e bonds (d) two 2c – 2e bonds and four 3c – 2e bonds [2005] 3. Which one of the following is the correct statement? (a) B2H6 . 2NH3 is known as inorganic benzene. (b) Boric acid is a protonic acid (c) Beryllium exhibits coordination number of six (d) Chlorides of both beryllium and aluminium have bridged chloride structures in solid base. [2008] 4. Boron cannot form which one of the following anions? (b) BF63– (c) BH4– (d) B(OH)4– (a) BO2– [2011 cancelled] 5. Which of these statements is not true ? (a) NO+ is isoelectronic with O2 (b) B is always covalent in its compounds (c) In aqueous solution, the Tl+ ion is much more stable than Tl (III) [2014, online] (d) LiAlH4 is a versatile reducing agent in organic synthesis. 6. In the following sets of reactants which two sets best exhibit the amphoteric character of Al2O3. xH2O ? Set 1: Al2O3 . xH2O (s) and OH- (aq) Set 2:

Al2O3 . xH2O (s)

and

H2O (l)

Set 3:

Al2O3 . xH2O (s)

and

H+ (aq)

Set 4: Al2O3 . xH2O (s) and NH3 (aq) (a) 1 and 2 (b) 1 and 3

(c) 2 and 4

(d) 3 and 4 [2014, online]

15.16 Complete Chemistry—JEE Main

7. Identify the reaction which does not liberate hydrogen: (a) Reaction of lithium hydride with B2H6 (c) Reaction of zinc with aqueous alkali (d) Allowing a solution of sodium in liquid ammonia and stand

[2016 online]

ANSWERS 1. (b) 7. (a)

2. (c)

3. (d)

4. (b)

5. (a)

6. (b)

HINTS AND SOLUTIONS 1. In water AlCl3 exists as [Al(H2O)6]3+ and 3Cl– ions. Therefore, the choice b is correct. H

2. The structure of B2H6 is

H

H

Be H

Be H

H

Terminal Be—H is 2c–2e bond Bridged Be—H—Be is 3c–2e bond 3. Inorganic benzene is borazine B3N3H6 with the structure

N

B

N

B

B

N

Boric acid is a Lewis acid. It does not liberate H+ but accepts a hydroxyl ion. Beryllium exhibits coordination number of four. Both BeCl2 and AlCl3 exist as bridged chlorides Cl

Be

Cl

Be — Cl ;

Cl 2

Cl Cl

2

Al

Cl Cl

Al

Cl Cl

2

(2s) (2p) . Its valence-shell has only four orbitals and thus it can show a maximum of four valency. Hence, the choice (b) is correct (because the formation of BF63– is not possible). 5. NO+ contains 14 electrons while O2 contains 16 electrons. These are not isoelectronic. 6. In acidic medium, aluminium exists as Al3+ and in alkaline medium, it exists as AlO2– (aluminate ion). Thus, Set 1 and Set 3 (choice b) represent amphoteric nature of aluminium. 7. The reaction of B2H6 with LiH is 2 LiH + B2H6 2Li(BH4)

The Group 14 Elements – Carbon Family Group 14 of the periodic table consists of elements carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb). The physical properties of these elements are recorded in Table 2.

Study of the p-Block Elements (Groups 13, 14 and 15)

Atomic and Physical Properties of Group 14 Elements

Table 2 Property

15.17

C

Si 2

Ge

Relative atomic mass

Covalent radius, r/pm

a

Ionic radius M4+, r/pmb 2+

Ionic radius M , r/pm

b

Melting point, T/K Boiling point, T/K

Density at 293 K, r/g cm

–3

2

2

Pb

Sn

[He](2s) (2p) [Ne](3s) (3p) [Ar](3d) (4s) (4p) [Kr](4d) (5s) (5p) [Xe](4f)14(5d)l0(6s)2(6p)2 Atomic number

2

10

2

2

10

2

2

6

14

32

50

82

12.011

28.086

72.59

118.69

207.2

77.2

117.6

122.3

140.5

146

—

40

53

69

78

—

—

73

118

119

4373

1693

1218

505

600

—

–3550

3123

2896 d

2024

3.51°

2.34

5.32

7.27

I

1086

786

761

708

715

II

2352

1577

1537

1411

1450

III

4619

3228

3301

2942

3081

IV

6221

4354

4409

3929

4082

2.5

1.8

1.8

1.8

1.9

1014 – 1016

~48

~47

1.1 ¥ 10–5

2 ¥ 10–5

Ionization energy, I/kJ mol

11.34

–1

Electronegativity Electrical resistivity at 293 K,

r/ohm cm

(a) tetracovalent, (b) six-coordinate, (c) for diamond form, (d) (b-form stable at room temperature.

Description of Physical Properties ns)2(np)2, where n varies from 2 to 6. The elements exhibit +4 covalent oxidation state. size, high electronegativity and nonavailability of d orbitals. 3. The group starts from a nonmetal and ends with a metal. There is increase in the metallic character on descending the group. C and Si are nonmetals. Ge is a nonmetal but also has some metallic characteristics Sn and Pb are metals. 4. The covalent and ionic radii increase on descending the group. (Fig. 8). The small difference in sizes between Si and Ge is due the intervening 3d electrons in Ge which shields the nuclear charge less effectively. Similarly, the small difference in sizes between Sn and Pb may be explained on the basis of intervening 4f electrons in case of Pb. 5. The melting points decrease on descending the group, with the exception of Pb whose melting point is slightly higher than that of Sn (Fig. 9).

15.18 Complete Chemistry—JEE Main

Fig. 8 Covalent and Ionic Radii of the Group 15

Fig. 9 Melting Points of the Group 14 Elements Elements

Carbon has extremely high melting point. This is due to the stronger C—C bonds in the network of carbon atoms which has face-centred cubic lattice (known as diamond lattice). The melting points of Si and Ge are also high. Both these elements have diamond type crystal lattice. The decrease in melting point from C to Si to Ge is due to the weakening of the M—M bond caused by the increase in covalent radii of the element. The melting points of Sn and Pb are low. They do not use all four outer electrons for metallic bonding. 6. The boiling points of the elements of Group 14 also decrease on descending the group. 7. The ionization energies decrease from C to Si, but then change in an irregular way because of the effects of 4+

ions is excluded.

oxidation state of these elements in the compounds is primarily +2. 8. The phenomenon of catenation (ability to like atoms to link one another through covalent bonds) decreases down the group. This phenomenon is linked with the M—M bond enthalpy, which decreases down the group. 9. Besides sigma bond, carbon forms pp—pp multiple bonding with itself and with other nonmetals; especially nitrogen and oxygen (e.g. C == C, C ∫∫ C, C ∫∫ N, C == O, C == S). The compounds of Si with multiple bond are rare. However, the element can form double bond through dp—pp bonding, in which the lone pair in p orbital of an atom is extended to an empty orbital of Si. This is known as back bonding. One of the examples of trisilylamine N(SiH3)3. 10. Maximum covalency of carbon is four, whereas the other elements of group can exhibit maximum valence of six due to the participation is d orbitals (e.g. [SiF6]2–, [Sn(OH)6]2–, [GeCl6]2– and [PbCl6]2–).

Trisilylamine

Fig. 10 The First Four Ionization Energies of the Group 14 Elements.

Study of the p-Block Elements (Groups 13, 14 and 15)

15.19

Chemical Properties of Group 14 Elements The reactivity of elements decreases down the group. The inert pair effect becomes increasingly effective down the group. The stability of the +4 oxidation state decreases while that of the +2 oxidation state increases on descending the group. For example, Sn2+ exists as simple ion and is strongly reducing. Sn4+ is covalent. Pb2+ is ionic, stable and more common than Pb4+.

Action of Water C, Si and Ge are unaffected by water. Sn reacts with steam to give SnO2 and H2. Pb is unaffected by water due to the formation of protective oxide layer at the surface. Action of Acids

C, Si and Ge are unaffected by dilute acids. Sn and Pb reacts with dilute nitric acid. 4Sn + 10HNO3 Æ 4Sn(NO3)2 + NH4NO3 + 3H2O (dilute)

3Pb + 8HNO3 Æ 3Pb(NO3)2 + 2NO + 4H2O. (dilute)

Action of Alkalis C is not affected by alkalis. Si reacts with alkalis forming silicates. Sn and Pb also react with alkalis forming stannate, [Sn(OH)6]2–, and plumbate [Pb(OH)6]2–. These reactions show that Sn and Pb are amphoteric.

Action of Halogens Graphite but not diamond is affected by F2 at higher temperatures giving (CF)n. Si and Ge react with all halogens, forming volatile SiX4 and GeX4. Sn and Pb are less reactive but do react giving SnX4 and PbX2.

Compounds of Group 14 Elements 1. Hydrides All the elements of Group 14 form covalent hydrides. Carbon forms a very large number of compounds which include alkanes (CnH2n+2), alkenes (CnH2n), alkynes (CnH2n–2) and aromatic compounds. Silicon forms a limited number of saturated hydrides (SinH2n+2) called silanes, which are strong reducing agents. One of the methods used in the preparation of silanes is by the reduction of silane halides by LiAlH4, LiH or NaH. SiCl4 + LiAlH4 Æ SiH4 + AlCI3 + LiCl Si2Cl6 + 6LiH Æ Si2H6 + 6LiCl Si3Cl8 + 8NaH Æ Si3H8 + 8NaCl Silanes are easily hydrolyzed in alkaline medium. of alkali Si2H6 + (4 + n) H2O ⎯trace ⎯⎯⎯⎯ → 2SiO2 ◊ nH2O + 7H2 Stannane (SnH4) and plumbane (PbH4) are also known but are less stable.

2. Halides All the elements of Group 14 form tetrahalides with the exception of PbI4 which is not known. Probably, this is due to the fact that PblV is a strong oxidizing agent and I– is a strong reducing agent, so both strong oxidizing and reducing agents cannot exist together. They are all covalent and volatile with the exceptions of SnF4 and PbF4 which have threedimensional structures and are high melting. The stability of the halides decreases down the group. CCl4 is stable while other halides are hydrolyzed due to the availability of d orbitals. The hydrolysis of SiCl4 may be represented as follows.

15.20 Complete Chemistry—JEE Main

The hydrolysis of SiF4 besides producing SiO2 also produces [SiF6]2– SiF4 + 2H2O Æ SiO2 + 4HF SiF4 + 2F– Æ [SiF6]2–

3. Oxides The elements of Group 14 form oxides of the type MO and MO2. The well-known oxides of carbon are CO and CO2. Besides these, less stable C3O2, C5O2 and C12O9 are also known. Carbon monoxide is extremely poisonous gas. If inhaled, it forms a complex with haemoglobin in the blood which is much more stable than oxy-haemoglobin complex. This prevents the haemoglobin in the red blood corpuscles from CO is a good reducing agent. It is also an important ligand which can donate or share a lone pair of electrons located on carbon atom. Carbon dioxide is a colourless, odourless gas, which play a vital role in the photosynthesis. Carbon dioxide is represented as O == C == O. Each double bond involves one s bond and one p bond. The formation p bonds is possible because of the small size of carbon and thus the overlapping between 2p(C) and 2p(O) are possible. Because of the double bonds, CO2 exists as discrete molecule and is a gas. Silicon also forms silicon dioxide which is solid and high melting. Silicon is not able to form pp-pp double bonds with oxygen due to its large size which prevents effective overlap between 2p(Si) and 2p(O) orbitals. Instead, a continuous tetrahedrally to four oxygen atoms and each oxygen atom is shared by two silicon atoms, as shown in Fig. 11. Acidic nature of the dioxides decreases down the group. CO2 and SiO2 are purely acidic. GeO2 is weakly acidic. SnO2 and PbO2 are amphoteric.

Fig. 11 Structure of Silicon Dioxide

The lower oxides GeO, SnO and PbO are also known. They are slightly more basic and ionic than the corresponding higher oxides. GeO is purely acidic SnO and PbO are amphoteric.

Study of the p-Block Elements (Groups 13, 14 and 15)

15.21

The stability of lower oxide increases down the group. Lead also forms a mixed oxide 2PbO ◊ PbO2 (i.e. Pb3O4). It is used as red pigment.

Silicates The earth’s crustal rocks and their breakdown products like clays, soils and sands, are mostly silicates and silica. Mica, asbestos, quartz, feldspar, zeolites, etc. are some of the important silicate minerals. Basically, all the silicates and silica contain SiO4– 4 tetrahedra. (Fig. 12). 4–

O

–

– –

= o

Si

= Si O

O

O –

Fig. 12

The basic out of SiO4– 4

Depending on the number of corners (0, 1, 2, 3 or 4) of SiO4– 4 tetrahedron shared with other tetrahedra, a variety of silicates are obtained. These are described below.

Orthosilicates These contain individual discrete SiO4– 4 tetrahedra. Examples are phenacite (Be2SiO4), willemite (Zn2SiO4), and zircon (ZnSiO4) minerals. Pyrosilicates 4– These contain discrete Si2O6– 7 ions formed by the sharing of one oxygen atom between be two SiO4 tetrahedra (Fig. 13). This structure is known as an island structure. Examples include thortveitite (Sc2Si2O7) and hemimorphite (Zn4(OH)2Si2O7). – = Oxygen = Silicon

–

Fig. 13

Island structure

Chain and Cyclic Silicates These silicates are formed when two oxygen atoms of SiO4– 4 tetrahedron are shared with two other tetrahedra. The resulting structure may be the chain type (Fig. 14) or the cyclic type (Fig. 15). The general formula of these silicates is (SiO3)2n– n, . Examples of cyclic silicates are benitoite (BaTiSi3O9), catapleite (Na2ZrSi3O9 ◊ 2H2O), dioptase (Cu6Si6O18 ◊ 6H2O) and beryl (Be3Al2Si6O18). –

–

–

–

= Oxygen –

–

–

–

–

Fig. 14

–

Chain structure

= Silicon

15.22 Complete Chemistry—JEE Main – –

– –

–

–

–

–

– –

= Oxygen

–

–

–

= Silicon

–

– =O = Si

–

–

Fig. 15

–

Cyclic structure 4– 4 tetrahedron

are shared with the adjacent tetrahedra. If further sharing of oxygen atoms occurs by half of the silicon atom, a double chain or band structure is formed. Examples of the former include enstatite MgSiO3 and diopside CaMg(SiO3)2. Asbestos also contains silicate chain. The band structure contains the basic Si4O6– 11 repeating unit. An example is Ca2Mg5(OH)2(Si4O11)2.

Layer and Sheet Silicates empirical formula (Si2O5)2n– n . This is shown in Fig. 16. The minerals kaolin [Al2(OH)4Si2O5], talc [Mg3(OH)2Si4O10] and the micas are common examples possessing layer and sheet structures.

= Oxygen = Silicon

Fig. 16

Layer and Sheet Structure

Three-dimensional Silicates In these silicates, all the four oxygen atoms of a SiO44 – tetrahedron are shared with other tetrahedra resulting in a threedimensional lattice. In case there is no replacement of silicon by other metals, the resulting formula is SiO2 (quartz). General Structure of Silicates The number of oxygen atoms shared in the general formula of silicate may be determined as follows.

(

)(

)(

Number of Charge on + Charge on Ê Number of ˆ Si atom silicate ion Á O atoms ˜ = Si atoms Ë shared ¯ (Number of Si atoms)

)

Study of the p-Block Elements (Groups 13, 14 and 15)

15.23

For example, Silicate anion

Number of O atoms shared

Expected structure

SiO4– 4

(1)(+ 4) + (-4) =0 1

Discrete

Si2O6– 7

(2)(+ 4) + (-6) =1 2

Island structure

(SiO3)2n– n

n(+4) + (-2n) =2 n

Chain or cyclic type

(Si2O5)2n– n

(2n)(+4) + (-2n) =3 ( 2n)

Two dimension sheet

SiO2

1(+4) + 0 =4 1

Three dimension lattice

Silicones Silicones form a group of organosilicon polymers of wide commercial use. They have general formula (R2SiO)n, where R may be methyl, ethyl or phenyl group. For the straight-chain polymer, dimethyldichlorosilane (CH3)2SiCl2 is used. Its hydrolysis followed by polymerization may be represented as follows

The above reaction continues because of active OH group at each end of the chain.

atoms.

Trimethylmonochlorosilane (CH3)3SiCl can be used to block the polymerization as it does not create the active OH group at the end of the chain.

15.24 Complete Chemistry—JEE Main

The hydrolysis of methyltrichlorosilane RSiCl3 produces a very complex cross-linked polymer. The durability and inertness of silicones is due to the silica-like arrangement Si—O—Si—O—Si which involves very strong Si—O bonds with bond enthalpy 502 kJ mol–1. The Si—C bond is also very strong. are excellent electrical insulators. Silicon resins are used in paints and varnishes.

4. Carbides These are prepared by direct combination of metals with carbon at elevated temperatures or indirectly, the heating of Ionic carbides are formed by metals of Group 1, 2 and 13. These compounds, in general occur as transparent crystals and in the solid state they are nonconductor of electric current. They give hydrocarbons when treated with water or acids. 4– 4– ), acetylides (C2– 2 ) and allylides. (C3 ). The examples are Be2C + 4H2O Æ 2Be(OH)2 + CH4 Al4C3 + 12H2O Æ 4Al(OH)3 + 3CH4 CaC2 + 2H2O Æ Ca(OH)2 + C2H2 Al2C6 + 6H2O Æ 2Al(OH)3 + 3C2H2 Mg2C3 + 4H2O Æ 2Mg(OH)2 + CH3C ∫ CH Covalent carbides formed by the elements of Group 16 and 17 are discrete molecules (CH4, CO2, CS2, etc.). Some covalent carbides are giant molecules, e.g. carborundrum (SiC) and boron carbide (B4C3). These are characterized by high decomposition temperature, chemical inertness and extreme hardness. Carborundum has a diamond like structure. Interstitial (or metallic) carbides are formed by the transition elements such as TiC, HfC, W2C, MoC, Mo2C, Cr3C2 and Cr4C. In these the small carbon atoms occupy interstitial position in the crystal lattices of the metals. These compounds are characterized by hardness, chemical inertness and high electric conductivity.

5. Zeolites In the framework where all four oxygen atoms of SiO4 tetrahedron are shared, if some of the silicon atoms are substituted by aluminium atoms, the aluminosilicate framework (known as zeolite with the general formula) x– M n+ x/n [(AlO2)x (SiO2)y] ◊ 3H2O) is formed. Here each AlO2 unit acquires one negative charge as the aluminium atom is trivalent, whereas the silicon atoms is tetravalent. To balance the electrical charge some other cations (such as Na+, Ca2+ and Mg2+) are absorbed in voids. Zeolites are characterized by their open structures that permit the exchange of cations and water molecules. Zeolite structures also contain tunnels or systems of interconnected cavities which have have been used as gaseous separator, cation exchanger, desiccants and catalysts.

Allotropes of Carbon Carbon exists in two true allotropic forms, namely, diamond and graphite. Diamond Diamond crystallizes in the face-centred cubic system. The diamond crystal is built up of a giant three dimensional structure with tetrahedral arrangement of carbon atoms which are at equal distance of 154 pm from one another (Fig. 17). The forces acting between atoms are very strong. The crystal of diamond is a nonconductor of electricity as all the electrons of carbon are held in the bonds between carbon atoms. It is the densest and purest variety of carbon. Also, it is the hardest natural substance known. Diamond has a high refractive index (= 2.45) because of which almost all chemical reagents. It burns in air at 900° C in oxygen at 700 °C to form carbon dioxide. Graphite Structurally graphite consists of a two-dimensional sheet-like network in which the carbon atoms are joined together in hexagonal rings (Fig. 18).

Study of the p-Block Elements (Groups 13, 14 and 15)

15.25

141.5 pm 154 pm

340 pm

Fig. 17

The crystal structure of diamond

Fig. 18

The layer structure of graphite

The two layer of carbon network in graphite are held together by weak van der Waals forces and are about 340 pm apart. Within the layer, each carbon atoms covalently bonded by sp2 hybrid orbitals to three cabron atoms with C—C distance equal to 141.5 pm. Thus, three of the four valence electrons are utilized in bonding with other carbons within the layer and the fourth electron is in the pz atomic orbital which is perpendicular to the layer and is involved in p bonding with such electrons in the same layer. Because of the highly delocalized nature of p electrons, graphite is a good conductor of heat and electricity. It is less dense as compared to diamond. Since the different layers are held by weak van der Waals forces, each layer can slide over the other and this leads to the softness and the lubrication properties of graphite Graphite is thermodynamically more stable than diamond. Besides graphite and diamond, there are several amorphous forms of carbon which resemble graphite in character. These include coal, coke, wood charcoal, carbon black, etc. Coal is believed to have been formed by the slow carbonization of vegetable matter buried underneath the earth centuries ago, in limited supply of air under high temperature and pressure prevailing there. The different varieties of coal available are peat (60% C), lignite (70 % C), bituminous (78 % C), semibituminous (83% C) and anthracite (90% C). The common variety of coal is bituminous. When coal is subjected to destructive distillation by heating in the absence of air, the residue left is known as coke. Wood charcoal is obtained by heating wood in the absence of air.

Fullerenes Heating of graphite is an electric arc in the presence of inert gases followed by condensation of sooty material results in the formation of fullerenes. The latter mainly consists of C60 with small quantity of C70. The molecule C60 is known as Bukminster fullerene membered rings. All C atoms are sp2 hybridized and occupy 60 vertices of ball-type molecule.

Anomalous Behaviour of Carbon Carbon differs from rest of the elements of the group because of its small size, high electronegativity and nonavailability of d-orbitals. It has the unique characteristics of forming compounds with multiple bonds such as C == C, C == O, C ∫∫ C, C ∫∫ N, etc. It also exhibits the property of catenation, i.e. forming chains of identical atoms. This property is related to the strength of the C—C bond. The higher the bond enthalpy, the greater will be the tendency to form chains (C—C bond enthalpy 348 kJ mol–1 , Si—Si bond enthalpy 222 kJ mol–1). The covalency of carbon is limited to four because of the nonavailability of d orbitals. The other elements of the group are capable of forming compounds in which they attain a covalency higher than four. The melting and boiling points of carbon are exceptionally high as compared to those of silicon and other elements of the group. Example

Which of the bond in each of the following pairs is stronger?

(a) C—C and Si—Si (b) C—O and Si—O

15.26 Complete Chemistry—JEE Main

Answer (a) The C—C single bond is stronger than the Si—Si bond. This is due to the smaller size of C leading to the good overlap between the orbitals involved in the bond formation. (b) The Si—O bond is much stronger than the C—O bond. It is due to the formation of pp—dp bond between Si and O atoms. Example

Describe the structural aspects of trimethylamine and trisilylamine.

Answer Trimethylamine has a pyramidal structure whereas trisilylamine has a planar structure. In trisilylamine, (H3Si)3N, nitrogen is sp2 a empty silicon 3dxz (or 3dyz) orbital. Thus, a dative pp—dp bond is established. This linkage provides additional bond strength in each Si—N bond. Because of this linkage the skelton NSi3 In trimethylamine, the above type of pp—dp bond is not possible as carbon does not have low lying d orbitals.

MULTIPLE CHOICE QUESTIONS 1. Which of the following elements does not belong to Group 14? (a) Carbon (b) Silicon (c) Germanium (d) Arsenic 2. The atomic number of the element tin which belongs to 5th period is (a) 49 (b) 50 (c) 81 (d) 82 3. Which of the following elements may be regarded as nonmetal? (a) Carbon (b) Silicon (c) Germanium (d) Tin 4. Which of the following elements may be regarded as semi-metals? (a) Carbon (b) Germanium (c) Tin (d) Lead 5. Which of the following elements behaves like a metal? (a) Carbon (b) Silicon (c) Germanium (d) Tin 6. Which of the following orders regarding the melting points is correct? (a) C > Si > Ge > Sn (b) C < Si < Ge < Sn (c) C > Si < Ge < Sn (d) C < Si > Ge > Sn 7. Which of the following orders regarding the boiling point is correct? (a) Si > Ge > Sn (b) Si < Ge < Sn (c) Si > Ge < Sn (d) Si < Ge > Sn 8. Which of the following statements is not correct? (a) The stability of the +4 oxidation state of elements of Group 14 decreases down the group (b) The stability of the +2 oxidation state of elements of Group 14 increases down the group (d) A majority of the compounds of elements of Group 14 are four covalent and involve sp3 hybridization 9. Which of the following statements is not correct? (a) All the elements of Group 14 form covalent hydrides (b) Carbon forms a very large number of compounds with hydrogen (c) Silicon like carbon forms a large number of compounds with hydrogen. These are collectively known as silanes (d) The general formula of silanes is SinH2n+2 and these are strong reducing agents 10. Which of the following statements is not correct? (a) All the elements of Group 14 form tetrahalides with the exception of PbI4 (b) All tetrahadies of elements of Group 14 are tetrahedral (c) The stability of the halides of elements of Group 14 increases down the group (d) SiCl4 exhibits the phenomenon of hydrolysis 11. Which of the following oxides is highly poisonous? (a) CO (b) CO2 (c) C3O2 (d) C5O2

Study of the p-Block Elements (Groups 13, 14 and 15)

15.27

12. Which of the following statements is not true? (a) Carbon produces carbon monoxide when it is burnt in a limited supply of air (b) Carbon monoxide is a good reducing agent (c) Carbon monoxide acts as a ligand through the donation of lone pair of oxygen (d) Carbon monoxide acts as a ligand through the donation of lone pair of carbon 13. Which of the following statements is not correct? (a) Carbon dioxide is a linear molecule with no dipole moment (b) Carbon dioxide and silicon dioxide have the same structures (c) Carbon dioxide is a gas whereas silicon dioxide is solid (d) In carbon dioxide, a double bond exists in carbon and oxygen whereas in silicon dioxide each silicon-oxygen bond is linked through a single sigma bond 14. Which of the following oxides is an acidic oxide? (a) CO2 (b) Ge3O2 (c) SnO3 (d) PbO2 15. Which of the following oxides is an amphoteric oxide? (a) CO2 (b) SiO2 (c) GeO2 (d) PbO2 16. Which of the following oxides is a basic oxide? (a) CO2 (b) SiO2 (c) GeO2 (d) PbO2 17. Which of the following statements is not correct? (a) Organosilicon polymers are known as silicones (b) Silicones have the general formula (R2SiO)n where R = – CH3, – C2H5, – C6H5, etc. (c) Hydrolysis of dialkyldichlorosilane produces cross-linked silicon polymer (d) Hydrolysis of alkyltrichlorosilane produces cross-linked silicon polymer 18. Which of the following statements is not correct? (a) The durability and inertness of silicones is due to the high bond enthalpy of Si —O bond

19. 20. 21. 22. 23. 24. 25.

26.

(c) Silicon rubbers are excellent electrical insulators (d) The silicones always involve cross-linked between Si and O atoms Carborundum is (a) CaC2 (b) Fe3C (c) CaCO3 (d) SiC Which of the following is expected to be conducting? (a) SiO2 (b) Graphite (c) Diamond (d) Methane Which of the following compounds of carbon may be considered a resonance hybrid? (a) C2H5OH (b) C6H6 (c) C4H10 (d) C2H2 Iron may be rendered passive by application of (a) PbO (b) Pb(OH)2 (c) PbCO3 (d) Pb3O4 The dissolution of stannous hydroxide in excess base produces (a) Sn(OH)4 (b) Sn(OH)62– (c) SnO32– (d) Sn(OH)–3 Synthetic zeolites have been prepared for use as (a) lubricants (b) molecular sieves (c) semiconductors (d) plastics Which of the following statements is correct? (a) Graphite is a bad conductor of electricity (b) Dry graphite in vacuum is not slippery (c) The adsorption of substances by graphite increases the friction as the layers slide past each other (d) The graphite does not possess metallic properties Which of the following statements is not correct regarding graphite crystal? (a) The carbon atoms are arranged in planar layers (b) Each carbon in graphite is sp2 hybridised (c) The atoms in each layer is not tightly bonded together (d) The binding force between layers is weak allowing the layers to slip over each other

15.28 Complete Chemistry—JEE Main

27. Which of the following carbides is an ionic carbide? (a) SiC (b) B4C (c) TiC (d) CaC2 28. Which of the following is white lead? (a) Pb(OH)2 · 2PbCO3 (b) Pb(OH)2 · Pb(CH3COO)2 (c) Pb(OH)2 (d) PbCO3 29. Which of the following is not hydrolysed? (a) CCl4 (b) SiCl4 (c) GeCl4 (d) SnCl4 30. Which of the following order regarding bond energies is correct? (a) e(C—H) > e(Si—H) > e(Ge—H) (b) e(C—H) < e(Si—H) < e(Ge—H) (c) e(C—H) > e(Si—H) < e(Ge—H) (d) e(C—H) < e(Si—H) > e(Ge—H) 31. Tin exists in (a) one allotropic form (b) two allotropic forms (c) three allotropic forms (d) four allotropic forms 32. Which of the following is red lead? (a) PbO (b) Pb3O4 (c) Pb2O (d) PbO2 33. Which of the following represents producer gas? (a) CO and N2 (b) CO2 and H2 (c) CO and H2 (d) CO2 and N2 34. Silicones contain (a) C, O and Si (b) C and Si (c) O and Si (d) C, N, O and Si 35. Silanes are compounds which contain (a) C, H and Si (b) C and Si (c) O and Si (d) H and Si 36. Which of the following statements is not correct? (a) Silicon is the second member of Group 14 (b) Silicon tetrahalide exhibits the phenomenon of hydrolysis (c) Because of large radius of silicon atom, the latter does not form the bonds of the type Si == Si, Si Si (d) Silicon dioxide molecules in solid phase are held together by van der Waals forces 37. Which of the following statements is not correct? (a) Silicon is extensively used as a semiconductor (b) Carborundum is SiC (c) Silicon occurs in free state in nature (d) Mica contains the element silicon 38. Which of the following represents pyrosilicate ion? (a) SiO4– (b) Si2O6–7 (c) (SiO8)21– (d) Si3O6– 4 n 9 39. Only one oxygen atom of two SiO4 tetrahedra is shared in (a) orthosilicate (b) pyrosilicates (c) chain silicates (d) cyclic silicates 40. In chain and cyclic structures of silicates (a) no oxygen atom is shared amongst SiO4 tetrahedra (b) one oxygen atom per tetrahedron is shared (c) two oxygen atoms per tetrahedron are shared (d) three oxygen atoms per tetrahedron are shared 41. In the sheet type silicates (a) no oxygen is shared amongst SiO4 tetrahedra (b) one oxygen atom per tetrahedron is shared (c) two oxygen atoms per tetrahedron are shared (d) three oxygen atoms per tetrahedron are shared 42. In the structure of quartz, (a) no oxygen atom is shared amongst SiO4 tetrahedra (b) one oxygen atom is shared between two tetrahedra (c) two oxygen atoms per tetrahedron are shared (d) all the four oxygen atoms of a SiO4 tetrahedron are shared with other tetrahedra 43. Which of the following statements regarding glass is not true? (b) Ordinary glass is a mixture of sodium and calcium silicates and may be represented as Na2SiO3 · CaSiO3 · 4SiO2 (c) Small amounts of Co(II), Cr(III), Fe(III) and Mn(IV) compounds impart respectively blue, green, brown and violet colours to the glass (d) Flint glass has a low refractive index

Study of the p-Block Elements (Groups 13, 14 and 15)

44. Which of the following statements is correct? (a) All compounds of silicon are collectively known as silicones (b) Silicon hydrides are known as silicones (c) Silicon halides are known as silicones (d) Silicones are polymeric organosilicon compounds 45. Which of the following statements regarding silicones is not correct? (a) Silicones have good thermal stability (b) Liquid silicones are lubricants (c) Silicones are chemical inert substances (d) Silicone rubber is not attack by ordinary chemical reagents except ozone 46. Lead dioxide is a/an (a) neutral oxide (b) acidic oxide (c) basic oxide 47. Plumbo-solvancy means dissolution of lead in (a) water (b) acids (c) alkalis 48. Red lead has the chemical formula (a) PbO (b) PbO2 (c) Pb2O3 49. Which of the following is not correct? (a) Lead is a true metal with + 2 electrovalency (b) Lead forms PbCl4 which is soluble in organic solvents (c) Lead reacts with concentrated HCl to form PbCl2 (d) Lead reacts with NaOH solution to form Pb(OH)4 50. Which of the following is known as butter of tin? (a) SnCl2·2H2O (b) SnCl2·3H2O (c) SnCl2· 4H2O 51. Which of the following halides does not exist? (a) PbF4 (b) PbCl4 (c) PbBr4 52. Which of the following halides is of yellow colour? (a) PbCl2 (b) PbBr2 (c) PbI2 53. The element tin and lead belong to Group (a) 11 (b) 12 (c) 13 54. Which is the chief ore of tin? (a) Cerrusite (b) Anglesite (c) Cassiterite 55. The chief ore of lead is (a) galena (b) cinnabar (c) cassiterite

(d) amphoteric oxide (d) organic solvents (d) Pb3O4

(d) SnCl2·5H2O (d) PbI4 (d) PbF2 (d) 14 (d) Cinnabar (d) zincite

(a) Crude lead acts as anode (b) Pure lead acts as cathode (c) Electrolytic solution contains PbSiF6 and H2SiF6 (d) Anode mud contains copper only 57. Heating lead in air at higher temperature (T 725 K), the oxide formed is (a) PbO (b) PbO2 (c) Pb2O3 (d) Pb3O4 58. Which of the following equations is not actually observed? (a) 4Sn + 10HNO3 (dilute)

(b) 3Sn + 8HNO3 (dilute)

(c) Sn + 2HCl (conc.)

(d) Sn + 2H2SO4 (conc.)

4Sn(NO3)2 + NH4NO3 + 3H2O 3Sn(NO3)2 + 2NO + 4H2O

SnCl2 + H2 SnSO4 + 2H2O + SO2

15.29

15.30 Complete Chemistry—JEE Main

59. Which of the following equations is not actually observed? 4Pb(NO3)2 + NH4NO3 + 3H2O (a) 4Pb + 10HNO3 (dilute)

(b) 3Pb + 8HNO3 (dilute)

(c) PbCl2 + 2HCl

(conc.)

(d) Pb + 4HNO3 (conc.)

3Pb(NO3)2 + 2NO + 4H2O H2PbCl4 Pb(NO3)2 + 2NO2 + 2H2O

60. Which of the following statements is not correct? (a) Divalent tin or lead compounds are essentially ionic in nature while tetravalent are generally covalent (b) Tin(II) compounds are invariably reducing agents (c) Lead(II) compounds are more stable than Pb(IV) compounds (d) (C2H5)2Pb acts as an antiknocking compound in petrol 61. Tin dioxide is a/an (a) neutral oxide (b) acidic oxide (c) basic oxide (d) amphoteric oxide 62. Which of the following is massicot? (a) PbO (b) PbO2 (c) SnO (d) SnO2 63. Lead oxide is a/an (a) neutral oxide (b) acidic oxide (c) basic oxide (d) amphoteric oxide 64. Which of the following halides will have minimum ionic character? (a) PbF2 (b) PbCl2 (c) PbBr2 (d) PbI2 65. Which of the following statements is not correct? (a) Lead(II) chloride and lead(II) iodide are soluble in hot water but crystallize out on cooling (b) Lead(IV) chloride cannot be separated from the solution as it readily decomposes to lead(II) chloride and chlorine (c) Lead(IV) chloride is not readily hydrolysed by water (d) Per cent of lead(II) halides decreases with the increase in atomic number of the halogen 66. Which of the following statements is not correct? (a) Tin(IV) chloride is an ionic compound (b) Tin(IV) chloride undergoes hydrolysis with water (c) With excess of hydrochloric acid, tin(IV) chloride forms hexachlorostannic acid (H2SnCl6) (d) Tin (II) chloride can reduce Fe(III) to Fe(II), Cu(II) to Cu(I), Hg(II) to Hg(0) and Au(III) to Au(0) 67. The only stable tetrahalide of lead is (a) PbF4 (b) PbCl4 (c) PbBr4 (d) PbI4 68. Which of the following oxides is used in lead accumulators? (a) PbO (b) Pb2O3 (c) PbO2 (d) Pb3O4 69. Which of the following elements does not show allotropy? (a) Carbon (b) Silicon (c) Sulphur (d) Lead 70. Mosaic gold is (a) SnS (b) PbS (c) SnS2 (d) CdS 71. Brown SnO is (a) an acidic oxide (b) basic oxide (c) an amphoteric oxide (d) neutral oxide 72. Lead sesquioxide is (a) PbO (b) Pb2O3 (c) PbO2 (d) Pb3O4 73. Galena is an ore of (a) Fe (b) Cu (c) Zn (d) Pb 74. Desilverization of lead is known as (a) Cyanide process (b) Parke’s process (c) MacArthur-Forest process (d) Parting process

Study of the p-Block Elements (Groups 13, 14 and 15)

15.31

75. Which of the following statements is not correct? (a) Zeolite contains aluminosilicate framework x– (b) The general formula of zeolite is Mn+ x/n [(AlO2)x (SiO2)y] · z H2O (c) Zeolites are characterized by their open structures that permit the exchange of anions and water molecules (d) Sodalite cage is formed by linking 24 SiO4 tetrahedra 76. Which of the following statements is not correct? (a) Zeolite A is formed by linking sodalite cages through double four-membered rings (b) Faujasite zeolite is formed by linking the sodalite cages through double six-membered rings

77. 78. 79. 80.

82. 83. 84. 85. 86. 87.

88. 89.

90. 91. 92.

dimensions on the atomic scales (d) Zeolites are anion exchanger Bond energy is highest amongst (a) C — C (b) Si — Si (c) Ge — Ge Lead pencil contains (a) PbS (b) PbO (c) FeS Amongst the following element, which one has the highest melting point? (a) Ge (b) Sn (c) Pb Hydrolysis of SiF4 in alkaline medium produces (a) SiO2 (b) H2SiO4 and H2SiF6 (c) H2SiF6

(d) Sn — Sn (d) graphite (d) Na (d) H2SiO4 and SiO2

(a) Water gas (b) Coal gas (c) Producer gas (d) Semi-water gas Which of the following compounds of lead is the most stable? (a) PbF4 (b) PbCl4 (c) PbBr4 (d) PbI4 Which of the following is a methanide? (a) Be2C (b) CaC2 (c) Mg2C3 (d) TiC Which of the following is an acetylide? (a) Be2C (b) Al4C3 (c) CaC2 (d) Mg2C3 Which of the following is an allylide? (a) Be2C (b) Al4C3 (c) CaC2 (d) Mg2C3 Which of the following bonds has the highest bond energy? (a) Si — Si (b) Si = Si (c) Si – O (d) Si = O Which of the following statements is correct? (a) Si6Cl14 exists as a cyclic chlorosilane +2 (b) Pyrolysis of SiCl4 at 1150 °C in the presence of Si produces members of the series SinCl2n (c) Si — F bond is a weaker bond than C — F bond (d) Si — Si bond is a stronger bond than C — C bond Which of the following elements is not the constituent of zeolite A. (a) Al (b) Si (c) O (d) Mg Each silicon atom in the trimer cyclic silicate has (a) one terminal oxygen and three bridging oxygens (b) two terminal oxygens and two bridging oxygens (c) three terminal oxygens and one bridging oxygen (d) all the four oxygens as the bridging oxygen Which one of the following is a covalent carbide? (a) Mg2C (b) Al4C3 (c) Li3C2 (d) SiC x The charge x of Si6O 18 will be (a) 4 – (b) 6 – (c) 8 – (d) 12 – The oxidation state of Sc in Sc2Si2O7 is (a) +2 (b) +3 (c) +4 (d) +6

15.32 Complete Chemistry—JEE Main

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79. 85. 91.

(d) (a) (b) (d) (b) (c) (c) (d) (d) (a) (d) (a) (d) (a) (d) (d)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 86. 92.

(b) (c) (a) (b) (c) (b) (b) (d) (d) (d) (a) (c) (b) (b) (c) (b)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75. 81. 87.

(a) (c) (c) (b) (d) (a) (b) (d) (d) (d) (d) (d) (c) (a) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76. 82. 88.

(b) (c) (d) (d) (a) (a) (c) (d) (c) (b) (d) (c) (d) (a) (d)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77. 83. 89.

(d) (a) (c) (d) (a) (d) (c) (a) (d) (a) (c) (a) (a) (a) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 90.

(c) (c) (d) (b) (a) (d) (d) (d) (c) (d) (a) (b) (d) (c) (d)

HINTS AND SOLUTIONS 6. Because of catenation of carbon, its melting point is greater than silicon. 7. Boiling point decreases from Si to Pb. 9. 10. 12. 13.

Silicon forms a limited number of saturated hydrides. The stability of the halides of elements of Group 14 decreases down the group. Carbon monoxide donates a lone pair of carbon. Carbon dioxide exists as individual molecule whereas silicon dioxide exists as giant molecule containing a network of C and O bonds. 14. Acidic nature of the oxides decreases down the group. 17. The hydrolysis of dialkyldichlorosilane produces linear silicon. R R R H O

2 Æ n R2SiCl2 æææ

- HCl

O

Si R

–

O

Si O

Si

R

R

O

Sn(OH)3–

23. Sn(OH)2 + OH 36. Each silicon atom in SiO2 is bounded to four oxygen atoms tetrahedrally. Each oxygen atom is shared by two silicon atoms. 43. Flint glass is a lead-potash-lime glass and it has high refractive index. 49. Pb + 2OH – æÆ PbO22 – + H2 plumbate

54. 55. 56. 58.

Cerrusite is PbCO3. Anglesite is PbSO4. Cainnabar is HgS. Cassiterite is SnO2. Galena is PbS. Anode mud contains many other metals such as Ag, Au, Sn, As and Sb. 4Sn(NO3)2 + NH4NO3 + 3H2O 4Sn + 10HNO3 (dilute)

59. Choice (b) is observed, i.e. 3Pb + 8HNO3 (dilute)

3Pb(NO3)2 + 2NO + 4H2O

Study of the p-Block Elements (Groups 13, 14 and 15)

15.33

60. (C2H5)4Pb acts as an antiknocking compound in petrol. 65. Lead(IV) chloride gets rapidly hydrolyzed by water. PbCl4 + 2H2O PbO2 + 4HCl 66. SnCl4 is a covalent compound. It is soluble in organic solvents. 67. PbF4 is yellow compound with melting point 873 K. PbCl4 is yellow oil stable below 273 K and decomposes to PbCl2 and Cl2 above 323 K. PbBr4 is even less stable. PbI4 does not exist. 77. C—C bond has the highest bond energy. As the size of atoms becomes larger, the bond energy decreases. 78. Lead pencil contains graphite 79. Melting point decreases on descending a group. 80. SiF4 + 4H2O Æ Si(OH)4 + 4HF and SiF4 + 2HF Æ2H+ + [SiF6]2– 81. Water gas is a mixture of CO and H2 both CO and H2 burn and evolve heat. 82. Stability decreases with increasing atomic number of halogen. 83. Be2C is a methanide as it produces CH4 on reacting with H2O. 84. CaC2 is acetylide as it produces actylene on reacting with H2O. 85. Mg2C3 is allylide as it produces propyne on reacting with H2O. 86. Amongst the given bonds, Si—O has the highest bond energy. 87. SiCl4 + SiÆSi2Cl6 + higher members of the series 5Si2Cl2+Si6Cl14+4SiCl4 88. Zeolites are the three-dimensional silicates and contain Na, Al and Si elements. 89. Trimeric cyclic silicate is

90. SiC is a covalent carbide, rest are ionic carbides. x 2– 12 – ∫ (SiO4– 91. Si6O 18 4 )6 – 6O ∫ Si6O 18 92. The charge of Si2O7x is

2– 6– Si2O 7x ∫ (SiO4– 4 )2 – O ∫ Si2O 7 . Hence,

Alternatively, 2x + 2(4+) + 7(–2) = 0

fi

the charge on Sc is (6+)/2 = 3+

x = 3+

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN graphite (a) has carbon atoms arranged in a large plates of rings of strongly bound carbon atoms with weak interplate bond (b) is a non-crystalline solid (c) is a allotropic form of diamond (d) has molecules of variable molecules masses like polymers [2003] 2. In curing cement plasters water is sprinkled from time to time. This helps in (a) converting sand into silicic acid (b) keeping it cool (c) developing interlocking needle-like crystals of hydrated silicates (d) hydrating sand and gravel mixed with cement [2003] 3. The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to

15.34 Complete Chemistry—JEE Main

(a) (b) (c) (d)

an interaction with nitrogen of the air at very low temperatures a change in the crystalline structure of tin a change in the partial pressure of oxygen in the air an interaction with water vapour contained in the humid air.

[2004]

4. In silicon dioxide (a) each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms (b) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms (c) silicon atoms are bonded to two oxygen atoms (d) there are double bonds between silicon and oxygen atoms [2005] 5. A metal, M forms chlorides in its + 2 and + 4 oxidation states. Which of the following statements about these chlorides is correct? (a) MCl2 is more easily hydrolysed than MCl4 (b) MCl2 is more volatile than MCl4 (c) MCl2 is more soluble in anhydrous ethanol than MCl4 (d) MCl2 is more ionic than MCl4 [2006] 6. Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is (b) R4Si (c) RSiCl3 (d) R2SiCl2 [2008] (a) R3SiCl 7. Which of the following exists as covalent crystals in the solid state? (a) Iodine (b) Silicon (c) Sulphur (d) Phosphorus [2013] 8. Global warming is due to increase of: (a) methane and nitrous oxide in atmosphere (b) methane and CO2 in atmosphere (c) methane and O3 in atmosphere (d) methane and CO in atmosphere [2014, online] 9. Example of a three-dimensional silicate is: (a) Zeolites (b) Ultramarines (c) Feldspars (d) Beryls [2014, online] 10. Match the items in column I with its main use listed in column II Column I Column II (A) silica gel

(i) Transistor

(B) Silicon

(ii) Ion-exchanger

(C) Silicone

(iii) Drying agent

(D) Silicate (iv) Sealant (a) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii) (c) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

(b) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii) (d) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

[2016 online]

11. Assertion: Among the carbon allotropes, diamond is an insulator, whereas, graphite is a good conductor of electricity Reason: Hybridization of carbon in diamond and graphite are sp3 and sp2, respectively (a) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion. (b) Both assertion and reason are correct, but the reason is the correct explanation for the assertion (c) Both assertion and reason are incorrect (d) Assertion is incorrect statement, but the reason is correct [2016 online]

ANSWERS 1. (c) 7. (b)

2. (c) 8. (b)

3. (b) 9. (d)

4. (a) 10. (a)

5. (d) 11. (b)

6. (c)

Study of the p-Block Elements (Groups 13, 14 and 15)

15.35

HINTS AND SOLUTIONS 3. At low temperature, tin undergoes a change in the crystalline structure; Grey Sn Grey tin is easily changed into a powder. Therefore, the choice b is correct. O O

4. The structure of SiO2 is

Si

291 K

white Sn.

O O

Si

O

O

O

5. The salt of lower oxidation state is ionic while that of higher oxidation state is covalent. 6. The hydrolysis of RSiCl3 produces a very complex cross-linked polymer. The hydrolysis of R2SiCl2 produces a straight-chain polymer. The compound R3SiCl is used to block the polymerization as it does not create the active OH group at the end of the chain. 7. Iodine, sulphur and phosphorus exist as I2, S6 and P4, respectively. Silicon does exist as covalent crystal in the solid state. Thus the choice (b) is correct. 8. Global warming is due to increase in the concentration of methane and carbon dioxide in atmosphere. 9. The replacement of Si4+ ions in the three-dimensional silicates (of general formula SiO2) by a combination of Al3+ and other cations (to maintain electrical neutrality) gives minerals known as feldspars, zeolites and ultramarines. 10. Silica gel is used as drying agent Silicon is used in transistor Silicone is used as sealant Silicate is used in ion-exchanger Therefore, the Choice (a) is the correct matching. 11. The choice (b) is correct.

The Group 15 Elements Group 15 of the periodic table consists of elements nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi). The physical properties of these elements are recorded in Table 3. Table 3

Property

N

P 2

Atomic number Relative atomic mass

Covalent radius, r/pma Ionic radius M3+, r/pmb M5+, r/pmb Melting point, T/K Boiling point, T/K

Atomic and Physical Properties of Group 15 Elements

As 2

1012 1903 2910 5825 11220 2.1

3

Sb 10

947 1798 2736 5481 10880 2.0

2

3

Bi

[He](2s) (2p) [Ne](3s) (3p) [Ar](3d) (4s) (4p) [Kr](4d) (5s) (5p) [Xe](4f)14(5d)10(6s)2(6p)3 7 15 33 51 83 14.01 30.97 74.92 121.75 208.98 152 74 110 120 140 — 44 58 76 103 — 38 46 60 76 63 317.1c 1089d 903.7 544.4 77.2 553.5 888 1860 1837 e f 0.879 1.823 5.778 6.697 9.808

Density at 298 K, r/g cm–3 Ionization energy, I/kJ mol–1 I 1402 II 2856 III 4577 –1 Sum (I + II + III)/kJ mol 8835 Sum (IV + V)/kJ mol–1 16920 Electronegativity 3.0

3

10

834 1595

2443 4872 9636 1.9

2

3

703 1610 2466 4779 9776 1.9

(a) tervalent, single bond, (b) 6-coordinate, (c) white form, (d) grey a-form at 38.6 atm, (e) at 63 K, (f) grey a-form.

15.36 Complete Chemistry—JEE Main

Description of Physical Properties ns)2(np)3, where n varies from 2 to 6. electronegativity and nonavailability of d orbitals. 3. The metallic character increases on descending the group. N and P are nonmetals. As and Sb are metalloids. Bi is a true metal 4. The covalent radii increase on descending the group (Fig. 12). The variation is according to the expectation. From phosphorus onwards, the difference becomes small due to poor shielding of nuclear charge by 3d and 4f electrons. The elements do not form M5+ 5. The melting point increases from N to P (Fig. 20). This is attributed to their structures. Nitrogen exists as diatomic molecule due to the multiple bond formation. Phosphorus exists in different allotropic forms, with white phosphorus existing as P4. participate in metallic bonding.

Fig. 19

Covalent Radii of the Group 15 Elements

6. The boiling points of elements increase down the group with the exception of the last member of the group (Fig. 21).

Fig. 20 Melting Points of the Elements of Group 15 Elements

Fig. 21 Boiling Points of the Group 15 Elements

Study of the p-Block Elements (Groups 13, 14 and 15)

15.37

7. Ionization energies of elements generally decrease down the group. high and they decrease down the group. The elements Sb and Bi comparatively have low values, so these elements form Sb3+ and Bi3+ in their salts with high electronegative element F. 5+ ions. 8. The densities of elements increase on descending the group.

Reactivity of Elements Nitrogen exists as diatomic molecules. It is a gas at room temperature and is relatively unreactive. The common allotropic forms of phosphorus are white and red. White phosphorus is soft, waxy and reactive. It is stored under water as it ignites spontaneously in air at about 35 °C to give P4O10. It exists as tetrahedral P4 molecules. The bond angle P—P—P is 60°, much smaller than the value expected for a tetrahedron structure. This indicates the highly strained structure which leads to the instability and hence reactivity of the molecule. Red phosphorus is formed when white phosphorus is heated to a high temperature. It is a polymeric solid which is less reactive then white phosphorus. Heating white phosphorus under high pressure results into a highly polymerized form of P called black phosphorus. It is the most stable form of phosphorus in which atoms are arranged in corrugated planes.

The formation of diatomic nitrogen is due to the fact the bond enthalpy e(N ∫∫ N) is greater than the bond enthalpy e(N—N) taken thrice. For phosphorus, the reverse is true, i.e. e(P ∫∫ P) < 3 e(P—P). The reactivity of other elements decreases down the group. The tendency to form oxides with +3 oxidation state of element decreases down the group.

Compounds of Group 15 Elements Hydrogen Compounds All the elements of Group 15 form volatile hydrogen compounds of the type MH3. The physical properties of these hydrogen compounds are recorded in Table 4. Table 4

Properties of Hydrogen Compounds of Group 15 Elements

Property

NH3

PH3

AsH3

SbH3

BiH3

Melting point, T/K

195.4

139.5

156.7

185

—

Boiling point, T/K

239.7

185.5

210.6

254.6

289.8

Density, r/g cm

0.68

0.75

1.64

2.20

(239 K)

(183 K)

(209 K)

(255 K)

(M—H) distance/pm

101.7

141.9

151.9

H—M—H angle

107.8°

93.6°

91.8°

91.3°

—

Bond enthalpy/kJ mol–1

391

322

247

255

—

–46.1

–9.6

66.4

145.1

277.8

–3

170.7

—

Enthalpy of formation, DtH°/kJ mol–1

15.38 Complete Chemistry—JEE Main

Temperature /K

Description of Properties 1. Ammonia, NH3, has exceptionally high melting 275 point. For other hydrides, the melting point of Bolling points hydrides increases on descending the group (Fig. 250 15). Explanation Because of the high electronegativity of 225 N, ammonia involves extensive hydrogen bondings. The other hydrides do not form hydrogen bonds. 200 2. The boiling point of NH3 is also higher than PH3. Melting points Beyond PH3, the boiling point increases down the 175 group (Fig. 15). Explanation In liquid phase, ammonia is hydrogen 150 bonded due to the high electronegativity of nitrogen. The other hydrides do not involve hydrogen bonding. 125 PH3 AsH3 SbH3 Their boiling point increases due to the increase in the van der Waals forces which increases with increase in size of molecule. Fig. 15 Melting and Boiling Points of Hydrides of Group 15 3. The densities recorded near the boiling point indicates the increasing trend on descending the group. 4. The M—H bond distance increases on descending the group. This is due to increasing size and decreasing electronegativity value of the element M. 5. The bond angle in NH3 is 107.8° indicating that sp3 hybrid orbitals of N are involved in the bondings with H atoms. N in ground state

N in NH3

The repulsion between a lone pair and bonded pairs is responsible for reducing bond angles from 109° 27¢ (the expected value) to 107° 48¢. For other hydrogen compounds, the angles are close to 90°, (and also decreases down the group) indicating the involvement of more and more p orbitals in the bonding. 6. The bond enthalpy M—H decreases down the group indicating that the strength of bond decreases on descending the group. Besides the above characteristics, the following characteristics are observed on descending the group. 2. The thermal stability of hydrogen compounds decreases. 3. The reducing property of hydrogen compounds increases. 4. The replacement of H by other groups such as Cl or CH3 5. The ability to form complexes by donating the lone pair of electrons decreases. 6. The basic character decreases. The above characteristics are explainable on the basis of bond enthalpy (which decreases down the group) and size of the atom (which increases down the group). The basic character is due to lone pair which becomes more diffuse on descending the group causing a decrease in basic nature.

Ammonia Ammonia is a colourless and pungent smelling gas. Its molecule is trigonal pyramidal with nitrogen at the apex and is highly polar. In laboratory NH3 may be produced on treating ammonium salt with NaOH or Ca(OH)2: NH4Cl + NaOH æDæ Æ NH3 + H2O + NaCl In industry, NH3 is prepared by Haber’s process:  N2(g) + 3H2(g)   2NH3(g)

DfH° = – 46. kJ mol–1

Study of the p-Block Elements (Groups 13, 14 and 15)

15.39

The condition used are about 200 atm and 700 K with a suitable catalyst (iron oxide with small amount of K2O and Al2O3, earlier iron with promoter molybdenum). Because of its ability to form hydrogen bond, it is highly soluble in water (53.1 g NH3 dissolves in 100 g water at 20 °C and 1 atm pressure). Ammonia dissolved in water behaves as a weak base. + –  NH3 + H2O   NH 4 + OH

Kb = 1.8 ¥ 10–5 mol dm–3

Ammonia forms salts with acids. In aqueous solution, ammonium salts are acidic due to the hydrolysis of NH+4 ions. Ammonium salts decompose on heating. The decomposition reaction depends on the nature of anion. 2– (a) If the anion is not oxidizing such as CI– , CO2– 3 or SO 4 . heat NH4Cl ⎯⎯→ NH3 + HC1

⎯→ 2NH3 + H2SO4 (NH4)2SO4 ⎯heat

(b) If the anion is oxidizing such as NO–2,NO–3 ClO–4 ,Cr2O2– 7 The salt decomposes to N2 or N2O. ⎯→ N2 + 2H2O NH4NO2 ⎯heat

⎯→ N2O + 2H2O NH4NO3 ⎯heat

⎯→ N2 + 4H2O + Cr2O3 (NH4)2Cr2O7 ⎯heat

Ammonia is used in the manufacture of nitric acid, various nitrogenous fertilizers and other compounds. It is also nonaqueous solvent. Its self ionization is + –  2NH3   NH 4 + NH 2 A substance producing NH+4 ion in liquid ammonia acts as an acid and that producing NH2– ion acts as a base. Liquid ammonia is an extremely good solvent for the alkali metals and the heavier metals Ca, Sr and Ba of Group 2. These solutions have conductivity comparable to that of pure metals. The solution of these metals are good reducing agents due to the presence of free electrons. NH3 Na ⎯liquid ⎯⎯⎯ → [Na(NH3)n]+ + e–

Hydrazine and Hydroxylamine Nitrogen also forms hydrazine (N2H4) and hydroxylamine (NH2OH). The oxidation state of nitrogen in these compounds are –2 and –1, respectively. Hydrazine is manufactured by the Raschig process, in which ammonia is oxidized by sodium hypochlorite in dilute aqueous solution. NH3 + NaOCl Æ NH2Cl + NaOH (fast) 2NH3 + NH2C1 Æ NH2NH2 + NH4Cl (slow) The side reactions occurring are (i) N2H4 + 2NH2Cl Æ N2 + 2NH4Cl (ii) 3NH2Cl + 2NH3 Æ N2 + 3NH4Cl The reaction (i) is catalyzed by heavy metal ions in solution. For this reason, distilled water along with glue or gelatin (which mask the metal) is used in this reaction. The reaction (ii) is minimized by using dilute solution. The structure of hydrazine is

situated tetrahedrally with respect to N atom. The bond length N—N is 145 pm. The N—N bond is quite weak as compared to the C—C bond in C2H6. This results due to the repulsion between nonbonding lone pairs. Due to this, nitrogen has little tendency to catenation. Hydrazine is a powerful reducing agent and is used to prevent the oxidation of the boiler and pipes. The compound and its derivatives are also used as rockel fuel. Hydroxylamine is manufactured by reducing nitrites or from nitromethane. NH4NO2 + NH4HSO3 + SO2 + 2H2O Æ [NH3OH]+ HSO–4 + (NH4)2SO4 CH3NO2 + H2SO4 Æ [NH3OH]+ HSO–4 + CO.

15.40 Complete Chemistry—JEE Main

Hydroxylamine is a weaker base than ammonia or hydrazine. It can acts as ligands. It undergoes addition reaction across a double bond. Hydroxylamine is used to obtain cyclohexanone oxime, which is an intermediate in the manufacture of nylon 6. N–OH

O

(

Hydrogen Compounds of Other Elements Phosphine (PH3), arsine (AsH3), and stibine (SbH3) are obtained by acid hydrolysis of phosphide, arsenide and antimonide, respectively. Zn3M2 + 6HCl Æ 3ZnCl2 + 2MH3 (M = P, As or Sb)

In the laboratory, PH3 is obtained by boiling white phosphrous with concentrated NaOH or KOH solution in an inert atmosphere. P4 + 3KOH + 3H2O Æ PH3 + 3KH2PO2. The compounds MH5 for the elements of Group 15 are not known. To exhibit pentavalency, the central atom should exhibit sp3 participate in sp3d hybridization. Such a hybridization is possible with chlorine which is highly electronegative. Thus, the pentahalides are known.

Halides Trihalides 3 is stable, NC13 is explosive, NBr3 and Nl3 are known only as their unstable ammoniates NBr3◊6NH3 and NI3◊6NH3). All the trihalides are predominately covalent with the exception of BiF3 which is ionic. All of them have a tetrahedral structure with one position occupied by a lone pair. All the trihalides with the exception of NF3 hydrolyze in aqueous solution, the products depend on the element. NCl3 + 3H2O Æ NH3 + 3HOCl PCl3 + 3H2O Æ H3PO3 + 3HCl AsCl3 + 3H2O Æ H3AsO3 + 3HCl SbCl3 + H2O Æ SbO+ + 2H+ + 3Cl– BiCl3 + H2O Æ BiO+ + 2H+ + 3Cl– NF3 reacts with H2O vapour if exposed to a spark. The reaction occurring is 2NF3 + 3H2O Æ 6HF + N2O3. NF3 has little tendency to form complexes while PF3 shows good tendency. The dipole moment of NF3 is very low (0.23 D). Explanation The high electronegative F atoms attract electrons, and the sum of the three N—F bond moments partly cancel the moment from the lone pair, and thus reduces both the dipole moment and its donor characteristics

Pentahalides PF5 PC15 PBr5 PI5

The following pentahalides are known. AsF5 SbF5 BiF5 (AsCl5) SbCl5

Study of the p-Block Elements (Groups 13, 14 and 15)

15.41

Arsenic pentachloride is highly reactive and unstable. Nitrogen does not form pentahalides as its valence shell has only four orbitals (2s and 2ps) and these can accommodate a maximum of eight electrons. All the pentahalides have a trigonal bipyramidal structure resulted from the sp3d hybridization of the central atom.

PCl5 undergoes complete hydrolysis in water. PCl5 + 4H2O Æ H3PO4

+ 5HCl

Phosphoric acid

Structures of Phosphorus Pentahalides PF5 is covalent both in gaseous, liquid and solid states. PC15 is covalent in gaseous and liquid states, but in solid it exists as [PCl4]+ [PCl6]–, cation has tetrahedral structure while anion has octahedral structure. PBr5 and PI5 exist as [PBr4]+Br– and [PI4]+I– in solid states, respectively. The compounds PHF4 and PH2F3 have been prepared. Oxides of Nitrogen Nitrogen forms many oxides, exhibiting all the oxidation states from +1 to +5. The negative oxidation state is not possible as oxygen is more electronegative than nitrogen. Nitrogen could exhibit larger number of oxidation states in its oxides due to the possibility of pp—pp multiple bonds between N and O. The other atoms of the Group 15 (i.e. P, As, Sb and Bi) are too large to allow such type of multiple bonding due to poor overlap between the orbitals. A brief description of oxides of nitrogen is in order. Nitrous Oxide (N2O)

N2O is a colourless gas. It is a neutral oxide. It can be prepared by heating a solution of NH4NO3 HCl NH4NO3 ⎯heat ⎯→ N2O + 2H2O

The oxidation state of N in N2O is +1. Its structure is N

≈

N

N

O

≈

N

O

The bond lengths N—N is 112.6 pm and N—O is 118.6 pm. These are shorter than the expected values. N2O is used as an anaesthetic. It is also known as laughing gas, because small amounts cause euphoria.

Nitric Oxide (NO)

NO is a colourless gas. It is a neutral oxide. It can be prepared by treating Cu with dilute HNO3. 3Cu + 8HNO3 Æ 2NO + 3Cu(NO3)2 + 4H2O (dilute)

The oxidation state of N in NO is +2. Its structure is N

O

≈

N

O

The bond length N—O is 115 pm which is intermediate between a double and a triple bond. Being an odd electron molecule (valence electrons are 11), it is paramagnetic. However in the liquid and solid states, nitric oxide becomes diamagnetic due to dimerization O—N—N—O. The asymmetrical dimer O—N—O—N has been isolated in the presence a of Lewis acid (say, HCl).

15.42 Complete Chemistry—JEE Main

Nitric oxide can act as a neutral ligand. The brown ring test for nitrates is due to the complex [Fe(H2O)5NO]+ ion. The other examples of complexes are sodium nitroprusside Na2[Fe(CN)5NO] ◊ 2H2O, [Fe(CO)2(NO)2] and [Cr(NO)4]. Problem Which species NO or NO+ has a stronger bond? Answer s 2s)2 (s*2s)2 (p2px)2 (p2py)2 (s 2pz)2(p*2p)1. The electron in the p* orbital is relatively easily lost to give the nitrosonium ion (NO+). Since the electron is removed from antibonding orbital, the bond is stronger in NO+ than in NO and the bond length decreases.

Nitrogen Sesquioxide (N2O3) Nitrogen sesquioxide is stable at temperatures below –30 °C. Above this temperature, it decomposes as N2O3 Æ NO + NO2. It can be prepared by mixing equimolar amounts of NO and NO2 at low temperatures. N2O3 is an acidic oxide and is the anhydride of nitrous acid (HNO2). N2O3 + H2O Æ 2HNO2. The average oxidation number of N in N2O3 is +3. Nitrogen sesquioxide exists in two form, which are interconvertible. O

O N

O

N O

(asymmetric form)

N

N

O

O (symmetric form)

The bond length N—N in asymmetric form is 186.4 pm, which is much longer than the expected value (147 pm). This indicates that the N—N bond in asymmetric form of N2O3 is weak.

Nitrogen Dioxide (NO2) nitrate.

Nitrogen dioxide is a brown gas. It is an acidic oxide. It can be prepared by heating lead 2Pb(NO3)2 Æ 2PbO + 4NO2 + O2

The action of concentrated nitric acid on Cu also produces NO2. Cu + 4HNO3 Æ 2NO2 + Cu(NO3)2 + 2H2O (concentrated)

The oxidation number of N in NO2 is +4. Its structure is

Being an odd electron molecule, it is paramagnetic. On cooling, paramagnetism disappears due to the dimerization of NO2. The structure of dimerized N02 (i.e. N2O4) is

The bond length N—N is 164 pm. It is much longer than the single bond length of 147 pm in hydrazine. This indicates that N—N bond is weak. The dimerization of NO2 is due to the fact that the unpaired electron is localized mainly on the N atom. This may be compared with NO and ClO2 molecules which show little tendency to dimerize, where the electron is not localized on the single atom. N2O4 is a mixed anhydride. N2O4 + H2O Æ HNO2 + HNO3

Study of the p-Block Elements (Groups 13, 14 and 15)

15.43

Liquid N2O4 serves as non-aqueous solvent. It ionizes as + –  N2O4   NO + NO 3

acid

Dinitrogen Pentoxide (N2O5)

base

N2O5 is a colourless ionic solid. It is an acidic oxide. It is an anhydride of nitric acid. H2O + N2O5 Æ 2HNO3

It can be prepared by dehydrating nitric acid with P2O5 at low temperatures. 4HNO3 + P4O10 Æ 2N2O5 + 4HPO3 The oxidation state of N in N2O5 is +5. Its structure is

In solid state, N2O5 exists NO+2NO–3 (nitronium nitrate).

Oxides of Phosphorus The most common oxides of phosphorus are P4O6 and P4O10 and less common oxides are P4O7, P4O8 and P4O9. P4O6 is a dimeric of phosphorus trioxide, P2O3. It is obtained when phosphorus is burnt in a limited supply of air: P4 + 3O2 Æ P4O6 P4O10 is a dimeric of phosphorus pentoxide, P2O5. It is obtained when phosphorus is burnt in excess of air. P4 + 5O2 Æ P4O10 The oxidation number of phosphorus in P4O6 is +3. The structure of P4O6 is P

165.6 pm

O

O O

P

127.0°

P O

O P

99.5°

O

O

P The four P atoms are at the corners of a tetrahedron, and each of the six O atoms are along each of the six edges. The bond angle POP is 127°. The bond length P—O is 165.6 pm. Each O atom is strictly above the edges, but it is more convenient to draw them on the edges. P4O6 is an acidic oxide. It is an anhydride of orthophosphorous (or simply phosphorous) acid. P4O6 + 6H2O Æ 4H3PO3

The schematic representation of hydrolysis is as follows. OH

OH P

P

P

OH

OH OH P

O O

O OH OH

O P

O P

O

OH O

P O

P O O OH

P

H OH O P O O P OH

OH

15.44 Complete Chemistry—JEE Main

2

OH OH

P

O 2H

P

OH O

P

OH

O H OH O P

OH

OH OH

P

OH O

P

OH

H O

OH P

OH O

P

OH

O 4H

H

OH

P

OH

OH

Pyrophosphorous acid

Orthophosphorous acid

P4O10 is a dimer of phosphorus pentoxide, P2O5. The oxidation state of phosphorus in P4O10 is +5. Its structure is O 143 pm

P 102°

O

O

123°

O O

O

P O

P P

160 pm

P O

O

O

O

In addition to the structure of P4O6, each phosphorus with its lone pair forms a coordinate bond with an oxygen atom. The bond length P—O along the edges is 160 pm and the bond length PÆO on the corners is 143 pm. The bonds on the corners are much shorter than a single bond. In fact, these are double bonds-one p bond (which is coordinatecovalent bond) and one is p bond formed by the back bonding of 2p electrons on oxygen atom to the empty 3d orbital on phosphorus (i.e. it represents pp—dp back bonding). When P4O10 P4O10 + 6H2O æÆ 4H3PO4 Tetrapolyphosphoric acid

H4P4O12 + H2O æÆ H6P4O13 Tetrapolyphosphoric acid

H6P4O3 + H2O æÆ 2H4P2O7 Pyrophosphroric acid

[H4P2O13 + H2O æÆ 2H3PO4] ¥ 2 Pyrophosphroric acid

overall reaction

P4O10 + 6H2O æÆ 4H3PO4

The above reactions are shown schematically in the following.

Study of the p-Block Elements (Groups 13, 14 and 15)

O

O

P O OH

O P

O

O

O

O

O H OH P O O O P

P

P O

O

P

O OH

P

OH

OH

Orthophosphoric acid

O

OH

P H OH O O P O OH

OH

Tetrametaphosphoric acid

O

P

OH P

HO

O H

OH

O

O O

not isolated

O

OH P

O

OH P

O O

O

O

OH P

15.45

OH

OH

P OH

Pyrophosphoric acid

O O

P

O H

OH

O

OH

P

OH

O O

P

OH

OH

Tetrapolyphosphoric acid

P4O10 is a strong dehydrating agent and removes water from many inorganic and organic compounds. 2H2SO4 + P4O10 Æ 4HPO3 + 2SO3; 4HNO3 + P4O10 Æ 4HPO3 + 2N2O5

Oxides of Other Elements Arsenic forms As4O6 and As4O10. Antimony forms Sb4O6 and Sb4O10. Bismuth forms only Bi2O3. It is not dimerized. The basicity of oxides (as well as hydroxides) increases on descending the group. N2O3 and P4O6 are acidic. As4O6 and Sb4O6 are both amphoteric. Bi2O3 is basic. The amphoteric reaction of As4O6 are As4O6 + 12NaOH Æ 4Na3AsO3 + 6H2O; As4O6 + 12HCl Æ 4AsCl3 + 6H2O The stability of pentoxide decreases on descending the group. This is illustrated from the fact that Bi does not form pentoxide. The pentoxides are more acidic than trioxides. Oxoacids of Nitrogen Serval oxacids of nitrogen are known (Table 5). Table 5 Oxoaicds of nitrogen Oxo-acid

Molecular formula

Description

Hyponitrous acid

H 2N 2O 2

Weak acid, salts known, HON = NOH

Hyponitric acid

H 2N 2O 3

Known in solution

Nitroxylic acid

H 2N 2O 4

Explosive

Nitrous acid

HNO2

Weak acid, unstable, salts known (called nitrites)

Nitric acid

HNO3

Most stable, strong acid, stable salts known (called nitrates)

Peroxonitrous acid

HOONO

Isomeric with nitric acid, unstable

Peroxonitric

HNO4

Unstable, HOONO2

Nitrous acid is unstable except in dilute solution. With bases, it forms nitrite salts. Both nitrous acid and nitrites are weak oxidizing agents. In their oxidizing actions (Fe2+ to Fe3+, I– to I2), they are reduced to either N2O or NO. By strong oxidizing agents such as MnO –4, HNO2 and NO –2 are oxidized to NO–3. Nitrous acid is used to make diazo compounds. Nitric acid is the only important oxoacids of nitrogen. Pure nitric acid, is colourless but on exposure to light it turns slightly brown because of slight decomposition into NO2 and O2. 4HNO3 Æ 4NO2 + O2 + 2H2O

15.46 Complete Chemistry—JEE Main

Nitric acid forms nitrate salts (eg. NaNO3, known as Chile saltpetre, and KNO3 known as Indian saltpetre). It is a strong oxidizing agent. Aqua regia contains 25% concentrated nitric acid and 75% concentrated HCl. Gold, which is not soluble in nitric acid, dissolves in aqua regia due to the oxidizing power of HNO3 coupled with the ability of Cl– to form complex with the metal ion. Laboratory grade nitric acid contains about 68% of HNO3 gravity of about 1.50. Nitric acid attacks most of metals except noble metals (gold and platinum). The products depends on temperature and the nature of metal. For example 3Cu + 8HNO3 ææ Æ 3Cu(NO3)2 + 2NO + 4H2O

Cu + 4HNO3 ææ Æ Cu(NO3)2 + 2NO2 + 2H2O

(dilute)

(conc.)

4Zn + 10HNO3 ææ Æ 4Zn(NO3)2 + 5H2O + N2O

Zn + 4HNO3 ææ Æ Zn(NO3)2 + 2H2O + 2NO2

(dilute)

(conc.)

Oxoacids of Phosphorus Phosphorus forms two series of oxoacids. These are phosphorous and phosphoric series. Phosphorous acids contain P in +3 (or +1) oxidation state and are reducing agents. Phosphoric acids contain P in + 5(4-4) oxidation state and are oxidizing agents. In all oxoacids, P is four-coordinate and tetrahedrally surrounded by other atoms. The oxoacid containing —OH group produce H+ on ionization and thus is acidic. The H atom directly attached to P is not ionizable and has reducing characteristic. Table 6 records the oxo-acids formed by phosphorus. Oxoacids of Phosphorus

Table 6 Oxoacid

Molecular formula

Oxidation, state of phosphorus

Hypophosphorous acid

H3PO2

+1

Orthophosphorous acid (or simply phosphorous acid)

H3PO3

+3

Hypophosphoric acid

H 4P 2O 6

+4

Orthophosphoric acid (or simply phosphoric acid)

H3PO4

+5

Dipolyphosphoric acid (also called pyrophosphoric acid)

H 4P 2O 7

+5

Cyclotrimetaphosphoric acid

(HPO3)3

+5

Polymetaphosphoric acid

(HPO3)n

+5

The structure of the acids recorded in Table 6 are shown in the following.

The Phosphorous Acid Series

The phosphorous acid series is obtained by the hydrolysis of phosphorus trioxide. O P O

O O

P

OH OH

-2H2O

4H

P

OH

O

O

Pyrophosphorous acid

Orthophosphorous acid

Both pyrophosphorous acid and orthophosphorous acid contain P in +3 oxidation state and are reducing agents since H is directly attached to P atom. Both are dibasic acids as each contains two —OH groups. Only H attached to O ionizes in solution. Two other acids belonging to the phosphorous acid series are metaphosphorous acid and hypophosphorous acid. These are obtained as follows. 25 mmHg

PH3 + O 2 ææææÆ H 2 +

HPO 2 Metaphosphorous acid H+

P4 + 3OH - + 3H 2 O Æ PH3 + 3H 2 PO 2- ææÆ PH3 +

3H3 PO 2 Hypophophorous acid

Study of the p-Block Elements (Groups 13, 14 and 15)

15.47

The above reaction may be shown schematically as follows, P OH P

H

H OH

HO

H OH P

P HO

PH3 +

OH P H OH P

H OH H

H PH3 + 3 HO

P

OH

OH

P

H PH3 + 3H

P

OH

O

The metaphosphorous acid is believed to exist as a polymer—similar to metaphosphoric acid. The oxidation state of P in hypophosphorous acid is +1. It is a monobasic acid and is strongly reducing agent.

The Phosphoric Acid Series The phosphoric acid series is obtained by the hydrolysis of phosphorus pentoxide. Orthophosphoric Acid (H3PO4) The oxidation state of P in H3PO4 is +5. Its structure is O

P HO

OH OH

It is a tribasic acid as it contains three replaceable H atoms. Its stepwise ionization is as follows. + –  H3PO4 K°a1 = 7.5 ¥ 10–3   H + H2PO 4 – + 2–  H2PO 4  K°a2 = 6.2 ¥ 10–8  H + HPO 4 2– + 3–    HPO 4 K°a3 = 1 ¥ 10–12   H + PO 4 Phosphoric acid forms three series of salts with alkalis, e.g. NaH2PO4, Na2HPO4 and Na3PO4. Dihydrogen phosphate is slightly acidic in nature. Monohydrogen phosphate is slightly basic in nature. Phosphate is appreciably basic in solution. Phosphoric acid is prepared by ‘the wet process’ in which phosphate rock is treated with sulphuric acid. Ca3(PO4)2 + 3H2SO4 Æ 2H3PO4 + 3CaSO4 [(3Ca3(PO4)2) ◊ CaF2] + 10H2SO4 Æ 6H3PO4 + 10CaSO4 + 2HF. Pure phosphoric acid is prepared by the ‘furnace process’ in which molten P is burnt to give P4O10. This is followed by the hydrolysis of P4O10. P4O10 + 6H2O Æ 4H3PO4. P4 + 5O2 Æ P4O10;

Another method of preparing phosphoric acid is to treat phosphorus with concentrated nitric acid. P4 + 20HNO3 Æ 4H3PO4 + 20NO2 + 4H2O. Phosphoric acid loses water steadily on heating. 2H3 PO 4 Orthophosphoric acid

gentle heat

ææææ Æ 220 ∞C

H 4 P2 O7 Pyrohosphoric acid

strong heat

ææææ Æ 320 ∞ C

The structure of polymetaphosphoric acid is

O

O

O

O

P

P

P

OH O

O

O

OH O

Polymetaphosphoric acid

(HPO3 )n Polymetaphosphoric acid

15.48 Complete Chemistry—JEE Main

Polyphosphoric Acids A large number of polyphosphoric acids are known. These consist of chains of tetrahedra [PO4] linked through the O atoms at one or two corners. A few such acids are as follows. O

O

P

P O

HO

Phosphoric acid is tetrabasic acid

OH

OH

OH

Pyrophosphoric acid (H4P2O7) O

O

O

P

P

P

O

HO

O

OH

Tripolyphosphoric acid is a pentabasic acid

OH OH

OH

O

O

O

O

P

P

P

P

O

HO

O

OH

O

OH

OH

Tetraphosphoric acid OH is a hexabasic acid OH

Tetrapolyphosphoric acid is one of the intermediate acids obtained in the hydrolysis of P4O10. The predecessor of this acid is tetrametaphosphoric acid which has a cyclic structure. O HO

P

O O

P

O HO

P

OH Tetrametaphosphoric acid is tetrabasic acid

O O

O

P

OH

O

Tetrametaphosphoric acid (H4P4O12)

The immediate hydrolysis product of tetrametaphosphoric acid is pyrophosphoric acid. The various phosphates having cyclic structures have been isolated. OO

O

O

P O O

O

O P

P O

O

P

P

O

O -

Dimetaphosphate ion

O

O

O

O O

O

P

O-

O-

Trimetaphosphate ion

O

P O-

P

O

O O

P

O

O-

Tetrametaphosphate ion

Study of the p-Block Elements (Groups 13, 14 and 15)

Hypophosphoric Acid

15.49

(H4P2O6) The structure of hypophosphoric acid is OH OH O

P

P

O

OH OH

It contains one atom less than pyrophosphoric acid and the oxidation state of P is +4. It is obtained by hydrolysis and oxidation of red phosphorus by NaOCl or yellow phosphorus by water and air. P H HO H HO H

P H

OH 2 P

H

H

P

P

OH OH 4O2

P

2O

OH OH

P

O

OH OH Hypophosphoric acid

OH

P Since no H atom directly attached to P is present, hypophosphoric acid is not a reducing agent. It is a tetrabasic acid. It is slowly hydrozysed as P—P bond is stronger than P—O—P.

O

OH

OH

P

P

OH

OH

OH O

O

P

OH H

+

OH Orthophosphorous acid

HO

P

O

OH Orthophosphoric acid

Anomalous Behaviour of Nitrogen Factors such as the small size, high electronegativity and nonavailability of d orbitals are responsible for the anomalous behaviour of nitrogen as compared to the rest of the elements of the group. Some of the anomalies are as follows. 1. Nitrogen is a gas and exists as a stable diatomic molecule with a triple bond. The remaining elements are solids and do not form stable diatomic molecules. The bond enthalpy of N2 is highest (941.4 kJ mol–1) O 2. Nitrogen form compounds with multiple bonding (like N N , N O 3. Nitrogen exhibits catenation as in H2N—NH2. 4. Nitrogen covalency does not exceed beyond four as in the NH4+ ion. 5. The hydrogen compounds of nitrogen have hydrogen bonding. 6. Except nitrogen, all elements of the group can make use of d orbitals forming compounds of the type PC15, AsF5, etc. 7. The single N–N bond is weaker than the single P–P bond because of high interelectronic repulsion of the nonbonding electrons owing to the small bond length. This results in low tendency of catenation in nitrogen.

MULTIPLE CHOICE QUESTIONS 1. Which of the following elements does not belong to Group 15? (a) Nitrogen (b) Phosphorus (c) Arsenic (d) 2. Atomic number of bismuth which belongs to 6th period is (a) 51 (b) 52 (c) 83 (d) 3. The number of elements in Group 15 is (a) 3 (b) 4 (c) 5 (d) 4. Which of the following orders regarding the melting points of hydrides of Group 15 is (a) NH3 > PH3 > AsH3 (b) NH3 < PH3 < AsH3 (c) NH3 > PH3 < AsH3 (d)

Tin 84 6 NH3 < PH3 > AsH3

15.50 Complete Chemistry—JEE Main

5. Coordination number of nitrogen never exceeds (a) 3 (b) 4 (c) 5 (d) 6 6. The ammonia molecule is (a) triangular planar (b) regular tetrahedron with bond angles equal to 109° 28 . (c) trigonal bipyramidal (d) trigonal pyramidal 7. Ammonia acts as a (a) neutral species (b) Lewis acid (c) Lewis base (d) amphoteric hydride 8. Which of the following orders regarding the bond distance M—H of the hydrides of Group 15 elements is correct? (a) N—H > P—H > As—H (b) N—H < P—H < As—H (c) N—H > P—H < As—H (d) N—H < P—H > As—H 9. Which of the following orders regarding the bond angle H—M—H of the hydrides of Group 15 elements is correct? (a) H—N—H > H—P—H > H—As—H (b) H—N—H < H—P—H < H—As—H (c) H—N—H > H—P—H < H—As—H (d) H—As—H < H—P—H > H—As—H 10. Which of the following orders regarding the bond enthalpy of M—H bond in the hydrides of Group 15 elements is correct? (a) N—H > P—H > As—H (b) N—H < P—H < As—H (c) N—H > P—H < As—H (d) N—H < P—H > As—H 11. Which of the following orders regarding the enthalpy of formation of hydrides MH3 of Group 15 is correct? (a) NH3 > PH3 > AsH3 (b) NH3 < PH3 < AsH3 (c) NH3 > PH3 < AsH3 (d) NH3 < PH3 > AsH3 12. Which of the following molecules includes nitrogen atom having oxidation state equal to – 2? (a) N2 (b) NH2OH (c) N2H4 (d) NH3 13. The molecule in which the oxidation state of nitrogen is –1 is (a) N2 (b) NH2OH (c) N2H4 (d) NH3 14. Which of the following trihalides is not known? (a) NCl3 (b) PCl3 (c) NI3 (d) PI3 15. Which of the following halides is not known? (a) NCl5 (b) PF5 (c) AsF5 (d) SbCl5 16. Which of the following chemical equations is correctly formulated? (a) NCl3 + 3H2O Æ NH3 + 3HOCl (b) PCl3 + 3H2O Æ PH3 + 3HOCl (d) SbCl3 + 3H2O Æ SbH3 + 3HOCl (c) AsCl3 + 3H2O Æ H3As + 3HOCl 17. Which of the following chemical equations is not correctly formulated? (b) AsCl3 + 3H2O H3AsO3 + 3HCl (a) PCl3 + 3H2O H3PO3 + 3HCl (c) SbCl3 + 3H2O H3SbO3 + 3HCl (d) BiCl3 + H2O BiO+ + 2H+ + 3Cl– 18. Which of the following chemical equations is not correctly formulated? (a) NCl3 + 3H2O NH3 + 3HOCl (b) PCl3 + 3H2O H3PO3 + 3HCl (c) AsCl3 + H2O AsO+ + 2H+ + 3Cl– (d) SbCl3 + H2O SbO+ + 2H+ + 3Cl– 19. Which of the following statements is not correct? (a) With the exception of nitrogen, all other elements of Group 15 form pentahalides (b) The pentahalides of elements of Group 15 are trigonal bipyramidal (c) The pentahalides of elements of Group 15 undergo incomplete hydrolysis to give the corresponding acids (d) The pentahalides of elements of Group 15 undergo complete hydrolysis to give the corresponding acids 20. Which of the nitrogen oxides is obtained when ammonium nitrate is heated? (a) N2O (b) NO (c) NO2 (d) N2O5 21. The nitrogen oxide obtained on heating lead nitrate is (a) N2O (b) NO (c) NO2 (d) N2O5 22. Which of the following is expected to be paramagnetic? (a) NH2OH (b) N2H6Cl2 (c) N2O3 (d) NO2

Study of the p-Block Elements (Groups 13, 14 and 15)

15.51

23. The oxidation of NO in air produces (a) N2O (b) N2O3 (c) N2O4 (d) N2O5 24. The dimerization of NO2 is accompanied with (a) an increase in paramagnetism (b) a decrease in paramagnetism (c) no change in paramagnetism (d) a darkening in colour 25. Lightning bolts in the atmosphere cause the formation of (a) NO (b) NH3 (c) NH3·H2O (d) NH2OH 26. Which of the nitrogen oxide is released when P4O10 is treated with HNO3? (a) N2O (b) NO (c) NO2 (d) N2O5 27. Given that 2NO2 N 2O 4 DH = negative The conditions at which the conversion of NO2 to N2O4 becomes more and more are (a) lower temperature and lower pressure (b) lower temperature and higher pressure (c) higher temperature and higher pressure (d) higher temperature and lower pressure 28. Which of the following acids of phosphorus is a reducing acid? (a) H3PO3 (b) H3PO4 (c) H4P2O7 (d) (HPO2)3 29. Which of the following acids is not an oxidizing agent? (a) H3PO4 (b) H4P2O6 (c) H4P2O7 (d) H3PO2 30. The structure of (HPO3)3 is (a) linear (b) tetrahedral (c) cyclic (d) octahedral 31. Orthophosphoric acid is a (a) monobasic acid (b) dibasic acid (c) tribasic acid (d) tetrabasic acid 32. Orthophosphorous acid is a (a) monobasic acid (b) dibasic acid (c) tribasic acid (d) tetrabasic acid 33. Which of the following acids is unstable? (a) HNO2 (b) HNO3 (c) H2N2O2 (d) HOONO 34. Which of the following acids is explosive? (a) Hyponitrous acid (b) Hyponitric acid (c) Nitroxylic acid (d) Peroxynitrous acid 35. Polymetaphosphoric acid has (a) linear structure of HPO3 units (b) branched structure of HPO3 units (c) cyclic structure of HPO3 units (d) discrete molecules of (HPO3)2, (HPO3)3, and so on 36. In which of the following acids, P—P bond is present? (a) Hypophosphoric acid (b) Pyrophosphoric acid (c) Orthophosphoric acid (d) Polymetaphosphoric acid 37. Which of the following formulae represents hypophosphoric acid? (a) H3PO3 (b) H4P2O6 (c) H3PO4 (d) H4P2O7 38. Which of the following formulae represents pyrophosphoric acid? (a) H3PO3 (b) H4P2O6 (c) H3PO4 (d) H4P2O7 39. With excess of water, PCl5 gives (a) H3PO4 + HCl (b) H3PO3 + HCl (c) H3PO2 + HCl (d) H4P2O7 + HCl 40. The oxide which on dissolving in water turns blue litmus red is (a) P2O5 (b) As2O3 (c) BaO (d) Sb2O3 41. Conc. HNO3 oxidizes iodine to (a) HI (b) HIO3 (c) NH4I (d) HIO4 42. Which of the following acids contains phosphorus in the +4 oxidation state? (a) Hypophosphorous acid (b) Orthophosphorous acid (c) Phosphoric acid (d) Hypophosphoric acid 43. Phosphine may be produced by adding to water some (a) Ca3P2 (b) P4O6 (c) P4O11 (d) HPO3

15.52 Complete Chemistry—JEE Main

44. In phosphorus oxide the number of oxygen atoms bonded to each phosphorus atom is (a) 1 (b) 2 (c) 3 (d) 4 45. Which of the following statements is not correct? (a) The molecule of N2O is linear (b) NO shows a strong tendency to form dimer N2O2 – (d) NO can oxidize SO2 in water to H2SO4 4 oxidizes NO to NO3 46. Which of the following statements is not correct? (a) The molecule of NO2 is angular (b) Low temperatures favours the dimerization of NO2 to N2O4 (c) NO2 is soluble in water giving a mixture of HNO2 and HNO3 (d) The structure of N2O4 is nonplanar 47. Which of the following statements is not correct?

48.

49.

50.

51.

52.

53. 54. 55. 56. 57. 58. 59. 60. 61.

(c) Nitrite ion has a linear structure (d) Thermal decomposition of NaNO3 gives NaNO2 Which of the following statements is not correct? (a) Phosphorus trioxide exists as dimer P4O6 (b) Phosphorus pentoxide exists as dimer P4O10 (c) Orthophosphorous acid, H3PO3, is a tribasic acid (d) Orthophosphoric acid, H3PO4, is a tribasic acid Which of the following statements is not correct? (a) Orthophosphoric acid has approximately tetrahedrally shaped molecule (b) Pyrophosphoric acid is represented as H4P2O7 (c) All the four sodium salts of pyrophosphoric acid are known (d) Metaphosphoric acid is a glassy polymeric solid with the empirical formula HPO3 Which of the following represents superphosphate of lime? (a) Ca(H2PO4)2 (b) Ca(H2PO4)2 + 2CaSO4 (c) Ca(H2PO4)2 + 2Ca(NO3)2 (d) Ca3(PO4)2 Which of the following represents triple superphosphate? (a) 3Ca(H2PO4)2 (b) Ca3(PO4)2 (c) Ca(H2PO4)2 + 2CaSO4 (d) Ca(H2PO4)2 + 2Ca(NO3)2 Which of the following represents nitrophos? (b) Ca3(PO4)2 (a) Ca(H2PO4)2 (c) Ca(H2PO4)2 + 2CaSO4 (d) Ca(H2PO4)2 + 2Ca(NO3)2 Which of the following halides does not exist? (b) PBr5 (c) PCl5 (d) PF5 (a) PI5 Which of the following halides does not exist? (b) AsF5 (c) PF5 (d) NF5 (a) SbF5 — — The P P P bond angle in white phosphorus is (a) 120° (b) 90° (c) 60° (d) 109° 28¢ Which of the following forms maximum P —H bonds? (b) H3PO3 (c) H3PO4 (d) H4P2O7 (a) H3PO2 The solid phosphorus pentachloride exists as (b) PCl+4 Cl– (c) PCl–6 (d) PCl+4 · PCl–6 (a) PCl5 The hydrolysis of PCl3 results into (b) H3PO4 (c) H3PO2 (d) H3PO3 (a) HPO3 The most reactive form of phosphorus is (a) red (b) yellow (c) violet (d) black Poisonous form of phosphorus is (a) white (b) red (c) black (d) violet Which of the following hydrides does not exist? (b) PH5 (c) AsH3 (d) N2H4 (a) NH3

Study of the p-Block Elements (Groups 13, 14 and 15)

15.53

62. Which of the following hydrides is thermally least stable? (b) PH3 (c) AsH3 (d) SbH3 (a) NH3 63. Which of the following halides does not hydrolyse? (b) PCl3 (c) AsCl3 (d) SbCl3 (a) NF3 64. Which of the following oxides does not contain N — N bond? (b) N2O3 (c) N2O4 (d) N2O5 (a) N2O 65. POCl3 is obtained when (a) PCl5 is heated in air (b) PCl5 is completely hydrolysed (c) PCl5 is hydrolysed to the limited extent (d) H2SO4 is added to PCl5 66. Which of the following hydrides has the highest melting point? (a) NH3 (b) PH3 (c) AsH3 (d) SbH3 67. In the reaction N2 + 3H2 2NH3, the yield of ammonia is expected to be maximum at (a) low temperature and low pressure (b) low temperature and high pressure (c) high temperature and low pressure (d) high temperature and high pressure 68. Nitrogen is in the lowest oxidation state in (a) nitrous oxide (b) nitric oxide (c) nitrogen dioxide (d) dinitrogen pentoxide 69. Highest oxidation state of nitrogen is achieved in (a) nitrogen dioxide (b) dinitrogen trioxide (c) dinitrogen tetroxide (d) dinitrogen pentoxide 70. Which of the nitrogen oxides is obtained when copper reacts with conc. HNO3? (c) H2O3 (d) N2O5 (a) NO (b) NO2 71. Which of the following orders is correct regarding the boiling points of elements of Group 15? (b) NH3 > PH3 < AsH3 (c) NH3 < PH3 > AsH3 (d) NH3 < PH3 < AsH3 (a) NH3 > PH3 > AsH3 72. Which of the following statements regarding nitrogen molecule is not correct? (a) Amongst the homonuclear diatomic molecules of second period, the bond dissociation enthalpy is maximum in case of N2 molecule (b) Amongst the homonuclear diatomic molecules of second period, the bond length is minimum in case of N2 molecule (c) Nitrogen molecule is paramagnetic in nature (d) Only at high temperatures, nitrogen molecule reacts with metals and nonmetals forming ionic and covalent nitrides 73. Which of the following statements regarding ammonia is not correct? (a) Ammonia is a colourless and pungent smelling gas

74. 75. 76. 77.

(c) Ammonia can act as Lewis acid (d) Ammonium chloride dissolves in liquid ammonia has acidic properties The gas liberated when copper reacts with dilute HNO3 is (a) NO (b) NO2 (c) N2O3 (d) N2O5 In the brown-ring test of nitrate ion, the compound formed is (a) [Fe(H2O)5NO]2+ (b) [Fe(H2O)5NO]3+ (c) [Fe(H2O)4(NO2)]2+ (d) [Fe(H2O)3(NO)3]2+ The treatment of glycerol with conc. HNO3 in the presence of conc. H2SO4 gives (a) mononitroglycerol (b) dinitroglycerol (c) nitroglycerine (d) CO2 and H2O Which of the following equations is not correctly formulated? (a) Conc. HNO3 acquires a yellow colour due to the following photo-decomposition reaction 4HNO3 Æ 4NO2 + 2H2O + O2 (b) Cu + 4HNO3

Cu(NO3)2 + 2H2O + 2NO2

dilute

(c) 4Zn + 10HNO3

4Zn(NO3)2 + 5H2O + N2O

dilute

(d) P4 + 20HNO3

4H3PO4 + 4H2O + 20NO2

15.54 Complete Chemistry—JEE Main

78. The aqua regia is (a) 3 parts conc. HNO3 + 1 part conc. HCl (b) 1 part conc. HNO3 + 3 parts conc. HCl (c) 2 parts conc. HNO3 + 2 parts conc. HCl (d) 2.5 parts conc. HNO3 + 0.5 part conc. HCl 79. Ammonia can be dried by (a) conc. H2SO4 (b) P4O10 (c) anhydrous CaCl2 (d) anhydrous CuSO4 80. Which of the following is the neutral oxide? (a) NO (b) N2O (c) N2O3 (d) NO2 81. Nitrogen dioxide is not obtained on heating (a) KNO3 (b) Pb(NO3)2 (c) Cu(NO3)2 (d) AgNO3 82. Which of the following halides of nitrogen is expected to be most stable? (a) NF3 (b) NCl3 (c) NBr3 (d) NI3 83. Which of the following halides of nitrogen is expected to have least dipole moment? (a) NF3 (b) NCl3 (c) NBr3 (d) NI3 84. Which of the following oxides of nitrogen occurs as a white solid? (a) NO (b) NO2 (c) N2O3 (d) N2O5 85. The pseudohalogen amongst the following is (a) NO (b) CO (c) Cl2 (d) (CN)2 86. Distillation of conc. HNO3 with P4O10 gives (a) NO (b) N2O (c) NO2 (d) N2O5 87. Which of the following hydrides is acidic in nature? (a) NH2OH (b) NH3 (c) N3H (d) N2H4 88. The order of stability of hydrides of Group 15 is (a) NH3 > PH3 > AsH3 (b) NH3 < PH3 < AsH3 (c) NH3 > PH3 < AsH3 (d) NH3 < PH3 > AsH3 89. On heating ammonium dichromate, the products obtained are (a) N2O and H2O (b) N2 and H2O (c) NO2 and H2O (d) NO and N2O 90. The laughing gas is (a) N2O (b) NO (c) NO2 (d) N2O3 91. Which of the following statements regarding phosphorus is not true? (a) Phosphorus belongs to the Group 15 of the periodic table (b) The element phosphorus is obtained by heating the rock phosphate with coke and sand in an electric furnace at about 1700 – 1800 K (c) The formula of phosphorus is P4 (d) Black phosphorus is the least stable form of the allotropes of phosphorus 92. Pyrophosphorous acid is a (a) monobasic acid (b) dibasic acid (c) tribasic acid (d) tetrabasic acid 93. Which of the following represents metaphosphoric acid? (a) HPO3 (b) H3PO3 (c) H3PO4 (d) H4P2O7 94. Which of the following represents orthophosphoric acid? (a) HPO3 (b) H3PO3 (c) H3PO4 (d) H4P2O7 95. Which of the following represents hypophosphorous acid? (a) HPO3 (b) H3PO3 (c) H3PO4 (d) H3PO2 96. The white phosphorus is stored (a) in air (b) under water (c) under kerosene (d) under CS2 97. Which of the following statements is true? (a) Both white and red phosphorus are soluble in water (b) Both white and red phosphorus are soluble in carbon disulphide (c) White phosphorus is soluble in carbon disulphide whereas red phosphorus is insoluble (d) White phosphorus is insoluble in carbon disulphide whereas red phosphorus is soluble

Study of the p-Block Elements (Groups 13, 14 and 15)

98. Which of the following statements is true? (a) Both white and red phosphorus are reactive (b) Both white and red phosphorus are inactive (c) White phosphorus is reactive whereas red phosphorus is inactive (d) White phosphorus is inactive whereas red phosphorus is reactive 99. Which of the following statements about phosphorus is not true? (a) Phosphorus does not occur in free state (b) Phosphorus is present in bones and teeth (c) Phosphorus exists in several allotropic forms (d) White phosphorus is much less active than red variety 100. When phosphorus is heated with conc. HNO3, it reduces the acid to (a) NO (b) NO2 (c) N2O3 (d) N2O5 101. When phosphorus is heated with conc. HNO3, it is oxidized to (a) H3PO2 (b) H3PO3 (c) H3PO4 (d) H4P2O7 102. P4O6 is the anhydride of (a) H3PO2 (b) H3PO3 (c) H3PO4 (d) H4P2O7 103. P4O10 is the anhydride of (a) H3PO2 (b) H3PO3 (c) H3PO4 (d) H4P2O7 104. Which of the following is orthophosphorous acid? (a) H3PO2 (b) H3PO3 (c) H3PO4 (d) H4P2O7 105. When P4O6 is dissolved in cold water, the acid obtained is (a) H3PO2 (b) H3PO3 (c) H3PO4 (d) H4P2O7 106. When P4O10 (a) H3PO2 (b) H3PO3 (c) H3PO4 (d) H4P2O7 107. Which of the following halogen compounds of nitrogen is not observed? (a) NF3 (b) NF5 (c) N2F2 (d) NCl3 108. Which of the following allotropes of phosphorus is most reactive? (a) Black (b) Brown (c) Red (d) White 109. Which of the following oxo acids of phosphorus is a reducing agent? (a) H3PO4 (b) H3PO3 (c) H4P2O7 (d) HPO3 110. The number of P—O—P bonds in cyclic metaphosphate ion is (a) 0 (b) 1 (c) 3 (d) 4 111. Which of the following is least basic? (a) NF3 (b) NH3 (c) NCl3 (d) NI3 112. Which of the following is least basic? (a) H3PO3 is diabasic and reducing acid (b) H3PO3 is diabasic and nonreducing acid (c) H3PO4 is diabasic and reducing acid (d) H3PO4 is tribasic and reducing acid 113. Strong heating of orthophosphoric acid yields (a) P2O5 (b) P4O10 (c) (HPO3)n (d) H3PO3+O2 114. Which of the following nitrogen oxide is a neutral oxide? (a) N2O3 (b) N2O5 (c) NO2 (d) N2O 115. Which of the following order regarding second ionization energy of N, P and As is correct? (a) N > P > As (b) N > As > P (c) P >N >As (d) P >As >N 116. Which of the following metaphosphate ion is not known to exist in free state? (a) PO3– (b) (PO3)22– (c) (PO3)33– (d) (PO3)4– 4 117. Which of the following facts regarding bond energy is not correct? (a) EP – F > EN – F (b) EP – Cl > EN – Cl (c) EP – O < EN – O (d) EP – C < EN – C 118. Which of the following order regarding stability of phosphorus is correct? (a) White > Red > Black (b) White > Black > Red (c) Black > Red > White (d) Black > White > Red

15.55

15.56 Complete Chemistry—JEE Main

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79. 85. 91. 97. 103. 109. 115.

(d) (c) (b) (c) (a) (c) (b) (a) (c) (c) (b) (b) (c) (c) (d) (d) (c) (c) (b) (b)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 86. 92. 98. 104. 110. 116.

(c) (b) (c) (a) (d) (b) (d) (c) (b) (a) (d) (a) (a) (b) (d) (b) (c) (b) (c) (a)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75. 81. 87. 93. 99. 105. 111. 117.

(c) (a) (a) (c) (b) (d) (a) (b) (a) (d) (a) (d) (a) (a) (c) (a) (d) (b) (a) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76. 82. 88. 94. 100. 106. 112. 118.

(c) (a) (a) (d) (a) (c) (a) (d) (d) (d) (d) (b) (c) (a) (a) (c) (b) (c) (a) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77. 83. 89. 95. 101. 107. 113.

(b) (b) (c) (c) (d) (a) (b) (c) (a) (b) (c) (b) (b) (d) (b) (d) (c) (b) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 90. 96. 102. 108. 114.

(d) (c) (c) (b) (c) (a) (d) (c) (d) (d) (a) (c) (b) (d) (a) (b) (b) (d) (d)

HINTS AND SOLUTIONS 1. 2. 4. 8. 9. 10. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 26.

Tin belongs to Group 14. Atomic number of bismuth will be 2 + 8 + 8 + 18 + 18 + 14 + 15 = 83 Because of hydrogen bonding, the melting point of NH3 is greater than PH3. As the size of the atom of Group 15 increases, the bond length between the atom and hydrogen also increases. Because of more positive charge on hydrogen and smaller size of nitrogen, the repulsion between H and H atoms makes the bond angle in NH3 bigger than in PH3 and AsH3. N — H bond is more stronger than P—H which, in turn, is stronger than As— H. Because of the bigger size of iodine, it cannot be accommodated around the smaller size nitrogen atom. The oxidation number of nitrogen does not exceed +3 because of the nonavailability of d orbitals. NCl3 + 3H2O NH3 + 3HOCl AsCl3 + 3H2O H3AsO3 + 3HCl H3PO3 + 3HCl SbCl3 + H2O SbO+ + 2H + + 3Cl– PCl3 + 3H2O SbCl3 + H2O SbO+ + 2H + + 3Cl– AsCl3 + 3H2O H3AsO3 + 3HCl The petahalides of elements of Group 15 undergo complete hydrolysis to give corresponding acids, e.g. 5HCl + H3PO4 PCl5 + 4H2O NH4NO3 N2O + 2H2O 2PbO + 4NO2 + O2 2PbNO3 NO2 is an odd-electron molecule. N 2O 4 NO + O NO2 N2O4 is diamagnetic whereas NO2 is paramagnetic. P4O10 + 4HNO3 4H3PO3 + 2N2O5

Study of the p-Block Elements (Groups 13, 14 and 15)

15.57

27. DH negative means exothermic. Hence, lower temperature and higher pressure (as D g = –1) favours the formation of dimer. 28. H3PO3 contains phosphorus in +3 oxidation state. It can be oxidized to +5 oxidation state (i.e. H3PO4). Hence, it is a reducing agent. 29. H3PO2 contains phosphorus in +1 oxidation state. Its oxidation state cannot be reduced further. Hence, it is not an oxidizing agent. 35. The structure of polymetaphosphoric acid is O O

P OH

OH O

P OH

O P

O

O

OH

36. Hypophosphoric acid is H4P2O6. Its structure is HO

O

O

P

P

OH

OH OH

38. 39. 40. 42. 43. 45. 46. 47.

Pyrophosphoric acid is H4P2O7. 5HCl + H3PO4 PCl5 + 4H2O P2O5 is an acidic oxide. Hypophosphoric acid is H4P2O6. The oxidation state of P is + 4. Ca3P2 + 6H2O 2PH3 + 3Ca(OH)2 NO is dimerized to a very little extent. The structure of N2O4 is planar. The shape of NO2– is bent. It is due to the presence of a lone pair of electrons on the nitrogen atom. N O

48. 49. 55. 56.

O

O

N O

H3PO3 is a diabasic acid. Pyrophosphoric is tetrabasic acid, but it forms only two series of salts, e.g. Na2H2P2O7 and Na4P2O7. White phosphorus has a molecular formula P4. The four atoms are present at the corners of a tetrahedron. The structures of acids are O O O O H

P H

OH, HO

P

OH HO

OH

P

OH, H

OH

P OH

O

P

OH

OH

62. The stability of hydrides decreases down the group. POCl3 + 2HCl 65. PCl5 + H2O calculated quantity

66. Because of hydrogen bondings, NH3 has the highest melting point. 67. The reaction is exothermic and is attended to by the decrease in the gaseous molecules. Hence, to increase the yield of NH3, the reaction should be carried out at low temperature and high pressure. Cu(NO3)2 + 2H2O + 2NO2 70. Reaction is Cu + 4HNO3 (conc.)

72. Nitrogen is a diamagnetic as it contains no unpaired electrons. 73. Ammonia contains a lone pair. Hence, it acts as Lewis base.

15.58 Complete Chemistry—JEE Main

74. Reaction is

3Cu + 8NHO3

3Cu(NO3)2 + 4H2O + 2NO

(dil.)

77. 81. 82. 86. 89. 91. 96. 99. 102. 103. 107. 108. 109. 110.

Same as Q. 74. Heating of KNO3 liberates O2. Fluorine has the smallest size amongst halogens. The reaction is 4HNO3 + P4O10 4HPO3 + 2N2O5 The reaction is (NH4)2Cr2O7 N2 + 4H2O + Cr2O3 Phosphorus exists in three allotropic forms—white, red and black. Black phosphorus exists in one amorphous and three crystalline forms. Black phosphorus is the most stable form of the element. White phosphorus is insoluble in water. Chemically, red phosphorus is much less reactive than the white form. 4H3PO3 P4O6 + 6H2O P4O10 + 6H2O 4H3PO4 The maximum valence of nitrogen is four as it has four valence orbitals. It has no d orbitals to show valence more than four White phosphorus is discrete P4 molecule. H3PO3 is HOP(OH)2. The bond P—H is responsible for reducing agent. The cyclic metaphosphate ion is – O O P O O O O P P – – O O O

111. The highly electronegative F atoms make NF3 a poor electron donor. 112. See Q.109 113. The reactions are

220∞ C 320∞ C æææ Æ H4P2O7 æstrong æææ Æ (HPO3)n H3PO3 ægentle heat heat

114. N2O3, N2O5 and NO2 are acidic oxide while N2O is a neutral oxide. 115. The effective nuclear charge of As is larger than P due to less shielding effect of 3d orbitals-making the second ionization energy of As greater than that of P. 116. The free monometaphosphate ions does not exist. The metaphosphates form a family of ring compounds

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. What may be expected to happen when phosphine gas is mixed with chlorine gas? (a) PH3◊Cl2 is formed with warming up (b) The mixture only cools down (c) PCl3 and HCl are formed and the mixture warms up (d) PCl5 and HCl are formed and the mixture cools down [2003] 2. The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is (a) zero (b) one (c) two (d) three [2005] 3. Regular use of which of the following fertilizers increases the acidity of soil? (a) Potassium nitrate (b) Urea (c) Superphosphate of lime (d) Ammonium sulphate [2007] 4. The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence (b) SiX2  GeX2  PbX2  SnX2 (a) GeX2  SiX2  SnX2  PbX2

Study of the p-Block Elements (Groups 13, 14 and 15)

(c) SiX2  GeX2  SnX2  PbX2 (d) PbX2  SnX2  GeX2  SiX2 5. Identify the incorrect statement among the following. (a) Ozone reacts with SO2 to give SO3 (b) Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O (c) Cl2 reacts with excess of NH3 to give N2 and HCl (d) Br2 reacts with hot and strong NaOH solution to give NaBr, NaBrO4 and H2O 6. Three reactions involving H2PO 4– are given below. (i) H3PO4 + H2O Æ H3O+ + H2PO 4–

15.59

[2007]

[2007]

(ii) H2PO4– + H2O Æ HPO 42– + H3O+

7.

8. 9.

10.

(iii) H2PO4– + OH– Æ H3PO4 + O2– In which of the above reactions does H2PO4– acts as an acid? (a) (i) only (b) (ii) only (c) (i) and (ii) only (d) (iii) only [2010] Which of the following statements is wrong? (a) N2O4 has two resonance structures. (b) The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table. (c) Nitrogen cannot form dp-pp bond. (d) Single N—N bond is weaker than the single P—P bond. [2011 cancelled] The molecule having smallest bond angle is (b) AsCl3 (c) SbCl3 (d) PCl3 [2012] (a) NCl3 Which one of the following properties is not shown by NO? (a) It’s bond order is 2.5 (b) It is diamagnetic in gaseous state (c) It is a neutral oxide (c) It combines with oxygen to form nitrogen dioxide [2014] Which one of the following does not have a pyramidal shape? (a) (CH3)3N

(b) (SiH3)3N

(c) P(CH3)3

(d) P(SiH3)3 [2014, online]

11. Which of the following statements is true? (a) HNO3 is a stronger acid than HNO2

(b) H3PO3 is a stronger acid than H2SO3

(c) In aqueous medium HF is a stronger acid than HCl (d) HClO4 is a weaker acid than HClO3. 12. Which of the following compounds has a P—P bond? (b) H4P2O6 (c) H4P2O7 (d) (HPO3)3 (a) H4PO2O5

[2006]

[2015, online] 13. Addition of phosphate fertilisers to water bodies causes: (a) enhanced growth of algae (b) increase in amount of dissolved oxygen in water [2015, online] 14. Assertion: Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen Reason: The reaction between nitrogen and oxygen requires high temperature. (a) Both assertion and reason are correct, and the reason is the correct explanation for the assertion (b) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion (c) The assertion is incorrect, but the reason is correct. (d) Both the assertion and reason are incorrect. [2015] 15. The pair in which phosphorus atoms have a formal oxidation state of +3 is (a) Pyrophosphorous and hypophosphoric acids (b) Orthophosphorous and hypophosphoric acids (c) Pyrophosphorous and pyrophosphoric acids (d) Orthophosphorous and pyrophosphorous acids [2016]

15.60 Complete Chemistry—JEE Main

1000 ppb, 40 ppb, 100 ppm and 0.2 ppm. This is unsuitable for drinking respectively due to high concentration of [2016]

ANSWERS 1. (c) 7. (b) 13. (a)

2. (c) 8. (c) 14. (a)

3. (d) 9. (b) 15. (d)

4. (c) 10. (b) 16. (b)

5. (d) 11. (a)

6. (b) 12. (b)

HINTS AND SOLUTIONS O

2. The structure of hypophosphorous acid (H3PO2) is

H

P

H

OH There are two hydrogen atoms attached to P atom. 3. The acidity in soil increases due to H+ and SO42– ions released from ammonium sulphate. 4. The stability of +2 oxidation state of elements of Group 14 increases on descending the group. 5. The correct reaction is 3Br2 + 6NaOH æÆ 5NaBr + NaBrO3 + 3H2O (hot) +

6. The reaction (ii) provides H3O . Hence, in this reaction H2PO –4 acts as an acid. 7. The stability of hydrides decreases from NH3 to BiH3 in group 15 due to decrease in bond energy M—H. 8. The bond angle Cl —A—Cl (where A is N, P, As or Sb) decreases with the decrease in the electronegativity or increase in the size of the central atom A. Thus, out of NCl3, PCl3, AsCl3 and SbCl3, the latter is expected to have the smallest bond angle. 9. (s 2s)2 (s*2s)2 (p 2px)2 (p 2py)2 (s 2p)2 (p* 2px)1 The molecule is paramagnetic as it contains one unpaired electron in p*2px orbital. Its bond order is 2.5 ( = (6 bonding electrons –1 antibonding electron)/2). It is a neutral oxide and combines with oxygen to give NO2. 10. Tetrasilylamine is planar due to back bonding of the lone pair in p orbital of N atom to an empty orbital of Si thereby form double bond

12. Hypophosphoric acid has P—P linkage OH OH Ω Ω O == P -- P == O Ω Ω OH OH (H4P2O6)

14. Nitrogen and oxygen in atmosphere do not combine unless high temperature is employed. Atmospheric lightening also helps in forming oxides of nitrogen. 15. Phosphorous series contains P in +3 state while phosphoric series contains P in +5 state. 16. The permissible limits are as follows. Iron Fluoride

0.2 ppm 1.5 ppm

Lead 50 ppb Nitrate 50 ppm

16 Study of the p-Block Elements (Groups 16, 17 and 18) The Group 16 Elements Group 16 of the periodic table consists of oxygen (O), sulphur (S), selenium (Se), tellurium (Te) and polonium (Po). These are known as chalcogens or ore-forming elements. The physical properties of these elements are recorded in Table 1. Atomic and Molecular Properties of Group 16 Elements

Table 1 Property

O

S

Se

Te

Po

[He](2s)2(2p)4 [Ne](3s)2(3p)4 [Ar](3d)10(4s)2(4p)4 [Kr](4d)10(5s)2(5p)4 [Xe](4f)14(5d)10(6s)2(6p)4 Atomic number

8

16

34

32.08

78.96

Relative atomic mass

16.00

Covalent radius, r/pma

74

104

Ionic radius M2–, r/pm,

140

184

Melting point, T/K

54

Boiling point, T/K Density at 298 K, r/g cm

393

90 –3

Ionization energy, I/kJ mol

1.32

d

718 e

2.06

52

210

114

137

168b

198

221

230c

490

725

519-527

1263

1235

958 f

4.34

g

6.25

1314

1000

941

869

II

3388

2251

2045

1790

E/kJ mol–1

9.14b

–1

I Electronegativity

84

127.60

3.5 –142

2.5 –200

813 –

2.4

2.1

2.0

–

–

–

(a) two-coordinate covalent radius, (b) six-coordinate metallic radius, (c) approximate, (d) monoclinic form, (e) at 54 K, (f) rhombic form, (g) a-monoclinic, (h) a-form

Description of Physical Properties ns)2(np)4, where n varies from 2 to 6. electronegativity and nonavailability of d orbitals. 3. The metallic character increase on descending the group. O and S are non-metallic Se and Te are weaker nonmetailic Po is metallic The above fact is supported by the chemical behaviour of these elements. The tendency to from M2+ ions increases down the group. The stability of M2– ions decreases down the group.

16.2 Complete Chemistry—JEE Main

4. The covalent radii increase on descending the group. The variation is as per expectation (Fig. 1). From S to Se, the difference in atomic radii is small due to the poor shielding of nuclear charge by 3d electrons. The ionic radii of M2– also increases on descending the group. 5. The melting and boiling points increase down the group with the exception of the last member for which a decrease is observed (Fig. 2). 6. The ionization energy decreases on descending the group. The ionization energy of an element in Group 16 is

incoming electron from the smaller, more compact electronic cloud of oxygen atom. 1200

b.pt

Atomic Radii 1000 150 800

r/pm

T/K

130

m.pt

110

400

90

70

Fig. 1

600

200

O

S

Se

Ta

O2

Po

Atomic Radii of the Group 16 Elements

Fig. 2

S8

Se8

Te8 Po

Melting and Boiling Points of Molecules of the Group 16

8. Polonium, the last member of he group, is a radioactive element. 9. Oxygen exists as diatomic molecules with two unpaired electrons and is paramagnetic. 10. Sulphur exists in two allotropic forms—rhombic sulphur (or a sulphur), monoclinic sulphur (or b g-monoclinic sulphur, is obtained by chilling hot concentrated solution of S in CS2 (or toluene or C2H5OH). All three forms contain puckered S8 rings. Rhombic sulphur is stable at room temperature while monoclinic sulphur is stable at temperature above 369 K. These two forms change reversibly with slow heating or slow cooling. In addition to S8, sulphur rings of 6, 7, 9–15, 18 and 20 atoms are known. In cyclo–S6, the ring adopts a chair form in the solid. At elevated temperatures (~ 1000 K), S2 is the dominant species with paramagnetic property like O2. Selenium exists in six allotropic forms. Four of these forms are red and two are grey. Three of the four red forms have Se8 molecules differing in packing in crystals. The fourth form is amorphous and contains polymeric chains. The stable form of grey selenium Tellurium exists only in one crystalline form. Polonium exists in two crystal forms—cubic and rhombohedral.

Structure of S8 molecule

205.7 pm 102.2°

Structure of S6 molecules

Study of the p-Block Elements (Groups 16, 17 and 18)

16.3

Reactivity of Elements ns)2(np)4. They may accept 2 electrons to attain noble-

2–

2 electrons, thus forming two covalent bonds. 2. Oxygen is the most electronegative element in this group. Thus, its compounds with metals are ionic in nature. Sulphur, selenium and tellurium also form sulphides, selenides and tellurides with the more electropositive metals of Groups I and II. The ionic nature of these compounds decreases down the group. 3. With hydrogen, covalent compounds such as H2O and H2S are formed. The compounds with more electronegative elements are also covalent with the oxidation state +II. Examples are F2O, Cl2O and SCI2. 4. S, Se and Te also exhibit +4 and +6 oxidation states due to the availability of d orbitals. These are more stable than +2 oxidation state. Formation of Two Bonds s

p

(i)

s

p

(ii) Two bonds, bond angle near to 90°

sp3 hybridization, bond angle near to 109° 28¢ Tetrahedral structure with two positions occupied by two lone pairs

Formation of Four Bonds s

p

d

sp3d hybridization Trigonal bipyramid structure with one position occupied by a lone pair

Formation of Six Bonds s

p

d

sp3d2 hybridization octahedral structure

5. The compounds of S, Se and Te with O are tetravalent with oxidation state +4. They have both oxidizing and reducing characteristics. 6. The compounds of S, Se and Te with F are hexavalent with oxidation state +6. They have oxidizing characteristics. The stability of higher oxidation states becomes less on descending the group. 7. Oxygen exists as diatomic gaseous molecules while other elements of the group catenate to form chains or rings. The tendency of catenation is maximum in sulphur and is displayed in its compounds like sulphanes H—Sn—H, polysulphuric acids HO3S—Sn—SO3H, polysulphides S2–n and various allotropic forms of elements. 8. The bonds between S and O, or Se and O, are much shorter than the expected single S O bond length between these atoms. This is due to the possibility of forming a double bond via pp — dp bonding in addition to the normal sigma bond. In pp — dp Sulphur forms stronger p bonds than selenium due to similar sizes of orbitals.

pp — dp sideways overlap

Comparison of Sideways Overlap Along a Period As one travels from Si to P to S to CI, the size of the 3d orbitals decreases due to the increase in the nuclear charge. This decrease in size of the 3d orbitals is associated with the progressively stronger pp — dp bonds, due to the achievement of comparable sizes of the 2p orbital and 3d orbitals. Thus pp— dp bonding does not exist between O and Si, the SiO4 units polymerize involving Si—O—Si s bonds. pp — dp bonding between O and P is stronger with large tendency of polymerization of oxoanions.

16.4 Complete Chemistry—JEE Main

pp — dp bonding between O and S is still more stronger with a small amount of polymerization of oxanions. pp — dp bonding between O and Cl is strongest with no polymerization of oxoanions.

Ozone through pure and dry oxygen or air. The silent discharge is carried out by packing insulating materials in the space between electrodes through which discharge passes. This prevents any local rise in temperature, which would lead to the decomposition of ozone. The reaction 2O3 3O2 is endothermic (DH = +284.5 kJ mol–1). Ozone is pale-blue gas with a characteristic strong smell. When inhaled in small amount, it causes headache and nausea. In larger quantities, it is poisonous. The main reactions of ozone are the oxidation reactions. 2KI + H2O + O3 Æ 2KOH + I2 + O2 PbS + 4O3 Æ PbSO4 + 4O2; (neutral medium)

2HCl + O3 Æ Cl2 + H2O + O2; 2I2 + 2H2O + 4O3 Æ 4HIO3 + O2;

2K2MnO4 + H2O + O3 Æ 2KMnO4 + 2KOH + O2 2K4Fe(CN)6 + O3 + H2O Æ 2K3Fe(CN)6 + 2KOH + O2

(moist)

Moist sulphur, phosphours and arsenic are also oxidized to higher acids. When ozone is passed through mercury, the latter loses its meniscus and sticks to glass due to the formation of Hg2O which dissolves in mercury. With peroxides, the reactions taking place are H2O2 + O3 Æ H2O + 2O2 BaO2 + O3 Æ BaO + 2O2; With organic compounds containing double and triple carbon-carbon bonds, ozone forms ozonides which gets decomposed by water or dilute acids. Ozone is formed at the height of about 20 km from oxygen by the absorption of sunlilght. There is a formation of ozone (such as CCl2F2, which is used as a refrigerant (Freon-12) and as a propellant in aerosol spray) contribute to the decomposition of ozone layer. Sunlight causes the decomposition of these compounds producing active chlorine which catalysed the decomposition of ozone. Emission of nitrogen oxides from supersonic jets also contribute in depleting the ozone layer. The structure of ozone is O 116° 49¢

O

128 pm O

The oxygen-oxygen bond length in ozone is 128 pm which is intermediate between a single bond (148 pm) and double bond (121 pm). It has been explained on the base on the base of the resonating structures shown below. -

-

O

O

O O

O

+

O

-

The structure of ozone is best described as follows. The oxygen are sp2 hybridized. Two of these orbitals in the central atom and one each of the terminal oxygens are involved in s bonds. The remaining sp2 orbitals contain lone pairs of electrons (= 18 – 4 – 10) occupy molecular orbitals formed from the three pz atomic orbital from the three oxygen atoms. Two of these electrons are in the bonding molecular orbital and the remaining two in the nonbonding molecular orbital. Thus, double bonds involving three oxygen atom is best described as four-electron three-centre bond. The bond order of each O —O bond is 1.5.

Compounds of Group 16 Elements Hydrogen Compounds All the elements of Group 16 form covalent compounds with hydrogen. These are water (H2O), hydrogen sulphide (H2S), hydrogen selenide (H2Se) hydrogen telluride (H2Te) and hydrogen polonide (H2Po).

Study of the p-Block Elements (Groups 16, 17 and 18)

16.5

A convenient method to prepare H2S, H2Se and H2Te is to treat the metal sulphides with acids. FeS + H2SO4 Æ H2S + FeSO4 FeSe + H2SO4 Æ H2Se + FeSO4 Al2Te3 + 3H2SO4 Æ Al2(SO4)3 + 3H2Te In the laboratory, H2S is generated by the action of dilute H2SO4 on iron(II) sulphide in a Kipp’s apparatus. Some of the physical properties of hydrogen compounds of Group 16 elements are recorded in Table 2. Table 2 Characteristic

Properties of Hydrogen Compounds of Group 16 H 2O

H 2S

H2Se

Melting point, T/K

273

187.5

207.5

Boiling point, T/K

373

212.3

231.5

1.8 ¥ 10

Dissociation constant at 298 K Bond angle H—M—H

104.5°

Bond length H—M/pm Bond energy H—M, E/kJ mol–1 Enthalpy of formation DfH°/kJ mol

–1

–16

1.3 ¥ 10

–7

92.5°

1.3 ¥ 10 90°

H2Te 221.9 271.2 –4

2.3 ¥ 10–3 89.5°

96

134

146

169

463

347

276

238

–242

–20.2

81

154.4

T/K

Water has exceptionally high melting and boiling points due to hydrogen bondings (Fig. 3). Beyond H2S, the melting and boiling 360 points increase on descending the group. The bond angle in H2O is 104.5° which is close to the tetrahedral angle 109°28¢. This indicates the sp3 hybridization of oxygen orbitals. 320 Two positions of a tetrahedron are occupied by H atoms and two are occupied by lone pair of electrons. The decrease of bond angle from 109°28¢ to 104.5° has been explained on the basis of lone pair-bonding 280 b.pt pair repulsion of electrons. The bond angle in other hydrides are more near to 90° indicating the involvement of p orbitals of element in bonding with hydrogen atoms. 240 The bond enthalpy H—M decreases down the group indicating the m.pt weakening of bond. This may be attributed to the increasing size of 200 bonding atomic orbital which overlap with 1s orbital of hydrogen atom. Also there occurs a decrease in ionic character of the bond which results into the decrease of electrostatic attraction. 160 The dissociation constant (K = [H+]2 [M2–]/[H2M]) increases on H2O H2S H2Se H2Te descending the group. This fact is in agreement with the weakening of Fig 3. Melting and Boiling Points of Hydrides of bond strength on descending the group. the Group 16 Elements All hydrogen compounds except water are poisonous and have unpleasant odour. The stability of hydrogen compounds decreases on descending the group. This is also shown by the decrease in their enthalpies of formation. Solid water is less dense than the liquid. Liquid water has maximum density of 1 g cm–3 at 4 °C. Ice has a rather open structure due to the hydrogen bondings (Fig. 4). There is a lot of unoccupied space in solid water. On melting, some free water molecules occupy some of this space resulting into the decrease in volume and hence increase in density. This effect is continued up to 4 °C, beyond which normal expansion occurs so the density decreases. In ice, hydrogen atoms are located between oxygen atoms, and by forming hydrogen bonds with the latter bind the whole structure together into a huge network of atoms. Throughout the network there are four hydrogen atoms arranged tetrahedrally about each oxygen atom, two oxygen atoms on either side of each hydrogen atom

Hard and Soft Water Hard Water If a sample of water contains bicarbonate, chloride and sulphate of calcium and magnesium, it is known as hard water. Such a sample of water does not form lather with soap. The Ca2+ and Mg2+ ions present in water precipitates out the soap. For examples,

16.6 Complete Chemistry—JEE Main

2C17H35COONa(aq) + M2+(aq) Æ (C17H35COO)2M Ø + 2Na+(aq) (soap)

(Ca2+ or Mg2+)

Soft water A sample of water free from soluble salts of calcium and magnesium is known as soft water.

Methods of Removing Hardness of Water and permanent. Temporary Hardness The temporary hardness in water is due to the presence of bicarbonates of calcium and magnesium. These can be removed by the following two methods (i) Boiling of Water On boiling temporary water, the soluble salts are precipitated as shown in the following. Æ Mg(OH)2Ø + 2 CO2 ≠ Mg(HCO3)2 æDæ Æ CaCO3Ø + H2O + CO2≠ Ca(HCO3)2 æDæ

O H H

H O H O H

H O HO H H O HO H H O H O H O

(ii) Clark’s Method To the temporary water, calculated quantity of lime is added. This precipitates out soluble calcium and magnesium bicarbonates: Ca(HCO3)2 + Ca(OH)2 Æ 2CaCO3Ø + 2H2O Mg (HCO3)2 + 2Ca(OH)2 Æ 2CaCO3Ø + Mg (OH)2Ø + 2H2O

H O H

O H

O H

Fig. 4

O H H O

H O H

H

O

H

O

H O

O

O H H H

O H O

HO H

O

H H

H OH H H

O

O H

H

H

O H H H O H O

O H H

H H O

H

H

H H O O H

H

O H

O

H

O

Schematic Arrangement of Water Molecules in Ice

Permanent Hardness The permanent hardness in water is due to the presence of chlorides and sulphates of calcium and magnesium. These cannot be removed by boiling water. The following methods are employed to remove these salts from water. (i) Treatment with Washing soda The addition of washing soda (sodium carbonate) causes the precipitation of calcium and magnesium. MCl2 + NaCO3 Æ MCO3Ø + 2NaCl (Mg, Ca)

MSO4 + Na2CO3 Æ MCO3Ø + Na2SO4 (ii) Calgon’s Method The addition of calgon (sodium hexametaphosphate, Na6P6O18) causes the following reactions 2– Na6P6O18 Æ 2Na+ + Na4P6O18 M2+

2– + Na4P6O18 Æ [Na2MP6O18]2– + 2Na+

(Ca2+ or Mg2+)

The Ca2+ and Mg2+ ions are retained in the complex ions and thus do not interfere the functioning of soda. (iii) Ion-Exchange Method Both natural and synthetic zeolites (having aluminosilicate framework) have been used for exchanging Ca2+ and Mg2+ ions in hard with Na+ ions. For example, for the synthetic zeolite permutit, we have Na6(Al6Si6(OH)12O18) + 3CaSO4 = 3Na2SO4 + Ca3(Al6Si6(OH)12O18) If the calcium zeolite is treated with a solution of sodium chloride, the action begins in reverse, and the original sodium zeolite is regenerated in a form ready to be used again. (iv) Using Ion Exchange Resins Giant organic molecules having acidic or basic groups are known as ion-exchage resin. Acid resins contain the acid group – COOH. Such resins exchange their H+ ions with cations Ca2+ and Mg2+ present in hard water. 2 RCOO– H+ + Ca2+ Æ (RCOO)2 Ca + 2H+ acid resin or base-exchange resin

Study of the p-Block Elements (Groups 16, 17 and 18)

16.7

Basic resins exchange their OH– ions with anions HCO–3, Cl–, and SO2– 4 present in hard water. – RNH+3 OH– + SO2– 4 Æ (RNH3)2 SO4 + 2OH

base resin or acid-exchange resin

Hard water is passed successively through a cation exchange and anion exchange resins. Cation-exchange resin makes water acidic due to the release of H+ and anion-exchange makes water alkaline due to the release of OH–. Liberated H+ and OH– combine to give neutral water. The exhausted cation and anion exchange resins are regenerated by treating them with dilute acid and alkali solutions, respectively. Heavy Water Heavy water is D2O. It may be obtained by exhaustive electrolysis of water. It is used as moderator in nuclear H H reactions and also in the preparation of deuterium compounds. 95.0 pm 98.8 pm 145.8 pm 147.5 pm 90.2° Peroxides and Polysulphides Oxygen and to a greater extent 111.5° sulphur form polyoxides and polysulphides, which are less stable 94.8° 101.9° H H than the normal salts. Examples are H2O2, H2S2, H2S3, H2S4, etc. The higher elements do not form such compounds. (b) (a) H2O2 and H2S2 can be prepared by the addition of acid to a Fig. 5 Nonplanar Dihydroxyl Structure of Hydrogen peroxide or a persulphide salt. Peroxide in (a) Gaseous Phase and (b) Crystalline Phase BaO2 + H2SO4 Æ BaSO4 + H2O2 Na2S2 + H2SO4 Æ Na2SO4 + H2S2 The structure of H2O2 molecules is as shown in Fig. 5. H2O2 is a strong oxidizing agent. In acidic solutions, oxidizing action is slow while in alkaline medium it is fast. 2– – Examples exhibiting oxidizing action are: Fe2+ to Fe3+, [Fe(CN)6]4– to [Fe(CN)6]3–, SO2– 3 to SO 4, I to I2 and NH2OH to HNO3. With a stronger oxidizing agent such KMnO4, KIO4, O3, H2O2 acts as a reducing agent. The reactions exhibiting oxidizing and reducing agents are H2O2 + 2H+ + 2e– Æ 2H2O E° = 1.77 V

H2O2 Æ O2 + 2H+ + 2e– H2O2 is a more powerful oxidizing agent in acidic medium as compared to alkaline medium. On the other hand, H2O2 is a more powerful reducing agent in alkaline medium as compared to acidic medium. These facts may be explained based on the above equations or on the basis of effects caused on the reduction potentials. For oxidizing action, RT 1 ln E = E° + 2 2F ([H ] / c∞) ([H 2 O 2 ] / c∞) E becomes more positive on increasing the concentration of H+, hence, its reduction tendency or causing oxidation of some other species is enhanced. For reducing action, E becomes more negative on decreasing the concentration of H+, hence, its reduction tendency or causing oxidation of some other species is lowered. Conversely, its oxidation tendency or causing reduction of some other species in enhanced. Kinetically, oxidation with H2O2 is slow in acidic medium but proceeds rapidly in alkaline medium.

Halides All the elements of Group 16 react with halogens to form several types of binary compounds such as MX6, MX4, MX2, M2X2 and M2X (Table 3).

16.8 Complete Chemistry—JEE Main

Table 3 MX6

MX4

O

Binary Compounds of Group 16 Elements with Halogens MX2

M 2X 2

M 2X

Others

OF2

O2F2

ClO2

O3F2, O4F2

Cl2O

Cl2O6 Cl2O7

Br2O

BrO2

BrO3 I2O4, I2O5, I4O9

S

SF6

SF4

SF2

S 2F 2

SCl4

SCl2

S2Cl2 S2Br2

Se

Te

Po

SeF6

TeF6

PoCl4 PoBr4 PoI4

SeF4 SeCl4 SeBr4 TeF4 TeCl4 TeBr4 TeI4 PoCl2 PoBr2

Se2Cl2 Se2Br2 TeCl2 TeBr2

Compounds of Oxygen

oxides. 3. In more common compounds with other halogens, oxygen has oxidation states of –2 and –1/2.

Compounds with Other Elements 1. Fluorine is the most electronegative element. It is able to bring out the maximum valency of six in S, Se and Te. 2. Other halogens are able to bring out the maximum valency of four in S, Se, Te and Po. 3. On descending the group, the higher oxidation-state halides are more stable than those of the lower oxidation states. 3 2 d hybridization of the central atom. 5. The structures of tetrahalides are trigonal bipyramidal with one of the equatorial Cl positions occupied by a lone pair of electrons. Cl 6 is extremely stable compound. It is used as a gaseous dielectric (insulator) in high voltage Te transformers. SeF6 is slightly reactive whereas TeF6 is more reactive. The latter is easily hydrolyzed as Cl TeF6 + 6H2O Æ 6HF + H6TeO6 Cl 7. Tetrahalides are more reactive than hexahalides but are more stable than the Structure of tetrahalides lower halides. Tetrahalides are very sensitive to water. SF4 + 2H2O Æ SO2 + 4HF

Study of the p-Block Elements (Groups 16, 17 and 18)

8. The dihalides form angular molecules, based on a tetrahedron with two positions occupied by lone pairs. The lone pairs distort the tetrahedral angle of 109°28¢ to 103° in SCl2, 101.5° in F2O and 98° in TeBr2. The tetrahedral structure indicates the sp3 hybridization of the central atom. 9. The structure of monohalides (M2X2) is similar to that of H2O2.

Oxides Practically, all of the elements react with oxygen to form oxides. Based on their structures,

16.9

S Cl

Normal Oxides In these oxides, the number of oxygen atoms in a molecule is in accordance with the oxidation state of the other element. Examples are H2O, MgO, Al2O3, etc.

Cl

Structure of dihalide

Peroxides In these oxides, the number of oxygen atoms is more than the expected number determined from the oxidation number of other elements. All of them contain —O—O— linkage attached ionically or covalently to the other element. Ionically bound — Na2O2, K2O2, BaO2 Covalently bound — H2O2, H2SO5 (peroxomonosulphuric acid) Peroxo compounds are strong oxidizing agents, and are hydrolyzed by water to give H2O2. H2SO5 + H2O Æ H2SO4 + H2O2 Suboxides In these oxides, the number of oxygen atoms in a molecule is less than the expected number determined from the oxidation number of other element. They involve M—M bonds in addition to M—O bonds. Example O == C == C == C == O. Basic Oxides The oxides of metals are generally basic Examples are Na2O, MgO, CrO, etc. The ionic basic oxides have high lattice energy. So, their melting points are high. If a metal can exist in more than one oxidation state, the oxide with the lowest oxidation state is the most ionic and the most basic. The basicity decreases (or acidity increases) with increase in the oxidation state of element. Examples CrO is basic, Cr2O3 is amphoteric and CrO3 is acidic. Amphoteric Oxides Oxides which can react with both acids and bases are amphoteric oxides. Examples BeO, Al2O3, GaO3, SnO, PbO and ZnO. Acidic Oxides The oxides of nonmetals are generally acidic. Examples CO2, NO, SO2, Cl2O, SO3, P2O5 Acidic oxides have low melting and boiling points. In cases where the element can exist in more than one oxidation state, the oxide of the highest oxidation state is the most acidic. Examples N2O5 is more acidic than N2O3 SO3 is more acidic than SO2. Neutral Oxides A few covalent oxides are neither acidic nor basic. These are said to be neutral oxides Examples N2O, NO, CO Oxides of Higher Elements of the Same Group as follows. Sulphur: SO2, SO3, S2O Selenium: SeO2, SeO3 Tellurium: TeO, TeO2, TeO3 Polonium: PoO, PoO2 The structures of SO2 and SO3 are

The more common oxides of the higher elements of group 16 are

16.10 Complete Chemistry—JEE Main

O

S

S

O

O

O O SO3 in gaseous phase

SO2 is a strong reducing agent: SO2 + 2H2O Æ SO42– + 4H+ + 2e– Its reducing characteristics is enhanced in alkaline medium. At room temperature, SO3 is solid and exists in three distinct forms, namely, a-SO3, b-SO3 and g-SO3. g-SO3 has a cyclic structure, (SO3)3. O

O

O

S O

O

S

S

O

O

O O

O

i.e.

O

O

S O

O

O

S

S

O

O

O ] units each sharing two corners. 4

b-SO3

O

O S

O O S O

O O

S O

O

O O

S

i.e.

O

S

O

O

O

a-SO3 is made up of chaions of cross-linked sheets.

Oxoacids 1. Sulphurous Acid Series HO

Di-or pyrosulphurous acid (H2S2O5)

S

O

HO

HO

O

O

S

S

OH

O

Dithionous acid (H2S2O4)

HO

O

O

S

S

O

O S

O

O

Sulphurous acid (H2SO3)

O

OH

S

O O

Study of the p-Block Elements (Groups 16, 17 and 18)

16.11

Comments 1. These acids contain one sulphur atom in +3 oxidation state. 2. Sulphurous acid is mostly present as SO2 2O. 3. Sulphurous acid form salts known as sulphites (SO32– ). 4. The structure of SO2– 3 ion is pyramidal, i.e. tetrahedral with one position occupied by a lone pair.

S O

5. 6. 7. 8.

O

bond length S—O 151 pm bond angle O—S—O 106° p bond is delocalized.

O Free di- or pyrosulphurous acid does not exist. The salts of disulphrous acid are known as disulphites. These are also known as pyrosulphites and metasulphites. These salts contain an S—S linkage. Dithionous acid does not exist in nature. The salts of dithionous acid are known as dithionites.

2. Sulphuric Acid Series O HO

Sulphuric acid (H2SO4)

S

OH

O O

Thiosulphuric acid (H2S2O3)

HO

S

OH

S O

Di- or pyrosulphuric acid (H2S2O7)

HO

S

O O

S

OH

O O Comments 1. The oxidation state of sulphur atom in H2SO4 is +6. In H2S2O3, the oxidation states are +6 and –2. 2. The salts of sulphuric acid are known as sulphates. The structure of sulphate is O

S O-

O

tetrahedral structure, bond length S—O 149 pm. The two p bonds are delocalized over the S and the four O atoms.

O-

3. The salts of thiosulphuric acid are known as thiosulphates. 4. The structure of thiosulphates is

16.12 Complete Chemistry—JEE Main

O

S O-

O

O5. Thiosulphuric acid cannot be formed by adding acid to thiosulphates because the free acid decompose in water in a mixture of S, H2S, H2Sn, SO2 and H2SO4. It can be prepared in the absence of water at low temperatures.

Sulphuric Acid Sulphuric acid is one of the most important industrial chemicals. Commercially, this is manufactured by the contact process. Sulphur dioxide obtained from the burning of sulphur or sulphide ores are made to combine with O2 to give SO3 where V2O5 is used as a catalyst. This is followed by the absorption of SO2 in H2SO4 to give oleum (H2S2O7). Æ SO2(g) S(s) + O2(g) æDæ V2 O5 2SO2(g) + O2(g) æææ Æ 2SO3 SO3 + H2SO4 Æ H2S2O7 The conversion of SO2 to SO3 is an exothermic process and is accompanied with decrease in volume. According to Le-Chatelier principle, low temperature and high pressure help increasing the production of SO3. A pressure of 2 bar and a temperature of 720 K are used in contact process. Dilution of oleum with water give sulphuric acid of desired concentration. Sulphuric acid ionizes as (a strong acid) H2SO4 + H2O Æ H3O+ + HSO4– K°a2 = 1.2 ¥ 10–2 HSO4– + H2O Æ H3O+ + SO42– Concentrated H2SO4 is a strong dehydrating agent. Hot concentrated H2SO4 acts as a moderately strong oxidizing agent: Cu + 2H2SO4 Æ CuSO4 + SO2 + 2H2O (conc.)

S + 2H2SO4 Æ 3SO2 + 2H2O (conc.)

C + 2H2SO4 Æ CO2 + 2SO2 + 2H2O (conc.)

3. Thionic Acid Series

Dithionic acid (H2S2O6)

HO

O

O

S

S

S

S

O

Polythionic acid (H2Sn+2O6)

HO

S O

OH

O (S)n

S

OH

O

Comments 1. Dithionic acid is a dibasic acid. It does not exist in free state. However, its salts are stables and are known as dithionates. 2. The average oxidation state of S in dithionic acid is +5.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.13

3. The dithionate ion has a structure similar to that of ethane in eclipsed conformation. The bond length S—S is 215 pm The bond length S—O is 143 pm The bond angle S—S—O is close to 103°. 4. The salts of polythionic acids are trithionate (S3O62–), tetrathionate (S4O62–), pentathionate (S5O62–) and hexathionate (S6O62–).

4. Perosoacid Series O H

Peroxomonosulphuric acid (H2SO5)

O

S

O

OH

O O HO

Peroxodisulphuric acid (H2S2O8)

S

O O

O

O

S

OH

O

Comments 1. Peroxoacid contains —O—O— linkage. 2. The oxidation state of S in peroxoacid is +6. 3. Peroxodisulphuric acid is a obtained by electrolysis of sulphates at high current density. 4. Peroxodisulphuric acid is a powerful oxidizing agents. It converts Mn2+ to MnO4– and Cr3+ to Cr2O72– 5. Peroxomonosulphuric acid is known as Caro’s acid. It may be obtained by the hydrolysis of peroxodisulphuric acid. O O O O HO

S O

O

O

S O

OH

HO

S

O

OH + HO

O

S

OH

O

Anomalous Behaviour of Oxygen Oxygen behaves differently from other elements in many respects. This is due to its smaller size, higher electronegativity, and the absence of d orbitals. Two of the anomalies may be mentioned here. 1. Oxygen forms strong hydrogen bonds in the compounds hydrogen attached to oxygen. 2. Oxygen has a tendency to form strong double bonds. The other elements also form double bonds. But they become weaker on descending the group. Thus O == C == O is a stable species S == C == S is less stable CSe2 polymerizes instead of forming double bonds CTe2 is unknown.

MULTIPLE CHOICE QUESTIONS 1. Which of the following elements does not belong to Group 16? (a) Oxygen (b) Sulphur (c) Selenium 2. Atomic number of tellurium which belongs to 5th period is (a) 50 (b) 51 (c) 52 3. The number of elements in Group 16 is (a) 3 (b) 4 (c) 5

(d) Bismuth (d) 84 (d) 6

16.14 Complete Chemistry—JEE Main

4. Which of the following orders of melting points of hydrides of Group 16 elements is true? (a) H2S > H2Se > H2Te (b) H2S < H2Se < H2Te (c) H2S > H2Se < H2Te (d) H2S < H2Se > H2Te 5. Which of the elements of Group 16 has the maximum tendency of catenation? (a) Oxygen (b) Sulphur (c) Selenium (d) Tellurium 6. Which of the following triatomic molecules does not exist? (a) CO2 (b) CS2 (c) CSe2 (d) CTe2 7. Which of the allotropes of sulphur represents a-sulphur? (a) Rhombic sulphur (b) Monoclinic sulphur (c) Plastic sulphur (d) Colloidal sulphur 8. Which of the following allotropes represents prismatic sulphur? (a) Rhombic sulphur (b) Monoclinic sulphur (c) Plastic sulphur (d) Colloidal sulphur 9. Which of the following hydrides of Group 16 has the lowest boiling point? (a) H2O (b) H2S (c) H2Se (d) H2Te 10. Which of the following statements is not correct? (a) The stability of hydrides of elements of Group 16 increases down the group (b) Oxygen cannot display oxidation states of +2, +4 and +6 (c) Oxygen exists as diatomic gaseous molecules while other elements of the Group 16 catenate to form chains or rings (d) Rhombic sulphur is stable at room temperature 11. Which of the following compounds undergoes hydrolysis? (a) SF6 (b) SCl6 (c) SeF6 (d) TeF6 12. Which of the following substances is used as the gaseous dielectric in high voltage transformers? (a) SF4 (b) SF6 (c) SCl6 (d) SeF6 13. Sulphur trioxide exists (a) in two polymorphic forms (b) in three polymorphic forms (c) in four polymorphic forms (d) only in one polymorphic form 14. The structure of SF4 is (a) square planar (b) tetrahedral (c) trigonal bipyramidal (d) octahedral 15. Which of the following acids does not exist in the free form? (a) H2SO2 (b) H2S2O3 (c) H2S2O7 (d) H2SO4 16. Which of the acids does not include S—S bond? (a) Dithionous acid, H2S2O4 (b) Dithionic acid, H2S2O6 (c) Polythionic acid, H2Sn+2O6 (d)Pyrosulphuric acid, H2S2O7 17. Which of the following acids exists in free form? (a) H2SO2 (b) H2SO3 (c) H2S2O3 (d) H2S2O4 18. Which of the following formulae represents peroxodisulphuric acid? O O O O (a) H O O S O S OH (b) H O O S O S O O H O

S

O

O S

O

O

O

O

(c) H O

O

OH

O O 19. The molecular formula of pyrosulphuric acid is (a) H2S2O3 (b) H2S2O4 20. The molecular formula of dithionous acid is (a) H2S2O3 (b) H2S2O4 21. The molecular formula of dithionic acid is (a) H2S2O3 (b) H2S2O4

(d) H O O

O

O

S

S OH

O

O

(c) H2S2O6

(d) H2S2O7

(c) H2S2O6

(d) H2S2O7

(c) H2S2O6

(d) H2S2O7

Study of the p-Block Elements (Groups 16, 17 and 18)

16.15

22. The molecular formula of thiosulphuric acid is (a) H2S2O3 (b) H2S2O4 (c) H2S2O6 (d) H2S2O7 23. The oxidation state of sulphur in peroxodisulphuric acid, H2S2O8, is (a) + 5 (b) + 6 (c) + 7 (d) + 8 24. The oxidation state of sulphur in peroxomonosulphuric acid, H2SO5, is (a) + 5 (b) + 6 (c) + 7 (d) + 8 25. Which of the following oxo-acids of sulphur has more than one oxidation state of sulphur? (a) H2S2O3 (b) H2S2O4 (c) H2S2O6 (d) H2S2O8 26. Which of the following oxo-acids of sulphur has more than one oxidation state of oxygen? (a) H2S2O3 (b) H2S2O4 (c) H2S2O6 (d) H2S2O8 27. When pure H2SO4 is electrolysed, the product at the anode is (a) H2S2O7 (b) H2S2O8 (c) H2S2O3 (d) H2S4O6 28. When aqua regia dissolves HgS, the function of the HCl in the mixture is to (a) oxidize the sulphur (b) oxidize the mercury (c) complex the sulphur (d) complex the mercury 29. In the manufacture of sulphuric acid by the contact process, SO3 is not added to water directly to form H2SO4 because (a) the reaction does not go to completion (b) the reaction is exothermic (c) the reaction is quite slow (d) SO3 is insoluble in water 30. The catalyst used in the contact process is (a) Fe2O3 (b) V2O5 (c) SO3 (d) Ag2O 31. Of the following acids, the strongest is (a) H2S (b) He2Se (c) H2Te (d) H2O 32. Pyrosulphuric acid may be synthesized by dissolving SO3 in (a) H2O (b) H2SO3 (c) H2SO4 (d) H2S2O8 33. The stable form of sulphur at room temperature and normal atmospheric pressure is (a) orthorhombic (b) monoclinic (c) hexagonal (d) amorphous 34. The hybridization of sulphur in SO2 is (a) sp (b) sp2 (c) sp3 (d) dsp2 35. Earth is protected from UV radiations by a layer of (a) N2 (b) O2 (c) O3 (d) CO2 36. H2S can be distinguished from SO2 by its action on (d) lead acetate solution 2Cr2O7 37. Marshall’s acid is (b) H2S2O5 (c) H2S2O7 (d) H2S2O8 (a) H2SO5 38. Caro’s acid is (a) H2SO5 (b) H2S2O5 (c) H2S2O7 (d) H2S2O8 39. Oxygen atoms in H2O2 has (a) sp hybrid orbitals (b) sp2 hybrid orbitals (c) sp3 hybrid orbitals (d) pure p orbitals 40. Which of the following is not true about H2O2? (a) It is used as a bleaching agent (b) It is used as a mild antiseptic (c) It is used as a rockel fuel (d) It is used for the manufacture of heavy water 41. The oxide that gives H2O2 on treating with a dilute acid is (a) PbO (b) Na2O2 (c) MnO2 (d) TiO2 42. Which of the following catalysts is used in the lead chamber process? (a) Pt (b) V2O5 (c) Oxides of nitrogen (d) Ni 43. Which of the following bonds has the highest bond energy? (a) O—O (b) S—S (c) Se—Se (d) Te—Te 44. Which of the following is the weakest acid? (a) H2S (b) H2Se (c) H2O (d) H2Te

16.16 Complete Chemistry—JEE Main

45. Which of the oxidation state(s) is/are shown by the element sulphur? (a) –2 only (b) +2 only (c) –2, +2 and +4 (d) –2, +2, +4 and +6 46. Which of the following is not the mineral of sulphur? (a) Chloroapatite (b) Fool’s gold (c) Sphalerite (d) Chalcopyrite 47. Which of the following statements regarding sulphur is not true? (a) Rhombic sulphur is stable at room temperature (b) Monoclinic sulphur is stable at room temperature (c) Both rhombic and monoclinic sulphur are soluble in CS2 (d) Both rhombic and monoclinic sulphur have a puckered-ring structure of eight sulphur atoms 48. Which allotropic form of sulphur is obtained on passing H2S gas through nitric acid? (a) Rhombic sulphur (b) Monoclinic sulphur (c) Plastic sulphur (d) Colloidal sulphur 49. Which of the following equations represents the oxidising action of sulphur dioxide? (a) 2Fe3+ + SO2 + 2H2O 2Fe2+ + SO42– + 4H + (b) 3Fe + SO2 2FeO + FeS – 2Mn2+ + 4H + + 5SO42– (c) 2MnO4 + 2H2O + 5SO2 (d) Cr2O72– + 2H+ + 3SO2 2Cr3+ + 3SO42– + H2O 50. The most favourable conditions for the formation of SO3 via the equation SO2 +

51.

52.

53. 54.

55.

56. 57.

1 2

O2

SO3

DH = – 95 kJ mol –1

are (a) low temperature and low pressure (b) low temperature and high pressure (c) high temperature and low pressure (d) high temperature and high pressure Which of the following statements regarding sulphur dioxide is not true? (a) SO2 is an angular molecule (b) SO2 may be regarded as the resonance hybrid of two canonical structures (c) SO2 is an anhydride of sulphuric acid (d) SO2 is an anhydride of sulphrous acid Which of the following statements regarding sulphur trioxide is not true? (a) Sulphur trioxide exists in three polymorphic forms (b) Sulphur trioxide is an anhydride of sulphuric acid (c) Vanadium pentoxide is used as a catalyst in converting sulphur dioxide into sulphur trioxide (d) sulphur trioxide has tetrahedral geometry The elemental sulphur is written as (a) S (b) S2 (c) S4 (d) S8 Which of the following statements is true? (a) Both rhombic and monoclinic sulphur are soluble in water (b) Both rhombic and monoclinic sulphur are soluble in carbon disulphide (c) Both rhombic and monoclinic sulphur are insoluble in carbon disulphide (d) Rhombic sulphur can be converted into monoclinic sulphur but the reverse is not possible Colloidal sulphur is obtained when (a) sulphur is heated in absence of air (b) sulphur is heated in presence of air (c) sodium thiosulphate is treated with iodine (d) H2S is passed through water containing HNO3 Which of the following forms of sulphur is stable at room temperature and 1 atmospheric pressure? (a) Amorphous (b) Hexagonal (c) Monoclinic (d) Orthorhombic Burning sulphur in air produces (a) H2S (b) H2S2 (c) SO2 (d) SO3

Study of the p-Block Elements (Groups 16, 17 and 18)

16.17

58. Sulphur dioxide is a strong reducing agent. However, it can also act as an oxidizing agent. Which of the following reactions shows its oxidizing nature? 4 solution (c) Reaction with H2 59. Which of the following statements is not true of sulphur? (a) Sulphur belongs to Group VIA of the periodic table (b) Sulphur exhibits allotropy (c) Sulphur occurs both in the native form and combined form (d) Sulphur is soluble in water 60. Which of the following gases is used for the qualitative analysis of metal ions? (a) CO2 (b) H2S (c) SO2 (d) SO3 61. Which of the following is an anhydride of H2SO4? (a) H2S (b) H2S2 (c) SO2 (d) SO3 62. Which of the following is an anhydride of H2SO3? (a) H2S (b) H2S2 (c) SO2 (d) SO3 63. Which of the following is not true for SO2? (a) SO2 is obtained by roasting metal sulphides (b) SO2 is acidic in nature (c) SO2 is used as a disinfectant(d)SO2 is an anhydride of H2SO4 64. Which of the following statements regarding sulphuric acid is not true? (a) Concentrated H2SO4 acts as a dehydrating agent (b) Hot concentrated H2SO4 acts as a powerful oxidizing agent (c) The reaction of PCl5 with sulphuric acid produces sulphuryl chloride (d) The structure of sulphuric acid in vapour phase is square planar 65. The formula of sodium thiosulphate is (a) Na2S2O3 · H2O (b) Na2S2O3 · 3H2O (c) Na2S2O3 · 5H2O (d) Na2S2O3 · 7H2O 66. Which of the following statements for sodium thiosulphate is not true? (a) Sodium thiosulphate is used as an antichlor in the textile industry

67. 68. 69.

70.

(c) Sodium thiosulphate acts as an oxidizing agent in its reaction with iodine (d) The structure of S2O 32– is tetrahedral The oxidation number of sulphur in sodium thiosulphate is (a) 0 (b) + 1 (c) + 2 (d) + 4 The dissolution of AgBr in S2O32– produces (b) Ag(S2O3)3– (c) Ag(S2O3)5– (d) Ag(S2O3)33– (a) Ag(S2O3)– 2 3 Which of the following statements regarding ozone is not true? (a) Ozone is an allotrope of oxygen (b) The ozone layer protects the earth’s surface from an excessive concentration of harmful ultraviolet radiation (c) The conversion of oxygen into ozone is an exothermic process (d) Ozone is much more powerful oxidizing agent than molecular oxygen Which of the following statements regarding ozone is not correct? (a) The ozone molecule is angular in shape (b) The ozone is a resonance hybrid of the two structures (c) The oxygen-oxygen bond length in ozone is identical with that of molecular oxygen

71. Which of the following oxidation state of oxygen is not observed in its compounds (a) –1/2 (b) –1 (c) +1 72. Which of the following facts regarding bond energies is correct? (a) eO=O < eS=S (b) eO – O < eS –S (c) eO–H < eS–H

(d) +2 (d) eO – C < eS –C

16.18 Complete Chemistry—JEE Main 10

73. The structure of TeCl4 Cl Te

(a) Cl

Cl

Cl

Cl

Te

(b) Cl

Cl

5s25p6)

(c)

Cl

Cl

Cl

Cl

Te

(d)

Cl

Te Cl

Cl

Cl Cl 74. The hybridization of S in SF4 is (b) sp3 (c) dsp3 (d) sp3d (a) dsp2 75. Silver bromide is soluble in sodium thiosulphate due to the formation of the complex (b) Na3[Ag(S2O3)2] (c) Na5[Ag(S2O3)3] (d) Na7[Ag(S2O3)4] (a) Na[Ag(S2O3)]

76. The structure of SO23– is (a) O

S

O

2–

O S

(b)

O

O

2–

O

(c)

(d)

O

S

2–

O

O

77. Given are the standard reduction potentials of a some species .51 V 0.084 V 0.44 V 2– SO2 æ0ææ Æ S4O2– Æ S8 6 æææÆ S2O3 æææ E° The value of E° will be about (a) 0.44 V (b) 0.54 V (c) 0.65 V (d) 0.76 V 78. Which of the following element will exhibit maximum tendency of forming pp–dp bonding in their oxoanions? (a) Si (b) P (c) S (d) Cl 79. Which of the following elements has a strongest tendency to form double bonds? (a) S (b) Se (c) P (d) Si 80. Which of the following oxide has acidic characteristics? (a) N2O (b) NO (c) CO (d) N2O3

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79.

(d) (a) (b) (d) (a) (c) (d) (b) (b) (d) (d) (c) (c) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80.

(c) (b) (c) (b) (d) (c) (a) (c) (b) (d) (c) (b) (d) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(c) (b) (a) (c) (b) (a) (c) (d) (c) (c) (d) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76.

(b) (a) (d) (a) (d) (b) (d) (a) (d) (c) (d) (c) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77.

(b) (d) (c) (b) (c) (c) (b) (b) (d) (d) (c) (c) (a)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78.

(d) (b) (c) (b) (b) (d) (c) (d) (b) (b) (c) (b) (d)

Study of the p-Block Elements (Groups 16, 17 and 18)

16.19

HINTS AND SOLUTIONS 1. Bismuth belongs to Group 15. 2. Atomic number of tellurium is 2 + 8 + 8 + 18 + 16 = 52. 6. The tendency of the elements of Group 16 to form multiple bonds to C, N and O decreases down the group. CO2 is very stable, CS2 is moderately stable, CSe2 decomposes readily and CTe2 does not exist. 9. Because of hydrogen bondings, the boiling point of H2O is greater than H2S. H2Se and H2Te have higher boiling points than H2S because of heavier atoms Se and Te. 10. The stability of hydrides of elements of Group 16 decreases down the group. H6TeO6 + 6HF 11. TeF6 is hydrolysed: TeF6 + 6H2O 13. SO3 exists in three forms: a-, b-, and g-sulphur trioxide. 14. The hybridization of S in SF4 is sp3d. Hence, it is trigonal bipyramidal. 16. H2S2O4 : H

O

O

S

S

O

O

O OH ;

H2Sn+2O6 HO

S

(S)n

O

H2S2O6 : HO

O

O

S

S

O

O

O

O OH ;

H2S2O7 HO

S

O O

O –2 O –1 +6

23. Peroxodisulphuric acid is

S

OH

O

–2 O –1

–1 +6

HO S O –2

S OH

O

–1

S OH

–2 O

O

The oxidation state of S is + 6. –2

O –1 +6

24. Peroxomonosulphuric acid is

HO

O

–1

S O

–2

–1

S 0

–2

O

–1 +4

25. Thiosulphuric acid (H2S2O3) is HO

OH

OH

S 26. See Q. 23. H2S2O8 + 2H + + 2e–. 27. 2H2SO4 31. Electronegativity decreases down the group. Hence, the hold of H in the bond H—M decreases down the group as the atom M becomes more and more metallic. H 2S 2O 7. 32. H2SO4 + SO3 40. Heavy water is obtained by the electrolysis of alkaline solution of water. 44. See Q. 31. 46. Fool’s gold is Fe3S2. Sphlaerite is ZnS. Chalcopyrite is CuFeS2. 47. Rhombic sulphur is stable at room temperature. 51. SO3 is the anhydride of sulphuric acid. 52. SO3 has a plane triangular structure. 64. The structure of H2SO4 is tetrahedral. 66. Sodium thiosulphate acts a reducing agent. – S4O2– 2S2O32– 6 + 2e

I2 + 2e–

2I –

16.20 Complete Chemistry—JEE Main

69. The conversion of O2 to O3 is an endothermic process. 70. The bond length in ozone is intermediate between those of single O —O and double O==O bonds. The bond angle is O3 is 117° with O —O distance 127.8 pm. 71. The oxidation state +1 is not observed in the compounds of oxygen. 72. Due to lone pair-lone pair repulsions on the neighbouring atoms, the single bond energies of O—O is smaller than that of S—S bond. 73. The structure of choice c represents the structure of TeCl4. 74. sp3d hybridization in S is involved in SF4. 75. Solubility of AgBr in Na2S2O3 is due to the formation of the complex Na5[Ag(S2O3)3]. 76. Sulphur in SO32– involves sp3 hybridization. There are a total of four pairs of electrons around S. 77. We have DG°1 = – 12F (0.44 V) 4SO2 + 4H+ + 6e– Æ S4O2– 6 + 2H2O] ¥ 2 S4O–6 + 2e– Æ 2S2O32–] ¥ 2 Add Hence

4S2O2– 3 +

24H +16e Æ S8 + 12H2O +

–

8SO2 + 32H+ + 32e– Æ S8 + 16H2O 32 E° = (12 ¥ 0.44 + 4 ¥ 0.08 + 16 ¥ 0.47) V E° = (14.12/32) V = 0.44 V

DG°2 = – 4F (0.08 V) DG°3 = – 16F (0.47 V) DG°4 = – 32F E°

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Which one of the following processes will produce hard water? (a) Addition of Na2SO4 to water (b) Saturation of water with CaCO3 (c) Saturation of water with MgCO3 (d) Saturation of water with CaSO4 2. Which of the following statements regarding sulphur is incorrect? (a) The oxidation state of sulphur is never less than +4 in its compounds. (b) S2 molecule is paramagnetic. (c) The vapour at 200 °C consists mainly of S8 ring. (d) At 600 °C, the gas mainly consists of S2. 3. Identify the incorrect statement from the following. (a) Oxides of nitrogen in the atmosphere can cause the depletion of ozone layer (b) Ozone absorbs the intense ultraviolet radiation of the sun

[2003]

[2011 cancelled]

(d) Ozone absorbs infrared radiation [2011] 4. Which of the following is the wrong statement? (b) O3 molecule is bent (a) ONCl and ONO– are not isoelectronic (c) Ozone is violet-black in solid state (d) Ozone is diamagnetic gas [2013] 5. Which of the following statements about the depletion of ozone layer is correct ? (a) The problem of ozone depletion is less serious at poles because NO2 consuming ClO radicals. (b) The problem of ozone depletion is more serious at poles because ice crystals in the clouds over poles act as catalyst for photochemical reactions involving the decomposition of ozone by Cl and ClO radicals. (c) (d) Oxides of nitrogen also do not react with ozone in stratosphere. [2014, online] 2+ 6. Consider the reaction: H2SO3(aq) + Sn4+(aq) + H2 (aq) + 3H+(aq) + HSO–4

Study of the p-Block Elements (Groups 16, 17 and 18)

7.

8.

9.

10.

16.21

Which of the following statements is correct ? (a) Sn4+ is the oxidizing agent because it undergoes oxidation (b) Sn4+ is the reducing agent because it undergoes oxidation (c) H2SO3 is the reducing agent because it undergoes oxidation [2014, online] (d) H2SO3 is the reducing agent because it undergoes reduction Photochemical smog consists of excessive amount of X, in addition to aldehydes, ketones, peroxy-acetyl nitrile (PAN), and so forth, X is: (b) CO (c) CO2 (d) O3 [2015, online] (a) CH4 Identify the incorrect statement regarding heavy water: (a) It reacts with SO3 to form deuterated sulphuric acid (D2SO4) (d) It is used as a coolant in nuclear reactors (c) It reacts with CaC2 to produce C2D2 and Ca(OD)2 (d) It reacts with Al4C3 to produce CD4 and Al(OD)3. [2016, online] Identify the incorrect statement: (a) The S — S — S bond angles in S8 and S6 rings are the same (b) Rhombic and monoclinic sulphur have S8 molecule (c) S2 is paramagnetic like oxygen [2016, online] (d) S8 ring has a crown shape. Which of the following statements about water is false? (a) Water can act both as an acid and as a base. (b) There is extensive intramolecular hydrogen bonding in the condensed phase (c) ice formed by heavy water sinks in normal water (d) Water is oxidized to oxygen during photo synthesis [2016]

ANSWERS 1. (c) 7. (d)

2. (a) 8. (b)

3. (d) 9. (a)

4. none 10. (b)

5. (b)

6. (c)

HINTS AND SOLUTIONS 2. The oxidation state of S in H2S is – 2, which is less than +4. Hence, the choice (a) is incorrect. 3. Ozone does not absorb infrared radiations. Hence, the choice (d) is correct. 4. The number of electrons in ONCl and ONO– are : 8 + 7 + 8 + 1=24 ONCl 8 + 7 + 17 = 32 ONO– Thus, these two species are not isoelectronic. Ozone is bent molecule, violet-black in colour in solid phase and diamagnetic in nature. Thus, all the four choices are correct and none of them is incorrect. 6. The reaction along with the oxidation number of species undergone oxidation and reduction is as follows.

7. Photochemical smong also contains O3.

16.22 Complete Chemistry—JEE Main

235

U.

9. The bond angle in S8 and S6 are different. S8 adopts a puckered ring or “Crown” structure. S6 adopts a chair form. 10. Water involves intermolecular hydrogen bondings.

The Group 17 Elements collectively known as halogens which means sea salts producers (from Greek, halos = sea salts, genes = born). Astatine is a short-lived radioactive element. The physical properties of these elements are recorded in Table 10.2. Table 2. Property

Atomic Properties of the Halogens

F

CI 2

[He](2s) (2p) Atomic number

5

9

Br 2

[Ne](3s) (3p)

5

10

I 2

[Ar](3d) (4s) (4p)

5

10

[Kr](4d) (5s)2(5p)5

17

35

Relative atomic mass

18.998

35.453

79.904

Covalent radius r/pm

72

99

114

133

133

184

196

220

Ionic radius, r/pm

4.0

Electronegativity

2.8

2.5

–1

1681

1256

1143

1009

–1

–333

–349

–325

–296

Ionization energy, l/kJ mol E/kJ mol

3.0

53 126.905

Description of Physical Properties ns)2(np)5, where n varies from 2 to 6. They have only one electron less than the number present in the adjacent noble gas. Thus, they show a strong tendency to complete metals to form ionic compounds or by sharing one electron with nonmetals to form covalent compounds. 2. The covalent and ionic radii increase down the group (Fig. 6). 3. The electronegativity and ionization energy decrease down the group (Fig. 7). The value of ionization energy

220

Radii

ionic

Ionization energies 1600

180

140

colvalent

100

60

Fig. 6

IE/kJ mol-1

r/pm

1400

1200

1000

F

Cl

Br

800

I

Covalent and Ionic Radii of the Elements of Group 17

Fig. 7

F

Cl

Br

I

Ionization Energies of the Elements of Group 17

Study of the p-Block Elements (Groups 16, 17 and 18) 350

EA/kJ mol-1

than Cl. This is due to the more repulsion experienced by the incoming atom. Molecular Properties All the halogens form diatomic molecules. Table 3 records some of the molecular properties of halogens.

16.23

Electron affinities

330

310

290

F

CI

Br

I

Fig. 8 Elements of Group 17

Table 3

Molecular Properties of the Halogens

Property

F2

Cl2

Br2

I2

Melting point, T/K

54

172

266

387

Boiling point, T/K

85

239

333

458

Density, liquid state, r/g cm

–3

1.513 (85 K)

Bond enthalpy, e/kJ mol

–1

Bond distance, l/pm Enthalpy of vaporization, DvapH/kJ mol

Enthalpy of fusion, DfusH/kJ mol E°/V

–1

–1

3.187

3.960

(203 K)

(273 K)

(493 K)

greenish yellow

reddish brown

violet

158.8

242.6

192.8

151.1

143

199

228

266

pale yellow

Colour

1.655

6.54

20.41

29.56

41.95

0.51

6.41

10.57

15.52

2.87

1.36

1.09

0.54

Description of Molecular Properties 1. The halogens are coloured beacuase they abosrb radiations in the visible region resulting in the excitation of outer electrons to higher energy levels. The smaller the atom, the more will be the energy needed for excitation. Gaseous F2 appears pale yellow as it absorbs high energy violet light. Gaseous Cl2 appears greenish yellow. Gaseous Br2 apperars reddish-brown. Gaseous I2 appears violet as it absorbs low energy yellow light 2. The density of halogens increases down the group. 3. The melting and boiling points increase down the group (Fig. 9). From the data on melting and boiling points, it follows that at room temperature F2 and Cl2 are gases Br2 exists in liquid form and I2 exists in solid form. 4. The bond enthalpy increases from F—F to Cl—Cl followed by a decrease (Fig. 10). Fluorine has exceptionally low bond enthalpy. This may be attributed to the lone pair repulsion which is maximum due to small size of atoms. Cl2 has a maximum value of bond enthalpy. This is because Cl—Cl bond acquires a double bond character due to the back bonding of p electrons on one chlorine atom to the empty d orbital of the second atom. 5. The bond distance increases down the group (Fig. 11). 6. The nonmetallic character of halogens decreases down the group. Iodine is associated with some metallic character. It has a metallic lusture and can exist as positive ion in ICl or ICN  ICl   I+ + Cl– + –  ICN   I + CN

16.24 Complete Chemistry—JEE Main Bond enthalpies

Melting and boiling points 500

240 b.pt

400

200

m.pt

Bond distances

200

100

0

Fig. 9

200

250

180

200

r/pm

ex-x/kJ mol-1

T/K

300

150

160

F2

Cl2

Br2

140

I2

Melting and Boiling Points of the Halogens

Fig. 10

F2

Cl2

Br2

100

I2

Bond Enthalpies of the Halogens

Fig. 11

F2

Cl2

Br2

I2

Bond Distances of Halogens

Reactivity of Halogens occurs as insoluble CaF2 3AIF6(cryolite) and Ca2(PO4 occur as chlorides and bromides in seawater and underground brines. Sodium chloride also occurs as rock salt. Some sea weeds and sponges contain iodine as iodides. Chile saltpeter (NaNO3) contains 0.02–1% iodine in the form of sodium iodate, NaIO3. The position of abundance of halogens and their concentrations in earth’s crust (in ppm) are given in Table 4. Table 4. Halogen Fluorine Chlorine Bromine Iodine

Distribution of Halogens in Earth’s Crust

Position of abundance in earth’s crust 13 20 46 60

Abundance in earth’s crust (ppm) 544 126 2.5 0.46

2. All the halogens exhibit –1 oxidation state. 3. Fluorine is the most electronegative element, it exhibits only –1 oxidation state. 4. The other halogens, besides exhibiting –1 oxidation state, also exhibit +1, +3, +5 and +7 oxidation states. This states are as follows. s +1 oxidation state ≠Ø

≠Ø

≠Ø

≠

≠Ø

≠Ø

≠

≠

+3 oxidation state

p

d

≠

 sp3d hybridization

+5 oxidation state

≠Ø

≠

≠

≠

≠

≠

≠

≠

 sp3d2 hybridization

+7 oxidation state

≠

≠

≠

≠

≠

 sp3d3 hybridization

5. The oxidizing ability of halogens follows the order

F2 > Cl2 > Br2 > I2.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.25

Fluorine has maximum oxidizing ability due to its low enthalpy of dissociation and maximum enthalpy of hydration. This fact also follows from their reduction potentials. The consequence of the above order is that a halogen of lower atomic number will oxidize halide ions of higher atomic number F2 + 2Cl– Æ 2F– + Cl2 Cl2 + 2Br– Æ 2C1– + Br2 Br2 + 2I– Æ 2Br– + I2 6. F2, being strongly oxidizing agent, it oxidizes water as 2F2 + 6H2O Æ 4H3O+ + 4F– + O2 A similar reaction is expected for chlorine, but the reaction is slow due to the high value of energy of activation. However, chlorine in water undergoes rapidly disproportionation reaction  Cl2 + H2O   HCl + HOCl The extent of disproportionation is still less for Br2 and almost negligible for I2. I2 is the weakest oxidizing agent. In this case, the reverse reaction is spontaneous, i.e. I– in acidic medium is oxidized by atmospheric oxygen 4I– + O2 + 4H+ Æ 2I2 + 2H2O

Compounds of Group 17 Elements 1. Hydrogen Halides All halogen react with hydrogen to give hydrogen halides. The reactivity of halogens towards hydrogen decreases down the group. Table 5 records some physical properties of hydrogen halides. Table 5.

Some Properties of Hydrogen Halides HF

HCl

HBr

HI

Melting point, T/K

190

158.9

186.1

222.2

Boiling point, T/K

293

188.1

206.4

237.6

Density/g cm

–3

0.99

1.19

3.2

pKa

Azeotropic composition (mass percent) Bond enthalpy/kJ mol

–1

35.37 566

Bond distance, l/pm

2.16

–7

–9

20.24

47.0

431

2.80 –11

57.0

366

299

100

170

Description 1. The melting and boiling points decrease from HF to HCl followed by an increase (Fig. 12). HF has exceptionally high melting and boiling points. This is attributed to the hydrogen bonds due to the very and the H atom of another molecule. The links the molecules together as (HF)n and they form zig-zag chains in both the liquid and the solid (Fig. 13). 300

260 b.pt

F

T/K

220

H

m.pt 180

140

Fig. 12

F

HF

HCI

HBr

F H

H F

F H

H F

HI

Melting and Boiling Points of Hydrogen Halides

Fig. 13

Hydrogen Bonding in HF

H F

16.26 Complete Chemistry—JEE Main

(X == Cl, Br, I)

Bond enthalpies 550

450

e/kJ mol-1

The hydrogen bonding even persists in gaseous HF (and not in HCl, HBr, HI) forming cyclic (HF)6, dimeric (HF)2, and monomeric HF. 2. In the gaseous state, all halides are covalent. In aqueous solutions they ionize as follows. + –  HF + H2O   H 3O + F HX + H2O æÆ H3O+ + X–

Thus, HF acts as a weak acid and HC1, HBr and HI behave as strong 350 acids. The aqueous solution of HX is known as hydrohalic acid or simply halogen acid. 3. In glacial acetic acid medium, the order of acid strength is 250 HF HCl HBr HI HI > HBr > HCl > HF This order is in agreement with the bond enthalpy data (Fig. 14). Fig. 14 Bond Enthalpies of Hydrogen 4. All halogen acids form azeotropic solutions with water. An azotropic Halides solution has a constant boiling point. Liquid HF has been used as a nonaqueous solvent. The mineral acids HNO3, H2SO4 and HCl behave as bases in HF solvent. The compounds SbF5, AsF5 and BF3 act as acids. The self ionization of HF is + –  2HF   [H2F] + F

2. Other Halides Halogens react with almost all elements of the periodic table, except helium, neon and argon, to form a vast variety of Ionic Halides Metals with low ionization energies (like alkali and alkaline earth metals) form ionic halides. When a metal exhibits more than one oxidation state, the halides in the lower oxidation states have more ionic character. For example, PbCl2 is ionic while PbCl4 is covalent. The solubilities of ionic halides usually increase from F– to Br– to Cl– to I–. This is due to the decrease in the value of lattice energy as the ionic radii increase. Covalent (or Molecular) Halides Metals with high ionization energies form covalent halides. When a metal exhibits more than one oxidation state, the halides in the higher oxidation states have more covalent character. For example, UF6 is covalent while UF4 is ionic. Most of electronegative elements also form covalent halides. Examples are BCl3, PCl3, PCl5, CCl4, etc. A large number of covalent halides formed from nonmetals are hydrolyzed in aqueous solutions. PCl3 + 3H2O Æ H3PO3 + 3H+ + 3Cl– BCl3 + 3H2O Æ H3BO3 + 3H+ + 3Cl– + – SiCl4 + 4H2O Æ Si(OH)4 + 4H + 4Cl PCl5 + 4H2O Æ H3PO4 + 5H+ + 5Cl– Most of molecular halides are gases or volatile liquids. This is because of the strong covalent bonds within the molecules and different molecules are held together by weak van der Waals forces. When a nonmetal, E, reacts with halogen, the bond strength decreases in the order E—F > E—Cl > E—Br > E—I. Bridging Halides In these halides, halogen forms bridge between two atoms. Examples are AlCl3, BeF2 and BeCl2. The bridge is formed due to one normal covalent bond with one of the atoms and one coordinate covalent bond with another atom. This coordinate bond is formed by sharing a lone pair of electrons on halogen with another atom. Both the bridging bonds are completely identical. 3. Halogen Oxides while the compounds formed by other elements are called halogen oxides. Table 6 describes the various compounds formed between halogens and oxygen.

Comments 1. Because of small difference in electronegativity between the halogens and oxygen, the bonds are largely covalent with the exception of I2O4 and I4O9 which are ionic.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.27

2. Most of the halogen oxides are unstable, and tend to explode when subjected to shock or sometimes even when exposed to light. 3. The stability of oxides increases down the group. Thus, the iodine oxides are the most stable, then the chlorine oxides and then bromine oxides (which decompose below room temperature). 4. The oxides with the higher oxidation states of halogens are more stable than the lower states. Table 6.

Compounds of Halogens with Oxygen

Oxidation state of halogen

Oxygen

Chlorine oxides

Bromine oxides

Iodine oxides

–1

OF2, O2F2

—

—

—

+1

—

Cl2O

Br2O

—

+3

—

Cl2O3

—

+4

—

ClO2

BrO2

I 2O 4

+5

—

—

—

I 2O 5

+6

—

Cl2O6

—

—

+7

O 4F 2

Cl2O7

—

I 4O 9

A Brief Description of Important Oxides 2F2 + 2NaOH Æ 2NaF + OF2 + H2O. It is a colourless gas and is a strong oxidizing agent. It has been used as a rocket fuel.

Dichlorine Monoxide diluted with dry air.

This is prepared by heating freshly precipitated (yellow) mercuric oxide with halogen gas 575 K

2Cl2 + 2HgO æææÆ HgCl2 ◊ HgO + Cl2O It is a yellow-brown gas which explodes in the presence of reducing agents or NH3, on heating. 3Cl2O + 10NH3 Æ 6NH4Cl + 2N2 +3H2O C12O is acidic anhydride of hypochlorous acid.  Cl2O + H2O   2HOCl Structures of F2O and Cl2O The Lewis dot structures of F2O and C12O (valence electrons: 20) are

There are four unpaired electrons around oxygen, so the geometry adopted by these electrons is tetrahedral as predicted from VSEPR theory. The same structures are predicted from the valence-bond theory. s Electronic structure of oxygen atom in free state Electronic structure of oxygen atom in F2O or Cl2O

p

≠Ø

≠Ø

≠

≠

≠Ø

≠Ø

≠

≠

 sp3 hybridization, tetrahedral geometry

Thus, in F2O and Cl2O two positions of a tetrahedron is occupied by halogens and the rest two positions by lone pairs. The expected bond angle X—O—X is 109°28¢. But, the observed values are F—O—F 103° Cl—O—Cl 111° The decrease in bond angle in F2O has been attributed to the repulsion between the lone pairs. The increase in bond angle in Cl2O has been attributed to the removal of steric crowding of the larger chlorine atoms.

O X X

16.28 Complete Chemistry—JEE Main

Chlorine Dioxide This is most important oxide of chlorine. It is a yellow gas. Being highly reactive, ClO2 is prepared in situ and is used after diluting with air or CO2. Laboratory Preparation H2O ææ Æ 2ClO2 + 2CO2 + (COONa)2 + 2H2O. 2NaClO3 + 2(COOH)2 æ90 ∞C Commercial Preparation trace of

Æ 2ClO2 + 2NaHSO4 2NaClO3 + SO2 + H2SO4 ææææ NaCl 2HClO3 + 2HC1 Æ 2ClO2 + Cl2 + 2H2O. ClO2 is a powerful oxidizing, bleaching and chlorinating agent. It is used for the The ClO2 is an odd electron molecule and is thus paramagnetic. The odd electron is delocalized over the entire molecule and thus ClO2 does not show any tendency for dimerization. The bond length Cl—O is 147 pm and it is shorter than for a single bond. This indicates the delocalization of odd electron on the entire molecule.

Dichlorine Hexoxide

This oxide exists in equilibrium with ClO3.  Cl2O6   2ClO3 Cl2O6 is made from ClO2 and O3. It is a strong oxidizing agent. It undergoes hydrolysis as shown in the following reactions. Cl2O6 + H2O Æ HClO3 + HClO4 Cl2O6 + 2NaOH Æ NaClO3 + NaClO4 + H2O Its structure is not known. Probable structures are O O O

Dichlorine Heptoxide

Cl

Cl

O O O

O

O Cl

or O

O Cl

or

O

ClO+2 ClO-4

O

It is obtained by dehydration of perchloric acid with P4O10. PO

2 5 2HClO4 æ-ææ Æ Cl2O7. H O 2

O

O 141 pm O 171 pm O 119° Cl 115°

O Cl O

O

Oxides of Bromine Two oxides are Br2O and BrO2. Br2O does not give HOBr on treating with water, but with NaOH it gives OBr–. It is a strong oxidizing agent, and oxidizes I2 to I2O5. BrO2 is hydrolyzed in alkaline solution giving Br– and BrO –3 ions. Oxides of Iodine

Iodine form stable oxides with oxygen. I2O5 is an anhydride of iodic acid. I2O5 + H2O Æ 2HIO3. It is an oxidizing agent. It is used to estimate carbon monoxide quantitatively based on the following reaction. I2O5 + 5CO Æ 5CO2 + I2 The liberated iodine is determined by titrating against sodium thiosulphate solution. I2O5 also oxidizes H2S to SO2 and NO to NO2 2, BrF3 or SF4 it forms IF5 2I2O5 + 10F2 Æ 4IF5 + 5O2 O

The structure of I2O5 is

O I

O

O

I O

The other oxides I2O4 and I4O9 are less stable. Their probable structures are IO+◊IO3– and I3+◊(IO3– )3, respectively.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.29

4. Oxoacids of Halogens Halogens form four series of oxoacids of the type HOX (hypohalous acid), HXO2 (halous acid), HXO3 (halic acid) and HXO4(perhalic acid). Table 7 records the oxoacids formed by halogens. Table 7 HOX

Oxoacids Formed by Halogens

HXO2

HXO3

HXO4

HOF HClO3

HClO4

HOBr

HBrO3

HBrO4

HOI

HIO3

HIO4

HOCI

HClO2

All oxoacids have tetrahedral structures as follows. -

X

-

-

X

X O

O

OX

-

X

O

O

O

O

O

OX -2

-

O

XO -3

O O XO -4

Hypohalous Acids Amongst the four hypohalous acids, HOCl is the most stable acid. The most unstable acid is HOF which has been made using special techniques. HOCl, HOBr and HOI can be obtained by shaking freshly precipitated HgO in water with appropriate halogen. 2HgO + H2O + 2X2 Æ HgO ◊ HgX2 + 2HOX All hypohalous acids are very weak acids. They are good oxidizing agents especially in acidic medium. They are known only in aqueous solution. Sodium salt of hypochlorous acid (NaOCl) is well known, and is used for bleaching cotton fabrics. It is also used as a disinfectant and sterilizing agent. The hypohalite ions can be generated by passing halogens in alkali solution: X2 + 2NaOH Æ NaX + NaOX + H2O – However, the OX ions tend to disproportionate in basic medium, rate of disproportionation increases with temperature. 3OX– Æ 2X– + XO3– Halous Acids Chlorous acid is the only halous acid known so far. This exists only in solution. Its salts are known as chlorites. It is a weak acid, but stronger than hypochlorous acid. Chlorus acid can be prepared by treating barium chlorite with H2SO4. Ba(ClO2)2 + H2SO4 Æ 2HClO2 + BaSO4. Sodium chlorite can be prepared by passing ClO2 through NaOH or Na2O2 solutions. 2ClO2 + 2NaOH Æ NaClO2 + NaClO3 + H2O Chlorite

Chlorate

2ClO2 + Na2O2 Æ 2NaClO2 + O2. Chlorite ions are stable in basic solution but undergo disproportionation in acidic solution. 5HClO2 Æ 4ClO2 + HCl + 2H2O

Halic Acids Three halic acids are known: HClO3, HBrO3 and HIO3. The acids HClO3 and HBrO3 are known only in solution. The acid HIO3 exists as a white solid. Both HClO3 and HBrO3 decompose if evaporated to dryness. 4HClO3 Æ 4ClO2(g) + 2H2O(g) + O2(g) The halic acids are strong oxidizing agents. HIO3 is made by oxidizing I2 with concentrated HNO3 or O3 I2 + 10HNO3 Æ 2HIO3+ 10NO2 + 4H2O

16.30 Complete Chemistry—JEE Main

HClO3 and HBrO3 are made by adding H2SO4 to barium halates. Ba(ClO3)2 + H2SO4 Æ BaSO4 + 2HClO3 The salt of chloric acid can be made by passing Cl2 into a hot solution of NaOH. ∞C 6NaOH + 3Cl2 æ80 ææ Æ NaClO3 + 5NaCl + 3H2O. Perhalic Acids Three perhalic acids are known: HClO4, HBrO4 and HIO4. They are the strongest acids, very powerful oxidizing agents, and explods in contact with organic matter. Perchloric acid can be obtained by the action of concentrated hydrochloric acid on sodium perchlorate. The latter is obtained by the electrolytic oxidation of chlorates. electrolysis NaClO3 + H2O ææææÆ NaClO4 + H2. Perbromates can be obtained by the action of powerful oxidizing agents such as F2 or XeF2 on bromates in aqueous solution. KBrO3 + F2 + 2KOH Æ KBrO4 + 2KF + H2O RbBrO3 + XeF2 + H2O Æ RbBrO4 + 2HF + Xe The common form of periodic acid is HIO4 ◊ 2H2O, i.e. H5IO6. This is prepared by oxidizing iodates by Cl2 in alkaline medium. – IO3– + 6OH– + Cl2 Æ IO 5– 6 + 3H2O + 2Cl This can also be prepared by electrolytic oxidation. – IO –3 + 6OH– Æ IO 5– 6 + 3H2O + 2e Heating H5IO6 at 100 °C gives HIO4. On strong heating, HIO4 decomposes giving I2O5 and O2. 100 ∞C

∞C Æ 2HIO4 æ200 ææÆ I2O5 + O2 + H2O. 2H5IO6 æ-æææ 4H O 2

Strengths of Oxoacids The strength of oxoacids depends on the number of oxygen atoms attached to the halogen. The more oxygen atoms that are bonded to the halogen, the more the electrons will be pulled away from the O—H bond and the more this bond will be weakend. Hence, the order of easiness with which the hydrogen ions can be removed from acids is O3Cl—OH > O2Cl—OH > OCl—OH > Cl—OH i.e.

HClO4 > HClO3 > HClO2 > HOCl.

Reason for not Existing HFO2, HFO3 and HFO4 All the oxoacids have tetrahedron structures. The sp3 hybrid orbitals of halogens form weak s bonds with oxygen atoms, because the energies of s and p orbitals of halogens differ appreciably. The ions of these acids are stabilized by strong pp—dp bonding between the full 2p orbitals on oxygen with emptly d orbitals on the halogen atoms. Despite this, many of the oxoacids are known only in solution. Fluorine does not have d orbitals and thus cannot form pp—dp of HOF, does not form oxoacids. 5.7 Interhalogen Compounds Each halogen reacts with every other halogen to form interhalogens or interhalogen compounds. Table 8 records the types of interhalogens formed by halogens. Table 8 F

Cl

ClF 1.0 ClF3 ClF5

1.2

Interhalogen Compounds†

BrF BrF3 BrF5 1.5

Br

(IF)* IF3 IF5 IF7

0.2

BrCl

ICl 0.5 (ICl3)2 0.3 IBr

I †

The numbers on vertical lines are the difference in electronegativity values.

* Unstable

Study of the p-Block Elements (Groups 16, 17 and 18)

16.31

Comments 1. There are four types of interhalogens, AX, AX3, AX5 and AX7. 2. The compounds AX and AX3 are formed where the electronegativity difference is not very large. 3. The compounds AX5 and AX7 are shown by large atoms Br and I surrounded by small atom F. Also, the difference of electronegativity is large. 4. The compound IF is not stable. It involves the maximum difference in electronegativity and the least number of F atom. On increasing F atom, the stability of IFx is increased. 5. There are never more than two different halogens in a molecule. Preparations The interhalogens can be prepared by direct combination of the halogens. The type of interhalogen formed depends on conditions. 200 ∞C I2 + Cl2 æÆ 2IC1 Cl2 + F2 æææÆ 2CIF (equal volumes)

(equimolar)

Cl2 + 3F2 æææÆ 2ClF3

I2 + 3Cl2 æÆ (ICl3)2

Br2 + 3F2 æÆ 2BrF3

Æ 2IF5 I2(s) + 5F2 æææ

300 ∞C

(excess)

(excess)

20 ∞C

250 - 300 ∞C

I2(s) + 7F2 æææææ Æ 2IF7

(diluted with nitrogen)

Br2 + 5F2 æÆ 2BrF5 (excess)

General Characteristics 1. The bonds between interhalogens are covalent because of the small difference in electronegativity values. 2. The melting and boiling points increase as the difference in electronegativity increases. 3. The interhalogen are more reactive than the halogens (except F2). This is because the bond A—X is weaker than A—A and X—X bonds. 4. Hydrolysis of interhalogen gives halide and oxohalide ions. The latter is formed from the larger halogen present in the interhalogen. 5. Liquids ICl3, BrF3, and IF5 have appreciable conductivity. This has been explained on the basis self-ionization. + –  2ICl3   [ICl2] + [ICl4] + –  2IF5   [IF4] + [IF6] + –  2BrF3   [BrF2] + [BrF4] Structures of Interhalogens In AX3, the number of valence electrons is 4 ¥ 7 = 28. The Lewis structure of AX3 is

X—A—X —

X orientations. Experimental studies have shown that AX3 is T-shaped. X In general, in trigonal bipyramidal orientation, lone paris are always present in equatorial positions. X Gaseous state Bond angle 87°40¢ A Equaterial A—X bonds (those in the triangle) = 159.8 pm Apical A—X bonds (those pointing up and down) = 169.8 pm The structure is distorted trigonal bipyramidal since bond angle is 87°40¢. The decrease of bond angle from 90° to 87°40¢ is attributed to the lone pair-lone pair repulsion. Also note that X the apical bond distance is larger than the equational bond distance. e molecule AX3 are as follows. A in free state A in the AX3 molecule

≠Ø

≠Ø

≠Ø

≠

≠Ø ≠Ø ≠ ≠ ≠  sp3d hybridization, trigonal bipyramidal geometry

16.32 Complete Chemistry—JEE Main

In AX5 molecules, there are 6 ¥ 7 = 42 valence electrons. In the Lewis structure there are six electron pairs around the central atom X. According to the VSEPR theory, these electrons will acquire octahedral orientation. Thus, the structure of AX5 is X

X

X

X

X

A

A

X

X X

X

X

Lewis structure In general, in octahedral orientation, loan pairs are present in axial positions. molecule are as follows. s A in free state A in AX5

p

d

≠Ø

≠Ø

≠Ø

≠

≠Ø

≠

≠

≠

≠

≠

 sp3d2 hybridization, octahedral geometry

In AX7, there are 7 ¥ 8 = 56 valence electrons. In the Lewis structure of the molecule there are seven unpaired electrons around the central atom A. These seven pairs will acquire pentagonal bipyramidal orientations. Thus, the structure of IF7 is F

X X A

X X

F

X X

F I

F

F F

X

F

follows s I in free state I in IF7

p

d

≠Ø

≠Ø

≠Ø

≠

≠

≠

≠

≠

≠

≠

≠

 sp3d3 hybridization, pentagonal bipyramidal structure

In the gaseous phase, ICl3 is decomposed to IC1 and Cl2. So, its structure is not known. In solid state, 1Cl3 dimerizes to (ICl3)2. Two T-shaped ICl3 molecules join together, forming a planar dimeric molecule (IC13)2. Cl

Cl I

Cl

Cl I

Cl

Cl

Anomalous Behaviour of Fluorine Fluorine differs from rest of the members of its group because of its small size, high electronegativity and nonavailability of d orbitals in the valence shell.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.33

1. The oxidation state of F in its compound is always –1 while other elements exhibit a large number of oxidation states (–1, +1. +3, +5, +6, +7). 2. Fluorine is the most reactive among halogens due to its low bond enthalpy. The latter arises due to the larger repulsion between the nonbonding electrons in the compact molecule. 4. Fluoride exhibits a large tendency to form complex ions ([AlF6]3–, [FeF6]3–) due to its small size. The other halides have much less tendency for complexation. 5. In its compound with hydrogen, it exhibits extensive hydrogen bondings. 7. HF is the weakest acid in aqueous solution, while HCl, HBr and HI are stronger acids. 8. In its salt with silver ions. AgF is sloble in water while other halides are insoluble.

MULTIPLE CHOICE QUESTIONS 1. The number of halogens known so far is (a) 4 (b) 5 (c) 6 2. The atomic number of astatine which belongs to six period is (a) 53 (b) 54 (c) 85 3. Which of the following is radioactive halogen? (a) Cl (b) Br (c) I

(d) 7 (d) 86 (d) At

(a) F < Cl < Br (b) F > Cl > Br (c) F > Cl < Br (d) F < Cl > Br 5. Which of the following orders of bond dissociation enthalpy of halogens is correct? (a) F—F > Cl—Cl > Br—Br (b) F—F < Cl—Cl < Br—Br (c) F—F > Cl—Cl < Br—Br (d) F—F < Cl—Cl > Br—Br 6. Which of the following halogens does not exhibit positive oxidation state? (a) F (b) Cl (c) Br (d) I 7. Which of the following orders of melting point of hydrides of halogens is correct? (a) HF > HCl > HBr (b) HF < HCl < HBr (c) HF > HCl < HBr (d) HF < HCl > HBr (a) Fluorspar (b) Cryolite (c) Fluoroapatite (d) Chile salt peter 9. Which of the halogen is most abundant in the earth’s crust? (a) F (b) Cl (c) Br (d) I 10. Which of the following orders of bond strengths between a nonmetal E and halogens is correct? (a) e (E—F) > e (E—Cl) > e (E—Br) (b) e (E—F) > e (E—Cl) < e (E—Br) (c) e (E—F) < e (E—Cl) < e (E—Br) (d) e (E—F) < e (E—Cl) > e (E—Br) 11. Which of the following reactions is quite violent in nature? (c) Between hydrogen and bromine (d) Between hydrogen and iodine 12. The strength of halogen acids in water follows the order (a) HF > HCl > HBr (b) HF < HCl < HBr (c) HF > HCl < HBr (d) HF < HCl > HBr 13. Which of the following statements is not correct? (a) Halides formed with nonmetals are covalent in nature (b) The reactivity of halogens decreases with increasing atomic number (c) Hydrogen halides are ionic molecules in the gaseous phase (d) Metals with low ionization energies form ionic halides while those of high ionization energies form covalent molecules

16.34 Complete Chemistry—JEE Main

14. Which of the following statements is not correct? (b) For a metal exhibiting more than one oxidation state, the halides in the lower oxidation state is more covalent than the one in the higher state (c) Halogens do not occur freely in nature (d) Most of binary compounds between oxygen and halogens are unstable 15. Which of the following halogen oxides are ionic in nature? (a) Cl2O (b) Br2O (c) BrO2 (d) I2O4 16. Which of the following interhalogens does not exist? (a) ClF (b) ClF2 (c) ClF3 (d) ClF5 (a) XeF2 (b) XeF4 18. Which of the following represents hypohalous acid? (a) HOX (b) HXO2 19. Which of the following represents halous acid? (a) HOX (b) HXO2 20. Which of the following represents halic acid? (a) HOX (b) HXO2 21. Which of the following represents perhalic acid? (a) HOX (b) HXO2 21. Which of the following statements is not correct?

23. 24. 25.

26. 27. 28. 29.

(c) XeF6

(d) XeF8

(c) HXO3

(d) HXO4

(c) HXO3

(d) HXO4

(c) HXO3

(d) HXO4

(c) HXO3

(d) HXO4

(b) Hypohalous acids HOCl, HOBr and HOI are all strong acids (c) Hypochlorous acid is the most stable among hypohalous acids (d) Chlorous acid is the only halous acid known Which of the interhalogens is unstable? (c) IF5 (d) (a) IF (b) IF3 Which of the following exhibits +7 oxidation state in interhalogen compounds? (a) F (b) Cl (c) Br (d) The structure of IF7 is (a) planar (b) distorted octahedral (c) pentagonal bipyrimadal (d) distorted trigonal pyrimidal Iodine in IF7 involves (a) s2p3d2 hybridization (b) sp3d3 hybridization (c) p3d4 hybridization (d) The anhydride of HClO4 is (a) ClO–4 (b) Cl2O7 (c) ClO2 (d) Which of the following molecules is not paramagnetic? (a) ClO2 (b) ClF3 (c) ClO3 (d) Hypobromous acid is (a) HOBr (b) HOBrO (c) HOBrO2 (d)

IF7 I

p2d5 hybridization ClO3 BrO2 HOBrO3

(a) is a weak acid (b) is a strong acid (c) does not exist (d) is a good oxidizing agent 31. Which of the following exhibits the lowest bond energy? (a) HF (b) HCl (c) HBr (d) HI 32. Which of the following has a pyramidal structure? (a) ClO2 (b) ClO2– (c) ClO–3 (d) ClO–4 33. Which of the following forms the most basic 0.1 M solution? (a) NaCl (b) NaOCl (c) NaClO2 (d) NaClO3

Study of the p-Block Elements (Groups 16, 17 and 18)

34. The shape of the XeF4 molecules is (a) square planar (b) tetrahedral

36. 37. 38. 39.

40. 41. 42. 43. 44.

(c) square pyramidal

(d) trigonal bipyramidal

(a) XePt F6 (b) Xe(RuF6)2 (c) XeOF4 (d) The shape of XeOF4 molecule is (a) square pyramidal (b) tetrahedral (c) distorted tetrahedral (d) The F—Xe—F bond angle in XeF4 is (a) 109° (b) 103° (c) 90° (d) Which of the following exhibits the largest electrical conductivity in the liquid state? (a) F2 (b) Cl2 (c) Br2 (d) Fluorine is highly reactive because (a) F—F bond energy is high (b) F—F bond energy is low (c) it is gaseous at room temperature (d) F has a smaller size Which of the following molecules has minimum dipole moment? (a) HF (b) HCl (c) HBr (d) The strongest reducing agents amongst the halides is (a) F– (b) Cl– (c) Br– (d) BrF3 molecule is (a) planar trigonal (b) tetrahedral (c) T-shaped (d) The correct order of stability of the halous acids in aqueous solution is (a) HClO < HBrO < HIO (b) HClO > HBrO < HIO (c) HClO > HBrO > HIO (d) The strongest Bronsted base is (a) ClO– (b) ClO–2 (c) ClO–3 (d)

XeOF2 irregular 60° I2

HI I– trigonal pyramidal HClO < HBrO > HIO ClO–4

(a) Cryolite (b) Feldspar (c) Fluorspar (d) Fluoroapatite 46. Which of the following contains chlorine with oxidation number +1? (a) ICl (b) Cl2O (c) HCl (d) HClO2 47. Which of the following is not a polar molecule? (a) HCl (b) NI3 (c) Cl2O (d) CCl4 48. Which of the following will liberate Br2 from KBr? (a) Cl2 (b) HI (c) I2 (d) SO2 49. The order of strength of oxyacids of chlorine is (a) HClO < HClO2 < HClO3 < HClO4 (b) HClO > HClO2 > HClO3 > HClO4 (c) HClO > HClO2 > HClO3 < HClO4 (d) HClO < HClO2 < HClO3 > HClO4 50. The shape of XeO3 is (a) triangular planar (b) tetrahedral (c) triangle pyramidal (d) square planar 51. The salts of halogens are extensively found in seawater with the exception of that of

(a) violet light (b) red light (c) green light (d) orange light 53. The gaseous iodine molecules absorb yellow light from the visible region. Its colour would be (a) violet (b) red (c) green (d) yellow (a) F2 > Cl2 > Br2

(b) F2 < Cl2 < Br2

(c) F2 > Cl2 < Br2

55. Which of the following statements regarding halogens is not true? (a) Ionization energy decreases with increase in atomic number (b) Electronegativity decreases with increase in atomic number (d) Enthalpy of fusion increases with increase in atomic number

16.35

(d) F2 < Cl2 > Br2

16.36 Complete Chemistry—JEE Main

56. Which of the following halogens does not show positive oxidation states? (a) Fluorine (b) Chlorine (c) Bromine (d) Iodine 57. Which of the following orders regarding the boiling point of halogens is true? (a) HF > HCl > HBr > HI (b) HF < HCl < HBr < HI (c) HF > HCl < HBr < HI (d) HF > HCl > HBr < HI 58. Which of the following statements regarding halogens is not true? (a) Halogens act as strong oxidizing agents. (b) Fluorine oxidizes water to oxygen and ozone (c) Chlorine oxidizes water to oxygen (d) Bromine disproportionates in water producing the reaction Br2 + H2O HBr + HOBr 59. Which of the following equations is not shown by the appropriate halogen? 2F – + H2O + OF2 (a) F2 + 2OH – 2% solution hot

(b) 2F2 + 4OH – ææÆ 4F– + 2H2O + O2 conc.

hot

(c) X2 + 2OH – ææÆ H2O + X– + XO–; where X is Cl, Br and I conc.

light

(d) Xe + F2 æææ Æ XeF2 60. Which of the following reactions would not proceed to right hand side? (b) Cl2 + Br– Æ (c) Br2 + I– Æ (d) I2 + Cl– Æ (a) F2 + Cl– Æ 61. Which of the following reactions regarding the preparation of hydrogen halide is not correct? (b) NaCl + H2SO4 Æ NaHSO4 + HCl (a) CaF2 + H2SO4 Æ CaSO4 + 2HF conc.

(c) NaBr + H2SO4 Æ NaHSO4 + HBr conc.

conc.

(d) PI3 + 3H2O Æ H3PO3 + 3HI

62. Which of the following orders regarding the acid character of hydrogen halides in the aqueous medium is correct? (a) HCl > HBr < HI (b) HCl < HBr > HI (c) HCl > HBr > HI (d) HCl < HBr < HI (a) Fluorine (b) Chlorine (c) Bromine (d) Iodine 64. Which of the halogens is found in DDT? (a) Fluorine (b) Chlorine (c) Bromine (d) Iodine 65. Which of the following interhalogens does not exist? (a) ClF (b) ClF2 (c) ClF3 (d) IF7 66. When chlorine reacts with cold and dilute solution of sodium hydroxide the products obtained are (a) Cl– + ClO– (b) Cl– + ClO–2 (c) Cl– + ClO–3 (d) Cl– + ClO–4 67. When iodine reacts with a hot and concentrated solution of sodium hydroxide, the products obtained are (a) I– + IO– (b) I– + IO3– (c) I– + IO–4 (d) I– + I2O94– 68. Which of the following halogens is the strongest oxidizing agent? (a) Fluorine (b) Chlorine (c) Bromine (d) Iodine 69. Which of the following interhalogen is not observed? (a) AX (b) AX3 (c) AX5 (d) AX6 70. The structure of interhalogen AX3 is (a) triangular planar (b) pyramidal (c) T-shaped with two lone pairs of electrons at the equilateral positions (d) tetrahedral with a single electron

Study of the p-Block Elements (Groups 16, 17 and 18)

71. Which of the following is anhydride of hypochlorous acid? (a) Cl2O (b) ClO2 (c) 72. Which of the following is anhydride of perchloric acid? (a) Cl2O (b) ClO2 (c) – 73. The hybridization of I in ICl 2 is (a) sp (b) sp3 (c) – 74. The hybridization of I in ICl 4 is (a) sp3 (b) dsp2 (c) 75. Which of the following is not a pseudohalogen? (a) CN– (b) SCN– (c) 3 3 76. The hybridization sp d leads to (a) octahedral geometry (b) (c) pentagonal bipyramid structure (d) 77. The structure of ClF3 is

F

F

(a)

F

Cl F

(d) Cl2O7

Cl2O7

(d) Cl2O7

d2sp3

(d) sp3d

sp3d2

(d) d2sp3

OCN–

(d) NO+

hexagonal pyramid structure square geometry F

F

Cl

(b)

Cl2O6

(c)

F

F

Cl

Cl

(d) F

F

78. Which of the following interhalogens is colourless? (a) ClF(g) (b) BrF(g) (c) BrCl(g) 79. Which of the following elements does not exhibit basic properties? (a) Fluorine (b) Chlorine (c) Bromine 80. In the reaction ClF + Cl2 + SbF5 Æ [Cl3] [SbF6] (a) Cl2 acts as a Lewis acid (b) Cl2 acts as a Lewis base (c) Cl2 is an amphoteric species (d)Cl2 acts as a reducing agent

F

F

(d) ICl(s) (d) Iodine

- 78 ∞C

81. In the reaction [Ph4As][Cl] + Cl2 æææÆ [Ph4As][Cl3] (a) Cl2 acts as a Lewis acid (c) Cl2 is an amphoteric species 82. Interhalogen compounds are (a) more reactive than halogens (c) equally reactive as halogens

(b) Cl2 acts as a Lewis base (d) Cl2 acts as an oxidizing agent. (b) less reactive than halogens (d) inert substances

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(b) (c) (c) (b) (c) (d) (c) (c) (a) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(c) (d) (b) (c) (b) (c) (d) (a) (c) (a)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(d) (a) (d) (d) (b) (b) (b) (b) (d) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(d) (a) (b) (b) (b) (a) (a) (b) (a) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(d) (a) (d) (a) (b) (a) (d) (d) (a) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(a) (b) (a) (d) (a) (a) (c) (a) (d) (d)

16.37

16.38 Complete Chemistry—JEE Main

61. 67. 73. 79.

(c) (b) (d) (a)

62. 68. 74. 80.

(d) (a) (c) (b)

63. 69. 75. 81.

(a) (d) (d) (a)

64. 70. 76. 82.

(b) (c) (c) (a)

65. (b) 71. (a) 77. (c)

66. (a) 72. (d) 78. (a)

HINTS AND SOLUTIONS 2. The atomic number of astatine is

2 + 8 + 8 + 18 + 18 + (14 + 17) = 85

5. Cl— Cl bond is the strongest. 6. F is the most electronegative atom. 7. Because of hydrogen bonding, HF has higher melting point than HCl, which, in turn, has smaller melting point than HBr. 8. Chile saltpeter is NaNO3. 13. Hydrogen halides are covalent molecules in the gaseous phase. 14. The halide in the lower oxidation state are less covalent than the one in the higher. 22. Hypohalous acids are all weak acids. H O

2 27. 2HClO4 æææ Æ Cl2O7

28. ClF3 contains even number of electrons whereas rest of the three contain odd number of electrons. 33. HOCl is the weakest acid, OCl– undergoes maximum hydrolysis. 56. Fluorine being most electronegative always exhibits the oxidation state of –1. 57. Because of small size and highly electronegative nature of F, there occurs extensive hydrogen bonding in HF. It is because of this, melting and boiling points of HF are greater than those of HCl. 58. The reaction of Cl2 with water is Cl2 + H2O HCl + HOCl 59. The reaction of halogens (except F2) with hot and concentrated alkali is hot

3X2 + 6OH – ææÆ 3H2O + 5X– + XO3– conc.

60. F2 can replace 2Cl – to Cl2, Cl2 can replace 2Br– to Br2 and Br2 can replace 2I– to I2. 61. Sulphuric acid oxidizes HBr (or HI) to Br2 (or I2). The halides HBr and HI are prepared by hydrolysis of phosphorus trihalides with cold water PX3 + 3H2O

H3PO3 + 3HX

(X = Br or I)

62. The order of electronegativity of halogens is F > Cl > Br > I Hence, acid strength follows the reverse order.

–

73. There are 3 ¥ 7 + 1 = 22 valence electrons in ICl2–. These are distributed as 3

Cl

d.

74. There are 5 ¥ 7 + 1=36 valence electrons in ICl4–. These are distributed as There are six pairs of electrons around I. It hybridization will be sp3d2 75. NO+ is not a pseudo halogen? 76. The hybridization sp3d3 leads to pentagonal bipyramid distribution of electrons.

I Cl

Study of the p-Block Elements (Groups 16, 17 and 18)

77. There are 4 ¥ 7 = 28 valence electrons in ClF3. These are distributed as

16.39

F Cl F

F Hence, its hybridization is sp3d which leads to trigonal bipyramid structure. The actual structure is T-shaped as shown by the choice (c). 78. ClF(g) is colourless. 79. Fluorine is the most electronegative element. It does not exhibit basic properties.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

(c) hydration enthalpy

(d) bond dissociation energy

[2004]

2. The smog is essentially caused by the presence of (a) O2 and O3 (b) O2 and N2 (c) oxides of sulphur and nitrogen (d) O3 and N2 [2004] 3. The correct order of the thermal stability of hydrogen halides (H—X) is (a) HI > HBr > HCl > HF (b) HF > HCl > HBr > HI (c) HCl < HF < HBr < HI (d) HI > HCl < HF < HBr [2005] 4. What products are expected from the disproportionation reaction of hypochlorous acid? (b) HClO3 and Cl2O (c) HClO2 and HClO4 (d) HCl and Cl2O (a) HCl and HClO3 [2006] 5. Which of the following reactions of xenon compounds is not feasible? (b) XeF6 + RbF Æ Rb [XeF7] (a) 2XeF2 + 2H2O Æ 2Xe + 4HF + O2 (d) 3XeF4 + 6H2O Æ 2Xe + XeO3 + 12HF + 1.5O2 (c) XeO3 + 6HF Æ XeF6 + 3H2O [2009] 6. Among the following oxoacids, the correct order of acid strength is (b) HOCl > HClO2 > HClO3 > HClO4 (a) HClO2 > HClO4 > HClO3 > HOCl (c) HClO4 > HOCl > HClO2 > HClO3

(d) HClO4 > HClO3 > HClO2 > HClO

[2014]

7. Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated? (a) IF7: pentagonal bipyramid (b) BrF5: trigonal bipyramid (d) ICl3: planar dimeric [2014, online] (c) BrF3: planar T-shaped 8. The least number of oxoacids are formed by (a) Nitrogen (b) Sulphur (c) Fluorine (d) Chlorine [2015, online] 9. Chlorine water on standing loses its colour and forms: [2015, online] (c) HCl and HOCl (d) HCl and HClO2 (a) HCl only (b) HOCl and HOCl2 10. The non-metal that does not exhibit positive oxidation state is: [2016, online] (a) Chlorine (b) Iodine (c) Fluorine (d) Oxygen 11. Which of the following is the most reactive? (a) Cl2 (b) Br2 (c) I2 (d) ICl

16.40 Complete Chemistry—JEE Main

ANSWERS 1. (c) 7. (b)

2. (c) 8. (c)

3. (b) 9. (c)

4. (a) 10. (c)

5. (c) 11. (d)

6. (d)

HINTS AND SOLUTIONS 1. The process F(g)– + aq Æ F(aq)– is accompanied with a high decrease in enthalpy as compared to other halogens.

2. 3. 4. 5.

6.

Therefore, the choice c is correct. The smog is caused by the oxides of sulphur and nitrogen. Therefore, the choice c is correct. The thermal stability of H—X (hydrogen halides) decreases with increase in the atomic number of X. The reaction is 3HClO Æ 2HCl + HClO3 Besides XeF2 and XeF4, the compound XeF6 also exhibit reaction with water. XeF6 + 3H2O Æ XeO3 + 6HF XeF6 also reacts with RbF forming Rb+[XeF7] –. The larger the number of oxygen atoms attached to chlorine, greater the electron pull towards oxygen, hence, more easy to remove hydrogen from the acids. The given acids are Cl — OH; OCl — OH; O2Cl — OH, O3Cl — OH (HOCl)

(HClO2)

(HClO3)

(HClO4)

7.

8. Flourine forms least number of oxoacids. 9. Cl2 + H2O Æ HCl + HOCl 10. Fluorine is the most electronegative atom. It does not have positive oxidation state. 11. Interhalogen compound is more reactive than halogens.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.41

The Group 18 Elements Group 18 of the periodic table consists of helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They are collectively known as noble gases because of their extremely low reactivity.

Discovery of Noble Gases In 1785, Henry Cavandish passed repeated electric sparks in a mixture of dry air (free from carbon dioxide) and oxygen in order to achieve complete conversion of nitrogen of the air to its oxides. The resulting nitrogen oxides were absorbed in caustic potash solution and the excess oxygen was removed by potassium pentasulphide. Still a very small volume of the gas (1/120th of the original volume of air) was left behind which neither combined with oxygen nor with any other element. This remarkable observation of Cavandish remained unnoticed for more than a century. In 1895, Lord Rayleigh found that the density of nitrogen obtained from atmosphere was about 0.47% higher than that prepared from the decomposition of ammonium nitrite. He concluded that the nitrogen obtained from atmosphere contains some constituent (or constituents) which is (or are) heavier than nitrogen. Ramsay and Rayleigh repeated Cavandish’s experiment and recorded the spectrum of the residual gas. It was found to be entirely different from that of any other element known till then. The vapour density of the gas was found to be 20 which gives its relative molar mass as 40. This gas was found to be chemically inactive, and therefore named ‘argon* (Greek word meaning lazy or inert). However, the new gas was soon found to be a mixture of a number of chemically The history of discovery of helium is also very interesting. As early as 1868, Lockyer concluded that the bright yellow line (which he called D3 line) observed in the solar spectrum during a total eclipse is due to the presence of a new element in the sun. He called this element helium (From the Greek word helios, the sun). Ramsay, in 1895, showed that the gas isolated by Hillebrand, from uranium mineral, had the same spectrum as helium discovered by Lockyer. Thus, In 1898, Ramsay and Travers carried out systematic fractional evaporation of a large volume of liquid gas obtained new). Its vapour density was found to be 10.1 and hence relative atomic mass 20.2. Its spectrum establishes it to be krypton (from Greek word krypto meaning hidden) and xenon (from Greek word which means stranger), respectively. The vapour densities of krypton and xenon were found to be 40 and 64 and hence their relative atomic masses were 80 and 128, respectively. In 1900, Dorn discovered the last member of noble gases as a disintegration product of radium and named it as radon 226 222 4 88 Ra Æ 86 Rn + 2 He

Physical Properties of Noble Gases The physical properties of the noble gases are recorded in Table 9. Table 9. Property

Physical Properties of Noble Gases He (1s)

Ne 2

2

(2s) (2p)

Ar 6

2

(3s) (3p)

Kr 6

2

(4s) (4p)

Atomic number

2

10

18

36

Relative atomic mass

4.002

20.179

39.098

83.80

Abundance in dry air by volume (ppm)

2

(5s) (5p)

Rn 6

54 131.29

(6s) (6p)6 86 222.018

5.24

18.18

93.40

*

24.6

83.8

115.9

161.3

202.0

Boiling point, T/K

4.2

27.1

87.3

119.7

165.0

211.0

Atomic radius, r/pm Enthalpy of vaporization,

0.09

2

Melting point, T/K

Ionization energy, I/kJ mol–1

1.14

Xe 6

93

112

154

167

190

2372

2080

1520

1351

1170

0.08

1.74

DvapH/kJ mol–1 * Helium is the only liquid that cannot be frozen without applying pressure.

6.52

9.05

12.65

traces

— 1037 18.1

16.42 Complete Chemistry—JEE Main

Description 1. All elements exist as monatomic gases. 2

ns)2(np)6, where n varies from .

inactive. 4. The radii of the elements are very large and increase on descending the group as expected (Fig. 15). These nonbonded radii should be compared with van der Waals radii of other elements and not with covalent bonded radii. 5. The ionization energies of the noble gases are very high order (Fig. 16)—helium has the highest value of all the elements. 6. Melting and boiling points are low indicating extremely weak interatomic forces (van der Waals types) between the atoms of noble gases (Fig. 17). This is supported by the low values of enthalpies of vaporization. b.pt.

200 200

Ionization energies

240°

van der Waals Radii

m.pt.

160 200° 120

120

160°

T/K

r/pm

IE/kJ mol-1

160

120°

80

Fig. 15

He

Ne

Ar

Kr

Molecular Radii of the Elements of Group 18

Xe

80°

Fig. 16

80

40

He

Ne

Ar

Kr

Xn

Ionization Energies of the

Rn

Fig. 17

He Ne Ar Kr

Xe Rn

Melting and Boiling Points of the

Elements of Group 18

Elements of Group 18

Special Properties of Helium When helium is cooled below 2.178 K at 1 atmospheric pressure, a remarkable liquid with abnormal properties is obtained. The liquid is referred to as helium II and the characteristic temperature is called the lamda (l) point. The

tension and compressibility.

Compounds of Noble Gas No compound of the noble gases was known before 1962. As highly oxidizing PtF6 could oxidize oxygen PtF6 + O2 æÆ O +2[PtF6]– Æ Xe+ + e–; 1170 kJ mol–1) is very similar to oxygen (O2 Æ O +2 + e–; 1165 kJ mol–1), should also react in the same way. In 1962, N. Bartlett could actually prepare a yellow-red powder corresponding to Xe+[PtF6]– by interaction of PtF6 with xenon. K Xe + PtF6 æ298 ææÆ Xe+[PtF6]– The reaction has since been shown to be more complicated, and the product is [XeF]+ [Pt2F11]–. Xe[PtF6] + PtF6 æ298 ææ Æ [XeF]+ [PtF6]– + PtF5 æ333 ææ Æ [XeF]+[Pt2F11]– K K Soon after this, a colourless volatile solid XeF4 was obtained when Xe and F2 reacted at 400 °C. The ionization energies of He, Ne and Ar are much higher than Xe and are high enough to enter into the chemical

Study of the p-Block Elements (Groups 16, 17 and 18)

16.43 2.

The

chemistry of radon is not well studied because of its short half-life.

Compounds of Xenon Xenon Fluorides The product depends on the ratio of Xe and F.

°C in a sealed nickel vessel. 2:1 ratio 1:5 ratio

Xe + F2

1:20 ratio

Oxidation Reactions

2

XeF2 XeF4 XeF6

to H+, Cl– to Cl2, I– to I2 and Ce(III) to Ce(IV).

XeF2 + H2 Æ 2HF + Xe XeF2 + 2HCl Æ 2HF + Xe + Cl2 XeF2 + 2KI Æ 2KF + Xe + I2 XeF2 + Ce2(SO4)3 + SO2– 4 Æ 2Ce(SO4)2 + Xe + F2 Similar reactions can be written for XeF4 and XeF6.

Fluorination Reaction XeF4 + 2SF4 Æ Xe + 2SF6 XeF4 + 2C6H6 Æ Xe + 2C6H5F + 2HF

Reaction with Water 2XeF2 + 2H2O Æ 2Xe + 4HF + O2

(slow reaction) 3 3XeF4 + 6H2O Æ 2Xe + XeO3 + 12HF + O2 (violent reaction) 2 XeF6 + 3H2O Æ XeO3 + 6HF (violent reaction) XeF6 undergoes partial hydrolysis reaction with small quantity of water. XeF6 + H2C Æ XeOF4 + 2HF XeOF4 is also obtained when XeF6 reacts with SiO2 2XeF6 + SiO2 Æ 2XeOF4 + SiF4 The reaction of XeF6 with XeO3 also produces XeOF4. 2XeF6 + XeO3 Æ 3XeOF4.

Formation of Complexes XeF2 forms complexes with PF5, AsF5. SbF5 structures of complexes are believed to be [XeF+] [MF6]– [XeF+] [M2F11]– [Xe2F3]+ [MF6]– XeF4 and XeF6 also form complexes, but their number are less. Structure of XeF2 The number of valence electrons in XeF2 are 8 + 2 ¥ 7 = 22. These are distributed as follows.

F—Xe—F bipyramidal orientations. Experimentally, XeF2 is found to be linear with bond distance X—F equal to 200 pm. Thus, F atoms in XeF2 occupy apical positions while the three lone pair of electrons occupy equatorial positions.

16.44 Complete Chemistry—JEE Main

F

Xe

Note:

F In triangular bipyramidal, lone pairs of electrons always occupy equatorial positions. 2

are as follows. s Xe in free atom Xe in FeX2

p

d

≠Ø

≠Ø

≠Ø

≠Ø

≠Ø

≠Ø

≠Ø

≠

≠

 sp3d hybridization, octahedral geometry

Structure of XeF4 There are 8 + 4 ¥ 7 = 36 valence electrons in XeF4. There are six electron pairs around Xe. According to the VSEPR theory, these will attain octahedral orientations. Experimentally, all the four X—F bonds in XeF4 have the identical bond distance of 195 pm. Thus, the four equatorial positions will be occupied by F atoms while the two lone pair of electrons will occupy apical positions. F F

Xe F

F

F Xe

F F

Note:

F

In octahedral orientation, lone pairs of electrons occupy apical positions. 2

s Xe in free state

p

are as follows.

d

≠Ø

≠Ø

≠Ø

≠Ø

≠Ø

≠Ø

≠

≠

≠

≠

 sp3d2 hybridization, trigonal bipyramidal geometry

Structure of XeF6 There are 8 + 6 ¥ 7 = 50 valence electrons. There are seven-electron pairs around Xe atom. Six of them attain octahedral orientations and the seventh electron pair of Xe causes distortion in the octahedral orientations. Thus, the structure of XeF6 is the distorted octahedral.

Study of the p-Block Elements (Groups 16, 17 and 18)

16.45

F F F

F

F Xe

F

F

F

Xe

F F

F F

slightly distorted octahedron

Structure of XeO3 There are 8 + 3 ¥ 6 = 26 valence electrons. There are four electron pairs around Xe. Hence, these electrons acquires tetrahedral orientations (VSEPR theory).

O

Xe

O

O

Xe O

O O

XeOF4 Valence electron = 8 + 6 + 4 ¥ 7 = 42 O F

F Xe

F

Number of electron pairs around Xe = 6 Arrangement of electron pairs around Xe = square bipyramidal O F

F

F

Lewis structure

Xe F

F

square bipyramidal (octahedral with one position unoccupied)

XeO2F2 Number of valence electron = 8 + 2 ¥ 6 + 2 ¥ 7 = 34 F O

Xe

O

Number of paired electrons around Xe = 5 Orientation of electron pairs around Xe is triangular bipyramidal. F O Xe

F

Lewis structure

O F triangular bipyramidal

16.46 Complete Chemistry—JEE Main

[XeO6]4– Number of valence electron = 8 + 6 ¥ 6 + 4 = 48

Number of paired electrons around Xe = 6 Orientation of electron pairs around Xe is octahedron O

4

O O O

O

O Xe

O

Xe

O O

O

Lewis structure

O O octahedral

XeO4 Number of valence electron = 8 + 4 ¥ 6 = 32

Number of paired electrons around Xe = 4 Arrangement of electron pairs around Xe is tetrahedron O

O Xe O

Xe

O

O

O

O

Lewis structure

O Tetrahedral

MULTIPLE CHOICE QUESTIONS 1. The number of elements in Group 18 is (a) 4 (b) 5 (c) 6 (d) 7 – 2. The ionization energy of Xe Xe+ + e is very close to that of (b) O2 Æ O+2 + e– (c) C2 Æ C+2 + e– (d) B2 Æ B+2 + e– (a) N2 Æ N+2 + e– 3. When a mixture of Xe and F2 in the ratio of 2 : 1 is heated at 400 C in a sealed nickel vessel, the compound formed is (c) XeF4 (d) XeF6 (a) XeF (b) XeF2 4. When a mixture of Xe and F2 in the ratio of 1 : 5 is heated at 400 °C in a sealed nickel vessel, the compound formed is (c) XeF4 (d) XeF6 (a) XeF (b) XeF2 5. When a mixture of Xe and F2 in the ratio of 1 : 20 is heated at 400 °C in a sealed nickel vessel, the compound formed is (a) XeF (b) XeF2 (c) XeF4 (d) XeF6 6. The number of paired electrons around Xe in XeF2 is (a) 3 (b) 4 (c) 5 (d) 6

Study of the p-Block Elements (Groups 16, 17 and 18)

16.47

7. The number of paired electrons around Xe in XeF4 is (a) 3 (b) 4 (c) 5 (d) 6 8. The number of paired electrons around Xe in XeF6 is (a) 4 (b) 5 (c) 6 (d) 7 9. The number of paired electrons around Xe in XeO3 is (a) 4 (b) 5 (c) 6 (d) 7 10. The structure of XeO4 is (a) square planar (b) tetrahedral (c) pentagonal with one paired electrons (d) octahedral with two paired electrons 11. Which of the following inert gases has the largest abundance (by volume) in air? (a) Helium (b) Neon (c) Argon (d) Krypton experimental conditions? (b) XeF3 (a) XeF2 13. Which of the following statements is correct? (a) The processes Xe(g) Xe+(g) + e– and O2(g) (b) XeF2 possesses an angular structure (c) XeF2 the three lone pairs occupying equatorial position (d) XeF4 14. The formula of freon – 12 is (b) CF2Cl2 (a) CFCl3

(c) XeF4

(d) XeF6

O2+ (g) + e– involve more or less identical ionization energy

(c) CF3Cl

(d) CF4

ANSWERS 1. (c) 7. (d) 13. (b)

2. (b) 8. (d) 14. (b)

3. (b) 9. (a)

4. (c) 10. (b)

5. (d) 11. (c)

6. (c) 12. (b)

HINTS AND SOLUTIONS 13. XeF2 molecule has a trigonal bipyramidal structure.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

(a) XeO2F2

ANSWERS 1. (d)

(b) XeOF4

(c) XeO3

(d) XeO4

[2014, online]

16.48 Complete Chemistry—JEE Main

HINTS AND SOLUTIONS XeF2 + H2

1 O2 + 2HF 2

XeF6 + 3 H2

3

XeOF4 + XEO3

+ 6HF 2

3 XeF4 + 6 H2 XeF6 + H 2 O

(

small quantity

3 O2 + XeO3 + 12 HF 2

)

4

+ 2HF

17 d– and f–Block Elements Transition Elements

Table 1

V Cr Mn Iron

Fe

Cobalt

Co Ni

Copper

6

17.2

Variable Oxidation States

Table 2

V

III

II III IV

II III IV V

Cr I II III IV V VI

Mn II III IV V VI

Fe II III IV

Co II III IV

Ni II III IV

I II III

II

VI

VII

Formation of Complexes

6]Cl

6

6

Size of Atoms

Y

17.3

Density

Melting and Boiling Points

Ionization Energies

Colour

Table 3 Colour absorbed Violet

Complimentary colou

Red

Table 4 Number of unpaired electrons

0 1 2 3 3 4 4 5 5

Metal ions +

Ni Co Cr Fe Cr Mn Fe

Colour observed

17.4

Magnetic Properties

n (n + 2)

B

B

Catalytic Activity

Table 5

V Fe Pd Cr Ni PdCl Pt Mn

Lanthanides and Actinides

n

n

17.5

MULTIPLE CHOICE QUESTIONS General Characteristics

Mn

is

6]

Oxidation State

17.6

Cr

to Cr

Magnetic Properties

B

n

B

B

n(n + 1)

B

n(n + 2)

B

B

Mass of a substancein a magnetic field Actual mass of the compound

Mass of a substancein a magnetic field Actual mass of the compound

is V

is is

B

B

B

is B

B

Cr

is

Fe

is

B

(n + 1)(n + 2)

17.7

6

V

Cr

Fe

Co

V

Fe

Mn

Colour

+

+

Characteristics of Compounds

17.8

]

Cr

is

Cr

Cl

Cr

Cl6 Cl6 is Cl Cl

Cl Fe … Fe

Cl

]

Cl

Cl

Cl Fe

Cl

]+

Cl

Cl Fe

Cl

Cl Cl

Cl

Cl Cl

Fe

Cl

F Cl

17.9

Cl

Inner Transition Elements

x x La

Ce

and Nb and

Sm

Mo and

66Dy

is

La

+

17.10

Ce

Pr

ANSWERS

HINTS AND SOLUTIONS

Mn

Sm

66Dy

17.11

Mn +7

Mn O -4 +7

Mn O -4 +7

Mn O -4

+ Mn

Mn +4

Mn O 2 +6

Mn O 24+

Cr

1(1 + 2)

B

V

B 6

Fe B

4( 6) 4( 6)

B

Cr

B

Fe

V

Cr

+

+

only V

Cr +

æDæ Æ

3(5)

B

5(7)

B

6(8)

Fe

and Co

2(2 + 2)

17.12

Sm

n

66Dy

La

n

La

and

n

and Dy

ANNEXURE

A Few Common Elements and Their Important Salts Iron 6

+C Fe

6

6

n

17.13

·

Copper, Silver and Gold n

n

S to

S · Sb S ]

] ]

]

] ]

]

17.14

]

+C N

]

exposure    

373 K    

423 K    

Zinc and Mercury n

Cl

Important Salt of Manganese Potassium Permanganate

Æ Æ Cl

+e

+ Cl Æ Æ

Æ Mn

n

17.15

Æ

+e Æ Æ Æ by

in

I to Æ

Important Salt of Chromium Potassium Dichromate

Æ

+

Æ Na Cr Æ

Na Cr +

Cr

+

Cr

+ 6e Æ

to Fe

to I

and S

Æ

Cr

Cr

Æ

¥

Æ

Cr

Cl Æ

Cr

Cl

Cl Cl

Æ

Cl

Æ Æ

Pb

Cr

Æ

+

17.16 +

Cr Cr and Cr O

O

2

Cr

O

Cr

O

O

O

2

Cr

O

O

O

O O

MULTIPLE CHOICE QUESTIONS Iron

is

17.17

Fe

6

6

C +

+

Copper, Silver and Gold

n

6

n

n

6

n

n

n

n

n

17.18

S · Sb S S

17.19

]

+C N ]

]

17.20

+

Æ

+

+e Æ

Zinc and Mercury

+N

]

smaller

+

+e Æ

+

+e Æ

17.21

Cl Cl ] ]

is

17.22

KMnO4 and K2Cr2O7 is

is

Cr

is

Cl

Cr

ANSWERS

17.23

HINTS AND SOLUTIONS

+

exposure    

373 K    

+N +N

423 K    

17.24

373 K

æææÆ -6 H 2 O

723 K

above 1000 K

æææÆ

æææææÆ

- H2O

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN [2003]

[2003] and Cr [2003] Z

[2004]

[2005]

[2005]

[2005]

[2006] mB [2006] a Ca [2006]

17.25

[2007]

[2007]

[2008]

[2009]

[2009]

[2011 cancelled]

[2012] O , heat

CO, 600∞C

CO, 700 °C

2 Fe ææææ Æ Fe3O 4 ææææÆ FeO æææææ Æ Fe

dil. H SO

H SO , O

heat

2 4 2 4 2 Fe ææææ æ Æ FeSO 4 æææææ Æ Fe 2 (SO 4 )3 æææ Æ Fe

O , heat

dil. H SO

heat

2 2 4 Fe ææææ Æ FeO ææææ Æ FeSO 4 æææ Æ Fe

Cl , heat

heat, air

Zn

2 Fe ææææ Æ FeCl3 ææææ Æ FeCl2 ææÆ Fe

[2014]

17.26

[2014, online]

[2014, online]

[2014, online] Cr [2014, online]

[2014] 6

[2015, online] false Cr

Cr

Cr [2015, online]

s

s* transition

[2015] +

and Cr

and Co

and Cr

[2016, online] and Co [2016, online] [2016]

and N

ANSWERS

[2016]

17.27

HINTS AND SOLUTIONS 6

+

6

6

Cr

mm = ation d

n

n ( n 2)

s

.

L

n (n + 2) mB

L

2 (4) mB

mB

n+l

and FeCl e

Cr2 O72- + 14H + + 6e- Æ 2Cr 3+ + 7H 2 O S2- Æ S + 2e- ] × 3 Cr2 O72- + 14H + + 3S2- Æ 2Cr 3+ + 7H 2 O + 3S S Æ Cr

Cr nt mB as V

n (n + )

17.28

and Cl Cr

Cr

+

Al

6

Æ Æ

+N

18 Coordination Chemistry and Organometallics 3) 6]

3+

ligands bound around a metal ion is called the coordination number Mono or Unidentate Ligand

–

–

–

– 2

– 2

3

Bidentate Tridentate Tetradentate Pentadentate Hexadentate

Nomenclature of Coordination Compounds

–

2

–

3 + 3

2

–

2–

– 2)

18.2 Complete Chemistry—JEE Main

— —

3) 6 3) 5]

2) 3 2

—

2

3 2+

3) 4

4

—

3 3) 3

4

K3

5 5

5) 2

6

6) 2

Explanation of Formation of Coordination Compounds

[Cr(NH3)6]3+ 5 3 2

3

hybrid orbitals

Coordination Chemistry and Organometallics 18.3

3d

4s

4p

Cr atom Cr3+ ion Cr3+ in [Cr(NH3)6]3+ d2sp3 hybridization, six pairs of electrons from six NH3 molecules

[CoF6]3– 3d

4s

4p

Co atom Co3+ ion 4d Co3+

in [CoF6]3 sp3d2 hybridization, six pairs of electrons from six F - ligands 6]

3–

[Ni(CN)4]2– and [NiCl4]2– 3d

4s

4p

Ni atom Ni2+ ion Ni2+ in [Ni(CN)4]2 dsp2 hybridization, four pairs of electrons from four CN - ligands

Ni2+ in [NiCl4]2 sp3 hybridization, four pairs of electrons from four Cl - ligands

3

4]

2–

2 –

4]

2–

is tetrahedron

18.4 Complete Chemistry—JEE Main

[Ni(NH3)6]2+ 3d

4s

4p

Ni atom Ni2+ ion 4d Ni2+

in [Ni(NH3)6]2+ sp3d2 hybridization, octahedral 3) 6]

2+

Ni(CO)4 3d

4s

4p

Ni atom Ni in Ni(CO)4 sp3 hybridization 4

Inner and Outer Orbital Octahedral Complexes 2

3

3 2

d

Example

Note

3) 6]

3+

2

3

6]

3–

3 2

d

–

Isomerism in Coordination Compounds Ionization Isomerism 3) 5

3) 5

4

4

Linkage Isomerism 3) 5

2

3) 5

2

2

Coordination Isomerism 3) 6]

3+

6]

3–

3) 6]

3+

6]

3–

Coordination Chemistry and Organometallics 18.5

Geometrical Isomerism cis isomer while those trans

Geometrical Isomerism in Square Planar Complexes 4

3

2b 2

cis – trans

2

cis

2

trans trans-

Example of Geometrical Isomers in Square Planar Complexes

H3N

H3 N

Cl Pt

Pt Cl

H 3N

Cl

cis-isomer

H3 N

Br

NH3 trans-isomer

H3 N

Pt Cl

Cl

Py

H3 N

Pt Py

Cl

Py Pt

Br

Br

Cl 1

Geometrical Isomerism in Octahedral Complexes cis

5

2 M

4

3

trans 6

Fig. 18.1

18.6 Complete Chemistry—JEE Main

Number of Isomers of a Complex trans trans

4

C2 trans

¥

2

6 4b 2

5

2b 2

2b 2c 2

3 3b 3

4

cis trans and 6)

cis trans

cis trans

cis

isomer trans ∑

2b 2c 2

trans

Trans positions; aa; bb; cc

aa; bc; bc

ab; ab; cc

(enantiomer) ab; ac; bc

ac; bb; ac

3b 3

Coordination Chemistry and Organometallics 18.7

∑ ∑

six

2b 2

trans

(and its enantiomer)

∑

6

∑

2b 2

cis and trans

∑

3

cis– and trans

2a 2

(and its enantiomer)

C2 2

cis

different donor

18.8 Complete Chemistry—JEE Main

Example of Geometrical Isomers in Octahedral Complexes cis

trans

H3N

Cl

2+

Cl

H3 N

Co H 3N

NH3 cis-isomer

2+

Cl

NH3

Co Cl

H3 N

NH3

Cl

trans-isomer

Optical Isomers

Stability of Coordination Compounds

–

–

–

Coordination Chemistry and Organometallics 18.9

–

–

2–

–

–

–

–

2– 2

– 2

3

– 2

–

– 2+

2+

and

2+ 2+

2+

2+

2+

2+

2+

2+

Crystal Fields Splitting in the Complexes xy

xz

z2

yz

dx2–y2 xy

xz

and dyz

x2–y2

eg

g

and dz2 2g

D

2g

followed by eg

Colour

18.10 Complete Chemistry—JEE Main

Table 18.1

The Relationship between the Colour Absorbed and Complimentary Colour Observed

Colour Absorbed Violet

Complimentary Colour Yellow–green Yellow Orange to red

Green to yellow green Orange

Colour of Some Transition Metal Ions

Table 18.2 Number of unpaired electrons

Metal ions

Colour observed

+

2+

2+

4+

2+

4+

2

2+

3+

3

2+

3

3+

Green

4

2+

Green

4

2+

Green

2+

5

3+

5

Yellow

Applications of Coordination Compounds

– 2+

3+

2+

Organometallic Compounds

s bonded p bonded s and p bonded p

2

3) 6

6

5) 6

3) 4

2

5) 4

2

h2 h5 and h6

MULTIPLE CHOICE QUESTIONS General Characteristics

5) 2

Coordination Chemistry and Organometallics 18.11

– 2

3) 6 3) 4

3) 3

3) 4

3

3

2

3) 5

2

2

3) 6

3

–

–

molecules

2

3) 4 · 4 Al2

4]

4 4

2

6]

4

3) 6] –

–

2+

is

–

is –

–

–

–

–

–

–

–

–

Geometry of Complex 3) 6 2

2

3

2

3

3) 6

3

3

is

–

18.12 Complete Chemistry—JEE Main

2

6]

3

3–

3 2

d

4] 4] 4] 4]

3) 6]

3

6]

3

2+

2–

4]

2–

4]

2–

4]

2–

2– 2– 2–

differ in the geometry

2– 4]

4

4]

4

4]

4

4]

4

4]

2– 2– 2– 2–

differ in the nature of ligands

3

2

3

3 2

3

2

3

3 2

d

4–

d

2

Nomenclature 4]

is

2) 6]

3

2

3) 4

3) 3

is

2

2

4

is

2

3) 4

4]

Magnetism of Complexes 6]

3–

4]

2–

3) 6]

2+

4]

2–

Coordination Chemistry and Organometallics 18.13 —

3) 6

—

÷

÷

is

2

—–

÷

6] 6]

—

4]

÷ 2– 4]

2–

2

3) 4] 2O)6]

2

4–

is

—–

3) 6

3+

3

2–

6]

4

2O)6]

2O 7

÷24 bohr magneton

2–

3

4]

is —–

÷ 4]

2+

4–

4) 2

2+

2O)6]

6

2+

4–

2O)6]

4

2+

6]

3

Isomerism in Complexes 2– 4

3

2

en+2

– 3

2

2

en+2

2 + 2en2 is 3) 3

3 – 2 2

2en 3) 5

4

3) 5

2

3) 2 3) 2

4

2 3) 6

6 2]

is

3) 5

2

3) 6

is

is

2 3) 5

6

can form 2]

2– 2

can form

3

2

3]

2–

ion is

2 + 3) 4

en 2

+ 3) 4

18.14 Complete Chemistry—JEE Main 2

2

3

2

2a 2

Additional Problems

) ion 4]

2–

4]

2–

3) 6]

3+

4]

3) 6

2

4]

3) 6] –

3] 4]

2–

–

3) 4]

2

2+

–

2+

3) 4

2–

2+

5) 2 3) 4

5

6

5) 2

5

6

4]

2–

4]

2–

4]

2–

4]

2–

4]

2–

4]

2–

are stable

2

6) 2

2 2

4]

2–

4]

2–

4

3

2

4) 2

2]

Coordination Chemistry and Organometallics 18.15

–

–

–

–

– 6]

–

–

–

–

–

–

–

3– 4

6]

3–

6]

3–

3+ 3) 6] 3+ 3) 6] 6]

3–

3) 6]

3–

6]

3–

3) 6]

3–

ANSWERS

HINTS AND SOLUTIONS

O

O 3 O 4 O

C C

CH2

1 N

CH2

CH2

O –

–

is strongest ligands

2– 4

2+

8

CH2

2 N

CH2

C

O 5

CH3

C

O 6

O

3+ 4]

–

4]

4] 2–

2–

3 2

d

–

–

18.16 Complete Chemistry—JEE Main 3d

4s

4d

4p

sp3d2 hybridization 2+

4

2

3

4s

3d

4p

d2sp3 hybridization

dsp2 hybridization 4]

m=

n m=

2–

n(n + 2) m =

n(n + 2) m =

8m

2(2 + 2) m =

8 m 2+

6

2+

3d

7

2+

4s

9

)

4p

d2sp3 hybridization 9 2

2

3

and d2

8

n(n + 2) m B

7

3

3) 4] 3+

2+

2+

2+

3

6

trans

trans trans

in three ways in the trans 2

cis

a2

Cis –

2+

–

–

trans

6

9

Coordination Chemistry and Organometallics 18.17

and p 2+ – –

–

–

–

–

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 4

4

[2003] 3) 5

3

3

is 3) 4 3) 3

2◊

3) 5

3

3] ◊ 2

2

3) 4

3

2

◊

[2003]

3

2+ 3) 4]

+ 4

3

molecule are not [2003]

2

p [2004] 6]

4–

6]

4–

3) 6]

3+

3) 6]

2+

[2004]

[2004] 3) 4

2]

+

3) 5

2+

2+

3) 2

2

2]

+

[2004] 4]

2– 6]

4] 4–

2– 4]

2–

6]

4–

4]

2–

4]

2– 6]

4–

4– 6] 2– 4]

4]

2–

4]

2–

[2004]

18.18 Complete Chemistry—JEE Main

6]

3–

6]

3–

6]

3–

6]

3–

[2005] 6]

3

is [2005]

3) 5

2

2

is [2006] 2– 4]

–

[2006] 5

s

p 4]

2–

s and p 4]

2–

4]

[2006]

2–

4]

2–

[2007] Do be the 3) 6]

3+

6]

3–

2O 4) 3]

3–

2O)6] 2O 4

2

3+

[2008]

2

[2008] 3) 5 3) 4

3

3) 5

4 3) 4

4

4

3

2

4

3) 4 ) 3 2

2

2 2

3) 4 3) 2

2 2]

[2009] 2O)4

3+ 2

3) 2]

3+

3) 3

+

3) 2]

2+

[2009] 2]

2+

3) 2]

2+

3]

3+

2O)4

3+

[2010] 2– 4] is

[2011 (cancelled)] 3) 6 2

3

3

[2011 (cancelled)] – 3 – 3

– 3

–

– 3

3

[2011]

Coordination Chemistry and Organometallics 18.19

3 3]

3

2

4]

2

3+ 3]

[2012]

2

2]

2

3) 3

–

+

3) 2

2]

+

[2013]

3+ 2

3

in gands

4

is 2

4

3

4

3

2

3

2

4

3

2

4

[2014] Æ K2

4

4]

Æ

2 3) 5] 2O)6]

+ K2

4

2O

+

+

2+

Æ

2+ 4–

+ 4

–

excess NaOH

æææææÆ

2+

2O

[2014]

AgCl Ø + 2NH3  [Ag ( NH3 )2 ] + Cl+

3

3

4

[2014]

K2

4] 5]

2-

3) 2 3) 4

4] 2

4

[2014, online] 3) 6 3) 6]

3+

-

3

-

3+ 3

2O)6]

D

-

3

3+

3) 6]

[2014, online] [2014, online]

4+

3) 6]

3+

3) 6]

2+

[2014, online] 3

4

3

2

[2014, online] 24] 3

2

2

3

[2014, online]

18.20 Complete Chemistry—JEE Main 3+ 3 2

2

d

d2

3 6]

6] 6]

3-

6]

3-

3

3-

6]

d

[2014, online]

3-

3-

6]

3-

[2014, online]

–

–

[2015, online]

4 2O)5

3]

2+

2O)4

3) 2]

2+

2O)3

3) 3]

2+

is

[2015, online] 2 6]

4–

6]

3–

4]

2–

2O)6]

2+

[2015, online] 2g 3– 6]

and eg

2+ 3) 6] 6]

4–

6]

3– 3

[2015, online] + is 2 [2015]

not 2

6

2) 6]

3

4) 3

3O

)4

[2015]

4 3

2

5

2

3

2O)5

6

2O)6

2

3

[2016, online] 2O)6]

2+

2O)6]

2+

2O)6]

3+

2O)6]

2+

2O)6]

2+

2O)6]

2+

2O)6]

3+

2O)6]

2+

2O)6]

2+

2O)6]

2+

2O)6]

3+

2O)6]

2+

2O)6]

2+

2O)6]

2+

2O)6]

3+

2O)6]

2+

3) 6

2O)6] 2– 4]

cis

2+

2 3) 4

3) 2

3

2 2

2O)6] 2+ 2O)6]

3) 4

2

2+

3) 5 2 [2016, online]

2

2+ 2O)6] 2+ 2O)6]

trans

2 3) 3

[2016, online]

3]

2O)6] 2– 4]

2+

[2015]

2

[2016]

Coordination Chemistry and Organometallics 18.21

ANSWERS

HINTS AND SOLUTIONS –

3d 26Fe

2+

+

4s

4p

d2sp3 (inner)

ion in complex ion in complex

d2sp3 (inner)

3+ 27Co

ion in complex

d2sp3 (inner)

2+

ion in complex

sp3d2 (outer)

25Mn

28Ni

2+

3) 6]

2+

2]

2

+

3d

4p

Hybridization

sp3

2+

sp3

25 Mn

27 Co

4s

2+

2+ 26 Fe

d2sp3

4] 3+ 26

5

2–

3+ 27

4]

2–

6

6] 3+

24

4–

3

3+ 25

4

18.22 Complete Chemistry—JEE Main

6]

3–

8

2

4]

3d

4p

4s

sp3 hybridization (tetrahedral arrangement)

Paramagnetic nature

M

2–

C

O

M

s-bonding

C

O

p-bonding

2+

2+

2+

6

7

8 2

3

2

4]

2+

4]

2–

2–

4]

Do –

–

–

–

– 2

3

– 2

–

Do

– 2– 2O 4

¥

– 2

2– 2O 4

2

3) 2]

3+

2–

4]

2–

are

Coordination Chemistry and Organometallics 18.23 3+

en

NH3

H3N

Co

2+

3]

H3N

NH3 3) 2]

3+

en

2+

2O)4en]

2+

has 3d

3+

2+

8

4]

28

4s

3d

4p

Æ Æ

Æ Æ Æ Æ Æ Æ

3]

3+

2]

3+

Co

en 3

en

sp3 hybridization Electrons from the four chlorines

m=

n(n + 2) m =

2(2 + 2) m = 3+

8 m

m

3

24

3d

4s

4p

Æ Æ Æ d2 sp3 hybridization Electrons from the six NH3 species 2

3 3 3

3+

– –

3) 3

3

2–

is

18.24 Complete Chemistry—JEE Main

4

2

3

3 4

2

3

4 4 2O

2

3 5]

23+

D -

3

5c3+

-

4+ 22

2

2+

)

23

-

È ˘ È ˘ ÍM (NH3 ) (SO 4 )˙ Cl and ÍM ( NH3 ) Cl˙ SO 4 ˙˚ ÍÎ ( B) (A) Î ˚ 2+ 28

3d

s

is 3d8

4s

4p 2+

2 3+

6

27

3d

4s

4p

3+ 3+

d2sp3 3+

3+ 27

2

3

in the

Coordination Chemistry and Organometallics 18.25 3+

6]

ion is 3d5

3d

4s

4p

3d

4s

4p

3-

d2sp3 hybridization 6]

3-

3d

4s

4p

4d

sp3d2 hybridization 4 2+ 3] 2+

2O)5 2O)4

3) 2]

2O)3

NH 3

H2O

NH3

Ni H 2O

H2O

OH2

cis

NH3

NH3

H2O

Ni

OH2

H2O

NH3

cis

OH2

cis OH2

NH3

NH3 trans

OH2

trans 2O)6] 2O)6]

3d6

NH3

H 2O

NH3

Ni

3+

do show geometrical isomerism

trans

H2O

27

2+

NH3

H2O

Ni

OH2

3) 3]

2+ 27

3d7

2+

25

and eg and e 2g g

2+ –

Æ

2+

3d5

4]

2–

3+ 26

2O

3d5 –

–

2g

2g)

3

g)

2

18.26 Complete Chemistry—JEE Main +

trans

a

c

d

b 2

a

b

a

b

d

c

c

d

6

– 3 3

3) 6

24 26 25 27

3d4 3d

6

3d

5

3d

7

≠

≠

≠Ø ≠

≠

≠

≠

≠

≠Ø ≠Ø ≠Ø 2O)6]

cis

≠

≠

3 2

≠

≠

3 2

≠

3 2

≠

d d d

3

≠

2+

2O)6]

2+

2a 2

–

N

N

Cl

N

Cl

Co

Co Cl

N N

N

Cl Mirror

N N

3

19 Nuclear Chemistry The natural nuclear changes are associated with the emission of three different types of radiations namely a-rays (i.e. 4 2+ 0 2He particles), b-rays (i.e.–1e) and -rays (i.e., electro-magnetic radiations of wavelength of the order of 0.1 nm). This phenomenon is known as radioactivity. The nuclear changes result into the following transformations. m-4 n-2 B

m nA m nA m * nA

m n +1 B m nA

+

+ 42 He 0 -1 e

+g

While expressing the nuclear reactions, the charges on the involved species are not mentioned. As a nuclear reaction proceeds, extra electrons within an atom lose energy so as to bring the atom to a stable state; and positive ions eventually pick up electrons. From the equations given above, it is obvious that the emission of an a-particle results in the formation of a daughter element which occupies a position two groups to the left of that of the parent element in the periodic table, whereas in b-particle emission, the daughter element occupies a position one group to the right of the parent element. This fact is known as the group displacement law. The stability of a nucleus depends upon the ratio of protons and neutrons. If this ratio of a nuclide lies away from the corresponding stability ratio, it results in an unstable nuclide which tries to attain stability by the spontaneous emission of an a or a b-particle along with the emission of g-radiations. The loss of mass when a given nucleus is formed, starting from the appropriate number of neutrons and protons, is known as mass defect. This loss of mass appears in the form of energy according to the Einstein equation E = mc2. A radioactive disintegration series involves the transformation of the parent radioactive element into the nonradioactive end product. There are four such series known as thorium (4n), neptunium (4n + 1), uranium (4n + 2) and actinium (4n + 3) series. Of these, neptunium or (4n + 1) series is man-made while others occur in nature. The mass number changes only when an -particle is emitted imparting a change of 4 units. In elements of 4n series, the mass numbers are divisible by 4. In (4n + 1), (4n + 2), and (4n + 3) series, the mass numbers give a remainder of 1, 2 and 3, respectively, when divided by 4. The starting and end elements of the four series are as follows. 232 208 ;(– 6a, – 4b) Thorium (4n) Series 90Th 82Pb Neptunium (4n + 1) Series

237 93Np

209 83Bi

;(– 7a, – 4b)

Uranium (4n + 2) Series

238 92U

206 82Pb

;(– 8a, – 6b)

Actinium (4n + 3) Series

235 92U

log N = log N0 –

l t 2.303

207 82Pb

;(– 7a, – 4b)

19.2 Complete Chemistry—JEE Main

where N and N0 are the number of radioactive atoms present at time t and 0, respectively, and l is the decay constant. The time required to reduce the initial number of atoms (or concentration) of a radioactive element to half its initial value is known as half-life period and is given by the expression 0.693 t1/2 = l Nuclear reactions are processes in which a nucleus is converted into one or more different nuclei as the result of an interaction with another nucleus or with an elementary particle. A nuclear reaction can be represented by an equation similar to those used for ordinary chemical reactions. For example, the transmutation of nitrogen into oxygen by the bombardment of a particles is represented as 14 4 17 1 7N + 2He 8O + 1H In shorthand notation, the above reaction is written as 147N( , p)178O.

MULTIPLE CHOICE QUESTIONS General Characteristics 1. Which of the following represents an alpha particle? (a) Proton (b) Neutron 2. Which of the following represents a beta particle? (a) Proton (b) Neutron 3. Which of the following represents gamma rays? (a) Stream of dipositive helium ions (c) Stream of electrons

5.

6.

7. 8. 9. 10.

11.

(c) Electron

(d) Dipositive helium ion

(c) Electron

(d) Dipositive helium ion

(b) Electromagnetic radiation (d) Cathode rays

(a) Alpha particles (b) Beta particles (c) Gamma rays (d) Protons The emission of beta particles is from (a) the valence shell of atom (b) the inner shell of atom (c) the nucleus due to the nuclear conversion proton neutron + electron (d) the nucleus due to the nuclear conversion neutron proton + electron The instability of a nucleus is due to (a) high proton : electron ratio (b)high proton : neutron ratio (c) low proton : electron ratio (d) low proton : neutrino ratio Which of the following is expected to have a negative charge? (a) Neutrino (b) Neutron (c) Positron (d) Antiproton Muon is heavier than electron by (a) 106 times (b) 207 times (c) 307 times (d) 407 times A positron has a mass (a) equal to proton (b) equal to electron (c) greater than electron (d) smaller than electron Mesons are responsible for (a) disintegration of the nucleus(b)repulsion between the protons within the nucleus (c) attraction between nucleons (d) attraction between neutrons only In a b decay, the ratio of p/n of the nucleus is (a) increased (b) decreased (c) not changed (d) changed but cannot be predicted

Nuclear Chemistry 19.3

Group Displacement Law 211 12. Group 15 element 211 83Bi is transformed to 84Po. To which group does the element Po belong? (a) 14 (b) 15 (c) 16 (d) 13 210 13. The element 84Po emits an alpha particle. The daughter element is (b) 206 (c) 208 (d) 210 (a) 204 82Pb 82Pb 82Pb 82Pb Mg emits a beta particle. The daughter element is 14. The element of 27 12 He (b) 23 (c) 27 (d) 23 (a) 23 10 11Mg 13Al 11Na 208 15. In a radioactive decay 232 90Th is transformed into 82Pb. The number of alpha and beta particles emitted respectively are (a) 4, 6 (b) 4, 4 (c) 6, 6 (d) 6, 4 16. Which of the following is man made disintegration series? (a) 4n series (b) 4n + 1 series (c) 4n + 2 series (d) 4n + 3 series 17. Which of the following is 4n series? (a) Thorium series (b) Neptunium series (c) Uranium series (d) Actinium series 18. Which of the following is 4n + 1 series? (a) Thorium series (b) Neptunium series (c) Uranium series (d) Actinium series 19. Which of the following is 4n + 2 series? (a) Thorium series (b) Neptunium series (c) Actinium series (d) Uranium series 20. Which of the following is 4n + 3 series? (a) Thorium series (b) Neptunium series (c) Actinium series (d) Uranium series 21. The end product of 4n + 1 disintegration series is (b) 209 (c) 206 (d) 297 (a) 208 82 Pb 83 Bi 82 Pb 82 Pb

22. For which radioactive series, lead is not the end product? (a) 4n (b) 4n + 1 (c) 4n + 2 23. After the emission of one a-particle followed by one b-particle from the atom daughter atom will be (a) 142 (b) 143 (c) 144 24. Lead 207 is produced as the end product in (a) 4n series (b) 4n + 1 series (c) 4n + 2 series

(d) 4n + 3 238 92 X,

the number of neutrons in the (d) 146 (d) 4n + 3 series

Kinetics of Radioactive Decay 25. The half-life of a radioactive decay is given as (b) t1/2 = log 2/l (c) t1/2 = l /log 2 (d) t1/2 = l/0.693 (a) t1/2 = 0.693/l where l is decay constant. 26. The disintegration rate of a radioactive element changes from an initial value of 10000 disintegrations per minute to 2500 disintegrations per minute in 50 days. The half-life of the element is (a) 25 days (b) 50 days (c) 75 days (d) 100 days 27. If N0 and N are the number of radioactive particles at t = 0 and at time t, then N N N 1 2.303 N t 1 log 0 (b) l = log (c) l = log 0 (d) l = ln 0 (a) l = N N N N0 2 t 2.303 t where l is the decay constant. 28. The activity of a sample has 40% as much radioactivity as present originally. If half life period of the radioactivity is 5000 y, the life of the sample undergoing disintegration is (a) 5000 y (b) 6000 y (c) 6667 y (d) 5667 y

19.4 Complete Chemistry—JEE Main

29. The activity of a sample of wood is due to the presence of 50% of 14C as compared to the original sample. If half life of 14C is 5760 y, the life of the sample of wood is (a) 5760 y (b) 5670 y (c) 6667 y (d) 5000 y 30. Carbon dating involves the counting of (b) 13C isotope (c) 14C isotope (d) 12C and 13C isotopes (a) 12C isotope 31. The radioactivity due to 14C isotope (t1/2 = 5760 y) of a sample of wood from a dead tree was found to be nearly one fourth of fresh wood. The tree died (a) 5760 y back (b) 9090 y back (c) 11520 y back (d) 2880 y back 32. The reciprocal of radioactive decay constant is called (a) half-life period (b) average life period (c) natural life period (d) root mean square life period

Mass Defect and Binding Energy 33. The energy equivalent to 1 atomic mass unit is (a) 921 MeV (b) 931 MeV 34. The expression of mass-energy conversion is (b) E = mc2 (a) E = m2c 35. MeV stands for (a) milli electron volt (b) milli electron velocity 36. Binding energy can be calculated from the formula 931 MeV ˆ (a) Binding energy = (Mass defect) ÊÁ Ë 1amu ˜¯ (c) Binding energy =

(mass defect/1amu) MeV 931.05

(c) 941 MeV

(d) 951 MeV

(c) E2 = mc

(d) E2 = mc2

(c) mill electron volume

(d) mega electron volt

9.31 MeV ˆ (b) Binding energy = (Mass defect) ÊÁ Ë 1amu ˜¯ (d) Binding energy =

(Mass defect/1amu) MeV 9.31

37. The mass defect for the formation of 12C is 0.10242 amu. Its binding energy would be (a) 95.35 MeV (b) 9.535 MeV (c) 95.35 eV (d) 9.535 eV 38. Energy released during the annihilation of one positron and one electron (me = 9.11 10–31 kg) is (a) 8.2 10–14 J (b) 1.64 10–13 J (c) 2.46 10–13 J (d) 4.92 10–13 J 10 0 39. The mass defect of the nuclear reaction 104Be 5B +–1e is (a) Dm = atomic mass of (105B – 104Be) (b) Dm = atomic mass of (105B – 104Be) + mass of one electron (c) Dm = atomic mass of (105B – 104Be) + mass of two electrons (d) Dm = atomic mass of (105B – 104Be) + mass of three electrons 40. The mass defect of the nuclear reaction 58B (a) Dm = atomic mass of (48Be – 85B)

8

4Be

++10 e is

(b) Dm = atomic mass of (48Be – 85B) + mass of one electron (c) Dm = atomic mass of (48Be – 85B) + mass of two electrons (d) Dm = atomic mass of (48Be – 85B) + mass of three electrons 41. The binding energy per nucleon of 37Li and 59 27 Co is 6.44 and 8.57 MeV. Which of the following statements is correct? (a) Li is more stable than Co (b) Li is less stable than Co (c) Both Li and Co are equally stable (d) The conversion of Co to Li is associated with release of energy

Nuclear Chemistry 19.5

42. If the binding energy per nucleon of 7Li and 4He are 5.60 MeV and 7.06 MeV, respectively, the energy associated with the reaction 7Li + p æÆ 242 He is (a) 17.3 MeV

(b) – 17.3 MeV

(c) 34.6 MeV

(d) – 34.6 MeV

Nuclear Reactions 43. Which one of the following is an ( , n) type nuclear transformation? 4 78 1 (b) 73Li + 11He (a) 75 33As + 2He 35Br + 0n (c)

45 21Sc

+ 01n

45 20Ca

+ 11H

(d)

44. In the nuclear reaction 14 4 7N + 2He the projectile is

17 8O

14 7N

+ 11H

7 4Be 15 8O

+ 10n +g

+ 11H

(b) 42H (a) 147N 45. Atom bombs are based on

(c) 178O

(d) 11H

(c) nuclear fusion 46. Hydrogen bomb is based on

(d) spontaneous chemical reaction

(c) nuclear fusion (d) spontaneous chemical reaction 47. The material used for absorbing neutrons in a nuclear reactor is (a) cadmium (b) radium (c) uranium (d) zinc 75 4 78 1 33As + 2He 35Br + 0n the projectile is (a)

14 7N

(b) 42He 9 4B

(a) 49Be

+ 11H

(b) 11H 10 5B

10 5B

+ 10n

1 0n

(b) (a) 51. Which of the following is capture reaction? 16 4 7 4 (a) 199F + 11H 8O + 2He (b) 3Li + 2He

(c) 84Be 7 4 3Li + 2He the target is (c) 73Li +g

(c)

32 16S

(b)

25 12Mg

(d)

235 92U

+ 01n

53. Which of the following is a spallation reaction? 14 4 (a) 168O + 21H 7N + 2He

(b)

63 29Cu

+ 24He

+g

(d)

238 92U

+ 01n

(a)

16 8O

+ 21H

(c)

75 33As

+ 24He

(c) 21H + 21H

14 7N

+ 42He

78 35Br

4 2He

11 5B

(c) 178O (d) 11H 8 2 4Be + 1H the subsidiary particle is

+ 01n

+ 13H +g

(d) 21H (d) 42He 34 17Cl

+ 01n

24 11Na

8 4Be

+ 10n

+ 11H

140 56Ba

1 + 94 36Kr + 20n

37 17Cl 239 92U

(d) 94Be + g

+ 1601n + 1411H

+g

54. Which of the following reactions represents a fusion reaction? (a)

238 92U

+ 01n

(c)

75 33As

+ 24He

239 92U 78 35Br

(b) 13H + 21H + 01n

55. For the conversion 238 92U (a) 4.14

109 J mol–1

(d)

234 90Th

32 16S

+ 13H

4 2He 34 17Cl

+ 10n + 01n

+ 24He, Dm = – 7.639 × 10–30 kg. The energy released during the process is

(b) 4.14

1010 J mol–1

(c) 4.14

1011 J mol–1

56. In a nuclear reactor, the speed of neutrons is slowed down by (a) ordinary water (b) heavy water (c) zinc rod

(d) 4.14

1012 J mol–1

(d) mercury

19.6 Complete Chemistry—JEE Main

57. The 146 C in upper atmosphere is generated by the nuclear reaction (a)

14 7N

+ 11 H æÆ 146 C + 01 e + 11 H

(b)

14 7N

æÆ 146 C + 01e

(c)

14 7N

+ 10 n æÆ 146 C + 11 H

(d)

14 7N

+ 11 H æÆ 116C + 42 He

(c)

238

58. In nuclear reactors, cadmium is used to (a) slow down neutron (b) absorb neutron (c) activate neutron (d) absorbed energy released during nuclear reaction. (b) 235U (a) 234U 60. In Breeder reactors, the coolant used is (a) an alloy or Na and K (b) cadmium

U

(d)

(c) graphite

239

U

(d) heavy water

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (d) (b) (d) (a) (c) (a) (a) (d) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(c) (b) (c) (c) (a) (b) (b) (b) (a) (b)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(b) (b) (d) (b) (d) (b) (a) (b) (b) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(c) (c) (b) (b) (c) (b) (c) (c) (d) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(d) (a) (a) (b) (a) (d) (b) (a) (b) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(b) (c) (b) (d) (c) (a) (a) (b) (b) (a)

HINTS AND SOLUTIONS 232 - 208 =6 4 Number of beta particles emitted = 82 – (90 – 6 ¥ 2) = 4 t0.5 26. 10,000 disintegrations ætæÆ 5000 disintegrations æÆ 2500 disintegrations 15. Number of alpha particles emitted =

0.5

Obviously, 2t0.5 = 50 days. 28.

l =

Hence, t0.5 = 25 days

0.693 5000 y

t = -

2.303 [A] 2.303 ¥ 5000 y 2.303 ¥ 5000 log = log (0.4) = ¥ 0.401 y = 6667 y. [A]0 l 0.693 0.693

29. 50% of 14C means half of 14C has disappeared. Therefore,

t = t0.5 = 5760 y.

-3

Ê 10 kg ˆ 33. E = mc2 = Á (3 ¥ 108 m s–1)2 = 1.49 ¥ 10–10 J Ë 6.023 ¥ 1023 ˜¯ 1 eV = (1e) (1 V) = (1.60 ¥ 10–19 C) (1 V) = 1.60 ¥ 10–19 J 1 eV ˆ (1.49 ¥ 10–10 J) Ê 1 amu 1.49 ¥ 10–10 J ÁË -19 ˜ 1.60 ¥ 10 J ¯

931.4 MeV

Nuclear Chemistry 19.7

Ê 931 MeV ˆ 37. Binding energy = (0.10242 amu) Á = 95.35 MeV Ë 1 amu ˜¯ 38. E = (2 me)c2 = (2 ¥ 9.11 ¥ 10–31 kg) (3 ¥ 108 m s–1)2 = 1.64 ¥ 10–13 J 39. The precise conversion is

10 4Be

(4p, 6n, 4e)

10 5B

(5p, 5n, 4e) + –10 e

Dm = {mass of (5p + 5n + 4e) + mass of an electron} – mass of (4p + 6n + 4e) = mass of (5p + 5n + 5e) – mass of (4p + 6n + 4e) = mass of (105B – 104Be) 40. The precise reaction is

8 5B(5p,

3n, 5e)

8 4Be(4p,

4n, 5e) + +10 e

Dm = {mass of (4p + 4n + 5e) + mass of a positron} – mass of (5p + 3n + 5e) = mass of (4p + 4n + 4e) + mass of (electron + positron) – mass (5p + 3n + 5e) = mass of (48Be – 85B) + mass of 2 electrons 42.

7 3Li

+

(3p, 4n, 3e)

1 1P

2 42He (2p, 2n, 2e)

Energy of reaction = Energy released in the formation of product – Energy require to separate n ??? of reactants = 8(7.06) MeV – 7(5.6) MeV = 17.3 MeV 55. Energy released per mole of reaction is E = (Dm)c2 NA = (7.64 ¥ 10–30 kg) (3 ¥ 108 m s–1)2 (6.022 ¥ 1023 mol–1) = 4.14 ¥ 1011 J mol–1

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of it remaining undercayed after 18 hours would be (a) 16.0 g (b) 4.0 g (c) 8.0 g (d) 12.0 g [2003] X 4 X A + 2. Consider the following nuclear reactions: 238 92 Y N + 2 2 He; Y B L + 2b The number of neutrons in the element L is (a) 142 (b) 144 (c) 140 (d) 146 [2004] 3. The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is (a) 1.042 g (b) 2.084 g (c) 3.125 g (d) 4.167 g [2004] 4. Hydrogen bomb is based on the principle of [2005] 24 5. A photon of hard gamma radiation knocks a proton out of 12 Mg to form 23 Na (b) the isobar of 23 (a) the nuclide 11 11Na (c) the isotope of parent nucleus (d) the isobar of parent nucleus [2005] 234 U to U, if one emission is an a-particle, what should be the other emission(s)? 6. In the transformation of 238 92 92 (b) Two b(c) Two b- and one b+ (d) One b- and one g (a) One b + and one b[2006]

ten times the permissible value, after how many days will it be safe to enter the room? (a) 1000 days (b) 300 days (c) 10 days (d) 100 days

[2007]

19.8 Complete Chemistry—JEE Main

8. Which of the following nuclear reactions will generate an isotope? (a) neutron particle emission (b) positron emission (c) a-particle emission (d) b-particle emission

[2007]

ANSWERS 1. (b) 7. (d)

2. (b) 8. (a)

3. (c)

4. (a)

5. (a)

6. (b)

HINTS AND SOLUTIONS 1. 18 hour implies six half-lives. Hence, the mass remaining undecayed will be 256 m = 6 = 4.0 g 2 2. The correct nuclear reactions are 238 92 M

Æ

230 88

N + 2 42 He

and

230 88

N

Æ

230 86

L + 2 +10 e

The number of neutrons in L is 144 (= 230 – 86). 3. Twenty four hours is equivalent to six half-lives. Mass of remaining radioisotope = (1/26) (200 g) = 3.125 g. 4. Hydrogen bomb is based on the fusion reactions involving the combination of small nuclei to form a larger nucles. During this process, energy is released because the binding energy per nucleon of lighter elements is smaller than those of the intermediate size. 5. The mass number and charge number both decrease due to the knocking out of proton. 6. The transformation reactions are 7. The fraction of radioactive material remaining undecayed after each successive half-life is as follows. 1 1 1 1 Æ Æ Æ 1 t Æ t t t 1/ 2 2 1/ 2 4 1/ 2 8 1/ 2 16 1/10th of the radioactive material will be present in between third and fourth half-lives, i.e. in between 90 days and 120 days. The minimum time after which it is safer to enter the room may be 100 days. This may be quantitatively

ln 2 ln 2 N0 = k t ; For a half-life; t1/2 = i.e. k = t1/ 2 k N N0 /N = 10, time required will be t1/ 2 ln N 0 / N 30 Ê ln 10 ˆ t1/ 2 = t= = Á = = 99.67 d ˜ Ë ln 2 ¯ log 2 0.3010 k ln

For

Thus, 100 days will be the minimum safer time to enter the room. 8. The emission of neutron does not change the atomic number of the element and thus generates an isotope.

20

Purification and Characterization of Organic Compounds

Quite often, one has to determine the molecular formula of a compound from the quantitative data on its elements. A brief review of this is described in the following. 1. Carbon is estimated in terms of carbon dioxide Mass per cent of carbon in the compound =

mCO2 MC · ¥ 100 = mcompound M CO2

Ê 12 ˆ Ê mCO2 ˆ ¥ 100 ÁË ˜¯ Á 44 Ë mcompound ˜¯

2. Hydrogen is estimated in terms of water Mass per cent of hydrogen in the compound =

mH 2O 2M H · ¥ 100 = M H 2O mcompound

Ê 2 ˆ Ê mH 2O ˆ ¥ 100 ÁË ˜¯ Á 18 Ë mcompound ˜¯

3. Nitrogen in the compound is estimated using either Dumas’ method or Kjeldahl’s method. In Dumas method, nitrogen is converted into nitrogen gas. From the volume of gas collected, we get Mass per cent of nitrogen in the compound =

mnitrogen mcompound

¥ 100 =

( pV / RT ) ( M N 2 ) mcompound

× 100

If the volume of gas is measured at STP, then we can use the following expression. Per cent of nitrogen in the compound =

{(VN 2 )STP / (22 400 cm3 mol-1 )}{M N 2 } mcompound

¥ 100

In Kjeldahl’s method, nitrogen is converted into ammonia which is absorbed in sulphuric acid. The remaining sulphuric acid is determined by titrating against a standard alkali solution. If Vacid and Nacid are the volume and normality of acid solution taken to start with and Valkali and Nalkali are the volume and normality of alkali solution used in the back titration, then we have Mass per cent of nitrogen in the compound =

where

mNH3 M NH3

M N . mNH3 × 100 M NH3 mcompound

= nNH3 = Vacid Nacid – Valkali Nalkali

4. Halogen is estimated in terms of AgX Mass per cent of X in the compound =

M X . mAgX × 100 M AgX mcompound

20.2 Complete Chemistry—JEE Main

5. Sulphur is estimated in terms of BaSO4 MS

Mass per cent of S in the compound =

.

mBaSO4

M BaSO4 mcompound

× 100

6. Phosphorus is estimated in terms of Mg2P2O7 (magnesium pyrophosphate) Mass per cent of P in the compound =

2M P

.

mMg 2 P2O7

M Mg 2 P2O7 mcompound

100

From the given percentages of elements, one can determine the empirical formula of the compound as described in the following. Divide each per cent by the respective relative molar mass. The resultant numbers give the relative ratio of atoms present in the molecule. If fractions appear in the ratio, then one can multiply by the smallest number so as to get integral ratio of atoms. After determining empirical formula, one has to determine molecular formula, which is an integral multiple of empirical formula. The integral multiple is given as Molar mass of the compound Empirical molar mass of the compoound Molar mass of the compound is determined by a suitable physico-chemical method, e.g. Victor-Meyer method, volumetrical method, and colligative properties. A few expressions to compute molar mass of a compound are as follows. m2 1. From relative lowering of vapour pressure M2 = (m1 / M1 ) (-D p1 / p1* ) n=

2. From elevation of boiling point 3. From depression of freezing point 4. From osmotic pressure

M2 =

K b m2 DTb m1

M2 =

K f m2 (-DTf ) m1

Ê m2 ˆ RT M2 = Á ˜ ËV ¯ P

In all the above expressions, the subscripts 1 and 2 stand for solvent and solute, respectively. 5. From vapour density hydrogen is 2 g mol–1, the molar mass of the gas will be Molar mass of gas = (2 g mol–1) (vapour density) 6. From the silver salt method for cabroxylic acids The carboxylic acid is converted into silver salt which on strongly heating produces silver. From the masses of silver salt and silver, the molar mass of acid can be determined as follows. mass of silver salt ¥ 108ˆ˜ g eq–1 Equivalent mass of silver salt, (Msalt)eq = ÊÁ ¯ Ë mass of silver Equivalent mass of acid,

(Macid)eq = (Msalt)eq – 107 g eq–1

Molar mass of acid,

Macid = (basicity) (Macid)eq

7. From the platinichloride method for organic bases Organic bases form solid platinichlorides (or chloroplatinates) with chloroplatinic acid (H 2PtCl6). These, on heating, leave behind platinum. Now, each chloroplatinic acid H2PtCl6 (==PtCl4 · 2HCl) contains 2 molecules of HCl. Hence, 1 1 equivalent of salt 1 equivalent of H2PtCl6 1 equivalent of base (molar mass of Pt) 2

20.3

195 g eq–1 2 Equivalent mass of platinum salt mass of platinum salt 195 (Msalt)eq = × g eq–1 mass of platinum 2 Equivalent mass of base 410 (Mbase)eq = (Msalt)eq – (MH2PtCl6)eq = ÈÍ( M salt )eq g eq -1 ˘˙ 2 Î ˚ Mbase = (acidity) (Mbase)eq

Detection of Elements (N, S and Halogens) extract of the given organic compound is prepared as described in the following. A pea sized dry piece of freshly cut sodium metal is taken in an ignition tube. This piece is melted to shining globule gently and then strongly till the lower end of the tube becomes red hot. The tube is then plunged and broken in about is known as sodium extract.

Test For Nitrogen To a little sodium extract (say, 2 mL), add a little solid ferrous sulphate. Boil the solution and add a little dilute sulphuric acid. Appearance of prussian blue or green coloration indicates the presence of nitrogen in the organic compound. Na +

C+N 

Fuse

æææ Æ Na+ + CN –

Ê from organicˆ ÁË compound ˜¯

Fe2+ + 6CN– 3[Fe(CN)6]4– + 4Fe3+

[Fe(CN)6]4– Fe4[Fe(CN)6]3 prussian blue

Test for Sulphur To a little sodium extract (say, 2 mL), add a few drops of acetic acid and lead acetate. Appearance of black precipitate indicates the presence of sulphur in the organic compound. 2Na + S 2Na+ + S2– (from organic compound)

S2– + Pb2+

PbS (black ppt)

Alternatively, a few drops of freshly prepared sodium nitroprusside is added to 2 mL sodium extract. Appearance of purple coloration indicates the presence of sulphur. [Fe(CN)5NOS]4– S2– + [Fe(CN)5NO]2–

Test for Nitrogen and Sulphur Present Together To a little sodium extract (say, 2 mL), add a little dilute sulphuric acid and ferric chloride solution. Appearance of red coloration indicates the presence of both nitrogen and sulphuric in the organic compound. Na + C + N + S Na+ + CNS– (from organic compound)

20.4 Complete Chemistry—JEE Main

Fe3+ + 3CNS–

Fe(CNS)3 red colour

Test for Halogens To about 5 mL of sodium extract, add equivalent volume of concentrated nitric acid. Boil the solution till volume is reduced to half. This ensures the evaporation of CN–, S2– or CNS– if present in the solution, since otherwise these ions also show the test carried out for halogens. Cool the solution. To a portion of this solution, add silver nitrate solution. Appearance of (a) white precipitates soluble in NH4OH indicates the presence of chlorine. (b) light yellow precipitates soluble in excess of NH4OH indicates the presence of bromine (c) yellow precipitates insoluble in NH4OH indicates the presence of iodine. Bromine and iodine may also be detected by carrying out the carbon tetrachloride (or carbon disulphide) test. To a little solution obtained after treating with concentrated nitric acid, add about 1 mL of CCl4 (or CS2). Add a little chlorine water or concentrated nitric acid. Shake the solution vigorously. Appearance of red (or violet) coloration in CCl4 layer indicates the presence of bromine (or iodine) in the organic compound.

MULTIPLE CHOICE QUESTIONS

(a) simple distillation (c) distillation under reduce pressure

4. 5. 6. 7.

8. 9.

(b) fractional distillation (d) steam distillation

(a) distillation (b) steam distillation (c) crystallization (d) Chromatorgraphy is based on the phenomenon of (a) adsorption (b) absorption (c) solubility (d) Essential oils can be isolated by (a) crystallization (b) steam distillation (c) sublimation (d) Sugar is decolourized by (a) coal (b) carbon black (c) charcoal (d) Two substances having slightly different solubility can be separated by (a) distillation (b) factional distillation (c) crystallization (d) fractional crystallization Liebig method is used for the estimation of (a) carbon only (b) hydrogen only (c) carbon and hydrogen (d) Per cent of carbon in an organic compound is given as M C . mCO2 M C mcompound ¥ 100 (b) ¥ 100 (a) M CO2 mcompound M CO2 mCO2 (c)

M CO2

mCO2

M C mcompound

¥ 100

(d)

M CO2 mcompound MC

mCO2

¥ 100

where m is the mass of the substance of M stands for its molar mass. 10. Per cent of hydrogen in an organic compound is given as mH 2O MH 2M H mH 2O ¥ 100 (b) ¥ 100 (a) M H 2O mcompound M H 2O mcompound

fractional distillation chemisorption distillation coke

nitrogen

20.5

(c)

M H 2O mcompound MH

mH 2O

¥ 100

(d)

M H 2O mcompound 2M H

mH 2O

¥ 100

11. Dumas method is used for the estimation of (a) carbon (b) hydrogen (c) sulphur (d) nitrogen 12. The expression to compute per cent of nitrogen in an organic compound by the Dumas method is (a)

(c)

(VN 2 )STP M N 2

¥ 100

(22414 cm mol ) mcompound (22414 cm3mol-1 ) M N ¥ 100 (VN )STP mcompound 3

-1

2

(b)

(d)

2

(VN )STP mcompound (22414 cm3 mol-1 ) M N

¥ 100

2

2

22414 cm3 mol-1 mcompound ¥ 100 M N2 VN

( )STP 2

13. Kjeldahl’s method is used to estimate the element (a) carbon (b) hydrogen (c) sulphur (d) nitrogen 14. The expression to compute per cent of nitrogen in an organic compound by the Kjeldahl’s method is Ê MN ˆ (a) Á ˜ (Vacid Nacid – Valkali Nalkali) ¥ 100 Ë mcompound ¯

Ê mcompound ˆ (b) Á (Vacid Nacid – Valkali Nalkali) ¥ 100 Ë M N ˜¯

Ê MN ˆ (c) Á ˜ (Vacid – Valkali) ¥ 100 Ë mcompound ¯

Ê mcompound ˆ (d) Á (Vacid – Valkali) ¥ 100 Ë M N ˜¯

15. The Carius method is used for the estimation of (a) carbon (b) hydrogen (c) halogens (d) nitrogen 16. The expression to compute the per cent of halogen X in an organic compound by the Carius method is Ê mAgX ˆ M Ag ¥ 100 (a) Á ˜ Ë M AgX ¯ mcompound (c)

M AgX mcompound MX

mAgX

¥ 100

(b)

mAgX MX ¥ 100 M AgX mcompound

(d)

M X mcompound ¥ 100 M AgX mAgX

17. Molar mass of volatile compounds can be determined by (a) volumetric method (b) Victor Meyer’s method (c) depression in freezing point (d) elevation in boiling point 18. Molar mass and vapour density are related to each other through the expression (a) Msubstance = (2 g mol–1) (Vapour density) Ê 1 g mol-1 ˆ (c) Msubstance = Á ˜¯ (Vapour density) 2 Ë

Ê ˆ 1 (b) Msubstance = Á (Vapour density) -1 ˜ Ë 2 g mol ¯ (d) Msubstance = Vapour density

19. 0.1890 g of an organic compound gives 0.287 g of silver chloride determined by Carius method. The per cent of Cl in the compound is (a) 18.7 (b) 37.5 (c) 56.2 (d) 75.0 20. An organic compound contains 90% C and 10% H by mass. Its empirical formula is (b) C2H4 (c) C3H8 (d) C3H4 (a) C2H6 3 21. 0.246 g of the organic compound gave 22.4 cm of nitrogen gas at STP as determined by Dumas method. The per cent of nitrogen in the compound is (a) 11.38 (b) 17.07 (c) 22.76 (d) 34.14

20.6 Complete Chemistry—JEE Main

22. An organic compound contains 58.53% C, 4.06% H and 11.38% N by mass. The empirical formula of the compound is (a) C6H5N2O (b) C6H5NO2 (c) C5H6N2O (d) C5H7NO2 3 23. A monoacid organic base (0.0915 g) requires 15 cm of N/20 HCl for complete neutralization. The molar mass of the base is (b) 122 g mol–1 (c) 244 g mol–1 (d) 183 g mol–1 (a) 61 g mol–1 24. An organic compound (0.156 g) on Kjeldahl’s method for nitrogen estimation requires 16.25 cm3 of 0.1 M NaOH for the remaining unneutralized acid in 60.25 cm3 of 0.1 N H2SO4 taken to start with. The per cent of nitrogen in the compound is (a) 39.26 (b) 78.52 (c) 49.26 (d) 59.26 3 25. A volatile liquid (0.146 g) displaced 30.5 cm of air at 305 K and 760 mmHg pressure. The molar mass of the liquid is (a) 117.8 g mol–1 (b) 118.8 g mol–1 (c) 119.8 g mol–1 (d) 120.8 g mol–1 26. A compound (0.244 g) containing sulphur yielded 0.9336 g of BaSO 4. The per cent of sulphur in the compound is (a) 25.46 (b) 52.46 (c) 64.25 (d) 46.52 27. An organic compound (0.45 g) on combustion yielded 1.1 g of CO2 and 0.3 g water. Besides C and H, nitrogen is also present in the compound. The empirical formula of the compound would be (a) CH2N

(b) C2H3N

(c) C3H4N

(d) C2H5N

28. 0.1 g of an organic monobasic acid on complete combustion gave 0.2545 g of CO2 and 0.04428 g of H2O. For complete neutralization 0.122 g of the acid required 10 cm3 of 0.1 M NaOH solution. The molecular formula of the acid would be (a) C7H6O2

(b) C6H7O2

(c) C7H7O2

(d) C6H6O2

29. Which of the following statements is true regarding the fractional distillation of a mixture of two liquids? (a) The distillate collected is the liquid of lower boiling point (b) The distillate collected is the liquid of higher boiling point

column 30. For which of the following compounds, Kjeldahl’s method can be used to estimate nitrogen? (a) Aniline (b) Methyl amine (c) Urea (d) Toluidine 31. Which of the following compounds is not expected to exhibit Lassaigne’s test of nitrogen? (b) CH3NH2 (c) CH3NO2 (d) NH2OH (a) CH3CN 32. Which of the following compounds is expected to give red colouration while performing Lassaigne’s test of nitrogen? (a) Urea (b) Thiourea (c) Aniline (d) Nitrobenzene

ANSWERS 1. 7. 13. 19. 25. 31.

(b) (d) (d) (b) (d) (d)

2. 8. 14. 20. 26. 32.

(c) (c) (a) (d) (b) (b)

3. 9. 15. 21. 27.

(b) (a) (c) (a) (c)

4. 10. 16. 22. 28.

(a) (b) (b) (b) (a)

5. 11. 17. 23. 29.

(b) (d) (b) (b) (a)

6. 12. 18. 24. 30.

(c) (a) (a) (a) (c)

20.7

HINTS AND SOLUTIONS Ê M Cl ˆ Ê 100 ˆ Ê 35.5 ˆ Ê 100 ˆ 19. Per cent of chlorine = Á (mAgCl) Á ˜ ˜ = ÁË 143.5 ˜¯ (0.287) ÁË 0.189 ˜¯ = 37.57 M m Ë AgCl ¯ Ë compound ¯ 90 10 :: 7.5 : 10 :: 3.4 ; Empirical formula C3H4 : 12 1 VN 2 Ê 100 ˆ Ê ˆ 22.4 ˆ 100 ˆ 21. Per cent of nitrogen = Á (MN2) Á = ÊÁ (28) ÊÁ = 11.38 ˜ ˜ -1 ˜ 3 Ë 0.246 ˜¯ Ë 22400 ¯ Ë 22400 cm mol ¯ Ë mcompound ¯ 20. C : H ::

22. C : H : N : O ::

58.53 4.08 11.38 26.01 :: 4.88 : 4.08 : 0.81 : 1.63 : : : 12 1 14 16

:: 6 : 5 : 1 : 2. Empirical formula = C6H5NO2 24. Initial amount of H+ = VM = (0.06025 dm3) (0.1 mol dm–3) = 0.006025 mol Remaining amount of H+ = (0.01625 dm3) (0.1 mol dm–3) = 0.001625 mol Amount of H+ reacted = (0.006025 – 0.001625) mol = 0.0044 mol Mass of NH3 produced = (Amount of H+) (MNH3) = (0.0044 mol) (17 g mol–1) = 0.0748 g Ê MN ˆ Ê 100 ˆ 14 100 ˆ (mNH3) Á = ÊÁ ˆ˜ (0.0748) ÊÁ = 39.5 Per cent of nitrogen= Á ˜ ˜ ¯ Ë 17 Ë 0.156 ˜¯ Ë M NH3 ¯ Ë mcompound ¯

(

)

(1 atm ) 30 . 5 ¥ 10- 3 L pV 25. n = = 0.00121 mol = RT 0.0825 L atm K - 1 mol- 1 (305 K )

(

M=

)

m 0.146 g = = 120.7 g mol–1 n 0.00121 mol

Ê MS ˆ Ê 100 ˆ 32 ˆ Ê 100 ˆ = 52.5 26. Per cent of sulphur = Á (mBaSO4) Á = ÊÁ ˜¯ (0.9336) ÁË ˜ ˜ ˜ Ë 233 0.244 ¯ Ë mcompound ¯ Ë M BaSO4 ¯ Ê MC ˆ Ê 100 ˆ Ê 12 ˆ Ê 100 ˆ (mCO2) Á 27. Per cent of C = Á ˜ ˜ = ÁË 44 ˜¯ (1.1) ÁË 0.45 ˜¯ = 66.67 M m Ë compound ¯ Ë CO2 ¯ Ê 2M H ˆ Ê 100 ˆ 2 100 ˆ Per cent of H = Á (mH2O) Á = ÊÁ ˆ˜ (0.3) ÊÁ ˜ = 7.4 ˜ ˜ ¯ Ë Ë 18 0.45 ¯ Ë mcompound ¯ Ë M H2O ¯ C:H:N

::

66.67 7.4 25.93 :: 5.56 : 7.4 : 1.85 :: 3 : 4 : 1 : : 12 1 14

;

Empirical formula = C3 H4 N

Ê MC ˆ Ê 100 ˆ 12 100 ˆ (mCO2) Á = ÊÁ ˆ˜ (0.2545) ÊÁ = 69.4 28. Per cent of C = Á ˜ ˜ Ë 44 ¯ Ë 0.1 ˜¯ Ë mcompound ¯ Ë M CO2 ¯ Ê 2M H ˆ Ê 100 ˆ 2 100 ˆ Per cent of H = Á (mH2O) Á = ÊÁ ˆ˜ (0.04428) ÊÁ = 4.9 ˜ ˜ Ë 18 ¯ Ë 0.1 ˜¯ Ë mcompound ¯ Ë M H2O ¯ C:H:O

::

69.4 4.9 25.7 :: 5.78 : 4.9 : 1.61 :: 3.6 : 3 : 1 :: 7 : 6 : 2 : : 12 1 16

Empirical formula = C7 H6 O2; Empirical molar mass = 122 g mol–1 Amount of acid = VM = (10 ¥ 10–3 dm3) (0.1 mol dm–3) = 10–3 mol

20.8 Complete Chemistry—JEE Main

Molar mass of compound =

m

n Molecular formula = C7H6O2.

=

0.122 g = 122 g mol–1 -3 10 mol

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is (b) Na3[Fe(CN)6] (c) Fe(CN)3 (d) Na4[Fe(CN)5NOS] (a) Fe4[Fe4(CN)6]3 [2004] 2. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is (a) acetamide (b) benzamide (c) urea (d) thiourea [2004] 3. 1.4 g of an organic compound was digested according to Kjeldahl’s method and the ammonia evolved was absorbed in 60 mL of M/10 H2SO4 solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution for neutralization. The percentage of nitrogen in the compound is: (a) 3 (b) 5 (c) 10 (d) 24 [2015, online] 4. Sodium extract is heated with concentrated HNO3 before testing for halogens because: (a) Ag2S and AgCN are soluble in acidic medium. (b) Silver halides are totally insoluble in nitric acid (c) S2– and CN–, if present, are decomposed by conc. HNO3 and hence do not interfere in the test. (d) Ag reacts faster with halides in acidic medium [2016, online]

ANSWERS 1. (a)

2. (c)

3. (c)

4. (c)

HINTS AND SOLUTIONS 1. The Lassaigne test involves the reactions Na + C + N Æ NaCN Fe2+ + 6CN– Fe3+ + [Fe(CN)6]4–

Æ [Fe(CN)6]4– Æ Fe4[Fe(CN)6]3

2. Amount of H+ used for the neutralization of NH3 n = 2 (0.1 mol L–1) (0.1 L) – (0.5 mol L–1) (0.02 L) = 0.01 mol Amount of N = Amount of NH3 = Amount of H+ used = 0.01 mol Mass of N = (0.01 mol) (14 g mol–1) = 0.14 g Mass per cent of N in the compound =

 0.14 g  (100) = 46.67%  0.3 g 

Mass per cent of N in the given compounds are as follows.

20.9

Acetamide (CH3CONH2) Urea (NH2CONH2)

14 × 100 = 23.73 59 28 × 100 = 46.67 60

Benzamide (C6H5CONH2)

It is less than 23.73

The compound is urea. 3. Amount of NaOH reacted with excess of H2SO4 is Ê1 ˆ n1 = VM = (20 ¥ 10–3 L) Á mol L-1 ˜ = 2 ¥ 10–3 mol Ë 10 ¯ Amount of H2SO4 reacted with NaOH is n n2 = 1 = 1 ¥ 10-3 mol 2 Amount of H2SO4 reacted with NH3 is Ê1 ˆ n3 = (60 ¥ 10–3 L) Á mol L-1 ˜ – 1 ¥ 10–3 mol = 5 ¥ 10–3 mol Ë 10 ¯ Amount of NH3 (also amount of N in the compound) liberated is n4 = 2n3 = 10 ¥ 10–3 mol percentage of N in the compound is n4 M N (1.0 ¥ 10-3 mol) (14 g mol-1 ) ¥ 100 = ¥ 100 = 10% mcompound (1.4 g ) 4. Both S2– and CN– are decomposed by conc. HNO3 and hence no precipitation of these ions occur with AgNO3.

21 Some Basic Principles Concept of Hybridization 2

2

2p2

hybridization

sp Hybridization diagonal or linear hybridization

Fig. 1

Structure of acetylene

sp2 Hybridization 2

hybridization

2

trigonal

21.2 Complete Chemistry—JEE Main

Fig. 2

Structure of ethylene

sp3 Hybridization ¢

3 3

tetrahedral hybridization 1s(H)

1s(H) H C H

1s(H) H

H sp3(C) 1s(H)

Fig. 3

Structure of methane

Comparison of sp, sp2 and sp3 Hybrid Orbitals of Carbon Number of s (or p) orbitals mixed Total number of orbitals mixed 2 3

Size of orbitals 3

2

Electronegativity of Orbitals 2

3

Some Basic Principles 21.3

Sigma and Pi bond s of pi (p π bond

σ bond

Fig. 4

Illustration of s and p bonds

bonding molecular orbital antibonding molecular orbital

σ* 1s

1s

1s

σ 1s

Fig. 5

s

—C in C2H6 C 2H 2

s and one p

2H

s and two p

∫∫ C bond in

Bond Lengths —C

∫∫ Molecule

—

CH 3

C 2H
CH3CH2+ > CH 3+ Cl3C + < Cl2CH+ < ClCH2+ < CH+3

CH2 == CH ⎯ CH2 ←→ CH2 ⎯ CH == CH2 CH2

CH2

←→

CH2

←→

CH2

CH2

←→

+

←→

primary (or 1°)

+

(CH3)2CH

secondary (or 2°)

←→

CH2

+

CH3CH2

CH2

+

(CH3)3C

tertiary (or 3°)

CH2

←→

21.16 Complete Chemistry—JEE Main

Electrophiles or Electrophilic Reagents

H+

+

+

+

+ 2

3C

+

+ 2

3

3

2

3

Nucleophiles or Nucleophilic Reagents

Cl

H2

3

2

Inductive Effect s bond moment p

d q) (d)

dq

d

C

—H

d+

d

CH3—CH2—

d

CH2CH3 |d+

d

d

CH3CH2—Pb—CH2CH3 | d CH2CH3

+I and I Groups I I I +I I

I

I

— CH3 —CH2CH3 —CH(CH3)2 —C(CH3)3 — C 6H — — — — 3 —

Resonance Effect

I

—

—

2

Some Basic Principles 21.17 O

O

R ⎯ CH ⎯ C ⎯ H ←→ R ⎯ CH == C ⎯ H

Fig. 12

—

Resonance stabilisation of a carbanion

2

O⎯H

O⎯H

NH2

←→

NH2 ←→

Fig. 13

Resonance donation of pi electrons

Acids and Bases +

HA +

B

 A

+

HB+

pair and a 3

3

3

2

Strengths of Acids HA + H2  A + H3 Ka Ka

K

+

[A - ] [H 3O + ] 2

[HA]

equilibrium constant standard euilibrium constant Ka Ka

Ka

Ka Ka

pKa

Ka

Ka

Ka (or

Ka Acids pKa Conjugate bases  CH3

3>

Cl + H+

3

> CH3 < CH3

2

2H

< C 2H 3

21.18 Complete Chemistry—JEE Main

I I I +I

3CH2

3) 2

3) 3

O H⎯C⎯O

O ←→

H ⎯ C == O

Resonance resonating or canonical structures

N == N == O ←→ N ≡≡ N ⎯ O

resonance energy

←→

Kekule’s structures

←→

←→

←→

Dewar’s structures

Some Basic Principles 21.19

Types of Organic Reactions

Substitution Reactions

Free-radical substitution

and Cl2 2

(a) Initiation step

Cl : Cl Æ 2 ·Cl

(b) Propagation steps

CH + ·Cl Æ CH 3· + HCl CH 3· + Cl2 Æ CH3Cl + Cl·

(c) Termination steps

Cl· + ·Cl Æ Cl2 CH 3· + ·Cl Æ CH3Cl

Electrophilic Substitution Reactions 3

C6H6 + Br2 æææÆ C6H Br + HBr (a) Formation of an electrophile Br : 3 Æ

Br +

(b) Electrophilic attack

H Br

+ Br+ →

(c) Elimination of proton

←→ Resonance hybrid

t

H Br

−

FeBr4

⎯⎯⎯→

Nucleophilic Substitution Reactions In S (CH3)3

H Br

Æ (CH3)3

(CH3)3CBr ææææææÆ (CH3)3C + + Br carbonium ion

Br + HBr + FeBr3

21.20 Complete Chemistry—JEE Main

Æ (CH3)3

(CH3)3C +

In S Æ CH3

CH3

H

H rate C ⎯ Br ⎯⎯⎯⎯→ HO− determining step

−

HO + H H

H

H + : Br −

Br → HO ⎯ C

C

H H

H

transition state

Addition Reactions

Electrophilic addition reaction CH2 2 + HBr Æ CH3CH2Br +

H—Br

H+

electrophile

H + + CH2 == CH2

CH3 ⎯ CH2 + Br−

+

Br –

nucleophile +

CH3—CH2

carbonium ion

CH3—CH2 — Br

Nucleophilic addition reaction C O CH3COCH3 + HCN

H—CN

H3C H3C

H+ + CN :–

H3C C == O + CN : → δ+

δ−

CN

C⎯O ; CH3

Elimination Reactions

OH | CH3 — C — CH3 | CN

H3C

H3C C⎯O + H → +

CN

CH3

CN

C ⎯ OH CH3

Some Basic Principles 21.21

— CH2— CH2—X ææææÆ — CH2

CH2

MULTIPLE CHOICE QUESTIONS Hybridization

2 2

2

3

∫∫C—C

2

2

2

2

3

·

3

 (a) H C

2

2

2

2





2C

2

2—CH2—C∫∫ 3

3

2

3



3) 3 C

3C

3

C —C 3

2

∫∫ 3

3

3

2 2

3

rect? 2

3

2

3

2

3

2

3

2

3

2

3

2

3

2

3

(a) CH > C2H > C2H2

(b) CH < C2H < C2H2

p

2H and C2H2 (c) CH > C2H < C2H2 (d) CH < C2H > C2H2 CH 2 2 2

s (a) C2H2

(b) C2H

(c) C2H6

2 3 2H

and C2H2

(d) C2H2Br2

p

21.22 Complete Chemistry—JEE Main

Nomenclature

d CH3 | H3C— C— CH | CH3

CH2

compound CH2

3) 2

d pound H CH3 C ⎯ CH NH2

CH3

(a

CH2 CH3 CH3 CH2 CH2 CH ⎯ CH ⎯ C ⎯ CH2 CH3 CH CH3

CH2 CH3

CH3 CH3

(a) CH3CH2CH2 CH | Cl (c) CH3CH2

3C∫∫C CH

| CH3

3

CH3 | CH3CH—CHCH2CH2CH—CH2 CH —CH3 | | | CH3 CH3 CH2CH2CH2CH3

3 CH CH2CH2

| CH3

Some Basic Principles 21.23

d

CH3 HO

CH3

(a) CH3C(CH3)2CH2CH2CH(CH3)2 (c) CH3C(CH3)2CH(CH3)CH2CH2CH3

(b) CH3C(CH3)2CH2CH(CH3)CH2CH3 (d) CH3C(CH3)2CH2C(CH3)2CH3

(a) — C ∫∫

— C— || C

(a) CH3CH(CH3)CH2CH2

3)3CCH2

(c) CH3(CH2)3

3CH2

6 octadien 2 6 diene 2H

)2

3

∫∫CCH3 octadien 6 (d) 6 diene

CH3 (c) 2

(d) —C ==O | H

2

N

N N N O CN || | CH3 — C — CH2—C—CH3 | CH3

CH3 HO

H

H

Cl C2H5

(a) (2R R (c) (2S R

R S S S 3

NO2 Cl

NO2

∫∫ CCH3

21.24 Complete Chemistry—JEE Main

OH CH2 CH

C(CH3)2

CH(CH3)2 3

|

2

CH3 3CH

CHCH CH3 CH2 C

CO

CH3

NH

N

CH3

O

NN

CH2

CH

CH2

CH2

C

N(CH2CH3)2

CH3 | 3C — N -- C -- CH 2 CH3 | | C 2 H5 CH

NN

3

Isomerism

C 6H cis trans B

Butanal

(a) n H

(a) n H

O || (a) H5C6—C—H

O || (b) H5C6—C—CH3

O || (c) H5C6— C —C6H5

O O || || (d) H5C6— C —CH2— C —CH3

Some Basic Principles 21.25

H

3H

Br3 3H

CH3 — CH

CH — CH3 and CH3— CH2— CH 3

H

3

COOCH3 OH

H

OH COOH (I)

(a)

CH == CH

O

(c)

O

3

OH

2

2

Br3

CH2

2

H

COOH OH

H

H

OH

HO

COOCH3 (II)

COOH OH H COOCH3 (III)

(b) O

O

O

(d)

O

21.26 Complete Chemistry—JEE Main

cis and trans

(a) A meso (b) A meso (c) A meso (d) A meso

(a) CH3

3CH(CH3

3CH(CH3)CH2

3CH(CH3)CH2Cl

Some Basic Principles 21.27

Cl Br | | (a) H—C—C—H | | H H

H H | | (d) H—C—C—CH3 | | Br OH

H Cl | | (c) H—C—C—H | | H H

H Cl | | (b) H—C—C—Cl | | H H

3

3

2

2

3

> CH2

3 > CH2

3 2

2 2

> C∫∫ 2 > C∫∫ 2

2 2

∫∫

2

2

2

2

and —CH2

> C∫∫ 2 > CH2

2

CH3 H

OH C2H5

> CH2

2

2

> CH2 2 > C∫∫

2

21.28 Complete Chemistry—JEE Main

S? CH3 H

OH OH

H

CH3

CH3

OH

H HO

CH == CH2

H

CH3

(I)

(II)

(III)

(IV)

(b) (I) and (III)

CH3 H

(c) (I) and (IV)

OH OH

H

CH == CH2 CH3

CH3

(d) (II) and (IV)

H

OH

HO

CH == CH2

CH == CH2

CH == CH2

H

CH3

(I)

(II)

(III)

(IV)

(a) (II) and (III)

(b) (I) and (IV) CH3

H

(c) (II) and (IV)

OH OH

H

CH == CH2 CH3

CH3

OH

(d) (III) and (IV) H

HO

CH == CH2

CH == CH2

CH == CH2

H

CH3

(I)

(II)

(III)

(IV)

(b) (I) and (III)

(c) (II) and (IV)

meso meso meso meso meso S racemic

(a) (R (c) meso S (a) (S (c) (R

CH == CH2

CH == CH2

(a) (I) and (II)

(a) (I) and (II)

CH == CH2

R R R

(d) none

Some Basic Principles 21.29

(a) CH3CH2

Æ CH3CH2

2

(b) CH3CH2

3) 2

(c) CH3

S S CH3

3)CH(CH3)2 +

2

3

+ Cl

3CHClCH(CH3)C2H

CH3

(b)

CH3

(c)

Cl

H

H

Cl

H

Cl

CH3

H

CH3

H

H

CH3

C2H5

(a)

CH(CH3)CH(D)CH2 (b) CH3

(c)

(d)

C2H5

H

CH3

H

CH3

H

D

H

D

D

H

CH2D

C2H5D

C2H5

CH(CH3)CH(D)CH2 (b) C2H5

H

H

CH3

H

D

H

D

(a) Cis

trans

H3C

H

D

H CH2D

CH2D

6H

(c)

C2H5

Cl2

(d)

C2H5

H3C

H

H

CH3

D

H

D

H

CH2D

CH2D R

meso

CH3

2H

H3C

CH2D

CH3

2H

H

(a)

H

C2H5

H3C

S R

H

C2H5

C2H5

C2H5

CH3

(d)

Cl

R R

+ HCl

+ H2

Æ CH3CH2CH2

(d) CH3CH2CH2CH(Cl)CH3 (a)

Æ CH3CH2

+ CH3 Æ CH3

2

2

CH2D

21.30 Complete Chemistry—JEE Main

2

E F and G H3C

H3C

O

H3C

CH3

H3C

(E)

H3C

OH CH3

CH3

H3C

OH

(F)

(G)

(a) E F and G (c) F and G

E F and E, G F and G

Inductive and Resonance Effects

in a molecule in a molecule I 3

3

I (a) —C6H

(a)

(b) —CH3

NH2

+

(b)

(a) CH 3·

(c)

NH3

+

(a) CH3

(c) —CH2CH3

(d) —C(CH3)3

(d)

OH

+ 2C

H2

·

+ 3C

2CH

2

(a) + I I

·

3C

I I I 2

2

2

2

I (a) — I < — Cl < — Br < — +

2

Cl

+ 3

C 2H < C 2H

(b) CH3 (d) CH3

3) 3

3

3) 3

3

< C 2H > C 2H

2

(a) CH3 (c) CH3 (a) (b) (c) (d)

CH3CH2 CH3CH2 CH3CH2 CH3CH2

(a) Cl3 (c) Cl3 (a) (b) (c) (d)

> (CH3)2 2 > (CH3)2

2

(CH3)3 (CH3)3 (CH3)3 (CH3)3

3CH(Cl)CH2 3CH(Cl)CH2 3CH(Cl)CH2 3CH(Cl)CH2

< (CH3)2 2 < (CH3)2 2

3) 3 3) 3

2CH2CH2 2CH2CH2 2CH2CH2 2CH2CH2

2

2

3

2

2

2

2

3

2

2

3) 2

3CH2

3) 2 3) 2 3) 2

3CH2 3CH2 3CH2

21.32 Complete Chemistry—JEE Main

(a) I

2 2

3C

3

3C

3

Ka

+ H+

CH3

3

3

3

(b) +I

Types of Organic Reactions 3

3

3

(a

2

+

(b

+

(c) H2

3

(c) H+

(d) Ca2+

(a) In S (b) In S centre

(a) In S

not correct? + 3

.

.

3) 3 C

3)2 CH 3) 3 3CH2

Æ (CH3)3 Æ CH3CH2

> CH3CH2+ > (CH3)2CH + > (CH3)3C+

Some Basic Principles 21.33 PCl5

HO2 CCH2CHCO2H

ether

HO2CCH2CHCO2 H

OH

Cl

(–)-malic acid

(+)-chlorosuccinic acid

HO2CCH2CHCO2H

PCl5 ether

HO2CCH2CHCO2H OH

Cl (+)-chlorosuccinic acid

(+)-malic acid

+

+

C

(d) S

ANSWERS

HINTS AND SOLUTIONS 1

2

3

4

CH 2 == C == CH -- CH3 2 2

2

3

21.34 Complete Chemistry—JEE Main 2

 in (CH ) C  C 3 3

3

3

C ∫∫ CH

2 2

3 3

2

2

p

H3C

CH3

C

H3C

CH2

H3C

CH2

CH3

CH3

CH3

C

H3C

CH2

CH2

CH3

CH3 CH3 H3C

C

CH

CH2 CH3

CH3 CH3

CH CH2

H3C

CH2

CH3

CH3 CH3 CH3

CH

CH3 4

H3C

3

C

2

1

CH

CH3

CH3 1

2

CH2

CH

3

CH CH3

CH3

CH3

C CH3

CH2OH

4

CH3

Some Basic Principles 21.35

1

2

3

4

5

6

7

CH3 CH == CH CH == CH C ∫∫ CCH3 1 5

4

3

CN

2

CH3COCH2C

CH3

CH3

R and R

R

R

N N

C C

C

C

C

C; C

C

C

C

C

C; C

C

C

C

C

3 2CH3

C

C

C

H

C; C

C

C

C

C

3CH2

2CH3

CH3

O

cis

H

trans 3 3 and (CH3)2 CH2ClCH2CH2CH3 and CH3CHClCH2CH3 —CH and CH3 2 3 2—CH3 cis trans

2—CH3

2

O

H5 C6 C

2

C

cis-trans

3

3

C; C

C

CH2

2

C

OH CH3

and cis

H5 C6 C

2

O CH

C

CH3

trans

CH 3CH 2CH 2CH 2 CH2CH 2CH3 * CH3CHCH2CH2CH2CH2 ; CH3CH2CHCH2CH2CH3 ; CH3CH CH3

* CHCH2CH3 ;

CH3 CH3

CH3 (two optical isomers)

CH3

(two optical isomers)

CH3

CH3

CH3 CHCH2CHCH3 ; CH3 CCH2CH2CH3 ; CH3CH2CCH2CH3 ; CH3C CH3

CH3CH2 Br3CCBr3

CH3 3CHBr2

CH3 3CBr3

CH3CH2CBr3 3CHBrCHBr2 CHBrCHBr 3 2

CH3 2CH2

3C(Br)2CH2

CHCH3

CH3 CH3 2CHBr2

2CH2CHBr2

2CBr3

2CHCHBr2

2CH(Br)CH2Br

2CHCBr3

21.36 Complete Chemistry—JEE Main

p H BrCH2 C

CH2CH3 and CH3

H

CH3

HO

OH

HO

H

H

CH3

H

C

C

H3C

CH3

CH3

H

CH3

Br

CH3

H

H

OH

OH

H

OH

CH3

plane of symmetry

CH3

optically active

optically inactive

meso * 3 CH

2

S

> CH3

R S and R S and R CH3

CH3

HO

H

H

H

Cl

Cl

C2H5 I

CH3

CH3

OH

HO

H

H

OH

H

Cl

H

H

Cl

C2H5 II

C2H5 III

C2H5 IV

Some Basic Principles 21.37 n

n

CH3

CH3 H HO

OH

OH

H

H

CH3 (I)

CH3

H

H

OH

OH

H

OH

CH3 (II)

CH3 (III)

CH3 OH

H

HO

H CH3

meso

CH3CHClCH2CH3

ClCH2CHClCH2CH3 + CH3C(Cl2)CH2CH3

( )

( )

achiral

+ CH3CHClCHClCH3 + CH3CHClCH 2CH2Cl ( )

CH3 H

Cl

CH3

H

H

CH3

H

H

D

H

H

D CH2 D

R

H C2H5

CH3

H3C

H

H CH3

C2 H 5

configuration

R

R

H

R H

configuration

C2 H5

C2H5

C2H 5 H 3C

H

CH3

CH2D

configuration

R

R

C2 H 5 H

CH3

R

C2 H 5

H3C

Cl

Cl

CH3 Cl

H

CH3

C2 H 5

H

( )

D CH2 D

D

R CH2D

configuration

21.38 Complete Chemistry—JEE Main

CH3 CH3

H

H

Cl CH3

CH2 Cl

CH3

CH3

CH3

CH3

H

CH2Cl

H

CH3

Cl

CH3

H

CH3

H

H

Cl

H

Cl

H

Cl

Cl

Cl

H

Cl

CH3

CH3

CH3

CH3

CH2Cl

(I)

(II)

(III)

(IV)

(V)

R R S meso

E and G cis

+

-- NH3

trans

R

Cl2

CH2Cl

H

Cl

H

CH3

H

Cl

H

CH3

C2 H 5

I I

S R

6H

C2 H 5

E and F

CH3

CH2Cl

CH3 H

Cl CH2

H

Cl CH3

Some Basic Principles 21.39

I I

d· d·

p

2C

CH2

d· æ H 2 C —CH CH— CH 2 ¨Æ

.

.

CH2 I I

K°a

(b) H2

3

2

2

p

(d) In Cl3

Cl

C

Cl

–

¨Æ æ

Cl

–

Cl

C

Cl

¨Æ æ

Cl

Cl

C

Cl

Cl – 3C

3C

3C

3

(b) +I

3

3) 3C

(b) In S

+

> (CH3)2CH+ > CH3CH+2 > CH+3

21.40 Complete Chemistry—JEE Main

2

(i) Delocalization of electrons X

X

←→

X

X

←→

←→

X

←→

(ii) Difference in hybridization of carbon 3

C

+ C

+ O

2

O

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN

[2003] H CH3 C 2H

CH (I)

H

CH3 C 3H

CH3

2H

(II)

C+ H (III)

CH3 C 2H

CH (IV)

C 2H

[2003] nH2n

2

[2003] 3

3) 2

[2003]

Some Basic Principles 21.41 2H

3

3

[2004] 2

(a) acetone

(b) acetic acid

(c) acetonitrile

(d) acetamide

[2004]

[2004] CH 3 | CH CH3

(b) CH3 CH2

(a) CH3—CH2—CH2—CH3 H

(c) CH3

(d) CH3—CH2—C∫∫

C

[2004]

C2H5

[2005]

(i) CH3C O –

3

O

[2005] 2

) 3]

6]

3)

]2+

]

[2005] [2005]]

CHO

(a) HO

(b)

H

SH

CH2OH

COOH H2 N (c) H Ph

NH2 H Ph

(d) H2N

H H

[2007]

21.42 Complete Chemistry—JEE Main

[2007] HOOC HO H

H

COOH OH

[2008] (a) CH3Cl

(b) (C2H )2 CHCl

2

(c) (CH3)3CCl

3

3

(d) (CH3)2 CHCl

3 2

[2008]

2 3

2

[2008]

CH3 [2009]

[2009] 3) 3C

3C

3)2CH

(a) Cl3C > C6H CH2 > (CH3)2CH > (CH3)3C (c) C6H CH2 > Cl3C > (CH3)3 C > (CH3)2CH

6H

CH2 (b) (CH3)3C > (CH3)2CH > C6H CH2 > Cl3C (d) (CH3)2CH > Cl3C > C6H CH2 > (CH3)3C [2009]

[2013]

(a)

(b) [2014, online] A

B

(b)

(a)

(c)

3C

∑

, (CH3)3 C∑

[2014, online]

[2015, online] 6H

[2015, online]

[2015]

Some Basic Principles 21.43

[2016, online]

CO2H H

OH

H

Cl CH3

[2016]

ANSWERS

HINTS AND SOLUTIONS

sp3

sp2

sp3

sp3

sp2

CH3

C

CH3 ;

CH3

C

OH ;

O

O

(acetone)

(acetic acid)

sp 3

sp

CH3

C

N;

sp3

sp2

CH3

C O

(aceonitrile)

(acetamide)

2

H3C

CH2

CH

CH3 ;

H 3C

Cl (2-chlorobutane)

H3C

CH2

CH

CH

Cl

Cl

CH3 ;

(2, 3-dichlorobutane)

CH

CH

Cl

Cl

CH3

H 3C

CH

COOH

OH

(2, 3-dichloropentane)

(2-hydropropanoic acid)

meso CH3

NH2

CH C2H5

CN– > CH3O– > CH3COO– > H3 C

SO– 3

21.44 Complete Chemistry—JEE Main 2

) 3]

CH3

CH3

CH3

H

Cl

H

Cl

Cl

H

H

Cl

Cl

H

H

Cl

CH3

CH3

(i)

(ii)

CH3 (iii)

CHO HO

H CH2OH

meso CH3 7

6

5

CH3 CH2 CH2

4

3

2

CH

C

1

CH2CH3

CH3 CH2CH3 1 COOH

1 COOH HOOC HO

COOH OH

H

HO H

H

HO

3 COOH 4

OH

R1

COOH R-configuration

H

3

2

H

R′

R′

H HO

H

2

H OH

4 COOH

HOOC

OH

R′ R-configuration

Some Basic Principles 21.45

cis-trans

CH3

C

H3C

CH3

CH3 3) 3C 3C

CH3 1

CH3

2

3

4

CH2

CH

CH

5

CH2

6

CH3

CH2 CH3

3C

3) 3C

CH3 -- CH -- CHO Ω Cl

;

2- chloropropanal

CH3 -- CH 2 -- CH -- CH3 Ω Cl 2- chlorobutane

CH3 -- CH -- CH 2 -- CH 2 -- CH3 Ω Cl 2- chloropentane

;

Cl Ω CH3 -- C -- CH 2 -- CH3 Ω CH3 2- chloro - 2- methylbutane

p

21.46 Complete Chemistry—JEE Main

C -- C -- C -- C -- C -- C

C -- C -- C -- C -- C Ω C

;

C Ω C -- C -- C -- C Ω C

C -- C -- C -- C -- C Ω C

Ê M Br ˆ Ê mAgBr ˆ ÁM ˜Á ˜ (100) Ë AgBr ¯ Ë mcompound ¯

C -- C -- C -- C Ω Ω C C

;

Ê 80 u ˆ Ê 141 mg ˆ ÁË 188 u ˜¯ ÁË 250 mg ˜¯ (100)

CH3 H3C

C

CH2 —

H2C

CH3

C

H

CH3 5

H3C

4

C

3

CH2

2

CH

1

CH2

CH3

3

CO2H

4

H

4

1

OH 2

H

Cl

Hold atom C-2 and rotate the other three clockwise

1

H

H

3

CO2H

HO 2

Cl

S Configuration

CH3

CH3

Priorities follow counter clockwise

CO2H H

2

H 4

3

CH3

Hold atom C-2 and OH rotate the other three counter clockwise Cl 1

CO2H H

2

CH3

Cl 1

OH

H 4

3

Priorities follow clockwise

R Configuration

22 Hydrocarbons

SECTION 1

Alkanes

Methods of Preparation Decarboxylation of Sodium Salt of Fatty Acids Sodium salt of a carboxylic acid produces alkane when it is dry distilled with sodalime (NaOH + CaO): RCO2Na + NaOH æCaO ææ Æ RH + Na2CO3 Wurtz Reaction An alkyl halide produces alkane when it is made to react with Na in dry ether: ææ Æ R—R + 2NaBr RBr + 2Na + BrR æether If R and R¢ are different alkyl groups, the product obtained is a mixture of R—R, R—R¢ and R¢—R¢. Wurtz reaction constitutes one of the methods to ascent the alkane series. Corey-House Alkane Synthesis To prepare alkane of the type R— R¢, this method can be used. This involves the following three steps: ææ Æ RLi + LiBr RBr + 2Li æether 2RLi + CuI Æ LiR 2 Cu + LiI Lithium dialkyl cuperate

LiR2Cu + R´Br Æ R—R¢ + RCu + LiBr

Kolbe Electrolytic Method Electrolysis of concentrated solution of sodium salt of a carboxylic acid produces alkane. O – O– ææÆ R— R + 2CO2 + 2e

Anodic reaction

2R

Cathodic reaction

2H2O + 2e– ææÆ 2OH– + H2

C

Hydrogenation of Alkenes and Alkynes Reduction of alkenes and alkynes with H2 in the presence of Ni catalyst at 200–300 °C produces alkanes: Ni CH2 ==CH2 + H2 ææÆ H3C—CH3 200-300∞C

22.2 Complete Chemistry—JEE Main Ni

CH + 2H2 ææÆ CH3— CH3

CH

200-300∞C

Reduction of Alkyl Halide This may be done by using any of the following reducing agents: Zn + HCl; Zn + CH3COOH; Zn–Cu couple in ethanol; LiA1H4 The reaction may be formulated as R— X + 2H Æ RH + HX Hydrolysis of Grignard Reagent RMgX + H2O Æ RH + XMgOH

Physical Properties 1. CH4 to C4H10 are gases, C5H12 to C17H36 are liquids and higher ones are solids. 2. Being nonpolar, these are insoluble in water and soluble in nonpolar solvents like benzene and carbon tetrachloride. 3. Boiling point increases with molecular mass; branched chain isomer has a lower boiling point than normal alkane as the former involves lesser van dar Waals interactions (dispertion forces) than the latter. 4. Liquid alkanes are lighter than water.

Chemical Properties Because of stronger C— H and C—C bonds, alkanes at room temperature are stable towards acids, bases, oxidizing and reducing agents and active metals. A few important reactions of alkanes are as follows.

Halogenation This involves the substitution of hydrogen atom by halogen atom. The order of reactivity is F2 > Cl2 > Br2(> I2). Fluorine reacts explosively, chlorination is vigorous, bormination is slow and iodination is reversible. The mechanism of chlorination and bromination involves free radicals. The reaction is initiated by the absorption of ultraviolet radiations or heating to about 250–400 °C. The following steps are involved in the mechanism of chlorination. (a) Chain-Initiation Step æÆ 2 Cl Cl2 æhv (b) Chain-Propagation step CH4 + Cl Æ CH3 + HCl Cl2 + CH3 Æ CH3Cl + Cl ------------------------------------------------(c) Chain-Terminal Step Cl + Cl Æ Cl2 CH3 + Cl Æ CH3Cl CH3 + CH3 Æ CH3—CH3 The chlorination is continued to produce CH2Cl2, CHCl3 and CCl4 depending upon the initial ratio of Cl2 and CH4. Only monochlorination takes place if methane is taken in a large excess in comparison to chlorine. In more complex alkanes, the abstraction of each different kind of H atoms gives a different molecule. The factors determining the relative yields of the isomeric product are as follows. 1. Probability factor This factor is based on the number of H atoms of different types. For example, CH2CH2CH3 contains six equivalent 1° H’s and two equivalent 2° H’s. Hence, the chance of removing 1° H relative to 2°H is 6 : 2, i.e., 3 : 1.

Hydrocarbons

22.3

2. Reactivity of H The order of reactivity of H is 3° > 2° > 1°. At room temperature, the relative rates of abstraction of hydrogen atoms by chlorine atom are 5.0 : 3.8 : 1.0. The relative rates of bromination of 1° : 2° : 3° H¢s are 1 : 82 : 1600. 3. Reactivity of X reactivity of H atoms. The less reactive Br For example, the chlorination of CH3CH2CH2CH3 produces two isomers:

CH3CH2CH2CH3 n-butane

Cl 2

Æ CH3CH2CH2CH2Cl + CH3CH2CHCH3

light, 25 C

n-butyl chloride

Cl sec-butyl chloride

The relative ratio of the two isomers is as follows. n-butyl chloride number of 1∞H reactivity of 1∞H 6 1.0 6 ¥ = = = ¥ sec-butyl chloride number of 2∞H reactivity of 2∞H H 4 3.8 15.2 6 ¥ 100 = 28.3 Percentage of n-butyl chloride = 21.2 Percentage of sec-butyl chloride = 71.7 in chain propagation steps, namely, Cl + CH4 Æ CH3 + HCl Br + CH4 Æ CH3 + HBr

DH° = 17 kJ mol–1 DH° = 75 kJ mol–1

Bromination of higher alkanes is decided by the reactivity H’s atoms and not by probability factor. For example, the major product (90%) in the bromination of n-butane is 2-bromobutane. ææææ Æ RNO2 + H2O Nitration RH + HNO3 æ400-500 ∞C Higher alkanes produce mainly nitrated products because of breaking of C— C bonds. D

RH + HO— SO3H ææ Æ R—SO3H + H2O

Sulphonation

fuming

where R is the larger alkyl group (≥C6H13). 3n + 1ˆ CnH2n+2 + ÊÁ O Æ nCO2 + (n + 1)H2O Ë 2 ˜¯ 2

Oxidation

Large quantity of heat is generated in the combustion of alkanes.

Isomerization Normal alkanes are converted into branched-chain isomer in the presence of AlCl3 and HCl. 3 CH3CH2CH2CH3 æAlCl ææ Æ CH3 CH

HCl

CH3

CH3

Aromatization For example,

CH3 H2C

CH3

H2C

CH2

Cr2O3-Al2O3 600°C

Æ

+ 4H2

CH2

A few facts about alkanes are as follows.

Conformations Ethane can exist in different conformations due to the rotation of methyl groups about C— C bond axis. The Newman projection formulae for the conformations of ethane are shown in Fig. 22.1.

22.4 Complete Chemistry—JEE Main

H

HH

H

H

H

H

H H

H H

H

H

Staggered

Eclipsed

Fig. 22.1

Newman projections of ethane

Staggered conformation has the lowest potential energy whereas eclipsed conformation has the highest potential energy. The difference between these two conformers is 11.3 kJ mol–1. The conformations of butane with reference to the central C—C bond are as follows.

Anti

The conformer in which methyl groups are far apart is known as anti conformer.

Gauche Eclipsed

The conformer in which methyl groups are closest to each other is known as gauche conformer The confer in which methyl group is near to either methyl or hydrogen are known as eclipsed conformers.

Isomerism The compound having different structural formulae but the same molecular formula are known as isomers. For example, the alkanes butane and pentane can have following carbon skeltons. Butane

C— C—C—C

and

C

C

C

C

Pentane

C— C—C—C—C,

C

C

C C

C

and

C

C

C

C

C

Optical Isomerism atom in a molecule is known as chiral atom. The spatial arrangement and its mirror image are not superimposable on each other and thus are two different isomeric compounds. They rotate the plane of polarised light by the same angle but in opposite directions. A mixture of equal amounts of these two isomers is known as racemic mixture. The net optical rotation of such a mixture is zero.

MULTIPLE CHOICE QUESTIONS ON SECTION 1 Preparation and Nomenclature 1. Ethane is produced during the electrolysis of potassium salt of (a) succinic acid (b) malonic acid (c) acetic acid 2. Ethane is produced when C2H5MgI is treated with (a) dilute HCl (b) aq KOH 3. Methane is produced during the hydrolysis of (a) CaC2

(b) Al4C3

(c) CO2

(d) H2O

(c) MgC2

(d) SiC

Zn - Hg

Æ C3H8 + H2O is known as 4. The reaction CH3COCH3 + 4H ææææ conc. HCl

(a) Rosenmund’s reduction (c) Wurtz reaction

(d) fumaric acid

(b) Clemmensen reduction (d) Kolbe’s reaction

Hydrocarbons

22.5

5. A concentrated solution of sodium butyrate is electrolysed, the product obtained at anode is (a) butane (b) propane (c) hexane (d) pentane 6. A hydrocarbon is obtained by heating (a) a sodium salt of fatty acid with alcohol (b) a calcium salt of fatty acid (c) an acid chloride of a fatty acid with sodium (d) a sodium salt of fatty acid with soda lime 7. The reaction

Ni

CH2== CH2 + H2 ææÆ C2H6

is an example of

600 K

(a) Sabatier and Senderens reaction (c) Sandmeyer’s reaction 8. The IUPAC name of the compound

(b) Kolbe reaction (d) Wurtz’s reaction CH3 CH3CH CH2CH2 CH CH is CH3 CH3 CH3

(a) 1, 1, 2, 5-tetramethylhexane (b)2, 5, 6, 6-tetramethylhexane (c) 2, 5, 6-trimethylheptane (d) 2, 3, 6-trimethylheptane

General Characteristics 9. The compound having one isopropyl group is (a) 2, 2, 3, 3-tetramethylpentane (c) 2, 2, 3-trimethylpentane 10. The compound with the highest boiling point is (a) n-hexane (b) n-pentane

(b) 2, 2-dimethylpentane (d) 2-methylpentane (c) 2, 2-dimethylpropane

(d) 2-methylbutane [IIT 1982]

11. Which of the following is expected to have minimum boiling point? (a) n-Pentane (b) n-Butane (c) 2-Methylbutane (d) 2, 2-Dimethylpropane 12. The highest boiling point is expected for (a) isooctane (b) n-octane (c) 2, 2, 3, 3-tetramethylbutane (d) n-butane [IIT 1986] 13. Which of the following statements is not correct? (a) Alkanes are nonpolar in nature (b) Alkanes are soluble in nonpolar solvents (c) Straight-chain alkanes have lesser boiling point than the corresponding branched-chain isomers (d) Alkanes exhibit alternations in melting point 14. Marsh gas is (a) methane (b) ethane (c) propane (d) butane 15. Which of the following molecules contains carbon atom of 2°? (b) (CH3)2CHCH3 (c) (CH3)3CH (d) (CH3)4C (a) CH3CH2CH2CH3 16. The correct order of stability of the given radicals is (a) tertiary > secondary > primary (b) tertiary > secondary < primary (c) tertiary < secondary > primary (d) tertiary < secondary < primary 17. Which of the following facts is correct? (a) C— D bond is slightly weaker than C—H bond (b) C—D bond is slightly stronger than C—H bond (c) Both C—H and C—D bonds are equally strong (d) Replacement of D in C—D by Cl is faster than the replacement of H in C—H 18. The most strained cycloalkane is (a) cyclopropane (b) cyclobutane (c) cyclopentane (d) cyclohexane

22.6 Complete Chemistry—JEE Main

19. Which of the following statements is correct? (a) The bond angle HCH in methyl radical is 120°. (b) The eclipsed conformation of ethane has lower potential energy as compared to staggered conformation. (c) The IUPAC name of CH3CH2CH2CH ⎯ CH ⎯ CHCH2CH3 is 3-ethyl-4-methyl-5-isopropyloctane.

CH CH3

CH3 CH2CH3 CH3

(d) Lower alkanes are soluble in water while higher ones are insoluble. 20. Which of the following statements is not correct? (a) Methane is the major constituent of natural gas. (b) In the reaction RMgX + NH3 Æ RH + Mg(NH2)X, ammonia acts as a base. (c) The reactivity of hydrogen atom in alkane towards replacement by a halogen follows the order 3° > 2° > 1°. (d) The enthalpy of formation of a free radical from the corresponding alkane follows the order 1° > 2° > 3°. 21. Which of the following statements is not correct? (a) The octane value of n-heptane is zero. (b) The octane value of isooctane is 100. (c) The octane number of a mixture containing 85% isooctane and 15% heptane is 15. (d) The 3° radicals react more slowly and smoothly with oxygen than the 1° and 2° radicals. 22. Which of the following statements is not correct? (a) Of the vinyl, allyl, 3°, 2°, 1° and benzyl radicals, the least stable one is vinyl. (b) Of the C6H5CH3, (CH3)3CH, (CH3)2CHCH3 and CH2==CHCH3, the maximum ease of abstraction of hydrogen atom is shown by C6H5CH3. (c) Free radical monochlorination of tert-butyl bromide gives 1-bromo-2-chloro-2 methyl propane. (d) Of Cl2, Br2, and I2, the most reactive halogen towards photohalogenation of alkanes is I2.

Chemical Reactions 23. Ethanol on treatment with concentrated HI and red phosphorus forms (a) C2H5I

(b) C2H4

(c) C2H6

(d) C3H8

(c) HIO3

(d) PI3

(c) Cr2O3/Al2O3 at 600 °C

(d) conc. H2SO4

24. Alkanes can be iodinated in the presence of (a) HI

(b) I2 and P

25. Isomerism in alkane can be brought about by using (a) Al2O3 at 100 °C

(b) Fe2O3

(a) CH3Cl (b) CH2Cl2 (c) CHCl3 (d) CCl4 27. Hexane when heated under pressure in the presence of Cr2O3 carried on Al2O3 support at 750–770 K produces (a) cyclohexane (b) hexene (c) cyclohexene (d) benzene –1 28. A single substitution of H atom in an alkane of molar mass 72 g mol by another substituent produces only one product. The alkane is (a) n-pentane (b) 2-methylbutane (c) 3-methylbutane (d) 2, 2-dimethylpropane 29. The reaction conditions leading to the best yield of C2H5Cl are uv light

(a) C2H6 (excess) + Cl2 ææææ Æ

dark

(b) C2H6 + Cl2 æææææ Æ room temperature uv light

(d) C2H6 + Cl2 ææææ Æ (c) C2H6 + Cl2 (excess) Æ 30. The relative reactivity of 1° : 2° : 3° hydrogens to chlorination is (a) 1 : 5 : 3.8 (b) 1 : 3.8 : 5 (c) 3.8 : 1 : 5 (d) 5 : 1 : 3.8

Hydrocarbons

22.7

31. The relative reactivity of 1° : 2° : 3° hydrogens to bromination is (a) 1 : 3.8 : 5 (b) 1 : 1600 : 82 (c) 1 : 82 : 1600 (d) 1600 : 82 : 1 32. Which of the following statements is not correct? (a) The order of reactivity of halogen towards alkanes is F2 > Cl2 > Br2. (b) The reaction of Cl2 and CH4 proceeds via ionic mechanism. (c) Oxygen acts as an inhibitor in the reaction between CH4 and Cl2. (d) Iodine does not react with CH4 at all. 33. Which of the following statements is correct? (a) The percentage of n-propyl chloride obtained in the chlorination of propane is about 56%. (b) The percentage of 1-chloro-2-methylpropane obtained in the chlorination of isobutane is about 64%. (c) The percentage of n-propyl bromide obtained in the bromination of propane is 44%. The relative reactivities of 3°, 2° and 1° H atoms are 1600:82:1. (d) The vinyl radical is more stable than 2° radical. 34. Which of the following statements is correct? Cl (a) The chlorination of CH3—CH—CH3 yields CH3 Br

C

CH3 .

Br (b) The number of isomers of dimethylcyclopropane is three. (c) The number of geometric isomers of 1,2,4-trimethylcyclohexane is three. (d) Substituted cycloalkanes do not show geometrical isomers. 35. Which of the following statements is correct? (a) The electrolysis of concentrated solution of sodium ethonate produces ethene. (b) If the relative rates of abstraction of 2° and 1° hydrogen atoms are 3.8 and 1.0, respectively, then the percentage ratios of the products 1-chloro-, 2–chloro–, and 3-chloro–, obtained in the photochlorination of n-hexane would be 16.48 : 41.76 : 41.76. (c) The relative rate of abstraction of 3°, 2° and 1° hydrogen atoms is maximum for 1° hydrogen atoms. (d) The relative rate of abstraction of 3°, 2°and 1° hydrogen atoms is minimum for 3° hydrogen atoms.

Isomers and Conformations 36. The minimum number of carbon atoms in alkane molecule to show isomerism is (a) 2 (b) 3 (c) 4 (d) 5 37. The maximum number of structural isomers for the alkane C4H10 is (a) 2 (b) 3 (c) 4 (d) 5 38. The maximum number of structural isomers for the alkane C5H10 is (a) 2 (b) 3 (c) 4 (d) 5 39. Which of the following is not found in alkanes? (a) Chain isomerism (b) Geometrical isomerism (c) Mesomerism (d) Positional isomerism 40. A molecule is said to be chiral if it (a) contains a centre of symmetry (b) contains a plane of symmetry (c) cannot be superimposed on its mirror image (d) exists as cis and trans-forms 41. The maximum number of structural isomers for an alkane hexane is (a) 2 (b) 3 (c) 4 (d) 5 42. Which of the following statements is not true? (a) In ethane, staggered conformation is more stable than eclipsed conformation (b) Cyclohexane exists in two conformations

22.8 Complete Chemistry—JEE Main

43. 44. 45.

46.

47. 48. 49.

50.

(c) The boat conformation of cyclohexane is more stable than chair conformation (d) The boat and chair conformations of cyclohexane do not exist as independent compounds An equimolar mixture of enantiomers is called (a) optical isomers (b) resolution (c) racemic mixture (d) dextroisomer Optical rotation of a racemic mixture is (a) positive (b) negative (c) zero (d) temperature dependent The energy barrier to rotation about the C— C bond in ethane is about 12.55 kJ/mol at room temperature (298 K). Assuming the contribution of entropy negligible, the ratio of staggered to eclipsed conforations is about (a) 150 (b) 159 (c) 165 (d) 175 The ratio of anti/gauche conforations of butane is 4.6. The anti conformation is more stable than the gauche conformation at room temperature (25 °C) by (a) 2.71 kJ/mol (b) 3.12 kJ/mol (c) 3.78 kJ/mol (d) 4.14 kJ/mol Which of the following is the stablest form of cyclohexane? (a) Boat form (b) Chair form (c) Half-chair form (d) Half-boat form Which of the following conformations of butane has the lowest energy? (a) Anti (b) Gauche (c) Eclipsed (d) Staggered Which of the following statements is not correct? (a) Cycloheptane and cyclooctane are more stable than cyclohexane. (b) The ring strain in cyclopropane is more than that in the cyclobutane. (c) The C—H bond in cyclopropane is shorter than in propane. (d) The H atoms in cyclopropane is more acidic than in propane. Which of the following statements is not correct? (a) The potential energy difference between eclipsed and staggered conformations of ethane is 12.552 kJ mol–1. Ignoring the entropy effect, the ratio of staggered to eclipsed conformation at 25 °C is about 155. (b) The potential energy difference between anti and gauche con formations of butane is about 3.8 kJ mol–1. Ignoring the entropy effect, the ratio of anti to gauche conformations at 25 °C is about 2.6. (c) The activation energy for the reaction 2 Cl◊ Æ Cl2 is zero. (d) The free radical chlorination of CH4 occurs faster than that of CD4.

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49.

(c) (a) (c) (a) (c) (c) (a) (c) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(d) (d) (a) (b) (d) (b) (b) (c) (b)

3. 9. 15. 21. 27. 33. 39. 45.

(b) (d) (a) (c) (d) (b) (c) (b)

4. 10. 16. 22. 28. 34. 40. 46.

(b) (a) (a) (d) (d) (b) (c) (c)

5. 11. 17. 23. 29. 35. 41. 47.

(c) (d) (b) (c) (a) (b) (d) (b)

HINTS AND SOLUTIONS 1. Kolbe’s electrolytic method 2CH3COOK + 2H2O Æ CH3CH3 + 2CO2 + 2KOH + H2 2. C2H5MgI + H2O Æ C2H6 + HOMgI

6. 12. 18. 24. 30. 36. 42. 48.

(d) (b) (a) (c) (b) (c) (c) (a)

Hydrocarbons

22.9

3. Al4C3 + 12H2O Æ 3CH4 + 4Al(OH)3 5. 2CH3CH2CH2COONa + 2H2O Æ C6H14 + 2CO2 + 2NaOH + H2 hexane

9. Isopropyl group is CH3 10. 17. 19.

20.

21.

CH

CH3 The boiling point gradually increases with increase in the molecular mass. Straight-chain alkanes have higher boiling point than the corresponding branched-chain isomer. The bond with the heavier isotope is slightly stronger than that with the lighter isotope. (a) The bond angle is 120° because carbon is sp2 hybridized. (b) In eclipsed conformation, hydrogen atoms of the two carbon atoms are closer to each other, hence more repulsive interactions resulting in the higher potential energy. (c) The correct name is 3-ethyl-5-isopropyl-4-methyloctane. Isopropyl comes before methyl. (d) Alkanes are souble in non-polar solvents such as benzene, ether and chloroform. (a) Natural gas contains about 97% of methane. (b) Here NH3 acts a stronger acid and RH as a weak acid. (d) The more stable the free radical, the lesser the enthalpy of formation of the free radical from the corresponding alkane. (c) The octane value is 85. HI P

Æ C 2H 6 + H 2O 23. C2H5OH æææ Cr O Al O

2 3 2 3 27. CH3CH2CH2CH2CH2CH3 æææææ Æ

750-770 K

Benzene

32. (b) (c) (d) 33. (a)

The reaction proceeds via free radical mechanism. This is due to the formation CH3—O—O· radical which is less reactive than CH3· radical. It is due to the fact, that the reaction I· + CH4 Æ HI + CH3· is highly endothermic (DH° = 138 kJ mol–1). The chlorination of propane gives two isomeric products: 1-chloropropane and 2-chloropropane. Their relative ratio would be given as 1-Chloropropane no. of 1∞ H reactivity of 1∞ H 1.0 6.0 = ¥ = 6 ¥ = 2-Chloropropane reactivity of 2∞ H 3.8 7.6 no. of 2∞ H 2 Percentage of 1° isomer =

(b) Isobutane is CH3

CH

6.0 ¥ 100 = 44% 6.0 + 7.6 CH3 . Its chlorination would give two isomers.

CH3

1∞ isomer no. of 1∞ H reactivity of 1∞H 9 1 9 = ¥ = ¥ = 3∞ isomer no. of 3∞ H reactivity of 3∞H 1 50 5 Percentage of 1° isomer =

9 ¥ 100 = 64.3% 9+5

(c) The propane is CH3CH2CH3. The relative ratio of the two isomers is 1∞ isomer no. of 1∞ H reactivity of 1∞ H 6 1 6 = ¥ = ¥ = reactivity of 2∞ H 2∞ isomer no. of 2∞ H 2 82 164 3 Percentage of 1° isomer = ¥ 100 4 3 + 164

22.10 Complete Chemistry—JEE Main

(d) The homolytic bond dissociation energy of vinylic H is greater than that of 2° H. Hence, vinylic radical is less stable than 2° radical. 34. (a) CH3

CH

∑

Cl

CH3

∑

Cl

Æ CH3CHCH 2 Br æææææ Æ CH3 C HCH 2 Br æææ CH2 ærearrangement | Br Cl 1° radical 1-bromo-2-chloropropane

CH3

CH

Br

Although alkyl free radicals seldom if ever rearrange by migration of hydrogen or alkyl, but they can rearrange by migration of halogen. (b) The number of isomers is three

Me

Me

Me

H

Me

Me

H

H

Me trans-1,2-dimethylcyclopropane

H

1,1-dimethylcyclopropane

cis-1,2-dimethylcyclopropane

Cis-and trans-isometers are due to the rigidity of the ring. The rotation about the ring C—C bonds is prohibitive. (c) There are four geometric isomers.

Me

Me

Me

Me

Me Me cis, cis-1,2,4-Trimethylcyclohexane

trans-2-cis-4-Trimethylcyclohexane

Me

Me Me

Me

Me

Me trans-2-trans-4-Trimethylcyclohexane

cis-2-trans-4-Trimethylcyclohexane

35. (a) (b) (c) (d)

ethane is the product of electrolysis 6 ¥ 1 : 4 ¥ 3.8 : 4 ¥ 3.8 :: 16.48% : 41.76% : 41.76% 3° 1°

37. C— C—C—C;

C

C

C

C

C

38. C—C—C—C—C;

C

C

C

C;

C

C

C

C

C C

41. C—C—C—C—C—C ;

C

C

C

C

C; C

C

C

C C

C

C; C

C

C

C

C

42. The boat conformation of cyclohexane is less than stable chair conformation. – 159 45. Keq = exp(–DG/RT) = exp(12550/8.314 ¥ 298) ~

C; C

C

C C

C

Hydrocarbons

22.11

46. Given is the fact that Keq = 4.6 for gauche anti. Hence ~ DH – DG = –RT ln Keq = [– (8.314) (298) ln 4.6] J/mol = –3781 J/mol 49. (a) Both are slightly less stable than cyclohexane. (c) The observed bond angle H—C—H in cyclopropane is 114°. To account for this angle, it is postulated that the hybrid orbital of C involves more s-character as compared to the sp3 hybrid orbital. (d) The C—H bonds in cyclopropane involves more s-character than those of an alkane. 50. (a) Here DG° DH°. Using DG° = – RT ln K°, we get ln K° = (b) Here DG°

DG∞ (12.552 ¥ 103 J mol-1 ) = = 5.066 RT (8.314 J K -1mol-1 ) (298 K)

fi

K° = 158.57

DH°. Using DG° = – RT ln K°, we get

ln K°= -

D G∞ 3.8 ¥ 103 J mol-1 = = 1.534 RT (8.314 J K -1mol-1 ) (298 K)

fi

K° = 4.64

(c) In this case, no bonds are broken, they are only formed. (d) The C—D bonds in CD4 are slightly stronger than C—H bonds in CH4. Thus, DH for abstraction of D is slightly greater than for H. The abstraction being the slow step, the removal of H will be faster.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloroalkane. The alkane could be (a) neopentane (b) propane (c) pentane (d) isopentance [2003] 2. Which of the following compounds is not chiral? (a) 1-chloropentane (b) 2-chloropentane (c) 1-chloro-2-methylpentane (d) 3-chloro-2-methylpentane [2004] 3. 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly (a) 2-bromo-3-methylbutane (b) 1-bromo-3-methylbutane (c) 1-bromo-2-methylbutane (d) 2-bromo-2-methylbutane [2005] 4. Alkyl halides react with dialkyl copper reagents to give (a) alkanes (b) alkenyl halides (c) alkenes (d) alkyl copper halides [2005] is (a) Anti, Gauche, Eclipse

(b) Eclipse, Gauche, Anti

(c) Gauche, Eclipse, Anti

(d) Eclipse, Anti, Gauche [2006]

6. Which one of the following conformations of cyclohexane is chiral? (a) Twist boat (b) Rigid (c) Chair (d) Boat 7. How many chiral compounds are possible on monochlorination of 2-methylbutane? (a) 8 (b) 2 (c) 4 (d) 6 8. The major product obtained in the photo catalysed bromination of 2-methylbutane is: (a) 1-bromo-2-methylbutane (b) 1-bromo-3-methylbutane (c) 2-bromo-3-methylbutane (d) 2-bromo-2-methylbutane

[2007] [2012]

[2014, online]

22.12 Complete Chemistry—JEE Main

ANSWERS 1. (a) 7. (b)

2. (a) 8. (d)

3. (d)

4. (a)

5. (a)

6. (a)

HINTS AND SOLUTIONS 2. The given compounds are CH3CH2CH2CH2CH2Cl ;

* CH3CH2CH2CHCH3 ;

1-chloropentane

* CH3CH2CH2CHCH2Cl ; CH3

Cl

2-chloropentane

1-chloro-2-methylpentane

* CH3CH2CHCHCH3 Cl CH3 3-chloro-2-methylpentane

The compound 1-chloropentane is not chiral as other compounds contain chiral carbon atoms. 3. The reaction is Br H3 C CH CH2

CH3 + Br2 Æ H3C

CH3

C CH2

CH3 + HBr

CH3 2-bromo-2-methylbutane

The reaction proceeds via free-radical mechanism. 3º radical being most stable is formed by the extraction of H from 2 position. 4. The reaction is

5. Because of hydrogen bonding, the order of stability is reversed when compared to ethane. Hence, the increasing order of stability is anti, gauche, eclipse. 7. The possible monochlorination of 2-methylbutane are as follows. Cl * * CH3 C CH2CH3 ; CH3CH CH CH3 ClCH2CHCH2CH3 ; CH3

CH3

(i)

(ii)

CH3 (iii)

Cl

CH3

CHCH2CH2Cl CH3 (iv)

Of the four compounds, the compounds (i) and (iii) are chiral compounds as each of the two has one asymmetric carbon atom. Each of these two will exhibit two enantiomers. . 8. 2- Methylbutane is

Photocatalysed bromination proceeds through the formation of radicals. Since 3° radical is most stable, the product obtained is 2-bromo-2-methyl butane.

Hydrocarbons

SECTION 2

22.13

Alkenes

Methods of Preparation Dehydrohalogenation of Alkyl Halides Heating of alkyl halides with alcoholic KOH produces alkanes: ethanol

Æ R— CH==CH2 + KX + H2O RCH2CH2X + KOH æææ D The ease of dehydrohalogenation is 3° alkyl halide > 2° alkyl halide > 1° alkyl halide. The dehydrobromination of optical isomers of 2, 3-dibromobutane depends whether the isomer is racemic or meso.

For the detailed analysis of these reactions, see the answers of Q.18c and Q.18d, respectively.

Dehydration of Alcohols Sulphuric acid is a dehydrating agent. The ease of dehydration is

H SO

2 4 R— CH2—CH2OH æææ Æ R— CH==CH2 + H2O D

3° alcohol > 2° alcohol > 1° alcohol.

Dehalogenation of Vicinal Dihalides This is carried out by heating with zinc dust in ethyl alcohol. RCH

alcohol

Æ R— CH==CH2 + ZnBr2 CH2 + Zn æææ D

Br Br The debromination of optical isomers of 2, 3-dibromobutane depends whether the isomer is meso- or racemic. CH3 CH3 CH3 H3C H H3C Br H Br H C C C C Br H H Br H H H CH3 CH 3 CH 3

cis-2-butene

racemic-2, 3-dibromobutane

trans-2-butene

meso-2, 3-dibromobutane

For the detailed analysis of these reactions, see the answers of Q.18a and Q.18b, respectively.

Partial Reduction of Alkynes R

H C

H

C R

Na or Li NH3

R

C

C

R

H2 Lindar catalyst

R

R C

C

H

H cis-alkene

trans-alkene

Kolbe’s Electrolytic Method The electrolysis of an aqueous solution of potassium salt of a saturated dicarboxylic acid produces alkene. Anodic Reaction

CH 2 COO - ææ Æ CH 2 + 2CO 2 + 2 | || CH 2 COO CH 2

-

Cathodic Reaction

2H2O + 2e– Æ 2OH– + H2

22.14 Complete Chemistry—JEE Main

Saytzeff Rule In dehydrohalogenation or dehydration, the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms. The alkene with the greater number of alkyl groups is the preferred product because it is formed faster than alternative alkenes. The sequence showing the relative rates of formation of alkenes is R2C==CR2 > R2C==CHR > R2C==CH2, RCH==CHR > RCH==CH2 The sequence of stability of alkenes also follows the above order. Hence, the more stable the alkene, the faster it is formed. For example, CH3CH 2 CHCH3 æKOH(alc) æææÆ | Br A carbocation formed as an intermediate is particulary prone to rearrangement. For example, the following reaction involves rearrangement. CH3

CH3

C

CH

CH3

EtOH

Æ CH3

CH3

C

CH3

C

CH3 + CH3

CH

C

CH3

CH3 Br

CH2

CH3

(main product)

This arrangement is due to the formation of more stable 3° carbocation from the lesser stable 2° carbocation: CH3 CH3

C

CH

CH3

CH3

–Br–

Æ CH3

CH3 Br

C

CH3

CH +

Æ CH3

CH3

C +

CH3

CH

Saytzeff rule

Æ CH3

CH3

CH3 C

CH3

CH3

CH3

(2°)

C

(3°)

The ease of dehydration of alcohols (3° > 2° > 1°) is in agreement with the formation of carbocation as the intermediate. Wherever possible, rearrangement takes place. For example CH3

CH2

CH

CH2OH

H+

Æ CH3CH

CCH3 + CH3CH2C CH3

CH3

CH2

CH3

(main product)

CH3 CH3CCHOHCH3

H+

CH3

Æ CH3C

CH3

CCH3 + CH2 | CH3

CH3 | CCHCH3 | CH3

(main product)

Hofmann Rule Heating of a quaternary ammonium hydroxide mainly produces the least substituted alkene. For example CH3CH2CHCH3 +

N(CH3)3

150 C

Æ CH3CH2CH (95%)

CH2 + CH3CH

CHCH3

(5%)

+ (CH3)3N: + H2O

A less substituted alkene is formed as a result of the loss of more acidic bH (which follows the trend 1° > 2° > 3°).

Chemical Properties Electrophilic Addition Reactions Because of loosely held p electrons, the carbon-carbon double bond serves as a source of electrons and thus addition reactions across the double bond are initiated by an electrophilic attack. The addition reactions exhibited by alkenes are as follows.

Hydrocarbons

22.15

1. Addition of hydrogen A solution of the alkene is shaken under a slight pressure of hydrogen gas in the presence of a small amount of the catalyst (Pt, Pd, or Ni) H , Ni

2 Æ RCH2CH3 RCH==CH2 æææ

The reaction is quantitative, the volume of hydrogen gas consumed provides the number of double bond in the compound. Heats of hydrogenation provide the information regarding the relative stabilities of unsaturated compounds. Of simple dialkylethylenes, the trans isomer is the more stable. It is also found that the greater the number of alkyl groups attached to the doubly bonded carbon atoms, lesser the heat of hydrogenation and thus more stable the alkene—a conclusion which is in agreement with the Saytzeff rule. The catalytic addition of hydrogen atoms across a double bond involves syn- addition, i.e. on the same side of the double bond. The hydrogenetaion of cis- and trans- isomers of 2, 3-dibromobutene yields different optical isomers of 2, 3-dibromobutane. CH3 CH3 Br CH3 CH3 H3C H Br H2/Ni H Br H2/Ni C C C C H Br Br H Br H3C Br Br CH3 CH3 cis-2, 3-dibromobutene

meso-2, 3-dibromobutane

racemic-2, 3-dibromobutane

For the detailed analysis of these reactions, see the answers of Q.43a and Q.43b, respectively. 2. Addition of hydrogen halide When dry gaseous hydrogen halide is directly passed into an alkene, the corresponding alkyl halide is formed. The addition of HX follows Markovnikov’s rule which states that the hydrogen of acid attaches itself to the carbon that already holds the greater number of hydrogens. For example, R— CH==CH2 + HCl Æ R

CH

CH3

Cl

In the presence of peroxide, the addition of HBr does not follow Markovnikov’s rule. The addition is just a reverse of Markovnikov’s rule and this fact is known as anti-Markovnikov’s rule. For example, peroxide

R— CH==CH2 + HBr ææææ Æ R—CH2—CH2Br

Mechanism of Addition of HBr in the Presence of Peroxide Anti-Markovnikov addition of HBr in the presence of peroxide occurs via free-radical mechanism. Chain-Initiating Steps Peroxide Æ Radical Radical + H : Br Æ Radical : H + Br Chain-Propagating Steps

......................... .........................

.......................... .........................

Factors that stabilize the free radical also stabilize the incipient free radical in the transition state. For example, secondary free radical, being more stable than primary radical, is formed faster in the reaction between R—CH==CH2 and HBr and thus the product obtained is RCH2CH2Br. Only for HBr, both the steps mentioned above are exothermic the reverse is true.

22.16 Complete Chemistry—JEE Main

3. Addition of sulphuric acid When a gaseous alkene is passed through concentrated sulphuric acid in cold or a liquid alkene is stirred with the acid, the addition of a molecule of H2SO4 takes place: R— CH==CH2 + H2SO4 Æ R— CH --CH3 | OSO3 H The addition of sulphuric acid follows Markovnikov’s rule. The resultant hydrogen sulphate when heated with water produces the corresponding alcohol: R— CH -- CH3 + H 2 O Æ R— CH -- CH3 + H 2SO 4 | OH

| OSO3 H

4. Addition of water In the presence of acids, water adds to the more reactive alkenes. The addition is in accordance with the Markovnikov’s rule. H+

R—CH==CH2 + H2O ææÆ R— CH -- CH3 | OH

Oxymercuration-Demercuration Process Alkenes in tetrahtydrofuran react with mercury acetate in the presence of water to give hydroxymercurial compounds which on reduction with sodium borohybride (NaBH4) yield alcohol. The net result is the addition of H2O to the double bond and the product obtained is in agreement with Markovnikov’s rule. For example, HgOAc

+ HgOAc CH3CH

Hg(OAc)2

CH2 ¾¾¾® CH CH —– CH 3 2

–

– OAc ¾¾® CH3CH—CH2 H2O

OH

NaBH4

¾¾® CH3CH—CH3 OH

The electrophile of Hg(OAc)2 is +HgOAc. No rearrangement occurs in this reaction. Hydroboration–Oxidation Process Alkenes undergo hydroboration with diborane in THF to yield alkylboranes which on oxidation with alkaline H2O2 give alcohols. The net result is the addition of H2O to the double bond and the product obtained is in accordance with the anti-Markovnikov’s rule. For example,

H2O2, OH –

==

OH

B2H6 in THF exists as BH3, 5. Addition of halogens Chlorine or bromine readily adds to double bond. Iodine does not react. R—CH==CH2 + X2 æ(æææææ Æ R -- CH -- CH 2 X 2 ∫∫ Cl2 , Br2 ) | | X X

Mechanism of Addition of Halogens Two steps involved are as follows. Step 1 Formation of halonium ion X

X+

C

C

X+

– Æ X +

C

C

halonium ion

Step 2 Addition of halide ion (or any other species similar to it) to halonium ions. This addition involves anti-addition. X+ C

C

X + Y– Æ

C

C Y

Hydrocarbons

22.17

For example, the addition of Br2 to ethylene in the presence of NaCl in aqueous medium, Y– may be Br–, Cl– or OH– and thus all the three compounds given below are formed. BrCH 2CH2Br, BrCH2CH2Cl, BrCH2CH2OH The addition of Cl2 or Br2 to double bond involves anti-addition, i.e. halogen atoms are attached at the opposite faces of the double bond. If the compound containing double bond exists in two stereoisomers, namely, cis and trans, then the addition of halogen may generate chiral carbon atoms and thus the product will exhibit optical isomers. The addition of halogen across cis alkene yields only racemic mixture whereas trans alkene produces meso compound. For example,

For the detailed analysis of these reactions, see the answers of Q.42c and Q.42d, respectively. 6. Addition of hypohalous acids (Chlorine or bromine in the presence of water) (X 2 , H 2O) æææææ Æ R -- CH -- CH 2 R— CH==CH2 æ(X 2 ∫∫ Cl 2 , Br2 ) | | OH X

The reaction in which two hydroxyl groups are added to the double bond is known as hydroxylation. This can be achieved by treating the compound containing double bond with cold alkaline potassium permanganate (Baeyer’s reagent). Mild conditions of oxidation are the essential requirement as otherwise the resultant compound may be oxidized further. The reaction is 3CH2==CH2 + 2KMnO4 + 4H2O Æ 3CH2

CH2 + 2MnO2+ 2KOH

OH OH Ozone can be used to break the double bond to give two smaller molecules. The reaction

takes place in two steps. Step 1 Formation of ozonide

O

RCH==CH2 + O3 Æ R

CH2

CH O

Step 2

O

Hydrolysis of ozonide in the presence of a reducing agent (frequently (CH3)2S or zinc dust and acetic acid) O R

H O

2 Æ RCHO + H2CO CH2 æææ

CH

Zn

O

O

The cleavage products are aldehydes or ketones depending upon the atom or group attached to carbons of the double bond. The function of reducing agent is to prevent the formation of H2O2 which would otherwise react with aldehydes and ketones. 9. Polymerization of alkenes Ethylene produces polyethylene when it is heated under pressure with oxygen. O , heat

2 Æ (— CH2—CH2—)n nCH2==CH2 ææææ

pressure

10. Degradation by sodium periodiate (NaIO4) in the presence of potassium permanganate (KMnO4). È | | KMnO 4 -- C == C -- ææææ Æ Í-Í ÍÎ

| C -| OH

˘ | C -- ˙ æææ Æ acids, ketones, CO 2 ˙ | OH ˙˚

22.18 Complete Chemistry—JEE Main

Carboxylic acid is obtained instead of aldehydes. The terminal == CH2 group is oxidized to CO2. CH3 CH3CH C

CH3

KMnO 4 NaIO 4

CH3

Æ CH3COOH + O

C

CH3

KMnO

4 CH3CH2CH==CH2 ææææ Æ CH3CH2COOH + CO2

NaIO 4

The rate of addition of an acid to a double bond depends upon the stability of the carbocation being formed. For a given acid, the order of reactivity of alkenes is as follows. CH3 CH3

C == CH2 > CH3CH==CHCH3, CH3CH2CH==CH2, CH3CH==CH2 > CH2==CH2 > CH2==CHCl

For the detailed analysis of these reactions, see answers of Q. 42c and Q. 42d, respectively.

Substitution Reactions An alkyl group attached to a double bond can undergo substitution reaction involving halogen—a typical reaction shown by alkanes. The working conditions are the high temperature (500–600°C) in the gaseous phase. These may be compared with those of addition reaction which are low temperature in the absence of light and the medium of the reaction is generally the liquid phase. For example,

Substitution reaction at alkyl group occurs via free-radical mechanism. It is also found that the low concentration of halogen can be used instead of high temperature to cause substitution at the allylic position. The compound N-bromosuccinimide has been used to brominate alkenes at the allylic position. As each molecule of HBr is formed by the bromination, N-bromosuccinimide generates Br2 according to the reaction O O H2C

C

HBr +

N H2C

Br Æ Br2 +

C

H2C H2 C

C N

H

C O

O N-bromosuccinimide

succinimide

Since substitution of H by X occurs only at allylic position and not at vinylic position, the following conclusions may be drawn. C

C

C

H

H

H

vinylic

Ease of abstraction of hydrogen atoms Ease of formation of free radical or Stability of free radical

allylic

allylic > 3° > 2° > 1° > CH4 > vinylic allyl > 3° > 2° > 1° > CH3 > vinyl

Since the allyl radical involves resonance (CH2==CH—CH2 ´ ∑CH2—CH==CH2), the allylic substitution may give two different products. For example, 1-octene on treating with N-bromosuccinimide gives the following two products:

The resonance in allylic radical makes it more stable than other alkene.

Hydrocarbons

22.19

MULTIPLE CHOICE QUESTIONS ON SECTION 2 General Characteristics (b) C2H4 (c) C2H6 (d) C3H6 (a) C2H2 2. Geometrical isomerism is shown by (a) alkanes (b) alkenes (c) alkynes (d) aromatic hydrocarbons 3. The geometric isomers (a) rotate the plane of polarised light (b) exhibit optical properties (c) contain atleast one carbon-carbon double bond (d) contain atleast one carbon-carbon triple bond 4. Which of the following compounds exhibits cis-trans isomerism? (a) 2-Butane (b) 2-Butyne (c) 2-Butanol (d) Butanal 5. The IUPAC name of the compound CH3 CH3

C

CH

CH2

CH3

6. 7.

8.

9.

is (a) 3, 3, 3-Trimethylpropane (b) 1, 1, 1-Trimethylprop-2-ene (c) 3, 3-Dimethylbut-1-ene (d) 2, 2-Dimethylbut-3-ene Propane and propene cannot be distinguished by using (b) Br2 (c) KMnO4 (d) O3 (a) Cl2 The compound CH2==C==CH2 (a) has linear geometry (b) has p bonds in a plane (c) has p bonds perpendicular to each other (d) has p electrons delocalized over the whole molecule Which of the following order regarding relative stability of the given alkenes is correct? (a) 1-Pentene < trans-2-pentene < 2-methyl-2-butene (b) 1-Pentene < trans-2-pentene > 2-methyl-2-butene (c) 1-Pentene > trans-2-pentene < 2-methyl-2-butene (d) 1-Pentene > trans-2-pentene > 2-methyl-2-butene Which of the following statements is not correct? (a) cis-Alkene can be converted into trans-alkene on heating with I2. (b) The compound

shows a faster rate of catalytic hydrogenation than the compound

.

(c) The larger the enthalpy of formation of an alkene, the more stable the alkene. (d) The compound C2H5CH(Cl)CH2 CH2 on hydrogenation produces an optically inactive compound. 10. Which of the following statements is not correct? (a) In general, cis isomer has higher boiling point and lower melting point as compared to its trans isomer. (b) The treatment of vicinal dihalides with zinc generates carbon-carbon triple bond. it by treating with zinc. (d) The ease of dehydrohalogenation of alkyl halides is 3° > 2° > 1°.

Preparation of Alkenes 11. The dehydrohalogenation of an alkyl halide to produce an alkene can be brought about by using (a) aqueous KOH (b) alcoholic KOH (c) conc. H2SO4 (d) AlCl3

22.20 Complete Chemistry—JEE Main

12. The treatment of 2, 3-dibromobutane with Zn followed by heating produces (b) CH3CH==CHCH3 (c) CH3C∫∫CCH3 (a) CH3C CH CH3

(d) CH3C == C CH3 | | Br Br

Br 13. The electrolysis of aqueous solution of potassium succinate produces (a) methane (b) ethene (c) acetylene (d) methyl alcohol 14. During debromination of meso-dibromobutane, the major compound formed is (a) n-butane (b) 1-butene (b) cis-2-butene (d) trans-2-butene 15. Which of the following statements is not correct? (a) The ease of dehydration of alcohols is 1° > 2° > 3°. (b) The dehydrohalogenation of alkyl halides is an example of 1,2-elimination and is brought about by action of a base. (c) 1-2 Elimination reaction involving E2 mechanism does not involve any rearrangement of carbon skeleton. (d) 1-2 Elimination reaction involving E1 mechanism may involve the rearrangement of carbon skeleton. 16. Which of the following statements is not correct? (a) In dehydrohalogenation, the more stable the alkene, the faster it is formed. (b) The order of reactivity of alkyl halides towards E2 dehydrohalogenation is 3° > 2° > 1°. (c) Hydrogenation of an alkene is an exothermic reaction. (d) n-propyl alcohol can be produced by treating propylene with 80% H2SO4 followed by heating with water. 17. Which of the following statements is correct? (a) The major product in the dehydrohalogenation of 2-chloro-2,3-dimethylbutane is 2,3-dimethyl-1-butene. (b) The order of reactivity of n-propyl bromide, isobutyl bromide and ethyl bromide towards E2 dehydrohalogenation is ethyl > n-propyl > isobutyl. (c) Of the simple dialkyl ethylenes, the cis isomer is more stable than the trans isomer. (d) If enthalpy of combustion of cis-2-butene is more negative than trans-2-butene, then the conversion of cis to trans isomer is exothermic. 18. Which of the following statements is correct? (a) The dehalogenation of meso-2, 3-dibromobutane produces cis-2-butene (b) The dehalogenation of (S, S)-2, 3-dibromobutane produces trans-2-butene (c) Dehydrobromination of (R, R)-2, 3-dibromobutane produces cis-2-bromobutene (d) Dehydrobromination of meso-2, 3-dibromobutane produces cis-2-bromo-2-butene

Reactions of Alkenes

20. 21.

22.

23.

(a) ethylene glycol (b) (c) ethyl alcohol (d) The Baeyer’s reagent is (c) (a) alkaline KMnO4 4 The addition of HOCl to but-1-ene produces (a) 1-hydroxy-2-chlorbutane (b) (c) 1-hydroxy-3-chlorobutane (d) The ozonolysis of 2, 4-dimethylpent-2-ene produces (a) two molecules of aldehydes (b) two molecules of ketones (c) one molecule of aldehyde and one molecule of ketone (d) neither aldehyde nor ketone The ozonolysis of 2, 3-dimethylpent-2-ene produces (a) two molecules of aldehydes

acetic acid carbon dioxide and water alkaline K2Cr2O7 2-hydroxy-1-chlorobutane 3-hydroxy-1-chlorobutane

2Cr2O7

Hydrocarbons

24.

25.

26.

27. 28. 29.

22.21

(b) two molecules of ketones (c) one molecule of aldehyde and one molecule of ketone (d) neither aldehyde nor ketone The ozonlysis of 2, 5-dimethylhex-3-ene produces (a) two molecules of aldehydes (b) two molecules of ketones (c) one molecule of aldehyde and one molecule of ketone (d) neither aldehyde nor ketone The addition of HI in the presence of peroxide does not follow anti-Markovnikov’s rule because (a) HI bond is too strong to be broken homolytically (b) I atom is not reactive enough to add on a double bond (c) I combines with H to give back HI (d) HI is a reducing agent The addition of HCl in the presence of peroxide does not follow anti-Markovnikov rule because (a) HCl bond is too strong to be broken homolytically (b) Cl atom is not reactive enough to add on a double bond (c) Cl combines with H to give HCl (d) HCl is a reducing agent Ozonlysis of 2-butene produces (a) ethanal only (b) methanal only (c) butanal (d) propanone Ozonlysis of 2-methylbut-2-ene produces (a) ethanal only (b) acetone only (c) ethanal and propanal (d) ethanal and acetone The peroxide effect involves (a) ionic mechanism (b) free-radical mechanism

30. Identify the product in the following reaction: heated with Cr O

2 3 æææææ Æ ‘product n-heptane æ supported on alumina

at 600∞C

(a) benzene (b) toluene (c) xylene (d) cyclohexane 31. The treatment of ethylene with Baeyer’s reagent produces (a) ethyl alcohol (b) acetaldehyde (c) ethylene glycol (d) a-hydroxyl acetaldehyde 32. In which of the following reactions, addition does not take place in accordance with the Markovnikov’s rule? RCOOR

ƺ (a) CH3CH == CH2 + HCl ææææ RCOOR

(c) CH3CH==CH2 + HI ææææ ƺ

RCOOR

(b) CH3CH==CH2 + HBr ææææ ƺ (d) CH3CH==CH2 + HOCl Æ …

(a) propene (b) 2-methyl-1-butene (c) 2, 3-dimethylbut-2-ene (d) 1-butene 34. The treatment of CH3CH==CHCH3 with NaIO4 or boiling KMnO4 produces (a) CH3CHO only (b) CH3COOH only (d) CH3COCH3 and HCOOH (c) CH3CHO and CH3COOH 35. The treatment of CH3C == CHCH3 with NaIO4 or boiling KMnO4 produces | CH3 (a) CH3COCH3 and CH3COOH (c) CH3CHO and CO2

(b) CH3COCH3 and HCHO (d) CH3COCH3 only

22.22 Complete Chemistry—JEE Main

36. When ethene reacts with bromine in aqueous sodium chloride solution, the product(s) obtained is (are) (a) ethylene dibromide only (b) ethylene dibromide and 1-bromo-2-chloroethane (b) 1-bromo-2-chloroethane only (d) ethylene dichloride only 37. Which of the following statements is not correct? (a) The electophilic addition across the double bond may be accompanied with the rearrangement of carbon skeleton. (b) Markovnikov’s rule is in agreement with Saytzeff’s rule. (c) The rate of addition of a hydrogen ion to a double bond depends upon the stability of the carbocation being formed. (d) The more stable the carbocation formed as an intermediate in the addition reaction of alkene, the slower it is formed. 38. Which of the following statements is correct? (a) During the addtion of halogens to alkenes, a carbocation is formed as an intermediate. (b) The addition of aqueous Br2 to ethylene in the presence of NaCl forms only dibromoethylene. (c) Polyethylene is essentially an alkane with a very long chain. (d) Treatment of an alkene with cold alkaline potassium permangate breakes the double bond. 39. Which of the following statements is correct? (a) 1,2-dichloroethene is more reactive towards addition of H2SO4 in comparison to vinyl chloride. (b) The function of the zinc dust in the ozonolysis of an alkene is to destroy H2O2 which is formed during the reaction. (c) Dehydration of 1-butanol mainly gives 2-butene. (d) Dehydrohalogenation of 2-bromopentane results in the formation of 1-pentene. 40. Which of the following statements is correct? (a) The addition of HBr to 1,3-butadiene at – 80 °C produces 1-bromo-2-butene as the major product. (b) The addition of HBr to 1,3-butadiene at 40 °C produces 3-bromo-1-butene as the major product. (c) The addition of HBr to 1,3-butadiene produces both 1-bromo-2-butene and 3-bromo-1-butene which are in equilibrium with each other. The relative amounts of the two are temperature dependent. (d) The treatment of gaseous propene with chlorine at 500-600 °C yields 1,2-dichloropropane. 41. Which of the following statements is correct? (a) The trans-2-butene is more stable than cis-2-butene. (b) The acid-catalysed dehydration of an alcohol follows E2 mechanism. (c) The dehydration of an alcohol can be carried out in the presence of a base. CH2

CH2OH

(d) The acid-catalysed dehydration of

produces

as the major product.

42. Which of the following statements is correct? (a) The addition of Br2 to cyclohexene produces cis-1, 2-dibromocyclohexane (b) The addition of Br2 to cyclohexene, in the presence of NaCl produces a mixture of trans- bromohexane, trans1-bromo-2-cholorohexane and trans-dibromohexane (c) The addition of Br2 to cis-2-butene produces racemic-2, 3-dibromobutane (d) The addition of Br2 to trans-2-butene produces racemic-2, 3-dibromobutane 43. Which of the following statements is correct? (a) The addition of H2 to cis-2, 3-dibromobutene produces meso-2, 3-dibromobutane (b) The addition of H2 to trans-2, 3-dibromobutene produces meso-2, 3-dibromobutane

Hydrocarbons

22.23

(c) The hydroxylation of cis-2, 3-dimethylbutene with Baeyer’s reagent produces racemic-2, 3-dihydroxybutane (d) The hydroxylation of trans-2, 3-dimethylbutene with Baeyer’s reagent produces meso-2, 3-dihydroxybutane.

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43.

(b) (c) (b) (d) (b) (c) (d) (a)

2. 8. 14. 20. 26. 32. 38.

(b) (a) (d) (d) (a) (b) (c)

3. 9. 15. 21. 27. 33. 39.

(c) (c) (a) (b) (a) (c) (b)

4. 10. 16. 22. 28. 34. 40.

(a) (b) (d) (c) (d) (b) (c)

5. 11. 17. 23. 29. 35. 41.

(c) (b) (d) (b) (b) (a) (a)

6. 12. 18. 24. 30. 36. 42.

(a) (b) (d) (a) (b) (b) (c)

HINTS AND SOLUTIONS CH3

5. CH3

C

3

2

1

CH

CH2

; 3, 3-Dimethylbut-1-ene

CH3

8. Increasing R groups attached to double bonds increases stability of alkene. 10. (a) In general, cis isomer has higher polarity and hence has higher boiling point. On the other hand, because of lower symmetry, cis (b) With zinc, dehalogenation occurs producing carbon-carbon double bond. The use of alcoholic KOH followed by KNH2 produces carbon-carbon triple bond. 12. CH3

CH

CH

Br

Br

CH3

Zn

Æ CH3CH==CHCH 3

13. CH 2 COOK + 2H 2 O Æ H2C==CH2 + 2CO2 + 2KOH + H2 | CH 2 COOK 16. (b) This is due to the fact that the intermediate carbocation has the stability in the order 3° > 2° > 1°. (d) The addition of H2SO4 to double bond follows Markovnikov’s rule H SO

H O

2 4 2 CH3CH == CH 2 æææ Æ CH3CHCH3 æheat ææ Æ CH3CHCH3 | | OSO3 H OH

Hence, n-propyl alcohol cannot be produced by this method. 17. (a) According to Saytzeff’s rule, the more substituted alkene is more stable and is thus a preferred product. Thus, the major product is H3C H CH3

C

C

CH3 Æ CH3

Cl CH3

C

C

CH3

CH3 CH3 2,3-dimethyl-2-butene

(b) Reactivity in E2 dehydrohalogenation depends mainly upon the stability of the alkenes being formed. CH 3

CH3

CH3 CH CH 2Br Æ CH 3 C CH 2 ; CH CH CH Br Æ H C—CH=CH ; CH CH Br Æ CH ==CH 3 2 2 3 2 3 2 2 2 isobutyl bromide disubstituted n-propyl bromide monosubstituted thyl bromide unsubstituted

22.24 Complete Chemistry—JEE Main

According to Saytzeff’s rule, the stability of alkenes increases with increase in the number of substituents on the doubly bonded carbons. Hence, the order of reactivity towards dehydrohalogenation is isobutyl > n-propyl > ethyl (c) A trans isomer is usually more stable than a cis isomer. 18. (a) The Product is trans-2-butene.

(b) The product is cis-2-butene.

(c)

(d)

Cr O 2- H +

2 7 19. CH2==CH2 æææææ Æ 2CO2 + 2H2O

HOCl

21. CH3—CH2—CH==CH2 æææÆ CH3CH2CH(OH)CH2Cl 3 Æ CHCHCHO + OC (CH3 )2 22. CH3CHCH == CCH 3 ææ | | | (ketone) CH3 CH3 CH3

O

(aldehyde)

Hydrocarbons

23. CH3CH2C

CCH3

O3

Æ CH3CH2C

CH3 CH3

O+O

C

CH3

22.25

CH3

CH3

(ketone)

(ketone)

3 24. CH3CHCH == CHCHCH3 æOæ Æ 2 (CH3 )2CHCHO | | (aldehyde) CH3 CH3 3 27. CH3CH == CHCH3 æOæ Æ 2CH3CHO

ethanol

O

3 28. CH3CH == C(CH3)2 ææ Æ CH3CHO + OC(CH3)2

ethanal

acetone

reagent 31. H2C==CH2 æBaeyer’s æææææ Æ HOCH2CH2OH RCOOR

Æ CH3CH2CH2Br 32. CH3CH==CH2 + HBr ææææ 33.

37. (b) Both Markovnikov and Saytzeff rules follows from the fact that there is an intermediate formation of the more stable carbocation. (c) The more stable the carbocation, the more quickly it is formed. 38. (a) A halonium ion is formed as the intermediate instead of a carbocation. (b) Besides dibromo compound, the bromochloro and bromoalcohol compounds are also formed. (c) The structure of polyethylene is (—CH2—CH2—)n. (d) A diol is formed. 39. (a) Two chlorine atoms attached to ethylene destabilizes the intermediate carbocation more than the monochloro substituted ethene. 40. (a) At lower temperature, 1,2-adduct predominates over the 1,4-adduct. Hence, the major component is 3-bromo1-butene. (b) At higher temperature, 1,4-adduct predominates over the 1,2-adduct. Hence, the major component is 1-bromo-2-butene. (d) At the given conditions, substitution at alkyl group takes place. This gives allyl chloride. 41. (b) The mechanism for the acid-catalysed dehydration of an alcohol (say, isopropyl alcohol) is as follows. Step 1

CH3

CH3 + H+

CH

+

OH

CH3

OH2

+

Step 2

CH3

CH +

Step 3

CH

CH3

from acid

CH3

CH

CH3

Æ CH3CHCH3 + H2O

OH2

CH2 + HSO4

CH3CH == CH2 + H2SO4

H Step 1 is fast equilibrium reaction. Step 2 is the slowest step as bond breaking involves high energy intermediate. It is also a rate determining step. Since it involves only one species, the elimination reaction follows E1 mechanism. (c) The base catalysed reaction will involve the reaction

22.26 Complete Chemistry—JEE Main +

CH3— C H—CH3 + OH –

CH 3 — CH —CH 3 | OH

which is not possible, as the leaving group OH– is a strong base and is a poor leaving group. In the presence of acid, the poor leaving group OH– is converted into a good leaving group H2O (a very weak base). (d) The product is 1-methylcyclohexene. An :H shift converts the less stable 1° to a more stable 3° carbocation.

+ H+ 42. (a) The addition of halogen to an alkene involves anti-addition. Thus, trans-1, 2-dibromocyclohexane is obtained. (b) Only a mixture of trans-dibromocyclohexane and trans-1-bromo-2-chlorocyclohexane is obtained. Transdichlorocyclohexane is not obtained as it will require the presence of Cl+ which is not available. (c) The addition of Br 2 across a double bond involves the formation of a bromonium ion intermidiate, followed by the anti addition of Br–. For cis-2-butene, the reaction proceeds so as to form racemic-2, 3-dibromobutane.

H CH3 Br2

(c) H

Br

+

H

H CH3

H

H Br

CH3

CH3

CH3 +

Br Br2

(d) H

H

CH3

CH3

H

Br

H H3C CH3

Br

Br

Br Br H

H3C

Br

CH3

Br

Br H CH3 (2S, 3S)

H

H

Br

H

H

Br

Br

CH3 (2S, 3R) meso CH3

Br CH3

H

H

CH3

H3C

H

Br CH3

H

Br

H

H

H

H H3C

CH3

H3C

Br CH3

H

Br

Br CH3 (2R, 3R)

Br

CH3

CH3

H

Br

H

Br CH3

H

H

H

H3C

CH3

H

Br

CH3

CH3

CH3 Br

H

Br Br H

H3C

CH3

H

CH3

CH3

Br

CH3

Br

H

H Br

H

Br CH3

H Br

CH3

Br

H

CH3

Br

H

CH3 (2R, 3S) meso

43. (a) Hydrogenation of an alkene is a heterogeneous process which occurs on the surface of a solid catalyst (PtO2 or Pd/C). The addition occurs with syn stereochemistry (both hydrogens add from the same side). For cis isomer, the product is a meso compound. H3C H H

H3C

CH3 Br

H3C

Br

Br

CH3

Br H

H2 CH3

Br

H

H CH3

Br

CH3

H

Br

Br

CH3

H

Br

H

CH3 (2S, 3R) meso CH3

H Br

H3C

CH3

H

Br

H Br

H3C

Br

Br

H

H3C

Br

Br

H

H

CH3 (2R, 3S) meso (same product)

Hydrocarbons

22.27

(b) The product is a racemic mixture. H3C H H H3C

Br Br

Br

Br

H2

CH3 H3C

CH3

H

H3C

CH3

H Br

Br H3C

CH3

H

Br

Br

Br

H

H

CH3 (2S, 3S) CH3

H Br

Br

CH3

H

Br

H H CH3

H3C Br

Br

Br

H

CH3

H

Br

H

CH3 (2R, 3R)

(c) The addition of two hydroxyl groups involves syn addition (on the same side of a double bond). The result is same as that of hydrogenation. The cis isomer produce meso compound. (d) The addition of two hydroxyl groups involves syn addition (on the same side of a double bond). The result is same as that of hydrogenation. The trans isomer produces racemic mixture.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Among the following structures CH3 C 2H 5

CH (I)

CH3 C 3H 7

CH3COCHC2H5 (II)

H H

C

CH3

+

H (III)

C 2H 5

CH (IV)

C 2H 5

it is true that (a) only II and IV are chiral compounds (b) all four are chiral compounds (c) only I and II are chiral compounds (d) only III is a chiral compound [2003] 2. Reaction of one molecule of HBr with one molecule of 1, 3-butadiene at 40ºC gives predominantly (a) 3-bromobutene under thermodynamically controlled conditions (b) 1-bromo-2-butene under kinetically controlled conditions (c) 3-bromobutene under kinetically controlled conditions (d) 1-bromo-2-butene under thermodynamically controlled conditions [2005] 3. HBr reacts with CH2 = CH—OCH3 under anhydrous conditions at room temperature to give (b) CH3CHO and CH3Br (a) H3C—CHBr—OCH3 (d) BrCH2—CH2–OCH3 [2006] (c) BrCH2CHO and CH3OH 4. Reaction of trans 2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces (a) 3-phenylcyclopentene (b) 4-phenycylclopentene (c) 2-phenylcyclopentene (d) 1-phenylcyclopentene [2006]

22.28 Complete Chemistry—JEE Main

Me

5. The alkene formed as a major product in the elimination reaction N OH

-Bu

D Me Et

Me

Me

(b) Me

(a)

(c) CH2 = CH2

is

(d)

[2006]

6. In the following sequence of reactions, the alkene forms the compound B.

CH3 CH == CHCH3 The compound B is (a) CH3CHO

(b) CH3CH2CHO

7. The alkene that exhibits geometrical isomerism is (a) 2-Butene (b) 2-Methyl-2-butene

O3

A

H2O B Zn

(c) CH3COCH3

(d) CH3CH2COCH3 [2008]

(c) Propene

(d) 2-Methylpropene [2009]

8. The main product of the following reaction conc. H SO

2 4 C6H5CH2CH(OH)CH(CH3)2 æææææ Æ?

is C6H5CH2CH2 (a)

(c)

CH3 C6H5CH2

C

C

C

CH2 CH3

C6H5

(b)

(d)

H C6H5

C

C

C

C

H CH(CH3)2 CH(CH3)2

[2010] H CH3 H H 9. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass 44 u. The alkene is (a) ethene (b) propene (c) 1-butene (d) 2-butene [2010] 10. Out of the following, the alkene that exhibits optical isomerism is (a) 2-methyl-2-pentene (b) 3-methyl-2-pentene (c) 4-methyl-1-pentene (d) 3-methyl-1-pentene [2010] (a) an acetylenic triple bond (b) two ethylenic double bonds (c) a vinyl group (d) an isopropyl group 12. Elimination of bromine from 2-bromobutane results in the formation of (a) predominantly 1-butene (b) predominantly 2-butyne (c) equimolar mixture of 1-butene and 2- butene (d) predominantly 2-butene 13. In allene (C3H4), the type (s) of hybridization of the carbon atoms is (are): (b) sp2 and sp (c) only sp2 (a) sp and sp3

[2011 cancelled]

[2005]

(d) sp2 and sp3 [2014, online]

14. The gas liberated by the electrolysis of dipotassium succinate solution is: (a) Ethane (b) Ethyne (c) Ethene (d) Propene [2014, online] 15. In the presence of peroxide, HCl and HI do not give anti-Markownikoff’s addition to alkenes because: (a) One of the steps is endothermic in HCl and HI (b) Both HCl and HI are strong acids (c) HCl is oxidizing and the HI is reducing (d) All the steps are exothermic in HCl anHI [2014, online]

Hydrocarbons

16. What is the major product expected from the following reaction? CH3

22.29

[2015, online]

D—Cl

where D is an isotope of hydrogen CH3 (b) (a) H

CH3

Cl

D

(c) H

Cl

Cl

CH3

Cl

H

D

CH3

(b)

CH3

H

17. Which of the following compounds will exhibit geometrical isomerism? (a) 1-phenyl-2-butene (b) 3-phenyl-1-butene (c) 2-phenyl-1-butene 18. Which compound would give 5-keto-2-methylhexanal upon ozonolysis? CH3 CH3 CH3 (a)

D

(d)

[2015] (d) 1, 1-Diphenyl-1-propane [2015] CH3

(d) H3C

(c) CH3

CH3 19. Bromination of cyclohexene under conditions given below yields: Br2/hn

Br Br

Br

Br

(a)

(b)

(c)

Br

(d)

[2016, online]

Br

Br 20. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate +

+

(a) CH3 – CH – CH2Cl

(b) CH3 – CH(OH) – CH 2

(c) CH3 – CHCl – CH 2

(d) CH3 – CH – CH2 – OH

+

+

[2016]

ANSWERS 1. 7. 13. 19.

(c) (a) (b) (d)

2. 8. 14. 20.

(d) (b) (c) (a)

3. (a) 9. (d) 15. (a)

4. (a) 10. (d) 16. (a)

5. (c) 11. (c) 17. (a)

HINTS AND SOLUTIONS 2. At 40°C, thermodynamically controlled, 1, 4-addition product predominates. CH2==CH — CH==CH2

HBr 40°C

CH3CH==CHCH2Br 1-Bromo-2-butene

At low temperature (0 ºC), kinetically controlled 1,2-addition product predominates. Br

4. The involved reaction is

1

C6H5 2

2

ethanolic KOH

1

C6H5 3

6. (a) 12. (d) 18. (b)

22.30 Complete Chemistry—JEE Main

5. The major product in Hofmann elimination is to give the less substituted alkene resulting from the loss of the more acidic b H (1° < 2° > 3°). 6. The given reactions are as follows. O Zn/H2O CH 3CH CHCH 3 O3 CH3 CH CHCH3 2 CH3CHO O

O

7. The given compounds are: (a) H3C—CH == CH—CH3

(b) H3C—CH == C — CH3 CH3

(d) H3C—C == CH2

(c) H3H—CH == CH3

CH3 Only the compound (a) can exists as cis-trans isomers. 8. The dehydration of the given alcohol can give two products. These are C6H5

C6H5CH2

H C

and

CH

C

CH(CH3)2

H

C

CH3 CH3

H

9. The aldehyde CH3CHO has molecular mass of 44 u. Hence, the symmetrical alkene is CH3CH

CHCH3.

O3

CH3CH==CHCH3 ææÆ 2CH3 CHO 2-Butene.

10. The structures of given alkenes are as follows. CH3 CH3 CH3C

CH CH2CH3

CH3CH

2-Methylpentene

CH3 CH2

C — CH2 CH3

3-Methyl-2-pentene

CH CH2CHCH3

4-Methyl-1-pentene

CH3 CH2

CH CHCH2CH3

3-Methyl-1-pentene

The molecule 3-methyl-1-pentene has one assymetrical carbon atom and thus it will exhibit optical isomerism. 11. A vinyl group will produce formaldehyde on ozonolysis. H 2C

O3

ææÆ HCHO + other fragment

CH

12. The predominant product is 2-butene (a Saytzeffs products). –HBr

CH3CHCH2CH3

CHCH3

2-butene

Br 1

CH3CH

2

1

13. The allene is C H 2 = C = C H 2 . In this compound, carbon labelled as (1) is sp2 hybridized while carbon labelled as (2) is sp hybridized. 14. The electrolysis of dipotassium succinate solution releases ethene. CH2COO CH2 + 2CO2 + 2e CH2COO CH2 ethene

15. In the chain propagation steps, we have Step 1

X+

C

C

C

C +

X

Step 2

C X

C +

+ HX

C X

C

+X

Hydrocarbons

22.31

With HCl, Step 2 is endothermic owing to strong H-Cl bond. With HI, Step 1 is endothermic owing to weak HI bond. In fact, HI is oxidised to I2 by peroxide. Cl 16. CH3 H H CH3 DCl + H + CH3 CH3 Cl D D D (minor) (major) 17. The compound 1-phenyl-2-butene will exhibit geometrical isomerism. The isomers are:

CH3

PhCH2 C == C H 6

6

CH3 Ozonolysis

1 3

2

C == C

and H

H

CH3

CH3

5

5

18. 4

H

PhCH2

O

4

CHO 3

CH3

1

2

CH3

5-keto-2-methyl-hexanal

19. In the presence of ultraviolet radiation, the allylic C—H bond undergoes bromination. Br Br2/hn

20. The reaction proceeds as follows CH3

CH

CH2 + Cl+

CH3

+

CH

CH2Cl OH CH3

CH

CH2Cl

OH

SECTION 3

Alkynes

Methods of Preparation Dehydrohalogenation of Vicinal Dihalides Dehydrohalogenation takes place in two stages

First stage requires mild conditions whereas second stage requires more vigorous conditions as vinylic halide is unreactive.

22.32 Complete Chemistry—JEE Main

Reaction of Metal Acetylides with Primary Alkyl Halides C

C

LiNH 2

H

Æ

– C : Li + + RX Æ

C

C

C

R + LiX

(R must be 1°)

With secondary and tertiary halides, elimination is the predominant reaction which results in the formation of alkenes. This method can be used to generate larger alkyne from the smaller one.

Hydrolysis of CaC2 and Mg2C3 Calcium carbide gives acetylene whereas Mg2C3 gives propyne. CaC2 + 2H2O Æ HC Mg2C3 + 4H2O Æ CH3C

CH + Ca(OH)2 CH + 2Mg(OH)2

Kolbe’s Electrolytic Method The electrolysis of an aqueous solution of potassium salt of an unsaturated dicarboxylic acid forms alkyne. CHCOO– Æ HC

Anodic reaction

CH + 2CO 2 + 2e–

CHCOO–

2H2O + 2e– Æ 2OH– + H2

Cathodic reaction

Dehalogenation of vic-Tetrahalogen Compounds C 2 H 5 OH

CH3— CBr2— CBr2— CH3 + 2Zn

heat

Æ CH3— C

C— CH3 + 2ZnBr2

Chemical Properties Addition of Hydrogen If the triple bond is not present at the end of the chain of the molecule, its reduction produces either a cis alkene or a trans alkene depending upon the choice of reducing agent. The use of a catalyst causes syn addition resulting into cisalkene whereas the reduction with a reducing agent causes anti addition resulting into trans-alkene. Na, NH3 (liq)

H R

R

C

C

C

C

C

C

R

anti addition

H

R Na or Li Lindlar catalyst (Pb-BaSO4)

R H

R

syn addition

H

Further reduction produces an alkane.

Addition of Halogens Chlorine and bromine add on alkyne: R

C

CH

X2

ÆR

C

CH

X

X

X2

ÆR

X

X

C

C

X

X

H

Alkynes are less reactive than alkenes. The mechanism of the reaction involves the initial formation of a cyclic halonium sulting in the lower activity of alkynes.

Hydrocarbons

22.33

Addition of Hydrogen Halides (HCl, HBr, HI) Cl R

C

CH

HCl

ÆR

C

HCl

ÆR

CH2

C

Cl

CH3

Cl

The addition takes place in accordance with Markovnikov’s rule. Peroxides have the same effect on addition of HBr to acetylenes as they have on alkenes.

Addition of Water In the presence of acid and HgSO4, a molecule of water adds to the triple bond. Initially enol is formed which is rapidly converted into an equilibrium mixture containing keto form in excess. C

C

H SO

2 4 R´ + H2O HgSOÆ R 4

C

C

H

OH

Æ Æ

R

R

R´

Keto-enol tautomerism

CH2

C

R´

O 2COR¢ and RCOCH2R¢ ).

Formation of Metal Acetylides A triply bonded carbon which is sp hybridized has a high electronegativity and hydrogen atom attached to it shows appreciable acidity. For example, acetylene reacts with Na and Li liberating H2 gas: C—H + Na Æ HC C–: Na+ + 12 H2 HC sodium acetylide

Compared to H2O, NH3 and alkane, the order of acidity of acetylene is as follows. CH > NH3 > RH H2O > HC The order of conjugate base will be OH– < HC

C– < NH2– < R–

Lithium acetylide has been used to convert lower alkynes into the bigger one using the reaction R—C C—Li + R¢ X Æ R—C C—R¢ + LiX (1° alkyl halide)

Heavy metal ions, mainly, Ag+ and Cu+, react with acetylinic hydrogen to form insoluble acetylides. This reaction can be used to differentiate between terminal and nonterminal alkynes: R —C

C —H

ammoniacal Ag+ Æ R— C C—Ag (Tollens reagent) precipitate

H —C

C —H

2Ag +

Æ Ag—C

R —C

C —H

Cu +

Æ R— C

ammoniacal medium

C—Ag C—Cu

Reaction with Grignard Reagent or Alkyllithium The acetylinic hydrogen on reacting with R´MgBr or R´Li produces the alkane R´H: RC CH + R¢ MgBr Æ R¢ H + RC CMgBr RC

CH + R¢ Li Æ R¢ H + RC

CLi

22.34 Complete Chemistry—JEE Main

Ozonolysis Alkynes produces carboxylic acids. O

CH3CH2C

O3

CCH3

Æ CH3CH2C

H 2O

CCH3

Æ CH3CH2COOH + CH3COOH

O O Polymerization Acetylene dimerizes in the presence of cuprous chloride and ammonium chloride to give vinyl acetylene which on reacting with HCl gives chloroprene. The latter polymerizes to give neoprene—a synthetic rubber. The reactions are:

HC

CH + HC

Cu Cl

2 2 CH NH ClÆ CH2 4

CH

C

CH

HCl

vinyl acetylene

Cl CH2CH C

Polymerization

CH2

n

Æ peroxide

CH2

CH

C

CH2

Cl chloproprene

Cyclic Polymerization The cyclic polymerization takes place when alkyne is passed through red hot iron tube at 400°C. 3HC

CH

hot iron tube 400 C

Æ benzene

CH3

3CH3C

CH

hot iron tube 400 C

Æ H3C

mesitylene

CH3

Isomerization The following reactions can be used to convert 1-alkyne to 2-alkyne and vice versa. CH3CH2C

KOH (alc)

Æ CH3CH==C==CH2

Æ

CH

CH3CH2C

CNa Æ

NaNH 2 in

inert solvent

Æ

allene

H2O

CH3C

CCH3

MULTIPLE CHOICE QUESTIONS ON SECTION 3 General Characteristics of Alkynes 1. Ethylene can be separated from acetylene by passing the mixture through (a) fuming H2SO4 (b) pyrogallol (c) ammoniacal CuCl2 2. Which of the following has the highest boiling point? (a) 1-Butyne (b) 2-Butyne (c) 1-Pentyne

(d) charcoal powder (d) 2-Hexyne

Hydrocarbons

22.35

3. Lindlar’s catalyst is (d) Na in liquid NH3 (a) Pt in ethanol (b) Ni in ethanol (c) Pd with BaSO4 4. The two hydrogen atoms in acetylene (a) are acidic in nature (b) are alkaline in nature (c) are neutral in nature (d) are acidic and alkaline in nature, respectively. 5. Which of the following is true? (a) Acetylene is more reactive than ethylene to an electrophilic attack (b) Acetylene is less reactive than ethylene to an electrophilic attack (c) Acetylene may show more reactivity or less reactivity towards electrophilic attack depending upon electrophilic reagent (d) Acetylene and ethylene show identical reactivities towards an electrophilic attack 6. Which of the following molecules has all the three types sp, sp2 and sp3 of carbon atoms? (b) CH3—CH==CH—CH3 (a) CH2==CH—CH==CH2 (c) CH3—CH==CH—C CH (d) CH3 CH2 CH2 CH3 7. The molecule CH3—CH==CH—C CH contains (a) one sp3, two sp2 and two sp carbon atoms (b) one sp3, one sp2 and three sp carbon atoms 2 (d) two sp3, two sp2 and one sp carbon atoms (c) two sp and three sp carbon atoms 8. The IUPAC name of the compound CH3 CH3 C

C

CH

CH3

9. 10. 11.

12.

13.

14.

is (a) 3, 3, 3-trimethylpropane (b) 1, 1, 1-trimethylpropyne (c) 3, 3-dimethylbut-1-yne (d) 2, 2-dimethyl but-3-yne CCH2CH==CH2 is The IUPAC name of the compound HC (a) 1-penten-4-yne (b) 1-pentyn-4-ene (c) allylacetylene (d) 1-penten-5-yne The compounds 1-butyne and 2-butyne can be distinguished by using (c) Tollens reagent (d) chlorine gas (a) bromine water (b) KMnO4 solution Which of the following statements is correct? (a) Alkynes are reduced less readily than alkenes (b) Alkynes are reduced more readily than alkenes (c) Both alkynes and alkenes are reduced with equal speeds (d) Alkynes cannot be reduced to the alkene stage Which of the following orders regarding acid strength is correct? (a) CH3COOH > CH3CH2OH > CH CH (b) CH3COOH > CH CH > CH3CH2OH CH > CH3COOH > CH3CH2OH (d) HC CH > CH3CH2OH> CH3COOH (c) HC Which of the following statements is correct? (a) Acetylene is more reactive than ethylene towards the addition of halogens (b) Acetylene is less reactive than ethylene towards the addition of halogens (c) Acetylene and ethylene show identical reactivities towards an electrophilic attack. (d) The reactivities of acetylene and ethylene towards electrophilic attack depend on the electrophilic reagent. Which of the following statements is correct? (a) Ethyne and its derivatives give white precipitate with ammoniacal silver nitrate solution. (b) The carbon-carbon triple bond in acetylene has a bond length of 121 pm. (c) Alkynes are more dense than water. (d) The hydrohalogenation of vinylic halide requires mild conditions.

22.36 Complete Chemistry—JEE Main

15. Which of the following statements is correct? (a) Tautomers differ in the arrangement of atoms and are also in rapid equilibrium with each other. (b) Keto-enol tautomerism is an equilibrium between two acids favouring the formation of the stronger acid. (c) A pair of electron in an sp hybrid orbital of carbon is less near to the nucleus as compared to that in an sp3 hybrid orbital. (d) 2-Butyne reacts with ammoniacal silver nitrate solution.

Preparation of Alkynes 16. The hydrolysis of Mg2C3 produces (a) acetylene (b) propyne (c) butyne (d) ethylene 17. The hydrolysis of calcium carbide produces (a) acetylene (b) propyne (c) butyne (d) ethylene 18. Acetylene is produced when C2H4Br2 is heated with (a) aqueous KOH (b) ethanolic KOH and NaNH2 (c) dilute HCl (d) Zn dust 19. Acetylene is produced by carrying out the electrolysis of potassium salt of (a) acetate (b) succinate (c) fumarate (d) oxalate 20. The product(s) obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be (b) CH3CH2CH2CHO (a) CH3CH2COCH3 (d) CH3CH2COOH + HCOOH (c) CH3CH2CHO + HCHO Reactions of Alkynes 21. The addition of H2 to CH3C∫∫CCH3 using Lindlar’s catalyst produces (a) cis-but-2-ene (b) trans-but-2-ene (c) a mixture of cis- and trans-but-2-ene (d) 2-methylprop-1-ene 22. The addition of H2 to CH3C∫∫CCH3 in the presence of Na or Li in liquid ammonia produces (a) cis-but-2-ene predominantly (b) trans-but-2-ene predominantly (c) an equal mixture of cis- and trans-but-2-ene (d) 2-methylprop-1-ene 23. Ozonolysis of propyne gives (a) acetic acid only (b) formic acid only (c) acetic and formic acids (d) acetaldehyde 24. When propyne is passed through red hot iron, it is converted into (a) benzene (b) toluene (c) mesitylene (d) xylene 25. The addition of hypohalous acid to acetylene produces OH Cl

(a)

H HO

C

C

Cl H

(b) H

C

C

OH OH

H

(c) H

C

C

H

(d)

Cl Cl

CHCHO

OH Cl Cl Cl 26. The hydration of propyne in the presence of HgSO4/H2SO4 produces (b) CH3CH2CHO (c) CH3COCH3 (d) HCHO (a) CH3CHO 27. Sulphuric acid reacts with acetylene in the presence of mercuric sulphate to give (a) acetic acid (b) acetaldehyde (c) ethyl hydrogen sulphate (d) diethyl sulphate 28. Which of the following compounds reacts with an aqueous solution of Ag(NH3)OH? (a) 2-Butyne (b) 1-Butyne (c) Ethene (d) Ethane 29. The cyclic polymerization of dimethylacetylene produces (a) benzene (b) o-xylene (c) 1, 3, 5-trimethylbenzene (d) 1, 3, 5-trimethylbenzene 30. The cyclic polymerization of acetylene produces (a) benzene (b) cyclohexane (c) toluene (d) 1, 3, 5-trimethylbenzene

Hydrocarbons

22.37

31. The order of reactivity of acetylene with halogen acids is (a) HI > HBr > HCl (b) HI < HBr < HCl (c) HI < HBr > HCl (d) HI > HBr < HCl 32. The ozonolysis of a triple bond produces (a) a mixture of aldehyde/ketone and carboxylic acid (b) a mixture of aldehydes/ketones (c) a mixture of carboxylic acids (d) CO2 and H2O 33. The reduction of an alkyne to alkene using Lindlar catalyst results into (a) cis addition of hydrogen atoms (b) trans addition of hydrogen atoms (c) a mixture obtained by cis and trans additions of hydrogen in the ratio 1 : 1 (d) a mixture obtained by cis and trans additions of hydrogen in ratio 1 : 2 34. Which of the following statements is correct? (a) The electrophilic addition of hydrogen halides to alkyne proceeds via the formation of an intermediate carbocation. (b) The addition of protic acids to alkynes takes place at very much the same rate as to alkenes. (c) The addition of halogens to alkynes takes place at very much the same rate as to alkenes. (d) The equilibrium

C

C

OH

C— C

O lies much in favour of enol form.

H

35. Which of the following statements is correct? (a) The addition of H2 to 2-butyne using Na in liquid NH3 gives cis-1,2-dimethylethene. (b) The addition of H2 to 2-butyne using H2 and Lindlar catalyst gives cis-1,2-dimethylethene. (c) The cis-1,2-dimethylethene is more stable than its trans isomer. (d) cis-trans mixture of 2-pentene cannot be converted to either cis or trans form. 36. Which of the following statements is correct? (a) The addition of HCl to 3,3-dimethyl-1-butyne exclusively gives 2,2-dichloro-3,3-dimethylbutane. (b) The number of isomers (including structural and stereoisomers) of alkynes C6H10 is eight. (c) The treatment of 1,1-dibromopentane with KOH(s) at 200 °C produces 1-pentyne. (d) The treatment of 1,1-dibromopentane with NaNH2 followed by acid gives 2-pentyne. 37. Which of the following statements is correct? (a) 2-Butyne is not soluble in concentrated H2SO4. (b) The reaction of one mole of HBr with 1-pentene-4-yne is H 2C

CHCH2C

CH

HBr Æ H C 2

CHCH2 C

CH2

Br (c) Alkynes are more reactive than alkenes toward addition of Br2. (d) Alkynes are more reactive than alkenes toward catalytic hydrogenation. 38. Which of the following statements is correct? (a) The treatment of 3-heptyne with KMnO4 under neutral conditions at room temperature gives CH3CH2COOH and CH3CH2CH2COOH. (b) The treatment of 3-heptyne with KMnO4 under alkaline or acidic conditions at higher temperatures gives CH3CH2COCOCH2CH2CH3. (c) The addition of HBr to 1,3-pentadiene proceeds at a slower rate than that to 1,4-pentadiene. (d) The addition of HBr to butadiene at –80 °C to give 1,2-adduct is kinetically controlled while that at 40 °C to give 1,4-adduct is thermodynamically controlled.

22.38 Complete Chemistry—JEE Main

39. Which of the following statements is not correct? (a) The addition of acetylene to lithium amide dissolved in ether produces NH3 and HC C Li + . (b) If water can generate HC∫∫CH from HC∫∫CLi then water must be a stronger acid than acetylene. (c) Acetylide ion (which has a pair of electrons in sp orbital of carbon) is more stronger base than ethide ion, C2H5– (which has a pair of electrons in sp3 orbital of carbon). (d) Lindlar catalyst is Pb/BaSO4. 40. Which of the following statements is correct? (a) The polymerization of chloroprene gives synthetic rubber known as neoprene. (b) The ozonlysis of propyne produces acetic acid. (c) The IUPAC name of HC∫∫C—CH2==CH2 is 4-pentyne-1-ene. (d) The number of isomer of alkyne C6H10 is six.

ANSWERS 1. 7. 13. 19. 25. 31. 37.

(c) (a) (b) (c) (d) (a) (d)

2. 8. 14. 20. 26. 32. 38.

(d) (c) (b) (a) (c) (c) (d)

3. 9. 15. 21. 27. 33. 39.

(c) (a) (a) (a) (b) (a) (c)

4. 10. 16. 22. 28. 34. 40.

(a) (b) (b) (b) (b) (a) (a)

5. 11. 17. 23. 29. 35.

(b) (b) (a) (c) (d) (b)

6. 12. 18. 24. 30. 36.

(c) (a) (b) (c) (a) (b)

HINTS AND SOLUTIONS 1. Hydrogen atoms in acetylene are acidic. These can be replaced by Cu+. 2. The boiling point increases with increase in carbon number. 5. For an electrophilic attack, acetylene is less reactive than ethylene. 6. CH3—CH==CH—C sp3

sp2

sp2

sp

CH sp

8.

9. C==C has priority over C C and gets the smaller number. 11. The adsorption of alkynes on the surface of a catalyst involves less strained transition state as compared to that of alkenes. Hence, alkynes react faster than alkenes. 14. (a) White precipitate is obtained only when —C∫∫C— functional group is present at the terminal(s) of the molecule, since only then it will have acidic hydrogen atom. (d) A vinylic halide is very unreactive. The use of vigorous conditions (such as the use of a strong base, NaNH2) are required to remove HX. 15. (b) Equilibrium reaction always favours the formation of weaker substances. In keto-enol tautomerism, we have C

C

O

stronger acid

H

C

C

O

H weaker acid

(c) An sp hybrid orbital has more s-character as compared to an sp3 hybrid orbital. (d) 2-butyne does not contain acetylinic hydrogen atom.

Hydrocarbons

16. Mg2C3 + 4H2O Æ CH3C

22.39

CH + 2Mg(OH)2

17. CaC2 + 2H2O Æ HC

CH + Ca(OH)2

19. HCCOOK + 2H2O Æ HC

CH + 2CO2 + 2KOH + H2

HCCOOK

21. The use of Lindlar’s catalyst causes cis-addition. 22. The use of Na or Li in liquid ammonia produces trans form predominantly. 23. CH3C CH O 3 Æ CH3COOH + HCOOH 25. HC

CH

HOCl

Æ HOCH==CHCl

HOCl

Æ HOCH CHCl2

–H 2 O

Æ OCH CHCl2

OH OH

26. CH3C

HgSO4 H 2SO 4

CH

H2O

Æ CH3C

HgSO 4 / H 2 SO 4 CH2 H 2O

CH3COCH3

–H2O

CH3C

CH3

OH

OH

OH

27. HC

CH

HgSO 4 H 2SO 4 H2O

Æ CH2==CHOH

HgSO 4 / H 2 SO 4 H 2O

CH3CHO

–H 2O

CH3

C

H

OH

28. A compound containing acidic hydrogen in —C ∫∫ CH group reacts with Ag(NH3)OH. CH3

29. 3CH3C

H3 C

CH3

H3 C

CH3

CH3 Æ CH3

30. 3HC

CH Æ

34. (c) Alkynes are considerably less reactive than alkenes as for as the addition of halogen is concerned. This is due (d) 35. (a) (c) (d)

36.(a)

It lies much in favour of keto form. The trans isomer is obtained. The trans isomer is more stable than its cis isomer. The conversion may be achieved as follows.

The product is a mixture of three isomers as explained in the following.

22.40 Complete Chemistry—JEE Main CH3 CH3

C

CH3 C

HCl

Æ CH3

CH

CH3

C

+ C

CH2

CH3 Cl

Cl–

C

CH3

CH3 vinylic cation

C

CH3 Cl CH2

HCl

Æ CH3

CH3

C

C

CH3

CH3 Cl 2,2-dichloro3,3-dimethylbutane

Besides this, the other products are obtained due to the methyl shift in vinylic cation. CH3 CH3

CH3

CH3 ⎯ C ⎯ C == CH2 → CH3 ⎯ C CH3

C

CH 2

Cl −

CH3 CH3

CH3 CH3 CH3 ⎯ C == C ⎯ CH2Cl

CH3 ⎯ C ⎯ C == CH2 Cl

HCl

HCl

CH3 CH3

CH3 CH3

CH3 ⎯ C ⎯ CH ⎯ CH2Cl

CH3 ⎯ C ⎯ C ⎯ CH3

Cl

Cl

1, 3-dichloro-2, 3-dimethylbutane

(b) There are eight isomers including stereoisomers. HC CCH2CH2CH2CH3; CH3C CCH2CH2CH3; 1-pentyne

Cl

2, 3-dichloro-2, 3-dimethylbutane

2-pentyne

CH3CH2C

CCH2CH3;

3-pentyne

CH 3 — CH—C ≡≡ CCH 3 | CH 3 4-methyl-2-pentyne

H

CH 3 — CH—CH 2 —C | CH 3

CH

H C

CH

CH3

4-methyl-1-pentyne

CH 3 | CH 3 — C—C | CH 3

CH3CH2

;

(S)-3-methyl-1-pentyne

HC

CH2CH3

C CH3

(R)-3-methyl-1-pentyne

CH

3,3-dimethyl-1-butyne

(c) The initially formed 1-pentyne is converted into 2-pentyne because the more highly R-substituted alkyne is more stable. (d) The very strong base NaNH2 removes the terminal proton from the dehydrohalogenation product and thus it cannot rearrange. CH3CH2CH2CH2CHBr2

NaNH 2

Æ CH3CH2CH2C

C–

H 3O

Æ CH3CH2CH2C

CH

37. (a) The vinyl carbocation from 2-butyne is more stable because of the presence of electron-repelling CH 3 + group attached to C+ (i.e. CH3—C==CH—CH3). Thus, 2-butyne is soluble in concentrated H2SO4 due to the + formation of the salt (CH3—C==CH—CH3)(HSO4–).

Hydrocarbons

22.41

(b) The addition of HBr proceeds through the formation of a carbocation. The alkyl carbocation formed from the + + alkene group (H3C—CHCH2C∫∫CH) is more stable than the vinyl carbocation (CH2==CHCH2C==CH2) from the alkyne group. The DH ‡ for the former is lesser than that of the latter resulting in the addition to HBr across the double bond, i.e. the alkene reacts at a faster rate than the alkyne towards electrophilic addition. Thus, the reaction taking place is H2C==CHCH2C

CH + HBr

H3CCHCH 2 C CH | Br

(c) The addition of Br2 across a double or triple bond occurs via the formation of the three membered ring of bromonium ion. CH == CH H2C CH2

Br

Br

(from alkyne)

(from alkene)

The brominium ion from alkyne is more strained due to the presence of double bond (shorter bond length). Moreover, its carbon atoms have more s-character. These factors make this brominium ion less stable than that from the alkene makaing alkynes less reactive than alkenes. (d) The hydrogenation proceeds via the adsorption on the surface of the catalyst. The adsorption of alkene occurs when the plane of the p-bond approaches perpendicular to the catalyst. The alkyne, any direction (parallel to C—C bond) can lead to the adsorption due to the cylindrical nature of the p-bonds. Thus, the transition state in case of alkynes is less constrained leading to the more positive value of DS ‡. This makes alkynes to react at a faster rate than alkenes. 38. (a) With neutral conditions at room temperature, a diketone is formed via the hydroxylation reaction. CH3CH2C

Æ CH3CH2C

C(CH2)2CH3

Æ

C(CH2)2CH3

OH OH

CH3CH2C—CH(CH2)2CH3

CH3CH2CH—C(CH2)2CH3

+

OH O

[O]

O OH

Æ

CH3CH2C—C(CH2)2CH3 O O

(b) Under neutral or alkaline conditions at higher temperatures, the triple bond is cleaved to give carboxylic acids. CH3CH2C

[O]

C(CH2)2CH3

Æ CH3CH2COOH + HOOC(CH2)2CH3

(c) The addition of HBr proceeds via the formation of carbocation. H + Æ CH — C H—CH==CH—CH 3 3

CH2==CH—CH==CH—CH3

allylic carbocation

CH2==CH—CH2—CH==CH2

H + Æ CH — C H—CH —CH==CH 3 2 2 2° carbocation

Allylic carbocation is more stable due to resonance

CH3

CH

CH == CH

CH3

+ or CH3

CH3

CH == CH

CH

CH

CH

CH

CH3

CH3

Its formation requires lesser DH‡ as compared to 2° carbocation and hence is formed at a faster rate.

22.42 Complete Chemistry—JEE Main

(d) The additions may be depicted as follows.

Intermediate complex

80 C H2C

CH

40 C

Br

H

CH == CH2

1, 2-adduct Br , 80 C

CH2 == CH

CH == CH2

H+

CH2 H

CH

CH

CH2

allylic carbocation Br , 40 C

Intermediate complex

H2C

CH == CH

H

CH2 Br

1, 4-adduct

The conversion allylic carbocation Æ 1,2-adduct involves a lower DH‡ as compared to carbocation Æ 1,4-adduct. At low temperature, 1,2-adduct is formed due to the lower DH‡. Thus, it formation is kinetically-controlled. As temperature is raised, 1,2-adduct passes over to the allylic carbocation and then to thermodynamically more stable 1,4-adduct. The 1,4-adduct is more stable as it is a more substituted alkene. Once, 1,4-adduct is formed, it cannot be converted to 1,2-adduct on lowering temperature as the conversion of 1,4-adduct to intermediate complex involves a larger DH‡ value. 39. (c) Acetylide is a weaker base than ethide ion. A pair of electrons is more tightly bound in acetylene than in ethide ion. (d) The Lindlar catalyst is Pd/BaSO4 O

3 Æ CH3COOH + HCOOH 40. (b) CH3C∫∫CH (c) The correct name is 1-penten-4-yne. C==C has priority over C∫∫C and gets the smaller number (d) There are seven isomers. 1-hexyne, 2-hexyne, 3-hexyne, 4-methyl-1-pentyne, 4-methyl-2-pentyne, 3-methyl-1-pentyne, 3, 3-dimethyl-1-butyne

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Which of the following reactions will yield 2, 2-dibromopropane? (b) CH3CH == CHBr + HBr æÆ (a) CH3—C ∫∫ CH + 2HBr æÆ [2007] (c) CH ∫∫ CH + 2HBr æÆ (d) CH3—CH == CH2 + HBr æÆ 2. The hydrocarbon which can react with sodium in liquid ammonia is (a) CH3CH2C∫∫CCH2CH3 (b) CH3CH2CH2C∫∫CCH2CH2CH3 (c) CH3CH2C∫∫CH (d) CH3CH∫∫CHCH3 [2008] 3. The treatment of CH3MgX with CH3C∫∫C—H produces (b) CH3CH==CH2 (c) CH3C∫∫CCH3 (d) CH3CH==CHCH3 (a) CH4 [2008] 4. The major organic compound formed by the reaction of 1, 1, 1– trichloroethane with silver powder is (a) 2-butene (b) acetylene (c) ethene (d) 2-butyne [2014]

Hydrocarbons

22.43

5. The reagent needed for converting H

Ph Ph

C

C Ph

C

C

H

Ph

is: (a) Cat. Hydrogenation

(b) H2/Lindlar Cat.

(c) Li/NH3

(d) LiAlH4

[2014, online]

ANSWERS 1. (a)

2. (c)

3. (a)

4. (d)

5. (c)

HINTS AND SOLUTIONS Br

CH + 2HBr Æ CH3

CH3—C

1. The reaction yielding 2, 2–dibromopropane is

C

CH3

Br

2. The compound CH3CH2C∫∫CH contains hydrogen atom attached to triply bonded sp hybridized carbon which acquires acidic nature due to high electronegativity of carbon. This acetylinic hydrogen is replaced by Na+. – + 1 CH3CH2C CH + Na Æ CH3CH2C C : Na + H2O. 2 3. The acetylinic hydrogen on reacting with CH3MgX produces the alkane CH4. CH3C∫∫CH + CH3MgX Æ CH4 + CH3C∫∫CMgX 4. The reaction is Cl CH3

C

Cl Cl

Cl + 6 Ag + Cl C Cl

CH3

CH3

C

C

CH3

2-Butyne

5. The anti addition of H2 to an acetylinic compound occurs by the use of Li/NH3. The use of catalytic hydrogenation causes syn. addition of hydrogen.

SECTION 4

Benzene

Structure of Benzene According to Kekule, benzene is a resonance hybrid of the following resonating structures.

The resonance energy of benzene is about 150.6 kJ mol–1, i.e. benzene is more stable than cyclohexatriene by 150.6 kJ mol–1. The above structure of benzene accounts for the following.

22.44 Complete Chemistry—JEE Main

1. 2. 3. 4.

Molecular formula of benzene is C6H6. Benzene yields only one monosubstitution product. Benzene yields three isomeric disubstitution products. All carbon-carbon bond distance in benzene are equal (139 pm) and are intermediate in length between single (154 pm) and double bonds (134 pm). 5. Benzene undergoes substitution rather than addition reactions. For example, the addition reaction of alkenes with cold dilute/alkaline KMnO4, Br2/CCl4 and HCl are not shown by benzene. However, under drastic conditions benzene is reduced to cyclohexane. Benzene undergoes substitution reactions in which H is replaced by another atom or group of atoms. For examples H 2SO 4 Æ C6H5—NO2 + H2O (i) C6H6 + HONO2 æææ

nitrobenzene

(ii) C6H6 + Cl2 æFeæÆ C6H5—Cl + HCl chlorobenzene

(iii) C6H6 + HOSO3H æSO æ3 Æ C6H5—SO3H + H2O benzenesulphonic acid

(iv) C6H6 + RCl æææ Æ C6H5—R + HCl AlCl3

alkylbenzene 3 (v) C6H6 + CH3COCl æAlCl ææ Æ C6H5—COCH3 + HCl

Reaction (iv) is known as Friedel-Crafts alkylation and reaction (v) is known as Friedel-Crafts acylation. 6. The enthalpies of hydrogenation and combustion of benzene are lower than those expected for cyclohexatriene.

Orbital Representation of Benzene Benzene is a planar molecule where each carbon is sp2 hybridized. Of the three hybrid orbitals, two are used in s-bonding with two other carbon atoms and the third is used in s-bonding with hydrogen atom. In addition to these orbitals, each carbon has one p orbital occupied by one electron. Since this orbital lies perpendicular to the plane of benzene ring, the electron in this orbital is of p-type. The p orbital of each carbon atom can overlap with the adjacent p orbitals of carbon atoms making additional bond of p-type. But this bond is not localized between two carbon atoms but form two continuous doughtnut-shaped electrons cloud one lying above and the other below the plane of cyclic carbon skelton. It is the overlap of the p orbitals in both directions that corresponds to the resonance hybrid of two Kekule structures. The delocalization of p-electrons gives rise to resonance energy and makes the molecule more stable. Substitution Reactions in Benzene The typical reactions of the benzene ring are those in which the p-electrons serve as a source of electrons for electrophilic (acidic) reagents. Because of delocalization of p-electrons, benzene does not show additional reactions as in the case of alkenes but undergoes substitution reactions in which hydrogen atom attached to carbon atom is replaced by another atom or group of atoms. The Hückel 4n + 2 Rule A molecule acquires aromatic characteristics provided it has cyclic clouds of delocalized p electrons above and below the plane of the molecule and this p clouds has a total of (4n + 2) p electrons. The requirement of 4n + 2 electrons is known as 4n + 2 rule or Hückle rule. Examples are benzene (n = 1), naphthalene (n = 2) and anthracene (n = 3). Besides and also there is a reasonably good overlap for p-bond formation in these two rings. In smaller or bigger rings, it is largely offset by angle strain or poor overlap of p orbitals or both.

Hydrocarbons

22.45

Mechanism of Electrophilic Substitution Reactions Nitration A mixture of nitric acid and sulphuric acid is the nitrating reagent. The steps involved in the mechanism are as follows. (a) Formation of nitronium ion H3O+ + 2HSO–4 + +NO2

HONO2 + 2H2SO4

(b) Electrophilic attack by nitronium ion H + +NO2 Æ

NO2 (slow)

+ benzenonium ion

(c) Removal of proton H

NO2

NO2 + HSO4– Æ

+

+ H2SO4

(fast)

The rate determining step is the formation of carbocation, which is the resonance hybrid of the following three structures. NO2

H

H

H

NO2

+

H

H

NO2

+

H

NO2 +

+

H

The dispersal of positive charge over the molecule by resonance makes this ion more stable than with a localized positive charge. Probably because of this, a carbocation is formed from the relatively stable benzene.

Sulphonation The involved steps are (a)

2H2SO4

H3O+ + HSO4– + SO3 SO3–

H

(b)

SO3 +

(slow)

+ –

H

(c)

SO3

–

SO3

+ HSO–4

+ –

SO3

(d)

+ H2SO4

(fast)

SO3H

+ H 3O +

+ H 2O

(equilibrium far to the left)

22.46 Complete Chemistry—JEE Main

In fact, the entire sulphonation process is an equilibrium process H+

C6H6 + H2SO4

C6H5SO3H + H2O

For sulphonation, we require excess of H2SO4 along with SO3 and for desulphonation we use excess of H2O along with heating (100 –175 °C).

Friedel-Crafts Alkylation The steps are: RCl + AlCl3

AlCl–4 + R+ H

R

R+ +

H

R

R

+ AlCl–4 Æ

+

(slow)

+

+ HCl + AlCl3

(fast)

Here AlCl3 acts as a Lewis acid.

Mechanism of Halogenation The steps are Cl2 + FeCl3

Cl3 –Fe — +Cl — Cl H

–

+

Cl3 Fe— Cl—Cl +

H

Æ

+

+ FeCl–4

(slow)

Cl

Cl +

Cl

+ FeCl–4 Æ

+ HCl + FeCl3

(fast)

Theory of Orientation

Activating Group A group that releases electrons to benzene ring is an activating group. It directs the incoming electrophile to ortho or para position. Examples include — NH2, —NHR, —NR2, —OH Strongly activating: Moderately activating: —OCH3, —NHCOCH3 —CH3, —C6H5 Weakly activating: Deactivating Group A group that withdraws electrons from benzene is a deactivating group. It directs the incoming electrophile to meta position. Examples include — NO2, —CN, —COOH, —SO3H, —CHO, —COR and —N(CH3)3+ A group attached to benzene also affects the rate of electrophilic substitution reaction. An electron-releasing group disperses the positive charge of carbocation and thus makes it more stable causing the reaction to proceed faster as

Hydrocarbons

22.47

positive charge of carbocation and thus makes it less stable causing the reaction to proceed slower as compared to the substitution reaction involving benzene. An activating group activates all the positions of a benzene ring; even the positions meta to it are more reactive than any single position in the benzene ring itself. It directs ortho and para simply because it activates the ortho and para positions much more than it does the meta. Alternatively, only in the ortho and para attack by an electrophile, a carbocation is generated in which positive charge is located on the carbon atom to which electron-releasing group is directly attached making this carbocation to be much more stable than the other resonating structures. Because of contribution from the stable structure, the hybrid carbocations resulting from an attack at ortho and para positions are more stable than the carbocation formed during the attack at meta position and since the stable hybrid carbocation is formed faster, it follows that the electron-releasing or activating group is ortho and para directing. The above facts are depicted in the following structures. CH3

CH3

CH3 +

Para attack

+

H

+

Y

H

Y

H

Y

stablest

CH3

CH3 H

Ortho attack

+

CH3 H

Y

H

+

Y

+

Y stablest

CH3

CH3 +

Meta attack

CH3

+

H

H

Y

H +

Y

Y

A deactivating group deactivates all positions in the ring, even the positions meta to it. It directs meta simply because it deactivates the ortho and para positions even more than it does the meta. Alternatively, only in the ortho and para attack by an electrophile a carbocation is generated in which positive charge is located on the carbon atom to which electron-attracting group is directly attached making this carbocation to be much more unstable than the other resonating structures. Because of contribution from the unstable structure, the hybrid resulting from attack at ortho and para positions are more unstable than the carbocation formed during the attack at meta position and since the stable carbocation is formed faster, it follows that the electron-attracting or deactivating group is meta directing. The above facts are depicted in the following structures. NO2

NO2

NO2

+

Para attack

+

H

+

Y

H

Y

H

Y

unstable

NO2

NO2 H

Ortho attack

+

Y

NO2 H

+

+

Y

H Y

unstable

22.48 Complete Chemistry—JEE Main

NO2

NO2

+

Meta attack

+

H

NO2 H

Y

H +

Y

Y

no unstable structure

Activation of Benzene via Resonance The groups —NH2, —OH, and —Cl can activate benzene ring. This is shown in the following structures. +

NH2

+

NH2

+

NH2

NH2

–

–

– +

OH

+

OH

+

OH

OH

–

–

–

+

Cl

+

Cl

+

Cl

Cl

–

–

–

For — NH2 and —OH, the resonance effect is much more important than the inductive effect. In the former benzene is activated whereas in the latter, it is deactivated. In —Cl, both the resonance and inductive effects are evenly matched. It is because of this that —Cl occupies the unusual position of being deactivating group but ortho and para Wdirector.

Effect of Activating and Deactivating Groups on Weak Organic Acids and Bases (a) Benzoic acid If an electron-releasing group is present at the para position (where resonance effect outweighs the inductive effect) of the — COOH group, the tendency to release proton from —COOH group is decreased making such a substituted benzoic acid weaker than benzoic acid itself. For example, p-toluic acid, p-hydroxybenzoic acid and p-anthranilic acid are weaker than benzoic acid. On the other hand, if such a group is present at the meta position (where inductive effect outweights the resonance effect), the electron-withdrawing tendency of the group makes the meta substituted acid stronger than benzoic acid. The presence of electron-withdrawing group present at para position also makes the substituted benzoic acid more strong than benzoic acid itself. For example, p-nitrobenzoic acid is a stronger acid than benzoic acid itself. (b) Phenol Phenol is a weaker acid. This is because of the resonating structures in which oxygen atom acquires positive charge making removal of proton from —OH group more easily. O

H

+

O

H

+

O

–

H

+

O

H

–

–

O

H

Hydrocarbons

22.49

The electron-withdrawing group present at the para or meta position makes oxygen atom more positive and thus helps in releasing the proton and thus such a substituted phenol is more acidic than phenol itself. For example, p-nitrophenol is more acidic than phenol. On the other hand, electron-releasing group at the para position (where resonance effect example, p-cresol is less acidic than phenol. Ortho meta and para isomers. Nearly all ortho substituents exert an effect of the same kind—acid strengthening—whether they are electron-withdrawing or electronreleasing, and the effect is a usually large. The ortho effect undoubtedly has to do with the nearness of the groups involved, but is more than just steric hindrance arising from their bulk. (c) Aniline Aniline is a weaker base. It is a Lewis base, i.e. the lone pair present on nitrogen can be given to the Lewis acids. The electron-releasing group present at the para position increases the electron density on nitrogen and thus makes it more basic. For example, para-toluidine is more basic than aniline. On the other hand, the presence of electron-withdrawing group decreases the electron density on nitrogen and hence makes it less basic. For example, p-nitroaniline is less basic than aniline.

Rules of Predicting Orientation in Distributed Benzenes The following rules are followed. 1.

CH3

If the groups reinforce each other, there is no problem. For example,

2.

CH3

3.

A strongly activating group, competing with a 4. weakly activating group, controls the orientation. For example, OH OH

In an ortho-para director and meta director are not reinforcing, the ortho-para director controls the orientation. The incoming group goes mainly to ortho to the meta director. For example,

CH3

When two weakly (or strongly) activating or deactivating groups compete, substantial amounts of both isomers are obtained for example,

Cl

NO2

or Cl

CH3

CH3

5.

Very little substitution occurs in the sterically hindered position between meta substituents.

CH3 very little CH3

Arenes (Aromatic-Aliphatic Hydrocarbons) Arenes contain both aliphatic and aromatic units. Examples include ethylbenzene, toluene, xylenes and ethyl toluenes. These compounds show two sets of chemical properties: aromatic ring shows electrophilic substitution and side chain undergoes free radical substitution. Each portion of the molecule affects the reactivity of the other portion and determines the orientation of attack. Reactions shown by alkylbenzenes are electrophilic substitution on the ring and free radical substitution at the alkyl group. For example,

22.50 Complete Chemistry—JEE Main

CH3 X

X2, FeX3

CH3

CH3 +

(X2 = Cl2, Br2)

X CH2Cl X

Cl2

CCl3

CHCl2

heat or light

Cl2

Cl2

heat or light

heat or light

benzyl chloride

benzal chloride

benzo-trichloride

Alkenyl Benzenes Alkenyl benzenes contain a double bond in the side chain of benzene ring. For example, CH

CH2;

styrene

CH

CH

CH3;

1-phenylpropene

Alkenylbenzenes show both substitution in the ring and addition to the double bond in the side chain. The reactivity of the double bond is greater than the benzene towards electrophilic reagents. Thus, mild conditions are required for the addition reaction across the double bond as compared to those required for the benzene ring. For example,

ÈCarbocation ˘ ˙ ææ Æ Product Reactant ææ Æ ÍÍ or ˙ ÍÎ free radical ˙˚ The more stability of benzyl cation or radical +

CHCH2CH3 or

. CHCH2CH3

relative to the conjugated alkenylbenzene makes the latter more reactive than simple alkenes. Also more stability of + + + benzyl cation or radical relative to other possibilities, such as C6H5CH2C HCH3 and C 6H5CH2CH2C H2, determines the

Hydrocarbons

22.51

In the presence of peroxides, the reaction taking place is +

MULTIPLE CHOICE QUESTIONS ON SECTION 4 Electrophilic Substitution 1. Which of the following substituents is ortho or para directing in the electrophilic substitution reactions in the benzene ring? (b) —NO2 (c) —CHO (d) —COOH (a) — NH2 2. Which of the following substituents is meta directing in the electrophilic substitution reactions in the benzene ring? (b) —CH3 (c) —CHO (d) —OH (a) — OCH3 3. An ortho-para directing group (a) activates ortho and para positions and deactivates the meta position (b) activates meta position and deactivates ortho and para positions (c) activates ortho and para positions more than the meta position (d) deactivates ortho and para positions less than the meta position 4. A meta directing group (a) deactivates ortho and para positions more than the meta position (b) activates ortho and para positions less than the meta position (c) activates meta position and deactivates ortho and para positions (d) deactive meta position and activates ortho and para positions 5. Among the following, the compound that can be readily sulphonated is (a) benzene (b) toluene (c) nitrobenzene (d) benzaldehyde 6. The reaction of toluene with chlorine in the presence of ferric chloride gives predominantly (a) benzyl chloride (b) benzoyl chloride (c) o- and p-chlorotoluene (d) m-chlorotoluene 7. Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives (a) o-cresol (b) p-cresol (c) 2, 4-dihydroxytoluene (d) benzoic acid 8. Which of the following reagents is weakly activating group? (b) —NR2 (c) —C6H5 (d) —NHCOCH3 (a) — NH2 9. In aniline, the —NH2 group (a) activates the benzene ring via both inductive and resonance effects (b) activates the benzene ring via resonance effect and deactivates via inductive effect (c) activates the benzene ring via inductive effect and deactivates via resonance effect (d) deactivates the benzene ring via both inductive and resonance effects 10. In the Friedel-Crafts acylation, the electrophile is (b) AlCl–3 (c) CH3CO+ (d) C6H5CH+2 (a) C6H5+ 11. In the nitrating mixture, HNO3 acts as (a) base (b) acid (c) reducing agent (d) catalyst 12. Which of the following species is expected to yield maximum percentage of meta substitution product? (b) ArCH2Cl (c) ArCHCl2 (d) ArCCl3 (a) ArCH3

22.52 Complete Chemistry—JEE Main

13. A direct iodination of benzene is not possible because (a) iodine is an oxidizing agent (b) resulting C6H5I is reduced to C6H6 by HI (c) HI is unstable (d) the ring gets deactivated 14. Which of the following statements is correct? (a) The principal mononitration product of o-cresol is 2-methyl-4-nitrophenol. (b) The carbon-hydrogen homolytic bond dissociation energy for benzene is considerable smaller than for cyclohexane. (c) The number of isomers of mononitration of o-dichlorobenzene is 4. (d) A tribromobenzene gives two isomers when it is mononitrated. The compound is 1,2,4-tribromobenzene. 15. Which of the following statements is correct? (a) Monobromination of p-toluenesulphonic acid followed by treatment with acid and superheated steam gives m-bromotoluene. O2N

(b) The mononitration of

CH2 NO2

gives the product

CH2 NO2

(c) The reactivity towards ring nitration of 2,4-dinitrotoluene is greater than that of m-dinitrobenzene. (d) The reactivity towards ring nitration of 2,4-dinitrophenol is smaller than that of 2,4-dinitrochlorobenzene. 16. Which of the following statements is not correct? (a) The order of reactivity towards ring nitration is p-xylene > toluene > p-toluic acid > terephthalic acid (b) The order of reactivity towards ring nitration is Ph(CH2)2NO2 > PhCH2NO2 > PhNO2 (c) The order of reactivity towards electrophilic substitution is C6H5CH3 < C6H5CH2COOC2H5 < C6H5CH(COOC2H5)2 (d) Each ring of biphenyl, C6H5 — C6H5, is more reactive than benzene towards electrophilic substitution, and the chief products are ortho and para isomers. 17. Which of the following statements is not correct? (a) In electrophilic aromatic substitution reaction, the formation of intermediate arenium cation is the ratedetermining step. (b) The C—H bond in benzene is slightly stronger than the C—D bond in deuterated benzene (C6D6). (c) The overall rate of an electrophilic substitution reaction, except sulphonation in benzene and deuterated benzene, are identical. (d) The sulphonation reaction involving benzene is a reversible reaction. 18. Which of the following statements is not correct? (a) All electrophilic substitution reactions except sulphonation involving benzene are irreversible. (b) The rates of sulphonation reactions involving benzene and deuterated benzene are slightly different. (c) Iodine solution in benzene is brown due to the formation of charge-transfer complex between iodine and benzene. (d) The reaction of benzene with n-propyl chloride in the presence of AlCl3 gives 1-phenylpropane as the only product.

Acid–Base Strengthening 19. Which of the following order regarding the acidity of aromatic acids is correct? (a) benzoic acid > p-toluic acid > p-hydroxybenzoic acid (b) benzoic acid > p-toluic acid < p-hydroxybenzoic acid (c) benzoic acid < p-toluic acid > p-hydroxybenzoic acid (d) benzoic acid < p-toluic acid < p-hydroxybenzoic acid 20. Which of the following orders regarding the base strength of aromatic bases is correct? (a) p-nitroaniline > aniline > p-toluidine (b) p-nitroaniline > aniline < p-toluidine

Hydrocarbons

22.53

(c) p-nitroniline < aniline > p-toluidine (d) p-nitroaniline < aniline < p-toluidine 21. Which of the following statements is not correct? (a) Aniline is a weaker base than methylamine (b) Phenol is a stronger acid than acetic acid (c) p-Anthranilic acid is weaker acid than benzoic acid (d) Phenol is a weaker acid than benzoic acid 22. Which of the following groups is acid-strengthening effect on benzoic acid? + (c) — NH3 group (d) — CH3 group (a) — OH group (b) — NH2 group 23. Which of the following groups is acid-weakening effect on benzoic acid? + (a) — OH group (b) — CHO group (c) — ON group (d) — NH3 group

Additional Problems 24. Which of the following carbocations is expected to be most stable? (a) O2N

H +

Y

+

(b) O2N

H Y

(c) O2N

+

(d) O2N

+

H Y H Y 25. The major product when 1 mol of H2 is added to C6H5CH==CHCH==CHCH3 using Pt catalyst is (b) C6H5CH2CH2CH==CHCH3 (a) C6H5CH2CH==CHCH2CH3 (c) C6H5CH==CHCH2CH2CH3 (d) C6H5CH2CH2CH2CH==CH2 26. Which of the following statements is correct? (a) Although benzene contains three double bonds, normally it does not undergo addition reaction. (b) m-chlorobromobenzene is an isomer of m-bromochlorobenzene. (c) In benzene, carbon uses all the three p orbitals for hybridization. (d) An electron donating substitutent in benzene orients the incoming electrophilic group to the meta position. 27. Which of the following statements is correct? (a) Benzene readily docolourises dilute KMnO4 solution. (b) The inductive effect is the only factor which decides the orientation effects in the monosubstituted benzene. (c) The number of isomers of aromatic compound tribromobenzene is four. (d) The dipole moment of p-nitrotoluene is expected to be larger than p-chloronitrobenzene. 28. Which of the following statements is correct?

(a) The compound O2N (b) The compound H3CH2C

NH2 is named 4-aminonitrobenzene. CH2CH2CH2CH3 is named 1-ethyl-4-butylbenzene.

(c) Cyclopentadiene is much more acidic than 1,3-cyclohexadiene. (d) The two phenyl groups in biphenyl lie in the same plane. 29. Which of the following statements is correct? (a) The oxidation of naphthalene with KMnO4 in acid gives phthalic acid. (b) The oxidation of naphthalene with O2/V2O5 gives 1, 4-naphthaquinone. (c) Treatment of p-chlorotoluene with NaOH(aq) at 340 °C exclusively gives p-hydroxytoluene. (d) Chlorine in p-chlorotoluene is replaced by OH when taken in NaOH(aq) at 340 °C. No such replacement occurs for 2,6-dimethylchlorobenzene. 30. Which of the following structures is correct? (a) The structure of the intermediate product, formed by the oxidation of toluene with CrO3 and acetic anhydride, whose hydrolysis gives benzaldehyde, is C6H5CH2OCOCH3. (b) The carbon-carbon bond length in benzene is 139 pm.

22.54 Complete Chemistry—JEE Main

(c) Cyclopentadienyl anion has unusually low stability. (d) Cycloheptatrienyl cation is also known as tropylium ion.

ANSWERS 1. 7. 13. 19. 25.

(a) (d) (b) (a) (c)

2. 8. 14. 20. 26.

(c) (c) (a) (d) (a)

3. 9. 15. 21. 27.

(c) (b) (c) (b) (d)

4. 10. 16. 22. 28.

(a) (c) (b) (c) (c)

5. 11. 17. 23. 29.

(b) (a) (b) (a) (d)

6. 12. 18. 24. 30.

(c) (d) (d) (d) (b)

HINTS AND SOLUTIONS 1. —NH2 group is electron-releasing group and is thus ortho or para directing. 2. —CHO group is electron-attracting group and is thus meta directing. 3. An ortho-para directing group activates all positions of benzene, the ortho and para positions are activated more than meta position. 4. A meta directing group deactivates all positions of benzene, the ortho and para positions are deactivated more than meta position. 5. —CH3 group in toluene activates benzene ring, its sulphonation is easier than rest of the three molecules. 6. In the presence of ferric chloride, electrophilic substitution at ortho and para positions take place. 7. In the presence of light and heat, methyl group is chlorinated. 8. Phenyl group is weakly activating group. 9. —NH2 group activates benzene ring via resonance and deactivates via inductive effect. 10. The electrophile is CH3CO+. 11. HNO3 acts as a base. It accepts proton from H2SO4. H3O+ + 2HSO4– + NO+2 HNO3 + 2H2 SO4 12. —CCl3 has a maximum electron attracting power. 13. C6H5I is reduced by HI to C6H6.

OH CH3 14. (a) The product is

NO2 A strongly activating group, competing with a weakly activating group, controls the orientation. The position 6 will be sterically hindered. (b) The C—H bond in benzene involves sp2–s character while in cyclohexane it is sp3–s. The bond C—H in benzene is expected to be shorter and thus stronger than in cyclohexane. (c) There are only two isomers.

Cl NO2

Cl

and

Cl

O2N Cl

Hydrocarbons

22.55

(d) 1,2,4-tribromobenzene gives three isomers.

Br

Br

Br

Br

Br

O2N

Br

and NO2 ,

O2N

Br

Br

Br

The compound 1, 2, 3-tribromobenzene gives two isomers.

Br

Br Br

Br and

Br

Br

O2N

NO2 15. (a) The product is o-bromotoluene.

CH3

CH3

CH3 Br

Br2

⎯⎯→

H2O, heat

Br

⎯⎯⎯⎯→

Fe

H2SO4

SO3H

SO3H

(b) Substitution is faster in the ring that is not deactivated by —NO2. Thus, the compounds obtained will be

NO2 CH2

CH2

O2N

NO2

O2N

(c) In 2,4-dinitrotoluene, there are two deactivating (NO2) groups and one activating (CH3) group while in m-dinitrobenzene, there are only two deactivating groups. (d) 2,4-dinitrophenol contains a strongly activating group (—OH) while 2,4 dinitrochlorobenzene contains weak deactivating group (—Cl). 16. (a) p-xylene contains two activating group, toluene contains one activating group, p-toluic acid contains one activating group (CH3) and one deactivating group (COOH) and terephthalic acid contains two deactivating groups (COOH). (b) The farther away a deactivating group from the ring, the less effective it is . (c) The more electron-withdrawing groups in benzene, the greater the deactivation. (d) The positive charge of the intermediate can be dispersed by the second phenyl group when the latter is attached to ortho or para, but not to meta.

H Y Intermediate

17. (a) The formation of intermediate arenium cation is slow while the elimination of H+ from this cation is fast. The former involves the removal of aromaticity while the latter restores aromaticity.

22.56 Complete Chemistry—JEE Main

(b) The C—D bonds are stronger than C—H bonds. In fact, the bond with the heavier isotope is slightly stronger than that with the lighter isotope. (c) In electrophilic substitution reactions, the rate-determining step is the formation of arenium cation.

H slow + E+

18.

19. 20. 21. 24.

E

E fast

+ H+

In this step, E+ bonds to the aromatic ring without cleavage of a C—H bond. The latter is broken in the second step, which is fast as it restores the stable aromatic system. Since C—H or C—D bond is not broken in the rate-determining step, the overall rates of reaction are identical. (b) In sulphonation reaction, both steps involving the formation of arenium cation and its conversion to the substituted benzene determine the overall rate of reaction. In the second step, the loss of D+ from the deuterated s complex is a little slower than the loss of H+ from the ordinary s complex. (d) The product cantains both PhCH2 CH2 CH3 and PhCH(CH3)2. Electron releasing group decreases the acidity of benzoic acid. Hydroxyl group is more powerful activator than methyl group. Electron releasing group increases the basicity of aniline while electron attracting group decreases the basicity. Phenol is a weaker acid than acetic acid. —NO2 group is benzene deactivator, it is meta directing.

25. Double bond attached to methyl group is reduced. 26. (a) Benzene gets stabilised because of resonance. (b) Both represent one and the same compound. (c) Each carbon in benzene is sp2 hybridized. (d) An electron attracting group directs the incoming electrophilic group to the meta position. 27. (c) There are three isomers.

Br

Br

Br

Br

Br ;

and Br

Br

Br

Br 1, 2, 3-tribromobenzene

1, 2, 4-tribromobenzene

1, 3, 5-tribromobenzene

(d) In p-nitrotoluene, both group moments act in the same direction while in p-chloronitrobenzene they act in the opposite directions:

CH3

ΝΟ2

Cl

ΝΟ2

28. (a) Amino group has priority over nitro; the correct name is 4-nitroaniline. (b) Substituents should be listed in alphabetical order. The correct name is 1-butyl-4-ethylbenzene. (c) The removal of H+ from cyclopentadiene gives the resonance stabilised aromatic cyclopentadienyl anion. This is not so in case of 1,3-cyclohexadiene. (d) The two phenyl groups are not in the same plane. They are joined by a single bond and have free rotation about it. 29. (a) The product is 1, 4-naphthaquinone. (b) The product is phthalic acid. (c) Two products namely p-hydroxytoluene and m-hydroxytoluene are obtained. This reaction proceeds through the formation of benzyne and is known as elimination-addition reaction

Hydrocarbons

CH3

CH3

CH3

OH

CH3 OH

22.57

CH3 + OH

Cl

Cl

OH

(d) There is no ortho-H, thus benzyne cannot form. 30. (a) The intermediate is C6H5CH(OCOCH3)2, benzylidene acetate (c) Cyclopentadienyl anion has high stability

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly: (a) benzoyl chloride (b) benzyl chloride (c) o-and p-chlorotoluene

(d) m-chlorotoluene [2007]

2. Presence of a nitro group in a benzene ring (a) activates the ring towards electrophilic substitution (b) renders the ring basic. (c) deactivates the ring towards nucleophilic substitution (d) deactivates the ring towards electrophilic substitution [2007] 3. The electrophile E+ attacks the benzene ring to generate the intermediate s – complex. Of the following, which s – complex is of lowest energy? NO2 NO2 NO2 H H (b) (c) (d) (a) + + E H + + E E [2008] H E 4. Toluene is nitrated and the resulting product is reduced with Sn and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains (a) mixture of o- and m- bromotoluenes (b) mixture of o- and p- bromotoluenes (c) mixture of o- and p- dibromobenzenes (d) mixture of o- and p- bromoanilines. [2008] 5. The nonaromatic compound along the following is (b)

(a)

(c)

S

(d)

6. Which compound exhibits maximum dipole moment among the following? NO2 (a) (b) (c) NO2 NO2

[2011] –

(d)

[2015, online]

NO2 NH2

NH2

NH2 7. In the following sequence of reaction: Toluene

The product C is (a) C6H5COOH

KMnO4

A

SOCl2

(b) C6H5CH3

B

H2/Pd BaSO4

C

(c) C6H5CH2OH

(d) C6H5CHO

[2015]

22.58 Complete Chemistry—JEE Main

ANSWERS 1. (c) 7. (d)

2. (d)

3. (c)

4. (b)

5. (a)

6. (b)

HINTS AND SOLUTIONS 1. Toluene undergoes electrophilic substitution reaction in ortho and para positions, since the methyl group attached to benzene, being ring activator, is ortho and para directing group. 2. Nitro group is a electron-withdrawing group, thereby it deactives the ring towards electrophilic substitution of benzene ring. 3. The nitro group is electron withdrawal group. This decreases the tendency of electrophilic substitution reaction. The intermediate s – complex of nitrobenzene has larger energy as compared to that of benzene. 4. The given reactions are as follows. CH3

CH3

NO2

1. HNO3 2. H2SO4

CH3

CH3 Sn/HCl

+ NO2

CH3 Br

NH2

1. Na NO2 / HCl (0 – 5°C)

+

CH3

CH3 +

2. CuBr

NH2

Br

5. An aromatic compound has (4n + 2) p-electons, where n is an integer. Cyclopentadiene has only 4p electrons and thus will be nonaromatic. NO2 NO2 NO2 NO2 6. NH2

NH2

NH2

The group moments of NO2 and NH2 act in the same direction, their vector addition will give maximum value. 7. The reaction are: CH3

COOH KMnO4

SECTION 5

COCl SOCl2

CHO H2/Pd BaSO4

Sources of Hydrocarbons

Coal and petroleum are the two main natural sources of hydrocarbons. The main source of aromatic hydrocarbons is coal while that of aliphatic hydrocarbons is petroleum.

Hydrocarbons

22.59

The fractional distillation of petroleum provides different fractions in different temperature ranges containing different chain lengths of carbon atoms. Table 1 describes the composition and uses of some of the fractions. Table 1 Fraction

Composition and uses of some of the fractions of petroleum Boiling point range T/K

Composition

Uses

Gaseous hydrocarbons

110–300

C1 to C5

As a fuel and in the manufacture of hydrogen and ammonia

Petroleum ether or naphtha

300–360

C5 to C7

As a solvent and in drycleaning

Gasoline (petrol)

340–470

C1 to C12

As fuel in motor-cars, etc.

Kerosene

450–550

C12 to C15

As fuel and for illumination

Diesel oil (gas oil or fuel oil)

520–670

C15 to C18

As fuel in trucks, buses, diesel engines, etc. fuel in electricity generation.

Lubricating oils, petroleum jelly

above 520

C16 and more

For lubricating of machines

m. pt 325–330

C20 and more

greases,

canvas etc. Petroleum coke (residue)

m.pt above 330

As fuel for making electrodes and road-building.

Petrol is the most important fraction of petroleum. It may also be produced by cracking process in which larger molecules are broken into smaller ones. This cracking may be carried out thermally or catalytically. For example, heating lubricating oil or diesel oil at a high temperature produces petrol of low octane rating. In catalytic cracking, silica and alumina are used. This cracking produces petrol of higher octane rating. Coal on pyrolysis (heating to a high temperature in the absence of air) produces coke. The volatile matter in coal volatilizes. It is passed through sulphuric acid to absorb basic compounds such as ammonia. The solution of the salt petrol. A foul-smelling, black, sticky liquid settles down at the bottom, and is known as coal tar. The latter on distillation produces aromatic hydrocarbons. The major fraction of coal tar distillation is shown in Table 2. Table 2 Name of fraction Crude light oil

Fractions of coal tar

Range of boiling point T/K

Approximate percentage

up to 440

2.2

Main Compounds benzene, toluene, xylene

Middle oil or carbolic oil

440–550

7.5

phenol

Heavy oil or creosote oil

500–540

16.5

cresols

Green oil or anthracene oil

540–630

20.0

naphthalene, anthracene, phenanthrene

Pitch

residue

56.0

carbon and other compounds

Octane Rating The quality of petrol is judged from its knocking behaviour in an engine. It is compared with two standards, namely, n-heptane and isooctane. The former burns explosively and has been given an octane rating of zero. Isooctane (2, 2, the mixture with n-heptane which will have the same performance as the sample of petrol.

Cetane Number In diesel engine, the straight-chain hydrocarbons generates more power than branched-chain ones. Cetane which is n-hexadecane ignites spontaneously and has been given a rating of 100. Zero rating is given to a-methylnaphthalene which ignites slowly in the diesel engines. The cetane number of diesel oil is the percentage of cetane in the mixture of cetane and a-methylnaphthalene which has the same ignition qualities as the sample of the diesel.

22.60 Complete Chemistry—JEE Main

Gasoline Additive The octane rating of a fuel is improved by adding tetraethyl lead to it. To avoid the more wear and tear of cylinder and piston of the engine, ethylene bromide is also added to the petrol.

MULTIPLE CHOICE QUESTIONS ON SECTION 5 1. The constituent obtained in the temperature range 340 K to 470 K during the fractional distillation of petroleum is (a) gaseous hydrocarbons (b) petroleum ether (c) gasoline (d) kerosene 2. The constituent obtained in the temperature range 450 K to 550 K during the fractional distillation of petroleum is (a) gaseous hydrocarbons (b) petroleum ether (c) gasoline (d) kerosene 3. The constituent obtained in the temperature range 520 to 670 K during the fractional distillation of petroleum is (a) gasoline (b) kerosene (c) diesel oil (d) lubricating oil 4. The range of number of carbon atoms in the gasoline fraction is (b) C7 to C12 (c) C12 to C15 (d) C15 to C18 (a) C1 to C7 5. The range of number of carbon atoms in the kerosene fraction is (b) C7 to C12 (c) C12 to C15 (d) C15 to C18 (a) C1 to C7 6. The range of number of carbon atoms in the diesel oil is (b) C7 to C12 (c) C12 to C15 (d) C15 to C18 (a) C1 to C7 7. The range of number of carbon atoms in the lubricating oils is (b) C12 to C15 (c) C15 to C18 (d) C18 and more (a) C7 to C12 8. Octane number is zero for (a) n-heptane (b) 2, 3, 3-trimethylpentane (c) n-octane (d) 2, 2, 4-trimethylpentane 9. Octane number is 100 for (a) n-pentane (b) 2, 3, 3-trimethylpentane (c) n-octane (d) 2, 2, 4-trimethylpentane 10. Which of the following is not true? (a) Petrol containing a larger amount of straight-chain hydrocarbons produces less knocking (b) Petrol containing a larger amount of branched-chain hydrocarbons produces less knocking (c) Octane rating is zero for n-heptane (d) Octane rating is 100 for isooctane 11. Which of the following is used as gasoline additive? (a) n-Heptane (b) Isooctane (c) Tetraethyl lead (d) Diethyl lead 12. Dry distillation of coal gives many valuable and important products. The solid residue left behind after dry distillation of coal is called (a) asphalt (b) coal pitch (c) coal tar (d) coke 13. Which of the following statements about petroleum is not correct? (a) Petroleum is a dark coloured oily liquid with a characteristic smell (b) Petroleum is heavier than water (c) Petroleum is insoluble in water (d) Petroleum is formed by decomposition of animal and plant remains inside the earth

Hydrocarbons

14. Which of the following gaseous fuels is obtained by dry distillation of coal? (a) Coal gas (b) Oil gas (c) Producer gas

(d) Water gas

ANSWERS 1. (c) 7. (d) 13. (b)

2. (d) 8. (a) 14. (a)

3. (c) 9. (d)

4. (b) 10. (a)

5. (c) 11. (c)

6. (d) 12. (d)

22.61

23

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes) SECTION 1

Haloalkanes

Methods of Preparation Replacement of Hydroxyl Group in Alcohols The reagents used are: For primary and secondary alcohol—hydrogen chloride in the presence of zinc chloride. For tertiary alcohol—concentrated hydrochloric acid in the presence of zinc chloride. The reaction proceeds as follows. SN1 mechanism; ROH + ZnCl2 Æ R

O

ZnCl2 Æ R+ + [HO—ZnCl2]– HClÆ RCl + H2O + ZnCl2

H SN2 mechanism; ROH + ZnCl2 + HCl Æ ZnCl3– + ROH +2 Æ ZnCl2 + RCl + H2O Tertiary alcohol follows SN1 mechanism. Straight chain primary alcohols favour SN2 mechanism. Secondary alcohol may react via both the mechanisms. For alkyl bromides, constant boiling hydrobromic acid (48%) is used. For primary alcohols, a little concentrated sulphuric acid is used as a catalyst. For secondary and tertiary alcohols, sulphuric acid is not used as these alcohols tend to dehydrate to give alkenes. HBr can be generated in situ by the action of concentrated sulphuric acid on potassium bromide. For alkyl iodides, constant boiling HI(57%) is used. It may be generated in situ by the action of 95% phosphoric acid on KI. The convenient reagents to replace — OH by —Cl are phosphorus chloride and thionyl chloride (SOCl2). The latter is preferred as the end products are gaseous products which escape in air giving almost pure alkyl chloride.

Addition of Halogen Acids to an Alkene The addition occurs in accordance with Markovnikov’s rule. HBr in the presence of peroxide adds according to antiMarkovnikov’s rule.

Physical Properties The C —X bond in alkyl halides is a polar bond. The dipole moment of methyl halides follows the order: CH3Cl > CH3F > CH3Br > CH3I The bond strength of C—X decreases with increasing size of halogen atom, i.e.

23.2 Complete Chemistry—JEE Main

e(C —F) > e(C—Cl) > e(C—Br) > e(C—I) The boiling points of alkyl halides are higher than those of the corresponding alkanes. The boiling point of alkyl halides increases with increasing size of the halogen atom. Alkyl chlorides are lighter than water while alkyl bromides and iodides are heavier than water.

Chemical Reactions The alkyl halides undergo nucleophilic substitution reactions in which a stronger necleophile replaces a weaker nucleophile. Some alkyl halide can undergo elimination reactions producing an alkene. Nucleophilic substitution reactions A few typical nucleophilic substitution reactions are as follows. 1. CH 3— Cl +

OH –

D

from aq KOH or Ag 2O in boiling water

Æ CH3OH + Cl–

2. CH3CH2—I + CH3O– Æ CH3CH2OCH3 + I– 3. CH3CH -- CH 2 I + CN Æ CH3CHCH 2 -- CN + IFrom alcoholic | | solution of CH3 KCN CH3 When silver cyanide is used, the product is isocyanide. CH3CHCH 2 -- I + AgCN Æ CH3CHCH 2 -- N ∫∫ C + Ag I | | CH3 CH3 4. CH3CH2I + :NH3 Æ CH3CH2NH+3 + I– 5. RX + KNO2 Æ R —O—N==O + RX alkyl nitrite

If silver nitrite is used, the product is mainly nitroalkane. O + AgX R — X + Ag—O—N==O Æ R N O The nucleophilic substitution reaction proceeds via either SN1 or SN2 mechanism.

SN1 Mechanism CH3

H3 C

... C H3 C .. CH3

Slow Br –Br

Fast

C+ H 3C

CH3

CH3

.. C Nu H3C ... CH3

Main characteristics 1. The more stable the carbocation intermediate, the faster the SN1 mechanism. 2. The nucleophilie plays no role in the mechanism. 3. The weaker bases are the better leaving groups and thus favour SN1 mechanism. 4. Polar solvents accelerate the SN1 reaction because it favours the formation of polar transition state. 5. The SN1 reaction on a chiral starting material ends up with the recemization of the product.

SN2 Mechanism

H

.

X

... .

H

C .....

H – Nu.....C.....X –

H

H

Nu

C

H .....

Nu–

H

H

H + X–

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes) 23.3

Main Characteristics 1. The conversion of reactants to transition state is the rate determining step. 2. The SN2 reaction involves the inversion of stereochemistry around carbon atom of the substrate. This inversion is known as Walden inversion. 3. The rate of reaction depends on the steric bulk of the alkyl group. The kinetic study have shown that the methyl halides are the most reactive in SN2 reactions. The increase in the length of chain of alkyl group decreases the rate of reaction. Alkyl branching next to the leaving group decreases the rate drastically. The reactivity order for SN2 reactions follows the following order. CH3 > 1° > 2° >> neopentyl > 3° 4. Aprotic solvents increase the rate of SN2 reactions. 5. The replacement of halide group follows the order

I– > Br– > Cl–.

Elimination Reaction In the presence of alcoholic KOH and heating, the elimination reaction occurs which results into a double bond. If more than one product is possible, the major product is of more substituted alkene (Saytzeff rule). CH3

CH2

CH

CH3

alcoholic KOH,

CH3CH CHCH3 + CH3 CH2CH 80%

Br

CH2

20%

Competition Between Substitution and Elimination Reactions The relative proportion of products depends on mainly three factors, namely, basicity of the nucleophile, hindrance in the haloalkane, and steric bulk around the nucleophilic atom. Factor 1. Base strength of the nucleophile Weak bases (H2O, ROH, halides, RS–, N–3, NC–, RCOO–) lead to more substitution. Strong bases (HO–, RO–, H2N–, R2N–) lead to more elimination. Factor 2. Steric hindrane around the reacting carbon Sterically unhindered (primary) haloalkanes lead to more substitution. Sterically hindered (branched primary, secondary, tertiary) haloalkanes lead to more elimination. Factor 3. Steric hindrane in the nucleophile Sterically unhindered (OH–, CH3O–, CH3CH2O–, H2N–) nucleophile lead to more substitution. Sterically hindered (CH3)3CO–, [(CH3)2CH2NH–] nucleophiles lead to more elimination. Formation Grignard Reagent Haloalkanes react with magnesium turnings in dry ether to form alkyl magnesium halide, known as Grignard reagent. The latter is most versatile compound as it can be used in the preparation of many different types of compounds. H2O or dilute acid dry oxygen HCHO

RX + Mg

ether

RMgX

R CHO

RH ROH ROH RCHOHR R R C(OH)

R COR Ethyl formate Cyanogen

R

RCHO RCN

Cyanides can also be prepared by treating alkyl halides with alcoholic solution of potassium cyanide. These, in turn, can be converted to amide, carboxylic acid and amine.

23.4

Complete Chemistry—JEE Main Conc. HCl

R -- C ∫∫ N + H 2 O æoræææææ Æ RCONH 2 H O -NaOH 2

2

Mineral acid

R -- C ∫∫ N + H 2 O æææææ Æ RCOOH + NH3 or alkali Na/C H OH

2 5 R -- C ∫∫ N + 4H æorææææææ Æ RCH 2 NH 2 H / Ni or LiA1H 2

4

MULTIPLE CHOICE QUESTION FROM AIEEE AND JEE MAIN 1. The major product formed when 1, 1, 1 - trichloro - propane is treated with aqueous potassium hydroxide is: (a) Propyne

(b) 1- Propanol

(c) 2- Propanol

(d) Propionic acid [2004]

2. Which of the following reagents is not suitable for the elimination reaction?

[2016, online]

Br

(a) NaI (b) NaOEt/EtOH (c) NaOH/H2O 3. 2-Chloro-2-methylpentane on reaction with sodium methoxide yields

(d) NaOH/H2O-EtOH

—

CH3

(a) (i) and (iii)

—

CH3

CH3

(b) (iii) only

(iii) C2H5CH — C—CH3 —

—

(i) C2H5CH2—C—OCH3 (ii) C2H5CH2—C — CH2

CH3

(c) (i) and (ii)

(d) all of these

[2016]

ANSWER 1. (d)

2. (a)

3. (d)

HINT AND SOLUTION -H O

2 1. CH3CH 2 CCl3 + 3KOH Æ [CH3CH 2 C (OH )3 ]+ 3KCl æææÆ CH3CH 2 COOH

Propionic acid

2. The reagent NaI cannot be used. 3. Strong base like CH3O– leads to more elimination. Sterically unhindred base like CH3O– leads to more substitution. Sterically hindrance around the reacting carbon leads to more elimination. In elimination, major product is of more substituted alkene (Saytzeff rule). In the given reaction CH3O– is unhindered whereas reacting carbon of 2-chloro-2-methylpentane is sterically hindered and thus all the three given compounds will be formed.

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes) 23.5

SECTION 2

Haloarenes

Methods of Preparation These can be prepared by Friedel-Crafts halogenation. Iodination cannot be carried out by this method as the reaction is endothermic. Aryl halides can also be prepared by Sandmeyer reaction. In this reaction, benzenediazonium chloride is treated with cuprous chloride or cuprous bromide dissolved in the corresponding acid. +

NCl–

N

Cl CuCl/HCl 60°C

+ N2

Iodobenzene is obtained by warming diazonium salt solution with aq. KI solution. +

N

N

Cl–

I + KI

+ N2 + KCl

warm

Fluorination is carried out by warming diazonium salt solution with HBF4. This reaction is known as Schiemann reaction. +

N

NCl–

+

N

NBF4–

HBF4

F + N2 + BF3

120°C

Chemical Reactions In general, aryl halides are less reactive than alkyl halides towards nucleophilic substitution reactions. This is due to the resonance effect in which lone pair of electron on halogen atom is delocalized to benzene ring imparting a partial double bond character to C—X bond. +

X

X

+

X

–

+

X

X –

–

In alkyl halide, the C—X bond involves sp3(C) whereas in aryl halide, sp2(C) is involved. Since the sp2(C) is more electronegative than the sp3(C), the C — X bond in aryl halide is shorter than in alkyl halides. This makes C—X bond more strong in aryl halides. Under normal conditions, halobenzenes are inert to nucleophiles. However, chlorobenzene can be made to react if the experimental conditions are drastic, i.e. high temperature and high pressure.

23.6

Complete Chemistry—JEE Main

O–Na+

aq. NaOH, 300 °C

Cl

H+

OH

NH2

aq. NH3, 200 °C Cu2O CuCN in

CN

Dimethylformamide

The presence of electron-withdrawing substituent at ortho and/or para positions is a favourable factor for the nucleophilic substitution reaction. More such substituents, the faster the reaction. Cl

Cl

OH

OH NO2

15% NaOH 160 °C

;

NO2

aq. NaOH 100 °C

NO2

NO2

NO2 Cl

NO2 OH

O 2N

NO2

O 2N

NO2

aq. NaOH Warm

NO2

NO2

The factors responsible for the nucleophilic substitution reaction is due to (1) decrease in electron density in the ring, thereby favouring the nucleophilic attack, and (2) the stabilisation of intermediate anion by resonance. No such resonance is possible if the electron-withdrawing substituent occupies meta position.

Wurtz-Fittig Reaction The combination of aryl halide with alkyl halide in an ethereal solution in the presence of sodium to give alkyl substituted benzene is known as Wurtz-Fittig reaction Cl + 2Na + ClCH3

Ethereal solution

CH3 + 2NaCl

Fittig Reaction Treatment of ethereal solution of bromobenzene with sodium gives biphenyl. Br + 2Na + Br

+ 2NaBr

Chloro substituent in chlorobenzene is ortho and para director, since resonance effect of Cl predominates over its inductive effect.

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes) 23.7

MULTIPLE CHOICE QUESTIONS FOR THE ENTIRE CHAPTER General Characteristics 1. Which of the following represents sec-butyl bromide? CH3 (a) CH3CH2CH2CH2Br (b) CHCH2Br CH3

CH3

(c) CH3

(a) CH3CH2CH2CH2Br

(b)

CH3

Br

(d)

CH3CH2CHCH3 Br

CH3

2. Which of the following represents tert-butyl bromide? CH3CHCH2Br

C

CH3

(c) CH3

C

Br

(d)

CH3CH2CHCH3 Br

CH3

3. Which of the following represents isobutyl bromide?

(a) CH3CH2CH2CH2Br

(b)

CH3 CHCH2Br CH3

CH3

(c) CH3

C

Br

(d) CH3CH2CHCH3

CH3

4. Which of the following order regarding bond enthalpy e(C—X) in an alkyl halide (RX) is correct? (a) e(C —Cl) < e(C—Br) < e(C— I) (b) e(C— Cl) > e(C—Br) > e(C— I) (c) e(C— Cl) > e(C—Br) < e(C— I) (d) e(C— Cl) < e(C—Br) > e(C— I) 5. The ease with which an alcohol can be halogenated by treating it with HX in the presence of anhydrous ZnCl2 is (a) (b) (c) (d)

tertiary alcohol < secondary alcohol < primary alcohol tertiary alcohol > secondary alcohol > primary alcohol tertiary alcohol > secondary alcohol < primary alcohol tertiary alcohol < secondary alcohol > primary alcohol

6. When toluene is treated with Cl2 at low temperature in the presence of a catalyst (FeCl3), the product obtained is

(a) only o-chlorotoluene (c) only p-chlorotoluene

(b) only m-chlorotoluene (d) a mixture of ortho-and para-chlorotoluenes

7. When toluene is treated with Cl2 at high temperature without the use of any catalyst, the product obtain is (a) o-chlorotoluene (b) m-chlorotoluene (c) a mixture of orhto- and para-chlorotoluenes (d) benzylchloride 8. For the same alkyl group, the order of boiling points of alkyl halides is (a) chloride < bromide < iodide (b)chloride > bromide > iodide (c) chloride < bromide > iodide (d)chloride > bromide < iodide

(a) CH4 (b) CH3Cl (c) CH2 = CHCl (d) CCl4 10. Chloroform is stored in dark coloured bottles because it is oxidized in the presence of light and air to a poisonous compound (a) CO (b) COCl2 (c) CO2 11. Which of the following is expected to have high density?

(d) CCl4

(a) CH3F (b) CH3Br 12. Which of the following is used as an antiseptic?

(c) CH3Cl

(d) CH3I

(c) CHI3

(d) CHCl3

(a) CH3I

(b) CH2I2

23.8

Complete Chemistry—JEE Main

13. Which of the following compounds represents gammexane? (a) Hexachlorobenzene (b) Benzene hexachloride (c) Dichlorobenzene (d) Trichlorobenzene 14. The dipole moment of CH3X (where X is a halogen) follows the order (a) CH3F > CH3Cl > CH3Br (b) CH3F > CH3Cl < CH3Br (d) CH3F < CH3Cl < CH3Br (c) CH3F < CH3Cl > CH3Br 15. Which of the following orders regarding the boiling points of the given alkyl chloride is correct? (a) CH3(CH2)3Cl > CH3CH2CHClCH3 > (CH3)3CCl (b) CH3(CH2)3Cl < CH3CH2CHClCH3 < (CH3)3CCl (d) CH3(CH3)3Cl < CH3CH2CHClCH3 < (CH3)3CCl (c) CH3(CH2)3Cl > CH3CH2CHClCH3 < (CH3)3CCl 16. The number of isomers of dibromobutane is (a) 4 (b) 5 (c) 6 (d) 8 17. Chlorination of benzene proceeds via (a) nucleophilic substitution mechanism (b) electrophilic substitution mechanism (c) elimination-addition mechanism (d) addition-elimination mechanism 18. Which of the following reactions is highly exothermic? (a) Fluorination of benzene (b) Chlorination of benzene (c) Bromination of benzene (d) Iodination of benzene 19. Which of the following reactions is an endothermic reaction? (a) Fluorination of benzene (b) Chlorination of benzene (c) Bromination of benzene (d) Iodination of benzene 20. Which of the following is an allylic halide? (b) CH2==CH—CH2X (c) CH2==CHX (d) C6H5Cl (a) CH3CH2X 21. Which of the following is a vinylic halide? (b) CH2==CH—CH2X (c) CH2==CHX (d) C6H5Cl (a) CH3CH2X 22. Carbylamine test is performed in alcoholic KOH by heating a mixture of (a) chloroform with silver powder (b) trihalogenated methane and a primary amine (c) an alkyl halide and primary amine (d) an alkyl cyanide and a primary amine 23. Haloform test is shown by the compound (a) methanol (b) methanal (c) 1-propanol (d) 2-propanol 24. Treating iodoform with silver powder gives (a) ethylene (b) acetylene (c) methane (d) diiodomethane 25. Amongst the following, a refrigerant is (b) CH2F2 (c) CCl2F2 (d) CCl4 (a) CHCl3 26. Heating bleaching powder with ethanol gives (a) acetone (b) methanol (c) chloroform (d) acetic acid

Nucleophilic Substitution Reactions 27. Which of the following order regarding SN2 reaction involving replacement of halogen atom in alkyl halide by a nucleophile is correct? (a) CH3 > 1° < 2° (b) CH3 < 1° > 2° (c) CH3 > 1° > 2° (d) CH3 < 1° < 2° 28. In the reaction anhydrous   ROH + HX    RX + H2O ZnCl 2

the order of reactivity of halogen acids is (a) HI > HBr > HCl (b) HI < HBr < HCl (c) HI > HBr < HCl (d) HI < HBr > HCl 29. Which of the following compounds gives formic acid when it is hydrolysed with boiling aqueous alkali?

(a) CH3Cl

(b) CH2Cl2

(c) CH3CH2Cl

(d) CHCl3

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes) 23.9

30. Which of the following is an example of Sandmeyer reaction? Cl

Cl

(a)

+ Cl2

FeCl 3

(b)

Æ

+ Cl2

low temperature

uv light

Æ

Cl

Cl

Cl

Cl

high temperature

Cl

CH3

N CH2Cl

(c)

+ Cl2

light, high temp. no catalyst

Æ

N

Cl

Cl CuCl, heat

(d)

Æ

+ N2

31. The major product with 2-boromobutane is treated with ethanolic KOH is (a) 2-hydroxybutane (b) butanol (c) 1-butene (d) 2-butene 32. Which of the following reagents is most effective in replacing —OH group of an alcohol by —Cl group? (a) HCl gas (b) Concentrated HCl (c) SOCl (d) SOCl2 33. The order of reactivity for the SN2 reaction involving replacement of — X by a nucleophile in alkyl halide is (a) CH3 < 1° < 2° (b) CH3 > 1° > 2° (c) CH3 < 1° > 2° (d) CH3 > 1° < 2° 34. Which of the following statements is correct? (a) Alkyl halides are more reactive than aryl halides towards nucleophilic substitution reactions (b) Alkyl halides are less reactive than aryl halides towards nucleophilic substitution reactions (c) The presence of an electron-withdrawing substituent at ortho and/or para position decreases the reactivity of nucleophilic substitution of chlorine in the substituted chlorobenzene (d) The replacement of chlorine in chlorobenzene by strong bases proceeds via elimination-addition reaction 35. The Wurtz-Fittig reactions involves (a) two molecules of aryl halides (b) two molecules of alkyl halides (c) one molecule of each of aryl halide and alkyl halide (d) one molecule of each of aryl halide and phenol 36. Which of the following compounds undergoes replacement of —Cl by —OH by merely warming the compound with aqueous NaOH? Cl Cl Cl Cl NO2 NO2 O2N NO2

(a)

(b)

(c)

(d)

NO2

NO2

37. Which of the following has the highest nucleophilicity? (b) OH– (c) CH–3 (d) NH–2 (a) F– 38. The order of reactivities of the following alkyl halides for a SN2 reaction is (a) RF > RCl > RBr > RI (b) RF > RBr > RCl > RI (c) RCl > RBr > RF > RI (d) RI > RBr > RCl > RF 39. Which of the following orders about nucleophilicity is correct? (b) MeO– < HO– < MeCOO– (a) MeO– > HO– > MeCOO– – – – (c) MeO > HO < MeCOO (d) MeO– < HO– > MeCOO– 40. Which of the following order about nucleophilicity is not correct? (b) I– > Cl– (c) HS– < HO– (d) HOO– > HO– (a) HO– > H2O

23.10

Complete Chemistry—JEE Main

41. Chlorine reacts with benzaldehyde to give (a) benzyl chloride (b) benzal chloride (c) benzoyl chloride (d) chlorobenzene 42. DDT is prepared from (b) C6H5Cl and chlorine (c) Chloral and C6H5Cl (d) Chloral and benzene (a) benzene and C6H5Cl 43. The correct sequence of the alkyl halides towards SN2 reactions is (b) CH3Br > CH3Cl < CH3F (a) CH3Br > CH3Cl > CH3F (d) CH3Br < CH3Cl < CH3F (c) CH3Br < CH3Cl > CH3F 44. Which of the following factors does not favour SN1 elimination in alkyl chloride? (b) Steric hindrance (c) Protic solvent (d) Aprotic solvent (a) Stability of R+ 45. Which of the following statements is not correct for SN1 reactions? (b) The nucliophile plays no role in the mechanism (c) The rate determining step is the conversion of transition state to the product. (d) The rate of reaction depends on the steric bulk of the alkyl group. 46. Which of the following statements is not correct for SN 2 reactions? (a) The conversion of reactant to transition state is the rate determining step. (b) The rate depends on the concentration and nature of nucleophile. (c) Protic solvents increase the rate of reaction. (d) The weaker the leaving base, faster the rate of replacement. 47. Which of the following factors does not favour elimination reaction than substitution reaction.... in an alkyl halide? (a) Low polarity of solvent (b) Strong base (c) High concentration of base (d) Sterically unhindered haloalkene.

Aryl Halides 48. The chlorine atom in chlorobenzene is ortho and para director because (a) resonance effect predominates over inductive effect (b) inductive effect predominates over resonance effect (c) both inductive and resonance effects are evenly matched (d) only resonance effect and not inductive effect is operating 49. Aryl halides are less reactive towards nucleophilic substitution reactions as compared to alkyl halides due to (a) the formation of less stable carbonium ion (b) resonance stabilization (c) longer carbon-halogen bond (d) inductive effect 50. Aryl halides are less reactive towards nucleophilic substitution as compared to alkyl halides. Which of the following factors is not responsible for this fact? (a) the formation of more stable carbonium ion (b) resonance stabilization (c) the inductive effect (d) sp2-hybridized carbon attached to the halogen 51. Which of the following reactions is highly exothermic? (a) Fluorniation of benzene (b) Chloroniation of benzene (c) Bromination of benzene (d) Iodination of benzene 52. In the reaction of p-chlorotoluene with KNH2 in liquor ammonia, the major product is (a) o-toluidine (b) m-toluidine (c) p-toluidine (d) p-chloroaniline 53. The chlorine atom in chlorobenzene is ortho and para director because (a) resonance effect predominates over inductive effect (b) inductive effect predominates over resonance effect (c) both inductive and resonance effects are evenly matched (d) only resonance effect and not inductive effect is operating

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes)

23.11

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49.

(b) (d) (b) (d) (c) (d) (c) (a) (b)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(c) (a) (c) (b) (c) (d) (d) (d) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51.

(b) (d) (a) (c) (c) (b) (a) (c) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52.

(b) (b) (c) (b) (a) (a) (c) (c) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53.

(b) (d) (b) (d) (d) (c) (c) (d) (a)

6. 12. 18. 24. 30. 36. 42. 48.

(d) (c) (a) (b) (d) (d) (c) (a)

HINTS AND SOLUTIONS 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

Bond enthalpy C—X (where X is Cl, Br, I) decreases with increase in atomic number of X. The decreasing order of reactivity is 3° > 2° > 1 > MeOH. At low temperature in the presence of a catalyst (FeCl3), electrophilic substitution at benzene ring takes place. At high temperature, substitution at methyl group takes place. The boiling point of alkyl halides increases with increasing size of halogen atom. CCl4 is known as pyrene. CHCl3 is oxidized to COCl2. Higher the size of halogen, higher the density. HCI3 is an antiseptic Benzene hexachloride is known as gammexane. The dipole moment order is CH3Cl >CH3F > CH3Br Boiling point decreases with increasing branching of the alkyl group. The skeletons of molecule are C

C

C

CBr ; C

C

C

C

Br Br

C ;

CBr ; C

Br Br

Br C

C

C

C

C

C Br

C

C Br ; C

C

C

CBr ;

Br

C

Br

17. 18. 19. 27. 28.

Chlorination of benzene proceeds via electrophilic substitution mechanism. Fluorination is highly exothermic reaction. Iodination is endothermic reaction. Replacement of halogen atom is alkyl halide by SN2 decreases with increase in steric hindrance. The stronger the acid, higher the reactivity.

29. 31. 32. 33. 34. 35.

H2O Æ HCOOH CHCl3 ææÆ CH(OH)3 æææ The major product is 2-butene (CH3CH==CHCH3). The most effective reagent is SOCl2. The rate decreases with increasing bulk of alkyl group. Alkyl halides are more reactive than aryl halides. Wurtz-Fitting reaction involves one molecule of each of aryl halide and aryl halide.

23.12

36. 37. 38. 39. 40. 44. 45.

47. 48. 49. 50. 52.

Complete Chemistry—JEE Main

The reactivity of Cl increases with increasing number of —NO2 group. In going from left to right across the periodic table, both basicity and nucleophilicity decrease. The reactivity increases with increasing size of halogen. When the nucleophilic and basic sites are the same atom, nucleophilicity parallels basicity. In going down a group in a period table, nucleophilicity increases. Stability of R+, steric hindrance and protic solvent favour SN1 reactions. (a) The rate expression is r = k [haloalkane] (b) Nucleophile is involved in the fast reaction. (c) Substrate Æ Transition state is the slow step. (d) Steric bulk favours SN1 mechanism. Æ Transition state is the slow step. (b) The reaction follows second-order kinetics; rate = k[haloalkane][nucleophile] (c) The weaker the base, faster the replacement. Low polarity of solvent, strong base and high concentration of base favour elimination reaction. Chlorine exhibits resonance effect more predominately than inductive effect. Aryl halides are less reactive due to resonance stabilization. The lesser reactivity of aryl halide is due to resonance stabilization and sp2-hybridized carbon attached to the halogen. The reaction proceeds via elimination-addition mechanism. The major product is m-toluidine.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The compound formed on heating chlorobenzene with chloral in the presence of concentrated sulphuric acid is (a) gammexene (b) DDT (c) freon (d) hexachloroethane [2004] 2. Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of (a) inductive effect (b) steric hindrance (c) insolubility (d) instability [2005] 3. Fluorobenzene (C6H5F) can be synthesized in the laboratory (a) by reacting bromobenzene with NaF solution (b) by heating phenol with HF and KF (c) from aniline by diazotisation followed by heating the diazonium salt with HBF4 [2006] 2 gas – – – 4. The decreasing order of the rate of the reaction CH3Br + Nu Æ CH3—Nu + Br with nucleophiles (Nu ) A to D [Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O–] is (a) B > D > C > A (b) D > C > A > B (c) D > C > B > A (d) A > B > C > D [2006] 5. The structure of the major product formed in the following reaction

CH2Cl NaCN DMF

I is

CH2CN (a)

(b)

CH2Cl

(c)

CH2Cl

(d)

CN

CN

I

CN

[2006]

Organic Compounds Containing Halogens (Haloalkanes and Haloarenes)

23.13

6. Which of the following is the correct order of decreasing SN2 reactivity? (b) RCH2X > R2CHX > R3CX (a) RCH2X > R3CX > R2CHX (d) R2CHX > R3CX > RCH2X (c) R3CX > R2CHX > RCH2X (X = a halogen) [2007] 7. Compound (A) C8H9Br, gives a white precipitate when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Indentify the compound (A).

[2013] 8. In SN2 reactions, the correct order of reactivity for the following compounds: CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is (a) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl (b) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl (c) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl (d) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl

[2014]

9. Chlorobenzne reacts with trichloro acetaldehyde in the presence of H2SO4 O H2SO4 Cl 2 + H C CCl3 The major product formed is: Cl

(a) Cl

C

Cl Cl

(b) Cl

C

Cl

(c) Cl

CH

Cl

CH2Cl

(d) Cl

Cl

CH

Cl

[2014, online]

Cl

CCl3

10. For the compounds CH3Cl, CH3Br, CH3I and CH3F, the correct order of increasing C- halogen bond length is: (b) CH3F < CH3Br < CH3Cl < CH3I (a) CH3F < CH3Cl < CH3Br < CH3I (d) CH3Cl < CH3Br < CH3F < CH3I (c) CH3F < CH3I < CH3Br < CH3Cl -

DMF

[2014, online]

-

11. In a nucleophilic substitution reaction: R - Br + Cl æææÆ R - Cl + Br , which one of the following undergoes (a) C6H5CHC6H5Br (c) C6H5CHCH3Br

(c) Finkelstein reaction

(b) C6H5CH2Br (d) C6H5CCH3C6H5Br

(d) Swarts reaction

[2014, online] [2015]

23.14 Complete Chemistry—JEE Main

ANSWERS 1. (b) 7. (d)

2. (b) 8. (b)

3. (c) 9. (c)

4. (b) 10. (a)

5. (a) 11. (b)

6. (b) 12. (d)

HINTS AND SOLUTIONS 1. The reaction is

2. The reactivity order for SN2 mechanism decreases with increase in the bulkiness of alkyl group. The order is CH3 > 1° > 2° >> neopentyl > 3° 4. The order acid-strength of the conjugate acids of the given bases is CH3OH < H2O < PhOH < AcOH The base-strength of the given bases will be CH3O– > OH– > PhO– > AcO– (D)

(C)

(A)

(B)

The stronger the base, stronger the nucleophile, greater the rate of nucleophilic displacement. 6. Bulkier the alkyl group, lesser the SN2 reactivity. 7. Aliphatic Br gives precipitate with AgNO3.The Oxidation of —CH2Br gives —COOH and also —CH3 gives —COOH. The two —COOH group must be in ortho positions to give anhydride. Hence,

8. Increasing bulkiness of alkyl group decreases the SN2 reaction. Thus, the correct choise is CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl 9.

10. As the size of halogen atom increases, the bond length also increases in the same order i.e. C - F < C - Cl < C - Br < C - I N2

CH3 > 1° > 2°

mechanism. The reactivity order is

neopentyl > 3°

C6H5CH2Br contains 1° C to which Br is attached and will undergo SN2 mechanism with complete inversion of For example, C2H5Cl + AgF Æ C2H5F + AgCl

3,

SbF3, AgF, Hg2F2, etc.

24 Organic Compounds

Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) SECTION 1

Alcohols

Methods of Preparation Oxymercuration-Demercuration Process Alkenes react with mercury acetate in the presence of water to give hydroxymercurial compounds which on reduction with sodium borohydride (NaBH4) yield alcohol. The net result is the addition of H2O to the double bond and the product obtained is in agreement with Markovnikov’s rule. For example, CH3CH

CH2 + H2O + Hg(OAc)2 Æ CH3

CH

CH2

NaBH4

CH

CH3

OH HgOAc

CH3

OH

No rearrangement occurs in this reaction.

Hydroboration-Oxidation Process Alkenes undergo hydroboration with diborane to yield alkylboranes which on oxidation with alkaline H2O2 give alcohols. The net result is the addition of H2O to the double bond and the product obtained is in accordance with the anti-Markovnikov’s rule. For example, CH3CH

CH2 + BH3 Æ CH3CH2CH2BH2

H2O2, OH–

CH3CH2CH2OH n-propyl alcohol

No rearrangement occurs during the hydroboration.

Grignard Synthesis of Alcohols Reaction of aldehyde or ketone with RMgX produces alcohol. H

H C H

O + R MgX Æ H

C R

H – +

OMgX

H2O

Æ

H

C

OH

R 1° alcohol

24.2 Complete Chemistry—JEE Main

H

H R¢

C

O + R MgX Æ R¢

H – +

H2O

Æ

OMgX

C

R¢

OH

C

R

R 2° alcohol

R≤

R≤ R¢

C

O + R MgX Æ R¢

R≤ – +

C

H2O

Æ

OMgX

R¢

C

R

OH

R 3° alcohol

Reaction of Ethylene Oxide with RMgX – +

H2O

CH2 + RMgX Æ RCH2CH2OMgX

CH2

Æ RCH2CH2OH

O

Reactions of Lithium Acetylides or Alkynyl Grignard Reagents with Aldehyde or Ketone R HC

CLi + O C

R R¢ Æ HC

C

R R¢

C

H+

Æ HC C C

OLi R HC

CMgBr + O

R¢

OH R

C R¢ Æ HC

C

C

R R¢

H+

Æ HC

C

C

OMgBr

R¢

OH

Reduction of Aldehydes and Ketones Sodium borohydride, NaBH4, is the convenient reagent to carry out the reduction of aldehydes or ketones into alcohol. RCHO

1.NaBH 4 in alcohol 2.H 3 O

Æ RCH2OH

LiAlH4 in ether can also be used for reduction. It is particularly useful for reducing a, b-unsaturated ketones. The reduction of such a ketone gives a mixture of both unsaturated and saturated alcohols. For example, HO

H

1. NaBH4, alcohol

O

HO +

2. H3O+ 2-cyclohexenol (59%)

HO

2-cyclohexenone

H

cyclohexanol (41%)

H

1. LiAlH4, ether

HO

H

+

2. H3O+ 94%

2%

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.3

Physical Properties Due to the hydrogen bonding, alcohols have higher boiling points than their structural isomers. The boiling point decreases with increase in branching, i.e. the order of boiling point is primary > secondary > tertiary.

Chemical Properties namely, R

OH bond cleavage and RO

H bond cleavage.

Reactions Exhibiting R···OH Bond Cleavage Reaction with Hydrogen Halides R—OH + HX Æ RX + H2O As such —OH is a poor leaving group. But its protonation converts into a good leaving group. There is formation of carbocation as the intermediate and thus the reaction may show rearrangement. The following are the reactivities of HX and ROH. HI > HBr > HCl;

allyl, benzyl > 3° > 2° > 1°

The reagents, used are concentrated HBr or NaBr + concentrated H2SO4, HCl + ZnCl2 and concentrated HCl.

Dehydration CH3 — CH2 — CH2— CH2OH

H

Æ CH3CH==CHCH 3 + CH3CH2CH ==CH2 2-butene (main product)

1-butene

The reaction may show rearrangement. Reactivity of ROH:

3° > 2° > 1°

The primary alcohol is dehydrated using concentrated H2SO4. The secondary and tertiary alcohols are dehydrated using dilue H2SO4 to avoid the polymerization of alkenes. Saytzeff’s rules is applicable so as to get more substituted alkene.

Reaction with Phosphorous Trihalides 3R— OH + PX3 Æ 3R— X + H3PO3 (PX3

PBr3, PI3)

Reactions Exhibiting RO···H Bond Cleavage Reaction with Active Metals RO— H + M Æ R— OM+ + (M=Na, K, Mg, Al, etc.)

1 H2 2

Reactivity of alcohol CH3OH > 1° > 2° > 3° The above reaction shows alcohol as an acid. It is worth comparing the acid strength of alcohol with other species. H2O > ROH > HC

CH > NH3 > RH

The relative order of basicity follows the reverse order, i.e. OH– < OR– < HC

C– < NH–2 < R–

24.4 Complete Chemistry—JEE Main

Reaction with Carboxylic Acid This results in the formation of an ester. O CH3

C

H+

OH + H OC2 H5

O CH3

C

OC2H5 + H2O

ethyl acetate

Instead of acetic acid, acetic anhydride or acetyl chloride can also be used.

Oxidation of Alcohols The oxidation of an alcohol involves the loss of one or more a-hydrogens. 1° alcohol is changed to an aldehyde by using the reagent pyridinium chlorochromate (C5H5NH+CrO3Cl–) formed by the reaction between chromic acid and pyridinium chloride. For example, H CH3CH2CH2OH

C 5 H 5 NH CrO 3 Cl

Æ CH3CH2 C

O

propanal

1-propanol

1° alcohol is directly converted into a carboxylic acid by the use of potassium permangnate. For example, O CH3CH2CH2OH

KMnO 4

1-propanol

Æ CH3CH2 C OH propanoic acid

2° alcohol is changed into a ketone by the use of potassium dichromate or CrO3 in glacial acetic acid or CrO3 in pyridine. For example, CH3CHOH CH3 isopropyl alcohol

K 2 Cr2 O 7

Æ CH3C

O

CH3 acetone

3° alcohol is not oxidizable as it does not contain a-hydrogen.

Characterization of Alcohols Oxidation Method

However, drastic oxidation will convert ketone (formed from 2° alcohol) and 3° alcohol into carboxylic acid containing fewer carbon atoms. When the vapours of alcohols are passed over hot metallic Cu at 570 K, limited oxidation takes place.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.5

Victor-Meyer Method The alcohols are treated with P4 and I2 to form an alkyliodide. This on treatment with AgNO2 yields nitroalkanes. On treating the latter with aqueous NaNO2 1° alcohol

P+12

CH3CH2OH

AgNO 2

Æ CH3CH2 I

Æ CH3CH2NO2

HNO2

Æ CH3

C

NO2

NOH

nitrolic acid (Blood red colour)

2° alcohol

P+12

(CH3)2CHOH

AgNO 2

Æ (CH3)2CHI

Æ (CH3)2CHNO2

HNO2

Æ (CH3)2CNO2 NO pseudo nitrol (Blue colour)

3° alcohol

P+12

(CH3)3 COH

AgNO 2

Æ (CH3)3CI

Æ (CH3)3CNO2

HNO2

Æ no reaction

Lucas Test Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride. It converts alcohol into chloride and the formation of the latter is indicated by the appearance of cloudiness which marks the separation of chloride from the solution. react appreciably at room temperature. Allyl alcohol reacts as rapidly as tertiary alcohols but it remains in the solution.

Characteristic Test of CH3CO— Group OH is oxidized to R

C CH3 which gives iodoform test. || O

CH3

|

CH

|

An alcohol of the type R

The reagent used is iodine and sodium hydroxide (sodium hypoiodite, NaOI). The reactions involved are H R

C

CH3 + NaOI

R

C

CH3 + 3NaOI

O R

CH3 + NaI + H2O

O

OH R

C

R

C

CI3 + 3NaOH

O

C

CI3 + NaOH

RCOO–Na+ + CHI3 Iodoform (Yellow ppt.)

O

Analysis of Molecules Containing —OH or ==O Group Attached to Adjacent Carbon Atoms Molecules containing —OH or ==O groups attached to adjacent carbon atoms undergo oxidation with cleavage of carbon-carbon bonds when treated with periodic acid. For example, RCH

CH

OH OH

R¢

HIO4

Æ RCH

OH + R¢CH

OH

OH

OH Æ RCHO + R¢COH

24.6 Complete Chemistry—JEE Main

RCH

C

R¢

HIO4

Æ RCH OH

OH O R

R

CH R¢

CH

CH

OH

OH OH

C

C

O

O

OH + HO

R¢

CR¢ Æ RCHO + R¢COOH O

2HIO4

Æ

RCHO + HCOOH + R¢CHO

HIO4

Æ RCOOH + R¢COOH

The amount of HIO4 consumed is equal to the amount of carbon-carbon bond broken in the molecule.

MULTIPLE CHOICE QUESTIONS ON SECTION 1 Preparation and Physical Properties 1. Which of the following is not characteristic of alcohols? (a) The boiling point increases with increasing molecular mass (b) They are lighter than water (c) Lower members have a pleasant smell and burning taste, higher members are odourless and tasteless (d) Lower members are insoluble in water and organic solvents but the solubility regularly increases with molecular mass 2. Oxymercuration-demercuration of CH3CH == CH2 produces (b) CH3CH(OH)CH3 (c) CH3CH(OH)CH2OH (d) CH3COCH3 (a) CH3CH2CH2OH 3. Hydroboration-oxidation of CH3CH = CH2 produces (b) CH3CH(OH)CH3 (c) CH3CH(OH)CH2OH (d) CH3COCH3 (a) CH3CH2CH2OH 4. Which of the following order regarding acidity of the following compounds is correct? (b) ROH < HC ∫∫ CH < NH3 (a) ROH > HC ∫∫ CH > NH3 (c) ROH > HC ∫∫ CH < NH3 (d) ROH < HC ∫∫ CH > NH3 5. The reaction of CH3CHO with RMgX produces (a) 1° alcohol (b) 2° alcohol (c) 3° alcohol (d) a carboxylic acid 6. The reaction CH3COCH3 with RMgX produces (a) 1° alcohol (b) 2° alcohol (c) 3° alcohol (d) a carboxylic acid 7. An industrial method of preparation of methanol is (a) catalytic reduction of carbon monoxide in presence of ZnO-Cr2O3. (b) by reacting methane with steam at 900 °C with a nickel catalyst (c) by reducing formaldehyde with lithium aluminium hydride (d) by reacting formaldehyde with aqueous sodium hydroxide solution 8. Hydrogen bonding is maximum in (a) ethanol (b) diethylether (c) ethyl chloride (d) triethylamine 9. Which of the following alcohols is expected to have minimum boiling point? (a) 1-Butanol (b) 2-Butanol (c) 2-Methyl-2-propanol (d) 1-Pentanol 10. Glycerine is a/an (a) secondary alcohol (b) tertiary alcohol (c) ester (d) trihydric alcohol 11. Power alcohol is (a) an alcohol of 95% purity (b) a mixture of petrol, ethanol and benzene

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.7

12. The enzyme which converts glucose to ethanol is (a) invertase (b) zymase (c) maltase (d) diastase 13. Which of the following statements is correct? (a) The branched isomer of an alcohol has lower boiling point than the unbranched alcohol. (b) Ethylene glycol boils at a temperature lower than that of ethanol. (c) The hydroboration-oxidation process gives product corresponding to Markovnikov addition of water to the carbon-carbon double bond. (d) The oxymercuration-demercuration process gives products corresponding to anti-Markovnikov addition of water to the carbon-carbon double bond. 14. Which of the following statements is not correct? (a) The addition of water to the carbon-carbon double bond via hydroboration-oxidation process does not involve any rearrangement of carbon skeleton. (b) The rearrangement of carbon skeleton may occur during the conversion of alcohol into alkene. (c) The rearrangement of carbon skeleton may occur during the conversion of alcohol into alkyl halide. (d) The cleavage of carbon-oxygen bond in alcohols is not catalyzed in the presence of an acid. 15. Which of the following statements is not correct? (a) The substitution of hydroxyl group by a halogen group in alcohol is an electrophilic substitution reaction. (b) Alcohols are weak acids as well as weak bases. (c) A secondary alcohol on oxidation gives a carboxylic acid containing the lesser number of carbon atoms. (d) A primary alcohol on oxidation gives a carboxylic acid containing the same number of carbon atoms. 16. Which of the following statements is correct? (a) Absolute alcohol can be obtained by distillation of ethanol and water mixture. (b) Cyclohexanol is more soluble in water than 1-hexanol. (c) The hydration of 3-phenyl-1-butene in dilute H2SO4 produces 3-phenyl-2-butanol. (d) The hydration of cyclobutylethene in dilute H2SO4 gives 1-cyclobutylethanol.

Cleavage of R...OH Bond 17. The ease of dehydration of alcohols is (a) tertiary > secondary > primary (b) tertiary < secondary < primary (c) tertiary > secondary < primary (d) tertiary < secondary > primary 18. HBr reacts faster with (a) 2-methylpropan-2-ol (b) propan-1-ol (c) propan-2-ol (d) 2-methylpropan-1-ol 19. Dehydration of 2-butanol gives (a) but-1-ene (b) but-2-ene (c) propene (d) 2-methylbut-2-ene 20. The order of reactivity of dehydration of alcohol is (a) 1° > 2° > 3° (b) 1° > 2° < 3° (c) 1° < 2° > 3° (d) 1° < 2° < 3° 21. The order of reactivity of hydrogen halides with an alcohol is (a) HCl > HBr > HI (b) HCl < HBr < HI (c) HCl > HBr < HI (d) HCl < HBr > HI 22. The iodoform test is not shown by the compound (b) CH3COCH2CH3 (c) CH3CH2COCH2CH3 (d) (CH3)2CHOH (a) CH3COCH3 23. The compound which gives the most stable carbonium ion on dehydration is

(a) CH3 CHCH 2 OH | CH 3

CH3 | (b) CH3 — C— OH | CH3

(c) CH3CH2CH2CH2OH

(d) CH3CHCH 2 CH3 | OH

24.8 Complete Chemistry—JEE Main

24. The order of reactivity of alcohols towards hydrogen halide is (a) benzyl > 3° > 2° > 1° (b) benzyl < 3° < 2° < 1° (c) 3° > 2° > 1° > benzyl (d) 3° > 2° > benzyl > 1° 25. The dehydration of 1-butanol gives (a) 1-butene as the main product (b) 2-butene as the main product (c) equal amounts of 1-butene and 2-butene (d) 2-methylpropene 26. Reaction of tertiary butyl alcohol with hot Cu at 350°C produces (a) butanol (b) butanal (c) 2-butene (d) 2-methylpropene 27. The reaction of ROH with R¢MgX produces (a) RH (b) R¢H (c) R — R (d) R¢ — R¢ 28. The product obtained when a limited amount of HI reacts with glycerol is (a) 1, 2, 3-Triiodopropane (b) Allyl iodide (c) 1, 2-Diiodopropane (d) Isopropyl iodide 29. The product obtained when excess of HI reacts with glycerol is (a) 1, 2, 3-Triiodopropane (b) Allyl iodide (c) 1, 2-Diiodopropane (d) Isopropyl iodide 30. HBr reacts fastest with (a) 2-methylbutan-2-ol (b) butan-1-ol (c) propan-2-ol (d) 2-methylpropan-1-ol

Cleavage of RO...H Bond 31. The order of reactivity of alcohols towards Na or K metal is (a) primary > secondar > tertiary (b) primary < secondary < tertiary (c) primary < secondary > tertiary (d) primary > secondary < tertiary 32. Which of the following compounds is oxidised to prepare methyl ethyl ketone? (a) 2-Propanol (b) 1-Butanol (c) 2-Butanol (d) tert-Butyl alcohol 33. The relative order of acidity of alcohols in comparison to H2O and HC∫∫CH is (b) H2O > HC∫∫CH > ROH (a) H2O > ROH > HC∫∫CH (d) ROH > HCOOH > H2O (c) ROH > H2O > HC∫∫CH 34. The relative order of basicity of conjugate bases is (b) OH– < HC∫∫C– < OR– (a) OH– < OR– < HC∫∫C– – – – (c) HC∫∫C < OH < OR (d) HC∫∫C– < OR– < OH– 35. When vapours of 2-propanol is passed over hot metallic Cu at 570 K, the product formed is (a) 1-propanol (b) propanone (c) propanal (d) propene 36. Among the following compounds, the strongest acid is (c) C2H6 (d) CH3OH (a) HC∫∫CH (b) C6H6 37. A mixture of methanol vapour and air is passed over heated copper. The products are (b) HCHO + H2 (c) HCOOH (d) CO + H2O (a) CO + H2O (a) CH3OH > CH3CH2OH > (CH3)2CHOH (c) CH3CH2OH > (CH3)2CHOH > CH3OH 39. Which of the following statements is correct?

(b) CH3OH > (CH3)2CHOH > CH3CH2OH (d) (CH3)2CHOH > CH3CH2OH > CH3OH

–  (a) The reaction ROH + OH–    RO + HOH lies far towards the left side.

(b) The reaction ROH + R¢S–Na+ æÆ RO– Na+ + R¢SH is feasible. (c) RS–, in a protic solvent, acts as a stronger nucleophile than RO–. (d) The bond angle R—O—H in methanol is smaller than that of R—S—H in methanethiol.

Characteristic Tests 40. In the Victor-Meyer test, blue colouration is shown by (a) primary alcohol (b) secondary alcohol

(c) tertiary alcohol

(d) diol

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.9

41. Which of the following alcohols is most reactive with Lucas reagent? (a) Methanol (b) Ethanol (c) Isopropyl alcohol (d) tert-Butyl alcohol 42. Lucas reagent is (b) anhydrous ZnCl2 with conc. H2SO4 (a) anhydrous AlCl3 with conc. HCl (c) anhydrous ZnCl2 with conc. HCl (d) anhydrous CaCl2 with conc. HCl 43. The compound that does not react with Lucas reagent is (a) n–butanol (b) sec-butyl alcohol (c) isobutyl alcohol (d) tert-butyl alcohol 44. The treatment of a compound with HIO4 produces CH3CHO and CH3CH2CHO. The probable compound is (b) CH3CH(OH)CH(OH)CH2CH3 (a) CH3CH == CHCH2CH3 (c) CH3CH == CHCH = CH2 (d) CH3CH(OH) CH2CHO 45. A molecule consumes 2 molecules of HIO4. The number of carbon-carbon bond broken by this acid is (a) one (b) two (c) three (d) none 46. A primary alcohol can be oxidized to an aldehyde by choosing the reagent (b) K2Cr2O7 (c) C5H5 NH+ CrO3 Cl– (d) O3 (a) KMnO4 47. The compound which reacts fastest with Lucas reagent at room temperature is (a) 1-butanol (b) 2-butanol (c) 2-methylpropanol (d) 2-methylpropan-2-ol 48. In the Victor-Meyer test, red colouration is shown by (a) 1° alcohol (b) 2° alcohol (c) 3° alcohol (d) phenol 49. In the Lucas test of alcohols, appearance of cloudiness is due to the formation of (a) aldehydes (b) ketones (c) acid chlorides (d) alkyl chlorides 50. An alcohol on oxidation gives CH3COOH and CH3CH2COOH. The alcohol is (b) CH3CH(OH)CH2CH3 (a) CH3CH2CH2OH (c) (CH3)2C(OH)CH2CH3 (d) CH3CH(OH)CH2CH2CH3 51. Which of the following statements is correct? (a) Tertiary butyl alcohol gives positive iodoform test. H (b) CH3CH2

C

CH2CH3 gives positive iodoform test.

OH

(c) The carbon-carbon bond in R are two aldehydes.

(d) The carbon-carbon bond in R

H

H

C

C

R can be broken by the use of periodic acid and the product obtained

OH OH R

H

C

C

R can be broken by the use of periodic acid giving two aldehydes.

OH OH

52. Which of the following statements is not correct? (a) The molecule RCHCH2CHR is cleaved by HIO4 giving RCHO and R¢CHO. OH OH (b) Tertiary alcohols are more readily dehydrated than the secondary alcohols. (c) Tertiary butyl alcohol when passed over hot metallic Cu at 570 K produces isobutene. (d) Primary alcohols do not show positive Lucas test. 53. Which of the following statements is correct? (a) Tertiary alcohols show positive Lucas test with faster speed than in the case of secondary alcohols. (b) The order of increasing acidity amongst 1°, 2° and 3° alcohols is 1° alcohol < 2° alcohol < 3° alcohol

24.10 Complete Chemistry—JEE Main

(c) The reaction of glycerol with small amount of HI produces 2–iodopropane. (d) The reaction of glycerol with excess of HI produces 1, 2, 3-triiodopropane. 54. Which of the following statements is not correct? (a) An organic compound on treating with HIO4 gives CH3COCH3 and HCHO. The compound is CH3 CH3

C

CH2

OH OH

(b) An organic compound on treating with HIO4 gives 5HCOOH and one HCHO. The compound is H2 C(CHOH)4CH2 . OH

OH

(c) Thiols are less soluble in water as compared to the corresponding alcohols. (d) Thiol CH3SH is more acidic than CH3OH. 55. Which of the following statements is correct? (a) The reduction of p-O2NC6H4CH2COOH with LiAlH4 gives p-O2NC6H5CH2CH2OH. (b) The reaction of Ph2C==CHCH3 with BH3 in tetrahydrofuran followed by H2O2/OH– gives Ph2C(OH)CH2CH3. (c) Cyclopentylmethylcarbinol does not give iodoform test. (d) An alcohol containing —CH(OH)CH3 gives iodoform test.

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (a) (a) (b) (b) (a) (b) (a) (d) (d)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(b) (a) (d) (d) (d) (c) (a) (b) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51.

(a) (c) (a) (b) (b) (a) (c) (b) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52.

(a) (d) (b) (c) (b) (a) (b) (c) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53.

(b) (b) (a) (b) (d) (b) (d) (d) (a)

6. 12. 18. 24. 30. 36. 42. 48. 54.

(c) (b) (a) (a) (a) (d) (c) (a) (b)

HINTS AND SOLUTIONS 1. 2. 3. 4. 5. 6. 8. 9. 10. 13.

Reverse of the choice d is true. Addition follows Markovnikov’s rule. Addition follows anti-Markovnikov’s rule. Alcohol is more acidic than acetylene. The latter is more acidic than NH3. CH3CHO will produce 2° alcohol with RMgX. CH3COCH3 produces 3° alcohol with RMgX. Compound containing —OH group will show maximum hydrogen bonding. Increasing branching causes lowering of boiling point. Glycerin is a trihydric alcohol (CH2OHCHOHCH2OH). (b) Ethylene glycol (HOCH2CH2OH) has two centres of hydrogen bonding while ethanol has only one.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.11

(c) Anti-Markovnikov addition of water takes place. (d) Markovnikov addition of water takes place. 14. (b) This is because of the formation of carbocation as the intermediate. (c) This is because of the formation of carbocation as the intermediate. (d) In the presence of an acid, alcohol gives protonated alcohol. the latter readily loses the weakly basic water molecule leaving behind the carbocation. The unprotonated alcohol would have to lose the strongly basic 15. (a) Replacement of —OH by a halogen in an alcohol is nucleophilic substitution reaction. It is the protonated alcohol which acts as a substrate. (b) Alcohols are acidic enough to react with active metals to liberate hydrogen gas. They are basic enough to accept a proton from strong acids. (c) It gives a ketone containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing lesser number of carbon atoms. 16. (a) Ethanol forms a lower boiling azeotropic mixture containing 95.6 and 4.4 by volume of alcohol and water, respectively. Water can be removed by adding Mg metal to azeotropic mixture which reacts with water to form Mg(OH)2 and releasing H2. (b) The OH group of cyclohexanol is more exposed and available for H-bonding with water due to the more compactness of the remaining R group. (c) The product is 2-phenyl-2-butanol.

(Note: Although phenyl is a better migrator than H, yet the hydride migrates so as to give a more stable 3° benzylic carbocation.) (d) The product is cis- and trans-2-methylcyclopentanol. In this reaction, addition of H+ to C1 gives a 2° carbocation which rearranges by ring expansion to give substituted cyclopentanol. H

CH == CH2

H+

CH

17. 18. 19. 20. 21. 22.

CH3

CH

H CH H Æ 2O

OH CH3 H

The ease of dehydration is tertiary > secondary > primary. Tertiary alcohol reacts faster with HBr. But-2-ene is more stable than but-1-ene (Saytzeff’s rule). The order of reactivity is 1° < 2° < 3°. HCl < HBr < HI. The order of reactivity is The iodoform test will be shown by a molecule containing —COCH3 group or by a molecule producing this group on oxidation. The compound CH3CH2COCH2CH3 will not show iodoform test. 23. 3° carbonium ion is most stable. 24. The order of reactivity of HX with ROH is benzyl > 3° > 2° > 1° 25. The reaction occur with rearrangement as 2° carbocation is more stable than 1° carbocation. CH3 26. CH3

C

OH

CH3

Cu 350°C

Æ CH3

C CH3

CH2

24.12 Complete Chemistry—JEE Main

27. ROH + R¢MgX æÆ ROMgX + R¢H 28.

29.

31. Reactivity of alcohol towards Na is 32. CH3CHCH2CH3

MeOH > 1° > 2° > 3°.

[O]

Æ CH3COCH2CH3 .

OH 33. The relative order of acidity is H2O > ROH > HC CH. 34. The relative order of basicity of conjugate base is opposite to that of acid.

35. CH3CHOHCH3

Cu

Æ CH3COCH3.

570 K

36. The strongest acid amongst the given compounds is alcohol. 39. (a) The differences in acidities and basicities are slight, all species are present at equilibrium. (b) ROH is a weaker acid than R¢SH and also R¢S– is a weaker base than RO–. Hence, the reaction will not be feasible as the reactants contain weaker acid and weaker base as compared to the products. (c) Because of the bigger size of S, the negative charge on it is more easily polarised which makes it easier for the nucleophilic attack. The nucleophilic nature of RO– is also diminished in a protic solvent because there will be more stronger hydrogen bonding due to the large negative charge density on the smaller sized O atom. (d) The bond angles are 109° and 96° for C—O—H and C—S—H, respectively. The oxygen involves sp3 hydrid orbitals while sulphur involves p orbitals for bonding with C and H atoms. 40. Blue colour is obtained with secondary alcohol. 41. Tertiary alcohol is most reactive towards Lucas reagent. 43. Primary alcohol does not react with Lucas reagent. 44. CH3CH CHCH2CH3

HIO4

Æ CH3CHO + OCHCH2CH3

OH OH

45. 46. 47. 48. 49.

The number of C—C bond broken is equal to the number of HIO4 molecules used. The reagent C6H5NH+ CrO3Cl– oxidizes alcohol to aldehydic stage. Tertiary alcohol reacts fastest with Lucas reagent. Red colouration is shown by 1° alcohol. The formation of alkyl chloride produces cloudiness. H

51. (a) tert-Butyl alcohol does not possess

CH3

C

OH

group, hence it does not give iodoform test.

H

(b) The molecule does not possess

CH3

C

OH

group.

H H

(c) The reaction is

R

C

C

OH OH

R¢

HIO4

Æ RCHO + R¢ CHO

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.13

R¢ H

(d) The products are one aldehyde and one ketone. R

C

C

R¢ R≤

HIO4

Æ R

C

O + R≤ CHO

OH OH 52. (a) The molecule does not contain —OH groups attached to adjacent carbon atoms. (b) Tertiary alcohol gives 3° carbocation and secondary alcohol gives 2° carbocation. Since 3° carbocation is more stable than 2° carbocation, it is formed more readily. 53. (a) Tertiary alcohol reacts faster than secondary alcohol with Lucas reagent. (b) The correct order is 1° > 2° > 3°. (c) With small amount of HI, allyl iodide is obtained. See Q. 28.

(d) With excess of HI, 2-iodopropane is obtained. See Q. 29. CH3

54. (a) CH3

CH2

C

HIO4

CH3COCH3 + HCHO

OH OH (b) The compound is . H2C(CHOH)4CHO.

OH H2 C

CH

CH

CH

CH

CH

O Æ HCHO + 5HCOOH

OH OH OH OH OH

(c) Thiols are less polar than the alcohols. (d) The negative charge on S (which is larger in size than oxygen) is more spread on its conjugate base RS– of the acid RSH as compared to that on O of RO–, Thus, RS– is less likely to attract H+ than RO–, making RSH more acidic than ROH. 55. (a) Both NO2 and COOH groups are reduced. The product is p-H2NC6H4CH2CH2OH. (b) The hydration involves anti-Markovanikov addition. 1. BH THF

Ph2C==CHCH3 2. H O3 /OHÆ Ph2CHCH(OH)CH3 2 2 (c) An alcohol containing a methyl carbinol with at least one H on the carbinol C gives iodoform test. Cyclopentylmethylcarbinol shows iodoform test and the product is cyclopentane-carboxylic acid.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. During dehydration of alcohols to alkenes by heating with concentrated H2SO4 the initiation step is (a) formation of an ester (b) protonation of alcohol molecules (c) formation of carbocation (d) elimination of water 2. Among the following compounds which can be dehydrated very easily is OH (a) CH3CH2CH2CH2CH2OH

[2003]

(b) CH3CH2CH2 CH CH3

CH3

(c) CH3CH2 CCH2CH3 OH

(d) CH3CH2 CH CH2CH2OH CH3

[2004]

24.14 Complete Chemistry—JEE Main

3. The best reagent to convert pent-3-en-2-ol into pent-3-en-2-one is (a) chromic anhydride in glacial acetic acid (b) pyridinium chlorochromate (c) acidic permanganate (d) acidic dichromate [2005] 4. A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was (b) CH3COOC2H5 (c) CH3OH (d) HCHO [2009] (a) CH3COCH3 5. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2 is (a) 1-butanol (b) 2-butanol (c) 2-methylpropan–2–ol (d) 2-methylpropanol [2010] 6. Consider the following reaction C2H5OH + H2SO4 Æ Product Among the following, which one cannot be formed as a product under any conditions? (a) Ethyl-hydrogen sulphate (b) Ethylene (c) Acetylene (d) Diethyl ether [2011] 7. Consider thiol anion (RS–) and alkoxy anion (RO–). Which of the following statements is correct? (b) RS– is less basic but more nucleophilic than RO– (a) RS– is less basic and less nucleophilic than RO– – – (c) RS is more basic and more nucleophilic than RO (d) RS– is more basic but less nucleophilic than RO– [2011] 8. Iodoform can be prepared from all except (a) Ethyl methyl ketone (b) Isopropyl alcohol (c) 3-Methyl-2-butanone (d) Isobutyl alcohol [2012] 9. An unknown alcohol is treated with the Lucas reagent to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism? (b) tertiary alcohol by SN1 (a) secondary alcohol by SN1 (d) tertiary alcohol by SN2 [2013] (c) secondary alcohol by SN2 10. Which one of the following statements is not correct ? (a) Alcohols are weaker acids than water (b) Acid strength of alcohols decreases in the following order RCH2OH > R2CHOH > R3COH (c) Carbon-oxygen bond length in methanol, CH3OH is shorter than that of C - O bond length in phenol. (d) The bond angle

in methanol is 108.9°

O C

CH2

11.

(a)

CH

CH2

H

CH2

CH

[2014]

on mercuration- demercuration produces the major product:

CH3

(b)

CH2

CH2

CH2

(d)

CH2

COOH

OH

OH

(c)

CH2

CH

CH2

OH

OH

12. In the Victor-Meyer’s test, the colour given by 1°, 2° and 3° alcohols are respectively: (a) Red, colourless, blue (b) Red, blue, colourless (c) Colourless, red, blue (d) Red, blue, violet

[2014]

[2014]

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.15

13. In the hydroboration - oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is: (a) CH3CH2OH (b) CH3CHOHCH3 (c) CH3CH2CH2OH (d) (CH3)3COH [2014] 14. The product of the reaction given below is [2016] 1. NBS/hn 2. H2O/K2CO3

OH

OH

(a)

CO2H

(b)

(c)

(d)

ANSWERS 1. (b) 7. (b) 13. (c)

2. (c) 8. (d) 14. (a)

3. (a) 9. (b)

4. (b) 10. (c)

5. (c) 11. (a)

6. (c) 12. (b)

HINTS AND SOLUTIONS 2. The compound CH3CH2 C (CH3)CH2CH3 produces most stable 3° carbocation and thus the compound will OH undergo dehydration most easily. Therefore, choice (c) is correct. 4. There is a formation of ester when ethanol reacts with a carboxylic acid in acidic medium.

CH3COOH + HOC2H5

H+

CH3COOC2H5 + H2O

Ester (fruity smell)

5. The given alcohols are: CH3 CH3CH2CH2CH2OH ; 1-Butanol

CH3CHCH2CH3 ; OH 2-Butanol

CH3

C

CH3

CH3

OH

CH3 CH CH2OH

2-Methylpropan-2-ol

2-Methylpropanol

3° alcohol reacts fastest with conc. HCl and anhydrous ZnCl2. 6. Acetylene is not formed in the reactions between C2H5OH and H2SO4 under different experimental conditions. 7. RSH is a stronger acid than ROH. Thus, the conjugate base of RSH, i.e. RS–, is a weaker base than the conjugate base of ROH, i.e. RO–. In going down a group in the periodic table, nucloephilicity increasses. Thus, RS– is more nucleophilic than RO–. 8. The given compounds are:

24.16 Complete Chemistry—JEE Main

A compound containing —COCH3 and

groups show iodoform test. Hence, isobutyl alcohol will not

show iodoform test. 9. Tertiary alcohol reacts fastest with lucas reagent. Because of bulkiness at the site of reaction the alcohol reacts with Lucas reagent via SN1 mechanism. 10. Due to the resonance effect in phenol, the C - O bond in phenol is shorter than the C - O bond in methanol. 11. Alkene on oxymercuration - demercuration produces an alcohol in which the net reaction is the addition of H2O to the double bond in accordance with Markovnikov’s rule (i.e. -OH group adds to carbon containing lesser number of hydrogens). Hence, we have CH2

CH

CH2

CH2

CH(OH)CH3

H2 O

Therefore, the choice (a) is correct. 12. 1° alcohol produces blood red colour. 2° alcohol produces blue colour. 3° alcohol produces no colour. 13. In the hydroboration - oxidation reaction, an alken reacts with BH3 to give alkyl boranes which on oxidation with alkaline H2O2 gives alcohol. The product obtained in the accordance with the anti– Markovnikov’s rule. CH3 - CH = CH2 + BH3 Æ CH3CH2CH2BH2 H2O2, OH- Æ CH3CH2CH2OH 14. N – Bromosuccinimide (NBS) causes the bromination of an alkene at the allylic position. Br

NBS/n

SECTION 2

OH

H2O/K2SO3

Phenols

Methods of Preparation N

N+HSO–4

OH H2O, H+

Æ

Hydrolysis of Diazonium Salt

SO3H

Alkali Fusion of Sulphonates

+ N2

OH NaOH

Æ

fusion

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.17

Synthesis from Cumene Cumene reacts with air at high temperature by a radical mechanism to form cumene hydroperoxide which on treating with acid gives phenol and acetone. H3 C

CH3

H 3C

CH O2

D

O

OH

C

CH3

OH H3O+

Æ

D

Æ

+ CH3 COCH3

Physical Properties Because of hydrogen bonding, phenols have quite high boiling points. Phenols themselves are colourless but become coloured due to the atmospheric oxidation. Comparison of physical properties of o-nitrophenol with those of p-nitrophenol and m-nitrophenol revealed wild variations. For example, o-nitrophenol has a much lower boiling point, much lower solubility in water than its isomers. It can be steam-distilled while other isomers cannot be. These variations are due to the fact that o-nitrophenol involves intramolecular hydrogen bonding whereas the other two isomers involve intermolecular hydrogen bonding:

N

O

O.... H

O

O

N

O

O H

(monomolecular)

O H

O

N

O

(multimolecular)

The boiling point of o-nitrophenol is low because it exists as a monomolecule. The solubility of m- and p-isomers is due to their ability to form hydrogen bond with water. Steam distillation depends upon the appreciable vapour pressure of the substance at the boiling point of water. Ortho isomer being monomolecular has an appreciable vapour pressure whereas meta and para isomers being multimolecular have low vapour pressures.

Chemical Reactions Acidic Nature of Phenols Phenols are weak acids. These are soluble in aqueous NaOH and not in aqueous bicarbonates. Phenol are stronger acids than water and alcohols but weaker than carboxylic acids. Phenols behaves as a weak acid because the resultant phenoxide ion is more stable (because of resonating effect) as compared to phenol. O H

O–

+ H+ more stable than phenol

The resonating structures of phenols are as follows.

24.18 Complete Chemistry—JEE Main +

OH

+

OH

+

OH

OH

OH

H

–

´ –

´

´ –

H (I)

(II)

H ´

(III)

(IV)

(V)

The structures (II) to (IV) involve separation of charges and thus have much higher energy than those of structures (I) and (V). In other words, the resonance hybrid includes major contribution from (I) and (V). The corresponding structures of phenoxide ion are as follows: O

–

O

O

–

O

H

O

–

´ –

´

´ –

H (I)

(II)

H ´

(III)

(IV)

(V)

hybrid structure is more stable than phenol.

Formation of Ethers (Williamson Synthesis) In alkaline medium, phenols react with alkyl halides to yield ethers. O–Na+

OH NaOH

OC2H5 C2H5I

Æ

Æ

O– Na+

OH NaOH

OCH2COONa

CICH2COOH Æ heat

Æ

OCH2COOH

HCl

Æ

Formation of Esters Phenols are usually converted into their esters by the action of acid, acid chloride or anhydride. O OH

O + RCOCI

C

R

Æ

Fries Rearrangement When esters of phenols are heated with AlCl3, the acyl group migrates from the phenolic oxygen to an ortho or para position of the ring producing a ketone.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.19

O OH

O

C

C2H5COCI

C2H5

OH

OH COC2H5

AlCl3

Æ

Æ

+

CS2

o-hydroxyphenyl ethyl ketone (volatile in steam)

COC2H5 p-hydroxyphenyl ethyl ketone

Ring Substitution The phenolic group is a powerful activating group and is thus ortho and para directing. The intermediates formed are oxonium ions (instead of carbocations) which are formed faster than the formation of carbocations from benzene, hence, phenols are more reactive towards electrophilic substitution than benzene. The structures of oxonium ions are: +

+

OH

OH

H and

Y

H

Y

In an aqueous solution, bromination of phenol yields tribromophenol. In lesser polar solvent, such as CHCl3, CCl4 or CS2, monobromination products are obtained. In aqueous medium, PhOH gives PhO– of which complete bromination takes place. The less reactive PhOH does not ionize in nonpolar solvent and thus monobromination occurs. OH OH OH OH Br

Br2 CHCl3 or CCl4 or CS2

Br Br2 aqueous medium

Br +

medium

Br Br Similarly, concentrated nitric acid gives trinitrophenol (picric acid) with low yield as nitration is accompanied by considerable oxidation. Use of dilute nitric acid gives monosubstituted nitrophenol. Nitrous acid converts phenol to nitrosophenol.

OH

OH NaNO2 + H2SO4

Æ N

O

p-nitrosophenol

Phenols can be sulphonated with concentrated H2SO4. OH

OH conc. H2SO4 100°C

SO3H

OH conc. H2SO4 15–20°C

SO3 H

24.20 Complete Chemistry—JEE Main

Phenols can undergo Friedel-Crafts alkylation and acylations: OH

OH + CH3Cl

OH CH3

AlCl3

Æ

+ o-cresol

CH3 p-cresol

OH

OH + CH3COCl

OH COCH3

AlCl3

Æ

+ o-hydroxyacetophenone

COCH3 p-hydroxyacetophenone

Kolbe Reaction When the salt of phenol is treated with carbon dioxide, a —COOH group is introduced in the ring: ONa

OH

OH COONa + CO2

Æ (main product)

H+

COOH

Æ salicyclic acid

Reimer-Tiemann Reaction When phenol is treated with chloroform and aqueous sodium hydroxide, a —CHO group is introduced in the ring: Reimer-Tiemann reaction is an electrophilic substitution on PhO–. The electrophile is dichlorocarbene which contains OH – – . a C with only six electrons. Æ Cl – + .. CCl2 HC Cl3 + Æ . CCl3 –H

The treatment of phenol with carbon tetrachloride in aqueous NaOH produces salicylic acid.

Coupling Reactions Phenol couples with benzenediazonium chloride in alkaline medium to form p-hydroxyazobenzene. N

N

Cl +

OH

OH–

Æ

N

N

OH

Condensation Reactions Phenol on combining with phthalic anhydride in the presence of concentrated sulphuric acid produces phenolphthalein:

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.21

Phenol condenses with formaldehyde in the presence of dilute acid or alkali producing bakelite.

Oxidation in Air Phenol on exposure to air produces a red coloured product called phenoquinone. OH

O excess of Æ phenol

airÆ

OH....O

O....HO phenoquinone

O

Schotten-Baumann Reaction The reaction of phenol with benzoyl chloride is known as Schotten-Baumann reaction. OCOC6H5

OH + ClCOC6H5

NaOH

Æ phenylbenzoate

Test for Phenol Phenols liberate hydrogen gas when treated with sodium. These are soluble in sodium hydroxide solution and not in sodium bicarbonate solution. Phenols give coloured complexes (usually green, violet, blue) with ferric chloride.

MULTIPLE CHOICE QUESTIONS ON SECTION 2 General Characteristics 1. Picric acid is (a) 2-nitrophenol (b) 4-nitrophenol (c) 2, 4-dinitrophenol (d) 2, 4, 6-trinitrophenol 2. Which of the following orders is true regarding the acidic nature of phenol? (a) Phenol > o-Cresol > o-Nitrophenol (b) Phenol > o-Cresol < o-Nitrophenol (c) Phenol < o-Cresol > o-Nitrophenol (d) Phenol < o-Cresol < o-Nitrophenol 3. Which of the following statements is not true? (a) Phenol is a weak acid (b) Phenol is soluble in NaOH (d) Phenol gives violet colouration with neutral FeCl3 (c) Phenol liberates CO2 from Na2CO3 solution 4. Which of the following statements is not true? (a) When vapours of phenol are passed over Zn dust, benzene is formed. (b) The phenolic —OH group is meta directing group

24.22 Complete Chemistry—JEE Main

5. 6.

7.

8.

9.

10.

11.

(c) The phenolic —OH group is ortho-and para directing group (d) o-Nitrophenol has a lower boiling point as compared to that of p-nitrophenol Phenol is less acidic than (a) acetic acid (b) water (c) p-nitrophenol (d) ethanol Which of the following statements regarding phenols is not correct? (a) o-Nitrophenol has a much lower boiling point as compared to p-nitrophenol. (b) o-Nitrophenol has much lower solubility in water as compared to p-nitrophenol. (c) o-Nitrophenol is stronger acid than p-nitrophenol (d) o-Nitrophenol can be steam distilled while p-nitrophenol cannot be. Which of the following statements regarding phenols is not correct? (a) Phenols are stronger acids than water and alcohols (b) Phenols are weaker acids than carboxylic acids (c) Phenols are soluble in both aqueous NaOH and aqueous NaHCO3. (d) Phenoxide ions are more stable than the corresponding phenols. Which of the following orders regarding the acid strength of phenols is correct? (a) p-aminophenol > p-chlorophenol > p-nitrophenol (b) p-aminophenol < p-chlorophenol < p-nitrophenol (c) p-aminophenol > p-nitrophenol > p-chlorophenol (d) p-nitrophenol > p-aminophenol > p-chlorophenol Which of the following orders regarding acid strengths is correct? (a) Benzyl alcohol > phenol > p-hydroxybenzoic acid (b) Benzyl alcohol < phenol < p-hydroxybenzoic acid (c) Benzyl alcohol < p-hydrobenzoic acid < phenol (d) phenol > Benzyl alcohol > p-hydroxybenzoic acid Phenol can be prepared by (a) Dow process (b) Claisen reaction (c) Cannizzaro reaction (d) Reimer-Tiemann reaction The order of activity in (I) phenol, (II) m-nitrophenol, (III), p-nitrophenol and (IV) p-methylphenol is (a) III > II > I > IV

(b) II > III > I > IV

(c) I > III > II > IV

(d) IV > I > II > III

12. Which of the following statements is not correct? (a) o-Nitrophenol is steam volatile whereas p-nitrophenol is not. (b) Phenol is partially miscible with water. Solubility increases with increase in temperature. (c) Phenol themselves are colourless. (d) o-Nitrophenol has a higher boiling point than p-nitrophenol. 13. Which of the following statements is not correct? (a) o-Nitrophenol has lower solubility in water than its para isomer. (b) Phenols and their salts have opposite solubility characteristics in ionic and nonionic solvents. (c) Phenols are stronger acids than carboxylic acids. (d) The —O– group in phenolate ion is more strongly electron-releasing than the —OH group in phenol. 14. Which of the following statements is not correct? (a) The resonance stabilization of phenol is more than that of phenoxide ion. (b) p-Aminophenol is less acidic than phenol. (c) Heating of an ester of a phenol with aluminium chloride causes rearrangement of acyl group from the phenolic oxygen to an ortho or para position of the ring. (d) If halogenation of phenol is carried out in a less polar medium such as CHCl3, CCl4 or CS2, only monohalogenated products are obtained. 15. Which of the following statements is not correct? (a) Unlike alkyl halide, aryl halides are not readily prepared from the corresponding hydroxyl compounds. (b) Most ethers are inert toward bases, but 2,4-dinitroanisole is readily cleaved to methanol and 2,4-dinitrophenol

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.23

(c) The bromination of p-methylanisole produces 2-bromo-3-methylanisole as the principal product. (d) The nitration of m-nitroanisole produces a mixture of 3,4-dinitro-anisole and 2,5 dinitroanisole. 16. Which of the following statements is not correct? (a) Phenol is a stronger acid than ROH. (b) PhO– is a stronger base than RO–. (c) NaHCO3 does not react with phenol because it is less acidic than phenol. (d) Na2CO3 reacts with phenol because it is more acidic than phenol. 17. Which of the following statements is not correct? (a) Picric acid is 2,4,6-trinitrophenol. (b) Kolbe reaction helps in converting phenol to salicyclic acid. (c) Reimer-Tiemann reaction involving phenol and chloroform converts the former into salicyclic acid. (d) Phenol gives characteristic colour with FeCl3. This test is due to enol structure.

Chemical Reactions 18. Salicyclic acid is produced when phenol in alcoholic KOH is treated with (b) CH3Cl (c) CCl4 (d) CH2Cl2 (a) CHCl3 19. The reaction of phenol with chloroform in alkaline medium producing salicyldehyde is known as (a) Cannizzaro reaction (b) Claisen reaction (c) Reimer-Tiemann reaction (d) Hell-Volhard-Zelinsky reaction 20. The bromination of phenol in aqueous medium produces (a) 2–bromophenol (b) 4–bromophenol (c) 2, 4, 6-tribromophenol (d) a mixture of 2- and 4-bromophenols 21. Phenol on treating with concentrated H2SO4 at 15–20 °C mainly produces (a) phenol-2-sulphonic acid (b) phenol-4-sulphonic acid (c) a 50% mixture of ortho and para phenol sulphonic acid (d) phenol-2, 4, 6-trisulphonic acid 22. Phenol on treating with concentrated H2SO4 at 100 °C mainly produces (a) ortho-phenolsulphonic acid (b) para-phenolsulphonic acid (c) a 50% mixture of ortho- and para-phenolsulphonic acid (d) phenol-2, 4, 6-trisulphonic acid 23. The treatment of phenol with phthalic anhydride in the presence of concentrated H2SO4 produces (a) aspirin (b) methyl red (c) methyl orange (d) phenolphthalein 24. Phenol reacts with bromine in CS2 medium at low temperature to give (a) m-bromophenol (b) p-bromophenol (c) o- and p-bromophenol (d) 2, 4, 6-tribromophenol 25. Phenol on exposure to air produces a red coloured product known as (a) benzoquinone (b) hydroquinone (c) phenoquinone (d) quinone OH

ONa

OH COONa

26. The reaction

+ CO2

(a) Wurtz reaction (c) Schotten-Baumann reaction

120-140°C 6-7 atm

Æ

COOH

HCl Æ –NaCl

(b) Riemer-Tiemann reaction (d) Kolbe reaction

is known as

24.24 Complete Chemistry—JEE Main

OCOC6H5

OH + ClCOC6H5 Æ

27. The reaction

is known as

(a) Wurtz reaction (c) Kolbe reaction

(b) Riemer-Tiemann reaction (d) Schotten-Baumann reaction OH

OH

OH COCH3

28. The reaction

2

+ CH3COCl

Æ

is known as

+

COCH3

29. 30. 31. 32.

33.

34.

(a) Wurtz reaction (b) Friedel-Crafts reaction (c) Kolbe reaction (d) Wurtz-Fitting reaction The reaction of PhOCH2CH3 with one equivalent of HI produces (a) PhI + CH3CH2OH (b) PhI + CH3CH3 (c) C6H6 + ICH2CH3 (d) PhOH + ICH2CH3 Benzenediazonium chloride on reacting with phenol in weakly basic medium gives (a) p-hydroxyazobenzene (b) benzene (c) diphenyl ether (d) chlorobenzene Phenol on distilling with zinc dust gives (a) benzene (b) diphenol (c) diphenyl ether (d) zinc phenoxide In the reaction of phenol with CHCl3 in aqueous sodium hydroxide at 70°C, the electrophile attacking the ring is (a) CHCl3 (b) :CCl2 (c) –:CHCl2 (d) –:CCl3 Which of the following statements is not correct? (a) The relative amounts of o-phenolsulphonic acid and p-phenolsulphonic acid obtained during sulphonation of phenol depends upon the temperature at which reaction is carried out. (b) o-Phenolsulphonic acid is converted into the para isomer by sulphuric acid at 100 °C. (c) Alkaline hydrolysis of CHCl3 is much more rapid than CH2Cl2. (d) It is not possible to separate the compounds in a mixture containing ArNO2, ArNH2, ArCOOH and ArOH. Which of the following statements is correct? (a) Sulphonation of phenol at low temperature is rate-controlled to give o-HOC6H4SO3H.

(b) Sulphonation of phenol at higher temperatures is thermodynamically controlled to give o-HOC6H4SO3H. (c) The —OH is more activating than —O– towards aromatic electrophilic substitution reactions. (d) Bromination of phenol in aqueous medium gives monobrominated phenol while in nonaqueous medium, tribrominated phenol is formed. 35. Which of the following statements is not correct? (a) Phenol reacts with CCl4 in the presence of NaOH solution at 70°C to give salicyclic acid. (b) Phenol reacts with phthalic anhydride in the presence of concentrated H2SO4 forming phenolphthalein which is used as acid-base indicator. (c) Phenol on exposure to air produces a red coloured product known as phenoquinone. (d) The treatment of concentrated H2SO4 at 15–20 °C with phenol produces p-phenolsulphonic acid whereas at 100 °C, o-phenolsulphonic acid is produced. 36. Which of the following statements is not correct? (a) When vapours of phenol are passed over Zn dust, benzene is formed. (b) The penolic —OH group is ortho- and para- directing. (c) o-Nitrophenol has a lower boiling point as compared to that of p-nitrophenol. (d) Phenol is less acidic than o-cresol.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.25

ANSWERS 1. 7. 13. 19. 25. 31.

(d) (c) (c) (c) (c) (a)

2. 8. 14. 20. 26. 32.

(b) (b) (a) (c) (d) (b)

3. 9. 15. 21. 27. 33.

(c) (b) (c) (a) (d) (d)

4. 10. 16. 22. 28. 34.

(b) (a) (b) (b) (b) (a)

5. 11. 17. 23. 29. 35.

(a) (a) (c) (d) (d) (d)

6. 12. 18. 24. 30. 36.

(c) (d) (c) (c) (a) (d)

HINTS AND SOLUTIONS 2. Methyl group is activating group (electron release to benzene) and nitro group is deactivating group (electron withdrawal group). Methyl group will make release of H+ 3. Phenol being a very weak acid does not liberate CO2 from Na2CO3. 4. Phenolic —OH group is ortho- and para- directing. 12. (a) o-Nitrophenol involves intramolecular hydrogen bond whereas p-nitrophenol involves intermolecular hydrogen bond. Consequently, the former exists as single molecule whereas the latter exists as multimolecular form. (d) Because of intramolecular hydrogen bonding in o-nitrophenol, the latter has lower boiling point as compared to p-nitrophenol (which has intermolecular hydrogen bonding). 13. (b) Phenol is essentially a covalent compound whereas its salt is ionic. 14. (a) The resonating structrues of phenol involves charge separation while those of phenoxide ion do not involve any charge separation. Since energy is needed to separate opposite charges, the structrues of phenol involve more energy and thus are less stable than the structures of phenoxide ion. (b) Electron-releasing substituent destabilizes the phenoxide ion, hence, makes the such substituted phenol less acidic. (c) This is known as Fries rearrangement. 15. (a) The bond Ar—O in phenols is strong, due to partial double-bond character and/or sp2 hybridization of aromatic carbon. (b) The reaction is nucleophilic aromatic substitution of —OH for —OCH3. This is achieved due to the deactivation of two nitro groups in the ortho and para positions. (c) The principal directing group is —OCH3. Therefore, the obtained compound would be 1-bromo-3-methylanisole.

OCH3

OCH3 Br

Br2

⎯⎯→ CH3

CH3

(d) The principal directing group is — OCH3.

OCH3

OCH3

OCH3 HNO3

⎯⎯⎯→ NO2

H2SO4

O2N + NO2

NO2 NO2

24.26 Complete Chemistry—JEE Main

16. (b) The negative charge on the alkoxide anion, RO–, is completely localised, but the negative charge on PhO– is delocalised by extended p-bonding to the ortho and para ring positions. This makes PhO– weaker base than RO– and PhOH a stronger acid than ROH. (c) The reaction of phenol with NaHCO3 would produce the following reaction PhOH

HCO3–

+

pK°a = 10 (weaker)

PhO–

pK°b = 7.6 (weaker)

+

pK°b = 4 (stronger)

H2CO3

pK°a = 6.4 (stronger)

Both the reactants are weaker than the corresponding species on the right side. Thus, the reaction would not proceed to the right side. (d) The reaction would be PhOH

CO32–

+

pK°a = 10 (stronger)

Æ

pK°b = 3.7 (stronger)

PhO–

HCO3–

+

pK°b = 4 (weaker)

pK°a = 10.3 (weaker)

Both the products are weaker than the corresponding species on the left side. Thus, the reaction proceeds to the right side. 17.(b) The reactions are OH

OH O 125°C

+ C

OH COONa

COOH

H+

4–7 atm

O

(c) The compound formed is salicylaldehyde. O–

OH

OH CHCl2

CHCl3

H+

CHO

Aq. NAOH, 70 °C

18.

+ H+ salicylic acid

20. 21. 22. 24. 29.

In aqueous medium, trisubstituted phenol is obtained At low temperature, ortho-product is obtained. At high temperature, para-product is obtained. In nonaqueous medium, monosubstituted ortho- and para-products are obtained. Aryl ethers do not cleave on the aromatic side, but only on the alkyl side.

C6H5OH

33. (a) H2SO4 15

−20

˚C

100

100 ˚C

˚C

H2SO4

o-phenolsulphonic acid ⎯⎯⎯→ p-phenolsulphonic acid

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.27

(b) Sulphonation is reversible. The ortho isomer is formed more rapidly; the para isomer is more stable. At 15—20 °C, there is rate control of product composition; at 100 °C, there is equilibrium control. (c) The hydrogen atom in CHCl3 is more acidic than in CH2Cl2. (d) These can be separated as follows.

ArNO2, ArNH2, ArCOOH, ArOH NaHCO3

ArCOO− Na+ H

residue ArNO2, ArNH2, ArOH

+

NaOH

ArCOOH

ArO− Na+

residue ArNO2, ArNH2

H+

HCl

ArOH ArNH3+ Cl− OH

ArNO2

−

ArNH2 34. (b) The reaction is reversible and thus at higher temperatures, thermodynamically controlled para isomer is formed. (c) PhO– reacts faster than PhOH with an electrophile because the intermediate formed with PhO– is a fairly stable uncharged species while in PhOH, a less stable positively-charged intermediate is formed. Thus, DH‡ for the reaction of PhO– is less than the DH‡ for the reaction of PhOH. +

OH

+

OH

+

E

+

O

O

OH

E+

H E

+ H

H Phenol

O

E

H E

E

unsaturated ketone (more stable uncharged species)

Oxonium ions

(d) In aqueous solution, some phenol is ionised to give more reactive PhO– species, which, in turn, forms a tribrominated phenol. In nonaqueous medium, the ionisation is not possible and the monobromination of less reactive PhOH occurs. 35. (d) At low temperature, o-phenolsulphonic acid is formed which is converted into p-phenolsulphonic acid at higher temperature.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. p-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is CH3 CH3 CH2COOH (b)

(a)

CH2COOH OH

OH

24.28 Complete Chemistry—JEE Main

CH3

CH3 CH(OH)COOH

(c)

(d) CH(OH)COOH OH

OH

O–Na+

OH + CHCl3 + NaOH

2.

[2005]

CHO

The electrophile involved in the above reaction is ≈

≈

(a) formyl cation (CHO)

(b) dichloromethyl cation (C HCl2 )

(c) dichlorocarbene (: CCl2)

(d) trichloromethyl anion (C HCl2 )

[2006]

3. The structure of the compound that gives a tribromo derivative on treatment with bromine water is CH3 CH3 CH2OH (b)

(a)

(c)

(d)

OH OH

5. 6.

7.

8.

(a) nitrobenzene (b) 2, 4, 6 – trinitrobenzene (c) o- nitrophenol (d) p- nitrophenol [2006] The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is (a) Salicylic acid (b) Phthalic acid (c) Benzoic acid (d) Salicyladehyde [2008] Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above reaction is (a) 2,4,6-tribromophenol (b) 2-bromophenol (c) 3-bromophenol (d) 4-bromophenol [2009] The correct order of acid strength of the following compounds A. Phenol B. p-Cresol C. m-Nitrophenol D. p- Nitrophenol is (a) C > B > A > D (b) D > C > A > B (c) B > D > A > C (d) A > B > D > C [2011 (cancelled)] Arrange the following compounds in order of decreasing acidity:

(a) II > IV > I > III (b) I > II > III > IV (c) III >I > II > IV (d) IV > III > I > II [2013] 9. Sodium phenoxide when heated with CO2 under pressure at 125 °C yields a product which on acetylation produces C.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.29

ONa + CO2

125°C 5 atm

H+ Ac2O

B

C

The major product C would be OCOCH3

OCOCH3 COOH

(a)

(b) COOH OH

OH COCH3

(c)

COOCH3

(d) COCH3

[2014] m π 0?

(a) (iii) and (iv)

Cl

CN

OH

SH

Cl

CN

OH

SH

(i)

(ii)

(iii)

(iv)

(b) only (i)

(c) (i) and (ii)

(d) only (iii)

[2014]

11. The following OH + HCl + HCN

OH

Anhyd. ZnCl2

CHO

is known as: (a) Perkin reaction (c) Kolbe¢s reaction

(b) Gattermann-Koch Formylation (d) Gattermann reaction

[2014]

12. Which one of the following substituents at para-position is most effective in stablizing the phenoxide ion ? (a) - CH3

(b) - OCH3

(c) - COCH3

(d) - CH2OH

O-

[2014]

ANSWERS 1. (d) 7. (b)

2. (c) 8. (c)

3. (b) 9. (b)

4. (c) 10. (a)

5. (a) 11. (d)

6. (a) 12. (c)

HINTS AND SOLUTIONS 1. The main substituent which decides the position of electrohilic substitution in p-cresol is —OH group. Therefore, the electrophilic substitution occurs at ortho position to the —OH group. The reactions involved are as follows:

24.30 Complete Chemistry—JEE Main

CH3

CH3

CH3

CH3

CHCl3

2H2O

(i) HCN

OH –

–2HCl

(ii) H+

CHCl2

CHO

OH

OH

CH(OH)COOH

OH

2. The electrophile is :CCl2 generated by the reaction

CH3

OH

OH– + CHCl3

CH3 Br

Br2(aq)

3. The reaction is

H2O + CCl3 | Æ Cl – + : CCl 2

Br

OH

OH

Br In other choices, meta position will be included in tribromo derivative. 4. With concentrated H2SO4, the reaction proceeds as follows. OH

OH

OH SO3H

H2SO4

SO3H (High temperature) 100 °C

(Low temperature) 15–20 °C

On treating with concentrated HNO3, —SO3H group is knocked off

SO3H

NO2 ;

At low temperature, o-nitrophenol is formed. In fact, the product formed may be 2, 4, 6 – trinitrophenol. OH

OH HNO3

SO3H

OH NO2

O2N

SO3H

HNO3

NO2

O2N

NO2

5. The main reaction is

6. In acidic medium, the reaction between Br– and BrO–3 produces Br2 which on reacting with phenol in aqueous medium produces 2,4,6-tribromophenol.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.31

7. Electron withdrawing —NO2 substituent has an acid-strengthening effect while the electron releasing —CH3 substituent has an acid-weakening effect on phenol. Thus, p-cresol is the weakest acid amongst the given compounds. The m- and p-nitrophenols are stronger acid than phenol (p-isomer is more strong than m-isomers). The given choice (b) 8. Electron-releasing group decreases the acidity while electron-attracting group increases the acidity. Thus, the molecule III will have maximum acidity and the molecule IV will have minimum acidity. Thus, the choice (c) is correct. 9. The given reaction is Kolbe reaction followed by acetylation of — OH group. ONa

OH + CO2

COONa

125°C 5 atm

H+

OCOCH3 COOH

Ac2O

10. Assuming planar structure, hydroquinone has two conformations as shown in the following. O

H

O

H

and O

O

(I)

(II)

H

H

The structure (I) is expected to have zero dipole moment whereas the structure (II) is expected to have dipole moment. This conclusion may be derived based on the vector addition of components of dipole moment. The above facts are also applicable to 1, 4-benzenedithiol (structure iv). 11. Preparation of phenolic aldehyde by the treatment of phenol with hydrogen cyanide and hydrogen chloride in the presence of Lewis acid (e. g. Zn Cl2) is known as Gattermann reaction. 12. Phenoxide ion is an anion. Electron- releasing group will destabilise this while electron- withdrawing group will stabilise this. –COCH3 is the substituent which is electron- withdrawing, hence, will stablise the anion.

SECTION 3

Ethers

Ethers Ethers in the common system of nomenclature are named by the alkyl groups attached to the oxygen atom followed by the word ether, e.g., dimethyl ether, methyl ethyl ether. In IUPAC, ethers are considered to be alkoxy derivatives of alkanes, e.g. methoxyethane (CH3OC2H5), ethoxyethane (C2H5OC2H5). If other substituents like —CHO, —OH, etc. are present, the compound is named as the derivative of that substance, e.g., 2-ethoxyethanol (C2H5OCH2CH2OH), methoxyethanal (CH3OCH2CHO), etc. Ethers can be prepared by the dehydration of alcohols for which concentrated sulphuric acid at 410 K or catalytic hydrogenation (Al2O3 at 520 K) can be used. Unsymmetrical ethers can be prepared by using Williamson synthesis in which alkyl halide is heated with sodium alkoxide. However, tertiary halide produces alkenes with the elimination of HCl. Ethers are chemically inert substances. With concentrated acids, they form oxonium salts. Ether linkage is cleaved by heating with HI producing alcohol and alkyl halide. Ether containing one aromatic and one aliphatic group produces phenol and alkyl halide on cleaving with HI. Ether linkage in aromatic compound is ortho and para directing group.

24.32 Complete Chemistry—JEE Main

Ether when exposed to air for a long time produces compound containing peroxide linkage. For example, CH3 CH3

CH

O

CH

CH3 CH3

air Æ

CH3 CH3

CH3 CH

O

C

O

O H

CH3

This perioxide linkage can be detected by treating the sample with ferrous ammonium sulphate followed by the addition of ammonium thiocynate. Ferrous ions is oxidized to ferric ions which produces red coloration with NH4CNS.

MULTIPLE CHOICE QUESTIONS ON SECTION 3 1. Which of the following is expected to have the lowest boiling point? (a) CH3CH2OH (b) CH3CHO (c) CH3COOH 2. Which of the following does not react with sodium metal? (a) CH3CH2OCH2CH3 (b) CH3OH (c) CH3COOH 3. The heating of phenyl methyl ether with HI produces (a) ethyl chloride (b) iodobenzene (c) phenol 4. The IUPAC name of CH3CH2OCH2CH2CH3 is (a) diethyl ether (b) ethyl propyl ether (c) ethoxypropane 5. The reaction CH3I + C2H5 3OC2H5 + NaI is an example of (a) Wurtz synthesis (b) Clemenson reaction (c) Williamson synthesis

(d) CH3OCH3 (d) HCOOH (d) benzene (d) propoxyethane (d) Dow reaction

(b) CH3OCH(CH3)2 (c) CH3OCH2CH2CH3 (d) CH3CH2CH2CH2OH (a) C2H5OC2H5 7. The alkoxy group in benzene is (a) ortho directing (b) para directing (c) ortho and para directing (d) meta directing 8. The formation of peroxide linkage in ether due to the exposure in air can be detected by treating it with (a) sodium (b) dilute hydrochloric acid (c) aqueous ferrous ammonium sulphate followed by the addition of ammonium thiocynate (d) dilute sodium hydroxide 9. The exposure of ether in air for a long time may cause (a) oxidation to carboxylic acid (b)the formation of peroxide linkage (c) oxidation to produce aldehyde or ketone (d) the degradation of the molecule 10. Ethers (a) are soluble in concentrated acids (b) are insoluble in concentrated acids (c) have unpleasant smell (d) have higher boiling point in comparison to the alcohol of comparable molecular mass 11. Given are the two cleavage reactions: (i) (CH3)3COCH3 3I + (CH3)3COH (ii) (CH3)3COCH3 3OH + (CH3)3CI Which of the following statements is correct? (a) The reagent used in reaction (i) is anhydrous HI in ether and in reaction (ii) is concentrated HI (b) The reagent used in reaction (i) is concentrated HI and in reaction (ii) is anhydrous HI in ether (c) The reagent used both in reactions (i) and (ii) is concentrated HI

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.33

(d) The reagent used both in reactions (i) and (ii) is anhydrous HI in ether O

12. The reaction of (CH3)2 C — CH2 with CH3OH in (i) acid H+, and (ii) base CH3O–, respectively, give (a) (CH3)2C(OCH3)CH2OH and (CH3)2CH(OH)CH2OCH3 (b) (CH3)2C(OCH3)CH2OH and (CH3)2C(OCH3)CH2OH (c) (CH3)2C(OCH3)CH2OCH3 and (CH3)2C(OH)CH2OH (d) (CH3)2C(OH)CH2OH and (CH3)2C(OCH3)CH2OCH3

ANSWERS 1. (d) 7. (c)

2. (a) 8. (c)

3. (c) 9. (b)

4. (c) 10. (a)

5. (c) 11. (a)

6. (d) 12. (a)

HINTS AND SOLUTIONS 3. SN2 attack on a C of a benzene ring does not occur. Also high energy C6H5+ is not formed by an SN1 reaction. Hence, ArI cannot be a product even in an excess of concentrated HI. 11. The low polarity of solvent ether favours the SN2 mechanism and high polarity of solvent water favours SN1 mechanism. The mechanism is

+

(CH3)3COCH3 + HI Æ (CH3)3C O CH3 + I– H

+

SN1: (CH3)3C O CH3 Æ CH3OH + (CH3)3C

+

I

Æ (CH3)3CI ;

H

+

SN2: I– + (CH3)3C O CH3 Æ CH3I + HOC(CH3)3 H

12. There occurs cleavage of ether linkage in oxiranes (cyclic three membered ring including oxygen). In acidcatalyzed reaction, nucleophile –OCH3 from CH3OH attaches to the more substituted carbon where as in basecatalyzed reaction, it is attached to the less substituted carbon.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Sodium ethoxide has reacted with ethanolyl chloride. The compound that is produced in the above reaction is (a) ethyl ethanoate (b) diethyl ether (c) 2-butanone (d) ethyl chloride [2011 (Cancelled)] 2. Williamson synthesis of ether is an example of: (a) Nucleophilic addition (c) Electrophilic substitution 3. Allyl phenyl ether can be prepared by heating: (a) C6H5Br + CH2 = CH - CH2 - ONa (c) C6H5 - CH = CH - Br + CH3 - ONa 4. Consider the reaction sequence below:

(b) Electrophilic addition (d) Nucleophilic substitution

[2014]

(b) CH2 = CH - CH2 - Br + C6H5ONa (d) CH2 = CH - Br + C6H5 - CH2 - ONa

[2014]

24.34 Complete Chemistry—JEE Main OCH 3 Succinic anhydride

A

AlCl 3

OCH3 OH

Clemmenson’s

OCH3

OH

(a)

(b)

X

reduction

H3CO

H3CO

(c)

(d) [2016, online]

OH

OH

ANSWERS 1. (a)

2. (d)

3. (b)

4. (a)

HINTS AND SOLUTIONS 1. The reaction is

CH3CH2ONa + ClCOCH3

CH3CH2 — O — C — CH3 O Ethyl ethanoate

2. Williamson synthesis involves nucleophilic substitution. 3. To prepare an ether containing a phenyl group, the only choice is to take sodium phenoxide and not phenyl bromide as the latter is much less reactive towards nucleophilic substitution reaction. 4. H3CO

O

O O

AlCl3

H3CO

Zn - Hg HCl

O

O

Succinic anhydride

SECTION 4

H3CO

Et Aldehydes and Ketones hers

Methods of Preparation of Aldehydes H

Oxidation of Primary Alcohol

R— CH2OH

pyridinium Æ R chlorochromate

C

O

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.35

CH3

CHCl2

CHO

Cl2

H2O

Æ

Oxidation of Methylbenzene

Æ

heat

CH(OOCCH3 )2

Reduction of Acid Chlorides

CHO

CrO3

H2O

acetic anhydride

H+

Æ

R— COCl

LiAlH(OBu t) 3

R— COCl

H 2 Pd BaSO 4 ethyl acetate

Æ RCHO Æ RCHO

CH(CH3)2

CH(CH3)2 HCl, AlCl 3

+ CO

Gatterman-Koch Reaction

CuCl

Æ CHO OH

OH 1. ZnCl 2 , ether

+ HCN + HCl

Gatterman Reaction

2. H 2 O

OH activating ring

+ NH4Cl

Æ OH CHO

Heating a Mixture of the Calcium Salts of Formic Acid and any One of its Homologues (RCOO)2Ca + (HCOO)2

3

Methods of Preparation of Ketones Oxidation of Secondary Alcohols

R

CH R¢

CrO 3

Æ R

K 2 Cr2 O 7

OH

Friedel-Crafts Acylation

C

R¢

O

+ RCOCl

AlCl 3

Æ

C

R + HCl

O

+ (CH3CO)2O

AlCl 3

Æ

C

CH3 + CH3COOH

O

While preparing a ketone of the type ArCOAr¢ containing deactivating group in one of the aromatic ring, it is important to select the proper combination of Ar and Ar¢. For example, the preparation of m-nitrobenzophenone, one should proceeds as shown in the following

24.36 Complete Chemistry—JEE Main

NO2 + Cl

NO2 AlCl3

Æ

C

C

O The reaction

O 2N

O

+ Cl

Æ no reaction

C O

does not take place as the strongly deactivating nitro group prevents the acylation reaction

Heating the Calcium Salt of any Fatty Acid other than Formic Acid (RCO2

3

Physical Properties Because of C==O group, aldehydes and ketones are polar compounds and their boiling points are higher than those of nonpolar compounds of comparable molar masses. However, they are not as polar as alcohols and carboxylic acids due to the absence of hydrogen bonding and thus their boiling points are lower than those of alcohols and carboxylic acids of comparable molar masses. Formaldehyde is a gas, rest are liquids or solids. A 40% aqueous solution of formaldehyde is known as formalin. The lower aldehydes and ketones are soluble in water due to hydrogen bonding between solute and solvent.

Chemical Properties +

–

The carbonyl group, C==O, plays a dominating role in the chemistry of aldehydes and ketones. Because of the C O polarization the addition reactions occurring across C==O bond is of the nucleophilic type. Aldehydes generally undergo nucleophilic addition more readily than ketones as the former involves lesser crowding at the transition state. An alkyl group releases electrons and thus destabilises the transition state by intensifying the negative charge developing on oxygen. On the other hand, an aryl group, which is electron-withdrawing group, is expected to stabilise the transition state and thus speed up the reaction. However, an aryl group stabilises the reactant more than the transition state due to resonance and thus lowers down the speed of the reaction. The resonating structures are R

R

The nucleophilic reaction involving

C

+

C

O

C

O

–

O group is catalyzed by acids. This is because carbon acquires more positive

charge as depicted in the following. Z

R¢ R

C

O

H

+

R¢ R

C

+

OH

R¢

+

C

OH

R

R¢

Z C

OH

R

(undergoes nucleophilic attack more readily)

Typical reactions shown by aldehydes and ketones are as follows.

Oxidation The hydrogen atom attached to the carbonyl carbon is abstracted during oxidation. Hence, only aldehydes are oxidized easily and not ketones:

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.37 KMnO 4

R— CHO

K 2 Cr2 O 7

Æ R— COOH

Ketones requires vigorous conditions and the acids obtained contain fewer carbon atoms. Cleavage occurs at the double bond of the enol form: OH C

O C

C C

H

OH

C

H

enol

C

C

C C

H

H

ketone

enol

Aldehydes are oxidized by even mild oxidizing reagents such as Tollens reagent (ammoniacal silver nitrate): CH3CHO + 2Ag(NH3)2+ + 3OH– Æ 2Ag + CH3COO– + 4NH3 + 2H2O silver mirror

O Ketones containing CH3C

are oxidized by hypohalite to give carboxylic acid and haloform. For example, C2 H 5 C

– – CH3 + 3OI– Æ C2H5COO + CHI3 + 2OH

O Hypohalite does not attack carbon-carbon double bond present in the molecule. For example, H C

CH3 C

C

CH3

OCl–

Æ

H

CH3

C

C

COOH + CHCl3

O

Reduction Aldehydes are reduced to 1° alcohols whereas ketones to 2° alcohols. This can be achieved either by catalytic hydrogenation (H2, Ni) or by the use of lithium aluminium hydride LiAlH4: H2 + Ni, Pt or Pd

C

O

C OH

LiAlH4 or NaBH4; then H+

If carbon-carbon double bond is also present in the carbonyl compound, it is also reduced alongwith. However, the use of the reagent 9-BBN (9-borabicyclo[3, 3, 1]nonane) prevents this and thus only the carbonyl group is reduced. For example, CH

CH

CHO

9-BBN

Æ

HOCH2CH2NH2

cinnamaldehyde

Æ

CH

CHCH2OH

cinnamal alcohol

Reduction to Hydrocarbons Aldehydes and ketones are reduced to hydrocarbons by (a) Clemmensen reduction (amalgamated zinc and concentrated hydrochloric acid) and (b) Wolff-Kishner reduction (hydrazine, NH2NH2, and a strong base like KOH or potassium tert-butoxide). For example, CH3CH2COCH 3

Zn(Hg)

Æ CH3CH2CH2CH3

conc. HCl

24.38 Complete Chemistry—JEE Main

The Wolff-Kishner reduction proceeds as follows.

O C

N CH3

C

H

NH2 CH3

H2NHH2

N

N

C

H CH3

KOH

Æ

H –N2

C

Æ

H CH3

KOH

KOH

H

Action of Grignard Reagents

Ketones produces 3° alcohol with Grignard reagent. C

Reactions with Sodium Hydrogen Sulphite

OH

O + NaHSO3

C

SO3Na bisulphite

Most aldehydes form bisulphite compounds. Ketones of the type CH3COR, where R is a primary alkyl group, form bisulphite compounds. If R is secondary or tertiary alkyl group, the formation of bisulphite is slow. Ketones of the type RCOR¢ does not form bisulphite provided R is C2H5 or higher alkyl group.

Addition of HCN (or sodium cyanide and mineral acid) O + CN–

C

C

+ O– H

CN

OH

C CN

cyanohydrin

Cyanohydrin undergoes hydrolysis to give a-hydroxy acid or unsaturated acid. For example, H C

O

H

CN– H+

Æ

C

OH

HCl heat

H

Æ

C

CN

OH

NO2

NO2

NO2

m-nitrobenzaldehyde

m-nitromandelic acid

CN CH3CH2 CCH3 O

COOH

H+

H+

heat

Æ CH3CH2CCH3 OH

COOH

COOH

CN–

Æ

CH3CH2CCH3

Æ CH3 CH = C

CH3

2-methyl-2-butenoic acid

OH

2-butanone

Addition of Ammonia Aldehydes react with ammonia to give unstable aldehyde-ammonia which on eliminating water gives imines.

R C HO + NH3

H+

OH R

CH NH2

–

H2O

RCH

NH

an imine

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.39

Unsubstituted imines formed from NH3 are unstable and polymerized on standing. R

3RCH == NH

NH

HN

R NH

R

If a primary amine (RNH2) is used instead of ammonia, a more stable, substituted imines (known as Schiff bases) is formed. Formaldehyde does not form imine but gives hexamethylenetetramine, which is used as a urinary antiseptic under the name urotropine. 6HCHO + 4NH3 æÆ (CH2)6N4 + 6H2O The structure of hexamethylenetetramine is N CH2 N

CH2 N

CH2

N

Aromatic aldehydes (such as benzaldehyde) or arylamines (such as aniline) gives the most stable imines, but other aldehydes, ketones or primary amines may be used.

Addition of Derivatives of Ammonia The derivatives of ammonia which add on the carbonyl group are hydroxylamine, hydrazine, phenylhydrazine and semicarbazide. Their stable forms are available in the form of salts; hydroxylamine hydrochloride (HONH +3Cl–), phenylhydrazine hydrochloride (C6H5NHNH3+ Cl–) and semicarbazide hydrochloride (NH2CONHNH3+Cl–). The addition of sodium acetate releases free base in solution which adds on the carbonyl group as follows. C

O + NH2OH

H+

Æ

C

hydroxylamine

C

oxime

H+

Æ

O + NH2NHC6H5

O + NH2NHCONH2

NHNHC6H5

C

Æ

H+

Æ

O + H2N

C

NHNHCONH2

Æ

C NNHCONH2 + H2O semicarbazone

OH

NH2 Æ

C NNHC6H5 + H2O phenylhydrazone

OH

semicarbazide

C

C NOH + H2O

OH

phenylhydrazine

C

Æ

NHOH

C

N

NH2

hydrazone

The adjustment of pH (about 3 to 4) is necessary as in the low enough pH, bases may form protonated species H3N—G which inhibits the addition as these are no longer nucleophilic reagents. On the other hand, protonation of carbonyl oxygen makes the molecule more susceptible to nucleophilic attack at positively charged carbon. The addition products, especially with 2,4–dinitrophenylhydrazine, are used to characterize the aldehyde or ketone as they have characteristic melting points.

+

24.40 Complete Chemistry—JEE Main

Many aromatic aldoximes and ketoximes exist in two isomeric forms—cis- and trans-forms: R

Ar

R

Ar

C

C

and

N

N OH Aromatic ketoximes undergo Beckmann rearrangement when treated with PCl5, H2SO4, etc. The exchange occurs in anti-position and the product obtained is N-substituted acetamide. Ar

OH

C N

R

C

O

PCl5

Æ

OH

R

NH Ar

Addition of Alcohols Aldehydes react with anhydrous ethanol in the presence of anhydrous acids (usually dry HCl) to form acetal. The reaction taking place is H R¢

C

H HCl

O + ROH

R¢

H

C

ROH

OR

R¢

C

OR + H2O

OH

OR

hemiacetal

acetal

Hemiacetals are too unstable to be isolated. Acetals are readily cleaved by acids and are stable towards bases: H R¢

H +

OR + H2O HÆ R¢

C

C

O + 2ROH

OR

Cannizzaro Reaction This reaction is shown by aldehydes containing no a-hydrogen. In the presence of concentrated alkali, two molecules of an aldehyde undergo self-oxidation-and-reduction to yield a mixture of an alcohol and a salt of a carboxylic acid. For example, 50% NaOH

2HCHO

Æ CH 3 OH methanol

HCOO Na + sodium formate

In the crossed Cannizzaro reaction, two different aldehydes are involved yielding all possible products. However, if one of the reactants is formaldehyde, it is always converted into sodium formate. For example, CHO

CH2OH + HCHO

OCH3 anisaldehyde (p-methyoxybenzaldehyde)

conc. NaOH

+ HCOO–Na+

Æ OCH3

p-methyoxybenzyl alcohol

The Cannizzaro reaction involved two successive additions as depicted in the following.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.41

The rate of reaction is decided by the ease with which H– is transferred. The molecule of the type PhCOCHO shows internal crossed-Cannizzaro reaction.

Aldol Condensation This reaction is shown by aldehydes or ketones containing a-hydrogen. In the presence of a dilute solution of a base, a-carbon of one molecule gets attached to the carbonyl carbon of the second molecule. For example, H CH3

H

C

O H

C

H C

OH–

Æ CH3

O

H

H

H

C

C

C

O

OH H

H

aldol

The mechanism involved is as follows. (a) Formation of carbanion—a nuclephilic reagent

–   CH3CHO + OH–    H2O + CH2CHO

H

(b) Attack of carbanion to the second molecule

CH3

C

H O + -CH2CHO

CH3

C

CH2CHO

OH (c) Transfer of H+ from water

CH3

H

C

CH3

C

O OH The product obtained in aldol reaction is b-hydroxyaldehyde or ketone. This is very easily dehydrated to form a double bond at a- and b-carbon atoms. For example, H CH3

C

H CH

C

H O

dil HCl

Æ CH3

warm

C

H CH

crotonaldehyde

C

O + H 2O

OH H The resultant molecule contains a double bond in conjugation with the carbonyl group and it thus acquires extra stability. If the double bond is also in conjugation with an aromatic ring, the product is so stable that the product obtained in aldol reaction is unsaturated aldehyde or ketone instead of b-hydroxyaldehyde or ketone. For example,

24.42 Complete Chemistry—JEE Main

C

CH3 + CH3

O

(base)

O

CH3 C

NaOC2H5

C

CH

CH3

–H2O

Æ

C

C

CH

C

OH H

O

O

1,3-diphenyl-2-butene-1-one

A crossed aldol condensation involves two different carbonyl compounds yielding a mixture of all the four possible products. However, a single product is obtained as shown in the following. Aldehyde containing no a-hydrogen (e.g. benzaldehyde or formaldehyde)

CHO + OH–

For example,

CH3CHO

Æ

H

H

H

C

C

C

O

cinnamaldehyde

Intramolecular Aldol Condensation Properly constituted diketons give cyclic intramolecular aldol addition products. O 6

5 4

3

2 1

CH3CCH2CH3CCH3 O

OH-H2O

CH3CCH2CH2CCH2

O

O

H2O -OH

O

-H2O

OH

O

O

H2O -OH

CH3

CH3 O

CH3CCH2CH2CH2CHCCH3

O

C OH CH3

O CH3

-H2O

COCH3

CH3

Keto-Enol Tautomerism A carbonyl compound with an acidic alpha hydrogen may exist in two forms called tautomers: a keto tautomer and an enol tautomer. The keto tautomer has the expected carbonyl group while the enol (from – ene + – ol) tautomer contains vinylic hydroxyl group which is formed by transfer of an acidic hydrogen from the a-carbon to the carbonyl oxygen. Most simple aldehydes and ketones exist primarily in their keto forms. However, 2-4-pentanedione exists in 80% enol from due to the formation of six membered ring involving the internal hydrogen bonding.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.43

Distinction between Aldehydes and Ketones Aldehydes show the following characteristic tests. Ketones do not show these tests. Tollens Test The Tollens reagent is an ammoniacal solution of silver nitrate. Aldehydes reduce Tollens reagent to a bright silver mirror. Fehling’s Test Fehling’s solution consists of an equimolar mixture of Fehling’s solution A and Fehling’s solution B. Fehling’s solution A is CuSO4 solution and Fehling’s solution B is a mixture of Rochelte salt (sodium potassium tartarate) and sodium hydroxide. Aliphatic aldehydes (but not aromatic) reduces Fehling’s solution to red brown cuprous oxide. Schiff’s Test Schiff’s reagent is rosaniline hydrochloride solution in water decolourized with SO2. Aldehydes when warmed with Schiff’s reagent restore pink colour of the reagent.

MULTIPLE CHOICE QUESTIONS ON SECTION 4 General Characteristics 1. Which of the following is used in preservation of biological specimens? (b) CH3COOH (c) CH3OCH3 (d) HCHO (a) CH3CHO 2. The dry distillation of calcium formate gives (b) CH3COCH3 (c) HCHO (d) HCOOH (a) CH3CHO 3. The dry heating of a mixture of calcium acetate and calcium formate produces (b) CH3COCH3 (c) CH3COOH (d) HCHO (a) CH3CHO 4. Which of the following order of boiling point is correct? (a) Butane > Butan-1-al > Butan-1-ol (b) Butane < Butan-1-al < Butan-1-ol (c) Butane > Butan-1-al < Butan-1-ol (d) Butane < Butan-1-al > Butan-1-ol 5. Formalin is an aqueous solution of (a) 40% formaldehyde (b) 50% formaldehyde (c) 60% formaldehyde (d) 70% formaldehyde 6. The structural formula of metaformaldehyde is (b) (CH2O)3 (c) (CH2O)4 (d) (CH2O)n (a) (CH2O)2 7. Polarization of electrons in acrolein may be written as d–

d+

(a) CH ==CH—CH==O 2

d–

d+

(b) CH2==CH—CH==O

d–

d+

(c) CH ==CH—CH==O 2

d+

d–

(d) CH2==CH—CH==O

8. Which of the following statements regarding aldehydes is not correct? (a) The carbonyl functional group is planar (b) The boiling points of aldehydes is greater than those of alkanes of comparable molar masses (c) The boiling points of aldehydes is greater than those of alcohols of comparable molar masses (d) Formaldehyde is gas at room temperature 9. Which of the following has the most acidic hydrogen? (a) 3-Hexanone (b) 2, 4-Hexanedione (c) 2, 5-Hexanedione (d) 2, 3-Hexanedione 10. The enol form of acetone, after treatment with D2O, gives (a) CH 3 —C CH 2 | OD

(b) CH 3 —C—CH 3 || O

(c) CH 2

C—CH 2 D | OH

(d) CD 2

C—CD 3 | OD

11. Which of the following statements regarding CO group is not correct? (a) The carbon atom of CO group in aldehydes is sp2 hybridized (b) The carbon atom of CO group in the transition state formed during the addition reaction across CO group is sp3 hybridized

24.44 Complete Chemistry—JEE Main

(c) The aryl group in aromatic aldehydes speeds up the addition reaction across the CO group (d) An aryl group stabilises an aldehyde more than the transition state 12. Which of the following statements is correct? (a) The structure of the enol form of CH 3COCH 2COCH 3 with intramolecular hydrogen bonding is CH3 C CH C CH3 O

O H

(b) Carbonyl carbon of an aldehyde or a ketone is sp hybridized. (c) Fomaldehyde is a liquid at room temperature. (d) Formaline is 30% aqueous solution of formaldehyde.

Oxidation and Reduction Reactions 13. The reaction

RCOR

Zn Hg HCl

Æ RCH2R

(a) Clemmensen reduction (c) Wolff-Kishner reduction 14. The reaction

RCOCl + H 2

(b) Rosenmund reduction (d) Catalytic reduction Pd-BaSO 4

(a) Clemmensen reduction (c) Wolff-Kishner reduction 15. The reaction

16. 17. 18.

19.

C6H5COCH3

is known as

Æ RCHO + HCl

is known as

(b) Rosenmund reduction (d) Grignard reduction NH 2 NH 2 KOH

Æ C6H5CH2CH3

is known as

(a) Clemmensen reduction (b) Rosenmund reduction (c) Wolf-Kishner reduction (d) Catalytic reduction When acetaldehyde is heated with Fehling’s solution, it gives a precipitate of (d) Cu + Cu2O + CuO (a) Cu (b) CuO (c) Cu2O Which of the following compounds will yield methyl ethyl ketone on oxidation? (a) 2-Propanol (b) 1-Butanol (c) 2-Butanol (d) tert-Butyl alcohol The reagent which can be used to reduce —CHO to —CH2OH without affecting the double bond in the unsaturated aldehydic molecule is (c) 9-BBN (d) NaBH4 (a) Zn/HCl (b) LiAlH4 Which of the following statements is not correct? (a) Toluene can be oxidized to benzaldehyde by using alkaline KMnO4. (b) Phenolic aldehydes can be prepared by Reimer-Tiemann reaction. SO–3 Na + | (c) NaHSO3 reacts with RCHO in ethanol to give solid adduct having structure R—C—OH. | H (d) The formation of oximes is hindered at the very low pH.

Nucleophilic Addition Reactions 20. Hydroxylamine reacts with aldehyde and ketone to give (a) aldehyde ammonia (b) oximes (c) carboxylic acid (d) alkyl cyanide 21. The reaction of formaldehyde with ammonia produces (a) an amine (b) an amide (c) aldehyde ammonia (d) hexamethylenetetramine 22. Which reagent on condensation with benzaldehyde produces phenylhydrazone? (b) C6H5NH2 (c) H2NNHCONH2 (d) C6H5NHNH2 (a) NH2OH

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.45

23. The pH of the solution for the formation of addition products of carbonyl group and ammonia derivatives should be about (a) 0 to 1 (b) 2 to 3 (c) 3 to 4 (d) 7 to 8 24. An oxime is formed when an aldehyde/ketone reacts with (a) hydroxylamine (b) hydrazine (c) phenylhydrazine (d) semicarbazide 25. The semicarbazone is formed when an aldehyde/ketone reacts with (b) H2NNH2 (c) H2NNHC6H5 (d) H2NNHCONH2 (a) H2NOH 26. Which of the following is an example of Perkin condensation? KOH

(a) C6H5CHO + CH3NO2 (b) 2CH3COCH3

dry HCl

Æ C6H5CH = CHNO2 + H2O

Æ (CH3)2C = CHCOCH3 3OH + HCOONa CH 3 COONa

(d) C6H5CHO + (CH3CO)2O

Æ C6H5CH = CHCOOH

27. Which of the following represents Claisen-Schmidt reaction? (a) CH3CHO + CH3CHO

NaOH

(b) C6H5CHO + CH3CHO (c) CH3CHO + CH3CHO

Æ CH3CH = CHCHO + H2O

NaOH NaOH

(d) C6H5CHO + C6H5CHO

Æ C6H5CH = CHCHO + H2O

Æ CH3CH = CHCH3 + O2

NaOH

Æ C6H5CH = CHC6H5 + O2

28. Which of the following reactions represents Tischenko reaction? (a) 2CH3

3COOH

+ CH3CH2OH

(b) 2CH3CHO

K 2 CO 3

Æ CH3CHCH2CHO OH

(c) 2CH3CHO

Al(OC 2 H 5 ) 3

Æ CH3CO2C2H5

(d) 2CH3CHO

NaOH

Æ CH3CH = CHCHO + H2O

29. Acetone on reacting with CH3MgBr followed by hydrolysis produces (a) n-butyl alcohol (b) sec-butyl alcohol (c) isobutyl alcohol (d) tert-butyl alcohol 30. Which of the following compounds will react with ethanolic KCN? (a) Ethyl chloride (b) Acetyl chloride (c) Chlorobenzene (d) Methane 31. The formation of cyanohydrin from a ketone is an example of (a) electrophilic addition (b) nucleophilic addition (c) nucleophilic substitution (d) electrophilic substitution 32. Which of the following statements regarding aldehydes is not correct? (a) The addition reactions occurring across the C==O bond is electrophilic type (b) The addition reactions occurring across the C==O bond is nucleophilic type (c) Aldehydes undergo addition reactions more readily than ketones (d) The addition reactions shown by CO group is catalyzed by acids 33. Which of the following statements regarding CO group is not correct? (a) The carbon atom of CO group in aldehydes is sp2 hybridized (b) The carbon atom of CO group in the transition state formed during the addition reaction across CO group is sp3 hybridized (c) The aryl group in aromatic aldehydes speeds up the addition reaction across the CO group (d) An aryl group stabilises an aldehyde more than the transition state | | | | | | Æ —C==C—C==O + HCN —C—C—C==O involves 34. The reaction | | CN H

24.46 Complete Chemistry—JEE Main

(a) electrophilic addition across double bond (b) nucleophilic addition across double bond (c) free radical addition across double bond (d) no effect of the CO group on the C==C bond 35. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is (b) CH3CHO (c) CH3COOCH3 (d) CH3COOCOCH3 (a) CH3COCl 36. Which of the following statements is not correct? (a) In pinacol-pinacolone rearrangement reaction involving, PhC(CH 3 )—CPh(CH 3 ) , the phenyl group migrates | | in preference to methyl group. HO OH (b) In pinacol-pinacolone rearrangement reaction involving Me2 C—CH 2 , the 3° OH is lost, followed by :H | | OH OH migration. (c) Schiff base is N-substituted imine.

R (d) In the Beckmann rearrangement reaction

OH C == N

conc. H2SO4

A , the product A is R¢CONHR.

R

Aldol and Cannizzaro Reactions 37. Aldol condensation is shown by an aldehyde containing (a) no a-carbon atom (b) no a-hydrogen atom (c) an a-hydrogen atom (d) 38. Cannizzaro’s reaction is not shown by an aldehyde containing (a) no a-carbon atom (b) an a-carbon atom (c) no a-hydrogen atom (d) 39. Base catalyzed aldol condensation occurs with (a) trimethylacetaldehyde (b) benzaldehyde (c) acetaldehyde (d) 40. The Cannizzaro reaction is not given by (a) trimethylactaldehyde (b) benzaldehyde (c) acetaldehyde (d) 41. Base catalysed aldol condensation occurs with (a) propanal (b) benzaldehyde (c) 2-methylpropanol (d) 42. m-Chlorobenzaldehyde on reaction with concentrated KOH at room temperature gives (a) potassium m-chlorobenzoate and m-hydroxybenzaldehyde (b) m-hydroxybenzaldehyde and m-chlorobenzyl alcohol (c) m-chlorobenzyl alcohol and m-hydroxybenzyl alcohol (d) potassium m-chlorobenzoate and m-chlorobenzyl alcohol 43. Which of the following compounds does not undergo aldol condensation? (b) C6H5CHO (c) CH3COCH3 (d) (a) C6H5CH2CHO 44. In the Cannizzaro reaction

2Ph—CHO

– OH

Æ Ph—CH2OH + PhCO–2 ,

two a-hydrogen atoms an a-hydrogen atom formaldehyde formaldehyde 2,2-dimethylpropanal

CH3CH2CHO

the slowest step is

–

(a) the attack of OH at the carbonyl group, (b) the transfer of hydride to the carbonyl group, (c) the abstraction of proton from the carboxylic acid, (d) the deprotonation of Ph—CH2OH 45. Which of the following will not undergo aldol condensation? (a) Acetaldehyde (b) Propanaldehyde (c) Benzaldehyde (d) Trideuteroacetaldehyde 46. Aldol condensation will be observed in (a) CH3CH2CH2CHO

(b)

CHO

(c)

CHO

(1998)

(d)

CHO

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.47

47. Which of the following statements is not correct? (a) Acetamide on heating with P2O5 produces methanenitrile. (b) The structural formula of meta formaldehyde is (CH2O)3. (c) Benzaldehyde undergoes Cannizzaro reaction whereas acetaldehyde does not. (d) The reaction of formaldehyde with ammonia produces hexamethylenetetramine. 48. Which of the following statements is correct? (a) Benzaldehyde undergoes aldol condensation in an alkaline medium. (b) Hydrazones of aldehydes and ketones are prepared in highly acidic medium. (c) The reaction of methyl magnesium iodide with acetone followed by hydrolysis gives secondary butanol. 49. Which of the following statements is not correct? (a) The equilibrium constant of the hydration reaction

H H

is lower than that of the reaction

CH3 CH3

Cl | (b) The equilibrium Cl—C—C | | Cl H

C == O + H2O

O + H2 O

H C == O + H2O

CH3 CH3

OH C

H

OH

OH C OH

Cl OH | | Cl—C—C — OH, lies more to the right side. | | Cl H

(c) H2CO is always oxidized to HCOOH in the crossed-Cannizzaro reactions. (d) The species PhCOCHO undergoes Cannizzaro reaction to give PhCH(OH)COO–.

Distinctive Tests 50. Which of the following gives a silver mirror with Tollens reagent? (b) CH3CHO (c) CH3COOH (d) CH3CH2OCH2CH3 (a) CH3CH2OH 51. Which of the following compounds does not exhibit positive iodoform test? (a) Acetaldehyde (b) Methanal (c) Propanone (d) Ethanol 52. Which of the following compounds undergoes the Cannizzaro reaction? (a) Acetaldehyde (b) Acetone (c) Benzaldehyde (d) Propanal 53. Benzaldehyde differs from acetaldehyde in this respect that the former does not react with (a) Tollens reagent (b) Fehling’s solution (c) HCN (d) NaHSO3 54. Which of the following tests distinguishes acetophenone from propionaldehyde? (a) Baeyer test (b) Iodoform test (c) Tollens test (d) Lucas test 55. The reagent which reacts both acetaldehyde and acetone is (a) Tollens reagent (b) Fehling’s reagent (c) Grignard reagent (d) Schiff’s reagent 56. Acetophenone differs from aliphatic ketones in that it does not react with (c) I2 and NaOH (d) NH2OH (a) HCN (b) NaHSO3 57. The compound that gives a positive iodoform test is (a) 1-pentanol (b) 2-pentanone (c) 3-pentanone (d) pentanal 58. Which of the following compounds will give a yellow precipitate with iodine and alkali? (a) 2-Hydroxypropane (b) Benzophenone (c) Methylacetate (d) Acetamide 59. The compound that will not give iodoform on treatment with alkali and iodine is (a) acetone (b) ethanol (c) diethyl ketone (d) ethyl methyl keton 60. Which of the following compounds does not exhibit(s) positive iodoform test? (a) Acetaldehyde (b) Methanol (c) Ethanol (d) Propanone

24.48 Complete Chemistry—JEE Main

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (d) (a) (a) (d) (b) (c) (b) (a) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(c) (c) (b) (b) (d) (a) (d) (b) (b) (b)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(a) (b) (c) (d) (b) (c) (c) (c) (b) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(b) (a) (c) (d) (c) (b) (c) (a) (c) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(a) (c) (c) (c) (d) (a) (a) (a) (b) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(b) (a) (c) (a) (a) (d) (d) (d) (b) (b)

HINTS AND SOLUTIONS 1. 2. 3. 4. 5. 6. 8.

HCHO is used as preservative. Dry distillation of calcium formate gives formaldehyde. (CH3COO)2Ca + (HCOO)2 3CHO + 2CaCO3 Boiling point of alkane is less than the corresponding aldehyde, which in turn, is less than the alcohol. 40% formaldehyde is known as formaline. Metaformaldehyde is (CH2O)3. Aldehydes have boiling points lower than those of alcohols of comparable molar masses.

9. 2,4-Hexanedione is H3C

O

O

C

CH2 C

CH2 CH3 . Methylene hydrogens attached in between two C==O

group will have maximum acidic nature. 10. CH3

C

CH3

CH3

C

CH2

D2O

Æ CH3

C

CH2

O OH OD 11. The aryl group in aromatic aldehyde retards the addition reaction across the CO group. 12. (b) Carbon is sp2 hybridized. (c) Formaldehyde is a gas at room temperature. (d) Formaline is 40% aqueous solution of formaldehyde. 13. The reduction with Zn/Hg-HCl is known as Clemmensen reduction. 14. The reduction with Pd-BaSO4 is known as Rosenmund reduction. 15. The reduction with NH2NH2/KOH is known as Wolf-Kishner reduction. 16. Red precipitates are of Cu2O.

17. CH3CHOHCH2CH3

[O]

Æ CH3COCH2CH3

18. 9-BBN (9-borabicyclo[3, 3, 1]nonane) can be used. 19. (a) With an alkaline KMnO4, toluene is converted into benzoic acid. However, the use of CrO3 in acetic anhydride followed by hydrolysis will convert toluene to benzaldehyde. (b) At low pH, protonated aldehydes, ketones and hydroxylamines are formed. 20. Hydroxylamine forms oximes. 21. The reaction of formaldehyde with ammonia produces hexamethylenetetramine. 6HCHO + 4NH3 æÆ (CH2)6N4 + 6H2O

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.49

22. Phenylhydrazine (C6H5NHNH2) produces phenylhydrazone. 24. Hydroxylamine produces oxime. 25. Semicarbazide (H2NNHCONH2) forms semicarbazone. CH3

29. CH3COCH3 + CH3MgBr

Æ CH3

C

OMgX

CH3

H2O

Æ CH3 C

CH3

OH

CH3 tert-Butyl alcohol

30. 31. 32. 33.

Ethyl chloride will give CH3CH2CN. Addition across C==O group occurs via nucleophilic mechanism. Same as Q. 31. Aryl group lowers the speed of addition reaction across the C==O group due to the stabilization of the reactant more than the transition state via resonant effect. 34. Same as Q. 31. 35. The positive charge on carbon is enhanced due to electron-with drawing nature of cholrine. CH 3CH 3 CH 3 | | | conc H 2 SO 4 Æ Ph—C—C—CH 3 36.(a) The reaction is Ph—C—C— Ph | | | || OH OH Ph O (b) The reaction is

CH 3 | CH 3 —C—CH 2 | | OH OH

CH 3 | Æ CH 3 —CH—CHO

(c) The general reaction for the formation of a Schiff base is RCHO + R¢NH2 (d) It is the trans R¢ group which migrates

– H 2O

Æ RCH

N—R¢

Aldol condensation is shown by an aldehyde containing an a-hydrogen atom. Cannizzaro’s reaction is not shown by an aldehyde containing an a-hydrogen atom. Acetaldehyde (CH3CHO) shows aldol condensation, as it contains a-hydrogen atoms. CH3CHO contains a-hydrogen atom. It will not show Cannizzaro reaction. Only propanal contains a-hydrogen atoms. It will show aldol condensation. m-Chlorobenzaldehyde undergoes Cannizzaro reaction. Transfer of hydride to carbonyl group is the slowest step. Compounds containing a-hydrogen (or deuterium) undergo aldol condensation. Acetaldehyde, propanaldehyde and trideuteroacetaldehyde will undergo aldol condensation. 48. (a) Benzaldehyde does not possess a-hydrogen atom. It does not undergo aldol condensation but undergoes Cannizzaro reaction giving benzoic acid and benzyl alcohol. (b) Hydrazine gets protonated in highly acidic medium. Due to this, its nucleophilicity is decreased. (c) The product is tertiary butanol.

37. 38. 39. 40. 41. 42. 44. 45.

(d) Carbonyl carbon is sp2 hybridized.

24.50 Complete Chemistry—JEE Main

49. (a) The equilibrium constant for the hydration of H2CO is greater than that of (CH3)2CO. Two factors responsible for this trend are steric and inductive. Firstly, the addition of H2O involves the change in the hybridization of C from sp2 to sp3. The alkyl groups in (CH3)2CO will involve more steric hinderance and thus its hydration will be less effective than that of H2CO. Secondly, the methyl group is an electron-releasing group. The presence of two methyl groups attached to C of CO group in (CH3)2CO diminishes the positive charge on C causing the decrease in the reactivity of nucleophilic addition of H2O across

.

(b) The chloral is less stable than chloral hydrate due to d+ charge on the C adjacent to the carbonyl C.

50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

(c) H2CO is the most reactive aldehyde. It exists in aqueous OH– solution mainly as the conjugate base of its hydrate H2C(OH)O–. The hydride shift from the anion occurs more readily giving formic acid. (d) PhCOCHO undergoes internal crossed-Cannizzaro reaction. Aldehyde (CH3CHO in this case) reduces Ag+ in Tollens reagent to give silver mirror. Iodoform test is shown by a compound CH3CO— group or by a compound which produces this group. Methanal (HCHO) does not show iodoform test. Benzaldehyde does not show Cannizzaro reaction. Benzaldehyde does not react with Fehling’s solution. CH3CH2CHO does not containing CH3CO— group. Thus, iodoform test can distinguish between CH3CH2CHO and CH3COC6H5. Grignard reagent react both with CH3CHO and CH3COCH3. CH3COC6H5 does not react with NaHSO3. 2-Pentanone contains CH3CO— group. It will show iodoform test. 2-Hydroxypropane will give CH3COCH3 which, in turn, will show iodoform test. Diethyl ketone will not show iodoform test as it does not contain CH3CO— group. Iodoform test will be shown if the compound has CH3CO— group.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. Which of the following on heating with aqueous KOH produces acetaldehyde? (b) CH3CHCl2 (c) CH3COCl (a) CH2Cl CH2Cl 2. In Cannizzaro reaction given below: 2PhCHO

– :

OH

(d) CH3CH2Cl

[2009]

Æ PhCH2OH + PhCO2–

The slowest step is (a) The abstraction of proton from the carboxylic group (b) The deprotonation of PhCH2OH (c) The attack of –: OH at the carboxylic group (d) The transfer of hydride to the carbonyl group 3. Identify the compound that exhibits tautomerism. (a) Phenol (b) 2-Butene (c) Lactic acid

[2009] (d) 2-Pentanone [2011 cancelled]

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.51

4. Trichloroacetaldehyde was subjected to Cannizaro’s reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is (a) chloroform (b) 2,2,2-trichloroethanol (c) trichloromethanol (d) 2,2,2-trichloropropanol [2011 cancelled] 5. In the given transformation, which of the following is the most appropriate reagent? CH CHCOCH3 CH CHCH2CH3 Reagent

HO

HO

(a) NH2NH2, OH– (b) Zn-Hg/HCl (c) Na, liq. NH3 6. The most stable reagent for the conversion of RCH2OH to RCHO is (a) PCC (pyridinium chlorochromate) (b) KMnO4 (c) K2Cr2O7 (d) CrO3

(d) NaBH4

[2012]

[2014]

7. An organic compound A, C5H8O; reacts with H2O, NH3 and CH3CHOOH as described below: O H2O CH C OH O NH3 A CH C NH2 O O CH3COOH CH C O C CH3 A is: (a) CH3CH

C

(b) CH2

CHO

CH3

(c) CH3

CH2

CHCH

CHO

CH3

C

C

O

(d) CH3

CH2

C

C

O

[2014]

CH2 H

CH3

(a) Aldol condensation (b) Claisen condensation (c) Cannizzaro reaction (d) Pinacol-pinacolon reaction [2014] 9. Which one of the following reactions will not result in the formation of carbon - carbon bond ? (a) Reimer-Tieman reation (b) Friedel Craft,¢s acylation (c) Wurtz reaction (d) Cannizzaro reaction [2014] 10. Which is the major product formed when acetone is heated with iodine and potassium hydroxide ? [2014, online] (a) Iodoacetone (b) Acetic acid (c) Iodoform (d) Acetophenone 11. A compound A with molecular formula C10H13Cl gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis gives Cannizaro reaction but not aldol condensation. D gives aldol condensation but not Cannizaro reaction. A is: CH3 (b) C6 H5 -- CH 2 -- CH 2 -- CH -- CH3 (a) C6H5 — CH2 — C CH3 Ω Cl Cl CH2 —CH2 —CH3 —

— —

CH2Cl

(c) C6H5 — CH2 — CH2 — CH2 — CH2 — Cl

(d)

[2015, online]

24.52 Complete Chemistry—JEE Main

12. Which of the following pairs of compounds are positional isomers? (a) CH3 -- CH 2 -- CH 2 -- CH 2 -- CHO and CH3 -- CH 2 -- CH 2 -- C -- CH3 ΩΩ O (b) CH3 -- CH 2 -- CH 2 -- C -- CH3 and CH3 -- CH -- CH 2 -- CHO ΩΩ Ω CH3 O (c) CH3 -- CH 2 -- CH 2 -- C -- CH3 and CH3 -- CH 2 -- C -- CH 2 -- CH3 ΩΩ ΩΩ O O (d) CH3 -- CH 2 -- C -- CH3 and CH3 --CH -- CH 2 -- CHO ΩΩ Ω O CH3 13. In the reaction sequence OH D 2CH3CHO æææ Æ A ææ Æ B; the product B is: (a) CH3 — CH2 — CH2 — CH2 — OH

[2015, online]

(b) CH3 — CH == CH — CHO

(c) CH3 -- C -- CH3 (d) CH3 — CH2 — CH2 — CH3 [2015, online] ΩΩ O 14. The correct statement about the synthesis of erythritol (C(CH2OH)4) used in the preparation of PETN is: (a) The synthesis requires three aldol condensations and one Cannizzaro reaction. (b) Alpha hydrogens of ethanol and methanol are involved in the reaction. (c) The synthesis requires two aldol condensations and two Cannizzaro reactions. (d) The synthesis requires four aldol condensations between methanol and ethanol [2016, online]

ANSWERS 1. (b) 7. (c) 13. (b)

2. (d) 8. (c) 14. (a)

3. (a) 9. (d)

4. (b) 10. (c)

HINTS AND SOLUTIONS 1. The reaction goes as follows.

2. The mechanism of the reaction involves the following steps.

5. (a) 11. (a)

6. (a) 12. (c)

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.53 H

Ph

H

H

C == O + Ph

C

–

O

Ph

C

OH

–

O

+ Ph

C == O

H

OH

fast + H+

fast – H

+

H C

Ph

OH

Ph

C == O –

O

H

The shifting of hydride ion to the carbonyl group is the slowest step. 3. The tautomerism in 2-pentanone is CH3 — C — CH2CH2CH3

CH3 — CH — CHCH2CH3

O

OH OH

The tautomerism in phenol is

H H

O

. However, phenol mainly exists in hydroxy form.

4. The reaction is 5. The Wolff-Kishner reduction (which uses hydrazine, NH2NH2, and a strong base OH–) reduces only —C = O group without affecting the —C = C— group. 6. Pyridinium chlorochromate is a suitable reagent to convert an alcoholic group to an aldehydic group. The rest of the three are stronger oxidizing agent. 7. The reactions shown are those of a ketene. (which have CH3

C

C

O group). The reactions.

CH2

C

C

O + H2O

CH3

CH2

CH3

CH

CO2H

CH3 (carboxylic acid)

CH3

CH2

C

C

O + NH3

CH3

CH2

CH3 CH3

CH2

C

CH

CONH2

CH3

(amide)

C

CH3

O + CH3COOH

CH3

CH2

O CH

C

O O

C

CH3

CH3 (anhydride)

8. reaction by treatment with aluminium ethoxide. Under these conditions the acid and alcohol are combined to give as ester. This reaction is known as Tischenko reaction. For example aluminium

2 CH3 CHo ææææ Æ CH3COOH + CH3CH 2 oH Æ CH3COOCH 2 CH3 ethyl acetate ethoxide 9. Reimer - Tiemann Reaction

24.54 Complete Chemistry—JEE Main

O-

OH

OH CHO

CHCl3

CHO

H+

aq. NaOH 70°C

Friedal - Craft’s acylation COCH3 CH3COCl Al Cl3

Wurtz reaction RBr + 2Na + BrR Æ R - R + 2NaBr Cannizzaro reaction 50% NaCH

2 HCHO æææææÆ CH3OH + HCOO- Na In Cannizzaro reaction, no C – C bond is formed 10. A compound containing CH3

or CH3CH (OH)

C

O - shows iodoform test. The reaction is CH3COCH3 + 3NaOI

RCOCI3 + 3NaOH

(from I2 and NaOH)

alc. KOH

—

C6H5 — CH2 — C — CH3 Cl

—

CH3

CH3

C6H5 — CH — C — CH3 (B)

(A)

Ozonolysis

Cl gives white precipitate with AgNO3

C6H5CHO (C)

CH3

—

—

(Iodoform)

O— — C — CH3 (D)

+

aldehyde no a-hydrogen shows Cannizzaro reaction

ketone contains a-hydrogen shows aldol condensation

12. The compounds CH3 -- CH 2 -- CH 2 -- C -- CH3 and CH3 -- CH 2 -- C -- CH 2 -- CH3 are positional isomers. ΩΩ ΩΩ O O 13. The reactions are: 2CH3 -- CHO Æ CH3 -- CH -- CH 2 -- CHO æDæ Æ CH3 -- CH == CH -- CHO Ω (B) OH (A) 14. Erythritol is synthesised as follows. This requires three aldol condensations and one Cannizzaro reaction. OH CH3CHO

3HCHO OH3 aldol condensation

OHC HO

C

OH

HCHO OHCannizzaro reaction

HO HO

OH C

OH

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.55

Additional Information PETN stands for pentaerythritol tetranitrate. It is one of the most powerful explosive materials. It is produced by treating erythritol with concentrated nitric acid. OH OH

HNO3

C

ONO2

O2NO C

ONO2

OH HO

O2NO

(PETN)

SECTION 5

Carboxylic Acids

Methods of Preparation Oxidation of Primary Alcohols

R — CH2OH

Oxidation of Alkylbenzenes

C6H 5— R

KMnO 4 KMnO 4

Æ R — COOH

Æ C6H5— COOH

or K 2 Cr2 O 7

+ Carbonation of Grignard Reagent R—MgX + O==C==O Æ R— COO — MgX HÆ R— COOH + Mg2+ + X– RMgX may be prepared from RX by adding Mg.

R —C

Hydrolysis of Nitriles

acid

N + 2H2O

Æ R— COOH + NH3

or base

Nitriles may be prepared by adding NaCN to the corresponding halides. For example, CH3CH2CH2Br + CN– Æ CH3CH2CH2CN

H2O, H +

Æ CH3CH2CH2COOH

The above substitution of X with CN is applicable only when the halide is a primary halide. Tertiary halides yield alkene and for secondary halide, the yield of substitution product is poor. CH3 CH3

C

CH3

Br + CN

–

Æ CH3

C

CH2 + HCN

isobutylene

CH3 tert-butyl bromide

Aryl halides are unreactive towards substitution. Thus, aromatic nitriles are prepared via diazonium salt. For example, NH2

N NaNO2 Æ HCl

N+Cl–

CN

CuCN

Æ

COOH H+

Æ

Physical Properties The molecules of carboxylic acids are polar and exhibit hydrogen bonding. The boiling point of a carboxylic acid is higher than that of an alcohol of comparable molar mass. This is due to the fact that the carboxylic acids exist as dimer.

24.56 Complete Chemistry—JEE Main

O ....H R

O

C

C

R

O H .... O

Chemical Reactions Acidity Carboxylic acids are weak acids and their carboxylate anions are strong conjugate bases. The aqueous solutions of carboxylate salts are slightly alkaline due to the hydrolysis of carboxylate anion. Compared to other species, the orders of acidity and basicity of corresponding conjugate bases are as follows. CCH > NH3 > RH

RCOOH > HOH > ROH > HC

Acidity

CC < NH–2 < R– Basicity RCOO < HO < RO < HC The carboxylic acids react with metals to liberate hydrogen and are soluble in both NaOH and NaHCO3 solutions. For example –

–

2CH3

3COO

CH3

3COO

CH3COOH + NaHCO3

–

–

–

–

Na+ + H2

Na + H2O

3COO

–

Na+ + H2O + CO2

The acidity of a carboxylic acid is due to the resonance stabilization of its anion: O–

O R

C

R –

O

O equivalent to

C

R

C O

O

Because of the resonance, both the carbon-oxygen bonds in the carboxylate anion have identical bond length. In the carboxylic acid, these bond lengths are no longer identical. For example, in formic acid, one carbon-oxygen bond length is 136 pm (single bond) and another of 123 pm (double bond), whereas in the formate anion, both carbon-oxygen bonds have length equal of 127 pm (in between 123 pm and 136 pm). The acidity of carboxylic acid depends very much on the substituent attached to —COOH group. Since acidity is due to the resonance stabilization of anion, any substituent which stabilizes the anion increases acidity whereas substituent causing destablization of anion decreases acidity. For example, electron-withdrawing group disperses the negative charge of the anion and hence makes it more stable causing the increase in the acidity of the corresponding acid. On the other hand, electron-releasing group increases the negative charge on the anion and hence makes it less stable causing the decrease in the acidity. In the light of this, the following are the decreasing order of a few substituted carboxylic acids. 1. Increase in the number of chlorine atoms on a-position increases the acidity, e.g. Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH 2. Increase in the distance of Cl from —COOH decreases the acidity, e.g. CH3CH2CHCOOH > CH3CHCH2COOH > CH2CH2CH2COOH Cl

Cl

Cl

This is due to the fact that inductive effect decreases with distance. 3. Increase in the electronegativity of halogen increases the acidity. FCH2COOH > BrCH2COOH > ICH2COOH . The aromatic acids are similarly affected by substituent. Electron-releasing groups (—CH3, —OH, —NH2) make benzoic acid weaker. Electron-withdrawing groups (—Cl, —NO2) make benzoic acid stronger.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.57

Conversion to Acid Chloride This may be carried out by using thionyl chloride (SOCl2), phosphorus trichloride (PCl3) and phosphorus pentachloride (PCl5). Thionyl chloride is more convenient as the side products are only gaseous and thus the acid chloride can be easily separated; any excess of SOCl2 can be easily removed as its boiling point is low (79°C). RCOOH + SOCl 2 RCOOH + PCl 5

Conversion into Esters

RCOOH

SOCl 2

acid

reflux heat

Æ RCOCl + SO 2 + HCl

Æ RCOCl + POCl 3 + HCl R¢OH

Æ RCOCl

Æ RCOOR

acid chloride

ester

A direct reaction between acid and alcohol is a reversible one. Ester can be obtained either by using one of the reactants in excess or by removing one of the products. H+

RCOOH + R¢OH

RCOOR¢ + H2O

+

H acts as a catalyst.

CH3OH > 1° > 2° > 3° HCOOH > CH3COOH > RCH 2COOH > R 2CHCOOH > R3CCOOH OH

O

Mechanism

R

C

OH + H

+

R

C

O R

C

OR¢ + H

+

R

C

+

R¢O

OH H

R

C

+

OR

OH

OH H

OH

OH + H2O

R

C

OR¢

+ OR ¢ OH2 Mineral acid speeds up both processes by protonating carbonyl oxygen and thus rendering carbonyl carbon more ¢OH and leaving group is water and in the hydrolysis, the roles are reversed. Note that all steps are reversible. Forward reaction is favoured by H+ and the reverse reaction by excess of water.

Reduction of Acids to Alcohol Lithium aluminium hydride, LiAlH4, is used to convert acids into alcohols. The initial product is an alkoxide which on hydrolysis gives an alcohol. 4R— COOH + 3LiAlH 4 Æ 4H2 + 2LiAlO2 + (RCH2O)4 AlLi H O

2 Æ 4RCH2OH (RCH2O)4AlLi Alternatively, an ester may be converted into alcohol by the use of sodium metal and alcohol or LiAlH4. For example,

CH3(CH2)14COOC2H 5

ethyl palmitate

(i) LiAlH 4 (ii) H

Æ CH3(CH2)14CH2OH + C2H5OH

The reagent BH3/THF followed by H3O+ also reduces —COOH to —CH2OH. If the compound contains —CO—, —NO2 and —Br groups, LiAlH4 cannot be used as these are also reduced alongwith —COOH group. However, BH3/THF can be used as this does not affect these groups. If the compound contains C==C group, then BH3/THF cannot be used as this is rapidly reduced. However, LiAlH4 can be used. Halogenation of Aliphatic Acids (Hell-Volhard-Zelinsky reaction) In the presence of phosphorus, chlorine or bromine replaces a-hydrogen of an acid by halogen atom. For example,

24.58 Complete Chemistry—JEE Main

Br | Br2 , P Br2 , P Æ CH3CHCOOH Æ CH3— C— COOH CH3CH2COOH | | Br Br The halogen atom in the a-position can be easily replaced by another nucleophilic reagent. This provides a method of converting carboxylic acid into many important substituted carboxylic acids. For example, NH3

RCHCOOH

(excess)

Br

RCHCOOH NH2 an a -amino acid

NaOH

RCHCOONa

H+

RCHCOOH

OH

OH an a -hydroxy acid

alc. KOH

R¢CH

CHCOO–

H+

R¢CH CHCOOH an a , b -unsaturated acid

KCN

RCHCOOH

H2O

COOH RCH

+

H

COOH

CN

a dicarboxylic acid

Decarboxylation Heating of sodium salt of carboxylic acid with soda lime (NaOH + CaO) produces alkane RCOONa

NaOH H

Æ RH + Na2CO3

Heating of calcium salt of carboxylic acid produces the compound containing CO group. RCO2(Ca/2) + R¢CO2(Ca/2) RCOR¢ + Ca2CO3 If R is H, then aldehyde R¢CHO is produced.

Functional Derivatives of Carboxylic Acids The functional derivatives of carboxylic acids are acid chlorides (RCOCl), anhydride R (RCONH2) and esters RCOR¢ . They all contain acyl group R O Nucleophilic Substitution Reactions mechanisms are as follows. Basic medium

Acidic medium

O C

C O

O

C

R , amides

O

.

The acyl derivatives undergo nucleophilic substitution reactions. Their

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.59

Reactivity of Acyl Derivatives The reactivity of RCOG depends on the resonance stabilization of the –CO—G group.

The greater the degree of delocalization, the lesser reactive is RCOG, NH2 has the greatest degree of delocalization, thus RCONH2 is least reactive, while X has little or no delocalization and thus RCOX is the most reactive. Thus, the reactivity follows the order: Acid chloride > anhydride > easter > amide A more reactive derivative may be used to prepare a less reactive derivative by reaction with appropriate nucleophile. Incidently, the reactivity decreases as the base strength of the leaving group increases i.e. Cl– < RCOO– < RO– < H2N–. We describe in brief the method of preparation and the main reactions of the above functional derivatives.

Acid Chlorides (SOCl2 is the best as the unwanted products are gases.)

LiAlH4

RCH2OH

alcohol

Acid Anhydrides

O COOH 200°C

C

Æ

COOH

O C O phthalic anhydride

24.60 Complete Chemistry—JEE Main H2O

Æ 2RCOOH

H+ 2NH3

Æ RCONH2 + RCOO–NH4+

(RCO)2 O

R¢OH

Æ 2RCOOR¢ + RCOOH

ArH Æ 2RCOAr + RCOOH AlCl3

Cyclic anhydrides with NH3, R¢OH and ArH/AlCl3 gives only one product. O C

H2C

O + 2NH3 Æ

H2C

C

CH2CONH2

H+

Æ

CH2COONH4

CH2CONH2 CH2COOH

ammonium succinamate

succinamic acid

O O C

CH3OH

OCH3

O COOH

C O

O

C

C

C6H6

O

AlCl3

COOH o-benzoylbenzoic acid

Amides O R

C

NH 3

O

Æ R

C NH2

Cl

LiAlH4

RCH2NH2

Esters O R

+ R¢OH

C

O R

C

+ R¢— OH Æ R

C

OH

O R

C Cl

OR¢ O OR¢

+ H2O

+ HCl

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.61

O NO2 Æ CH3C

(CH3CO)2 + HO

NO2 + CH3COOH

O

p-nitrophenyl acetate

H2O

O

Æ R

C

+ R¢OH

OH (reversible reaction) OH–

NH3

O

Æ R

C

Æ R

C

O

O R

O–

+ R¢OH + R¢OH

NH2

C OR¢

R≤OH

O

Æ R

C

OR≤

+ R OH

(reversible reaction) transesterification 2R≤MgX

R≤

Æ R

C

R≤

OH H2

Æ RCH2OH + R¢OH

CuO-CuCr2O4 or LiAlH4

The reaction of RCOOR¢ with R¢¢MgX proceeds as follows -O(MgX)+

O R

C

OR¢

R¢¢MgX

R

C

-O(MgX)+

O

OR¢

R

C

R¢¢

R¢¢MgX

R

R¢¢

C

R¢¢

R¢¢

H+-(MgX)+

OH R

C

R¢¢

R¢¢

The reaction of RCN with R¢MgX gives a ketone: N -(MgX)+ R

C

N + R¢MgX

R

C

R¢

O hydrolysis

R

C

R¢

Here, a second equivalent of R¢MgX does not react because the intermediate imine salt bears a negative charge. Unlike the ester intermediate, where elimination of OR¢– leads to ketone, loss of the charge by elimination of R¢ merely reverse the reaction.

24.62 Complete Chemistry—JEE Main

MULTIPLE CHOICE QUESTIONS ON SECTION 5 General Characteristics and Preparations 1. The Hell-Volhard-Zelinsky reaction is used to synthesise (a) aldehydes (b) a-haloacids (c) acid halides (d) nitriles 2. The reagent used in the Hell-Volhard-Zelinsky reaction is (a) zinc (b) nickel (c) phosphorus (d) aluminium chloride 3. The —COOH group in benzene ring is (a) ortho directing (b) para directing (c) ortho and para directing (d) meta directing 4. Which of the following orders is true regarding the acidic nature of COOH? (a) Formic acid > Acetic acid > Propanoic acid (b) Formic acid > Acetic acid < Propanoic acid (c) Formic acid < Acetic acid < propanoic acid (d) Formic acid < Acetic acid > Propanoic acid 5. Which of the following orders is true regarding the acidic nature of COOH? (a) pK°a(HCOOH) > pK°a(CH3COOH) > pK°a(CH3CH2COOH) (b) pK°a(HCOOH) > pK°a(CH3COOH) < pK°a(CH3CH2COOH) (c) pK°a(HCOOH) < pK°a(CH3COOH) > pK°a(CH3CH2COOH) (d) pK°a(HCOOH) < pK°a(CH3COOH) < pK°a(CH3CH2COOH) 6. Which of the following belongs to – I group? (b) —CH2CH3 (c) —COOH (d) —CH(CH3)2 (a) —CH3 7. Which of the following orders is true regarding the acidic nature of acetic acid? (a) Acetic acid > Monochloroacetic acid > Dichloroacetic acid (b) Acetic acid < Monochloroacetic acid < Dichloroacetic acid (c) Acetic acid > Monochloroacetic acid < Dichloroacetic acid (d) Acetic acid < Monochloroacetic acid > Dichloroacetic acid 8. Which of the following orders is true regarding the acetic nature of monosubstituted acetic acid? (a) Fluoroacetic acid > Chloroacetic acid > Bromoacetic acid (b) Fluoroacetic acid < Chloroacetic acid < Bromoacetic acid (c) Fluoroacetic acid < Chloroacetic acid > Bromoacetic acid (d) Fluoroacetic acid < Chloroacetic acid < Bromoacetic acid 9. Which of the following orders is true regarding the boiling point of organic substances? (b) CH3COCl < CH3COOH < (CH3CO)2O (a) CH3COCl > CH3COOH > (CH3CO)2O (d) CH3COCl < CH3COOH > (CH3CO)2O (c) CH3COCl > CH3COOH < (CH3CO)2O 10. Which of the following is present in vinegar? (c) CH3CHO (d) CH3OH (a) HCOOH (b) CH3COOH 11. The oxidation products of 1-nitronaphthalene and a-naphthylamine respectively are (a) phthalic acid and 3-aminophthalic acid (b) 3-nitrophthalic acid and phthalic acid (c) phthalic acid and phthalic acid (d) 3-nitrophthalic acid and 3-aminophthalic acid 12. Which of the following orders of acid strength is correct? (b) RCOOH > HOH > ROH > HC CH (a) RCOOH > ROH > HOH > HC CH (c) RCOOH > HOH > HC CH > ROH (d) RCOOH > HC CH > HOH > ROH 13. Which of the following orders of base strength is correct? – – (a) R– > NH2– > HC C > RCOO– (b) R– > NH2– > RCOO– > HC C – – (c) R– > RCOO– > NH2– > HC C (d) HC C > NH2– > RCOO– > R–

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.63

14. Which of the following orders of acid strength is correct? (a) CH3CH2CHCOOH < CH3CHCH2COOH < CH2CH2CH2COOH | | | Cl Cl Cl (b) CH3CH2CHCOOH > CH3CHCH2COOH > CH2CH2CH2COOH | | | Cl Cl Cl (c) CH3CH2CHCOOH > CH2CH2CH2COOH > CH3CHCH2COOH | | | Cl Cl Cl (d) CH3CH2CHCOOH < CH2CH2CH2COOH < CH3CHCH2COOH | | | Cl Cl Cl 15. Acetic acid differs from formic acid in that (a) acetic acid is stable to heat (b) formic acid is stable to heat (c) acetic acid acts as a reducing agent (d) acetic acid shows positive test with Tollens reagent 16. Which of the following sequences is correct? (a) pK°a(p-O2NC6H4COOH) > pK°a(C6H5COOH) > pK°a (p-HOC6H4COOH) (b) pK°a(p-O2NC6H4COOH) < pK°a(C6H5COOH) < pK°a (p-HOC6H4COOH) (c) pK°a(p-O2NC6H4COOH) > pK°a(C6H5COOH) < pK°a (p-HOC6H4COOH) (d) pK°a(p-O2NC6H4COOH) < pK°a(C6H5COOH) > pK°a (p-HOC6H4COOH) 17. Which of the following statements is correct? (a) Formic acid like acetic acid can be halogenated in the presence of red phosphorus and Cl2. (b) Formic acid is a stronger acid than acetic acid. (c) The boiling point of propanoic acid is less than that of n-butyl alcohol, an alcohol of comparable molar mass. (d) The IUPAC name of CH3CH2CH2CN is propanenitrile. 18. Which of the following statements is not correct? (a) The melting point of a trans fatty acid is higher than its cis isomer. (b) The IUPAC name of CH3CH2 CH2 CH2 (CH2)7COOH C == C C == C C == C H H H H H H is cis,cis,cis-3,6,9-octadecatrien-18-oic acid. (c) Vinegar is a 5% aqueous solution of acetic acid. (d) The C==O bond in RCOOH is longer than in RCHO. 19. Which of the following statements is not correct? (a) The reactivity of C==O towards nucleophiles in RCOOH is less than that in RCHO. (b) Soft soaps are K+ salts of fatty acids. (c) Hard soaps are Na+ salts of fatty acids. (d) Peroxy acids, R COOH , are much stronger than R COH . || || O O 20. Which of the following statements is not correct? (a) H2C==CHCH2COOH is more acidic than CH3CH2COOH. (b) The acid (CH3)3CCH2C(CH3)2COOH is less acidic than CH3(CH2)7COOH. (c) Trans-4-t-butylcyclohexanecarboxylic acid is stronger than the cis-isomer (d) The pK°a1 of a dicarboxylic acid is usually greater than the pK°a of the comparable monoalkylcarboxylic acid.

24.64 Complete Chemistry—JEE Main

21. Which of the following statements is not correct? (a) Maleic acid is a weaker acid than fumaric acid. (b) Maleate monoanion is a weaker acid than fumarate monoanion. (c) Salicyclic acid, o-HOC6H4COOH, is a stronger acid than o-H3COC6H4COOH. (d) A more reactive acid derivative may be used to prepare a less reactive one by reaction with the appropriate nucleophile.

Chemical Reactions 22. The reaction of formic acid with concentrated H2SO4 gives (b) CO2 (c) HCHO (d) CO (a) CH3COOH 23. Which of the following reagent solutions can be used to distinguish between methanoic acid and ethanoic acid? (b) FeCl3 solution (a) Ammoniacal AgNO3 solution (d) NaOH solution (c) Na2CO3 solution 24. Acetic acid on reacting with ethanol in the presence of traces of sulphuric acid produces (b) C2H4 (c) CH3COOC2H5 (d) (CH3CO)2O (a) C2H6 (a) 1° < 2° < 3°

27. 28.

29.

30. 31.

32. 33.

34.

(b) 1° > 2° > 3°

(c) 1° > 3° > 2°

(d) 1° < 3° < 2°

(b) RCH2COOH < R2CHCOOH < R3CCOOH (a) RCH2COOH > R2CHCOOH > R3CCOOH (d) R3CCOOH > RCH2COOH > R2CHCOOH (c) RCH2COOH < R3CCOOH < R2CHCOOH The treatment of CH3CH2COOH with chlorine in the presence of phosphorus gives (b) CH3CH2CH2Cl (c) CH3CH(Cl)COOH (d) CH2(Cl)CH2COOH (a) CH3CH2COCl Which of the following reactants produces benzophenone on heating? (a) Calcium formate + Calcium benzoate (b) Calcium acetate + Calcium benzoate (c) Calcium benzoate (d) Calcium acetate + Calcium formate Which of the following reactants produces an aldehyde on heating? (a) Calcium formate + Calcium benzoate (b) Calcium acetate + Calcium benzoate (c) Calcium acetate (d) Calcium benzoate Which of the following reagent/solution can be used to distinguish between methanoic acid and ethanoic acid? (a) Tollens reagent (b) HCl solution (c) NaOH solution (d) NaHCO3 solution Which of the following reactants would produce acetophenone on heating? (a) Calcium acetate + Calcium formate (b) Calcium formate + Calcium benzoate (c) Calcium acetate + Calcium benzoate (d) Calcium acetate When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The C from CO2 comes from (a) methyl group (b) carboxylic acid group (c) methylene group (d) bicarbonate Which of the following statements is not correct? (a) The reduction of RCOOH with LiAlH4 produces RCH3. (b) The —COOH group attached to benzene ring is meta director. O || (c) In R—C— O –, both the carbon-oxygen bond lengths are identical. It is more than carbon-oxygen double bond length, and less than carbon-oxygen single bond length. (d) Electron-withdrawing substituents stabilises the carboxylate anion and hence increase the acidity of carboxylic acid. Which of the following statements is correct? (a) Alkaline hydrolysis of an ester is a reversible reaction. (b) In the alkaline hydrolysis of an ester, the bond broken is the bond between oxygen and alkyl group.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.65

(c) Alkaline hydrolysis of an ester follows second order kinetics and is thus SN2. (d) An electron-withdrawing group G in p-GC6H4COOC2H5 decreases the rate of hydrolysis as compared to C6H5COOC2H5.

Carboxylic Acid Derivatives 35. Acetamide on heating with phosphorus pentoxide gives (b) CH3NH2 (c) CH3CN (d) CH3CH2NH2 (a) CH3COOH 36. Acetamide reacts with Br2/NaOH to give (b) CH3NH2 (c) CH3Br (d) CH3CH2NH2 (a) CH3COOH 37. The treatment of an ester with LiAlH4 followed by acid hydrolysis produces (a) two alcohols (b) two acids (c) two aldehydes (d) one molecule of alcohol and another of carboxylic acid 38. Which of the following sequences of relative rates of alkaline hydrolysis of esters is correct? (a) CH3COOCH3 > CH3COOC2H5 > CH3COOC3H7 (b) CH3COOCH3 < CH3COOC2H5 < CH3COOC3H7 (c) CH3COOCH3 < CH3COOC2H5 > CH3COOC3H7 (d) CH3COOCH3 > CH3COOC2H5 < CH3COOC3H7 39. Which of the following sequences of relative rates of alkaline hydrolysis of esters is correct? (a) CH3COOC3H7 > CH3COOCH(CH3)2 > CH3COOC(CH3)3 (b) CH3COOC3H7 < CH3COOCH(CH3)2 < CH3COOC(CH3)3 (c) CH3COOC3H7 < CH3COOCH(CH3)2 > CH3COOC(CH3)3 (d) CH3COOC3H7 > CH3COOCH(CH3)2 < CH3COOC(CH3)3 40. Which of the following sequences of relative rates of alkaline hydrolysis of esters is correct? (a) HCOOCH3 < CH3COOCH3 < (CH3)2CHCOOCH3 (b) HCOOCH3 > CH3COOCH3 < (CH3)2CHCOOCH3 (c) HCOOCH3 < CH3COOCH3 > (CH3)2CHCOOCH3 (d) HCOOCH3 > CH3COOCH3 > (CH3)2CHCOOCH3 41. Which of the following sequences of rates of alkaline hydrolysis of esters is correct? (a) CH3CH2COOCH3 < (CH3)2CHCOOCH3 < (CH3)3CCOOCH3 (b) CH3CH2COOCH3 > (CH3)2CHCOOCH3 > (CH3)3CCOOCH3 (c) CH3CH2COOCH3 > (CH3)2CHCOOCH3 < (CH3)3CCOOCH3 (d) CH3CH2COOCH3 < (CH3)2CHCOOCH3 > (CH3)3CCOOCH3 42. Which of the following statements is not correct? (a) Carboxylic acids and their derivatives are collectively known as acyl compounds. (b) Acyl compounds (RCOW) undergo electrophilic substitution reaction in which W is replaced by some other basic group. (c) Acyl compounds undergo nucleophilic substitution reactions more readily than the compounds containing no carbonyl group. (d) The acyl compounds differ from aldehydes or ketones in the respect that the former undergo nucleophilic substitution reactions while the latter undergo nucleophilic addition reactions. 43. Which of the following statements is not correct? (a) The ease with which :W of the acyl compound RCOW is lost depends upon its basicity; the weaker the base, the better the leaving group. (b) Aldehydes or ketones undergo nucleophilic addition reactions instead of substitution reactions because the leaving group (:H– or :R–) are the strongest bases of all.

24.66 Complete Chemistry—JEE Main

followed by its reaction with ammonia or an alcohol. (d) Adipic acid HOOC(CH2)4COOH on heating produces cyclic adipic anhydride having structure. O

O C

O C CH2

CH2 CH2

CH2

44. Which of the following statements is correct? (a) An electron-releasing group G in p-GC6H4COOC2H5 decreases the rate of hydrolysis as compared to C6H5COOC2H5. (b) The rate of alkaline hydrolysis of RCOOR¢ increases with the increase in the carbon atoms in the alkyl group R¢. (c) The rate of alkaline hydrolysis of RCOOR¢ increases with the increase in the carbon atoms in the alkyl group R. (d) The rate of alkaline hydrolysis of RCOOR¢ decreases with the increase in the branching of carbon skeleton of the alkyl group R.

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43.

(b) (b) (a) (d) (b) (c) (a) (d)

2. 8. 14. 20. 26. 32. 38. 44.

(c) (a) (b) (d) (a) (d) (a) (a)

3. 9. 15. 21. 27. 33. 39.

(d) (b) (a) (a) (c) (a) (a)

4. 10. 16. 22. 28. 34. 40.

(a) (b) (b) (d) (c) (c) (d)

5. 11. 17. 23. 29. 35. 41.

(d) (b) (b) (a) (a) (c) (b)

6. 12. 18. 24. 30. 36. 42.

(c) (b) (b) (c) (a) (b) (b)

HINTS AND SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

The Hell-Volhard-Zelinsky reaction is used to synthesise a-haloacids. The reagent used is phosphorus. —COOH is electron-attracting group. It is thus meta director. Electron-releasing group weakens the acidic nature of COOH. The stronger the acid, the lesser the value of pK°a. —I group is electron-attracting group. Acidity increases with increase in the number of chlorine (electron-attracting group) atoms attached to methyl group. The larger the electronegativity of halogen, the larger the acidity of substituted acetic acid. CH3COCl does not involve hydrogen bonding. Its boiling point is lesser than CH3COOH. The boiling point of the latter is lesser than that of (CH3CO)2O due to larger molar mass of (CH3CO)2O. Vinegar contains acetic acid. Nitro bearing benzene ring is resistant to oxidation and amino bearing benzene is more susceptible to oxidation. The products are 3-nitrophthalic acid and phthalic acid, respectively.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.67

12. 13. 14. 15. 16.

Carboxylic acid is the strongest acid. R group in ROH makes it less acidic than H2O. Acetylene is the least acidic. The base strengths of the conjugate bases follow the reverse order of the acid strengths of acids. Farther Cl atom from COOH group, weaker the acid. Acetic acid is thermally more stable than formic acid. Electron-attracting group makes benzoic acid more strong acidic and electron-releasing group makes it less acidic. The pK°a follows the reverse order. 17. (a) Formic acid does not contain a-hydrogen atom. (c) The boiling point of carboxylic acid is higher than that of alcohol of comparable molar mass because of more extensive hydrogen bonding. (d) The —CN group is also counted in the carbon chain. Hence, its name is butanenitrile. 18. (a) A trans double-bond does not interfere with the uniform zig-zag and close packing is permitted. This makes its melting point higher than its cis isomer. (b) Counting starts with COOH group. The correct name is cis, cis, cis-9,12,15-octadecatrienoic acid. (d) Because of resonating structures the C==O bond acquires some single-bond character, making its length longer than in RCHO.

19. (a) In RCOOH there is positive charge on the C==O carbon causing this C to be less electrophilic and less reactive toward nucleophiles. (d) The conjugate base of RCOOH is stabilised due to the resonance effect while no such stabilisation occurs in the conjugate base of peroxy acid.

20. (a) The sp2 hybridised carbon of —C C— attracts the bonded electrons more than the sp3 hybridised carbon C— is an acid-strengthening electron-withdrawing group. does. Consequently. —C – (b) The —COO of the branched acid is shielded from solvent molecules and cannot be stabilised as effectively as can an unhindered anion. (c) In trans isomer, the —COO– is less shielded and is thus more stabilised by solvent. electron-withdrawing COOH group of dicarboxylic acid. Hence, pK°a1 of dicarboxylic acid is smaller than pK°a of monoalkylcarboxylic acid. 21. (a) The maleic acid is a stronger acid because of the stabilisation of its monoanion due to H-bonding between the cis COOH and COO–. This is not present in the trans isomer (fumarate anion).

where H-bonding is not present.

24.68 Complete Chemistry—JEE Main

(c) H-bonding in salicylate ion makes the salicyclic acid more acidic than o-H3COC6H4COOH.

–H O

2 Æ CO. 22. HCOOH 23. Methanoic acid (i.e. formic acid) reduces ammoniacal AgNO3 solution (Tollens reagent) while ethanoic acid does not. 24. There is formation of the ester ethyl acetate.

27. CH3CH2COOH

P4 Æ Cl2 CH3CH(Cl)COOH. This is Hell-Volhard-Zlinsky reaction.

28. 29. 30. 31. 32. 33.

Heating of calcium benzoate produces benzophenone. The use of calcium formate results into an aldehyde. Same as Q. 23. Acetophenone is produced on heating calcium acetate and calcium benzoate. CO2 comes from HCO3–. (a) The product obtained is RCH2OH. (c) This is due to the resonating effect amongst the two CO groups. (d) The stabilization is due to the dispersal of negative charge of the carboxylate anion. 34. (a) The reaction is

The reaction is irreversible because the resonance-stabilised carboxylate anion shows little tendency to react with an alcohol. (b) The bond broken is the bond between oxygen and acyl carbon. (d) An electron-withdrawing group stabilises the intermediate carboxylate anion, hence speeds up the rate of hydrolysis. 35. CH3CONH2 36. RCONH2 37. 38. 39. 40. 41. 42.

P2O5 Æ CH3CN. –H2O

OBr–

Æ RNH2 + CO32–. LiAlH

4 Æ Two alcohols are produced. RCOOR¢ RCH2OH + R¢OH. Increasing bulkiness of R¢ group in RCOOR¢ decreases the rate of alkaline hydrolysis of ester. Increasing crowdiness of R¢ in RCOOR¢ decreases the rate of alkaline hydrolysis of ester. Increasing bulkiness of R group in RCOOR¢ decreases the rate of alkaline hydrolysis of ester. Increasing crowdiness of R group in RCOOR¢ decreases the rate of alkaline hydrolysis of ester. (a) The acyl compounds contain RCO group.

(b) Because of

group, acyl compounds undergo nucleophilic substitution reactions because carbonyl

carbon is slightly positively charged.

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.69

(d)

43. (c) As both the steps are fast and essentially irreversible. Instead of producing cyclic anhydride, adipic acid loses carbon dioxide and form cyclopentanone. 44. (a) An electron-releasing group destabilises the intermediate carboxylate anion, hence decreases the rate of hydrolysis. (b) The steric effect and the electron-releasing effect of alkyl group decreases the rate of hydrolysis. (c) The steric effect will cause the decrease in the rate of hydrolysis.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. When CH2 CH—COOH is reduced with LiAlH4, the compound obtained will be (a) CH3CH2CHO (b) CH3CH2COOH (c) CH2=CHCH2OH (d) CH3CH2CH2OH [2003] 2. Rate of the reaction O O Æ R C R C + Nu + Z Z Nu is fastest when Z is (a) Cl (b) NH2 3. Consider the acidity of the carboxylic acids: (i) PhCOOH (ii) o-NO2C6H4COOH Which of the following order is correct? (a) i > ii > iii > iv (b) ii > iv > iii > i

(c) OC2H5 (iii) p-NO2C6H4COOH (c) ii > iv > i > iii

(d) OCOCH3

[2004]

(iv) m-NO2C6H4COOH (d) ii > iii > iv > i [2004]

4. On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is (b) CH3COONa + C2H5OH (a) CH3COOC2H5 + NaCl (c) CH3COCl + C2H5OH + NaOH (d) CH3Cl + C2H5COONa [2004] 5. Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl gives (a) acetone (b) acetamide (c) 2-methyl-2-propanol (d) acetyl iodide [2004] 6. Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon? (a) ethyl acetate (b) acetic acid (c) acetamide (d) butan-2-one [2004] 7. Which one of the following undergoes reaction with 50% sodium hydroxide solution to give the corresponding alcohol and acid? (a) phenol (b) benzaldehye (c) butanol (d) benzoic acid [2004] 8. Among the following acids which has the lowest pKa value? (b) CH3CH2COOH (c) CH3COOH (d) HCOOH [2005] (a) (CH3)2CHCOOH

24.70 Complete Chemistry—JEE Main

9. The reaction RCOX + Nu

Æ RCONu + X –

is fastest when X is (b) –OCOR (c) Cl– (a) –OC2H5 10. The increasing order of the rate of HCN addition to compounds (A) HCHO, (B) CH3COCH3, (C) PhCOCH3, and (D) PhCOPh is (a) C < D < B < A (b) A < B < C < D (c) D < B < C < A 11. The correct order of increasing acid strength of the compounds Me (A) CH3CO2H, (B) MeOCH2CO2H, (C) CF3CO2H, and (D) Me (a) A < D < C < B (b) B < D < A < C (c) D < A < C < B 12. In the following sequence of reactions, CH3CH2OH

P + I2

Æ A

Mg Æ B ether

HCHO

Æ C

H 2O

(d) –NH2

[2005]

(d) D < C < B < A [2006] CO2H is

(d) D < A < B < C [2006] Æ D

the compound ‘D’ is: (a) butanal (b) n-butyl alcohol (c) n-propyl alcohol 13. The compound formed as a result of oxidation of ethyl benzene by KMnO4 is: (a) benzophenone (b) acetophenone (c) benzoic acid

(d) propanal

[2007]

(d) benzyl alcohol

[2007] 14. Which of the following reagents may be used to distinguish between phenol and benzoic acid? (b) Aqueous NaOH (c) Tollens reagent (d) Molish reagent (a) Neutral FeCl3 [2011 cancelled] 15. The strongest acid amongst the following compound is (c) HCOOH (d) CH3CH2CH(Cl)CO2H (a) ClCH2CH2CH2COOH (b) CH3COOH [2011 cancelled] 16. An organic compound A upon reacting with NH3 gives B. On heating B gives C. The compound C in presence of KOH reacts with Br2 to give CH3CH2NH2. The compound A is (b) CH3CH2CH2COOH (a) CH3COOH (d) CH3CH2 COOH

(c) LiAlH

PCl

alc.KOH

5 4 Æ A æææ Æ B ææææ Æ C , the product C is 17. In the reaction CH3COOH ææææ (a) acetyl chloride (b) acetaldehyde (c) acetylene 18. Phthalic acid reacts with resorcinol in the presence of concentrated H2SO4 to give: (a) Phenolphthalein (b) Alizarin (c) Coumarin 19. Among the following organic acids, the acid present in rancid butter is: (a) Pyruvic acid (b) Lactic acid (c) Butyric acid 20. COOK

Electrolysis COOK A is:

A

[2013]

(d) ethylene

[2014]

(d) Fluorescein

[2014]

(d) Acetic acid

[2014]

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.71

(b)

(a)

(d) (c) n

[2015, online] 21. In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with chlorine or bromine to yield a compound in which a-hydrogen has been replaced by halogen. The reaction is known as (a) Wolff-kischner reaction (b) Etard reaction (b) Hell-Volhard-zelinsky reaction (d) Rosenmund reaction [2015, online]

ANSWERS 1. 7. 13. 19.

(c) (b) (c) (c)

2. 8. 14. 20.

(a) (d) (a) (b)

3. 9. 15. 21.

(d) (c) (d) (c)

4. (a) 10. (d) 16. (d)

5. (c) 11. (d) 17. (d)

6. (d) 12. (c) 18. (d)

HINTS AND SOLUTIONS 2. The weaker the base replaced, faster the rate of reaction. Of the given leaving species, Cl– is the weakest base and thus show the fastest rate of reaction. Therefore, the choice a is correct. 3. Electron withdrawal group is acid strengthening group. Due to ortho effect, o-NO2C6H4COOH is the strongest acid. p-NO2C6H4COOH is more acidic than m-NO2C6H4COOH, because the former generates positive charge on carbon bearing —COOH group whereas the latter generates positive charge on the next carbon atoms. –

O

+

N

O

O +

C

–

O

C O

More effective in displacing H+

+

O

N+

–O

O– Less effective in displacing H+

o-NO2C6H4COOH > p-NO2C6H4COOH > m-NO2C6H4COOH > C6H5COOH (ii) (iii) (iv) (i) 4. The choice a is correct (i.e. CH3COOC2H5 + NaCl) as no reaction occurs when ethyl acetate is mixed with aqueous sodium chloride. 5. The reaction is OH Thus, the correct order is

CH3COBr

(i) excess of CH 3 MgI (ii) saturated solution of NH 4 Cl

Æ CH3

C

CH3

CH3 2-methyl-2-propanol

24.72 Complete Chemistry—JEE Main

6. The reaction is Zn HCl

Æ CH3CH2CH2CH3

CH3COCH2CH3 butan-2-one

n-butane

7. An aldehyde containing no a hydrogen atom undergoes Cannizzaro reaction: 2C6H5CHO

50% NaOH

Æ C6H5CH2OH + C6H5COOH

benzaldehyde

benzyl alcohol

benzoic acid

8. The lowest pKa implies the highest Ka, which, in turn, implies the strongest acid. Of the given acids, HCOOH is the strongest acid. 9. The weaker the base-leaving group, the faster the nucleophilic displacement. 10. Increase in crowding at the transition state decreases the rate of addition of HCN 11. Electron releasing group decreases the acid strength while electron attracting group increases the acid strength. 12. The reactions are as follows. P + I2

CH3CH2OH

Mg Æ CH3CH2MgI ether (B)

Æ CH3CH2I (A)

H

HCHO + CH3CH2MgI

Æ H

C

OMg I

H2O

Æ CH3CH2CH2OH

CH2CH3 (C)

13. The reaction is CH2CH3

COOH KMnO4 [O] (Benzoic acid)

14. Neutral FeCl3 gives violet colouration with phenol while pale dull yellow precipitates are obtained with benzoic acid. 15. The electron-attracting Cl atom attached to carbon at a-position in carboxylic acid enhances the stability of its conjugate base and thus increases the acidicity of carboxylic acid. 16. We have

17. The reaction are CH3COOH

LiAlH4

CH3CH2OH

PCl5

CH3CH2Cl

alc. KOH

H2C CH2 ethylene

Organic Compounds Containing Oxygen (Alcohols, Phenols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 24.73

18. HO

OH H

HO O C

OH

H

HO H

O

OH O

CO

O CO

Fluorescein

19. Butyric acid is present in the rancid butter. 20. Anodic reaction during electrolysis is COO– + 2CO2 + 2e– COO–

21. Reaction is known as Hell-Volhard-Zelinsky reaction.

25 Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines) SECTION 1

Cyanides and Isocyanides

The general formula of alkyl/aryl cyanide is R—C ∫∫ N while that of isocyanide is R—N Æ C. These are known as nitriles and isonitriles, respectively. In the IUPAC system of nomenclature, the cyanides are named as alkanenitriles. The chain is numbered with number 1 is assigned to —C ∫∫ N carbon. Isocyanides are named according to the common system of nomenclature by placing the word isocyanide or carbylamine after the name of akyl/aryl group. Alkyl cyanides are prepared by SN2 reactions involving alkyl halide and an aqueous ethanolic solution of sodium or potassium cyanide. These are also prepared by the dehydration of acid amides by using P2O5 or POCl3 or SOCl2. Aryl cyanides are prepared via Sandmeyer reaction in which arenediazonium chloride is treated with KCN in presence of cuprous cyanide. Nitriles are hydrolyzed by aqueous acid or base to produce carboxylic acid. Reduction with LiAlH4 produces amine. Treatment with diisobutylaluminium hydride produces an imine derivative which on hydrolysis produces aldehyde. The —CN or —NC group is meta director in aromatic substitution reactions.

MULTIPLE CHOICE QUESTIONS ON SECTION 1 Cyanide and Isocyanide 1. The IUPAC name of CH3CH2CN is (a) 1-cyanoethane (b) ethyl cyanide 2. The IUPAC name of CH3CH2NC is (a) isocyanoethane (b) ethyl isonitrile 3. The hydrolysis of isocyanide produces (a) acid (b) amide 4. The partial hydrolysis of a nitrile produces (a) acid (b) amide 5. The complete hydrolysis of a nitrile produces (a) acid (b) amide

(c) ethylnitrile

(d) propanenitrile

(c) ethyl isocyanide

(d) propanenitrile

(c) amine

(d) ester

(c) amine

(d) ester

(c) amine

(d) ester

25.2 Complete Chemistry—JEE Main

6. The reduction of a nitrile by LiAlH4 produces (a) primary amine (b) secondary amine 7. The reduction of an isocyanide by LiAlH4 produces (a) primary amine (b) secondary amine

(c) tertiary amine

(d) amide

(c) tertiary amine

(d) amide

ANSWERS 1. (d) 7. (b)

2. (b)

3. (c)

4. (b)

5. (a)

6. (a)

MULTIPLE CHOICE QUESTION FROM AIEEE AND JEE MAIN 1. Ethyl isocyanide on hydrolysis in acidic medium generates (a) methylamine salt and ethanoic acid (b) ethylamine salt and methanoic acid (c) propanoic acid and ammonium salt (d) ethanoic acid and ammonium salt

[2003]

ANSWER 1. (b)

HINT AND SOLUTION H O

2 Æ C2H5NH2 + HCOOH 1. C2H5NC æææ +

H

SECTION 2

Nitro Compounds

The nitro compounds are named as nitro derivatives of the corresponding alkane with the lowest number assigned to the carbon bearing nitro group. Nitroalkane may be prepared by treating alkyl halide with silver nitrite in aqueous ethanolic solution. This method is useful for primary nitroalkanes. The yield is poor for sec- and tert-nitroalkanes. Aromatic nitro compounds are prepared by nitration of benzene ring with a mixture of concentrated HNO3 and concentrated H2SO4. The reduction of nitro group in nitroalkanes depends on the condition of reduction. LiAlH4

RCH2NO2

Zn dust

RCH2NH2 RCH2NHOH

NH4Cl soln. SnCl2/HCl

RCH2NHOH + RCH

The reduction of nitrobenzenes proceeds through the following stages.

NOH

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.3

– O NaOH/CH3OH

[PhNO + PhNHOH] Fe

PhNO2

Zn/CH3OH/NaOH

Ph N = N Ph

Zn/NaOH Zn/NaOH

PhN = N+Ph

CH3CO2H

SnCl2 HCl

NaIO 3

2PhNH2

Ph NHNHPh

In strongly acidic medium, the product is p-aminophenol formed by the rearrangement of phenylhydroxylamine. NO2

NHOH metal/conc. HCl

OH rearrangement

NH2

Nitroalkanes reacts with nitrous acid, the end product depends on the nature of nitroalkane.

Nitrolic acid

Nitro group is meta directing in aromatic substitution reactions.

MULTIPLE CHOICE QUESTIONS ON SECTION 2 1. The reduction of nitrobenzene with zinc in the presence of NaOH produces (a) aniline (b) hydrazobenzene (c) azobenzene (d) phenylhydroxylamine 2. The reduction of nitrobenzene with zinc in the presence of methanolic alkali produces (a) aniline (b) hydrazobenzene (c) azobenzene (d) phenylhydroxylamine 3. The reduction of nitrobenzene using zinc/tin in the presence of hydrochloric acid produces (a) aniline (b) hydrazobenzene (c) azobenzene (d) phenylhydroxylamine 4. The reduction of nitrobenzene in a neutral medium (by a reducing agent like zinc in aqueous NH4Cl) produces (a) aniline (b) hydrazobenzene (c) azobenzene (d) phenylhydroxylamine 5. The hydrolysis of CH3CH2NO2 with 85% H2SO4 gives (b) CH3COOH (c) CH3CH = NOH (d) C2H6 (a) CH3CH2OH 6. The major product of the reaction between m-dinitrobenzene with NH4SH is

25.4 Complete Chemistry—JEE Main

NH2

NO2

(a)

NHSH

(b)

NHSH

(c)

(d)

NH2

NH2

NO2

NHSH

7. In strongly acidic medium, phenylhydroxyl amine is converted into (a) aniline (b) phenol (c) p-aminophenol 8. Primary nitro compound on reacting with nitrous acid gives (a) nitrolic acid (b) pseudonitroles (c) no new compound

(d) anilinium salt (a) a dintro compound

ANSWERS 1. (b) 7. (c)

2. (c) 8. (a)

3. (a)

4. (d)

SECTION 3

5. (b)

6. (a)

Amines

Methods of Preparation Reduction of Nitro Compounds Reduction can be done by using either molecular hydrogen and a catalyst (Ni or Pt) or a metal (usually granulated tin) and an acid (hydrochloric acid). For example, NO2

NH2 H2/Ni or Pt or Sn/HCl

Reaction of Halides with Ammonia + – 3X

RX + NH3 RNH+3 X– + OH–

2

+ H 2O + X –

Replacement of X by NH2 is a nucleophilic substitution reaction. Tertiary alkyl halide generally produces alkene. Aryl halides show low reactivity. These can be converted into amine if a strong deactivating group, such as —NO2, is also attached to the ring at ortho or para position to halogen. During the ammonolysis of halides, ammonia is taken in excess so as to avoid the formation of other amines (2° and 3°) RX

NH3

Æ RNH2

RX

Æ R2NH

(1°)

RX

Æ R3N

(2°)

(3°)

Hofmann Degradation of Amides O R

C

OBr–

NH2

R

NH2 + CO 2– 3 (1°)

:

The reaction involves migration of a group from carbonyl carbon to the adjacent nitrogen atom. The reaction is believed to proceed by the following steps. O O – + OBr Æ + OH – R C R C (a) Halogenation of amide N Br NH2 H

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

O R

C

:

(b) Abstraction of H+ by OH–

N

Br

25.5

O

+ OH– Æ R

C

N

Br

+ H2O

H

O

(c) Separation of halide ion

Æ R

R

C

(d) Rearrangement to produce isocyanate

R

C

(e) Hydrolysis of isocyanate

R—N==C==O + 2OH –

N

O N

Br

Æ R

N

O C

N

+ Br –

O

C

H2 O

Æ R— NH2 + CO 2– 3

Reduction of Nitriles and Amides R— X

R — COOH

NaCN

1. SOCl 2. NH

1. LiAlH4 , ether

Æ R— CN

2. H 2 O

Æ RCH2NH2

O

Æ R

2

1. LiAlH 4 , ether Æ

C NH2

2. H 2 O

3

RCH2NH2

Physical Properties Amines are polar compounds. Except tertiary amines, primary and secondary amines form intermolecular hydrogen bonding. Amines have lower boiling points than alcohols and carboxylic acids of comparable molar masses.

Chemical Reactions Basic Nature of Amines Nitrogen of amines contains lone pair of electrons which can be shared with other species and thus these act as Lewis bases. Amines form salts with acids. For example, NH+3 Cl–

NH2 + HCl aniline

anilinium chloride

Aliphatic amines are somewhat stronger bases than ammonia. This is due to the fact that alkyl group is electronreleasing group which, in turn, increases the availability of lone pair of electrons. On the other hand, aromatic amines are considerably weaker than ammonia. This is due to the electron-withdrawing nature of phenyl group which decreases the availability of lone pair of electrons. Alternatively, the more stable the ion relative to the amine from which it is formed, the more basic the amines. For example, the ion formed in the reaction H H R

N + H+

R

N+ H

H H is stablized by alkyl group and hence increases basicity of RNH2 as compared to NH3. Two factors operate in deciding the basicity of alkylamines in aqueous medium. (i) Inductive Effect The alkyl group being electron releasing increases the charge density on nitrogen. This in turn increases the basicity of amines. The expected order of basicity is R3N > R2NH > RNH2 > NH3

(ii) Solvation Effect Because of the positive charge carried by the conjugate acid of an amine, it is stabilized by the hydrogen bonding with the solvent water. The larger the number of hydrogens attached to the nitrogen in the

25.6 Complete Chemistry—JEE Main

conjugate acid, the larger is its stability and thus larger is the basicity of the corresponding base. The expected order of basicity of the alkylamines will be NH3 > RNH2 > R2NH > R3N The inductive and solvent effects predict the opposite trend in the basiscity of alkylamines. In going from R2NH to R3N, the solvation effect plays a more dominating role as compared to the inductive effect making R2NH more basic than R3N. In solution, the order is R2NH > RNH2 > R3N > NH3. A simple explanation is that the steric factor in R3N makes the availability of a lone pair of electrons on nitrogen poor than in the dialkylamine, predicting R2NH a stronger base than R3N. In aromatic amines, the resonance stabilization of ArNH2 is larger than ArNH3+ , because there are more resonating structures of ArNH2 as compared to those of ArNH+3: +

+

NH2

NH2

+

NH2

NH2

–

+

NH2 –

NH3

+

NH3

;

–

Effect of substituents on basicity of aromatic amines Electron-releasing group pushes electrons toward nitrogen and hence increases the availability of lone pair of electrons and thus increases its basicity. On the other hand, electronwithdrawing group decreases the availability hence makes the amine less basic. In terms of the stability of ion, electronreleasing group stabilizes cation and hence increases basicity whereas electron-withdrawing group destabilizes cation making the amine less basic. For example G

NH2 is more basic if G is —NH2 or —OCH3 or —CH3 and

is less basic if G is —NH+3 or —NO2 or —SO–3 or —COOH or —X. It may be pointed out that the basicity of an amine and acidity of a carboxylic acid are affected by a substituent in the opposite manner. This is to be expected, since basicity depends upon ability to accommodate a positive charge (or ability to share lone pair) and acidity depends upon ability to accommodate a negative charge (or to accept lone pair of electrons).

Alkylation The alkylation of RNH2 produces quaternary ammonium salt. RNH2 (1°)

RX

Æ R2NH

RX

Æ R3N

(2°)

RX

Æ R4N+X –

(3°)

quaternary ammonium halide

Treatment of quaternary ammonium halide with aqueous Ag2O produces 4° ammonium hydroxides (which is very strong base like NaOH). 2R4N+X– + Ag2O + H2

4N

+

OH– + 2AgX

On heating quaternary hydroxide, less substituted alkene (Holfmann product) is obtained. Æ (CH3)3N + H2C== CHCH2CH3 + H2O [(CH3)3 NCH(CH3)CH2CH3]+OH– æDæ A less substituted alkene is formed as a result of the loss of more acidic bH (which follows the trend 1° > 2° > 3°).

Conversion of Amines into Substituted Amides Primary and secondary amines react with acid chloride to give substituted amides.

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.7

Tertiary amines do not react this way as they contain no replaceable hydrogen. Schotten-Baumann Technique to prepare substituted amides of aromatic carboxylic acid. In this method acid chloride is added to the amine in the presence of a base (aqueous NaOH or pyridine).

NH2 + Cl

base

Æ

C

NH

O

C O

benzanilide

Substituted amides undergo hydrolysis on heating in the presence of acid or base to give amine and carboxylic acid. Acetylation of aniline is generally carried out by using acetic anhydride instead of acetyl chloride.

NH2 + (CH3CO)2 O

CH3COONa

NHCOCH3 + CH3COOH

Ring Substitution in Aromatic Amines The —NH2, —NHR and —NR2 are benzene activating groups through resonance effect of nitrogen where the lone pair of nitrogen is shifted to the benzene ring making ortho and para positions available for electrophilic attack. +

NH2

+

NH2

NH2

+

H

NH2

H

H

The carbocation formed as intermediate are +

+

NH2

NH2

H Y

and H

Y

In these structures, nitrogen atom carries a positive charge and is joined to the benzene ring by a double bond. Such structures are quite stable as each atom (except hydrogen) has a complete octet of electrons. The group —NHCOCH3 is less powerful ortho and para director because of the electron-withdrawing characteristic of oxygen makes nitrogen a poor source of electrons. This fact is made use in preparing mono substituted aniline. The —NH2 group is such a powerful activator, that substitution occurs at all available ortho and para positions of aniline. If, however, —NH2 group is converted to —NHCOCH3, the molecule becomes less powerful activator. Hence, only mono substitution products are obtained. Finally, —NHCOCH3 is converted back to —NH2 by hydrolyzing with acid. This technique is especially used while nitrating aniline as strong oxidizing agent destroys the highly reactive ring. NH2

NHCOCH3 Ac2O

NHCOCH3

NH2

H2O, H+

HNO3 H2SO4

NO2

NO2

25.8 Complete Chemistry—JEE Main

If acetylation is not done, nitration of aniline gives about two-thirds meta and one-third para products, besides loss of aniline. This is due to the fact that the strong acid converts aniline to anilinium ion: +

NH2

NH3

+ H+

Now the group —+NH3 is the meta director because of its positive charge.

Reactions of Amine with Nitrous Acid In the reacting mixture, nitrous acid is generated by the action of mineral acid on sodium nitrite. The products obtained depend on the type of amine: Aliphatic primary amine It forms unstable diazonium salt which on decomposing liberates nitrogen and mixture of alcohols and alkenes. + – H2O R—NH2 + NaNO2 Æ N2 + mixture of alcohols and alkenes 2 X ] æææ Nitrogen obtained is quantitative and hence can be used to determine the strength of amine. Aromatic primary amine It forms stable diazonium salt. NH2 + NaNO2 + 2HX

cold

+ – N X + NaX + 2H2O

N

Aliphatic and Aromatic Secondary Amines Both of these produce N-nitrosoamines. CH3

CH3

N H + NaNO2 + HCl Æ

N

N

O + NaCl + H2O

N-nitroso-N-methylaniline

Tertiary Aromatic Amine This undergoes ring substitution, to yield a compound in which in nitroso group, —N = O, is joined to carbon: NaNO2, HCl

(CH3)2N

0 – 10° C

(CH3)2 N

N

O

In this reaction, the electrophile —+NO (or H2O+—NO) replaces H+ of benzene.

Synthetic Use of Diazonium Salt One of the best general ways of introducing F, Cl, Br, I, CN, OH and H into an aromatic ring is via diazonium salt. The freshly prepared diazonium salt (primary aromatic amine in cold aqueous mineral acid + sodium nitrite) is treated with other reagents to get the required product. (a) The replacement by —Cl or —Br can be carried out by either of the following ways: Sandmeyer Reaction The diazonium salt is treated with cuprous chloride or cuprous bromide

Ar -- N +2 X - æææ Æ Ar -- X + N 2 CuX

Gattermann Reaction In this, copper powder and hydrogen halide are used in place of cuprous halide. (b) The replacement by —I is achieved by simply adding KI to diazonium salt Ar—N+2 X– + I–

2

(c) The replacement by —F is achieved as follows

+ X–

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

Ar — N2+ X –

HBF4

Æ Ar — N2+ BF –4

heat

25.9

Æ Ar — F + BF3 + N2

stable salt, filtered and dried

(d) The replacement by —CN is carried out by adding cuprous cyanide to diazonium salt Ar -- N +2 X - æææÆ Ar -- CN + N 2 CuCN

Hydrolysis of nitrile produces carboxylic acid. (e) The replacement by —OH is carried out by adding water to diazonium salt. + Ar—N+2 X– + H2 2+H To avoid coupling of Ar—OH with Ar—N+2 X– (which form azo compounds by coupling), the latter is slowly added to large volume of water. (f) The replacement by —H is achieved by the reducing agent hypophosphorous acid, H3PO2. Ar—N+2X– + H3PO2 + H2

2

+ H3PO3 + HX

Hofmann Elimination Amines can be made to undergo an elimination reaction under suitable conditions to yield alkenes. The reaction is known as elimination reaction. The amine is methylated with excess of methyl iodide, yielding a quaternary ammonium iodide. The latter produces an alkene when heated with silver oxide. The less substituted alkene in contrast to the more substituted stable alkene predominates in this reaction. For example, 1-butene is formed from butylamine. +

CH I

3 → CH CH CH CH N (CH ) I– CH3CH2CH2CH2NH2 ⎯(excess) ⎯⎯ 3 2 2 2 3 3

Ag2O, H2O –Agl +

CH3CH2CH = CH2 ←⎯ CH3CH2CH2CH2N (CH3)3– OH– Δ⎯ 1-butene

Hofmann Rule In Hofmann elimination, an alkene is formed as a result of the loss of more acidic bH (which follows the trend 1° > 2° > 3°). Since the amine is relatively poor leaving group, the transition state involves more C—Hb bond breaking. Thus, the acidity of the bH becomes the deciding factor in the formation of the double bond.

Separation of a Mixture of Amines The mixture of the amine salt and quaternary salt is distilled with KOH solution. The three amines distill over leaving the quaternary salt behind. The three amines can be separated by either of the two methods given below.

Hinsberg Method The mixture of three amines is treated with Hinsberg reagent (benzenesulphonyl chloride, C6H5SO2Cl). The primary and secondary amines react with Hinsberg reagent while tertiary does not. The reactions are

After the reactions are over, the solution is made alkaline with NaOH. The sulphonamide of primary amine is soluble in NaOH (due to the acidic hydrogen attached with the nitrogen) while that of secondary amine is insoluble (as there is no acidic hydrogen attached to nitrogen). So after treating with NaOH, two layers are formed; aqueous phase containing soluble salt of primary amine

25.10 Complete Chemistry—JEE Main

(C6H5SO2NR) and second layer containing insoluble salt of secondary amine and tertiary amine as such. The two Na layers are separated with the help of ether. The aqueous phase is hydrolysed with concentrated HCl when the primary amine is regenerated. The ether layer is distilled and tertiary amine is distilled over. The residue is hydrolysed with concentrated HCl to recover secondary amine.

Hofmann’s Method The mixture of amines is treated with diethyloxalate which forms a solid oxamide with primary amine, a liquid oxime

remaining liquid is treated with KOH and distilled when the secondary amine is distilled over. The reactions occurring are as follows. COOC2H5 COOC2H5

+ 2RNH2

Æ CONHR CONHR

+ 2C2H5OH;

COOC2H5 COOC2H5

(dialkyloxamide) solid

+ R2NH

Æ CONR2

+ C2H5OH COOC2H5

(oxamic ester) liquid

Test for Primary Amine (Carbylamine Reaction) The treatment of a primary amine with chloroform and alcoholic potash produces carbylamine (isocyanide) which has most offensive smell: RNH2 + CHCl3 2O This reaction is not exhibited by secondary and tertiary amines. Dye Test for aromatic primary amines Cool in ice 0.2 g (or 4-5 drops) of the organic compound with 1 mL concentrated HCl. Add 2 mL of iced-cold dilute NaNO2 solution. Shake and add 2-3 mL of alkaline b-naphthol solution. A red or orange dye indicates the presence of aromatic —NH2 group

Test for Secondary Amine (Libermann Reaction) The secondary amine is converted into nitrosoamine by treating the amine with nitrous acid. The resultant solution is warmed with phenol and concentrated H2SO4 to green. The colour changes to red on dilution and further changes to greenish-blue or violet on treating with alkali.

MULTIPLE CHOICE QUESTIONS ON SECTION 3 General Characteristics of Amines 1. Which of the following is a primary amine? (b) CH3NHCH3 (c) (CH3)3N (d) (CH3)4N+ Cl– (a) CH3CH2NH2 2. Activation of benzene by —NH2 group can be reduced by treating the compound with (a) acetic acid (b) acetyl chloride (c) dilute HCl (d) methyl alcohol 3. Which of the following orders is true regarding the basic nature of NH2 group? (a) o-Toluidine > Aniline > o-Nitroaniline (b) o-Toluidine < Aniline > o-Nitroaniline (c) o-Toluidine < Aniline < o-Nitroaniline (d) o-Toluidine > Aniline < o-Nitroaniline 4. Which of the following order of basic strength is correct for the amines in aqueous phase? (b) (CH3)2NH > (CH3)3N > CH3NH (a) (CH3)2NH > CH3NH > (CH3)3N (d) (CH3)3N > CH3NH2 > (CH3)2NH (c) (CH3)3N > (CH3)2NH > CH3NH2 5. Which of the following orders amongst amines in the gaseous state is true regarding the basic nature of NH2 group?

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.11

(a) CH3NH2 > (CH3)2 NH > (CH3)3 N (b) CH3NH2 > (CH3)2 NH < (CH3)3 N (d) CH3NH2 < (CH3)2 NH < (CH3)3 N (c) CH3NH2 < (CH3)2 NH > (CH3)3 N 6. Which of the following order is true regarding the acidic nature of COOH group? (a) Benzoic acid > o-Aminobenzoic acid > o-Nitrobenzoic acid (b) Benzoic acid < o-Aminobenzoic acid < o-Nitrobenzoic acid (c) Benzoic acid > o-Aminobenzoic acid < o-Nitrobenzoic acid (d) Benzoic acid < o-Aminobenzoic acid > o-Nitrobenzoic acid 7. Among the following phenols, the most acidic is (a) p-aminophenol (b) p-nitrophenol (c) o-chlorophenol (d) m-nitrophenol 8. The correct order of basicity of the given compounds is (b) CH3CH2NH2 < HO(CH2)3NH2 > HO(CH2)2NH2 (a) CH3CH2NH2 < HO(CH2)3NH2 < HO(CH2)2NH2 (c) CH3CH2NH2 > HO(CH2)3NH2 > HO(CH2)2NH2

(d) CH3CH2NH2 > HO(CH2)3NH2 < HO(CH2)2NH2

9. Acetamide is treated separately with the following reagents. Which one of these would give methylamine? (b) sodalime (c) NaOH + Br2 (d) hot concentrated H2SO4 (a) PCl5 10. Chlorobenzene can be prepared by reacting aniline with (a) hydrochloric acid (b) cuprous chloride (c) chlorine in the presence of anhydrous aluminium chloride (d) nitrous acid followed by heating with cuprous chloride 11. Amongst the following, the most basic compound is (a) benzylamine (b) aniline (c) acetanilide (d) p-nitroaniline 12. Examine the following two structures for the anilinium ion and chose the correct statement from the ones given below. + NH NH3 3 ←→ (I)

(II)

(a) II is not acceptable canonical structure because carbonium ions are less stable than ammonium ions (b) II is not an acceptable canonical structure because it is nonaromatic (c) II is not an acceptable canonical structure because nitrogen has 10 valence electrons (d) II is an acceptable canonical structure 13. Which of the following statements is not correct? (a) Primary amines show intermolecular hydrogen bonding. (b) Secondary amines show intermolecular hydrogen bonding. (c) Tertiary amines show intermolecular hydrogen bonding. (d) Amines have lower boiling points as compared to those of alcohols and carboxylic acids of comparable molar masses. 14. Which of the following order is correct regarding the relative basicity of amines? (a) CH3O

NH2 >

NH2 > O2N

NH2

(b) CH3O

NH2
p–aminobenzaldehyde < p–bromoaniline 16. Which of the following orders regarding the basic strength of substituted aniline is correct? (a) p–methylaniline > p–chloroaniline > p–aminoacetophenone (b) p–methylaniline > p–aminoacetophenone > p–chloroaniline (c) p–aminoacetophenone > p–methylaniline > p–chloroaniline (d) p–aminoacetophenone > p–chloroaniline > p–methylaniline 17. The number of resonating structures of arylammonium ion is (a) 2 (b) 3 (c) 4 (d) 5 18. The number of resonating structures of aniline is (a) 2 (b) 3 (c) 4 (d) 5 19. Among the following, the strongest base is (b) p-NO2C6H4NH2 (c) m-NO2C6H4NH2 (d) C6H5CH2NH2 (a) C6H5NH2 20. Which of the following orders is correct regarding basicity of indicated molecules? (a) N, N-Dimethyltoluidine > p-toluidine > aniline > p-nitroaniline (b) Aniline > N, N-dimethyl-p-toluidine > p-toluidine > aniline (c) p-Toluidine > N, N-dimethyl-p-toluidine > aniline > p-nitroaniline (d) N, N-Dimethyltoluidine > aniline > p-toluidine > p-nitroaniline 21. Which of the following statements is correct? (a) Guanidine (H2N)2CN == H is expected to be a strong base. (b) Aniline is more basic than pyridine. N , has two basic sites. (c) Imidizole,

N H

(d) Purine,

22. Which of the following statements is correct? (a) Amines have lower boiling points than nonpolar compounds of the same molar mass. (b) Amines have higher boiling points than alcohols or carboxylic acids of comparable molar mass. (c) Aliphatic amines are as basic as ammonia but aromatic amines are more basic. (d) Although nitrogen in R ⎯ N

is chiral and also the mirror image of the molecule is not superimposable on

R′′ R′ the molecule itself, yet the molecule does not show optical activity. 23. Which of the following statements is not correct? (a) Quaternary ammonium salts in which nitrogen holds four different groups show optical activity.

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.13

(b) Aliphatic amines can be prepared by ammonolysis of alkyl halides but not aromatic amines. (c) Ammonolysis of alkyl halides gives highest yield with tertiary halides and is worthless for primary halides. (d) The more stable the substituted ammonium ion relative to the amine from where it is formed, the more basic is the amine. 24. Which of the following statements is correct? (a) Base-weakening substituents are the one that activate an aromatic ring toward electrophilic substitution. (b) Aliphatic amine forms stable diazonium salt with nitrous acid. (c) All amino compounds can be distinguished by using the reagents chloroform and potassium hydroxide. (d) The sulphonamide formed from secondary amine does not form sodium salt and is also insoluble in water. 25. Which of the following statements are is correct? (a) The order of acid strength in the given compounds is NH3 > CH3CONH2 > (CH3CO)2NH. (b) p-Aminobenzoic acid is a dipolar ion. (c) Sulphanilic acid can exist as a dipolar ion. (d) In Hoffmann elimination, a more substituted alkene is produced. 26. Which of the following statements is not correct? (a) NF3 is less basic than NH3. (b) Guanidine, HN==C(NH2)2, is a weak base. (c) The bond angle in NH3 is greater than in NF3. (d) The decreasing base strength of the amines CH3(CH2)2NH2, CH2==CHCH2NH2 and HC∫∫CCH2NH2 is CH3(CH2)2NH2 > H2C==CHCH2NH2 > HC∫∫CCH2NH2. 27. Which of the following statements is not correct? (a) Amides are much stronger bases than amines. (b) PhCONH2 is a stronger base than CH3CONH2. (c) N, N-Dimethylaniline is only slightly more basic than aniline whereas 2,6-dimethyl-N,N-dimethylaniline is much more basic than 2,6-dimethylaniline. (d) p-Nitroaniline is a weaker base than p-cyanoaniline. 28. Which of the following statements is correct? (a) 3,4,5-Trinitroaniline is less basic than 4-cyano-3,5-dinitroaniline. (b) The increasing order of basicities of pyrrole Pyrrole < Pyridine < Piperidine.

, pyridine

and piperidine

H is

(c) Pyrrole is soluble in dilute HCl. involving two equivalents of H2 produces

(d) The catalytic reduction of quinoline

N

N

Chemical Reactions of Amines 29. The primary amine on treating with NaNO2 and HCl produces (a) diazonium salt (b) nitrosamine (c) nitrite salt 30. The secondary amine on treating with NaNO2 and HCl produces (a) diazonium salt (b) nitrosamine (c) nitrite salt 31. Which of the following is Sandmeyer reaction? Cu Cl / HCl

2 2 Æ C6 H 6 + Cl2 (a) 2C6 H5 Cl æææææ

(c) C6H5OH

Zn dust

Æ C6H6 + ZnO

32. Aniline reacts with bromine water to give

(b) C6H5N2Cl

(d) nitrophenol (d) isonitrite salt Cu 2 Cl 2 / HCl

(d) C6H5NO2 + 6[H]

Æ C6H5Cl + N2

Sn / HCl

Æ C6H5NH2 + 2H2O

25.14 Complete Chemistry—JEE Main

(a) p-bromoaniline (b) o-bromoaniline (c) mixture of o-and p-bromaniline (d) 2, 4, 6-tribromoaniline 33. Benzenediazonium salt when heated in aqueous solution the compound obtained is (a) phenol (b) nitrobenzene (c) chlorobenzene (d) benzene 34. Aniline reacts with concentrated sulphuric acid at 450-475 K to give (a) phenol (b) sulphanilic acid (c) aniline hydrogen sulphate (d) azo dye 35. Heating of [CH3CH2CH2CH(CH3)N+(CH3)3]OH– produces (b) CH3CH2CH2CH==CH2 (a) CH2==CH2 (c) CH3CH2CH==CHCH3 (d) CH2==CHCH2CH==CH2 36. The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosoamine is (a) methylamine (b) ethylamine (c) diethylamine (d) triethylamine 37. Which of the following statements is not correct? (a) Replacement of halogen by NH2 in alkyl halide is a nucleophilic substitution reaction (b) Aryl halides show more reactivity as compared to alkyl halides in the replacements of halogen by the NH2 group (c) During the replacement of halogen by —NH2 group, ammonia is taken in large excess so as to avoid the formation of 2° and 3° amines (d) Tertiary alkyl halide generally produces alkene instead of the replacement of halogen by NH2 group 38. Which of the following statements is not correct? (a) Aliphatic amines are stronger bases than ammonia. (b) Aromatic amines are stronger bases than ammonia. (c) The alkyl group in alkyl ammonium ion more stabilizes the ion relative to the amine. (d) The aryl group in aryl ammonium ion less stabilizes the ion relative to the amine. 39. Which of the following amines form N–nitroso derivative when treated with NaNO2 and HCl. R (b) (c) (d) (a) CH3NH2 NH2 NH2 N⎯H 40. On adding KI to benzenediazonium chloride, the product obtained is (a) benzene (b) 1-4-diiodobenzene (c) iodobenzene (d) 1,3,5-triiodobenzene 41. Which of the following reagents can convert benzenediazonium chloride into benzene? (a) Water (b) Acid (c) Hypophosphorous acid (d) HCl 42. Benzenediazonium chloride on reaction with phenol in weakly basic medium gives (a) diphenyl ether (b) p-hydroxyazobenzene (c) chlorobenzene (d) benzene 43. Towards electrophilic substitution, the most reactive is (a) anilinium chloride (b) aniline (c) N-acetylaniline (d) nitrobenzene 44. Which of the followings undergoes diazotization? (b) C6H5NH2 (c) CH3CONH2 (d) CH3N(CH3)2 (a) CH3NH2 45. Which of the following statements is not correct? (a) The main product obtained in the reaction of chloroform with aniline in the presence of excess of alkali is benzeneisonitrile. (b) Amines are more basic than alcohols, ethers and esters. (c) Nitrogen involves sp2 orbitals in bonding with other atoms or groups in amines. (d) Amines are less basic than water and also less basic than hydroxide ion.

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.15

(a) Kolbe’s reaction (b) Riemer-Tiemann reaction (c) Carbylamine reaction (d) Haloform reaction 47. Hinsberg’s reagent is (a) phenylisocyanide (b) benzenesulphonyl chloride (c) p-toluenesulphonic acid (d) o-dichlorobenzene 48. Which of the following compounds will dissolve in an alkali solution after it has undergone reaction with Hinsberg reagent? (b) (CH3)3N (c) CH3NH2 (d) C6H5NHC6H5 (a) (C2H5)2NH 49. Which of the following can distinguish the three amines, viz., primary, secondary and tertiary? (a) Azo-dye test (b) Hinsberg reagent (c) Carbylamine test (d) Acetyl chloride 50. Carbylamine test is performed in alcoholic KOH by heating a mixture of (a) chloroform and silver powder (b) trihalogenated methane and a primary amine (c) an alkyl halide and a primary amine (d) an alkyl cyanide and a primary amine 51. Hofmann’s method to separate amines in a mixture uses the reagent (a) benzenesulphonyl chloride (b) diethyl oxalate (c) benzeneisocyanide (d) p-toluenesulphonic acid 52. A positive carbylamine test is given by (a) N, N-dimethylaniline (b) 2, 4-dimethylaniline (c) N-methyl-o-methylaniline (d) p-methylbenzylamine 53. Which of the following statements is correct? (a) Hinsgberg test involves the reagent benzenesulphonyl chloride, in the presence of aqueous potassium hydroxide to distinguish primary, secondary and tertiary amines. (b) Libermann reaction is shown only by primary amines. (c) Primary amine gives liquid oxamide when treated with diethyloxalate. (d) Secondary amine gives solid oxamide when treated with diethyloxalate.

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49.

(a) (b) (c) (d) (c) (b) (b) (b) (b)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(b) (c) (a) (a) (b) (d) (b) (b) (b)

3. 9. 15. 21. 27. 33. 39. 45. 51.

(a) (c) (c) (a) (a) (a) (c) (d) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52.

(a) (d) (a) (d) (b) (b) (c) (c) (d)

5. 11. 17. 23. 29. 35. 41. 47. 53.

(d) (a) (a) (c) (a) (b) (c) (b) (a)

6. 12. 18. 24. 30. 36. 42. 48.

(d) (c) (d) (d) (b) (c) (b) (c)

HINTS AND SOLUTIONS 1. The primary amine is RNH2. 2. Acetyl chloride converts —NH2 to —NHCOCH3, which reduces the activation of benzene ring. 3. Electron-releasing group increases the availability of lone pair of electrons on nitrogen while electron attracting group decreases this availability.

25.16 Complete Chemistry—JEE Main

4. The solution effect makes (CH3)3N poor base as compared to CH3NH2 and (CH3)2NH. 5. More the number of electron-releasing groups attached to —NH2, the larger is its basic nature. 6. The substituent at ortho position has the effect of increasing acidity. Intramolecular hydrogen bonding in o-nitrobenzoic acid makes it less acidic than o-aminobenzoic acid. 7. Electron-withdrawal group at para position makes phenol more acidic. 8. The —OH group causes decrease in the basicity of —NH2 group. 9. CH3CONH2

Br2, NaOH

Æ CH3NH2.

10. 11. 12. 13. 14.

Chlorobenzene is prepared via diazonium salt. Benzylamine is an aliphatic amine. It is most basic amongst the given compounds. N has 10 electrons around it. Tertiary amine does not contain H attached to N. —OCH3 is electro-releasing group, it enhances the basicity. —NO2 is electron-attracting group, it decreases the basicity. 15. —NO2 is more electron attracting then —CHO; —Br is more electron-releasing than its inductive effect. 16. —CH3 is more electron releasing than —Cl; —COCH3 is electron-attracting group. 17. There are only two resonating structures. +

NH2

+

+

NH3

NH3

+

NH

+

NH

NH

NH2

19. C6H5CH2NH2 is an aliphatic amine. It is most basic amongst the given compounds. 20. Aniline is a weaker base. It is a Lewis base, i.e. the lone pair present on nitrogen can be given to the Lewis acids. The electron-releasing group present at the para position increases the electron density on nitrogen and thus makes it more basic. For example, para-toluidine is more basic than aniline. On the other hand, the presence of electron-withdrawing group decreases the electron density on nitrogen and hence makes it less basic. For example, p-nitroaniline is less basic than aniline. 21. (a) Guanidine is a strong base as there is a greater resonance stabilization of the cation [C(NH2)3]+. This is due to the contribution from three equivalent structures with the accommodation of positive charge by three nitrogens. NH2

H2N +

+

+

NH2

H2N

NH2

H2N

NH2

H2N

C

C

C

C

NH2

NH2

NH2

NH2

+

(b) The lone pair on nitrogen in aniline is considerably delocalised to the benzene ring, making the molecule quite stable. This stability is destroyed when H+ adds to the nitrogen, thereby decreasing the basicity of aniline. In pyridine, lone pair is completely localized to nitrogen atom making it stronger base than aniline. (c) The electron pair of H—N: is delocalised over the ring to provide the aromatic sextet. and thus is not a basic site. The nitrogen of :N==C retains electron density and is a basic site. (d) Purine has three basic sites. These are the three double-bonded nitrogen. The lone pair of NH2 and NH are involved in delocalisation.

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.17

22. (a) Being more polar than nonpolar compounds, amines have higher boiling point. There also exist hydrogen bonds amongst amines. (b) Amines are less polar than alcohols and carboxylic acids. (c) Aromatic amines are less basic than ammonia and aliphatic amines. Phenyl ring being electron attractor makes lone pair on nitrogen less readily available to an acid. Aliphatic amines are more basic than ammonia. (d) The molecule and its mirror image is rapidly interconvertible. The energy barrier between the two is very less. 23. (b) Aryl halides have low reactivity toward nucleophilic substitution. (c) Ammonolysis gives good yield with primary amines as nucleophilic substitution predominates but is worthless for tertiary amines as elimination reaction predominates to give alkenes. CH3

NH3

CH3CH2CH2Br

Æ

CH3CH2CH2NH3+

–

Br ;

CH3

C

CH3

Br

NH3

Æ CH3— C ==CH2 + NH4Br

CH3

24. (a) Electron-releasing groups are aromatic-ring activators and they also push electrons toward nitrogen making lone pair of electrons more readily available for an acid. Hence, they are base-strengthening groups. Electronwithdrawing groups act in the opposite direction, hence, they are base-weakening group. (b) Aliphatic amines form unstable diazonium salt which decompose to give nitrogen and aliphatic alcohols. (c) Only primary amine can be distinguished as it produces carbylamine having most offensive smell RNH2 + CHCl3 2O 25. (a) The correct order is NH3 < CH3CONH2 < (CH3CO)2NH In the anion from ammonia, the negative charge is localized on nitrogen. In the anion from acetamide, the negative charge is shared by nitrogen and one oxygen. In the anion from diacetamide, the negative charge is shared by nitrogen and two oxygens.

least stable

most stable

(b) p-Aminobenzoic acid does not exist as dipolar ion. An aromatic —NH2 group is a weak base and it cannot neutralize the —COOH group. It can, however, neutralize the stronger acidic —SO3H group. Thus, sulphanilic acid can exist as dipolar ion. (d) In Hofmann elimination, a less substituted alkene is produced. 26. (a) In NF3, nitrogen acquires d+ nitrogen, the lone pair of electrons are more tightly held in NF3. (b) In guanidine, the imino (==NH) nitrogen is sp2 hybridised while amino (—NH2) is sp3 hybridised. The proton is likely to be attached to —NH2 as its nitrogen has less s-character. Actually the imino nitrogen is protonated because this leads to the stable symmetrical resonating cation with very high delocalisation energy as there are three contributing structures. H2N C H2N

NH

H+

H2N C H2N

+

H2N +

NH2

H2N C

H2N

NH2

H2N +

C

NH2

Because of the large delocalisation energy, the cation is very much stabilised leading the guanidine to act as a strong base. (c) Ammonia involves more bonding pair-bonding pair electronic repulsion as the pairs are located near the nitrogen atom due to its more electronegative character. In NF3, F is more electronegative and thus bonding pair lies more near F leading to less electronic repulsion.

25.18 Complete Chemistry—JEE Main

(d) The larger the s-character of a hybrid orbital of carbon, the larger electron-withdrawing (by induction) effect it has, consequently, the larger is its base-weakening effect. 27. (a) The adjacent C==O weakens the basicity by delocalisation of electron density from N to O. (b) The ability of C==O to cause base-weakening is attenuated by the phenyl group, making benzamide a stronger base than acetamide. (c) Phenyl group has a base-weakening effect because of its ability to form extended p-bonding between the amino N and the phenyl ring.

H NH2

N H

(I)

(II)

In the structure II, N—H bonds are in the plane of benzene ring. In 2,6-dimethyl-N,N-dimethylaniline, the formation of extended p-bonding is sterically hindered and thus interfere with the base-weakening extended p-bonding making this compound much more basic than 2, 6-dimethylaniline. (d) The nitro group is more effective than cyano group in weakening the basic nature of aniline.

O

H N

N

O

H N == C

H

N H

In nitroaniline, electron delocalisation ends up with the negative charge on the more electronegative oxygen atom whereas in cyanoaniline, the negative charge ends up in the nitrogen atom (which is less electronegative than oxygen). 28. (a) The base-weakening via electron delocalisation is achieved when N—O bonds of nitro group in nitroaniline are in the same plane of benzene ring. In 3,4,5-trimethylaniline, the attainment of coplanarity is sterically hindered and is thus not effective in base-weakening compared to 4-cyano-3-5-dinitroaniline. Cyano group, being linear, does not encounter this hinderance. (b) In pyrrole,

the lone pair on nitrogen is involved in the delocalisation over the ring and thus there is N , K°b = 2 ¥ 10–3)

because it has sp3 hybrid orbital (less s-character) while the N of pyridine (K°b = 2.3 ¥ 10–9) has sp2 hybrid orbital (more s-character).

withdrawing N. Thus, the product is

N H

29. A diazonium salt is produced. 30. A nitrosamine is produced. 31. Sandmeyer reaction is

C6H5N2Cl

Cu2Cl2/HCl

Æ C6H5Cl + N2

32. In aqueous medium, tribromo derivative is obtained. 33. C H N Cl 6 5 2

H2 O

Æ C6H5OH + HCl + N2.

34. Sulphanilic acid is produced. 35. In Hofmann elimination, a less substituted alkene is formed as a result of the loss of more acidic BH (which follows the trend 1° > 2° > 3°). The reaction is Æ CH3CH2CH2CH == CH2 CH3CH2CH2CH(CH3) N+(CH3)3 æDæ 36. 2° amine produces oily nitrosoamine. 37. Aryl halides is less reactive than alkyl halide.

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

25.19

38. 39. 40. 41. 42. 43. 44. 45.

Aromatic amines are lesser basic than ammonia due to resonance effect. 2° Amine produces N-nitroso derivative. Treating with KI produces iodobenzene. Hypophosphorous acid is used. The product is p-hydroxyazobenzene. —NH +3Cl, —NHCOCH3 and —NO2 groups deactivate benzene ring towards electrophilic substitution reaction. Aromatic Amine undergoes diazotization. (d) Amines are more basic than water.

47. 48. 49. 50.

Æ R—NC + 3KCl + 3H2O. RNH2 + CHCl3 + 3KOH ææ Hinsberg reagent is benzenesulphonyl chloride (C6H5SO2Cl). RNH2 gives RSO2NHR. Due to H atom attached to N, it dissolves in alkali solution. Hinsberg reagent can distinguish the three amines. Carbylamine test is given by primary amine by using CHCl3 in alcoholic KOH medium.

51. Hofmann’s method uses diethyl oxalate as a separating reagent. 52. Primary amine exhibits carbylamine test. 53. (b) Secondary amines show Libermann reaction. (c) Solid oxamide is obtained. (d) Liquid oxamide is obtained.

MULTIPLE CHOICE QUESTIONS FROM AIEEE AND JEE MAIN 1. The reaction of chloroform with alcoholic KOH and p-toluidine forms

[2003] 2. The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2NH is (b) CH3NH2 < NH3 < (CH3)2NH (a) CH3NH2 < (CH3)2NH < NH3 (d) NH3 < CH3NH2 < (CH3)2NH (c) (CH3)2NH < NH3 < CH3NH2 3. Which of the following is the strongest base? (a)

NH2

(b)

NHCH3

(c)

NH2 CH3

(d)

[2003]

CH2NH2

[2004]

4. Which one of the following methods is neither meant for the synthesis nor for separation of amines? (a) Wurtz reaction (b) Curtius reaction (c) Hinsberg method (d) Hofmann method [2005] 5. An organic compound having molecular mass 60 amu is found to contain 20% C, 6.67% H and 46.67% N, while rest is oxygen. On heating it gives NH3 alongwith a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is (b) CH3CH2CONH2 (c) CH3NCO (d) CH3CONH2 (a) (NH2)2CO [2005] 6. In the chemical reaction,

25.20 Complete Chemistry—JEE Main

7. 8.

9.

10.

CH3CH2NH2 + CHCl3 + 3KOH æÆ (A) + (B) + 3H2O, the compounds (A) and (B) respectively are: (b) CH3CH2CONH2 and 3KCl (a) C2H5CN and 3KCl (d) C2H5NC and 3KCl [2007] (c) C2H5NC and K2CO3 Which one of the following is the strongest base in aqueous solution? (a) Trimethylamine (b) Aniline (c) Dimethylamine (d) Methylamine [2007] A compound with molecular mass 180 u is acylated with CH3COCl to get a compound with molecular mass 390 u. The number of amino groups present per molecule of the former compound is (a) 2 (b) 5 (c) 4 (d) 6 [2013] On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is (a) an alkyl isocyanide (b) an alkanol (c) an alkanediol (d) an alkyl cyanide. [2014] Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value? (b) (CH3)2NH (c) CH3NH2 (d) (CH3)3N [2014] (a) C6H5NH2

11. Complete reduction of benzene-diazonium chloride with Zn/HCl gives: (a) Aniline (b) Phenylhydrazine (c) Azobenzene

(d) Hydrazobenzene [2014] 12. Conversion of benzene diazonium chloride to chloro benzene is an example of which of the following reactions? (a) Claisen (b) Friedel-craft (c) Sandmeyer (d) Wurtz [2014] 2

(a) Diazomethane

(b) Methylalcohol

and HCl is:

(c) Methylcyanide

(d) Nitromethane

[2014]

14. Which one of the following compounds will not be soluble in sodium bicarbonate ? (a) 2, 4, 6-Trinitrophenol

(b) Benzoic acid

(c) o-Nitrophenol

(d) Benzene sulphonic acid

[2014]

15. In a set of reactions p-nitrotoluene yielded a product E CH3 Br2 B FeBr3

Sn/HCl

C

NaNO2 D HCl

CuBr HBr

E

NO2

The product E would be:

(a)

Br

Br

Br

(b)

CH2Br

CH3

CH3

CH3

Br

(c)

(d)

[2014]

Br Br

Br

Br

16. The major product of the reaction NaNO2/H2SO4

NH2

OH

is: (b)

(a) H

OH

O

Organic Compounds Containing Nitrogen (Cyanides, Isocyanides, Nitrocompounds and Amines)

(c)

(d)

25.21

[2014]

N H

17. Arrange the following amines in the order of increasing basicity.