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08-Nov-13

Bonding in Ceramics

CERAMIC STRUCTURES

Dr. Mohsin Ali Raza Department of Metallurgy and Materials Engineering University of the Punjab

How to calculate % Ionic Character ? • Dipole moment of molecule is used to calculate ionic character • Covalent bonds - Equally shared electrons --- non-polar - Unequal sharing of electrons--- polar • Ionic bonds Complete transfer of electrons when one becomes +ve (cation) and other becomes –ve (anion)

Ceramics Crystal structures • If the atomic bonding is mainly ionic the crystal structures are composed of electrically charged ions instead of atoms. • Cation (+vely charged, Na+, Fe+2, Fe+3, Mg+2, NH4+) and Anion (-vely charged, O-2, Cl-1, F-1, CN-1,) • Ions do not exists on their own • Ions binds with ions of opposite charge to form a crystal lattice.

• Atomic bonding ranges from purely ionic to totally covalent • Many ceramics possess combination of both types of bonding • Ionic character depends on Electronegativities of atoms • We can calculate % ionic character of ceramic materials

Dipole moment • The measure of net polarity of a bond is called dipole moment, (µ) • Mathematically, 𝜇 =𝑄×𝑟 Q = Charge r= distance between charges Units: debye 𝐷 1 D = 3.336 x 10-30 C.m

Formation of ions • Removal of electron gives Cation (+vely charged ions), has more number of protons than electrons (All metals form cations) • Addition of electron gives anion (-vely charged), has more electrons than protons

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08-Nov-13

Criteria that influence crystal structure of ceramics First Criteria The magnitude of the electrical charge on each of the component ions. - The crystal must be electrically neutral - Charge must be balanced Bonding in MgCl2 and CaF2 Ca+2 and F-1 (Two flouride ions are required for ionic bonding and for charge neutrality)

Second Criteria • Each Cation prefers to have as many nearest neighbours anions as possible and likewise. • Stable ceramic structures form when anions surrounding a cation are in contact with that cation. • Size or ionic radii of the cations and anions is important – Cation ion radius (rc) and anion radius (rA) • Cations are smaller than anions, the relative size is quantified by the radius ratio (ρ) 𝑟

• 𝜌 = 𝑟𝐶 < 1

Requirement for stable structure • Anions surrounding a cation are all in contact with that cation. • Closed packing of anions and cations necessary • The condition for minimum energy implies maximum attraction and minimum repulsion. This leads to contact, configurations where the anions have highest number of cations neighbours and vice versa.

𝐴

Concept of stable structure

Coordination number (CN) • The number of nearest neighbours anions or cations (In crystallography)

UNSTABLE

• For a specific CN, there is a critical or minimum radius ratio (rC/rA), for which the cation-anion contact develops and this can be found by geometrical analysis

STABLE

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Concept of Ionic and Atomic Radii • • •

Measure of size of atom (distance from the nucleus to the boundary of surrounding electrons) Ionic radii can’t be determined absolutely Determined on the probability of encountering electron

Tetrahedron

Octahedron

Bond length: Average distance between nuclei of two bonded atoms

Cube

Atomic and Ionic Radii Pure element first Native Cu. Atomic radius = 1/2 bond length

Bonds between identical atoms atomic radius is half the interatomic distance

X-ray d100  a

a

Ionic radius =

a

2 4

a

In the bonds between different ions, distance is controlled by attractive and repulsive forces between the charges

Charge and Attractive Force Control on Effective Ionic Radii

Approach until Repulsive and Attractive Forces the same

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08-Nov-13

Interionic forces for an ion pair

Fnet = Fattractive + Frepulsive

Interatomic Energies for an ion pair 𝐸𝑛𝑒𝑡 = +

𝑍1 𝑍2 𝑒 2 𝑏 + 4𝜋𝜖0 𝑎 𝑎𝑛 Repulsive energy

Attractive energy

Attractive energy : Released as the ions come close together (-ve) Repulsive energy : absorbed as the ions come close together (+ve)

Minimum energy is achieved when ions are at their equilibrium separation distance (ao)

Ref. Principles of Materials Science Engineering by William F. Smith

Homework problems

Lattice Energy • Amount of Energy released on the formation of bonds (-ve) • Indicator of bond strength

Example Problem 2.10

Table: Lattice Energies and Melting points of selected ionic solids

Problems 2.5.4-2.5.9 (HW) Example 2.11

Ionic radii of selected ions

Book Principles of Materials Science and Engineering by W. F. Smith 3rd Edition

• As the size of ion increases in a group, lattice energy decreases and melting points decreases because bonding electrons are far from the nucleus • Multiple bonding electrons in an ionic solid increase lattice energy

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08-Nov-13

Bond Length • • • •

Average distance between the nuclei of the two bonded atoms Shorter bond length means stronger bonds Bond length is related to bond order Bond order is number of bonding pair of electrons between two atoms • More electrons involved means stronger bond

Bond strength and melting temperature MgO

NaCl

LiF

Structure

Melting point

2852 ºC

801 ºC

848 ºC

Rocksalt

Inter-ionic equilibrium distance

212 pm

283 pm

209 pm

Rocksalt

Factors on which Bond strength depends

Bond energies and length of selected covalent bonds

Comparison of Potential-interatomic distance curves

Melting points of ionic compounds

• Bond strength depends on valency and ionic radii Higher the valency stronger will be the bond • Bond strength is directly proportional to multiplication of ionic charges and inversely proportional to equilibrium interionic distance MgO

NaCl

LiF

Structure

Melting point

2852 ºC

801 ºC

848 ºC

Rocksalt

Inter-ionic equilibrium distance

212 pm

283 pm

209 pm

Rocksalt

Ionic character of the bond is > 60 % Why MgO has higher melting point than LiF despite having larger interionic distance?

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08-Nov-13

Improve the criteria (Look for other factors)

Covalent Character of a Bond depends on EN difference

Melting point of Al2O3 = 2054 ᵒC Melting point of MgO = 2852 ᵒC Look at the bonding : Al2O3 has higher covalent character than MgO

How we can qualitatively judge the COVALENCY of the Higher bonds? electronegativity Look at the Electronegativity (EN) differences difference, Higher will be the ionic Al2O3 ∆EN = EN of O – EN of Al = 3.44-1.61= 1.83 character and the MgO ∆EN = EN of O – EN of Mg = 3.44-1.31= 2.13 melting point

% Ionic Character vs EN

Covalent character increases Melting point decreases Structure also changes

Calculate % ionic character of mixed ionic-covalent bonding % 𝐼𝑜𝑛𝑖𝑐 𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟 = (1 − 𝑒

1 4

−

𝑋𝐴−𝑋𝐵

2

)(100%)

Where, XA = EN of atom A XB = EN of atom B

Polarisation

Polarization

Cations tend to attract outer electrons of the anion and at the same time it tends to repel the positively charged nucleus of the anions. This result in distortion of the anion (anion is polarised). The effect is called polarising effect.

Polarising power of Cation The tendency of metal cation to distort the electron cloud of anion. It is directly proportional to charge on cation and inversely proportional to size of cation. Polarising power of Anion

Polarisation arise due to the asymmertical distribution of the charges

The tendency of anion to get distorted because of the polarising power of cation is called polarisibilty power of anion.

Due to high polarizing power of Al ion, Al2O3 develops some covalent bonding and hence has lower melting point than MgO

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08-Nov-13

Factors Effecting Covalency of Ionic Bonds • Polarizing power of the Cation It is the ratio of the charge to the ionic radius Check for Φ (Al+3) = 60 nm-1 for Mg+2 = 31 nm-1 • Polarizing power of the anion Same for oxide anions • Electron configuration of the Cation no of electrons in d-orbital Both Al2O3 and MgO have no d shell electrons

• Smaller the cation, larger the anion greater is tendency for the covalency • AgCl is more covalent than NaCl because the former has 18 electrons in outermost shell while in the latter Na+ has 8 electrons in outermost shell.

Factors Effecting Covalency of Ionic Bonds • If polarisation is quite small, bonding is ionic. • If polarisation is very high, bonding is covalent. • If polarisation is in between the two values (above mentioned), the bond is partial covalent.

AlI3 (melting point 189 °C and AlF3 (1291 °C)

Crystal Structures Simple Cubic structure Packing of spheres in a simple cubic array

Interstitial site surrounded by eight atoms

CN = 8

Close-packed cubic structures

Close-packed cubic structures 1

1

Octahedral site (Type 2)

Tetrahedral Sites (Type 1)

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08-Nov-13

Add next layer of atoms

Add third layer of atoms (if it sits exactly on the atoms in layer 1 resulting structure is hexagonal closed pack with a layer ordering of A-B-A-B)

Type 1 are now covered But type 2 are still left vacant

Hexagonal close-packed structure

Cubic Closest packing

Add next layer of atoms Type 1 are now covered But type 2 are still left vacant

Interstitial site 3

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08-Nov-13

Add next layer on site 3 Ordering is A-B-C

FCC lattice

Octahedral hole

Octahedral hole is surrounded by 6 atoms

Tetrahedral hole

Tetrahedral hole is surrounded by 4 atoms

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08-Nov-13

Limiting Conditions for different Coordination numbers

For CN = 8

• Ratio of Rc/Ra defines the limiting situation in which all atoms touch contact each other

𝑑𝐴 + 𝑑𝐶 = 1.732

• Determine the limiting condition i.e., Rc/Ra for the cation to remain in 8 CN

dA = 1 Then

𝟐

dC = 0.732

1 (arbitrary) Cation will rattle if Relative Radius goes below 0.732

1 (arbitrary)

For CN = 6

𝑑𝐶 𝑑𝐴

𝑟

= 𝑟𝐶 = 0.732 𝐴

𝑑𝐴 + 𝑑𝐶 = 2 dA = 1 Then dC = 0.414

Use this plane

𝑑𝐶 𝑑𝐴

𝑟

= 𝑟𝐶 = 0.414 𝐴

Cation will rattle if Relative Radius goes below 0.414

Home work problem 1. Calculate the critical (minimum) radius ratio (rc/ra)

Example problem 10.1 Calculate the critical (minimum) radius ration rC/RA for the triangular coordination (CN =3) of three anions of radii R A surrounding a central cation of radius r in an ionic solid.

for the tetrahedral coordination (CN=4) of 4 anions of radii R surrounding a central cation of radius r in an ionic solid (consider FCC unit cell). 2. Prove that the tetrahedral site is bigger in BCC unit cell than FCC unit cell.

3. Calculate the octahedral vacancy for BCC unit cell. Cation will rattle if Relative Radius goes below 0.155

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Example Problem 10.2 Predict the CN for the ionic solids CsCl and NaCl. Use the following ionic radii for the prediction: Cs+ = 0.170 nm , Na+ = 0.102 nm and Cl- = 0.181 What will be the CN if the ionic radii of Na+ ions = 0.118 nm

AX Types crystal structures Type 1 when CN= 8 Cesium Chloride (CsCl)

Cl Cl-1 Cs

1 1 1

( , , ) 2 2 2

Position of Cs cation in unit cell

Cs+1 Lattice of CsCl Hard sphere model to represent unit cell of CsCl

Examples of AX type ionic compounds which have CsCl like structure • CsBr • TlCl • TlBr Intermetallic compounds

Ball and Stick model for CsCl

Example 10.3 Calculate the ionic packing factor for CsCl. Ionic radii are Cs+ = 0.170 nm and Cl-= 0.181 nm Ionic packing factor =

3𝑎 = 2𝑟 + 2𝑅 𝑎 = 0.405 𝑛𝑚

CsCl ionic packing factor = =

• AlNi

𝑉𝑜𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙

R = Cl-1 ion and r = Cs+1

• AgMg • LiMg

𝑉𝑜𝑙 𝑜𝑓 𝑖𝑜𝑛𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙

4 𝜋𝑟 3 3

+

4 3

−

1 𝐶𝑠 𝑖𝑜𝑛 + 𝜋𝑅3 (1 𝐶𝑙 𝑖𝑜𝑛) 𝑎3

0.68

• β-Cu-Zn

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08-Nov-13

Positions of octahedral site in FCC unit cell

Formation of Octahedrons

Octahedron sites are Bound by 6 ions

Eight

In FCC unit cell, there are 13 octahedral sites, and octahedrons can be formed

Rocksalt Structure

AX type crystal structures Type 2 When CN = 6, rc/Ra is between 0.414 and 0.732 Example NaCl (Rock salt structure), MgO, MnS, LiF,etc. Rock salt structure can be visualised as two interpenetrating FCC lattices, one composed of cations and the other of anion

ClNa+

Positions of Na+ and Cl- in NaCl unit cell

Charge Balance in NaCl (Halite)

In Halite, Na+ has CN 6 and valence +1 Interpretation: Each Na+ has 6 Cl- neighbors, so each Cl- contributes a charge of -1/6 to the Na+ 6 x -1/6 = -1, so a charge balance results between the Na+ cation and the six polyhedral Cl- anions with which it bonded. NEUTRALITY IS ACHIEVED

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08-Nov-13

Positions of octahedral sites

Example Problem 13.2

• The octahedral sites are located at the centre of the unit cell (1/2, 1/2, 1/2) as well as at the edges of the unit cell (1/2, 0, 0), etc.

On the basis of ionic radii, what crystal structure would you predict for FeO? rFe2+ = 0.077 nm RO-2 = 0.140 nm

r/R = 0.550 , It is between 0.414 and 0.732 (consult Table 13.2) , the CN number for Fe+2 ion is 6 ; this is also the CN of O-2, since there are equal number of cations and anions. The predicted structure will be Rock Salt, which is AX type structure having CN = 6

Density calculation from the knowledge of crystal structure

Example Problem 10.5 Calculate the linear density of Ca+2 and O-2 ions in ions per nm in the [110] direction of CaO which has the NaCl structure. (ionic radii: Ca+2= 0.106 nm and O-2 = 0.132 nm

Calculate the planar density of Ca+2 and O-2 ions in ions/nm2 on the (111) plane of CaO.

Same for Ca+2 ions 2.14 g/cm3 The handbook value of for the density of NaCl is 2.16 g/cm 3

Positions of tetrahedral sites in FCC unit cell

Positions of tetrahedral sites in unit cell

• 8 tetrahedral positions in FCC lattice and 24 in BCC lattice 4 Sites available for FCC unit cell

FCC

(¼ ,¼ , ¼ )

(¾, ¾, ¼ )

FCC BCC 12 Sites available for BCC unit cell

BCC

In FCC unit cell, 8 tetrahedral sites are positioned within the cell, at the general fractional Coordinates (n/4,n/4,n/4) where n = 1 or 3, e.g., (¼,, ¼, ¼ ) or (¼, ¼, ¾)

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08-Nov-13

Positions of tetrahedral sites in unit cell

AX type crystal structures Type III When CN = 4 Ions are tetrahedrally Coordinated

Zinc Sulphide (ZnS) Crystal Structure (Zinc Blende or Sphalerite)

Zinc Sulphide (ZnS) Crystal Structure (Zinc Blende or Sphalerite)

rZn/RS=0.60/1.84=0.32 (tetrahedral) S-2

Zn+2

4 S atom per unit cell so only 4 tetrahedral sites can be occupied to achieve charge balance

• 4 tetrahedral sites are occupied per unit cell to maintain charge neutrality • Bonding is typically covalent

Examples of other “Zinc blende” structures • Important Ceramic material which form such structure is SiC • Electronic materials are GaAS, ZnTe, etc. • Diamond also form “Zinc blende” structure

Diamond

Example 3.16 Calculating the Theoretical Density of GaAs The lattice constant of gallium arsenide (GaAs) is 5.65 Å. Show that the theoretical density of GaAs is 5.33 g/cm3. For the ‘‘Zinc blende’’ GaAs unit cell, there are four Ga and four As atoms per unit cell. Each mole (6.023  1023 atoms) of Ga has a mass of 69.7 g. Therefore, the mass of four Ga atoms will be (4 * 69.7/6.023  1023) g.

Silicon carbide (carborundum)

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08-Nov-13

Each mole (6.023  1023 atoms) of As has a mass of 74.9 g. Therefore, the mass of four As atoms will be (4 * 74.9/6.023  1023) g. These atoms occupy a volume of (5.65  10-8)3 cm3.

density 

mass 4(69.7  74.9) / 6.023  10  8 3 volume (5.65  10 cm)

Check the covalent character of ceramic compounds

23

Hence, the theoretical density of GaAs is 5.33 g/cm3.

Example Problem 10.7

AmXp Type Crystal Structures

Calculate the density of zinc blende (r(Zn2+) = 0.060 nm, r(S2-) = 0.174 nm)

• If the charges on the cations and anions are not same, a compound can exist with the chemical formula AmXp ,where m and/or p ≠ 1 • AX2 type compounds • Examples CaF2 (fluorite), UO2, PuO2, ThO2, BaF2, PbMg2

Fluorite Structure (AX2)

Fluorite Structure (AX2) rca+2/RF- = 0.8 CN = 8

rca+2/RF- = 0.8 CN = 8 for

Ca+2

What is CN for F-?

Half of the centre cube positions are occupied by Ca+2 ions

Ca+2 occupy FCC lattice

8 tetrahedral sites are occupied by Fions to maintain charge neutrality

Many octahedral sites are vacant. In UO2 this allows to accommodate fission products

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08-Nov-13

Fluorite Structure (AX2)

Fluorite Structure (AX2)

Examples: some Halides (CaF2, BaCl2...); Oxides (ZrO2...)

Calculate the density of UO2 (r (U+4) = 0.105 nm and r(O-2) = 0.132 nm

Anti Fluorite Crystal Structure • The reverse of the fluorite structure • Anions (F-) goes on the FCC lattice sites • Cations will fill up the tetrahedral sites

3 + − 𝑎 = 𝑟𝑈 4 + 𝑅𝑂 2 4

• A2X type structures (A is cation and X is anion) • Examples Li2O, Na2O, K2O and Mg2Si The handbook value of for the density of UO2 is 10.96

Structure of Li2O

g/cm3

Connections of tetrahedrons in Antifluorite structures

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08-Nov-13

Bond strength and coordination number

Octahedral sites in HCP unit cell Two atoms shared by 2 cells

12 atoms shared by six cells

Bond strength of ionic bond donated from a cation to anion is defined as “ Formal charge on the cation divided by its coordination number”

Octahedral sites

Example: Silicon with a valence of 4 has tetrahedral coordiantion (CN=4) , the bond strength is 4/4 = 1 Al+3 ions in octahedral coordiantion (CN=6) , the bond strength is 3/6 = 1/2

A2X3 structures • Hexagonal closed pack structures • Coordination number of cation is 6 and anion is 4 • 2/3 octahedral sites are filled • Structure name is Corundum • Examples Al2O3, Fe2O3, Cr2O3, Ti2O3, V2O3, Ga2O, Rh2O3

Corundum (Structure of Al2O3) • CN= 6 • Bond strength is ½ • 4 Al+3 ions are near to O-2 • There are 6 O-2 ions available per unit cell so for charge neutrality we need 4Al+3 ions) • 6 octahedral sites are available in HCP • Only 4 (2/3 sites will be occupied in HCP unit cell) • Such charge balance can only be possible in HCP structure.

O-2

Al+3

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08-Nov-13

Formation of Octahedrons in Corundum structure

Perovskite structures • ABO3 type structures (have two types of cations) A is a divalent cation (La, Ba, Sr, Ca, etc) B is tetravalent cation (Fe, Co, Ni, Mn, Ti, etc) O is oxide anion • Examples BaTiO3, CaTiO3, LaMnO3, etc.

Perovskite structures

Formation of octahedrons in pervoskite structures How many O-2 ions for Ti+4?

Grey sphere = B cation (Ti) (0,0,0) (corners of cube)

Ti+4 is surrounded by 6 O-2 ions

Green, blue, cyan sphere = O anion (occupy octahedral sites at the edges)

How many O-2 ions for Ca+2? Ca+2 is surrounded by 12 O-2 ions

Purple sphere= A, Ca (Cation) (occupy octahedral site at the centre of Cube, (½, ½, ½)

How many O-2 ions are near to each O-2? Each O-2 ions are near to 8 O-2 ions

Calculate bond strength in CaTiO3 pervoskite Bond strength of Ti-O bond =?

Effect of size of A and B cations on the Perovskite structures SrTiO3 an ideal Perovskite structure

CN for Ti is 6 , the bond strength is = 4/6= 2/3

Bond strength of Ca-O bond =? CN for Ca is 12 , the bond strength is = 2/12= 1/6 Also, How many O-2 ions are coordinated with Ti+4 and Ca+2 ?

2 Ti+4 , 4 Ca+2 Total bond strength of Ti-O bond = 2(2/3) = 4/3 And Total bond strength of Ca-O bond = 4(1/6)

What is a in terms of cation A and O-2 ion radius?

a = 𝟐(rO+rA)

Total bond strength in structure = 4/3+4/6= 2 It is equal to the oxygen valency.

What is a in terms of cation B and O-2 ion radius?

a = 2(rO+rB)

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08-Nov-13

GoldSchmidt’s Tolerance factor 𝑮𝒐𝒍𝒅𝑺𝒄𝒉𝒎𝒊𝒅𝒕′ 𝒔 𝒕𝒐𝒍𝒆𝒓𝒂𝒏𝒄𝒆 𝒇𝒂𝒄𝒕𝒐𝒓 (𝒕) =

What happens when tolerance factor deviates from ideal value?

𝒓𝑶 + 𝒓𝑨

• Cubic structure exists if 0.89 < t ≤ 1

𝟐 𝒓𝑶 + 𝒓𝑩

• If A ion is smaller than ideal then t will be smaller than ideal value.

Problem

• [BO6] octahedra will tilt in order to fill space.

Calculate GoldShmidt’s tolerance factor for SrTiO3? Ionic radii for Sr+2 = 0.144 nm ,Ti+4 = 0.0605 nm ,O-2 = 0.140 nm • For ideal cubic pervoskite t= 1, such as SrTiO3 • Higher or lower values than ideal value (1) changes the structure symmetry

Doubled perovskite structure

Changing the composition from ideal ABO3 •

• Non-stoichometry in perovskite structures exist • SrFeOx (where 2.5≤x≤3) • By oxidizing or reducing , oxygen content can be varied

By doubling all the edges of cubic unit cell it is possible to occupy equivalent positions with atoms of different elements.

• Such structures are called perovskite superstructues and also called as doubled perovskite.

• In SrFeO2.875 , there will be some Fe+3 and the other Fe+4 ions. Oxygen vacancies ordered so that FeO5 square pyramids formed. Oxygen vacancies provide oxide ion conductivity to the pervoskites, hence used in Solid oxide fuel cells

Barium titanate

Perovskite Superconductor • High temperature superconductor (YBa2Cu3O6.96) is superstructure of pervoskite.

• • • •

Perovskite structure Piezoelectricity High dielectric strength Ferroelectricity (exhibition of spontaneous polarisation and re-orientation of polarisation) • Curie temperature (120 ᵒC). It is tetragonal below 120 ᵒC (ferroelectric) but above 120 ᵒC it is cubic a perfect pervoskite (paraelectric) Uses: Ultrasonic equipment, capacitors, piezoelectric devices, doped titanate have applications in semiconductors.

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08-Nov-13

Ferroelectricity in BaTiO3

Ferroelectricity in BaTiO3

Ferroelectricity in BaTiO3

Ferroelectricity in BaTiO3

Ferroelectricity in BaTiO3

Ferroelectricity in BaTiO3

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08-Nov-13

Spinel structures (AB2X4)

Home work problems

The spinels have general formula AB2O4

10.2.5-10.2.24

AII is divalent cation e.g, Fe, Cr, Mn, Mg, Co, Ni, Cu, Zn, Cd, Sn

Principles of Materials Science and Engg. 3rd Edition By W. F. Smith

BIII is trivalent cation e.g, Al, Ga, In, Ti, V, Cr, Mn, Fe, Co, Ni

X is O, S, Se Named after the mineral spinel (MgAl2O4)

Spinel structure • Spinel structure is a cubic structure which is considered as combination of the rock salt and zinc blende structures.

Spinel structure

• O-2 ions are at face centred cubic position • For a subshell of this structure there are 4 atoms, 4 octahedral sites and 8 tetrahedral sites. • 12 total interstices to be filled up by three cations, one divalent and 2 trivalent. • In each elementary cell 2 octahedral sites are filled and one tetrahedral site • Each of these elementary cell are arranged so as to form a unit cell containing 32 oxygen ions, 16 octahedral cations and 8 tetrahedral cations

Example : MO.Fe2O3 (M is a divalent metal ion)

Examples of Spinel structures • In normal spinel structures the A+2 (8M+2) ions occupy 8 tetrahedral sites and B+3 (16 Fe+3) ions are on 16 octahedral sites. • Examples ZnFe2O4, CdFe2O4, MgAl2O4, FeAl2O4, CoAl2O4, NiAl2O4, etc.

Inverse Spinel • Most common structure • The A+2 ions and half B+3 ions are on octahedral sites, the other half of B+3 are on tetrahedral sites. • Examples Ferrites (these have magnetic properties) e.g., Fe3O4, FeMgFeO4, FeTiFeO4, ZnSnZnO4, etc.

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08-Nov-13

Pauling rules

Pauling rules

First Rule

Second Rule

 A Coordination polyhedron of anions is formed about each cation in the structure.

 Stable structures only form in ionic compounds by achieving local electrical neutrality.

 The cation-anion distance is determined by the sum of radii.

In a stable structure the total strength of the bonds reaching an anion from all surrounding cations should be equal to the charge of the anion.

 The coordination number (i.e, the number of anions surrounding the cation ), is determined by the ratio of radius of the two ions. (cation should be in contact with anion to achieve stability)

Linus Pauling

Pauling rules

Formation of Anionic groups

Second Rule The electrostatic valency (E.V) principle E.V = charge on cation/C.N

Charge is exactly balanced on anion and cation , this is called isodesmic bond and the bond strength are equal from all directions.

In this case, E.V > ½ the charge on anion i.e., 4/3>1/2*2, the oxygen are more strongly bonded to the central coordination ion (Anisodesmic bond)

Formation of Anionic groups

Pauling rules Third Rule

E.N difference is also less than 2 bond is covalent (Si = 1.9 and C = 2.55)

 Cation coordination polyhedra form links at corner, faces and edges.

In the case, E.V = ½ the charge on anion, the oxygen can bound tightly to the central coordination ion and to the ions outside the group (mesodesmic bond)

Sharing corners will put the cations at the greatest distance from each other.

In the stable structures, corners and faces of the coordination polyhedra tend to be shared.

What is isodesmic bond?

E.V = 4/4= 1 SiO4-4 a building block of silicate minerals

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08-Nov-13

Pauling rules

Pauling rules Fourth Rule

Third Rule

 Cations with high valency and low coordination numbers tend to not share their coordinating polyhedra. Pauling’s fourth rule says that highly charged cations will tend to be as far apart as possible to minimize electrostatic repulsion.

Fifth Rule  The number of different kinds of constituents in a structure tends to be small. Stability decreases

There are only few different kinds of anions and cations sites in a crystal. There are only octahedral, tetrahedral and cubic sites, most crystal will be formed by occupying these small number of sites.

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