UNIT I Syllabus : THE SOLID STATE Marks-4Classification of Solids based on different binding forces : molecular, Ionic,

covalent and metallic solids, Amorphous and crystalline solids (elementary

idea), unit cell in two dimensional and three dimensional lattices, calculation

of density of unit cell, packing in solids, voids number of atoms per unit cell

in a cubic unit cell, point defects, electrical and magnetic properties.

1. Gystalline and Amorphous solids (comparison)

Property Crystalline Solids Amorphous Solids

* Shape

* Melting

point

*Cleavage

property

* Heat of

fusion

*

Anisotropy

* Nature

* Order in

the

arrangement

of

constituent

particles

Examples

Definite characteristic shape.

Melt at a sharp and characteristic

temperature

When cut with a sharpedged tool,

they split into two pieces and the

newly generated surfaces are

plane and smooth.

They have a definite and

characteristic heat of fusion.

They are anisotropic in nature.

True-solids

They have long range order.

All solid elements and solid

compds : eg. Cu, Al, NaCl, Sugar

etc.

Irregular shape

Gradually soften over a

range of temperature

When cut with a sharp

edged tool, they cut into

two pieces with Irregular

surfaces.

They do not have definite

heat of fusion.

Isotropic in nature.

Pseudo solids or super

cooled liquids

They have only short range

order.

Glass, plastics, Rubber,

charcoal, lamp black etc.

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2. Classification of solids on the basis of different binding forces.

Type of

solid

Constituent

particles

Nature of

bond

Examples Physical

Nature

Electrical

Conductivity

Melting

point

1.

Molecular

Solids

(i) Non

Polar

(ii) Polar

(iii)

Hydrogen

bonded

Non polar

molecules

Polar

molecules

H-bonded

molecules

Dispersion

or London

forces

Dipole-

dipole

force

Hydrogen

bonding

Ar., CCl4,

H2, Solid

CO2, I2

HCl, HBr,

SO2

H2o (ice)

NH3(S)

HF (S)

Soft

Soft

Hard

Insulator

Insulator

Insulator

Very

low

Low

Low

2. Ionic

Solids

Ions Coulomic

forces

NaCl, Mgo,

ZnS, CaF2

Hard but

brittle

Insulator in

solid state but

conductors in

molten state

and aqueous

state

High

3. Metallic solids

Positive ions in a sea of delocalized electrons

Metallic bonding.

Fe, Cu, Ag, Mg

Hard but malleable and ductile

Conductor in solid as well as in molten state.

Fairly high.

4. Covalent or Network solids

Atoms Covalent bonding

SiO2(quartz) SiC ; C (diamond) graphite

Hard

Soft

Insulator

Conductor (exception)

Very high

3. *Some Important Terminology.

* Space lattice/Crystal lattice : A well defined ordered and regular

arrangement of atoms, molecules or ions in the three dimensional

space is called crystal or space lattice. There are fourteen types of

space lattices possible, they are also called bravais lattices.

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* Lattice Point : The atoms, molecules or ions in a crystalline

substance are shown by points in its space lattice and are called lattice

points.

* Lattice Site : The position occupied by a constituent particle in a

space lattice is called lattice site. Eg. Corner, Body centre, Face centre

etc.

* Unit Cell : The smallest part of the crystal lattice which when

repeated over and over again produces the complete crystal is called

unit cell.

4. Calculation of number of atoms (z) per unit cell/rank of Unit Cell

of a Cubic crystal system :

Unit Cell No. of

atoms at

corners

No. of atoms

at face centres

No.of atoms at body

centre

Total

Primitive/Simple

Cubic U.C.

8x1/8 = 1 0 0 1

Body centered

Cubic U.C.

8x1/8=1 0 1 2

Face Centred

Cubic U.C.

8x1/8=1 6x1/2=3 0 4

5. Seven Primitive unit cells and their possible variations as centred unit

cells:

Crystal Possible Axial Axial angles Examples

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System Variations distances or

Edge

Lengths

* Cubic Primitive,

Body

centred, face

centred

a=b=c £=ß=γ=900 NaCl, ZnS, Cu

* Tetragonal

*Orthorhom

-bic

* Hexagonal

*Rhombohe

d

-ral or

Trigonal

*Monoclinic

Tridinic

Primitive,

Body

centred

Primitive,

Body

Centred,

Face

Centred, End

Centred

Primitive

Primitive

Primitive

End Centred

Primitive

a=b=c

a=b = c

a=b = c

a=b=c

a=b=c

a=b=c

£=ß=r=900

£= ß=r=900

£=ß=900

r=1200

£=ß= r=900

£=r=900

ß = 1200

£=ß=r=900

White Tin, SnO2,

TiO2, CaSO4

Rhombic-

Sulphur, KNO3,

BaSO4

Graphite,

ZnO, CdS

Calcite

(CaCO3), HgS

Mono Clinic

Sulphur,

Na2SO4.10 H2O

K2Cr207,

CuSO4.5 H2O,

H3BO3

6. Close Packing in Crystals :

Two dimensional close packing in metallic crystals.

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Square close Packing Hexagonal Close Packing

* The particles of adjacent rows

show vertical as well as horizontal

alignment.

* Coordination number of each

particle is four.

* Packing fraction or packing

efficiency is about 52.4%

* The shape of void is tetraangular

The particles in every alternate row are

vertically aligned

Coordination number of each particle

is six.

Packing fraction or packing efficiency

is about 60.4%

The shape of void is triangular.

As the packing efficiency of hexagonal close packing is more than square

close packing. Hence H.C.P. is more stable arrangement in two dimensions.

Three dimensional close packing is metallic crystal.

(i) Three dimensional close packing from two dimensional square Close-

packed layers :-

In this packing the square close-packed layers of particles are stacked

one over the other so that all the atoms are vertically as well as

horizontally aligned.

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The lattice thus generated is the simple cubic lattice, and its unit cell is

the primitive cubic unit cell.

(ii) Three dimensional close packing from two dimensional hexagonal

close-packed layers :

In three dimentional close packed structures the arrangement of hexagonal layers

can be done in two ways :

ABAB….Packing ABC ABC….Packing

ABAB….or H.C.P. packing is generated by

stacking H.C.P.-layers one over the other

such that spheres in each alternate layer are

vertically aligned

Hexagonal close packing, e.g.

Mo, Mg, Be etc.

Packing efficiency is 74%

ABC ABC or C.C.P. is

generated by stacking

H.C.P. layers one over

the other such that

spheres in every fourth

layer are vertically

aligned

Cubic close packing,

e.g.

Cu, Ag, Au, Fe, Ni, Al,

etc.

Packing efficiency is

74%

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7. VOIDS : The hollow or vacant spaces among the constituent particles in

close packed structures are called voids or interstitials.

Two types of voids in crystals :

Tetrahedral void Octa-hedral void

*The void is surrounded by four

spheres arranged tetrahedral

*Coordination number of letra

hadral void is 4.

* The number of tetra hedral voids

per particle is 2.

Radius ratio in tetrahedral void is

0.225.

The void is surrounded by size spheres

arranged octahedrally

Coordination number of octahedral void

is 6.

The number of octahedral voids per

particle is one.

Radius ratio in octahedral void is 0.414.

8. Relation ship between density and edge length of a unit Cell of a

crystal.

Donsity ‘d’ of unit cell Mass of Unit cell =

Volume of unit cell

No. of atoms x Mass of each atom =

Volume of Unit cell

d = Z x M /a3 x NA

Z – No. of atom/U.C., M – Molar mass of element

a—edge length of U.C. and NA—Avogadro’s No.

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9. Packing efficiency in case of metal crystals with the assumption that

atoms are touching each other.

Packing efficiency = Volume occupied by all the spheres

in the unit cell x 100 / volume of U.C.

For simple cubic system :-

a = 2 r

No. of spheres/U.C. = one

Packing fraction = 1 x 4/3 kk r3x100 ---------------------- (2r)3

= 52.4%

For body centered Cubic system :-

BC = 2 a

CD = 4 r = 3 a or a = 4r/3

No. of spheres per U.C. = 2

Rocking efficiency = 2 x 4/3 kk r3 x 100 ----------------------

(4r/ 3)3

= 68%

For Face centered cubic system :

AC = 2 a

AC = 4r = 2 a or a = 2 2 r

No. of spheres per U.C. = 4

Packing efficiency = 4x4/3 kk r3 x 100 ------------------- (2 2 r)3

= 74%

10. Important Questions (Sold)

1. Some glass objects from ancient civilizations are found to become

milky in appearance, why ?

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Ans. At some temperature they under go crystallization, which make them

milky in appearance.

2. Glass panes fixed to windows or doors of old buildings are found to be

thicker at bottom than at the top. Why ?

Ans. Glass is an amorphous solid, which has property of fluidity.

3. A solid has a cubic structure in which ‘x’ atoms are located at the

corners of the cube ‘y’ atoms are at the cube centres and ‘o’ atoms are

at the edge centres. What is the formula of the compound ?

Ans. No. of ‘x’ atoms = 8 x 1/8 = 1

No. of ‘y’ atoms = 1

No. of ‘o’ atoms = 12 x ¼ = 3

Formula of compound = XYO3

4. Potassium crystallizes in a body centred cubic lattice. What is the

approximate number of unit cells in 11.7 g of potassium ? [ At. Mass

of k = 39 U ]

Ans. No. of moles of ‘k’ = 11.7/39 = 0.3

No. of atoms of ‘k’ = 0.3 x 6.022 x 1023

No. of atoms of ‘k’ per B.C.C. Unit Cell = 2

No. of Unit cells = 0.3 x 6.022x1023 = 9.03 x 1022

25. A compound is formed by two elements M and N. The element N

forms CCP and atoms of M occupy 1/3rd of tetrahedral voids, What is

the formula of the compound ?

Ans. No. of ‘N’ atoms = 8 x 1/8 + 6 x ½ = 4

No. of ‘M’ atoms = 8 x 1/3 = 8/3

Formula of compound = M8/3 N4 = M2N3

6. Niobium Crystallizes in body centered cubic structure of density 8.55

g cm-3. Calculate atomic radius of niobium.

[Atomic mass of Nb = 92.91 U ]

Ans. d = Z x M / a3 x NA or a3 = 2 x 92.91

6.0222 x 1023x8.55

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= 3.609 x 10-23 = 36.09 x 10-24

a = 3.305 x 10-8 cm = 330.5 pm

4r = 3 a or r= 1.732 x 330.5 ----------------

4 = 143 pm.

7. An element crystallizes in a structure having a FCC unit cell of an

edge 200 pm. Calculate the density, if 200 g of this element contains

2.4 x 1024 atoms.

Ans. Z x M d = a3 x NA

Here we can take M/NA = 200/2.4 x 1024

4 x 200 d = = 41.67 g/cc (200)3 x (10-10)3 x 2.4 x 1024

8. The density of chromium metal is 7.2 g/cc. If the unit cell is cubic

with an edge length of 289 pm, determine the type of the unit cell

present in its crystals.

[Atomic mass of Cr = 52 U ; NA = 6.02 x 1023 ]

Ans. Z x M or d x a3 x NA d = Z = a3 x NA M

7.2 x (289)3 x (10-10)3 x 6.02 x 1023

or Z = 52

= 2.011 = 2

Unit Cell is B.C.C.

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Important Questions (Un sold)

1. What is unit cell ? calculate the number of atoms in a f.c.c. unit cell of

an element.

2. Two elements P and Q have B.C.C. and F.C.C. structures. What will

be the number of tetrahendral and octahedral voids per unit cell ?

3. Give the coordination number of tetrahedral and octahedral voids.

4. An element exists as hexagonal close packed structures as well as

cubic close packed structure. In which case the element would have

higher density ?

5. In a compound, oxide ions have CCP arrangement. Cations A are

present in one eighth of the tetrahedral voids and cations B occupy

half the octahedral voids. What is the simplest formula of the

compound?

6. If the radius of bromide ion is 0.182 pm, how large a cation can fit in

each of the letrahedral void ?

7. Iron has body centred cubic structure. The edge length of the unit cell

is found to be 286 pm. What is the radius of an iron atom ?

8. Calculate the density of silver which crystallizes in the F.C.C.

structure. The distance between the nearest silver atoms is 287 pm.

(Molar mass of silver = 107.87 g/mol).

Imperfections in Crystals

11. Any deviation from perfectly ordered arrangement of constituent

particles in a crystal is called imperfection or defect.

A defect in a crystal arise due to Irregularity in the arrangement of

particles or presence of impurities or rise in temperature of the crystal.

Defects in crystals modify the existing properties or even introduce

new properties in a crystal.

Broadly, we can classify the defects into two categories :

(i) Point defects or atomic defects (ii) Line defects.

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Point defects arise due to irregularities from ideal arrangement

around a point or an atom in a crystal, where as line defects arise due

to deviation from ideal arrangement in the entire rows of lattice point.

Point defects are of three types :

(i) Stoichiometric defect

(ii) Non stoichiometric defect and (iii) Impurity defects.

Stolchiometric defects :

A defect in a crystal due to which the stoichiometry of the crystal

remains unaltered. These defects are of two types.

Property Schott Ky defect Frenkel defect

Definition

Density

Conductivity

Mechanical

Stress

Type of

Crystal in

which the

defect arises

Examples

A point defect which arises due to

complete missing of ion pairs.

It decreases

Increases

Decreases

Ionic crystals with high

coordination number and having

comparable ionic radii.

NaCl, KCl, CsCl and Ag Br

Ag Br shows both type of defects

A point defect which arises

due to displacement of a

smaller ion from its lattice

site to interstitial hole.

It remains the same.

Increases

Decreases

Ionic crystals with low

coordination number and

having large difference in

ionic radii

Zn S, Agcl, Ag Br, AgI

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Exeption -

Non stoichiometric defects : A defect in a crystal due to which

stoichiometry (ie ratio of cation to anion) of the compound is changed.

These defects arise in two ways :-

(i) By metal excess

(ii) Metal deficiency

* Metal Excess defect can arise in a crytal by anion vacancy or by extra

cation occupying Interstitial hole.

Metal excess defect due to anion

vacancy

Metal excess defect due to interstitial

cations

* A compound may have excess

metal ion if an anion is absent from

its lattice site leaving a hole which is

occupied by the electron to maintain

the electrical neutrality.

* This defect is generally developed

when an alkali metal halide (NaCl,

KCl) is heated in an atmosphere of

alkali metal vapours. Some of the

anions diffuse in to the surface of the

crystal and combine with alkali metal

atoms to give NaCl. The electron

released are trapped in anionic

vacancy

If an extra cation is present in an

interstitial site and electrical neutrality

is maintained by the presence of

electron in the nearby interstitial sites.

This defect arises in ZnO, when it is

heated, it loses oxygen and Zn2+ ions

are trapped is interstitial holes along

with electrons in the neighboring sites

to maintain electrical neutrality.

*These defects are found in the crystals

which is likely to possesses frenkel

defect.

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* These types of defects are found is

the crystal which is likely to possess

schottky defect.

* F-Centre : An electron trapped in

an anion vacancy is called F-centre.

F-centre in a crystal make it coloured

& para-magnetic. E.g. NaCl becomes

yellow, KCl – Violet and LiCl—

pink.

* Zn O becomes yellow other wise it is

white. The formula of Zinc oxide

becomes Zn(1+x) O.

* Metal deficiency defect :

There are many solids which are difficult to prepare in stoichiometric

composition and contain less amount of metal as compared to the

stoichiometric proportion.

This defect generally found in compounds in which metal exhibits

variable valency e.g. FeO, FeS and NiO. Incase of FeO some of Fe2+ ions are

missing and loss of +ve charge is balanced by the presence of required

number of Fe3+ ions. For the loss three Fe2+ ions in a crystal there will be two

Fe3+ ions in the cation vacancies and one cation vacancy will be unoccupied.

The formula of nonstoichiometric FeO is fe0.95 O1.0

Impurity defects :

These defects in the Ionic solids may be introduced by adding

impurity ions having different valence state than that of the host ions and

creating cation vacancy. For example, addition of SrCl2 to NaCl or CdCl2 to

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AgCl yields solid solutions in which there will be one cation vacancy for

every impurity ion added.

(Na+) (Cl-) (Na+) (Cl-) (Na+) Cation vacancy(Cl) (Na+) (Cl-) (Cl-)

(Sr2+) (Cl-) (Na+) (Cl-) (Na+)

(Cl-) (Na+) (Cl-) (Na+) (Cl-)

12. Properties of Solids :

(a) Electrical properties of Solids

Solids can be classified into three types on the basis of their

conductivities.

(i) Conductors : The solids with conductivities ranging between 104 to

107 ohm-1 m-1 are called conductors. Metals have conductivities in the

order of 107 ohm-1 m-1 are good conductors.

(ii) Semiconductors : These are the solids with conductivities ranging

between 10-6 to 104 ohm-1 m-1.

(iii) Insulators : These are the solids with very low conductivities ranging

from 10-20 to 10-10 ohm-1 m-1

Size of energy gap in conductors, semiconductoss and Insulators:-

In metal (conductors) there in no energy gap due to overlapping

between valence band and conduction band and it decreases with

increase in temperature. Sn (Tin) Eg = 8 kg/mol .

In case of semi conductors like germanium and silicon, the energy gap

is very small and conductivity increase with increase in temperature

eg. Si and Ge (Eg = 111 kj/mol).

In case of Insulators, the energy gap is very large and therefore the

vacant conduction bond is not available to the electrons of the

completely filled valence band. E.g. Diamond (Eg=511 kg/mol has

large energy gap. It is an Insulator.

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Diagrammatical representation of energy gap (Eg) :

Electrical conductivity of semiconductors increase with rise in

temperature, sincemore electons can jump to the conduction bond. Si and Ge

show this type of behavior and are called intrinsic semi conductors. The

conductivity of Semi conductors can be increased by adding an appropriate

amount of suitable impurity. This process is called doping.

Doping can be done with an imparity which is electron rich or electron

deficient as compared to the intrinsic semi conductor Si or Ge. Such

impurities introduce electronic defects in them.

(a) Electron-rich impurities : Si/Ge (Group -14) doped with electron-rich

impurity P/As (Group -15) is called n-type semi conductor. Fig. (b).

(b) Electron-deficient impurities : Si/Ge (Group-14) doped with electron-

deficient impurity B/Al (Group-13) is called p-type semi conductor. Fig.

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(c).

Creation of n-type and p-type semi conductors by doping groups 13 and 15

elements.

(b) Magnetic properties :

Every substance has some magnetic property associated with it. The

origion of these properties lies in the motion of an electron in a atom. Each

electron therefore behave like a tiny magnet. The magnetic moments which

originate from two types of motions.

(i) Orbital motion of electron around the nucleus

(ii) Spin motion of electron around its own axis.

Magnetic moment Magnetic moment

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On the basis of magnetic properties, substances can be classified into

five categories :-

(a) Diamagnetic solids : These are weakly repelled by external magnetic

field and do not have unpaired electron e.g. TiO2, V2O3, NaCl, C6H6.

(b) Paramagnetic solids : These are weakly attracted by external magnetic

field and have unpaired electrons. They lose magnetism in the absence of

external magnetic field e.g. Cu0, 02, Fe3+, TiO.

(c) Ferromagnetic solids : These are strongly attracted by the external

magnetic field and show permanent magnetism even if external magnetic

field is removed e.g. Fe, Co, Ni, CrO2.

(d) Anti ferromagnetism :- In these type of solids the domain (A group of

metal ions in a small region) orient in anti parallel direction in equal

number so that the net magnetic moment is zero eg. MnO.

(e) Ferromagnetic solids :- In these type of solids the magnetic moment of

domains are oriented in anti parallel direction in unequal number so that

the net magnetic moment is very small e.g. Fe3O4 (Magnetite) and all

ferrites like MgFe2O4 and Zn Fe2O4.

(a)

(b)

(c)

Schematic alignment of magnetic moments in

(a) Ferromagnetic (b) anti ferromagnetic and

(c) Ferromagnetic substances

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Important questions (Un solved)

1. State difference between schottky and frenkel defects.

2. Cite one example each of following : (i) Solid which exhibit schottky

defect and (ii) Frenkel defect. (iii) Schottky and frenkel defects both.

3. What is F-centre ? What are its effect on the properties of solid.

4. What happens when Zn O is heated strongly ? Point out the Kind of defect

developed in ZnO as a result of heating.

5. Which of these two CdCl2 and NaCl will produce schottky defect ?

6. What happens when Fe3o4 is heated to 850 K and why ?

7. Analysis shows that Nickel oxide has formula Ni0.98 O1.0

What fraction of Nickel exists as Ni2+ and Ni3+ ions ?

8. Give at least two important application of n and p-type semi-conductors.

9. Define the terms “ Valence band, conduction band, forbiddon zone,

ferrimagnetic solids and ferro magnetic solids.

10. If NaCl is doped with 10-2 mol% of SrCl2. What is the concentration of

cation vacancy and actual number of cation vacancies ?

For Bright Students (Add to your Knwolwedge)

1. Effect of temperature and pressure on crystal structure : with the

increase in pressure the corrdination number increases resulting is change

in the crystal structure e.g. when NaCl (6 : 6 coordination) crystal is

subjected to high pressure it changes to CsCl-type structure (8:8-

coordinations)

If temperature is increased, the coordination number decreases e.g.. When

CsCl (8:8-coordination) crystal is heated to 760 K it changes to NaCl

type structure (6:6 – coordination)

2. Relation between radius ratio and coordination number for Ionic

crystals :

Radius ratio : It is the ratio of radius of cation to radius of anion in an

ionic crystal.

Radius ratio r+/r- Structure Coordination No. Examples

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0-0.155

0.155-0.225

0.225-0.414

0.414-0.732

0.732-1.00

Linear

Triangular

Tetra hedral

O ctahedral

Cubic (b.c.c.)

2

3

4

6

8

HF2

B2O3, BN

SiO44-; ZnS

MgO, NaCl

CsCl, Cs Br.

3. Structure of simple cubic Ionic compounds (AB, AB2, A2B-Types)

Type Structure Example Co.No. No. of formula units per U.C.

Calculation of no. of cations and onions per unit cell.

(i) AB Rock Salt-type (NaCl)

F.C.C. Halides of Li, Na, K and Rb Mgo, AgF, NH4Cl

Na+=6Cl- = 6 4

Na+= 12x1/4+Body centre = edge centre+Body centre = 3+1 = 4Cl = 8x1/8+6x1/2 corners + face centres = 1 + 3 = 4

(ii) CsCl B.C.C. Cs Br, Cs I, TlCl,Te Br, TlI, TlCN

Cs+=8Cl-=8 1

Cs+ = Body centre = 1Cl- = Corners = 8x1/8=1

(iii) Zn S(Zinc blende)

F.C.C. CuCl, Cu Br, CuI AgI, BeS.

Zn2+=4S2-=4 4

Zn2+=4 (50% of Tetra hedral vids)S2-=4 (at corners and face cntre)

AB2-TypeCaF2

(Fluorite type)

F.C.C. BaF2, BaCl2,SrF2, SrCl2

CdF2, PbF2

Ca2+=8F-=4 4

Ca2+=4(one at corner + three At face centres)F = 8 (at 100% Tetra hedral voids)

A2B- F.C.C. K2O, Li2O, Na+=4 Na+=8 (one at each

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Type Na2O(Anti fluo rite)

Rb2O, Rb2S O2-=8 4 Tetra hedral void)O2-=4 (One at corner three at Face centres.)

Important Question (Un solved)

1. What are the coordination no. of Cation and anions in CaF2 and K2O.

2. Mgo has NaCl-type structure. What is the coordination no. of oxide

ion?

3. The ionic radii of Mg2+ and O2 ions are 66 pm and 140 pm

respectively. What type of interstial void Mg2+ is likely to occupy ?

What is the probable coordination number of Mg2+ ion ?

4. If the radius of Bromide ion is 0.182 nm, how large a cation can fit in

each of the Tetra hedral void ?

5. Lead II sulphide crystal has NaCl structure. What is its density ? The

edge length of the uniti cell of PbS crystal is 500 pm.

(NA=6.0222x1023 mol-1, at masses : Pb = 207, S = 32)

6. Caesium chloride crystallizes as a body centred cubic lattice and has a

density 4.0 g/cc. Calculate the edge length of unit cell. Also distance

between cesium and chloride ions. [Molar Mass of CsCl = 168.5,

NA=6.022 x 1023]

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