1 solid state1

8/9/2019 1 Solid State1

http://slidepdf.com/reader/full/1-solid-state1 1/56

Solid State

1

Solid State

General Characteristics of SolidsCrystalline and Amorphous SolidsClassification of Crystalline Solids Basedon Interparticle Binding Forces

Lecture-1

Crystal Lattices and Unit CellsCalculation of Number of Particles in a Unit Cell

Lecture-2

Lecture-3Close Packed Structures, Packing EfficiencyCalculations Involving Unit Cell Dimensions

Lecture-4 Interstitial Voids

Lecture-5 Imperfections in Solids

Lecture-6Electrical Properties of SolidsMagnetic Properties of Solids



100 Hours 100Marks

2

Believe where others doubt . Work where others refuse.

Save where others waste. Stay where others quit and

you will win where others lose.



Solid State

3

Polymers

Solid State 1

Lecture-1

1. General characteristics of states(1) Gas State(2) Liquid State(3) Solid State

Gas State :(i) In this state no definite shape and volume.(ii) Very less attractive force between the gaseous particles.

Liquid State :(i) In this state definite volume but indefinite shape is

there. Shape is according to the vessel size and shape.

(ii) Force of attraction between the particles is more ascompare to gas particles. Solid State :

(i) In this state shape and volume both are definite.(ii) Force of attraction between the particles is maximum.

Solids are having following properties(i) The intermolecular distances in solids are short and

intermolecular forces are strong.(ii) The constituent particles (atoms, ions or molecule

of solids have fixed postions and can only oscillaabout their mean positions.

(iii) Solids are almost incompressible.



100 Hours 100Marks

4

Distinction between crystalline & amorphous solidsDistinction between crystalline & amorphous solidsDistinction between crystalline & amorphous solidsDistinction between crystalline & amorphous solidsDistinction between crystalline & amorphous solids Properties Crystalline solids Amorphous solids 1. Shape The crystalline solids The amorphous solids

have definite shape. have characteristic shapeirregular shapes.

2. Order in They have regular arrange- They do not have any arrangement ment of constituent particles. regular arrangement of of constituent They are said to exhibt the constituent particles. particles long range order. They may have short range order.3. Melting Point They have sharp and They do not have sharp m.p

characteristic melting point. They gradually soften over arange of temperature.

4. Cleavage When cut with a sharp When cut with a sharp edged Property edged tool,they split into tool they cut into two pieces

two pieces and the newly with irregular surfaces.generated surfaces are plain and smooth.

5. Enthalphy They have a definite and chara- They do not have definitecteristic enthalpy of fusion enthalpy of fusion.

6. Anisotropy They are anisotropic and They are isotropic and havehave different physical same physical properties in properties in different all directionsdirections.

7. Nature They are true solids. They are isopseudo solidsand supercooled liquids.

8.Common Copper, silver, iron, common Glass, rubber, plastics, etc. examples salt, zincsulphide

potassium nitrate etc.

(iv) The density of solids is greater than that of liquidsand gases.

(v) Solids diffuse very slowly as compared toliquids and gases.

On the basis of different properties.We are having following types of solids :(a) Crystalline solids (b) Amorphous solids



Solid State

5

Depending upon the nature of intermolecular forces operating in them

Note Isotropic in nature means their electrical and mechnical properties are same in all the directions. Anisotropic in nature means, their electrical and mechanical propertiesare different in different directions.Now we come to classification of crystalline solids.

1. Type of solid :√√√√√ Ionic: NaCl, MgO, KCl, CsCl, ZnS, Na 2 SO4 , KNO3 ,

CuSO4.5H 2O.√√√√√ Covalent or network: Diamond(C), Graphite(C),

Carborundum(SiC), Quartz(SiO 2), Boron-nitride(BN).√√√√√ Molecular: Ice, Solid CO 2 , Sulphur (S8).Glucose,

Napthalene etc.√√√√√ Metallic: s-block and d-block metals and some p-block

elements, Na, CU, Ag, Au, Ni, Pt, Cr, etc.

Structure of quartz (crystalline) and quartzglass (amorphous)Quartz is a form of SiO 2 (silica). It has tetrahedral SiO4

– – units which are orderly arranged in crystalline quartz as shownin figure. When SiO 2 is melted and the melt is cooled, forms quartz glass which is amorphous. In this state, th SiO4 units are randomly joinded.

(a) crystal Quartz (b) Amorphous Glass



Solid State

7

Questions Problems based on this Lecture

Q.1 Why do solids have a definite volume ? Ans. The constituent particles in solids are bound to their mean position

strong cohesive forces of attraction. The interparticle distances reuncharged at a given temperature and thus solids have a definite vo

Q.2 Why are solids rigid ? Ans. In solids, the constituent atoms or molecules or ions are not free to

but can only oscillate about their mean positions due to strong interator intermolecular or interionic forces.

Q.3 What makes a glass different from a solid such as quartz ? Under whaconditions could quartz be converted into glass ? Ans. Glass is an amorphous solid in which the constituent particles (4

–

– tetrahedral) have only a short range order and there is no long rorder. In quartz the constituent particles (SiO4

– – tetrahedral) have bothshort range as well as long range orders. On melting quartz and cooling it rapidly is converted into glass.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................



100 Hours 100Marks

8

Q.4 Why does the window glass of the old building look milky ? Ans. It is due to heating during the day and cooling at night, i.e., annealing

a slow cooling process due to this after some years. Glass acquires somecrystalline character and these character look milky.

Q.5 How can a material be made amorphous ? Ans. First melting the material and then cooling it rapidly from the liquid state.Q.6 How does amorphous silica differ from quartz ? Ans. In amorphous silica SiO4

– – tetrahedral are randomly joined to each other while

in quartz they are joined in a regular manner.Q.7 Ionic solids conduct electricity in the molten state but not in the solid

state. Explain. Ans. In the molten state, ionic solids dissociate to give free ions and hence

can conduct electricity. However, in the solids state, since the ions arenot free to move but remain held together by strong electrostatic forcesof attraction. Therefore, they are electrically insulator.

Q.8 (a) ‘Stability of a crystal is reflected in the magnitude of its meltingpoints’.Comments.(b) The melting points of some compounds are given below :Water = 273K, Ethyl alcohol = 155.7K, Diethyl ether = 156.8K,Methane = 90.5K.What can you say about the intermolecular forces between thesemolecules ?

Ans. (a) Higher the melting point, stronger are the forces holding the constituentparticles together and hence greater is the stability.(b) The intermolecular forces in water and ethyl alcohol are mainly thehydrogen bonding. Higher melting point of water as compared to alcoholshows that hydrogen bonding in ethyl alcohol molecules is not as strongas in water molecules. Diethyl ether is a polar molecule. The intermolecularforces present in them are dipole-dipole attraction.Methane is a non polarmolecule. The only forces present in them are the weak van der Waals’forces.

Q.9 Refractive index of a solid is observed to have the same value along thedirections. Comment on the nature of the solid. would it show cleavageproperty ?

Ans. Since the solid has the same value of refractive index along all directions, itis isotropic and hence, amorphous. Being an amorphous solid, it would notshow a clean cleavege when cut with a knife. Instead, it would break intopieces with irregular surfaces.

Q.10 Classify the following as amorphous or crystalline solids:Polyurethane, naphthalene, benzoic, acid, teflon, potassium nitrate,cellophone, polyvinyl chloride, fibre glass, copper.

Ans. Amorphous sol ids : Polyurethane, teflon, cellophone, polyvinylchloride and fiber glass



Solid State

9

Crystalline solids : Naphthalene, benzoic acid, potassium nitrate acopper.

Q.11 What type of solids are electrical conductors, malleable and ductile ? Ans. Metallic solidsQ.12 Classify the following solids in different categories based on the nature o

intermolecular forces operating in them: Potassium sulphate, tin, benzeneurea, ammonia, water, zinc sulphide, rubidium, argon, silicon carbide.

Ans. Ionic solids : Potassium sulphate, zinc sulphateCovalent solids : Graphite, silicon carbideMolecular solids : Benzene, urea, ammonia, water, aMetallic solids : Rubidium, tin.

Q.13 Solids A is a very hard electrical insulator in solid as well as molten staand melts at an extremely high temperature. What type of solid is it ?

Ans. Covalent.Q.14 Urea has a sharp melting point but glass does not. Explain. Ans. Urea is a crystalline solid, so it has a sharp melting point. On the

hand, glass is a molecular solid, so its melting point is not sharp.Q.16 Why is glass considered a super-cooled liquid ? Ans. Glass has characteristics similar to liquids. It can flow extremely

so it is considered to be a super-cooled liquid.Q.17 Solid A is very hard electrical insulator in solid as well as in molten sta

and melts at extremely high temperature. What type of solid is it ? Ans. It is a covalent or network solid.Q.18 Write a distinguishing feature of metallic solids ? Ans. Electrical conduction in solid state.Q.19 How do metallic and ionic substance differ in conducting electricity ?

Ans. Metallic solids Ionic solids1. Conduct electricity both in 1. Conduct electricity only in solid as well as in molten state. molten state or when dissolv

in some polar solvent like wate2. Conductivity is due to the 2. Conductivity is due to the migration of electrons. migration of cations and anio

Q.20 (a) Define liquid crystals.(b) What are Semectic and Nematic liquid crystals ?

Ans. (a) In a temperature range just above the melting point, some substaare able to exist in a define pattern as in a solid but can flowa liquid. These substances are known as liquid crystals.

(b) Semectic liquid crystals are having soap like phase and Nemcrystals are those which are having needle or thread-like ph



100 Hours 100Marks

10

Lecture-2

Characteristics of a Crystal Lattice

Let us sum up the charateristics of a cystal lattice. The followingare the characteristics of a crystal lattices.(i) Each point in a crystal lattice is called lattices points

or lattices site.(ii) Each point in a crystal lattices represents one constituent

particles which may be an atom, a molecule or an ion.(iii) Lattics points are joined by straight lines to bring out

the geometry of the lattices.

Type of Crystal Systems :—There are basically two types of unit cells constituting differentcrystal systems. These are :(i) Primitive unit cells(ii) Centred unit cells

(i) Primitive unit cells:These are unit cells which have points (or particles)only at the corners. These are also called simple unit cells.

(ii) Centred unit cells:These are unit cells which have points (or particles) at the corners as well as at some other positions.(a) Face centred unit cells in which the points are present

at the corners as well as the centre of each face.(b) Body centred unit cells in which the points are present at all the corners as well as the body centre

of the unit cell.(c) End centred unit cells in which the points are present at all

the corners and at the centre of any two opposite faces.



Solid State

11

Space lattice or Lattice Points:—

Now we will find about details of crystal solid. Space Lattice :It is an array of points showing how ions, at-oms or molecules are situated (arranged) in different site in three dimensional space is called space lattice.Lattice Point :It is the point where atom, ion or molecule placed.Example – In simple cubic total numbers of lattice point are 8.

Total no. of atoms are requiredFor simple cubic = 8

For triangle arrangement no. of Lattice points three.

Space lattice = Triangle Lattice points = 8 Lattice points = 3 Space lattice = cubic

Unit cell :It is the smallest unit which on repeatation gives the requiredcrystal lattice and gives almost all the information about thesolid.

Example – In two dimensional:

Square closed structure Hexagonal closed structure Unit cell is square unit cell is hexagon



100 Hours 100Marks

12

In three dimensional:If cubic unit cell is there then we have as , , are axial angles. Parameter of Unit cell : A unit cell is characterized by :—(i) Its dimensions along the three edges as ‘a’, ‘b’ and ‘c’.

These edges may or may not be mutually perpendicular.(ii) Angles , and between the angle pair of edges,the

angle is between the edges ‘b’ and ‘c’, angle isbetween the edges ‘c’ and ‘a’ and angle is between theedges ‘d’ and ‘b’. These are known asinterstial angleor axial angles.

Crystal System : On the basis of boundaries of unit cell.We are having seven types of crystal systems.Example– Cubic, Tetragonal, orthorhombic, hexagonalrhombohedral or trigonal, monoclinic, tirclinic. Seven Primitive Unit Cells and their Possible arrangement.

Variation as Centred Unit Cells CrystalCrystalCrystalCrystalCrystal systemsystemsystemsystemsystem Possible variations AxialPossible variations AxialPossible variations AxialPossible variations AxialPossible variations Axial distances Axial anglesdistances Axial anglesdistances Axial anglesdistances Axial anglesdistances Axial angles ExamplesExamplesExamplesExamplesExamples

or edge le or edge le or edge le or edge le or edge lengthsngthsngthsngthsngths CubicCubicCubicCubicCubic Primitive, BodyPrimitive, BodyPrimitive, BodyPrimitive, BodyPrimitive, Body a=b=ca=b=ca=b=ca=b=ca=b=c NaCl, Zinc NaCl, Zinc NaCl, Zinc NaCl, Zinc NaCl, Zinc

centred, Face-centredcentred, Face-centredcentred, Face-centredcentred, Face-centredcentred, Face-centred blende, Cu blende, Cu blende, Cu blende, Cu blende, Cu TetragonalTetragonalTetragonalTetragonalTetragonal Primitive,Primitive,Primitive,Primitive,Primitive, a=ba=ba=ba=ba=b ccccc White tin, SnO White tin, SnO White tin, SnO White tin, SnO White tin, SnO22222,,,,,

Body-centred, Body-centred, Body-centred, Body-centred, Body-centred, TiO TiO TiO TiO TiO22222, CaSo, CaSo, CaSo, CaSo, CaSo44444

OrthorhombicOrthorhombicOrthorhombicOrthorhombicOrthorhombic Primitive,Body-Primitive,Body-Primitive,Body-Primitive,Body-Primitive,Body- aaaaa bbbbb ccccc Rhombic sulphur, Rhombic sulphur, Rhombic sulphur, Rhombic sulphur, Rhombic sulphur,centred, Face-centcentred, Face-centcentred, Face-centcentred, Face-centcentred, Face-cent KNO KNO KNO KNO KNO33333, BaSO, BaSO, BaSO, BaSO, BaSO44444

-red, End-centred-red, End-centred-red, End-centred-red, End-centred-red, End-centred HexagonalHexagonalHexagonalHexagonalHexagonal PrimitivePrimitivePrimitivePrimitivePrimitive a=ba=ba=ba=ba=b ccccc Graphite, ZnO, Graphite, ZnO, Graphite, ZnO, Graphite, ZnO, Graphite, ZnO,

CdSCdSCdSCdSCdS RhombohedralRhombohedralRhombohedralRhombohedralRhombohedral PrimitivePrimitivePrimitivePrimitivePrimitive a=b=ca=b=ca=b=ca=b=ca=b=c Calcite (CaCO Calcite (CaCO Calcite (CaCO Calcite (CaCO Calcite (CaCO33333),),),),), or Trigonalor Trigonalor Trigonalor Trigonalor Trigonal HgS (Cinnabar) HgS (Cinnabar) HgS (Cinnabar) HgS (Cinnabar) HgS (Cinnabar) MonoclinicMonoclinicMonoclinicMonoclinicMonoclinic Primitive,Primitive,Primitive,Primitive,Primitive, aaaaa bbbbb ccccc Monoclinic Monoclinic Monoclinic Monoclinic Monoclinic

End-centedEnd-centedEnd-centedEnd-centedEnd-cented sulphur, sulphur, sulphur, sulphur, sulphur, Na Na Na Na Na22222SOSOSOSOSO44444 , 10H , 10H , 10H , 10H , 10H 22222OOOOO

TriclinicTriclinicTriclinicTriclinicTriclinic PrimitivePrimitivePrimitivePrimitivePrimitive aaaaa bbbbb ccccc K K K K K22222Cr Cr Cr Cr Cr 22222OOOOO 7 7 7 7 7 ,CuSO,CuSO,CuSO,CuSO,CuSO44444



Solid State

13

All crystals do not have simple lattices. There can be 14different ways in which similar points can be arranged in a three dimensional space.

Bravais Lattices :On these crystal system, arrangement of the atom or ions which gives space lattices known as Bravai

lattices. We are having 14 types of bravais latticeExample–In case of cubic system we are having three types oarrangements.

(1) Simple Cubic (SC)

Total lattice point = 8

(2) Body Centred Cubic (BCC)

Total lattice points = 9

Eight at corners and one at centre.(3) Face centred Cubic (FCC)

Total lattice points = 14

Type of cellType of cellType of cellType of cellType of cell No. of atoms Number ofNo. of atoms Number ofNo. of atoms Number ofNo. of atoms Number ofNo. of atoms Number of No. of faces in TotalNo. of faces in TotalNo. of faces in TotalNo. of faces in TotalNo. of faces in Total at corners at corners at corners at corners at corners atoms in faces the body atoms in faces the body atoms in faces the body atoms in faces the body atoms in faces the body of cubeof cubeof cubeof cubeof cube

S ingle or primitive cubicingle or primitive cubicingle or primitive cubicingle or primitive cubicingle or primitive cubic 8×(1/8) = 18×(1/8) = 18×(1/8) = 18×(1/8) = 18×(1/8) = 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 Body-centred cubic (bcc) Body-centred cubic (bcc) Body-centred cubic (bcc) Body-centred cubic (bcc) Body-centred cubic (bcc) 8×(1/8) = 18×(1/8) = 18×(1/8) = 18×(1/8) = 18×(1/8) = 1 0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 Face-centred cubic (fcc)Face-centred cubic (fcc)Face-centred cubic (fcc)Face-centred cubic (fcc)Face-centred cubic (fcc) 8×(1/8) = 1 6×(1/2) = 38×(1/8) = 1 6×(1/2) = 38×(1/8) = 1 6×(1/2) = 38×(1/8) = 1 6×(1/2) = 38×(1/8) = 1 6×(1/2) = 3 0 0 0 0 0 4 4 4 4 4



100 Hours 100Marks

14

Eight at corners and six at face. So, in this we are having cube as crystal system and those arrangement of atoms as Bravais lattice. So, it isone crystal system is cubic and there bravais lattices s.c.,b.c.c. and f.c.c.

Calculation of number of particles in a Unit cell:Lattice point in the arrangement (Cubic) SC — 8 BCC — 9 FCC — 14 HCP — 17

In this we will calculate sharing of atoms, ions, molecules with the unit cell and accrording to the positions of the given

atom, ion, molecule.CUbic System

At corner, sharing of atom = 18

At face, sharing of atom = 1 2

At centre, sharing of atom = 1

At edge, sharing of atom = 14



Solid State

15

In case of hexagonal

At corner = 16 At face = 1

2 At centre = 3

Symmetry in the solid Solids are having three types of symmetries.(1) Plane of symmetry(2) Axis of symmetry(3) Centre of symmetry

Example :— In cubic Plan of symmetry = 9 Axis of symmetry = 13Centre of symmetry = 1

Formula of unit cell :When we combine the shared atoms, ions in the unit cell the formula is developped.

Ex.-1 (i) If in the cubic system A at corners B at all the faceThen, find out formula of compound.(ii) If one A is remove from one corner means, one cor

is empty only seven corners are occupied by atoms(iii) If one B is remove from face

Sol. (i) Each corner is having sharing= 18Each face is having sharing= 1

2‘A’ at a corner so each corner is having ‘A’ as = A

8

‘B’ at a face so each face is having ‘B’ as =6



100 Hours 100Marks

16

So total ‘A’ used= A8

Only one ‘A’ is used in the unit cellTotal face = 6

So total ‘ B’ used = B 26× =3B So in this case in the unit cell 1A and 3BTherefore, the formula is AB3 .

(ii) If one A is remove from one corner means, one corner is empty only seven corners areoccupied by atoms

So total ‘A’ used = A

8 7× = 7A

8( A

8 at each corner)

total ‘B’ used = 26× =3B

( B 2 at each face)

So formula 7A

: 3B or 7 A : 24B8Therefore, formula is A 7 B 24 .

(iii) If one B is removed then from ‘six’ face one isempty and ‘five’ are occupied.

So, total ‘B’ used = 5 5

2 2

total ‘A’ used = A

88× =A

So formula 5B A : or 2A : 5 B 2

Therefore, formula is A 2 B 5 .



Solid State

17

Ex.2 If in the above case A at corner B at face C at centre and Dedge, then find out formula of unit cell if followingconditions are given.(a) If one A from one corner is removed.

(b) If one B from one face is removed.(c) If one D from the edge is removed.(d) A, B, C are each are removed.

Sol. At corner sharing of atom = 18 At face sharing of atom= 1 2

At edge sharing of atom =

1

4If no condition : 3 3

A D8, 6, 12 AB CD8 2 4

1,C

(a) If one A is removed from one corner then total occupicorners are 7 which are having ‘A’Total ‘A’ = No. of corner occupied × Sharing of one atomat corner

Total ‘A’ used = A

7 × 8

B at face= B6 3B 2 (6 faces with

B 2 at each face)

C at centre = 1 × C = C(one centre is there)

D at edge= D 12 3D4

(12-edge with D4 at each edge)

Formula is

7A , 3B, C, 3D8 So simple formula is 7A 24B 8C 24D

A 7 B 24C 8 D 24



100 Hours 100Marks

18

(b) If one B is removed then total face occupied are = 5 Total A used A=8 A

8

Total B used B=5 2

5B

2Total C used = 1 × C = C

Total D used D=12 3D4

Therefore, 5B 2

A, , C, 3D and Simple formula is A 2 B 5 C 2 D6 .

(c) If one D is removedTotal A A=8 A

8Total B=6 3

2Total C = 1 × C = C

Total D D 114

11D4

Therefore, , 3B, C, 11D4 and Simple formula is

A4 B 12C 4 D 11(d) If one each removed

Total A A 7 8

7A8

Total B 5 2

5B 2

Total C=0 × C=0 (only one C at centre which is removed)

Total D D

11 4 11D

4Therefore, 11D

4 5B 2

7A8 and Simple formula is A 7 B 20 D 22



Solid State

19

QuestionsProblems based on this Lecture

Q.1 In NaCl crystal, Cl — ions are in fcc arrangement. Calculate the numberof Cl — ions in its unit cell ?

Ans. Cl — ion per unit cell 188 (from corners) + 16 2 (from face corners) = 4.

Q.2 Explain how much portion of an atom located at (i) corner and (ii) bodycentreof a cubic unit cell is part of its neighbouring unit cell.

Ans. (i) A point lying at the corner of a unit cell is shared equally by eigh

cells and therefore, only one-eight (1/8) of each such point belongs given unit cell.(ii) A body centred point belongs entirely to one unit cell since it ishared by any other unit cell.

Q.3 Define face-centred cubic structure. Ans. There is a structural particle at the centre of each face as well as each c

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................



Solid State

21

COORDINATION NUMBER (CN)

Coordination number is defined as number of atoms or

ions surrounded by the given atom or ion.

Example :X

for X CN is 4.

in three dimension coordination number is as;HCP = 12 for BCC, CN = 8FCC (salt like) = 6CCP = 12 , SC = 6


Q.1 What is the 2-dimensional coordination number of a molecule in squarclose packed layer ?

Ans. FourQ.2 (a) What is meant by the term ‘coordination number’ ?

(b) What is the coordination number of atoms(i) In a cubic-close packed structure ?(ii) In a body-centred cubic structure ?

Ans. (a) Coordination number defined as the number of nearst neighboursclose packing.In ionic crystal, coordination number of an ion incrystal is the number of oppositely charged ions surrounding thparticular ion.

(b) (i) 12 (ii) 8.Q.3 In a body-centred cubic (bcc) unit cell, a metal atom at the center of th

cell issurrounded by how many other metal atoms ? Ans. A metal atom at the center of the cell is surrounded by 8 atoms.



100 Hours 100Marks

22

Lecture-3Relation between radius of atom, ion and edgelength. Density, Packing fraction, Formula of

unit cell.(i) For Simple Cubic Unit Cell(SCC) :

In this figure we are having as.

r a

If edge length of unit cell is ‘a’ and the radius of theatom is r.Then from the figure.

(a) Radius r + r = a a 2r =

(b) Total no. of atoms in the unit cell at each corner

one atom is there with 18 sharing. So, total atoms = No. of corner × Part of sharing

188× = 1

One atom is shared in the unit cell.(c) If A and B atoms are palced at lattice points in the

unit cell alternativily



Solid State

23

B

B

B

B

A

A

A

A

Sharing of A 18=

Total lattice points of A = 4 So, A

8 ×4= Total sharing of A A 2=

Sharing of B 18=

Total lattice points of B = 4 So, B

8 ×4= Total sharing of B B 2=

Formula, A 2

B 2: AB

(d) Density :

3Mass of unit cellMassd

Volume a where a3 volume of unit cell

Mass of unit cell is due to atoms in the unit cell. Totalatoms in the unit cell=16.023×10 23 atoms having mass = atomic weight

23

23o

3 3

Atomic weightMass of one atom

6.023 10 Atomic weight Atomic weight

6.023 10 N So, densitya a



Solid State

25

(c) If A at corner and B at face then formula total A used in unit cell

total B used in unit cell A 3B, So AB3 is formula

(d) Density of the unit cellMass of unit cell

dVolume of unit cell

Mass is due to four atom

o3

Atomic weight4

Nd a(e) Packing fraction of the unit cell

Volume packedd Total Vol. of unit cellVolume is packed is due to four atoms. Which areinvolved in the unit cell.

Volume of one atomTotal volume packed by the atom

Total volume of unit cell = a3

So, Packing fraction

3

3

4 a43 2 2 74%

a So, empty space is 26%.



100 Hours 100Marks

28

Volume occupied in hcp arrangement = 74%

Hexagonal Closed Packing(h.c.p.)Main points in this are :(a) It AB AB AB types

(b) Packing fraction is 74%. So empty space is 26%.(c) Coordination number is 12.(d) Total atoms used in this = 6(e) Volume of h.c.p. = 3 24 2r

Cubic Closed Packing(c.c.p.)

Main points in this are :(a) ABC ABC ABC type.(b) Packing fraction is 74%.

So empty space is 26%.(c) Coordination number is 12.(d) CCP is same as that of FCC for the calculation (e) Volume of c.c.p. = 3 16 2r

Note : Volume of h.c.p. is more than CCP.Note: In case of f.c.c. salt like

Four formula in unit cell



Solid State

29

o3

M4Nd

aEx. NaCl, Cl at corner and Faces, Na+ at edge centre and at

centre.Edge length a = r – + 2r + + r – = 2(r ++r –) where r– = for Cl– and r + = for Na+

Na+

Cl— Cl—

In case of b.c.c.One formula in unit cell

o3

M 1Nd

aEx. CsCl, Cs+ at centre, Cl– at corner, analysis is as,

r 2r r a 3 2(r r ) a 3

aa 2

Cl—

Cl—Cs+

Note:— c.c.p. is equivalent to f.c.c. for calculation.



100 Hours 100Marks

30


Q.1 Which of the following lattices has the highest packing efficiency ?(i) Simple cubic(ii) Body-centred cubic(iii) Hexagonal close-packed lattice

Ans. Hexagonal close-packed lattice has the highest packing efficiency (74%)Q.2 A metal crystallizes into two cubic phases, face-centred cubic (FCC) and

body-centred cubic (BCC) whose unit cell lengths are 3.5Å and 3.0Å respectively. Calculate the ratio of the densities of FCC and BCC.

Ans. 3 A

3 A

z MDensity of unit cell(d) N a4 MDensity of f.c.c. unit cell N (3.5 )

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................



100 Hours 100Marks

34

Lecture-4VOIDS

Voids : It is three dimension space in the solid which is empty.

In two dimension s0me of the voids are as,

Square void Triangular voidIn three dimension we are having two types of voids.

(1) Tetrahedral Void (2) Octahedral Void

Tetrahedral Void : In case of tetrahedral void four atoms or ions form tetrahedral. In which three atoms/ions are on the base and oneis on the top.

It is having following properties :

(a) Coordination number = 4(b) If radius of void is r

void and radius of the atom is r

atom then,

voidatom

r 0.225 r (c) In the void generally cations are placed so,

+r 0.225 r —

(d) w.r.t. the given atom number of tetrahedral voids are two.Octahedral Void : In case of octahedral void we are having four

atoms on the base and two atom on the top and bottom of the four atoms.It is having following points.(a) Coordination number is six for octahedral void.(b) If radius of the void is r void and radius of atom is r atom then,



100 Hours 100Marks

38

2 2

2 2

AD C CD

2a a 3aThe tetrahderal void is present at the centre of the body diagonal AD show that half the length of this diagonal is equal to thesum of the radii of R and r. Thus,

AD 3.aR r .............(ii) 2 2

Dividing equation (ii) by (i), we get,

R r 3.a 2 3

R 2 2a 2r 3 1R 2

r 3 3 2or 1R 2 2 1.732 1.414 0.225

1.414

r 0.225 R Thus, for an atom to occupy a tetrahedral void, its radius mustbe 0.225 times the radius of the sphere.For a given atom number of octahedral void = No. of atomsNumber of tetrahedral void = 2×Number of atoms

B o d y d i a

g on

al

D

B

C

Face diagonal

a2r

a A



Solid State

39


Q.1 Define void. Ans. The empty spaces present between the metal atoms or the ions when

are packed within the crystal are called voids.Q.2 What is number of tetrahedral voids in an unit cell of a cubic close-packe

structure ?

Ans. There are 8 tetrahedral voids in a unit cell.Q.3 What is the coordination number of an octahedral void ? Ans. Six.Q.4 A compound forms hexagonal close-packed structure. What is the tot

number of voids in 0.5 mol of it ? How many of these are tetrahedara voids ?

Ans. No. of atoms in the close packing 0.5 mol = 0.5×6.022×1023 = 3.011×1023

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................



100 Hours 100Marks

40

No. of octahedral voids = 1×No. of atoms in the packing = 3.011×1023

No. of tetrahedral voids = 2×No. of atoms in the packing = 2×3.011×1023

= 6.022×1023

Total No. of voids = 3.011×10 + 6.022×1023 = 9.33×1023.Q.5 What is the relation of number of voids in terms of number of atoms ?

OrIn a close paked arrangement of N-spheres, how many (i) tetrahedral and(ii) octahedral sites are present?

Ans. Number of octahedral voids = No. of atoms in the close paked arrangement.Number of tetrahedral voids = 2 × No. of atoms

Q.6 If the radius of the tetrahedral void is ‘r’ and radius of the atom closepacking is R. What is the relation between ‘r’ and R ?

Ans. ‘r’ = 0.225R Q.7 If the radius of octahedral void is ‘r’ and radius of the atom in close

packing is R. what is the relation between ‘r’ and R ? Ans. ‘r’ = 0.414 R.Q.8 In a solid XY, ‘X’ atoms are in ccp arrangement and ‘Y’ atoms occupy all

the octahedral sites. If all the face centred atoms along one of the axesare removed, then what will be the resultant stoichiometry of the compound?

Ans. In a ccp types structure, 8 X atoms are at the corners and 6 X atoms are atthe face centres of the cube. Removal of all the face centred atoms along oneof axes means the removal of 2 X atoms. Thus, only 4 X atoms will be lefton the centred of the faces.

1 1No. of X atoms per unit cell 8(corner) (face centre) 38 2 Y atoms are present in all the octahedral sites i.e. 12 at edge centres and 1at body centre.

1No. of Y atoms per unit cell 12 1 44Therefore, Stoichiometry of the compound = X 3 Y 4 .

Q.9 Atoms of element B from hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formedby the element A and B ?

Ans. Suppose, the number of B atoms in the packing =nNumber of tetrahedral voids = 2n As only 2/3rd of tetrahedral voids are occupied by atoms A,

2 4No. of atoms A 2n n3 34Ratioof A andB n n 4 33

The formula of the compound is A 4B3.



Solid State

45

2x + 27 – 3x – 20 = 0 x = 7

So, % of Fe+2 = 7 100 77.7% 9

Fe+3 = 100 – 77.7 = 22.3%Ex. Find out % of Cu+ and Cu++ in Cu 1.5 O. Ans. In Cu 1.5 O or Cu 15 O 10

Let Cu++ is x and Cu+ is (15–x) in Cu 15 O 10 then Total charge (+2)x+(15–x)×(+1) + 10×–2

2x + 15 – x – 20 = 0 or x = 5

% of Cu++ = 5 100 33.33% 15 % of Cu+ = 100 – 33.33 = 66.67

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................



100 Hours 100Marks

46


Q.1 What is the meaning of the term imperfection in solids ? Ans. Inperfection refers to the departure from the perfect periodic arrangementof atoms, ions or molecules in the structure of crystalline substances.Q.2 What is the type of lattice imperfections found in crystals ? Ans. (a) Stoichiometric defects, viz., Schottky defect and Frenkel defect and

(b) Non-stoichiometric defects, viz., metal excess, metal deficiency andimpurity defects.

Q.3 What type of defect can arise when a solid heated ? Which physicalproperty is affected by it and in what way ?

Ans. On heating a solid vacancy defect is produced in the crystal. This isbecause on heating. some lattice sites become vacant. As a result of this

defect, the density of the substance decreases because some atoms or ionsleave the crystal completely.Q.4 CaCl2 will introduce Schottky defect if added to AgCl crystal. Explain. Ans. Two Ag+ ions will be replaced by one Ca2+ ions to preserve electrical

neutrality. Thus, a hole is created at the lattice site for every Ca2+ ionintroduced.

Q.5 Why is Frenkel defect not found in pure alkali metal halides ? Ans. This is because of the fact that alkali metal ions have large size which

cannot fit into the interstitial sites.Q.6 Name the non-stoichiometric point defect responsible for colour in

alkali halides. Ans. Excess of metal ions and formation of F-centres as a result of trappedelectrons.

Q.7 What are F-centres ? Ans. The free electrons trapped in the anion vacancies are called as F-centres.Q.8 Name one solid in which both Frenkel and Schoottky defects occur. Ans. AgBrQ.9 Why does Frenkel defect not change the density of AgCl crystals ? Ans. Because of the Frenkel defect, no ions are missing from the crystal as a

whole, therefore, there is no change in density.

Q.10 What is Schottky defect ? Ans. If equal number of cations and anions are missing from their lattice sites,the defect is known as defect.

Q.11 What are non-stoichiometric compounds ? Ans. If the actual ratio of the cations and anions is different as represented by

the ideal chemical formula of the compound, it is called a non-stoichiometric compound. Ex. Fe0.9O, Cu1.5O



100 Hours 100Marks

50

this electron leaves an electron hole at its original position. If ithappens, it would appear as if the electron hole has moved in thedirection opposite to that of the electron that filled it. On applying the electric fields, electrons moves towards the positively charged plate through electronic holes. But it seems that electrons holesare positively charged and are moving towards negativelycharged plate. This is called p-type semi-conductor.

Group-13 Group-14 Grou p-15 B C N(gas) Al Si P Ga Ge AsIn Sn SbTh Pb Bi

Application of n-type and p-type(1) For making electronic components(2) For making npn and pnp type transistors.

Magnetic Properties :—The magnetic properties of the substance depend on electrons.Every electrons revolves around the nucleus and also rotatesaround its own axis and thus every electron behaviours like amagnet. On the basis of magnetic properties the substance may be ;

(i) Paramagnetism :The substances which are weakly attracted by magnetic field arecalled paramagnetic substances. They are magnetised in amagnetic field in the same direction This is due to presence of one or more unpaired electrons. These electrons attracted by mag-netic field.



100 Hours 100Marks

52

Mechanism of Electrical Conduction

The conduction in most of the solids is through electron movementunder an electrical field. However, in some ionic solids, theconduction is by ions. Therefore in the solids where the conduction is by the movement of electrons, the electrical conductivitydepends on the number of electrons availiable to participate in the conduction process.The spaces between valence band and conduction band representenergies forbidden to electrons and we called energy gap or forbidden zone.(i) In metals, the conduction band is close to valence bandand therefore, the electrons can easily go into theconduction band. Therefore, metals are good conductros.

(ii) In insulators, the energy gap between valence band andconduction and therefore, electrons from valence bandcannot move into the conduction band.(iii) Several solids have properties intermediate between metalsand insulators. These are called semi-metals or semiconductors.



Solid State

53

Conduction of Electricity in Semiconductors

Ge and Si are the most important commercial examples

semiconductors. The crystal structures of Ge and Si are simila to that of diamond. Atoms of both Ge and Si have four electroin the outermost shell. Therefore each atoms is covalently bon with neighbouring atoms through sp3 hybrid.

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................

............................................................................................................................



100 Hours 100Marks

54

QuestionsProblems based on this concept

Q.1 What is piezoelectricity ? What type of crystals exhibit piezoelectricity ? Ans. The crystals, which on applying a mechanical stress produce electricity, are

called piezoelectric crystals and the phenomenon is called piezoelectricity.Polar crystals in which the dipoles align themselves in an ordered mannerunder the influence of an electric field exhibit piezoelectricity.

Q.2 Define the ‘forbidden zone’ of an insulator. Ans. The large energy gap between valence band and the conduction band in an

insulator called forbidden zone.Q.3 What is superconductivity ? Who discoverd it ?

OrDefine superconductivity of a substance.

Ans. A substance is said to be superconductor when it offers no resistance to flow of electricity. This property is called superconductivity. Most metals becomesuperconducting at very low temperature (2-5K). Kammerling Onnes discovered it.

Q.4 Define photovoltaic compound. Ans. A compound which generates current when exposed to light is called

photovoltaic compound.Q.5 A group 14 elements is to be converted into n-type semiconductor by

doping it with a suitable impurity. To which group should this impurity belong ?

Ans. n-type semiconductor means conduction due to presence of excess of electrons.Therefore, to convert group 14 elements inton-type semiconductor, it shouldbe doped with group 15 elements.

Q.6 Classify each of the following as either a p-type or n-type semiconductor:(a) Ge doped with In (b) B doped with Si

Ans. (a) Ge is Group 14 element and In is Group 13 element. Thus, anelectron deficit hole is created and therefore, it is p-type.

(b) B is Group 13 element and Si is Group 14 element, there will be afree electron. Thus, it is n-type.

Q.7 What is doping ? Why is it done ? Ans. It is process of adding impurities in a crystal lattice. Dopping is done by

adding calculated amount of impurities. It increases the conductivity.Q.8 What is the difference in the semi-conductors obtained by doping silicon

with Al or with P ?



Solid State

55

Ans. Silicon doped with Al forms p-type semiconductors, i.e., flow of curbecause of creation of positive holes while silicon doped with P pron-type semiconductorsi.e., flow of current is due to extra electrons havnegative charge.

Q.9 What type of crystal defect is produced when sodium chloride is dope

with MgCl2 ? Ans. It is called impurity defect. A cation vacancy is produced. A substitu

solids solution is formed (because 2Na+ ions are replaced by one Sr2+ ion thelattice site).

Q.10 How does the elecrical conductivity of metallic conductors vary witemperature?

Ans. Electrical conductivity decreases with rise in temperature because kbegan to vibrate and create hindrance in the flow of electrons.

Q.11 What is meant by ‘doping’ in a semiconductor ? Ans. Doping means incorporating small amount of forgien impurity in cry

Doping of group 14 elements with group 15 elements give rise to electron (n-type semiconductors) whereas with group 13 elements give rholes ( p-type semiconductors).

Q.12 Account for the following :(i) Phosphorus doped with silicon is a semiconductor.(ii) Some of the glass objects recovered from ancient monuments

look milky instead of being transparent.(iii) Schottky defect lower the density of a solid.

Ans. (i) Phosphorus doped with silicon has holes and hence acts as n-semiconductor.

(ii) Glass objects recover from ancient monuments look milky becaof some crystallization taken place in them.

(iii) This because some of the cations and anions are missing from crystal lattice.

Q.13 How does the electrical resistivity of (i) a semiconductor (ii) metallconductors vary with temperature ?

Ans. Electrical resistivity of metallic conductors increases with temperature whof semiconductors and super conductors decreases with temperature.



100 Hours 100Marks

Extra ShootQ.1. Explain why amorphous solids are isotropic.

Q.2. How many different types of unit cells are possible for two dimensionallattices ?

Q.3. If three elements X, Y and crystallize in a cubic solid lattice with X atomsat the corners, Y atoms at cube center and atoms at edges then what is theformula of the solid ?

Q.4. What is Frankel defect ?

Q.5. What is the coordination number of an atom present in an octahedral void ?

Q.6. What is F-center ?Q.7. Compare the size of energy gap in conductors, semiconductors and insulators?

Q.8. Define the term amorphous. Give four examples of amorphous solids.

Q.9. A compound is formed by two elements M and N, The element N formshcp and atoms M occupy 2/3rd of octahedral voids. What is the formulaof the compound?

Q.10. A compound contains two types of atoms- X and Y. It crystallises in acubic lattice with atoms X at the corners of the unit cell and atoms Y at the body centres. What is the simplest possible formula of this compound?What is the coordination of Y ?

Q.11. If three elements, P, Q and R crystallize in a cubic solid lattice with Patoms at the corners, Q atoms at the cube centre and R atoms at the cen-tre of the faces of the cube, then write the formula of the compound.

Q.12. Describe the following :

(i) Hexagonal close packing(ii) Cubic close packing(iii)Body centred cubic packingD i ll f h

1 solid state1

Documents