Solid State

S.C.F. 23 & 29 (Top Floor), Phase -7, Mohali. Ph. 9417009068

FUSION: S.C.F. 23 & 29, Phase -7 & SCF 21 PHASE 2 Mohali. Ph. 9417009068, SOLIDS STATE (Theory) (1)

General Introduction:- Solids are the substances having definite

volume and shape. Solids have a regular arrangement of their

constituent particles. The interparticle forces are very strong. The particle motion is restricted to

vibratory motion only. Basically the solids are of two types:- 1) Amorphous solids 2) Crystalline Solids 1) Amorphous solids:- Solids in which the constituent particles (ions, atom) do not possess an orderly arrangement are called as amorphous solids. E.g.:- Glass, rubber etc. Properties of Amorphous solids:-

a) They don’t possess sharp melting and boiling points. b) They undergo irregular cleavage when cut

with a knife. c) They are isotropic in nature i.e. they have

same physical properties in all directions d) They are pseudo solids. e) In amorphous solids the regular arrangement

of constituent ion is present over a very small region i.e. they are supposed to have a short range order.

2) Crystalline Solids:- The solids in which the constituent particles are arranged in a regular pattern are known as crystalline solids e.g:- NaCl, ZnS etc. Properties of Crystalline solids:-

a) They possess sharp melting & boiling points i.e. they abruptly change into liquid state at its melting point.

b) They undergo regular cleavage when cut with a knife.

c) They are anisotropic in nature i.e. they have different physical properties in different directions.

d) They are true solids. e) They have regular arrangement of constituent

particle. The crystalline solids can be further subdivided depending on the nature of interparticle forces

Types of Crystalline solids i) Ionic Solids ii) Metallic Solids iii) Covalent Solids iv) Molecular Solids

i) Ionic Solids:- In these the constituent particles are positively & negatively charged ions which are held together by strong electrostatics forces of attraction. e.g.:- NaCl In NaCl crystal each Na+ ion is surrounded by six Cl- ions & Vice versa.

Properties:-

1. Due to the fact that the ions are held together and are not free to move they are poor conductors of electricity in the solid state

2. They are hard and brittle. 3. High melting and boiling points. 4. Soluble in water and other polar solvents. 5. High enthalpies of vaporization.

Metallic Solids:- In these solids the constituent particles are positively charged kernel present in a sea of mobile electrons. The structure is also called as electron sea model.

Properties:-

1. They have high electrical and thermal conductivity due to the presence of mobile electrons

2. They can be hard as well as soft. 3. They are malleable and ductile in nature. 4. They possess metallic lusture.



Covalent Solids:- Covlent solids are also known as Network Solids.In these the constituent particles are atoms which are held together by system of covalent bonds. e.g.:- Diamond and graphite etc. These covalent bonds form a network resulting in

giant molecules. Properties:-

1. Covalent solids are directional in nature. 2. They are hard. 3. They have high melting points 4. They are poor conductor of electricity. 5. They have high enthalpies of fusion.

Example are diamond and graphite are: Structure of Diamond

Rigid tetrahedral structure of diamond Features of structures of Diamonds:-

Each carbon atom is sp3 hybridized and linked to four other carbon atoms by single covalent bonds.

There is a three dimensional network of covalent bonds this makes diamond extremely hard in nature.

Diamond is a poor conductor of electricity due to the absence of free electrons.

Structure of graphite

Hexagonal layered type structure of Graphite

Features of structure of Graphite:-

Each carbon atom is sp2 hybridized It is bonded to three carbon atoms by covalent

bond. Fourth e- is used for π bond formation Graphite has layered structure consisting of

hexagonal rings. Graphite is a good conductor of electricity

due to delocalized nature of πe.- Molecular Solids:- In these Solids, The Constituents particles are Molecule. The molecular solids are furthers divided in two types. Polar Molecule Solids Non Polar Molecule Solids Polar Molecule Solids:- These are those crystalline solids in which the constituents particle are polar molecule like HCl , SO2 etc. The forces holding these molecule together are dipole-dipole forces of attraction.(Which is again a type of Vander Waals forces) Properties:-

1. They are soft. 2. They are non-conductors of electricity. 3. The Mp and Bp are comparatively higher

than Non –polar Molecular. 4. Their Mp and Bp are not so high, they

even exits liquids or gases at room temperature.

Non-Polar Molecule Solids:- These are those crystalline solids in which the constituents particle either atoms of Noble gases or non polar molecule like H2 ,Cl2 ,Br2 Etc. The forces holding these molecule together weak Vander Waals forces. Properties:-

1. They are soft. 2. They are non-conductors of electricity

because they do not have any ions. 3. Their Mp and Bp are quite low. 4. They Generally exits liquids or gases at

room temperature.



Q1) What is the difference between glass and Quartz when both are made up of SiO4 tetrahedral? Q2) What are Pseudo solids? Why the window panes of old buildings are thick at the bottom? Q3) Write the difference between Amorphous

Solids and Crystalline solids . (Home Work)

Some Basic Definitions:- The smallest repeating unit which when

repeated again and again result in the formation of space lattice is called as Unit cell.

Diagram of unit cell The perfect regular arrangement of atoms,

molecules in crystalline solids is called as space lattice

The position occupied by the constituent atom or ions in the space lattice are known as lattice points.

Each unit cell is characterized by distances a, b and c along the three edges and angles α, β, γ between these edges.

Types of Unit Cells: i) Primitive unit cell. A unit cell is called primitive unit cell if it has particles only at the corners. It is also called simple unit cell. (ii) Centered unit cells: In this type of unit cells, particles are present not only at the corners but also at some other positions. These are of three types: i) Face Centred: - Particles are located at the corners

and also in the center of each face. ii) Body Centred: -Particles are located at the corners

and also at the center within the body.

iii) End Centred. Particles are located at the corners and also at the center of the end faces.

Calculation of the Number of points in a unit cell:- The following points should be taken into consideration for calculating the contributions.

1. A point present at the corner of a unit cell is shared by eight other unit cells. Therefore, each such point has contribution of 1/8 to the unit cell.

2. A point present on the face of a unit cell is shared by the two unit cells. Therefore, each such points has a contribution of 1/2 to the unit cell.

3. A point present on the edge of the unit cell is shared by four unit cells. Therefore, each such point has a contribution of 1/4 to the unit cell.

4. A point present within the body of the unit cell is not shared by any other unit cell. Its contribution to the unit cell is 1.

Simple or Primitive cubic cell: In a simple cubic cell, points are present only at the eight corners. Since a point present at each corner is shared by eight other unit cell, therefore, Contribution by one point present at corner to the unit cell = 1/8 Contribution by points present at eight corners to the unit cell = 8 1/8 = 1 Face centred cubic cell: In this type of unit cell, the points are located at the eight corners as well at the center of each face of the unit cell. Contribution by eight points present at the corners = 1



Contribution by each point present on the center of each face = 1/2. Contribution by six points present on the center of six faces = 1/2 6 = 3 Total no. of points present in a face centred cubic unit tell (fcc) = 1 + 3 = 4 Body Centred Cubic unit cell: In the body centred cubic unit cell (bcc) points (atoms or particles ) are located at all the eight corners and also in the body of the unit cell. Contribution by eight points present at the corners = 1 Contribution by the point present in the body of unit cell = 1 Total number of points present in the body centred cubic unit cell (bcc) = 1 + 1 = 2 End Centred Cubic: the points are located at the eight corners as well as one pair of opposite faces : Contribution by eight points present at the corners = 1 Contribution by point present at two faces= ½ x 2=1 Total contribution = 1 +1 = 2 Seven Crystal Systems:

Cubic Axial distance

Axial angles

Examples

Cubic a = b = c Α = β = γ = 90°

Diamond, Cu, NaCl, CsCl, CaO

Tetragonal a = b c Α = β = γ = 90°

White tin, SnO2, TiO2, (rutile), NH4Br

Hexagonal a = b c Α = β = 90° γ = 120°

Zn, Cd, Graphite HgS, ZnS, BN,Sb, Ice

Rhombohedral a = b = c Α = β = γ 90°

Sb, Ice, Calcite, CaCO3

Orthorhombic a b c Α = β = γ = 90°

Rhombic sulphur, BaSO4

Monoclinic a b c Α = γ = 90° β 90°

Monoclinic sulhpur, Gypsum

Triclinic a b c

Α β γ 90°

K2Cr2O7, CuSO4.H2O

Bravias Lattices: The seven crystal system listed above can be further classified on the basis of the unit cell present.

Although each crystal system is expected to have four different unit cells, but actually all of them

cannot exist in each case. fourteen different types of lattice called Barvias lattices are formed according to the arrangement of the points in the different unit cells involved. These are listed below:

Crystal System Types of Lattices Cubic Tetragonal Orthorhombic Monoclinic Rhombohedral Triclinic Hexagonal

Simple, Face centred, Body centred Simple, Body centred Simple, Face centred, Body centred, End centred Simple, End centred Simple Simple Simple



Closed Packing:- An arrangement of atoms in which the vacant space is minimized is called as closed packing. The closed packing can be present in i) One Dimension, ii) Two Dimensions iii) 3 Dimensions. i) Packing in One Dimension:- Only way of arranging the particles is in the form of a straight line. Now we can introduce a new term co-ordination number which can be defined as the number of sphere which are in contact with a given sphere. e.g.:- In 1 dimensional packing. The arrangement is shown above obviosuly coordination number is 2.

Packing in 1-D

Packing in 2D:- The arrangement can be done in 2 ways.

i) Square close packing ii) Hexagonal close packing

i) Square close packing In the first arrangement, the second layer of sphere lies exactly above the first row. So there will be horizontal as well as vertical

alignment. This can be shown as In this arrangement 60.4% of available space is

occupied. The co-ordination number of each sphere in

square close packing is 4.

Four neighbouring spheres are touching the given sphere.

This arrangement is called as square close packing.

ii) Hexagonal close packing in this arrangement the second row are different

from the first row, but the arrangement of atoms in third row is same as that of first row.

In second arrangement, the spheres of second layer are placed in the depressions of the spheres of first row

Six neighboring spheres are touching the

given sphere. In this case 52.4% of available space is

occupied. Close packing in 3 D can be done in two ways

i) Hexagonal close packing ii) Cubic close packing

i) Hexagonal close packing

Closed packing in 3D through extension of square close packing in 2D:- If the arrangement of sphere obtained in

square close packing in 2D is further extended by increasing the number of layers we obtain an arrange of type ‘ABA’.

This is called as Hexagonal close packing.

i) Hexagonal close packing ii) Cubic close Packing ii) Cubic Close packing When we extend the 2-D hexagonal close packing into 3D. The packing can be shown as The spheres in the first layer are labeled as A

on placing these sphere two types of voids are formed which are labeled as B & C.

Now the spheres of second layer can be placed on voids B.



When we placed the spheres on the void marked as B than the void marked as ‘A’ and ‘C’ remain unoccupied.

Whenever a sphere of second layer is placed above the void of the first layer a tetrahedral void is formed. The void ‘C’ represent the tetrahedral voids.

The third layer of sphere can be placed on the tetrahedral voids marked as ‘C’ on the second layer. The repeated layer labeled ABCABC…

Tetrahedral Voids: A tetrahedral voids is formed when the triangular void made by three spheres of a particular layer and touching each other has a contact with one sphere either in the upper layer or the layer below it. Thus a tetrahedral void is formed void is formed when four spheres placed at the corners of a tetrahedron each other as shown in this figure.

Calculation of the Ratio. A tetrahedral void has been shown in the figure. Here three spheres from the triangle base while the fourth sphere lies at the top. The shaded sphere occupies the tetrahedral void. Let the radius of void and sphere be r and R respectively. Let the edge length ‘a’. From the right hand triangle ABC AB = AC2 + BC2

AB is the face diagonal while AC and BC are the edge length (a). Therefore AB2 = a2 + a2 (AB)2 = (2a2)

AB = a 2 . As the spheres A and B are actually touching each other face diagonal AB = 2R

2R = a 2 or R = 2

2a

or 12

a ………..(i)

In the right angle triangle ABD, in which AD represent the body diagonal

AD = 2 2 2 2AB + BD = ( 2a) + a = 3a AD = 2R + 2r ( R=> radius of atom )

( r = radius of tetrahedral void)

2R + 2r = 3 a or R + r = 32

a …..(ii)

Dividing equation (ii) by equation (i) R r

R =

3 22

aa

=32

or r1+R

= 32

rR

= 32

-1 = 1.7321.414

- 1 = 1.732-1.4141.414

= 0.225

Thus, for tetrahedral void, the radius ratio = 0.225. Characteristics of Tetrahedral Void (hole): - Size of the void is much smaller than that of the

spheres. Bigger the size of the spheres, more will be the size of the void.

For tetrahedral void, the ratio of the radius of void (r) to that of the spheres (R) is 0.225 i.e. r/R = 0.225.

Since in ccp and fcc arrangement, each spheres is in contact with three spheres in the layer above it and three spheres of the layer below it, there are two tetrahedral voids per sphere.

*In packing consisting of N spheres no. of tetrahedral voids is 2N. Example

S. No Type of

packing Number of Atoms

per unit Cell Number of

tetrahedral voids per unit Cell

1 SCC 1 2 2 BCC 2 4 3 FCC 4 8 4 End

Center 2 4

Co-ordination number of tetrahedral void is four.

Octahedral Voids: - An octahedral void or site is formed when three spheres arranged at the corners of an

C

D

A

B



equilateral triangle are placed over another set of spheres pointing in the opposite directions.

Calculation of Radius Ratios Radius Ratio of Octahedral Void If the length of the unit cell is a cm, then ‘R’ is the radius of atom, ‘r’s radius of Octahedral void Considering the triangle ABC BC2 = AB2 + AC2

(2R)2 = 2 (R+r)2

r = 0.414 R

Radius ratio of the octahedral void = 0.414 R Note:-

S.No Type of Packing

Number of Atoms per unit Cell

Number of Octahedral voids

per unit Cell 1 SCC 1 1 2 BCC 2 2 3 FCC 4 4 4 End

Center 2 2

Packing Fraction:- Packing fraction is the ratio of volume occoupied by atoms in a cube by volume of cube. Packing fraction=(Volume of atoms in a cube/volume of cube). Efficiency of Packing:-

a) In Simple cubic:- Edge length = a , Radius of sphere = r Now edge length a = 2r

Number of spheres per unit cell = 8 × 1 18

Volume of sphere = 34 πr3

Volume of cube = a3 = (2r)3 = 8 r3 Volumeof onespherePacking efficiency = ×100

Volumeof cubic unit cell

= 3

3

4 π r π3 ×100= ×1008r 6

= 3.142×100 = 52.4%6

Volume occupied = 52.4%

b) Cubic close packing(FCC) :- Let edge length of unit cell = a Radius of sphere = r Now there are and sphere at the corners and six sphere at the faces. Number of spheres in unit cell =

8 × 18

+ 6 × 12

= 4

Consider Δ ABC AC2 = AB2 + BC2 = a2 + a2 = 2a2 AC = Face diagonal = 2 a From figure AC = 4 r

2 . A = 4 r a = 4r = 2 2r2

Volume of four spheres = 4 × 43

πr3 = 163

πr3

Packing efficiency =

Vol.of four spheresin a unit cell ×100

Total volumeofunit cell



= 3

3

16 πr3 ×100

16 2r = π

3 2 = 3.142

3×1.414 = 74%

c) Body centered cube (BCC ) :- Let edge length of unit cell = a Radius of sphere = r Now there are eight sphere at the corners and one sphere at the body center position. Number of spheres in unit cell =

(8 × 18

)+ (1×1) = 2

AC = Body diagonal = 3 a From figure AC = 4 r

3 . a = 4 r a = 4r3

Volume of two spheres = 2 × 43

πr3 = 83

πr3

Packing efficiency =

Vol.of fourspheresin a unit cell ×100

Total volumeofunit cell

=3

3

83 ×100643 3

r

r

= 68%

=

3

3

8π r π×100 = ×10024 2 r 3 2 = 74%

Relation Between Edge Length and Density in a unit Cell: Let us derive a general relation between the edge length and the density in a unit cell. Consider a unit cell with edge length=a pm = a 10-10 cm. Volume of the unit cell = -10 3(a×10 cm)

= 3 -30a ×10 cm3

Density of unit cell = Massof unit cell

Volumeof unit cell

Now, mass of unit cell = No. of atoms in a unit cell mass of each atom

= Z Atomic mass(M)

Avoadro'snumber(No.)

By substituting the value of mass of unit cell

Density of unit cell = 33 -30

0

Z×M (gcm )N ×a ×10

The value of Z is different for different types of cubic unit cell. PRACTICE:-

1. An Element has BCC structure with cell edge length of 288pm . The density of the element is 7.2gm/cm3.how many atoms are present in 208 gm of the elements.

(NCERT) (Ans:-2.4 x 1023) 2. X-ray Diffraction studies shows that copper

crystallizes in an fcc unit cell with edge 3.608 x 10-8cm.in a separate experiment, Copper is determined to have a density of 8.92 gm/cm3.Calculate the atomic mass of Cu.

(NCERT, Ans-63.1 U) 3. Ag forms CCP lattice and X-ray studies of its

Crystals shows that the edge length of its unit cell length is 408.6 pm. Calculate the density of silver (Atomic Mass = 107.9 u)

(NCERT, Ans- 10.5 x 103 g/cm3) 4. An element E forms bcc lattice. The edge

length of cell is 1.469 × 10-10 m and density is 19.3 g cm-3. Calculate the atomic mass of element, also calculate radius of an atom of element. (H.W)

5. An element A crystalline in fcc lattice 200 g of this element has 4.12 × 1024 atoms. The density is 7.2 g/cm3. Calculate the edge length. (H.W)

6. Crystalline CsBr has cubic structure, calculate the edge length if density of CsBr crystal is 4.24g/cm3(atomic mass:Cs = 133,Br = 80)(HW)

Radius ratio:- Ratio of radius of cation to radius of anion

is called as Radius ratios. Radius Ratio= RR

The

allowed size of the cation is determined by the critical radius ratio. If the cation is too small, then it will attract the anions into each other and they will collide hence the compound would collapse, this case occurs when the radius ratio drops below 0.155. At the stability limit the cation is touching all the anions and the anions are just touching at their edges (radius ratio = 0.155).



Beyond this (radius ratio > 0.155) stability limit the compound will be stable. The below table gives the relation between radius ratio and co-ordination nuber

Radius Ratio

Co-ordination number Type of void Example

0.155-0.225 3 Triangular Planar B2CO3

0.225-0.414 4 Tetrahedral ZnS,CuCl

0.414-0.732 6 Octahedral NaCl,

MgO

0.732-1.000 8 Cubic CsCl,

NH4Br

Structures of substances related to close packed lattice. Structure of NaCl: The salient features of this structure are: (a) The Cl- ions form cubic close packed arrangement, i.e. they are located at all the eight corners of the cube and also at centre of each face. (b) The Na+ ions occupy all the octahedral holes. In the cubic unit cell, the Na+ ions are present at the centre of each edge of cube and one Na+ ion is located at the body centre of the cube.

Number of formula units of sodium chloride per unit cell (i) Number of Cl- ions per unit cell: = [8Cl-(at corners)

81 ]+

[6Cl-(at the face centers)21 ]

= Cl- + 3Cl- = 4Cl- ions (ii) Number of Na+ ions per unit cell: =[12 Na+ ions (at the centre of each edge)

41 ]

+[1 Na+ (at the body centre)] = 3Na+ ions + 1 Na+ ion = 4Na+ ions So, the number of formula units of sodium chloride per unit cell = 4 Structure of CsCl: It is body centred cubic unit cell.

Salient features of the structure are: (a) The Cl- ions form the simple cubic arrangement, i.e., the Cl- ions are occupying all the eight corners of the cube. (b) The Cs+ ion is present at the body centre of cube. (c) Each Cs+ ion is surrounded by eight Cl - ions. Cl- ion is surrounded by eight Cs+ ion. (d) Number of formula units of cesium chloride per unit cell is: * Number of Cl- ions per unit cell = 8 Cl- (at the corners) 1

81

* Number of Cs+ ions per unit cell = 1 (present at the body centre) So, there is only one Cl- for one Cs+ ion in the unit cell. Structure of ZnS (Zinc Blende) In this, each of the Zn ion is surrounded by four sulphide ions which are disposed towards the four corners of a regular octahedron. Also each sulphide ion is surrounded by four zinc ions. The sulphide



ions are in cubic close packed type of arrangement and the Zn atoms are occupying octahedral sites. The co-ordination number of both Zn and sulphide ions is 4 : 4.

Fluorite Structure (Calcium Fluoride, CaF2) (1) In this, Ca++ ions are arranged in ccp arrangement, i.e., there are Ca++ present at all the eight corners of the cube and also at the centre of each face. The fluoride ions occupy all the tetrahedral voids in the lattice of Ca++ ions. (2) Since, there are two tetrahedral holes for each Ca++ ion and F- ions occupy all the tetrahedral there are two F- ions for Ca++ ion. Thus, the stoichiometry of crystal is 1 : 2 (3) In this arrangement co-ordination number of Ca++ ions is 8 and that of F- ion is 4. Thus, CaF2 has 8 : 4 coordination. (4) In CaF2, each unit cell has four Ca2+ ions and 8F - ions. It can be explained as follows: Number of Ca2+ ions = 8(at the corners)

81 +6(at

the face centered) 21 = 1 + 3 = 4

Number of F- ions=8 (within the body of unit cell) 1 = 8 Anti fluorite structure:(Na2O) The anions are arranged in cubic close packing

while the cations occupy all the tetrahedral voids. Magnetite also possess normal spinel structure

The O2- ions is in contact with 8 Na+ ions. Where as each Na+ ions is contact with 4O2- ions.

Thus coordination number is 4 and 8 respectively for each ions.

Structure Na2O

Spinel structure(MgAl2O4) The spinel structure (sometimes called garnet

structure) is named after the mineral spinel (MgAl2O4); the general composition is AB2O4. It is essentially cubic, with the O - ions forming a fcc lattice. The cations (usually metals) occupy 1/8 of the tetrahedral sites and 1/2 of the octahedral sites and there are 32 O-ions in the unit cell

Many Ferrites (ZnFe2O4) processes normal spinel structure.

Imperfections or Defects in Solids Defects in Solids are of two types

1) Point Defects 2) Line Defects

1) Point Defects:- Point defects are irregularities or deviation from ideal arrangements around the point or atom in the crystalline substance. 2)Line defects:-line defects are irregularities or deviation from ideal arrangements in entire rows of lattice points. Point Defects: can be classified into three categories * Stoichiometric defect * Non-Stoichiometric defects * Impurity defects(Doping) Stoichiometric defect: - This defect arises because certain ions are missing from the crystal lattice and vacancies or holes are created at their respective positions. Since a crystal in electrically neutral, the number of such missing cations (A+) and anions (B-) must be the same. In Stoichiometric defect the cation and anion ratio will remain same as represented in the molecular formula even after the defect. They are also known as Intrinsic or Thermodynamic defects. * Types of Stoichiometric defect i) Vacancy Defects ii) Interstitial Defects iii) Schottky Defects iv) Frenkel Defect i) Vacancy Defects:-When some of the lattice sites are

vacant, the crystal is said to have vacancy defects this results in decrease in density of the substance. This



defects is caused due to heating of substance. This substance

ii) Interstitial Defects:- When some constituent particles occupy an interstitial sites. This defects causes increases the density of solids.this defects also shown by non-polar compounds.

iii) Schottky Defect:-This is mainly a Vaccancy defects shown by ionic compounds Condition of favoring Schottky defect: Schottky defect is prevalent in lattice with With high co-ordination numbers In which cations and anions have almost identical

size Alkali metal halides such as NaCl, KCl, CsCl, KBr etc. normally show this defect.

Diagram of Schottky Defect

Consequence of Schottky defect: The presence of holes because of the missing ions

lowers the density of the crystal. Due to the presence of the large number of holes,

the stability of the crystal decreases and it also lowers its lattice energy.

The defect increases the electrical conductivity of the crystalline solid. When electrical field is applied, under its influence ions from the nearby positions occupy these holes and their positions become vacant. The process continues and the crystal is in a position to conduct electric current.

vi) Frenkel Defect:- Frenkel Defects is shown by ionic compounds. It results when certain ions leave their normal sites and occupy positions elsewhere in the crystal lattice. Holes are created at their respective positions. Since cations are smaller in size as compared to anions normally these are involved in Frenkel defect.

Diagram of Frenkel Defect Conditions Favoring Frenkel Defect: - Frenkel defect is prevalent in lattices with With low co-ordination numbers In which the anions are bigger in size as compared to

the cations. The defect has been found to be present in silver halides such as AgCl, AgBr, AgI etc, because the size of Ag+ ion is quite small. It is rather easy for the ion to leave its position and occupy a new position in the crystal lattice. Consequence of Frenkel Defect: The electrical conductivity of the crystal is expected

to increase because certain cations from the neighbouring positions will occupy these holes.

The presence of large number of holes is bound to decrease the stability of the crystal. However, its density remains unchanged since the number of ions per unit volume car the same.

The defect is expected to increase the dielectric constant of the crystalline solid

Difference between Schottky and Frenkel Defects: SCHOTTKY DEFECT FRENKEL DEFECT

1. Some ions are missing from their normal sites.

2. It lowers the density of

the crystal. 3. There is not change in the

dielectric constant value. 4. It is noticed in the solids

with high co-ordination numbers of ions.

Ions do not leave the lattice but occupy positions elsewhere in it. The density of the crystal remain unaffected The value of dielectric constant increases. It is noticed in the solids with low-ordination number of ions.

Non-Stoichiometric Defect: - The defects in the non-Stoichiometric crystalline solids are of two types. 1. Metal excess Defects:- This defect arises if the metal

ions or positive ions are in excess. The defect can arise in two ways:

By anion vacancies: In this case one or more



anions are missing from the lattice site and in order to maintain the electrical neutrality of the crystal, the electrons are actually trapped in these positions. This type of defect is Known as F-Center. Ex :-The yellow colour and the electrical conductivity of the non-stoichiometric ZnO is due to these trapped electrons.

By the presence of extra cations in the interstitial sites: The metal excess defect can also be created when there are extra positive ions occupying some of the interstitial sites and the electrons at some other sites to maintain the electrical neutral character of the crystal. Consequence of Metal Excess Defects:

The crystals with metal excess defects are conducting due to presence of free electrons. However, the conductivity is very low since the defects are small in number. They act as semiconductor and known as n-type semiconductors since the current is carried by the movement of the electrons.,

The crystals with metal excess defects are paramagnetic in nature due to the presence of free electrons.

Crystals with metal excess defects are generally coloured.

Metal Deficiency Defects: - By cation vacancies: In some of the crystals, certain cations are missing from the lattice sites and their positive charges are balanced by the presence of extra charges on the neighbouring or adjacent cations. 2. By the presence of extra anions at the interstitial

sites: These defects arise when certain extra anions or negative ions are present at the lattice sites and their charges are balanced by the neighbouring cations which are in higher oxidation state than the normal cations.

Consequence of Metal Deficiency Defects: Crystals with metal deficiency defect also act as

semiconductors. Since the conductivity is due to the holes, these are known as p-type semiconductors. Impurity defects: - If we add certain elements like phosphorus (P) or arsenic (As) of group 15 to these covalent crystals, their atoms will also get linked with those of group 14 elements by covalent bonds but will have the extra electron which is not involved in the bonding( atoms have five valence electrons). These extra electron will lead to electrical conductivity resulting in n-type of semi-conductors as these are conducting due to the movement of electrons. The introduction of defects in a particular crystalline solid by the addition of impurities of other elements is known as doping. The doping in the covalent crystal of group 14 elements can also be caused by the addition of small amounts of the element aluminium (Al) or gallium (Ga) belonging to group 13. The atoms of such elements can share only three electrons with the atoms of group 14. Thus, the holes will be created in the lattice since there is no fourth electron available for sharing. The holes will lead to electrical conductivity and the crystals thus formed are also semiconductors. These are known as p-type of semi-conductors. Paramagnetic Substances :-

These are the substances weakly attracted by a magnet and this property is known as Para magnetism

Elements have unpaired electrons. Their spins or magnetic moments may lead to

magnetic character. However, the electron spins may mutually cancel due to random orientation under normal conditions and the substances may not have any magnetic character.

Under the influence of an external magnetic field or magnet, the spins of certain electrons may allign to produce temporary magnetism they may be attracted by the magnet. However, they may lose their orientation and also magnetic character once the magnet is removed. Example Na, Li etc

Diamagnetic Substance:- These substances have all electrons paired Thus, no resultant magnetic moment They may be weakly repelled by the external

magnetic field. For example zinc metal is diamagnetic as it does not have any unpaired electrons. Some other examples are of TiO2, NaCl, N2 etc.

Ferromagnetic Substances:



Certain paramagnetic substance become permanently magnetized under the influence of

the applied magnetic field even in the absence of the magnetic field. Example Fe, Cr etc Anti-Ferromagnetic Substances: Certain paramagnetic substances allign the magnetic moments due to the unpaired electrons under the influence of external magnetic field in such a way that the net magnetic moment is zero Ferrimagnetic substances: In certain paramagnetic substances, the magnetic moments of electron spins allign in parallel and anti-parallel direction in unequal numbers so that they have a net magnetic moment.

Piezoelectricity: Thoses crystal which produced electricity on mechanical stress. Such a crystal is known as piezoelectric crystal and this property is called piezoelectricity or pressure electricity. Pyroelectricity: Some polar crystals upon heating produce small electric current. This phenomenon is known as pyroelectricity. Actually certain atoms Ferroelectricity and the effect involved is called ferroelectric effect. The examples of some ferroelectric solids are: Potassium dihydrogen phosphate (KH2PO4) Potassium tartarate or Rochelle’s salt (KOOCCH(OH)CH(OH)COOK) and barium titanate (BaTiO2). Anti –Ferroelectricity: When the dipoles in alternate lattice point up and down, the two dipoles cancel one another so that there is not resultant dipole moment. Such crystals are known as antiferroelectric crystals and the phenomenon is called antiferroelectricity. For example, lead zirconate (PbZrO3) is an antiferroelectric solid. Low Temperature supper conductivity (LTSC) : He found that when the temperature of mercury was decreased to 4.15 K, the resistance felt closely to zero. This is called critical temperature (Tc). At very low temperature ranging from 0.1 K to 10 K, many metals, alloys and certain compound become super conducting. This means that they can conduct electricity with out losing energy or the current can flow for ever. It may be noted that super conductors are diamagnetic in

nature i.e. they do not allow magnetic field to pass trough them.

High temperature super conductivity (HTSC): a cuprate which is a mixed oxide consisting of Ba-La-Cu-O system and was super conducting at 35 K. Later on a few more cooperates were synthesised and were conducting even at higher temperatures. A few important our of these are listed.

Ba0.7 K0.3 BiO3 30 K La1.8 ST0.2 CuO4 40K YBa2Cu3O7 90 K Bi2Ca2Sr2Cu3O10 105 K Tl2Ca2Ba2Cu3O10 125 K The phenomenon of super conductivity has wide range of application in the field of electronics

Super conducting cables have been used for transmitting electric current underground.

Some electronic devices and speacilised instruments such as cryogenic gyro operate at liquid helium temperature i.e. approximately 4K.

Super conductors also find applications in lazers and some computer circuits.

N.C.E.R.T. –Assignment

1. Define the term ‘amorphous’. Give a few examples of amorphous solids.

2. What makes a glass different from a solid such as quartz? Under what conditions could quartz? Under what conditions quartz could be converted into glass?

3. (i) What is meant by the term ‘coordination number’? (ii) What is the coordination number of atoms:

(a) in a cubic close-packed structure? (b) in a body –centered cubic structure?

4. How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

5. ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?



6. How will you distinguish between the following pairs of terms: (i) HCP and CCP? (ii) Crystal lattice and unit cell? (iii) Tetrahedral void and octahedral void?

7. How many lattice points are there in one unit cell of each of the following lattice? (i) Face centered cubic (ii) Face –centered tetragonal (iii) Body-centred

8. Explain (i) The basis of similarities and differences between metallic and ionic crystals. (ii) Ionic solids are hard and brittle.

9. Calculate the efficiency of packing in case of a metal crystal for (i) Simple cubic (ii) body-centered cubic (iii) Face-centered cubic (with the assumptions that atoms are touching each other)

10. Silver crystallizes in fcc lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 10.5 g cm-3, calculate the At Mass of Ag.

11. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-center. What is the formula of the compound? What are the coordination number of P and Q?

12. Niobium crystallizes in body-centred cubic structure. If density is 8.55 g cm-3. Calculate atomic radius of niobium using its atomic mass 93 u.

13. If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

14. Copper crystallizes into a fcc lattice with edge length 3.61 × 10-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm-3.

15. Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

16. What is a semiconductor? Describe the two main types of semiconductors.

17. Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 :

1. Can you account for the fact that this substances is a p-type semiconductor?

18. Ferric oxide crystallizes in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula.

19. Classify each of the following as being either a p-type or a n-type semiconductor: (i) Ge doped with In (ii) B doped with Si.

20. Gold (atomic radius = 0.144 nm) crystallizes in a face-centered unit cell. What is the length of a side of the cell?

21. In terms of band theory, what is the difference (i) between a conductor and an insulator (ii) between a conductor and a semiconductor?

22. Explain the following terms with suitable examples: (i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centers

23. Al crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in 1.00 cm3 of aluminium?

24. If NaCl is doped with 10-3 mol % of SrCl2. What is the concentration of cation vacancies?

25. Explain the following with suitable examples: (i) Ferromagnetism (ii) Paramagnetism (iii)Ferrimagnetism (iv) Antiferromagnetism (v) 12- 16 and 13-15 group compounds.

Previous Years Questions “2001”

1. What is non-stoichiometry defect in crystals? 2. Why does Frenkel defect not change the

density of AgCl crystals? [1] 3. If the close packed cations in an AB type

solid have a radius of 75pm, what would be the maximum and minimum sizes of the anions filling the voids? [2]

4. Sodium crystallizes in the cubic lattice and the edge of the unit cell is 430 pm. Calculate the number of atoms in a unit cell. [Atomic mass of Na = 23.0 amu, Density of sodium = 0.9623 g cm-3, NA= 6.023× 1023 mol-1] [2]



5. How the Schottky defect differs from Frenkel defect? [3]

“2002” 1. How many Na+ ions occupy second nearest

neighbor locations of a Na+ ion in the structure of sodium chloride crystals? [1]

2. State the difference between Schottky and Frenkel defects. Which of these two changes the density of the solid? [2]

“2003” 1. Mention one property which is caused due to

the presence of F-center in a solid. [1] 2. Cesium chloride crystallizes as a body

centered cubic lattice and has a density of 4.0 g cm-3. Calculate the length of the edge of the unit cell of cesium chloride crystal. [Molar mass of CsCl = 168.5 g mol-1] [2]

3. Br- ions form close packed structure. If the radius of Br- ion is 195 pm. Calculate the radius of the cation that just fits in the tetrahedral hole. Can a cation Ag+ having a radium of 82 pm be slipped into the octahedral hole of the crystal Ag+Br-? [2]

“2004” 1. Define the term ‘amorphous’. [1] 2. Which point detect lowers the density of a

crystal? [1] 3. What is meant by coordination number in an

ionic crystal? [1] 4. Calculate the value of Avogadro constant

from the following data: Density of NaCl is 2.165 g cm-3,Distance bt. Na+ and Cl- in NaCl = 281 pm(M.M of NaCl = 58.5 g mol-1) [2]

“2005” 1. Why is potassium chloride sometimes violet

instead of pure white? [1] 2. What happens when a ferromagnetic

substance is subjected to high temperature?[1] 3. How many atoms can be assigned to its unit

cell if an element forms (i) BCC (ii) FCC [1] 4. Aluminium crystallizes in a face centered

cubic close –packed structure. Its atomic radius is 125 × 10-12 m. (a) What is the length of the edge of the unit cell?

(b) How many such unit cells are there in a 1.00 m3 piece of aluminium? [2]

5. In the compound AX, the radius of A+ ions is 95 pm and that of X- ion is 181 pm. Predict the crystal structure of AX and write the coordination numbers of each of the ions. [2]

“2006” 1. A cubic solid is made of two elements X and

Y. Atoms Y are the corners of the cube and X at the body center. What is the formula of the compound? [1]

2. Name the non-stiochiometric point defect responsible for colour in alkali halides. [1]

3. Of 0.1 molal solutions of glucose and sodium chloride respectively, which one will have a higher boiling point? [1]

4. What makes the crystal of KCl appear sometimes violet? [1]

5. Name the crystal defect which lowers the density of an ionic crystal. [1]

6. What are the point’s defects in the crystals? Name the main point defects. [2]

7. An element has a body centered cubic structure with a cell edge of 288 pm. The density of the element is 7.2 g cm-3. Calculate the number of atoms present in 208 g of the element. [2]

8. Calculate the distance between Na+ and Cl- ions in NaCl crystal if its density is 2.165 g cm-3. [Molar mass of NaCl = 58.5 g mol-1; NA = 6.02 × 1023 mol-1] [2]

9. Calculate the efficiency of packing in case of a metal crystal in simple cube lattice. [2]

“2007” 1. Find out the number of atoms per unit cell in

a face-centered cubic structure having only single atoms at its lattice points. [1]

2. What is semiconductor? Describe the two main types of semiconductors and explain mechanisms for their conduction. [3]

“2008” 1. What is total number of atoms per unit cell in

a face-centered cubic (fcc) structure? [1] 2. What is the coordination number of each ion

in a rock –salt type structure? [1]



3. Name an element with which silicon should be doped to give n-type of semiconductor. [1]

4. What is meant by negative deviation from Raoult’s Law? Draw a diagram to illustrate the relationship between vapour pressure and mole fractions of components in a solution to represent negative deviation. [2]

5. Account for the following: (i) Fe3O4 is a ferromagnetic at room temperature but becomes paramagnetic at 850 K. ii) Zinc oxide on heating becomes yellow. iii) Frenkel defect does not change the density of AgCl crystals. [3]

6. With the help of suitable diagrams, on the basis of band theory, explain the difference between i) a conductor and an insulator ii) a conductor and a semiconductor [3]

7. What is a semiconductor? Name the two main types of semiconductors and explain their conduction mechanisms. [3]

8. Silver crystallizes in a fcc lattice. The edge length of its unit cell is 4.077 × 10-8 cm and its density is 10.5 g cm-3. Calculate on this basis the atomic mass of silver. [3]

9. Explain the following terms with suitable examples: (i) Schottky defect (ii) ferromagnetism [3]

10. a) State Henry’s law and mention its two important applications. b) Henry’s Law constant for CO2 dissolving in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 1 L of Soda Water when packed under 2.5 atm. CO2 pressure at 298 K. [3]

“2009” 1. How do metallic and ionic substances differ

in conducting electricity? [1] 2. Which point defect of its crystals decreases 3. the density of a solid? [1] 4. What is the number of atoms in a unit cell of

a face-centered cubic crystal? [1]

5. Silver crystallizes with face-centered cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver ? (Assume that each face atom is touching the four corner atoms). [3]

6. Silver crystallizes with face-centered cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching the four corner atoms). [3]

7. Iron has a body –centered cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number (At. Mass of iron =56 g mol-1) [3]

‘2010’ 1. Which point defect in crystals of a solid

decreases the density of the solid? [1] 2. The density of copper metal is 8.95 g cm–3. If

the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body-centered cubic or a face centered cubic structure? (M.M of Cu = 63.54 g mol–1) [3]

‘2011’ 1. Calculate the packing efficiency of a metal

crystal for a simple cubic lattice. [2] 2. Explain how you can determine the At mass of an unknown metal if you know its density and its edge length when its type of unit cell is kwown. The dimensions of units cell of its crystal. [2]

‘2012’ 1) Copper crystallizes with face centered cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. ( Atomic mass of Cu = 63.55 u and Avogadro’s number NA = 6.02 1023 mol-1) [2] or 2) Iron has a body centered cubic unit cell with the cell dimension of 286.65 pm. Density or iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 56.0 u) [2]

Solid State

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