pmt class xii chemistry solid state

5/24/2018 PMT Class XII Chemistry Solid State

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MEDICAL

THE SOLID STATE

C H E M I S T R Y S T U D Y M A T E R I A L

NARAYANA INSTITUTE OF CORRESPONDENCE COURSESF N S H O U S E , 6 3 K A L U S A R A I M A R K E T

S A R V A P R I Y A V I H A R , N E W D E L H I - 1 1 0 0 1 6PH.: (011) 32001131/32/50 FAX : (011) 41828320Websi te : w w w . n a r a y a n a i c c . c o mE-mai l : i n f o @ n a r a y a n a i c c . c o m


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2004 NARAYANA GROUP

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PREFACE

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CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS

THE SOLID STATE

Theory

Exercises

True and False Statements

Fill in the Blanks

Assertion-Reason Type Questions

Multiple Choice Questions

MCQs Asked in Competitive Examinations

Subjective Questions

Answers

C

O

N

T

E

N

T

S

C

O

N

T

E

N

T

S

C

O

N

T

E

N

T

S

C

O

N

T

E

N

T

S

C

O

N

T

E

N

T

S


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1

Chemistry : The Solid State NARAYANAINSTITUTE OF CORRESPONDENCE COURSES

CONTENTS

!!!!! Introduction

!!!!! Classification of Solids

!!!!! Types of Symmetry

!!!!! Types of Unit Cells

!!!!! Crystal Defects

!!!!! Radius Ratio

!!!!! Structure of Simple Ionic

Compounds

!!!!! Closest Packing

!!!!! Properties of Solids

!!!!! Metal Silicates

!!!!! Exercise

!!!!! Answers

THE SOLID STATE

M ost of the things around us are solidshaving different shapes and size. In a solid,

the molecules have fixed positions and their

motion is restricted to just vibrations. The

constituent particles in a solid are closely

packed and this leads to the properties like

incompressibility, rigidity, non-fluidity, slow

diffusion and mechanical strength. In this

unit, we will classify solids on the basis of

binding forces, understand structure of solids

describe the imperfections in solids and their

effects on properties and their application in

industries.


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Chemistry : The Solid State

FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 !Ph.: (011) 32001131/32 Fax : ( 011) 41828320

2

NARAYANAINSTITUTE OF CORRESPONDENCE COURSES

A substance is said to be solid if its melting point is above the room temperature. Some of the importantcharacteristics of solids are as follows :

(i) Solids have definite shapes, size and volume

(ii) Solids are almost incompressible and generally have high density

(iii) The diffusion of solid is negligible or rather very slow as the particles have permanent positions fromwhich they do not move easily.

CLASSIFICATION OF SOLIDS

Solids can be classified broadly into two types.

1. Crystalline Solids 2. Amorphous Solids

Difference Between Crystalline and Amorphous Solids

Crystalline Solids Amorphous Solids

1.They have definite and regular geometry due to definite andorderly arrangement of atoms, ions or molecules in threedimensional space

They do not have any pattern of arrangementof atoms, ions or molecules and thus, do nothave any definite geometrical shape.

2. They have sharp melting points and change abruptly into liquidsAmorphous solids do not have sharp melting

points and do not change abruptly into liquids.

3. Crystalline solids are anisotropic. Some of their physicalproperties are different in different directions

Amorphous solids are isotropic. Their physicalproperties are same in all directions.

4. These are considered as true solids.These are considered pseudosolids orsupercooled liquids.

5.Crystalline solids are rigid and their shape is not distorted bymild distoring forces.

Amorphous solids are not very rigid. These canbe distorted by bending or compressing forces.

6.

Crystals are bound by plane faces. The angle between any twofaces is called interfacial angle. For a given crystalline solid, it isdefinite angle and remains always constant no matter how thefaces develop. When a crystalline solid is hammered, it breaksup into smaller crystals of the same geometrical shape.

Amorphous solids do not have well definedplanes.When an amorphous solid is broken, thesurfaces of the broken pieces are generally notflat and intersect at random angles.

7.Crystals have some sort of symmetry. (i) plane of symmetry,(ii) axis of symmetry or (iii) centre of symmetry

Amorphous solids do not have any symmetry.

INTRODUCTION

Gases and liquids belong to fluid state. The fluidity of liquids and

gases is due to relative free motion of their molecules. In solids,

molecules or atoms or ions oscillate around their fixed positions

due to strong intermolecular or interatomic or interionic forces. This brings

rigidity and long range order in solids. Solids are classified as crystalline

and amorphous. In crystalline solids atoms, ions or molecules are held in

an orderly array. In amorphous solid there is only short range order.

Industrial applications of solids in based on their electrical, magnetic and

dielectric properties.


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3


TYPES OF CRYSTALLINE SOLIDS

Crystalline Solid

Ionic crystal Covalent crystal Metallic crystal Molecular crystal

Characteristics of the Different types of crystalline Solids

Characteris tics Ionic Crys tal Covalent Crys tal Metallic Crytsal Molecular crys tal

1Constituentparticles

Cations (positiveions) and anions(negative ions)

Atoms Atoms Small molecules

2Bindingforces

Strong electrostaticforces

Covalent forcesElectrostatic attraction

between positive ions andmobile electrons

Weak van der Waal'sforces

3Physical

properties

Brittle, high m.p. andb.p., good conductorsof electricity in moltenstate and aqueoussolution, high heats offusion.

Generally hard,high m.p. and b.p.,

poor conductors ofheat, high heats offusion.

Hard and high m.p., verygood conductors of heatand electricity, exhibitproperties such as lustre.malleability ductility etc.,moderate heats of fusion.

Generally soft, low m.p.volatile, poorconductors of electricity(insulators) , poorconductors of heat, lowheats of fusion.

4 ExamplesNaCl, KF, CuSO

4,

NaNO3etc.

Diamond, siliconcarbide (SiC)quartz (SiO

2) etc.

Cu, Ag, Fe, Na, K etc.Solid CO

2, Iodine,

Naphthalene, Argonice etc.

TYPES OF SYMMETRY IN CRYSTALS

(i) Centre of Symmetry : It is such an imaginary point within the crystal that any line drawn throughit intersects the surface of the crystal at equal distances in both directions. A crystal always possesses

only one centre of symmetry.

(ii) Plane of symmetry :It is an imaginary plane which passes through the centre of a crystal and

divides it into two equal parts such that one part is exactly the mirror image of the other.

A cubical crystal like NaCl possesses, in all, nine planes of symmetry; three rectangular planes

of symmetry and six diagonal planes of symmetry. One plane of symmetry of each of the above

is shown in fig.(a) and (b).

(iii) Axis of symmetry : It is an imaginary straight line about which, if the crystal is rotated, it will

present the same appearance more than once during the complete revolution. The axes ofsymmetry are called diad, triad, tetrad and hexad, respectively, if the original appearance is

repeated twice (after an angle of 180), thrice (after an angle of 120), four times (after an

angle of 90) and six times (after an angle of 60) in one rotation. These axes of symmetry are

also called two-fold, three-fold, four fold and six-fold, respectively.

In general if the same appearance of a crystal is repeated on rotating through an angle of360

,n

around an imaginary axis, the axis is called an n-fold axis.

In all, these are 13 axes of symmetry possessed by a cubical crystal like NaCl as shown in fig.(c),

(d) and (e).


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(iv) Elements of symmetry:The total number of planes, axes and centre of symmetry possessed by acrystal are termed as elements of symmetry. A cubic crystal possesse a total of 23 elements ofsymmetry.

Planes of symmetry = (3 + 6) = 9 Fig (a) and (b)

Axes of symmetry = (3 + 4 + 6) = 13 Fig (c), (d) and (e)

Centre of symmetry = 1 Fig. (f)

The total number of symmetry elements = 23

CRYSTAL LATTICE

The regular three dimensional arrangement of the points in a crystal is called space lattice. The spacelattice is also called crystal lattice. The locations of the points in the space lattice are called lattice pointsor lattice sites.

UNIT CELL

The smallest but complete unit in the space lattice which when repeated over and over again in the threedimensions, generates the crystal of the given substance.

TYPES OF UNIT CELLS

Unit cells are basically of two types. These are primitive and non primitive.

Primitive unit cells :A unit cell is called primitive unit cell if it has particles (or points) only at thecorners. It is also called simple unit cell.

Non-primitive unit cells :In this type of unit cells, particles (or points) are present not only at thecorners but also at some other positions. These are of three types :

(i) Face centred :Particles (or points) are located at the corners and also in the centre of each face.

(ii) Body centred : Particles (or points) are located at the corners and also at the centre within thebody.

(iii) End centred :Particles (or points) are located at the corners and also at the centres of the endfaces.


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5


Simple or Primitive unit cell Body centred unit cell Face centred unit cell End centred unit cell

Fig.Primitive and non-primitive unit cells

SEVEN CRYSTAL SYSTEMS

It we take into consideration the symmetry of the axial distances (a, b, c) and also the axial anglesbetween the edges ( , , ) , the various crystals can be divided into seven systems. These are alsocalled crystal habits. These crystals alongwith suitable examples are listed in the table

The seven crystal systems

System Axial distance Axial angles Examples

Cubic a = b = c === 90 NaCl, KCl, Diamond, Cu, Ag

Tetragonal a = b c === 90 White tin, SnO2Orthorhombic a b c === 90 Rhombic sulphur, KNO3, K2SO4, BaSO4, PbCO3Monoclinic a b c == 90,90 Monoclinic sulphur, Na2SO4.10H2OHexagonal a = b c 90 , 120 = = = Graphite, ZnO, CdS

Rhombohedral a = b = c 90 = = Calcite (CaCO3), quartz, NaNO3Triclinic a

b

c

=90 CuSO

4.5H

2O, K

2Cr

2O

7

BRAVAIS LATTICES

The seven crystal systems listed above can be further classified on the basis of the unit cells present.Although each crystal system is expected to have four different unit cells, but actually all of them cannotexist in each case. Bravaisin 1848 has established fourteen different types of lattices called Bravaislattices according to the arrangement of the points in the different unit cells involved listed below:

Crystal System Types of Lattices

Cubic Simple, Face centred, Body centred

Tetragonal Simple, Body centred

Orthorhombic Simple, Face centred, Body centred, End centred

Monoclinic Simple, End centred

Rhombohedral Simple

Triclinic Simple

Hexagonal Simple

In various unit cells, there are three kinds of lattice points : points located at the corners, points in the facecentres and points that lie entirely within the unit cell. In a crystal, atoms located at the corner and face-centre of a unit cell are shared by other cells and only a portion of such an atom actually lies within agiven unit cell.


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(i) A point that lies at the corner of a unit cell is shared among eight unit cells and, therefore, only one-eighth of each such point lies within the given unit cell.

(ii) A point along an edge is shared by four unit cells and only one-fourth of it lies within any one cell.

(iii) A face-centred point is shared by two unit cells and only one half of it is present in a given unit cell.

(iv) A body-centred point lies entirely within the unit cell and contributes one complete point to the cell.

Types of lattice point Contribution to one unit cell

Corner 1/8

Edge 1/4

Face-centre 1/2

Body-centre 1

Total number of constitutent units per unit cell

81= occupied corners + 41 occupied edge centres + 21 occupied face centres + occupiedbody centre.

Determination of Number of Constitutent units per unit cell :

Let edge length of cube = a cm

Density of substances = d g cm3

Volume of unit cell = a3cm3

Mass of unit cells = volume density = (a3 d) g

Number of mole per unit cell =M

da3 ;

where M = molar massNumber of molecules per unit cell = Number of mole Avogadros Number

3a d NZ

M

=

EXAMPLES - 1

The density of KCl is 1.99 g cm3and the length of a side of unit cell is 6.29 as determined by X-raydiffraction. Calculate the value of Avogadros number.

SOLUTION :

KCl has face-centred cubic structure,

i.e., Z = 4

Avogadros number =Vd

MZ

Given, d = 1.99; M = 74.5; V = (6.29 108)3cm3

Avogadros number =23

8 3

4 74.56.01 10

1.99 (6.29 10 )

=

EXAMPLES - 2

An element occurs in bcc structure with a cell edge of 288 pm. The density of element is 7.2 g cm3. Howmany atoms does 208 g of the element contain?


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SOLUTION :

Volume of the unit cell = (288 1010)3= 23.9 1024cm3

Volume of 208 g of the element =3cm88.28

2.7

208=

Number of unit cell in 28.88 cm3= 24109.23

88.28

= 12.08 1023unit cells

So, total no. of atoms in 12.08 1023unit cells = 2 12.08 1023 [since each bcc unit cell contain 2atoms]

= 24.16 1023

EXAMPLES : 3

The density of potassium bromide crystal is 2.75 g cm3and the length of an edge of a unit cell is 654 pm.The unit cell of KBr is one of three types of cubic unit cells. How many formula units of KBr are there

in a unit cell? Does the unit cell have a NaCl or CsCl structure?SOLUTION :

We know that

=

A3 N

M

a

N

M

NaN A

3 =

10 3 232.75 (654 10 ) 6.023 103.89 4

119

= = !

Number of mass points per unit cell = 4

Thus ,, has NaCl type crystal is fcc structure.

EXAMPLES : 4

A cubic solid is made up of two atoms A and B. Atoms A are present at the corners and B at the centreof the body. What is the formula of the unit cell?

SOLUTION :

Contribution by the atoms A present at eight corners = 1/8 8 = 1

Contribution by the atom B present in the centre of the body = 1

Thus, the ratio of atoms of A : B = 1 : 1

Formula of unit cell = AB

EXAMPLES : 5If three elements P, Q and R crystallise in a cubic unit cell with P atoms at the corners. Q atoms at thecubic centre and R atoms at the centre of each face of the cube, then write the formula of the compound.

SOLUTION :

Contribution made by the atoms of P present at all the eight corners of the tube = 1/8 8 = 1

Contribution made by the atoms of Q present in the centre of the body = 1

Contribution made by the atoms of R present at the centre of all the six faces = 1/2 6 = 3

Formula of the compound = PQR3


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8


RELATION BETWEEN EDGE LENGTH (A) AND THE RADIUS OF SPHERE (R)

We have calculated the number of atoms (or spheres since an atom or particle may be regarded as thesphere) in different types of cube unit cells. Now, let us calculate the relation between the edge length (a)and the radius (r) of the spheres present in these unit cells in case of pure elements i.e. elements composedof the same type of atoms.

(a) Distance between the centres of the spheres present on the corners of the edges of the cubic iscalled edge length (a).

(b) Distance between the centres of the two nearest spheres is called nearest neighbour distance (d)and is equal to r + r = 2r. (Here r is the radius of the spheres or the atomic radius)

FOR SIMPLE CUBIC LATTICE

In a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere,

volume occupied by one sphere present in the unit cell = 343r

Since the spheres at the corners touch each other;

Edge length (a) = r + r = 2r = d

No of atom on the corners = 188

1=

aTotal no. of atoms in a unit cell = 1

Radius of an atom = r

Distance between nearest neighbour (d) = a

Volume of cube = a3= (2r)3= 8r3

Packing fraction =cubeofVolume

sphereofVolume = 524.06r8

r3/4 3

3

==

% volume occupied by atom = 52.4

% volume unoccupied = 100 52.4 = 47.6

FOR FACE CENTRED CUBIC LATTICE

No. of atoms on the corners = 188

1=

a

B

C

A

No. of atoms on the faces = 362

1=

Total no. of atoms in a unit cell = 1 + 3 = 4

Volume occupied by atoms =3r

3

44

From ,ABC AC2 = AB2+ BC2

AC2 = a2+ a2= 2a2

AC = 2 a.

Distance between nearest neighbour (d) =AC a 2 a

2 2 2= =


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9


Radius of the atom (r) =d

2=

22

a

Volume occupied by atoms in a unit cell

3

22

a

3

4

4

=

Thus packing fraction 74.0a

22

a

3

44

3

3

=

=

% volume occupied by unit cell = 74

% volume unoccupied by unit cell = 100 74 = 28

FOR BODY CENTRED CUBIC LATTICE

No. of atoms on the corners = 188

1

=No. of atoms in the body = 1

Total no. of atoms in a unit cell = 1 + 1 = 2

a

B

C

A

DFrom ADC , AD2= AC2+ CD2= ( ) 222 a3aa2 =+

AD = 3a

Distance between nearest neighbour (d) =2

3a

2

AD=

Radius of an atom (r) = d/2 = 4

3a

Volume occupied by atoms =

34 a 3

23 2

Packing fraction =cubeof.Vol

atomsbyoccupied.Vol

3

3

4 a 32

3 20.68

a

= =

% volume occupied in a unit cell = 68

% volume unoccupied in a unit cell = 32

Lattice type Radius (r) Distance between Nearest neighbour Packing fraction

Simple cube 2/ar= d = 2r = a 0.524

Face centred r =22

a2d a= 0.74

Body centred

4

3ar=

2

3ad= 0.68


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EXAMPLES : 6

The radius of copper atom is 128 pm. If it crystallises in a face centred cubic lattice (fcc). what will bethe length of the edge of the unit cell?

SOLUTION :

For face centred cubic crystal

Edge length (a) = pm362)pm128()4142.1(2r22 ==EXAMPLES : 7

Xenon crystallises in a face centred cubic lattice and the edge of the unit cell is 620 pm. Calculate thenearest neighbouring distance and the radius of the xenon atom?

SOLUTION :

Nearest neighbour distance (d) = pm4.4384142.1

)pm620(

2

a==

Radius of xenon atom (r) = pm2.2194142.12

)pm620(

22

a

==

EXAMPLES : 8

The face diagonal of a cubic closed packed unit cell is 4. What is its face length?

SOLUTION :

Face diagonal = 2 2a a a 2+ = ; 83.24142.1

)4(

2

4

2

diagonalFacea ====

EXAMPLES : 9

An element A crystallises in face-centred cubic lattice with edge length of 100 pm. What is the radius ofits atom A and the nearest neighbour distance in the lattice? What is the length of its face-diagonal?

SOLUTION :

For a face centred cubic lattice (fcc)

Radius (r) = pm4.354142.12

)pm100(

22

a=

=

Nearest neighbour distance (d) = 2r = 2 35.4 pm = 70.8 pmLength of face diagonal = 4r = 4 35.4 pm = 141.6 pm

CRYSTAL DEFECTS OR IMPERFECTIONS IN SOLID

Crystal defects

Electronicimperfection

Atomic imperfectionor

point defect

Stoichiometric defect Non-stoichiometric defect Impurity defect

Schottky defect Frenkel defect Metal excess Metal deficiency

Anion vacancy defect Extra cation Impurity defect

in covalent solid

Impurity defect

in Ionic solid


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SCHOTTKY DEFECT

It consists of a pair of holes in the crystal lattice i.e one positive and one negative ion are absent fromtheir proper position. This sort of defect occurs mainly in highly ionic compounds where the +ve and -ve

ions are of similar size and hence co-ordination no. is high e.g.NaCl, CsCl, KCl, KBr etc.

No of Schottky defect formed for cm3(ns) is given by

ns=swN.exp.

2kT

where N = No. of sites per cm3that could be left vacant

A B A B A

A+

B B A B

A+

B AA A

A+

B AA B B

+ + +

+

+ + +

+ + +

WS= work necessary to form a schottky defect.

K = gas constantT = absolute temp.

At room temp. NaCl has one defect in 1015lattice site.At 500C, the value rising to one in 106sites.

FRANKEL DEFECT

It consists of a vacant lattice site (a hole) and the ionwhich ideally should have occupied the site nowoccupies an interstitial position.

In this type of defect metal ions are generaly smaller

than the anion i.e. large difference in size and co-ordination number is low. e.g. AgCl, AgBr, AgI.

The no. of Frenkel defect formed per cm3(Nf) isgiven by

ff

wN NN ' exp

2kT

=

A+

B

A+

A

+

B

B

A

+

A+

B

A+

A+

A+

B

A+

A+

B

B

B A

+A

+

A+

B

A+

B

A+

where N is the number of sites per cm3that could be left vacantNis the no. of alternative interstitial per cm3.wfis the work necessary to form a Frenkel defect, K is the gas constant and T is absolute

temp.

METAL EXCESS DEFECT

F-centres :This arises due to absence of anion from its Latticesite, leaving a hole which is occupied by an electron therebymaintaining the electrical balance. When compound such as

NaCl, KCl LiH or TiO are heated with excess of theirconstituent metal vapours or treated with high energy radiation,they become deficient in negative ion and their formula may be

1AX , where = small fraction.

Anion sites occupied by electrons in this way are called Fcentres.

More the no. of F-centres, more is the colour of compound.

A+

B A

A+

B B A

A+

AA

A+

B AA B

B

B

A

+

+ +

++

+ +

e


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12


Interstitial ions and electrons :Metal excess defect also occur when an extra +ve ion occupies oninterstitial position in the lattice, the electrical neutrality is maintained by the inclusion of an interstitialelectron. The composition may be represented by general formula XA1 + . The example include ZnO,CdO, Fe2O3and Cr2O3.

METAL DEFICIENCY DEFECT

This type of compound is represented by the formula XA1 . This deficiency can occur in two ways :

(i) Positive ion absent : If a positive ion is absent from its lattice site, the charges can be balanced by

an adjacent metal ion having an extra +ve charge. e.g. FeO, NiO, -TiO, FeS and CuI.

A B A

A+B B A

A+B AA

A+B AA B

B

B

+ +

+

2++

+ +

A B A

A+

B B A

A+

B AA

A+

B AA B

B

B

A

+ +

+ +

+ 2+

+ +

B

Positive ion absent Extra interstitial negative ion

(ii) Extra interstitial negative ion : It is possible to have an extra negative ion in an interstitialposition and to balance the charges by means of an extra charge on an adjacent metal ion.

RADIUS RATIO

The ratio of the radius of cation to that of anion in an ionic crystalline solid, is called radius ratio.

Radius ratio =)r(anionofRadius

)r(cationofRadius

+

The relation between the radius ratio (r+ / r) and co-ordination number is quite often known as Radiusratio rule.This helps in predicting the structures of ionic crystals. The same has been depicted in thetable.

Radius ratio Possible Co-ordination Structural Examples

(r+ / r) number of cation Arrangement

155.0 2 Linear

0.155 0.225 3 Trigonal planar B2O30.225 0.414 4 Tetrahedral ZnS, CuCl, CuBr, HgS

0.414 0.732 6 Octahedral NaCl, KBr, MgO, CaO, CaS, NaBr

0.732 1 8 Cubic CsCl, CsBr, CsI, NH4Br

Larger the cation more will be its co-ordination number due to increase of its radius ratio. As the radius ratio increases more and more beyond 0.732, anion more farther and farther apart and

cubic void are generated.


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EXAMPLES : 10

In the crystal of an ionic compound, the anions B2form a closed packed lattice and cations A+occupy allthe tetrahedral voids. What is the simplest formula of the compound ?

SOLUTION :

We know that in a close packed lattice involving anions B2having tetrahedral voids, there are twotetrahedral sites for each anion and both are occupied by the cations A+.

Thus, for one anion (B2), there are two cations (A+)

Formula of the compound = A2B.

EXAMPLES : 11

A solid AB has the NaCl structure. If radius of the cation A+is 120 pm. Calculate the maximum valueof the radius of the anion B.

SOLUTION :Since NaCl has octahedral structure.

The limiting ratioA

B

r0.414

r

+

=

or pm290414.0

120

414.0

rr AB ===

+

STRUCTURE OF SIMPLE IONIC COMPOUNDS

Anions (large in size) form a close packed arrangement but cations (smaller in size) occupy the interstitialsites resulting from the close packing of the anions e.g. NaCl, ZnS, CsCl etc.

In CaF2, Ca2+ions from the close packed arrangement while the Fions fit in the interstitial sites.

In a close packed arrangement, the number of octahedral sites is half the number of tetrahedral sites. For the radius ratio more than a minimum value or limiting value, close packed arrangement may open

slightly to accommodate the anions that are of bigger size.

Greater the co-ordination number (CN) more is the stability of ionic crystal due to greater forces ofattraction.

The different ionic compounds can have the following type of structures.

(i) Ionic compounds of AB type

(ii) Ionic compounds of AB2type

(iii) Ionic compounds of A2B type

STRUCTURES OF IONIC COMPOUNDS OF AB TYPE

Ionic compounds of the type AB suggest that in their crystal lattice A+and Bions are present in the ratio of1 : 1. These compounds have following three types of structures.

Rock salt (NaCl) type structures.

Cesium chloride (CsCl) type structure.

Zinc blende (ZnS) type structure.


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ROCK SALT (NaCl) TYPE STRUCTURE

Cl ion surrounded

octahedrally by six Na ions

+

Na surrounded

octahedrally by six Cl ions

+

ion

Fig.Sodium Chloride Structure

Characteristics:

NaCl is composed of Na+ion and Clions. The ionic radius of Na+ and Cl ion are 95 pm and 181 pm respectively. Thus radius ratio is

Na

Cl

r 950.525

r 181

+

= =

The structure is octahedral in which Clion constitute a cubic closed packed and Na+ions occupythe octahedral voids.

The coordination number is 6 : 6 ie. each Na+ion is surrounded by six Clions and vice versa.

A unit cell of NaCl consists of four NaCl units. Other examples having similar structure are

(i) Halide of Li, Na, K, Cs.

(ii) Halide of Ammonium.

(iii) Oxides and sulphides of Mg, Ca, Sr & Ba.

(iv) Halides of Ag (Except AgI)

(v) CaCO3and CaC2

CESIUM CHLORIDE (CsCl) TYPE STRUCTURES

Characteristics :

(i) The crystal is composed of Cs+and Clions and their ionic radii are 160 pm and 181 pm respectively.The radius ratio for the crystal is :

r Cs 160 pm0.889

r Cl 181pm

+

= =

The radius ratio suggests a body centred cubic structure. However, it is more than the ideal ratio which is0.732.

(ii) The Clions are present at the eight corners of the unit cell while one Cs+ion is present in the centreof the body as shown in the fig.(a)


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(iii) Each Cs+ion is surrounded by 8 Clions and each Clion by 8 Cs+ions. Thus, the coordinationnumber of each ion is 8 or these are in 8 : 8 co-ordination.

(iv) The contribution by all the 8 Clions at the corners is 1 while by the Cs+ion present in the centre ofbody is also 1. Therefore, the unit cell of CsCl consist of one Cs+ion and one Clion and the formula

of cesium chloride is CsCl.

= Cs+

= Cl

(a) (b)

Fig.Cesium chloride structure

ZINC BLENDE (ZnS) TYPE STRUCTURE

Fig.Structure of Zinc Blende (ZnS)

Characteristics :

(i) The sulphide ions (S2) with radius = 184 pm adopt cubic close packed arrangement (ccp) i.e. S2

ions are present at all the corners as well as at the centre of each face.

(ii) The radii of Zn2+and S2ions are 74 pm and 184 pm respectively. Therefore, the radius ratio is :

2

2

Zn

S

r 74pm0.402

r 184 pm

+

= =

(iii) Each Zn2+ion present occupies a tetrahedral site. As there are two tetrahedral sites per atom in aclose packed arrangement one half of these sites are occupied by Zn2+ions whereas the remaininghalf are vacant. Thus Zn2+ions are present at the alternative sites as shown in fig.

(iv) Each Zn2+ion is surrounding by four S2ions which are directed towards the four corners of atetrahedron. Similarly each S2ion is also surrounding by four Zn2+ions. Thus, coordination numberof both these ions is 4 and these are also present in the ratio of 4 : 4.

(v) Other examples of this type are : Chloride, bromides and iodides of copper, silver iodide, beryllium sulphideetc.

STRUCTURES OF IONIC COMPOUNDS OF AB2TYPE

These are the crystalline ionic solids in which cations and anions are in the ratio 1 : 2. Most of these

compounds have calcium fluoride (CaF2) type structure which is also knows as fluorite structure.


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= Ca2+

= F

Fig.Structure of Calcium Fluoride (Fluorite Structure)

Characteristics

(i) The number of Fions is twice the number of Ca2+ions i.e. Ca2+and Fions are in the ratio 1 : 2.The radii of Ca2+and Fions are 99 pm and 136 pm respectively. Therefore, radius ratio is :

Ca

F

r (99pm)0.73

r (136pm)= =

2+

(ii) The radius ratio suggests a body centred cubic structure in which the Ca2+ions are present at thecorners and at the centre of each face of the cube.

(iii) The radius ratio suggests that each Zn2+ion present occupies a tetrahedral site. As there are twotetrahedral sites per atom in a close packed arrangement, one half of these sites are occupied byZn2+ions whereas the remaining half are vacant. Thus, Zn2+ions are present at the alternative sites.

(iv) Each Zn2+ ion is surrounded by four S2 ions which are directed towards the four corners of atetrahedron. Similarly each S2ion is also surrounded by four Zn2+ions. Thus, coordination numberof both these ions is 4 and these are also present in the ratio of 4 : 4.

The other example of this type of Lattice are: BaF2, BaCl2, SrF2, SrCl2, CdCl2, PbF2etc.

ANTIFLUORITE STRUCTURE IN IONIC COMPOUNDS OF A2B TYPE

= Na

O ion(in fcc arrangement)

2

Na ion

(at tetrahedral hole)

+

= O

+

2

Fig.Structure of Na2O (Antifluorite Structure)

(a) The compounds of A2B type structure is known as antifluorite structure.

(b) In this structure, the positions of the cations and anions as compared to fluorite structure get reversedi.e. the smaller cations occupy the position of fluoride ions while the anions with bigger size occupythe positions of calcium ions.

(c) Sodium oxide (Na2O) is an example of antifluorite structure.

(d) The oxide ions (O2) have ccp arrangement and Na+ions occupy the tetrahedral voids.


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(e) Thus there are two tetrahedral sites per atom in close packed arrangement and these are occupiedby Na+ions.

(f) Each Na+ion is surrounded by four O2ions and each O2ions is surrounded by eight Na+ions. TheC.N. is 4 : 8.

(g) There are several oxides and sulphides of the metals with antifluorite structure e.g. Li2O, K2O,Rb2O and Rb2S.

EFFECT OF PRESSURE AND TEMPERATURE ON CRYSTALSTRUCTURE

At ordinary temperature and pressure, chlorides, bromide and iodides of alkali metals lithium, sodium,potassium and rubidium and halides of silver have NaCl type structure with 6 : 6 co-ordination.

NaCl structure transform to CsCl structure at high pressure in which the ions are in the ratio 8 : 8. When high temperature of 760 K is applied on cesium chloride (CsCl), with co-ordination of 8 : 8, it

changes sodium to chloride with co-ordination of 6 : 6.

BRAGGS LAW

When X-rays are incident on a crystal face, they penetrate into the crystal and strike the atoms indifferent planes. From each of these planes, X-rays are deflected. Bragg presented a relationship betweenthe wavelength of the X-rays and the distance between the planes, such as

= sind2nwhere n is an integer such as, 1, 2, 3, ..., is the wavelength, d is the distance between repeating planesof particles and the angle of deflection or glancing angle. Two parallel layers of the crystals are shownwith the help of horizontal lines separated from each other by distance d. Two waves comprising X-raysthat are in plane strike against the crystal and after reflection emerge from the crystal.

C

DB

A

d

E

Fig.X-ray diffraction by regularly spaced constituents in a crystal

Further, the wave emerging along path AE is ahead of the wave that follows the path CD. The differencein the distance travelled by the two rays after reflection also called path difference is :

Path difference = BC + CD = AC sin + AC sin = 2d sin This must be an integral multiple of wave length i.e. n . Therefore.

n 2d sin = (Bragg Equation)

EXAMPLES : 12

X-rays of wavelength 1.54 strike a crystal and are observed to be deflected at an angle of 22.5A.Assuming that n = 1. Calculate the spacing between the planes of atoms that are responsible for thisreflection.


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SOLUTION :

According to Braggs law,

= sind2n

Given n = 1, ,54.1= = 22.5 A01.2

383.02

54.1

5.22sin2

54.1d =

=

=

EXAMPLES : 13

If NaCl is doped with 103mol percent of SrCl2what is the concentration of cation vacancy?

SOLUTION :

Na+ Cl Na+ Cl

Cl Cl Na+

Sr2+ Cl Na+ Cl

Cl

Na

+

Cl

Na

+

Number of cationic vacancies per mol 18233

10023.6100

10023.610=

=

vacancies per mol.

CLOSEST PACKING

In crystals the atoms, ions or molecules are arranged in aregular way in a three-dimensional space. The arrangementhas minimum energy and hence maximum stability. Formaximum stability, a constituent in the aggregate must besurrounded by the maximum number of neighbours. Formaximum number of contacts, each constituent of the

crystal must be packed as closely as possible. In a two-dimensional plane the closest arrangement being that inwhich each sphere is in contact with six other spheres.This layer

is denoted by A.

To obtain the closest packing in space, the spheres areplaced over the layer obtain the closest packing in space,the spheres are placed over the layer A in a regular manner.There are six vacant sites or triangular pockets aroundany sphere in layer A. In fig around a sphere X these sitesare labelled as 1, 2, 3, 4, 5 and 6. We can place only three

spheres touching each other in alternate vacant sites, e.g.either in 1, 3 and 5 or in 2, 4 and 6.

Suppose the second layer is constructed by placing thespheres in the vacant sites labelled as 1, 3 and 5 then thesites marked 2, 4 and 6 are left unoccupied. This layer isdenoted by B.

A

B

A

C

B

A

B

A

B

A

Y

6

5

4

3

2

1

Close packed structures(a) Hexagonal closest packing(b) Cubical closest packing

(a)(b)

x

The second layer again has two types of vacant sites.Around any atom in the second layer, one set of vacantsites lies just above the vacant sites 2, 4 and 6 of the firstlayer and the other set lies above


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the centres of the spheres of the first layer. Thus after the second layer is complete, there are twodifferent ways of placing the spheres in the third layer. If the spheres in the third layer are placed in thevacant sites which are above the centres of the spheres of the first layer, then the third layer repeats thearrangement of the first layer and we have only two types of layers, viz., ABAB .... etc. This is called a

hexagonal closest packing, hcp. fig (a) On the other hand, if the spheres are placed in the vacant sitesabove 2, 4 and 6, a new arrangement (C) of the spheres in produced. Fig (b) The entire arrangement isnow of the type ABCABC ........... etc., and is referred to as a cubical closest packing, ccp or facecentred cubical closest packing, fcc.

The number of nearest neighbours (known as the coordination number) in each arrangement is twelve;six in the same layer three in the layer above and three in the layer below it.

In the fcc structure,there are two types of vacant sites or holes : tetrahedral and octahedral. A tetrahedralhole is surrounded by four spheres while an octahedral hole is the empty space surrounded by six spheres.These vacant sites can accommodate other smaller atoms or molecules giving rise to a variety of differentstructures.

(a) (b)

Fig.(a) Tetrahedral holes and (b) Octahedral holes in an fcc structure

In an fcc structure, there are eight tetrahedral holes per unit cell as shown in fig.(a) The number ofoctahedral holes in the unit cell of the fcc structure is four as can be deduced from the followingconsiderations. In fig(b) each (X) mark represent a vacant site and there are twelve such vacant sites atthe edges of unit cubic lattice. The vacancy at an edge is common to four unit cells and hence the number

of such vacant sites per unit cell is12

34

= . In addition to these vacancies there is one octahedral hole at

the centre of the unit cell. Thus the total number of octahedral holes per unit cell is four.

Fig.Body centered cubical close-packed structure

Another less packed arrangement of spheres is the body centered cubic arrangement as shown in figureabove in which each sphere has eight nearest neighbours; four in the same plane, two above and two

below it in the adjacent layers. Since the coordination number in fcc or hcp structure is greater than thatin the bcc structures, the latter are therefore less denser.

PROPERTIES OF SOLIDS

There are three main types of properties.

1. ELECTRICAL PROPERTIES

Based on their conductivity, solids can be divided into three categories :


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(i) Conductors

(ii) Insulators

(iii) Semi-conductors

(i) Conductors : The solids through which electricity can flow to a large extent are called conductors.They are either metallic conductors or electrolytic conductors. In the metallic conductors, the flowof electricity is due to flow of electrons (electronic conductors) without any chemical change occurringin the metal. While in electrolytic conductors like NaCl, KCl etc, the flow of electricity takes placeto a good extent only when they are taken in the molten state or in aqueous solution. The flow ofelectricity is due to flow of ions.

(ii) Insulators: The solids which almost do not allow the electricity to pass through them are calledinsulators e.g. S, P, plastics, wood, rubber etc.

(iii) Semi-conductors :The solids whose conductivity lies between those of metallic conductors andinsulators are called semi-conductors. The electrical conductivity of semi-conductors and insulators

is due to the presence of impurities and defects. Their conductivity increases with increase oftemperature as the defects (such as holes) increase with increase of temperature.

(a) Pure substances that show conductivity similar to that of silicon and germanium are calledintrinsic semiconductors.

(b) Electrical conductivity of semiconductors increases with increase of temperature becausewith increase of temperature, large number of electrons can jump from valence band toconduction band.

(c) Electrical conductivity of silicon and germanium is very low at room temperature. It can beincreased by doping with elements of Group 15 or Group 13.

(d) Similar to combination of Group 14 elements with those of Group 15 or Group 13, semiconductors

have been prepared by combination of elements of Group 13 and 15 (e.g. InSb, AlP and GaAs)or Group 12 and 16 (e.g. ZnS, CdS, CdSe and HgTe)

(e) n- and p-type semiconductors are combined suitably to form a number of electronic components.

2. MAGNETIC PROPERTIES

Solids are divided into following categories on the basis of magnetic properties.

(i) Diamagnetic substances : Substances which are weakly repelled by the external magnetic fieldare called diamagnetic substances e.g. TiO2, NaCl, benzene etc. The property thus exhibited iscalled diamagnetism. This property is shown only by those substances which contain fully filledorbitals i.e. no unpaired electron is present.

(ii) Paramagnetic substances : Substances which are attracted by the external magnetic field arecalled paramagnetic substances. The property thus exhibited is called paramagnetism. This propertyis shown by those substances whose atoms, ions or molecules contain unpaired electrons e.g. O2,Cu2+, Fe3+etc. These substances, however, lose their magnetism in the absence of the magneticfield.

(iii) Ferromagnetic substances : Substances which show permanent magnetism even in the absenceof the magnetic field are called ferromagnetic substances, e.g. Fe, Ni, Co, CrO2show ferromagnetism.Such substances remain permanently magnetised, once they have been magnetised. This type ofmagnetism arises due to spontaneous alignment of magnetic moments due to unpaired electrons inthe same direction, as shown in fig.(a)


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(c) Ferrimagnetism(a) Ferromagnetism (b) Anti-ferromagnetism(iv) Anti-Ferromagnetic substances : Substances which are expected to possess paramagnetism or

ferromagnetism on the basis of unpaired electrons but actually they possess zero net magneticmoment are called anti-ferromagnetic substances, e.g. MnO. Antiferromagnetism is due to the

presence of equal number of magnetic moments in the opposite directions as shown in fig.(b)

(v) Ferrimagnetic Substances : Substances which are expected to possess large magnetism on thebasis of the unpaired electrons but acutally have small net magnetic moment are called ferrimagneticsubstances e.g. Fe3O4, ferrites of the formula M

2+Fe2O4where M = Mg, Cu, Zn etc. Ferrimagnetismarises due to the unequal number of magnetic moments in opposite direction resulting in some netmagnetic moment, as shown in fig(c). Each ferromagnetic substance has a characteristic temperature

above which no ferromagnetism is observed. This is known as curie temperature.

3. DIELECTRIC PROPERTY

(i) Piezoelectricity or Pressure electricity : Electricity is produced due to displacement of ions,when mechanical stress is applied on such crystal. The electricity thus produced is calledpiezoelectricity and the crystals are called piezoelectric crystals.

+ + + + + + +

+ + +

+ + +

Electric field

Dipoles aligned inan ordered manner

Voltmeter

Mechanical stress

Mechanical stressFig.Production of electricity on applying mechanical stress on piezoelectric crystals

A few examples of piezoelectric crystals include titanates of barium and lead, lead zirconate (PbZrO3),ammonium dihydrogen phosphate (NH4H2PO4) and quartz. These crystals are used as pick-ups inrecord players where they produce electrical signals by application of pressure. They are also usedin microphones, ultrasonic generators and sonar detectors.

(ii) Pyroelectricity : Some piezoelectric crystals when heated produce a small electric current. Theelectricity thus produced is called pyroelectricity (pyre means heat).

(iii) Ferroelectricity : In some of the piezoelectric crystals, the dipoles are permanently polarized evenin the absence of the electric field. However on applying electric field, the direction of polarizationchanges. This phenomenon is called ferroelectricity due to analogy with ferromagnetism. Someexamples of the ferroelectric solids are barium titanate (BaTiO3), sodium potassium tartarate (Rochellesalt) and potassium dihydrogen phosphate (KH

2

PO4

). It may be pointed out here that all ferroelectricsolids are piezoelectric but the reverse is not true.


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(iv) Anti-ferroelectricity : In some crystals, the dipoles align themselves in such a way that alternately,they point up and down so that the crystal does not possess any net dipole moment. Such crystalsare said to be anti-ferroelectric. A typical example of such crystals is lead zirconate (PbZrO3).

METAL SILICATESMetal silicates have wide range of structures ranging from simple monomers to complex polymers,however, the basic unit of all the silicate ions is tetrahedron nature. These tetrahedral occur singly, or bysharing oxygen atoms in small groups, in small cyclic groups, in infinite chains or infinite sheets givingvarious structures.

1. Simple orthosilicates : In these silicates SiO44tetrahedra do not share oxygen atoms with one

another and exist as discrete SiO44orthosilicate anions e.g. Mg2SiO4, Fe2SiO4and Zn2SiO4. In all

these compounds SiO44ion is linked to M2+ions through co-ordinate bond. [See Fig. (a)]

2. Pyrosilicates or islands : In these two SiO44tetrahedral share a common oxygen, island structures

having formula (Si2O7)6

e.g. Pyrosilicates. [See Fig. (b)]

3. Ring or cyclic silicate anions : Ring anions are obtained when two oxygen atoms of eachtetrahedron are shared with others. (Si3O9)

6and (Si6O18)12. General formula of any such anion

must be (SinO3n)2n. [See Fig. (c)&(d)]

(a) SiO44

(b) (Si O )2 75

(c) (Si O )3 96

(d) (Si O )6 1812

(e) (SiO ) chain3 n2

(f) (Si O ) layer 2 5 n2


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4. Infinite chain silicates : These are of two types the pyroxenes which contain single strand chainsof composition (SiO3

2)nand amphiboles which contain double strand cross-linked chains of composition(Si4O11

6)n. General formula of the anion in a pyroxene is the same as in a silicate with cyclic anione.g. Diopside (MgCaSi2O6). Various asbestos minerals are examples of chain silicates (amphiboles).

5. Infinite sheet Silicates : When three oxygen atoms of each SiO4tetrahedron are shared, the twodimensional sheet structures with general formula of (Si2O5

2)n are obtained e.g. clay.[See Fig. (e)]

6. Framework silicate : Three dimensional network is obtained when all the four oxygen atoms ofeach of SiO4tetrahedron are shared. e.g. Quartz and zeolites. The general formula of frameworksilicate is SiO2. [See Fig. (f)]


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EXERCISE

Section A : (i) True and False Statements

(ii) Fill in the Blanks

(iii) Assertion-Reason Type Questions

Section B : (i) Multiple Choice Questions

(ii) MCQs asked in Competitive Examinations

Section C : Subjective Questions


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EXERCISES

SECTION - A

TRUE AND FALSE STATEMENTS

1. The number of atoms in a unit cell having f.c.c. structure is 2

2. The electricity produced by applying mechanical stress on a crystal is called piezoelectricity.

3. Quartz is amorphous whereas silica is crystalline.

4. Frankel defect not found in pure alkali metal halides.

5. Fe0.95O is called stoichiometric oxide of iron.

6. Crystalline solids are anisotropic.

7. Silicon carbide is a covalent crystal.

8. In (hcp) systems, each sphere is surrounded by eight other spheres.

9. The number of atoms per unit cell of body-centred cube is 2.

10. Schottky defect in crystal results in appreciable increase in its density.

FILL IN THE BLANKS

1. On applying __________________ pressure, NaCl structure changes to CsCl type strucgture.

2. As the radius of cation increases, C.N. __________________

3. No. of tetrahdral voids is __________________ the no. of octahidal void.

4. Iron, Nickel, copper, silver and gold crystallises in __________________ structure.

5. It is observed that NaCl crystal at room temperature there are about 10

22

ions and 10

6

schottky pairs percm3. It means that there is one schottky pair per __________________ ions.

6. The crystal structure of CsCl is __________________ .

7. A face-centred point is shared by __________________ unit cells.

8. In face-centred cubic structure the empty space is __________________ %.

9. In hcp mode of packing, a sphere has coordination number __________________ .

10. The presence of a trace of arsenic in germanium makes it __________________ .


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ASSERTION-REASON TYPE QUESTIONS

I n each of the foll owing questions two statements are given as Asserti on A and Reason R. Examine the

statements carefull y and answer the questions according to the instructions given below :

"

Mark(1)if both Aand R are correct and Ris the correct reason of A." Mark (2)if both Aand Rare correct and Ris not the correct reason of A." Mark (3)if Ais correct and Ris wrong." Mark (4)if both Aand Rare wrong.

1. Assert ion A : In crystal lattice the size of the cation is larger in octahedral hole than in tetrahedral hole.Reason R: Cations occupy more space than anions in crystal packing.(1) (2)(3) (4)

2. Assert ion A : In a ccp lattice made up of Yions, all the tetrahedral voids are occupied by X+ions;hence the formula of the solid is X2Y.Reason R: In a ccp lattice number of tetrahedral voids formed is twice the number of ions making thelattice.

(1) (2)(3) (4)

3. Assert ion A : NaCl crystal unit cell has fcc arrangement.Reason R: There are 4 units of NaCl present per unit cell.(1) (2)(3) (4)

4. Assert ion A : A crystal having fcc structure is more closely packed than a crystal having bcc structure.Reason R: Packing fraction for fcc structure is double that of bcc structure.(1) (2)(3) (4)

5. Assert ion A : The two ions A+and Bhave radii 88 and 200 pm respectively. Coordination number ofA+will be 6.Reason R: When r+/r= 0.414 0.732, the coordination number is 6.(1) (2)(3) (4)

6. Asser tion A : Covalent crystals have the highest melting point.Reason R: Covalent bonds are stronger than ionic bonds.(1) (2)(3) (4)

7. Assert ion A : CsCl as body-centred cubic arrangement.Reason R: CsCl has one Cs+ion and 8 Clions in its unit cell.(1) (2)(3) (4)

8. Assert ion A : Triclinic system is the most unsymmetrical system.Reason R: No axial angle is equal to 90 in triclinic system.(1) (2)(3) (4)

9. Asser tion A : Antiferromagnetic substances on heating to high temperature become paramagnetic.Reason R: On heating, randomisation of spins occur.(1) (2)(3) (4)

10. Assert ion A : In any ionic solid, [MX], with Schottky defects, the number of positive and negative ionsare same.Reason R: Equal number of cation and anion vacancies are present.(1) (2)

(3) (4)


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SECTION - B

MULTIPLE CHOICE QUESTIONS

1. A mineral having the formula AB2crystallizes in the c.c.p. lattice, with A atoms occupying the latticepoints. The CN of A is 8 and that of B is 4. What percentage of the tetrahedral sites is occupied by Batoms?

(1) 25% (2) 50%

(3) 75% (4) 100%

2. CsBr has b.c.c. structure with edge length 4.3. The shortest inter ionic distance in between Cs+and Bris

(1) 3.72 (2) 1.86

(3) 7.44 (4) 4.3

3. The formula for determination of density of cubic unit cell is

(1) 303

cmgMz

Na

(2)3

30 cmg

aMNz

(3)3

0

3

cmgNz

Ma

(4)3

30

cmgaN

Mz

4. The closest-packing sequence ABAB ..... represents

(1) Primitive cubic packing (2) Body-centred cubic packing

(3) Face-centred cubic packing (4) hexagonal packing

5. In Braggs equation for diffraction of X-rays n represents

(1) The number of mole (2) Quantum number (3) The order of reflection (4) Avogadros number

6. Tetragonal crystal system has the following unit cell dimensions

(1) a = b = c and === 90(2) ==== 90andcba(3) === 90andcba

(4) ==== 120,90andcba

7. In a solid lattice, the cation has left a lattice site and is located at interstitial position, the lattice defect is

(1) Interstitial defect (2) Vacancy defect

(3) Frenkel defect (4) Schottky defect

8. Each unit cell of NaCl consists of 13 chloride ions and _______.

(1) 13 Na+ions (2) 14 Na+ions

(3) 6 Na+ions (4) 8 Na+ions

9. Germanium or silicon becomes semi-conductor due to

(1) Schottky defect

(2) Chemical impurity

(3) Frenkel defect

(4) None of these


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10. Which one of the following is correct?

(1) Schottky defect lowers the density

(2) Frenkel defect increases the dielectric constant of the crystals

(3) Stoichiometric defects make the crystals good electrical conductors(4) All of these

11. The electrons trapped in anion vacancies in metal excess defects are called

(1) Mobile electrons (2) Trapped electrons

(3) Valence electrons (4) F-centres

12. The correct statement regarding F-centre is

(1) Electrons are held in the voids of crystals

(2) F-centre provides colour to the crystals

(3) Conductivity of the crystal increases due to F-centre

(4) All of these13. Crystals where dipoles may align themselves in an ordered manner, so that there is a net dipole moment,

exhibit

(1) Pyro-electricity (2) Piezo-electricity

(3) Ferro-electricity (4) Anitferro electricity

14. Some of the polar crystals, on heating produce a small electric current called

(1) Pyro-electricity (2) Piezo-electricity

(3) Ferro-electricity (4) Anitferro-electricity

15. Germanium is an example of

(1) An intrinsic semiconductor (2) An n-type semiconductor (3) A p-type semiconductor (4) Insulator

16. Which type of semiconductor is obtained on mixing the arsenic into the silicon?

(1) n-type (2) p-type

(3) Internal (4) both (1) and (2)

17. If indium is added in small quantity of Ge metal, we get

(1) An n-type semiconductor (2) A p-type semiconductor

(3) Rectifier (4) Insulator

18. When n- and p-type semiconductors are allowed to come into contact

(1) Some electrons will flow from n to p (2) Some electrons will flow from p to n

(3) The impurity element will flow from n to p (4) The impurity element will flow from p to n

19. The electrical conductivity of semiconductors

(1) Increases with temperature (2) Decreases with temperature

(3) Remains constants on heating (4) None of the above

20. Super conductors are substances which

(1) Conduct electricity at low temperature (2) Conduct electricity at high temperatures

(3) Offer high resistance to the flow of current (4) Offer no resistance to the flow of current


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21. The phenomenon of superconductivity was first discovered by

(1) Einstein (2) Soddy

(3) Hund and Mulliken (4) Kammerlingh Onnes

22. Which substances possess zero resistance at 0 K?(1) Conductors (2) Semiconductors

(3) Superconductors (4) Insulators

23. Amorphous solids are

(1) Solid substances in real sense (2) Liquids in real sense

(3) Supercooled liquids (4) Substances with definite M.P.

24. Which of the following is not an example of molecular crystal?

(1) Hydrogen (2) Iodine

(3) Ice (4) Sodium chloride

25. The pure crystalline substance on being heated gradually first forms a turbid liquid at constant temperatureand still at higher temperature turbidity completely disappears. The behaviour is a characteristic ofsubstance forming

(1) Allotropic crystals (2) Liquid crystals

(3) Isomeric crystals (4) Isomorphous crystals

26. The co-ordination number of a body-centred atom in cubic structure is

(1) 4 (2) 6

(3) 8 (4) 12

27. The radius of an ion in a body centred cube of edge ais

(1)2a (2) 2

4a

(3)3

4

a(4) a

28. The fraction of the total volume occupied by atoms in a simple cube is

(1)2

(2)

8

3

(3) 6

2

(4) 6

29. A substance AxBycrystallizes in an f.c.c. lattice in which atoms of A occupy each corner of the cubeand atoms of B occupy the centres of each face of the cube. Identify the correct composition of thesubstance AxBy.

(1) AB3 (2) A4B3(3) A3B (4) Composition cannot be specified

30. In a closed packed lattice, the number of octahedral sites as compared to tetrahedral ones will be

(1) Equal (2) Half

(3) Double (4) None of these


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31. CsCl on heating to 760 K changes into

(1) Liquid (2) NaCl structure

(3) ZnS structure (4) None of these

32. Compounds having NaCl type structure when subjected to high pressure changes into(1) Unstable compound (2) CsCl type structure

(3) ZnS type structure (4) None of the three

33. In a face centred cubic cell, an atom at the face centre is shared by

(1) 4 unit cells (2) 2 unit cells

(3) 1 unit cell (4) 6 unit cells

34. Ionic solids are characterised by

(1) Good conductivity in solid state (2) High vapour pressure

(3) Low melting point (4) Solubility in polar solvents

35. Silicon is a(1) Conductor (2) Semiconductor

(3) Non conductor (4) Metal complex

36. On adding a little phosphorus to silicon, we get a/an

(1) n-Type semiconductor (2) p-Type semiconductor

(3) Metallic conductor (4) Insulator

37. Piezoelectric crystals are used in

(1) Radio (2) TV

(3) Record player (4) Refrizerator

38. Glass is

(1) Supercooled liquid (2) Crystalline solid

(3) Liquid crystal (4) None of these

39. In a crystal pair of ions are missing from normal sites. This is an example of

(1) F-centres (2) Interstitial defect


40. Frenkel defect generally appears in

(1) AgBr (2) ZnS

(3) AgI (4) All of these

41. Missing of one cation and one anion from the crystal lattice is called :

(1) Ionic defect (2) Crystal defect

(3) Schottky defect (4) Frenkel defect

42. In a closed packed array of N spheres, the number of tetrahedral holes are

(1) N/2 (2) N

(3) 4N (4) 2N

43. For an ionic crystal of general formula AX and co-ordination number 6, the value of radius ratio willbe

(1) Greater than 0.73 (2) In between 0.732 and 0.414

(3) In between 0.41 and 0.22 (4) Less than 0.22


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44. Close packing is maximum in the crystal lattice of:

(1) Face-centred cubic (2) Body-centred cubic

(3) Simple-centred cubic (4) None of these

45. In the crystal of CsCl, the nearest neighbours of each Cs ion are(1) Six chloride ions (2) Eight Cs ions

(3) Six Cs ions (4) Eight chloride ions

46. Schottky defect in crystals is observed when

(1) An ion leaves its normal site and occupies an interstitial site

(2) Equal number of cations and anions are missing from the lattice

(3) Unequal number of cations and on ions are missing from the lattice

(4) Density of the crystal is increased

47. Pick up the correct statement

(1) The ionic crystal of AgBr does not have Schottky defect

(2) The unit cell having crystal parameters, a=b c, = = 90, =120 is hexagonal

(3) In ionic compounds having Frenkel defect, the ratio r+/ris high

(4) The coordination number of Na+ion in NaCl is 4

48. When NaCl is dopped with MgCl2the nature of defect produced is

(1) Interstitial defect (2) Frenkel defect

(3) Schottky defect (4) None of these

49. The second order Bragg diffraction of X-rays with 1.00 = from a set of parallel planes in a metaloccurs at an angle of 60. The distance between the scattering planes in the crystal is

(1) 0.575 (2) 1.00

(3) 2.00 (4) 1.15

50. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formationwill occur?

(1) p-Type (2) n-Type

(3) Both (1) and (2) (4) None of the two


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32


MCQS ASKED IN COMPETITIVE EXAMINATIONS

1. The fraction of total volume occcupied by the atoms present in a simple cube is: [CBSE PMT 2007]

(1) 4

(2) 6

(3)3 2

(4)

4 2

2. The appearance of colour in solid alkali metal halides is generally due to [CBSE PMT 2006]

(1) Interstitial positions (2) F-centres

(3) Schottky defect (4) Frenkel defect

3. CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomicmass of Cs=133 and that of Br = 80 amu and Avogadro number being 6.02 1023mol1, the density ofCsBr is

(1) 4.25 g/cm3 (2) 42.5 g/cm3

(3) 0.425 g/cm3 (4) 8.25 g/cm3 [CBSE PMT 2006]

4. In face-centred cubic unit cell, edge length is [DPMT 2005]

(1)4

r3

(2)4

r2

(3) 2r (4)3

r2

5. In a face-centred cubic lattice,unit cell is shared equally by how many unit cells?[CBSE P MT 2005]

(1) 4 (2) 2

(3) 6 (4) 86. If Z is the number of atoms in the unit cell that represents the closest packing sequence.... ABC

ABC....., the number of tetrahedral voids in the unit cell is equal to [A.I.I.M.S. 2005]

(1) Z (2) 2Z

(3) Z/2 (4) Z/4

7. What type of crystal defect is indicated in the diagram below? [A.I.I.M.S. 2003]

Na Cl Na Cl Na Cl+ + +

Cl Cl Na NaNa Cl Cl Na ClCl Na Cl Na Na

+ +

+ +

+ + +


(3) Interstitial defect (4) Frenkel and Schottky defects

8. The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and c = 0.504 nm and= = 90 and = 120 is [A.I.I.M.S 2004](1) cubic (2) hexagonal

(3) orthorhombic (4) rhombohedral

9. The liquefied metal expanding on solidification is [A.I.I.M.S 2004]

(1) Ga (2) Al

(3) Zn (4) Cu


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10. The pyknometric density of sodium chloride crystal is 2.165 103kg m3while its X-ray density is2.178 103kg m3. The fraction of the unoccupied sites in sodium chloride crystal is

[C.B.S.E. P.M.T 2002]

(1) 5.96 (2) 5.96 102

(3) 5.96 101 (4) 5.96 103

11. Which of the following metal oxides is anti-ferromagnetic in nature? [M.P.P.E.T. 2002]

(1) MnO2 (2) TiO2(3) VO2 (4) CrO2

12. To get n-type doped semiconductor, impurity to be added to silicon should have the following number ofvalence electrons [K.C.E.T. 2001]

(1) 1 (2) 2

(3) 3 (4) 5

13. Addition of arsenic to germanium makes the latter a [Tamilnadu, CET 2002]

(1) Metallic conductor (2) Intrinsic semiconductor

(3) Mixed conductor (4) Extrinsic semiconductor 14. Semiconductors are derived from compounds of [Kerala MEE 2002]

(1) p-block elements (2) Lanthanides

(3) Actinides (4) Transition elements

15. A semiconductor of Ge can be made p-type by adding [Manipal PMT 2002]

(1) Trivalent impurity (2) Tetravalent impurity

(3) Pentavalent impurity (4) Divalent impurity

16. Due to Frenkel defect the density of ionic solids [M.P.P.E.T. 2002]

(1) Decreases (2) Increases

(3) Does not change (4) Changes

17. Which of the following crystals does not exhibit Frenkel defect? [M.P.P.E.T. 2000](1) AgBr (2) AgCl

(3) KBr (4) ZnS

18. The packing fraction for a body centred cubic is [M.P.P.M.T. 2000]

(1) 0.42 (2) 0.53

(3) 0.68 (4) 0.82

19. The number of octahedral sites per sphere infcc structure is [M.P.P.M.T. 2000]

(1) 1 (2) 2

(3) 4 (4) 8

20. The interionic distance for cesium chloride crystal will be [M.P.P.E.T. 2002]

(1) a (2) a/2(3) 2/a3 (4) 3/a2


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34


SECTION - C

SUBJECTIVE QUESTIONS

1. What is the contribution of the atom when it is placed at the centre of the edge in a unit cell of a cube ?

2. What is the atomic radius of the atom in body-centred cubic structure ?

3. What is the coordination number in hcp and ccp arrangements ?

4. Explain, why ionic crystals do not conduct electricity in solid state ?

5. Name the two important kinds of holes normally encountered in a closed packed structure. How many

such holes are present per sphere in a close packed structure ?

6. Explain the term dislocation in relation to crystal.

7. Which type of crystals exhibit piezo-electricity.

8. What is the coordination no of each sphere in ccp of spheres in three dimensions.9. Mention one property which is caused due to presence of F-centre in solid.

10. What is the effect of pressure on NaCl type crystals.

11, A given solid can belong to one of the four crystal types, i.e., ionic, molecular, covalent and metallic.

Indicate the crystal type of the following :

(a) Diamond (b) Sodium chloride (c) Ice (d) Copper

(e) Boron nitride (f) Zinc oxide (g) Paraffin wax.

12. X-rays of wavelength 1.54 strike a crystal and are observed to be deflected at an angle of 22.5.

Assuming that n = 1, calculate the spacing between the planes of atoms that are responsible for this

reflection.

13. At room temperature, sodium crystallises in body-centred cubic lattice with a = 4.24. Calculate the

theoretical density of sodium (at. mass of Na = 23.0)

14. A compound formed by elements A and B crystallises in cubic structure where A atoms are at the

corners of a cube and B atoms are at the face centre. What is the formula of the compound ?

15. A solid AB has the NaCl structure. If radius of the cation A+is 120 pm, calculate the maximum value of

the radius of the anion B.

16. In a cubic lattice, the closed packed structure of mixed oxides of the lattice is made up of oxide ions; one

eighth of the tetrahedral voids are occupied by divalent ions (A2+) while one half of the octahedral voids

are occupied by trivalent ions (B3+). What is the formula of the oxides?

17. If NaCl is doped with 103mol percent of SrCl2, what is the concentration of cation vacancy?

18. A metal crystallises into two cubic phases, face-centred cubic (f.c.c.) and body-centred cubic (b.c.c.)

whose unit lengths are 3.5 and 3.0 respectively. Calculate the ratio of densities of f.c.c. and b.c.c.

19. Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a,

b, c) (iii) (6a, 3b, 3c), (iv) (2a, 2b, )

20. The unit cell length of NaCl is observed to be 0.5627 nm by X-ray diffraction studies; the measured

density of NaCl is 2.164 g cm3. Correlate the difference of observed and calculated density and calculate

% of missing Na+and Clions.


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35


ANSWERS

SECTION - A

(True and False Statements)

1. False 2. True

3. False 4. True

5. False 6. True

7. True 8. False

9. True 10. False

(Fill in the Blanks)

1. High 2. increases

3. double 4. ccp

5. 1016 6. B.c.c.

7. Two 8. 26

9. 12 10. Semi-conductor

(Assertion-Reason Type Questions)

1. (3) 2. (1)3. (1) 4. (3)

5. (1) 6. (3)

7. (3) 8. (2)

9. (1) 10. (1)


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36


SECTION - B

(Multiple Choice Questions)

1. (4) 2. (1) 3. (4) 4. (4) 5. (3)6. (2) 7. (3) 8. (2) 9. (2) 10. (4)

11. (4) 12. (4) 13. (2) 14. (1) 15. (1)

16. (1) 17. (2) 18. (1) 19. (1) 20. (4)

21. (4) 22. (3) 23. (3) 24. (4) 25. (2)

26. (3) 27. (3) 28. (4) 29. (1) 30. (2)

31. (2) 32. (2) 33. (2) 34. (4) 35. (2)

36. (1) 37. (3) 38. (1) 39. (4) 40. (4)

41. (3) 42. (4) 43. (2) 44. (1) 45. (4)46. (2) 47. (2) 48. (3) 49. (4) 50. (2)

(MCQs asked in Competitive Examinations)

1. (2) 2. (2) 3. (1) 4. (2) 5. (3)

6. (2) 7. (2) 8. (2) 9. (1) 10. (4)

11. (1) 12. (4) 13. (4) 14. (1) 15. (1)

16. (3) 17. (3) 18. (3) 19. (1) 20. (3)

SECTION - C

(Subjective Questions)

Answers are given in the separate booklet.

pmt class xii chemistry solid state

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