PMT_Class_XII_Chemistry_Solid State.pdf

 C H E M I S T R Y S T U D Y M A T E R I A L         THE SOLID STATE   MEDICAL                NARAYANA INSTITUTE OF CORRESPONDENCE COURSES

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CONTENTS

THE SOLID STATE Theory Exercises •

True and False Statements

•

Fill in the Blanks

•

Assertion-Reason Type Questions

•

Multiple Choice Questions

•

MCQ’s Asked in Competitive Examinations

•

Subjective Questions

Answers

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THE SOLID STATE CONTENTS

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

!

Introduction

!

Classification of Solids

!

Types of Symmetry

!

Types of Unit Cells

!

Crystal Defects

!

Radius Ratio

!

Structure of Simple Ionic Compounds

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

!

Closest Packing

binding forces, understand structure of solids

!

Properties of Solids

describe the imperfections in solids and their

!

Metal Silicates

!

Exercise

!

Answers

effects on properties and their application in industries.

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INTRODUCTION

G

ases 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.

A substance is said to be solid if its melting point is above the room temperature. Some of the important characteristics 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 from which 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 and They do not have any pattern of arrangement orderly arrangement of atoms, ions or molecules in three of atoms, ions or molecules and thus, do not dimensional space have any definite geometrical shape.

2.

They have sharp melting points and change abruptly into liquids

3.

Crystalline solids are anisotropic. Some of their physical Amorphous solids are isotropic. Their physical properties are same in all directions. properties are different in different directions

4. These are considered as true solids.

Amorphous solids do not have sharp melting points and do not change abruptly into liquids.

The se a re c o nsid e re d p se ud o so lid s o r supercooled liquids.

Crystalline solids are rigid and their shape is not distorted by mild distoring forces.

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

Crystals are bound by plane faces. The angle between any two faces is called interfacial angle. For a given crystalline solid, it is 6. definite angle and remains always constant no matter how the faces develop. When a crystalline solid is hammered, it breaks up into smaller crystals of the same geometrical shape.

Amorphous solids do not have well defined planes. When an amorphous solid is broken, the surfaces of the broken pieces are generally not flat and intersect at random angles.

5.

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.

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TYPES OF CRYSTALLINE SOLIDS Crystalline Solid Ionic crystal

Covalent crystal

Metallic crystal

Molecular crystal

Characteristics of the Different types of crystalline Solids Characteristics

Ionic Crystal Covalent Crystal C a tio ns (p o sitive io ns ) a nd a nio ns Atoms (negative ions)

Metallic Crytsal

Molecular crystal

Atoms

Small molecules

1

Constituent particles

2

Binding forces

Strong electrostatic forces

Covalent forces

3

Physical properties

Brittle, high m.p. and b.p., good conductors of electricity in molten state and aqueous solution, high heats of fusion.

Ge ne ra lly ha rd , high m.p. and b.p., poor conductors of heat, high heats of fusion.

4

Examples

Diamond, silicon NaCl, KF, CuSO 4 , carbide (SiC) Cu, Ag, Fe, Na, K etc. NaNO 3 etc. quartz (SiO 2) etc.

Electrostatic attraction between positive ions and mobile electrons Hard and high m.p., very good conductors of heat and electricity, exhibit properties such as lustre. malleability ductility etc., moderate heats of fusion.

Weak van der Waal's forces Generally soft, low m.p. volatile, poor conductors of electricity ( ins ula t o r s ) , p o o r conductors of heat, low heats of fusion. Solid C O 2 , Iod ine, Naphthalene, Argon ice etc.

TYPES OF SYMMETRY IN CRYSTALS (i)

Centre of Symmetry : It is such an imaginary point within the crystal that any line drawn through it 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 of symmetry 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 of

360° , 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). 3 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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(iv) Elements of symmetry: The total number of planes, axes and centre of symmetry possessed by a crystal are termed as elements of symmetry. A cubic crystal possesse a total of 23 elements of symmetry. 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 space lattice is also called crystal lattice. The locations of the points in the space lattice are called lattice points or lattice sites.

UNIT CELL The smallest but complete unit in the space lattice which when repeated over and over again in the three dimensions, 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 the corners. 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 the corners 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 the body. (iii) End centred : Particles (or points) are located at the corners and also at the centres of the end faces.

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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 angles between the edges (α, β, γ ) , the various crystals can be divided into seven systems. These are also called 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

α = β = γ = 90°

NaCl, KCl, Diamond, Cu, Ag

Tetragonal

a = b ≠c

α = β = γ = 90°

White tin, SnO2

Orthorhombic

a≠b≠c

α = β = γ = 90°

Rhombic sulphur, KNO3, K2SO4, BaSO4, PbCO3

Monoclinic

α = γ = 90°, β ≠ 90° Monoclinic sulphur, Na2SO4.10H2O

Hexagonal

a ≠ b≠c a = b≠c

α = β = 90°, γ = 120° Graphite, ZnO, CdS

Rhombohedral

a=b=c

α = β = γ ≠ 90°

Calcite (CaCO3), quartz, NaNO3

Triclinic

a ≠ b≠c

α ≠ β ≠ γ = 90°

CuSO4.5H2O, K2Cr2O7

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 cannot exist in each case. Bravais in 1848 has established fourteen different types of lattices called Bravais lattices 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 face centres and points that lie entirely within the unit cell. In a crystal, atoms located at the corner and facecentre of a unit cell are shared by other cells and only a portion of such an atom actually lies within a given unit cell. 5 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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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 oneeighth 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

1 1 1 = × occupied corners + × occupied edge centres + × occupied face centres + occupied 8 4 2 body centre. Determination of Number of Constitutent units per unit cell : Let edge length of cube = a cm Density of substances = d g cm–3 Volume of unit cell = a3 cm3 Mass of unit cells = volume × density = (a3 × d) g a3 × d Number of mole per unit cell = ; M where M = molar mass Number of molecules per unit cell = Number of mole × Avogadro’s Number Z=

a3 × d × N M

EXAMPLES - 1 The density of KCl is 1.99 g cm–3 and the length of a side of unit cell is 6.29 Å as determined by X-ray diffraction. Calculate the value of Avogadro’s number. SOLUTION : KCl has face-centred cubic structure, i.e., Z = 4 Avogadro’s number = Given, d = 1.99;

Z× M d×V

M = 74.5;

Avogadro’s number =

V = (6.29 × 10–8)3 cm3

4 × 74.5 = 6.01 × 1023 −8 3 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 cm–3. How many atoms does 208 g of the element contain? 6 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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SOLUTION : Volume of the unit cell = (288 × 10–10)3 = 23.9 × 10–24 cm3 Volume of 208 g of the element = Number of unit cell in 28.88 cm3 =

208 = 28.88 cm3 7.2 28.88 = 12.08 × 1023 unit cells 23.9 × 10 − 24

So, total no. of atoms in 12.08 × 1023 unit cells = 2 × 12.08 × 1023 [since each bcc unit cell contain 2 atoms] = 24.16 × 1023 EXAMPLES : 3 The density of potassium bromide crystal is 2.75 g cm–3 and 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 ρ=

NM   a3  NA 

ρ × a3 × NA 2.75 × (654 × 10−10 )3 × 6.023 × 10 23 = = 3.89 ! 4 M 119 Number of mass points per unit cell = 4 Thus α,β, γ has NaCl type crystal is fcc structure. N=

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 centre of 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 : 5 If three elements P, Q and R crystallise in a cubic unit cell with P atoms at the corners. Q atoms at the cubic 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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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 the sphere) 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 composed of the same type of atoms. (a) Distance between the centres of the spheres present on the corners of the edges of the cubic is called 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 =

4

3

πr 3

Since the spheres at the corners touch each other; Edge length (a) = r + r = 2r = d

1 ×8 =1 8 Total no. of atoms in a unit cell = 1 Radius of an atom = r Distance between nearest neighbour (d) = a No of atom on the corners =

a

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

4 / 3 πr 3 π Volume of sphere = = 0.524 Packing fraction = = 6 Volume of cube 8r 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

=

No. of atoms on the faces Total no. of atoms in a unit cell Volume occupied by atoms From ∆ABC,

AC

1 ×8 =1 8 1 = ×6 = 3 2 =1+3=4

C

2

4 3 = 4 × πr 3 2 = AB + BC2

⇒

AC2

= a2 + a2 = 2a2

∴

AC

=

A

B

a

2 a.

Distance between nearest neighbour (d) =

AC a 2 a = = 2 2 2

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Radius of the atom (r) = ∴

a d = 2 2 2

4  a   Volume occupied by atoms in a unit cell = 4 × π  3  2 2 

3

3

4  a  4 × π  3 2 2 = 0.74 Thus packing fraction = a3 % volume occupied by unit cell = 74 % volume unoccupied by unit cell = 100 – 74 = 28

FOR BODY CENTRED CUBIC LATTICE C

1 ×8 =1 8

No. of atoms on the corners =

No. of atoms in the body = 1 Total no. of atoms in a unit cell = 1 + 1 = 2 From ∆A DC , AD2 = AC2 + CD2 =

A

( 2a ) + a 2

∴

AD = a 3

∴

Distance between nearest neighbour (d) =

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

∴

= 3a 2

Vol. occupied by atoms Vol. of cube

3

=

4a 3  2×   3 2 

= 68

% volume unoccupied in a unit cell

= 32

Lattice type

Radius (r)

Simple cube

r = a/2

r=

Body centred

r=

a

AD a 3 = 2 2

% volume occupied in a unit cell

Face centred

D

a 3 4

4 a 3  Volume occupied by atoms = 2 × π  3  2 

Packing fraction =

2

B

a 2 2 a 3 4

3

a3

= 0.68

Distance between Nearest neighbour Packing fraction d = 2r = a

0.524

d =a

0.74

d=

2

a 3 2

0.68 9

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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 be the length of the edge of the unit cell? SOLUTION : For face centred cubic crystal Edge length (a) = 2 2 r = 2(1.4142) × (128 pm) = 362 pm EXAMPLES : 7 Xenon crystallises in a face centred cubic lattice and the edge of the unit cell is 620 pm. Calculate the nearest neighbouring distance and the radius of the xenon atom? SOLUTION : (620 pm) = 438.4 pm 1.4142 2 a (620 pm) = = 219.2 pm Radius of xenon atom (r) = 2 2 2 × 1.4142

Nearest neighbour distance (d) =

a

=

EXAMPLES : 8 The face diagonal of a cubic closed packed unit cell is 4Å. What is its face length? SOLUTION : Face diagonal =

a2 + a2 = a 2 ; a =

Face diagonal 2

=

4Å 2

=

( 4Å ) = 2.83Å 1.4142

EXAMPLES : 9 An element A crystallises in face-centred cubic lattice with edge length of 100 pm. What is the radius of its 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) a (100 pm) = = 35.4 pm Radius (r) = 2 2 2 × 1.4142 Nearest neighbour distance (d) = 2r = 2 × 35.4 pm = 70.8 pm Length of face diagonal = 4r = 4 × 35.4 pm = 141.6 pm

CRYSTAL DEFECTS OR IMPERFECTIONS IN SOLID Crystal defects Electronic imperfection

Atomic imperfection or point defect

Stoichiometric defect Schottky defect

Frenkel defect

Non-stoichiometric defect Metal excess

Anion vacancy defect

Impurity defect

Metal deficiency

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 from their 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

 −ws  ns = N.exp.   2kT  where

A+

B

B

A+

A+

B

B

A A++

B

3

N = No. of sites per cm that could be left vacant A+

A+

B

A+

A

B

A+

B

A++

B+

+

B–

A+

B–

A+

B–

A++

B

A+

B+– A

A++ A A

+

WS = work necessary to form a schottky defect. K = gas constant T = absolute temp. At room temp. NaCl has one defect in 1015 lattice site. At 500°C, the value rising to one in 106 sites.

FRANKEL DEFECT It consists of a vacant lattice site (a hole) and the ion which ideally should have occupied the site now occupies an interstitial position.

A

A+

In this type of defect metal ions are generaly smaller than the anion i.e. large difference in size and coordination number is low. e.g. AgCl, AgBr, AgI.

B

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

A+

where

–

B–

+ –  −w f  B B B– A A++ N f = NN ' exp   2 kT   N is the number of sites per cm3 that could be left vacant N’ is the no. of alternative interstitial per cm3. wf is 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 Lattice site, leaving a hole which is occupied by an electron thereby maintaining the electrical balance. When compound such as NaCl, KCl LiH or δ – TiO are heated with excess of their constituent metal vapours or treated with high energy radiation, they become deficient in negative ion and their formula may be AX1− δ , where δ = small fraction.

A+

B

A+

B

A+

B

A A+

A+

e

–

A+

B+ A

B

A+

B

+ A+

B +

Anion sites occupied by electrons in this way are called F– centres. More the no. of F-centres, more is the colour of compound.

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Interstitial ions and electrons : Metal excess defect also occur when an extra +ve ion occupies on interstitial position in the lattice, the electrical neutrality is maintained by the inclusion of an interstitial electron. The composition may be represented by general formula A1+ δ X . The example include ZnO, CdO, Fe2O3 and Cr2O3.

METAL DEFICIENCY DEFECT This type of compound is represented by the formula A1− δ X . 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, δ -TiO, FeS and CuI. A+

B

B

A+

B

B

A++

B

A+

A+

B

A+

B

A+

B B

+

A+

B

A2+

B+ A

A+

B

A2+

A B+

B

A+

B

A++

B

A+

B

+ A+

Positive ion absent

Extra interstitial negative ion

(ii) Extra interstitial negative ion : It is possible to have an extra negative ion in an interstitial position 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 of cation (r + ) Radius ratio = Radius of anion (r − )

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

• •

Radius ratio (r+ / r–)

Possible Co-ordination number of cation

Structural Arrangement

∠0.155 0.155 – 0.225 0.225 – 0.414 0.414 – 0.732 0.732 – 1

2

Linear

3 4 6 8

Trigonal planar Tetrahedral Octahedral Cubic

Examples

B2O3 ZnS, CuCl, CuBr, HgS NaCl, KBr, MgO, CaO, CaS, NaBr 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 B2– form a closed packed lattice and cations A+ occupy all the tetrahedral voids. What is the simplest formula of the compound ?

SOLUTION : We know that in a close packed lattice involving anions B2– having tetrahedral voids, there are two tetrahedral 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 value of the radius of the anion B.

SOLUTION : Since NaCl has octahedral structure.  rA+ The limiting ratio  −  rB

or

rB− =

  = 0.414 

rA+ 120 = = 290 pm 0.414 0.414

STRUCTURE OF SIMPLE IONIC COMPOUNDS • Anions (large in size) form a close packed arrangement but cations (smaller in size) occupy the interstitial sites 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 F– ions 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 of attraction. The different ionic compounds can have the following type of structures. (i)

Ionic compounds of AB type

(ii) Ionic compounds of AB2 type (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 B– ions are present in the ratio of 1 : 1. These compounds have following three types of structures. Rock salt (NaCl) type structures. Cesium chloride (CsCl) type structure. Zinc blende (ZnS) type structure. 13 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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ROCK SALT (NaCl) TYPE STRUCTURE

–

Cl ion surrounded + octahedrally by six Na ions +

Na ion surrounded – octahedrally by six Cl ions Fig. Sodium Chloride Structure

Characteristics:

• •

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

rNa + rCl−

=

95 = 0.525 181

•

The structure is octahedral in which Cl– ion constitute a cubic closed packed and Na+ ions occupy the octahedral voids.

• • •

The coordination number is 6 : 6 ie. each Na+ ion is surrounded by six Cl– ions 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) CaCO3 and CaC2

CESIUM CHLORIDE (CsCl) TYPE STRUCTURES Characteristics : (i) The crystal is composed of Cs+ and Cl– ions and their ionic radii are 160 pm and 181 pm respectively. The radius ratio for the crystal is :

r Cs + 160 pm = = 0.889 r Cl − 181pm The radius ratio suggests a body centred cubic structure. However, it is more than the ideal ratio which is 0.732. (ii) The Cl– ions are present at the eight corners of the unit cell while one Cs+ ion is present in the centre of the body as shown in the fig.(a) 14 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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(iii) Each Cs+ ion is surrounded by 8 Cl– ions and each Cl– ion by 8 Cs+ ions. Thus, the coordination number of each ion is 8 or these are in 8 : 8 co-ordination. (iv) The contribution by all the 8 Cl– ions at the corners is 1 while by the Cs+ ion present in the centre of body is also 1. Therefore, the unit cell of CsCl consist of one Cs+ ion and one Cl– ion 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 S2– ions are 74 pm and 184 pm respectively. Therefore, the radius ratio is :

rZn 2+ rS2−

=

74 pm = 0.402 184 pm

(iii) Each Zn2+ ion present occupies a tetrahedral site. As there are two tetrahedral sites per atom in a close packed arrangement one half of these sites are occupied by Zn2+ ions whereas the remaining half are vacant. Thus Zn2+ ions are present at the alternative sites as shown in fig. (iv) Each Zn2+ ion is surrounding by four S2– ions which are directed towards the four corners of a tetrahedron. Similarly each S2– ion is also surrounding by four Zn2+ ions. Thus, coordination number of 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 sulphide etc.

STRUCTURES OF IONIC COMPOUNDS OF AB2 TYPE 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. 15 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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= Ca – =F

2+

Fig. Structure of Calcium Fluoride (Fluorite Structure)

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

rCa 2+ rF−

=

(99 pm) = 0.73 (136 pm)

(ii) The radius ratio suggests a body centred cubic structure in which the Ca2+ ions are present at the corners 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 two tetrahedral sites per atom in a close packed arrangement, one half of these sites are occupied by Zn2+ 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 a tetrahedron. Similarly each S2– ion is also surrounded by four Zn2+ ions. Thus, coordination number of 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, PbF2 etc.

ANTIFLUORITE STRUCTURE IN IONIC COMPOUNDS OF A2B TYPE = Na+ = O2−

2–

O ion (in fcc arrangement)

+

Na ion (at tetrahedral hole)

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 reversed i.e. the smaller cations occupy the position of fluoride ions while the anions with bigger size occupy the 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. 16 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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(e) Thus there are two tetrahedral sites per atom in close packed arrangement and these are occupied by Na+ ions. (f) Each Na+ ion is surrounded by four O2– ions and each O2– ions is surrounded by eight Na+ ions. The C.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 CRYSTAL STRUCTURE • 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.

BRAGG’S LAW When X-rays are incident on a crystal face, they penetrate into the crystal and strike the atoms in different planes. From each of these planes, X-rays are deflected. Bragg presented a relationship between the wavelength of the X-rays and the distance between the planes, such as nλ = 2d sin θ where n is an integer such as, 1, 2, 3, ..., λ is the wavelength, d is the distance between repeating planes of particles and θ the angle of deflection or glancing angle. Two parallel layers of the crystals are shown with the help of horizontal lines separated from each other by distance d. Two waves comprising X-rays that are in plane strike against the crystal and after reflection emerge from the crystal.

E

λ

θ A

θ

B θ θ D

d

C 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 difference in the distance travelled by the two rays after reflection also called path difference is : Path difference = BC + CD = AC sin θ + AC sin θ = 2d sin θ This must be an integral multiple of wave length i.e. nλ . Therefore.

nλ = 2d sin θ

(Bragg Equation)

EXAMPLES : 12 X-rays of wavelength 1.54 Å strike a crystal and are observed to be deflected at an angle of 22.5°A. Assuming that n = 1. Calculate the spacing between the planes of atoms that are responsible for this reflection. 17 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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SOLUTION : According to Bragg’s law, nλ = 2d sin θ Given n = 1,

λ = 1.54 Å, θ = 22.5° A 1.54 1.54 d= = = 2.01 Å 2 sin 22.5° 2 × 0.383

EXAMPLES : 13 If NaCl is doped with 10–3 mol percent of SrCl2 what is the concentration of cation vacancy? SOLUTION : Na + Cl– Cl– Sr2+ Cl– Cl– Na +

Na + Cl– Na + Cl–

Cl– Na + Cl– Na +

Number of cationic vacancies per mol =

10 −3 × 6.023 × 10 23 = 6.023 × 1018 vacancies per mol. 100

CLOSEST PACKING In crystals the atoms, ions or molecules are arranged in a regular way in a three-dimensional space. The arrangement has minimum energy and hence maximum stability. For maximum stability, a constituent in the aggregate must be surrounded by the maximum number of neighbours. For maximum number of contacts, each constituent of the crystal must be packed as closely as possible. In a twodimensional plane the closest arrangement being that in which each sphere is in contact with six other spheres. This layer is denoted by A. To obtain the closest packing in space, the spheres are placed 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 around any sphere in layer A. In fig around a sphere X these sites are 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 the spheres in the vacant sites labelled as 1, 3 and 5 then the sites marked 2, 4 and 6 are left unoccupied. This layer is denoted by B. The second layer again has two types of vacant sites. Around any atom in the second layer, one set of vacant sites lies just above the vacant sites 2, 4 and 6 of the first layer and the other set lies above

B A C B A

A B A (a) (b) Y B

x A 2 3

1 ×

4

6 5

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

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the centres of the spheres of the first layer. Thus after the second layer is complete, there are two different ways of placing the spheres in the third layer. If the spheres in the third layer are placed in the vacant sites which are above the centres of the spheres of the first layer, then the third layer repeats the arrangement 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 sites above 2, 4 and 6, a new arrangement (C) of the spheres in produced. Fig (b) The entire arrangement is now of the type ABCABC ........... etc., and is referred to as a cubical closest packing, ccp or face centred 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 tetrahedral hole 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 different structures. × ×

× × ×

×

×

× ×

×

× ×

(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 of octahedral holes in the unit cell of the fcc structure is four as can be deduced from the following considerations. In fig(b) each (X) mark represent a vacant site and there are twelve such vacant sites at the edges of unit cubic lattice. The vacancy at an edge is common to four unit cells and hence the number 12 = 3 . In addition to these vacancies there is one octahedral hole at of such vacant sites per unit cell is 4 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 figure above 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 that in 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 : 19 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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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 flow of electricity is due to flow of electrons (electronic conductors) without any chemical change occurring in the metal. While in electrolytic conductors like NaCl, KCl etc, the flow of electricity takes place to a good extent only when they are taken in the molten state or in aqueous solution. The flow of electricity is due to flow of ions.

(ii) Insulators: The solids which almost do not allow the electricity to pass through them are called insulators e.g. S, P, plastics, wood, rubber etc. (iii) Semi-conductors : The solids whose conductivity lies between those of metallic conductors and insulators 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 of temperature 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 called intrinsic semiconductors. (b) Electrical conductivity of semiconductors increases with increase of temperature because with increase of temperature, large number of electrons can jump from valence band to conduction band. (c) Electrical conductivity of silicon and germanium is very low at room temperature. It can be increased 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 field are called diamagnetic substances e.g. TiO2, NaCl, benzene etc. The property thus exhibited is called diamagnetism. This property is shown only by those substances which contain fully filled orbitals i.e. no unpaired electron is present.

(ii) Paramagnetic substances : Substances which are attracted by the external magnetic field are called paramagnetic substances. The property thus exhibited is called paramagnetism. This property is 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 magnetic field. (iii) Ferromagnetic substances : Substances which show permanent magnetism even in the absence of the magnetic field are called ferromagnetic substances, e.g. Fe, Ni, Co, CrO2 show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetism arises due to spontaneous alignment of magnetic moments due to unpaired electrons in the same direction, as shown in fig.(a) 20 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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(a) Ferromagnetism (b) Anti-ferromagnetism (c) Ferrimagnetism (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 magnetic moment 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 the basis of the unpaired electrons but acutally have small net magnetic moment are called ferrimagnetic substances e.g. Fe3O4, ferrites of the formula M2+ Fe2O4 where M = Mg, Cu, Zn etc. Ferrimagnetism arises due to the unequal number of magnetic moments in opposite direction resulting in some net magnetic 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 called ‘piezoelectricity’ and the crystals are called piezoelectric crystals. Mechanical stress

Dipoles aligned in an ordered manner

+ + + + + + +

+

+

+

Voltmeter

Electric field +

+

+

Mechanical stress Fig. 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 in record players where they produce electrical signals by application of pressure. They are also used in microphones, ultrasonic generators and sonar detectors. (ii) Pyroelectricity : Some piezoelectric crystals when heated produce a small electric current. The electricity thus produced is called pyroelectricity (pyre means heat). (iii) Ferroelectricity : In some of the piezoelectric crystals, the dipoles are permanently polarized even in the absence of the electric field. However on applying electric field, the direction of polarization changes. This phenomenon is called ferroelectricity due to analogy with ferromagnetism. Some examples of the ferroelectric solids are barium titanate (BaTiO3), sodium potassium tartarate (Rochelle salt) and potassium dihydrogen phosphate (KH2PO4). It may be pointed out here that all ferroelectric solids are piezoelectric but the reverse is not true. 21 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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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 crystals are said to be anti-ferroelectric. A typical example of such crystals is lead zirconate (PbZrO3).

METAL SILICATES Metal 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 by sharing oxygen atoms in small groups, in small cyclic groups, in infinite chains or infinite sheets giving various structures. 1.

Simple orthosilicates : In these silicates SiO44– tetrahedra do not share oxygen atoms with one another and exist as discrete SiO44– orthosilicate anions e.g. Mg2SiO4, Fe2SiO4 and Zn2SiO4. In all these compounds SiO44– ion is linked to M2+ ions through co-ordinate bond. [See Fig. (a)]

2.

Pyrosilicates or islands : In these two SiO44– tetrahedral 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 each tetrahedron are shared with others. (Si3O9)6– and (Si6O18)12–. General formula of any such anion must be (SinO3n)2n–. [See Fig. (c)&(d)]

(a) SiO4

4–

(b) (Si2O 7)

5–

(c)

(Si3O9)

6–

(d) (Si6O18)

12–

2–

(e) (SiO3 ) n chain 2–

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4.

Infinite chain silicates : These are of two types the pyroxenes which contain single strand chains of composition (SiO32–)n and amphiboles which contain double strand cross-linked chains of composition (Si4O116–)n. General formula of the anion in a pyroxene is the same as in a silicate with cyclic anion e.g. Diopside (MgCaSi2O6). Various asbestos minerals are examples of chain silicates (amphiboles).

5.

Infinite sheet Silicates : When three oxygen atoms of each SiO4 tetrahedron are shared, the two dimensional sheet structures with general formula of (Si2 O 5 2– ) n are obtained e.g. clay. [See Fig. (e)]

6.

Framework silicate : Three dimensional network is obtained when all the four oxygen atoms of each of SiO4 tetrahedron are shared. e.g. Quartz and zeolites. The general formula of framework silicate is SiO2. [See Fig. (f)]

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EXERCISE

Section A :

Section B :

Section C :

(i)

True and False Statements

(ii)

Fill in the Blanks

(iii)

Assertion-Reason Type Questions

(i)

Multiple Choice Questions

(ii)

MCQ’s asked in Competitive Examinations

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 1022 ions and 106 schottky pairs per cm3. 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 In each of the following questions two statements are given as Assertion A and Reason R. Examine the statements carefully and answer the questions according to the instructions given below : " Mark (1) if both A and R are correct and R is the correct reason of A. " Mark (2) if both A and R are correct and R is not the correct reason of A. " Mark (3) if A is correct and R is wrong. " Mark (4) if both A and R are wrong. 1. Assertion 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. Assertion A : In a ccp lattice made up of Y– ions, 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 the lattice. (1) (2) (3) (4) 3. Assertion 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. Assertion 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. Assertion A : The two ions A+ and B– have radii 88 and 200 pm respectively. Coordination number of A+ will be 6. Reason R : When r+/r– = 0.414 – 0.732, the coordination number is 6. (2) (1) (3) (4) 6. Assertion A : Covalent crystals have the highest melting point. Reason R : Covalent bonds are stronger than ionic bonds. (2) (1) (3) (4) 7. Assertion A : CsCl as body-centred cubic arrangement. Reason R : CsCl has one Cs+ ion and 8 Cl– ions in its unit cell. (2) (1) (3) (4) 8. Assertion A : Triclinic system is the most unsymmetrical system. Reason R : No axial angle is equal to 90° in triclinic system. (2) (1) (3) (4) 9. Assertion A : Antiferromagnetic substances on heating to high temperature become paramagnetic. Reason R : On heating, randomisation of spins occur. (2) (1) (3) (4) 10. Assertion A : In any ionic solid, [MX], with Schottky defects, the number of positive and negative ions are same. Reason R : Equal number of cation and anion vacancies are present. (2) (1) (3) (4) 26 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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SECTION - B MULTIPLE CHOICE QUESTIONS 1.

2.

3.

A mineral having the formula AB2 crystallizes in the c.c.p. lattice, with A atoms occupying the lattice points. The CN of A is 8 and that of B is 4. What percentage of the tetrahedral sites is occupied by B atoms? (1) 25%

(2) 50%

(3) 75%

(4) 100%

CsBr has b.c.c. structure with edge length 4.3. The shortest inter ionic distance in between Cs+ and Br– is (1) 3.72

(2) 1.86

(3) 7.44

(4) 4.3

The formula for determination of density of cubic unit cell is (1)

a3N0 g cm −3 z×M

a3 × M g cm −3 (3) z × N0

4.

5.

6.

(2)

z × N0 g cm −3 3 M×a

(4)

z×M g cm −3 N0 × a3

The closest-packing sequence ABAB ..... represents (1) Primitive cubic packing

(2) Body-centred cubic packing

(3) Face-centred cubic packing

(4) hexagonal packing

In Bragg’s equation for diffraction of X-rays ‘n’ represents (1) The number of mole

(2) Quantum number

(3) The order of reflection

(4) Avogadro’s number

Tetragonal crystal system has the following unit cell dimensions (1) a = b = c and α = β = γ = 90° (2) a = b ≠ c and α = β = γ = 90° (3) a ≠ b ≠ c and α = β = γ = 90° (4) a = b ≠ c and α = β = 90°, γ = 120°

7.

8.

9.

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

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

Germanium or silicon becomes semi-conductor due to (1) Schottky defect (2) Chemical impurity (3) Frenkel defect (4) None of these 27

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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 these 13. 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 temperature and still at higher temperature turbidity completely disappears. The behaviour is a characteristic of substance 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 a is (1)

a 2 3a 4

(3)

2a 4

(2) (4) a

28. The fraction of the total volume occupied by atoms in a simple cube is (1)

(3)

π 2

3π 8

(2) 2π 6

(4)

π 6

29. A substance AxBy crystallizes in an f.c.c. lattice in which atoms of ‘A’ occupy each corner of the cube and atoms of ‘B’ occupy the centres of each face of the cube. Identify the correct composition of the substance 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 29

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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 (3) Frenkel defect (4) Schottky 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 will be (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 30 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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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, α = β = 90°, γ =120° is hexagonal (3) In ionic compounds having Frenkel defect, the ratio r+ /r– is high (4) The coordination number of Na+ ion in NaCl is 4 48. When NaCl is dopped with MgCl2 the 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 λ = 1.00 Å from a set of parallel planes in a metal occurs at an angle of 60°. The distance between the scattering planes in the crystal is (1) 0.575 Å

(2) 1.00 Å

(3) 2.00 Å

(4) 1.15 Å

50. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formation will occur? (1) p-Type

(2) n-Type

(3) Both (1) and (2)

(4) None of the two

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MCQ’S ASKED IN COMPETITIVE EXAMINATIONS 1.

The fraction of total volume occcupied by the atoms present in a simple cube is: [CBSE PMT 2007] (1) (3)

2.

3.

4.

π 4

(2) π

3 2

6.

7.

π 4 2

The appearance of colour in solid alkali metal halides is generally due to (1) Interstitial positions

(2) F-centres

(3) Schottky defect

(4) Frenkel defect

[CBSE PMT 2006]

CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs=133 and that of Br = 80 amu and Avogadro number being 6.02 × 1023 mol–1, the density of CsBr is (1) 4.25 g/cm3

(2) 42.5 g/cm3

(3) 0.425 g/cm3

(4) 8.25 g/cm3

In face-centred cubic unit cell, edge length is 4 r (1) 3

[CBSE PMT 2006] [DPMT 2005]

(2)

4 r 2

3 r 2 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) 8 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 What type of crystal defect is indicated in the diagram below? [A.I.I.M.S. 2003]

(3) 2r

5.

(4)

π 6

(4)

Na + Cl − Na + Cl − Na + Cl −

8.

9.

Cl− Cl− Na + Na + Na + Cl− Cl− Na + Cl− Cl− Na + Cl− Na + Na + (1) Frenkel defect (2) Schottky defect (3) Interstitial defect (4) Frenkel and Schottky defects The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and c = 0.504 nm and α = β = 90° and γ = 120° is [A.I.I.M.S 2004] (1) cubic (2) hexagonal (3) orthorhombic (4) rhombohedral 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 × 103 kg m–3 while its X-ray density is 2.178 × 103 kg m–3. The fraction of the unoccupied sites in sodium chloride crystal is [C.B.S.E. P.M.T 2002] –2 (1) 5.96 (2) 5.96 × 10 –1 (3) 5.96 × 10 (4) 5.96 × 10–3 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 of valence 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 in fcc 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)

3 a/2

(4) 2a / 3

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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

(e)

Boron nitride

(f)

Zinc oxide

(g)

Paraffin wax.

(d)

Copper

12. X-rays of wavelength 1.54Å 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 = 4.24Å. 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 10–3 mol 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 3.0Å 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 cm–3. Correlate the difference of observed and calculated density and calculate % of missing Na+ and Cl– ions. 34 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

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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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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)

(MCQ’s 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.

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