SOLID STATEBy MUKESH SHARMA ,DPS JODHPUR

Introduction :

Solids are characterised by incompressibility, rigidity and mechanical strength. This indicates that the molecules,atoms or ions that make up a solid are closely packed, they are held together by strong cohesive forces andcannot move at random. Thus, in solids we have well ordered molecular arrangements.

Moreover solids are also characterised by a definite geometrical arrangement. Substances which satisfy all thecharacteristics of a solid except the definite geometrical arrangement are called amorphous substances (e.g.glass, rubber etc.).The definite geometrical arrangement of atoms, molecules or ions in a solid extends over theentire structure of the solid. This is termed as long range order.Classification of Solids

is classified according to the regularity with which atoms or ions are arranged

Crystalline Solids atoms are organized in a well ordered and regular geometrical pattern

Amorphous Solids atoms are organized in a disordered and irregular geometrical pattern

.. Although amorphous solids possess rigidity, incompressibility, refractive index etc., but do not have

Difference B/W Crystalline and Amorphous Solids:

Cause of anisotropy in crystals For instance, atoms along the edge of givenstructure are more separated than along the facediagonal. This causes anisotropy

Types of crystalline solids

Space Lattice and Unit Cell: A crystal can be considered to be generatedfrom the repetition of some basic unit of pattern such as atoms, molecules or ions. In other words, a wellordered and regular arrangement of constituents in the three dimensional space is called crystal lattice. A spacelattice can be subdivided into a number of small cells known as unit cells. It can be defined as the smallestblock from which entire crystal can be built up by its translational repetition in three dimensions.

Unit cell: is the smallest unit of a space lattice which repeats itself to form the lattice.

In other words space-lattice is formed by face to face packing of unit cells

Characterizations of an unit cell

1. Axial property:

1 angle b/w two adjacent faces (sides)

Length of sides (edges)

2. Position of lattice point (constituent particle)

Types of Lattices

Seven Crystal Systems: The seven crystal systems are given below.

Crystal System Bravais LatticesParameters of Unit Cell

ExampleIntercepts Interfacial angle

1. Cubic Primitive, Face Centered, BodyCentered = 3

a = b = c = = = 90o Pb,Hg,Ag,AuDiamond, NaCl, ZnS

2. Orthorhombic Primitive, Face Centered, BodyCentered, End Centered = 4

a b c = = = 90o KNO2, K2SO4

3. Tetragonal Primitive, Body Centered =2 a = b c = = = 90o TiO2,SnO2

4. Monoclinic Primitive, End Centered = 2 a b c = = 90o, 90o

CaSO4,2H2O

5. Triclinic Primitive = 1 a b c 900 K2Cr2O7,CaSO45H2O

6. Hexagonal Primitive = 1 a = b c = = 900, = 120o

Mg, SiO2, Zn, Cd

7. Rhombohedral Primitive = 1 a = b = c = = 90o, 90o As, Sb, Bi, CaCO3

Total = 14

Seven Crystal Systems and Fourteen Bravais Lattice

Simple

TETRAGONAL

Body Centred

c

aa

Simple End Centred

MONOCLINIC

c

abb

a

aaSimple Face Centred Body Centred

CUBIC

Simple End Centred

ORTHORHOMBIC

c

ab

Body Centred Face Centred

Types of cubic unit cells [cubic bravaise lattice]

1 Primitive Unit Cells: In a primitive unit cell, the same type of particles are present at all the cornersof the unit cell.

However, it has been observed that the particles may be present not only at the corners butalso at some other special positions within the unit cells. Such unit cells are called ‘nonprimitive unitcells’. There are three types of nonprimitive unit cells as follows:

2 Face Centred: When atoms are present in all 8-corners and six face centres in a cubic unit cell thenthis arrangement is known as FCC.

3 Body Centred: When atoms are present at 8 corners as well as in the body centre in a cubic unit cellthen this arrangement is known as BCC.

Calculation of number of effective atoms [Z] in a Unit Cell.primitive BCC FCC

In a crystal atoms located at the corner and face center of a unit cell are shared by other cells and only a portionof such an atom actually lies within a given unit cell.

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 onefourth of it lies within any one cell.

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

iv) A bodycentred point lies entirely within the unit cell and contributes one complete point to the cell.

Close Packing in Crystals

RHOMBOHEDRAL

aa a

a

a

c

b

ab g

TRICLINIC

a120o

c

a

HEXAGONAL

In order to understand the packing of the constituent particles in a crystal, it is assumed that these particles arehard spheres of identical size. The packing of the crystals is done such that they occupy the maximum availablespace and hence the crystal has maximum density.

There are two common ways in which spheres of each size can be packed as shown below:

Arrnagement (I) Arrnagement (II)

In arrangement (I), the spheres are packed in such away that their centres are at the centres of anequilateral triangle. Each sphere is surrounded by sixother similar spheres. This arrangement is calledhexagonal close packing.

This arrangement can be extended in three dimensions by adjusting spheres on the top of hollows or voids ofthe two dimensional layers.

In the base layer shown in fig. (a), the spheres are marked as A and the two types of voids B/W the spheres aremarked as ‘a’ and ‘b’. The efficient way of placing the spheres in second layer is to place them in the ‘a’ voidsof the first layer, the ‘b’ voids remaining unoccupied.

There are two types of voids in the second layer i.e. ‘c’ and ‘d’. The ‘a’ and ‘b’ voids in the first layer aretriangular while only ‘c’ voids of the second layer are triangular. The ‘d’ voids are combination of twotriangular voids (one each of first and second layer) with the vertex of one triangle upwards and the vertex ofother triangle downwards.

The void surrounded by four spheres and placed at an angle of 109° 28 is known as tetrahedral voids.

Now, there are two ways to build up the third layer:i) When the third layer is placed over the second layer so as to cover the tetrahedral or ‘c’ voids, a

threedimensional closest packing is obtained where the spheres in every third layer are verticallyaligned to the first layer. This arrangement is called ABAB.,… pattern or hexagonal (HCP) closepacking (calling first layer as A and second layer B).

ii) When the third layer is placed over the second layer such that the spheres cover the octahedral or ‘d’voids, a layer C different from A and B is formed. This pattern is called ABCABC……pattern or cubicclose packing (CCP).

A A A A A

AAAAA

A A A A

a

b b b b

b b b

a a

a a a a

Fig. (a)

A A A A A

AAAAA

a

d

c

a

a a a a

d d

a

c

Fig. (b)

In both HCP and CCP methods of packing, a sphere is in contact with six other spheres in its own layer. Ittouches three spheres in the layer above and three spheres in the layer below. Thus a sphere is in direct contactwith 12 other spheres. In other words, the coordination number of that sphere is 12. Coordination number is thenumber of closest neighbours of any constituent particle.

The HCP and CCP arrangements can be also be shown as below.

Hexagonal closest packing of spheres: (a) normal and(b) exploded view

a

b b

(a) (b)

Fig. (X): Cubic closest packing of spheres: (a) generation of unit from closestpacked layers,and (b) rotation to show cubic symmetry

(a) (b)

BY MUKESH SHARMA ,DPS JODHPURFEATURES OF HEXAGONAL CLOSE PACKING

this is an A-B-, A- B –type packing in which every 3rd layer is similar to 1st layer in pattern ofpacking

it is resulted from packing of 3rd layer point in tetrahedral void of 2nd layer

no of atom [point] in one unit cell is 6

co . no . of a lattice point[atom] is 12

6 from same plane and 3 from above and 3 from below the plane

packing efficiency is 74%

CUBIC CLOSE PACKING

this is an A-B-C, A- B –C type packing in which every 4th layer is similar to 1st layer in patternof packing

it is resulted from packing of 3rd layer point in octahredral void of 2nd layer

no of atom [point] in one unit cell is 4

co . no . of a lattice point[atom] is 12

4 from same plane and 4 from above and 4 from below the plane

packing efficiency is 74%

Calculation of Spaces occupied i.e. Packing Fraction

a) In a simple cubicEdge length = aRadius of sphere = rAs the spheres are touching each other, a = 2r

No. of spheres per unit cell = 1 88 =1

Volume of the sphere = 34/3 r , Volume of the cube = a3 = 3 3(2r) 8r

fraction occupied (or packing fraction) =3

3

4 r38r

= 0.524% occupied = 52.4%

b) In face centred cubicPR = 4rIn right angled triangle PQR

PR = = = 4r 4a r2

Volume of the unit cell = a =3

34 32r r2 2

No. of spheres in the unit cell = 1 18 68 2 = 4

Volume of 4 spheres = 3 34 164 r r3 3

Fraction occupied (or packing fraction) =3

3

16 r332r

2

= 0.74 or % occupied = 74%

c) In body centred cubic: The atom at the body centre centretouches the spheres at the corners, Body diagonal, PS = 4rFace diagonal PR = 2 2PR QR = 2a

and Body diagonal PS = 2 2PR RS = 2 22a a 3a

Now, 3a 4r a = 4 r3

, Volume of unit cell = a3 = 364 r3 3

No. of spheres per unit cell =18 1 18

= 2

Volume of two spheres =3 34 82 r

3 3

Fraction occupied (or packing fraction) =

3

3

8 r364 r

3 3

= 0.68 or % occupied = 68%

ar

2r

r

a

a

B

C

a

2 2PQ QR 2 2a a 2a 2a4r

aa

a

PQ

S

R

Hexagonal Close Packing: Each corner atom would becommon to 6 other unit cells, therefore their contribution to one unit

cell would be 1/6. Total number of atom in 1 hcp unit cell = 126

(from 12 corners) + 22

(from 2 face centred + 31

(from body centre)

ABCD is the base of hexagonal unit cell

AD=AB=a. The sphere in the next layer has its centre F vertically above E it touches the three spheres whosecentres are A,B and D.

Hence FE = h2

=2

2 2r(2r)3

The height of unit cell (h) = 24r3

The area of the base is equal to the area of six equilateral triangles, = 236 (2r)4

. The volume of the unit cell =

23 26 (2r) 4r4 3

.

Therefore,

PF =3

2

46

π r

33 26 (2r) 4r

4 3

0.74;VF 0.26 BY MUKESH SHARMA ,DPS JODHPUR

Type of Lattice point Contribution to one unit cellCorner of cube 1/8Edge centre of cube 1/4

Facecenter of cube 1/2Body Center of cube 1

Type of Lattice point Contribution to one unit cellCorner of Hexagonal unit cell 1/6Facecenter of 1/2

Body Center of cube 3Edge centre of hexagonal [only voids] 1/3

C

B

a

Ar

D

GE

a

A

F

h/2

3/a E

a 2r

c

Type ofUNITCELL

Latticepoint

Effective no. Atoms[Z]

Coordination number

Relationb/w a and

r

Packingeffeicency

Atoms incontact

Primitive 8 [1/8X8]=1 6 2r =a 52 Oncorner

Bodycentred

9 [1/8X8]+[1X1]=2 8 4r=√ 68 On bodydigonal

Facecentred

14 [1/8X8]+[1/4 X12]=4 12 4r=√ 74 On facedigonal

Hexagonalprimitive

17 [1/6X12]+[1/2X2]+3=6 12 Shortesta=2r

74 Inhexagonal

plane

Different distances

Distance B/W nearest neighbour in primitive unit cell [two –corner atom ]2r =a

Distance B/W 2nd nearst neighbour in primitive unit cell [two face diagonal atom ] √Distance B/W 3rd nearst neighbour in primitive unit cell [twocube diagonal atom ] √Distance B/W nearest neighbour in FCC [corner to face cetre atom]

2r=√Distance B/W nearest neighbour in BCC [corner to body cetre atom] = √DistanceB/W 1stand4th layer of FCC [corner to corner on body diogonal] √DistanceB/W1stand2thlayerofFCCanytwosuccessivelayer √Distance B/W atom and tetrahedral void in FCC √Distance B/W octahedral void and tetrahedral void in FCC √Distance B/W atom and octahedral void in FCC

BY MUKESH SHARMA ,DPS JODHPUR

Illustration 1: A solid has a cubic structure in which X atoms are located at the corners of the cube, Y atomsare at the cube centres and O atoms are at the edge centres. What is the formula of thecompound?

Solution: Atoms of X are present at all the eight corners of the cube. Therefore, each atom of X at thecorner makes 1/8 contribution towards the unit cell.

Number of atoms of X per unit cell = 18 18

Y atom is present at the body centre, thus contribution of Y towards unit cell= 1 1 = 1O atom is present at each of the edge centre (number of edges of cube = 12)And each O atom present at edge centre will make 1/4 contribution towards the unit cell.

The number of O atoms per unit cell = 1124 = 3

The formula of the compound is, therefore XYO3

Illustration 2: Potassium crystallizes in a body centred cubic lattice. What is the approximate number of unitcells in 4.0g of potassium? Atomic mass of potassium = 39.

Solution: There will be eight atoms at corners of the cube and one atom at the body centre.

Number of atoms per unit cell = 188 + 1 1 = 2

Number of atoms in 4.0g of potassium = 234 6.023 1039

Number of unit cells in 4.0g of potassium =234 6.023 10

39 2

= 3.09 1022

IQ 1:In a solid, oxide ions are arranged in ccp. One sixth of the tetrahedral voids are occupied by the cations Awhile one third of the octahedral voids are occupied by the cation B. What is the formula of the compound?

Voids and Radius Ratio Rules

Octahedral void tetrahedral void

Radius Ratio Rules: The structure of many ionic solids can be accounted by considering the relative sizesof the cation and anion, and their relative numbers. By simple calculations, we can work out as how many ionsof a given size can be in contact with a smaller ion. Thus, we can predict the coordination number from therelative size of the ions.

Calculation of some limiting radius ratio valuesa) Coordination Number 3: The adjacent figure shows the

smaller positive ion of radius r+ in contact with three largeranion of radius r–.In this figure, AB = BC = AC = 2r–

BE = r–

BD = r+ + r–

ABC = 60° BDC = 120° or BDE = 60° DBC = 30°

A

B CE

D

BD = BEcos30

r+ + r– = r rcos30 0.866

= 1.155r–

r+ = 1.155r– – r+ = 0.155r–

rr

= 0.155

b) Coordination Number 4: (tetrahedron)Angle ABC is the tetrahedral angle

ABC = 109° 28’ ABD = 12

(109° 28) = 54° 44

sin (ABD) = 0.8164 = AD rAB r r

r r 1

r 0.8164

= 1.225 r

r

= 0.225

A

B

CD

c) Coordination Number 6: (Octahedron): A crosssection through an octahedral site is shown in the adjacentfigure and the smaller positive ion touches six larger negativeions. (Only four negative ions are shown in the figure but thereis one sphere above and the below the plane of paper).In this figureAB = r+ + r– and BD = r–

ABC = 45° cos(ABD) = 0.7071 = BD rAB r r

Solving, we get rr

= 0.414

A

B CD

Limiting radius ratiorr

Coordinationnumber Shape Examples

0.155 2 Linear BeF2

0.155 – 0.225 3 Trigonal planar B2O3

0.225 – 0.414 4 Tetrahedral ZnS, SiO44–

0.414 – 0.732 4 Square planar PtCl42–

0.414 – 0.732 6 Octahedral NaCl0.732 – 0.999 8 Body centred cubic CsCl

Illustration 3: Iron crystallises in a body centred cubic structure. Calculate the radius of Fe atom if edgelength of unit cell is 286 pm.

Solution: Edge length, a = 286 pm

For BCC, radius of atom, r = 3 a4

= 3 2864 = 1.732 286

4 = 123.8 pm

IQ2:Define coordination number, what is the coordination number of each sphere ina) Cubic closepacked structuresb) Bodycentred closepacked structures

c) c) Hexagonal closepacked structures

Calculation of density of a cubic crystal from its edge: If we know the type of crystalstructure possessed by the cubic crystal, so that the number of particles per unit cell are known and the edgelength for it is known by XRay studies, the density of the crystal can be determined.

Case I: For cubic crystals of elements:Let the edge of the unit cell = a pmNo. of atoms present per unit cell = ZAtomic mass of the element = M Volume of the unit cell = (a pm)3 = a3 pm3 = a3 10–30 cm3

Density of the unit cell,

= Mass of the unit cellVolume of the unit cell

a

= 3 30

No. of atoms in its unit cell Mass of each atoma 10

= 3A3 30

MZN g / cm

a 10

(NA – Avogadro’s number) = 3 30

A

Z MN a 10

g/cm3

Case II: For cubic crystals of ionic compounds: Here, Z is the number of formula units present in oneunit cell and M is the formula mass. The formula will remain the same viz.

Illustration 4: A facecentred cubic element (atomic mass 60) has a cell edge of 400 pm. What is its density?

Solution: For the facecentred cubic, Z = 4

= 33 30

0

Z M g / cma N 10

= 3 23 30

4 60(400) (6.02 10 ) 10

= 6.23 g/cm3

IQ3:The unit cell in a crystal of diamond belongs to a cubic crystal system but doesn’t correspond to the Bravaislattice. The volume of unit cell of diamond is 0.0454 nm3 and the density of diamond is 3.52 g/cc. Find thenumber of carbon atoms in a unit cell of diamond?

Classification of Ionic StructuresSimple ionic compounds are of the type AB or AB2 where A and B represent the positively and negativelycharged ions respectively. (In any solid of the type AxBy, the ratio of coordination number of A to B would bey :x.

Structures of Type AB: Ionic compounds of the type AB means compounds having the positively andnegatively charged ions in the ratio 1:1. These compounds can have following three type of structures.

1. Rock salt (NaCl) type2. Cesium chloride (CsCl) type3. Zinc blende (ZnS) type

330

A

Z M g./ cmN 10

1. Rock Salt (NaCl) type Structure: Cl– isforming a FCC unit cell in which Na+ is in the octahedralvoids. The coordination number of Na+ is 6 and that ofCl– would also be 6. Moreover, there are 4 Na+ ions and4 Cl– ions per unit cell. The formula is Na4Cl4 i.e. NaCl.Other examples for this type of structure are all halides ofalkali metals except CsCl and all oxides of alkaline earthmetals except BeO.

Na+

Cl-

2. Zinc Blende Structure (ZnS): Sulphite ions areface centred and zinc is present in alternate tetrahedralvoids. Formula is Zn4S4, i.e. ZnS. Coordination numberof Zn is 4 and that of sulphide is 4. Other substance thatexists in this kind of a structure is BeO.

3. Cesium Chloride (CsCl) type structure:CsCl has bodycentred cubic (bcc) arrangement. EachCs+ ion is surrounded by 8 Cl– ions and each Cl– ion issurrounded by 8 Cs+ ions i.e. this structure has 8:8coordination. A unit cell of CsCl consists of only oneunit of CsCl i.e. One Cs+ ion and one Cl– ion. Fewexamples of compounds having CsCl structure are CsBr,CsI, CsCN, TlCl, TlBr, TlI and TlCN.

Cs+ ion surrounded bby 8 Cl- ions

Structure of Ionic Compounds of the Type AB2: These are the ionic compounds havingcations and anions in the ratio 1:2. Most of these compounds have calciumfluorite (CaF2) type structure. Thesecompounds have cubic close packing (CCP) arrangement in which Ca2+ ions are present at the corners and thecentre of each face of the cube.Each Ca2+ ion is surrounded by 8F– ions i.e. it has acoordination number of 8 whereas each F– ion is surroundedby 4 Ca2+ ions i.e. has a coordination number of 4. Thus thisstructure has 8:4 coordination. Few examples of suchcompounds having CaF2 structure are BaF2, BaCl2, SrF2,SrCl2, CdF2, PbF2 and ThO2.

2CaF

In Na2O each oxide ion is coordinated to 8 Na+ ions and each Na+ ion to 4 oxide ions. Hence it has 4:8coordination. This is called antifluorite structure. Others examples being Cl2O, K2O, Li2O, K2S, Na2S etc.

Ionic Radii: The internuclear distance B/W the ions at the adjacent corners of a unit cell can be taken as thesum of the radius of the cation and the radius of the anion.

But it’s not simple to assign different values to constituting ions because it’s not possible to calculate the radiusof one ion without knowing the radius of other one.

It has been observed that radius of a cation is smaller than that of the corresponding atom. The reason is thatwith the removal of one (or more) electrons from the valence shell of the atom, the effective nuclear chargeincreases because now, it is acting on a smaller number of electrons, making the electron cloud pulled moreinward towards the nucleus than before.

On the other hand, the radius of an anion is larger than that of the corresponding atom, reason being, withaddition of one (or more) electrons to the valence shell, effective nuclear charge decreases because the same

2Zn2S

ionsZnfourbyllytetrahedra

surroundedionS

2

2

nuclear charge is now acting on a larger number of electrons. Consequently, the force of attraction on theelectron cloud decreases and hence the ionic radius increases.

Ionic radius increases in a group with increase in atomic number as:Li+ Na+ K+ Rb+ Cs+

Similarly, in the halogens, the ionic radius increase as:F– Cl– Br– I–

The reason for this increase is the increase in the principal quantum number though the number of electrons inthe valence shell remains the same. In the same period, the radius of isoelectronic ions (having same number ofelectrons) decreases with the increase in charge of the ions. e.g. Na+Mg2+ Al3+.

Illustration 5: If the close packed cations in an AB type solid have a radius of 75 pm, what would be themaximum and minimum sizes of the anions filling the voids?

Solution: For closed packed AB type solidrr

= 0.414 – 0.732

Minimum value of r– = r 750.732 0.732

pm = 102.5 pm

Maximum value of r– = r 750.414 0.414

pm = 181.2 pm

IQ 4:i) The radius of Na+ ion is 95 pm and that of Cl– ion is 181 pm. Predict whether the coordination number

of Na+ ion is 6 or 4.ii) How many unit cells are there in a 1.0g cube shaped ideal crystal of NaCl?iii) The ionic radii of K+ and F– are 133 and 136 pm, respectively. Calculate length of the unit cell of KF,

KF has rocksalt structure.

Imperfection in Solids

An ionic crystal which has the same unit cell containing the same lattice points throughout the crystal is knownas ideal crystal but crystals tend to have a perfectly ordered arrangement at only absolute zero. Thisarrangement corresponds to state of lowest energy. And as the temperature increases, the crystals start deviatingfrom the perfectly ordered arrangement. This defect may appear at a point, along a line or over a surface.

A point defect could arise due to the absence of a particle (vacancy), presence of some foreign particle at alattice site, presence of a foreign particle at the interstitial site or displacement of a particle to the interstitial site.

Two main defects in crystals which are discussed as follows are Schottky and Frenkel defects

Schottky Defect: This defect is caused when some of the lattice points are unoccupied and those pointsare called vacancies or holes fig. (a). The number of missing positive and negative ions is the same so that thecrystal remains neutral in all. Schottky defects are more common in ionic compounds with high coordinationnumber, and where the sizes of positive and negative ions are almost equal for example, NaCl, KCl, CsCl andKBr.

The number of defects increases with increase in temperature. The number of defects increases to one in 106sites at 775 K and one in 104 sites at 1075K. The presence of large number of Schottky defects in crystal resultsin significant decrease in its density.

Frenkel Defect: This defect is caused when some of the ions leave their lattice sites to occupy aninterstitial site fig. (b). Frenkel defects are more common in ionic compounds with low coordination numberand where there is large difference in size B/W positive and negative ions for example, ZnS, AgCl, AgBr andAgI.

In pure alkali metal halides, these defectsare not very common because the ionscannot get into interstitial positions due totheir large sizes.

Fig. (a): Schottky Defect

B- A+A+

B- B-

B-B-

B- B-

A+

A+ A+

A+

A+A+

A+B-

A+

Fig. (b): Frenkel Defect

B- B-

B-B-

B- B-

A+A+

A+

A+

B- A+B-A+

A+

Metal Excess Defect: This may occur in two different ways

F-Centres: A negative ion may be absent from its lattice site leaving a hole which is occupied by an electron,thereby maintaining the electrical balance. This type of defect is formed by crystals which would be expected toform Schottky defects. When compounds such as NaCl, KCl, are heated with excess of their constituent metalvapours, or treated with high energy radiation, they become deficient in the negative ions and their formulaemay be represented by AX1–, where is a small fraction. The crystal lattice has vacant anion sites which areoccupied by electrons. Anion sites occupied by electrons in this way are called F centres (F is an abbreviationFarbe, the German word for colour).

Interstitial ions and electrons: Metal excess defects also occur when an extra positive ion occupies aninterstitial position in the lattice and electrical neutrality is maintained by the inclusion of an interstitial electron.Their composition may be represented by general formula A1+X. This kind of metal excess defect is muchmore common than the first and is formed in crystals which would be expected to form Frenkel defects.Examples include ZnO, CdO, Fe2O3.

Crystals with either type of metal excess defect contain free electrons, and if these migrate they conduct anelectric current. These free electrons may be excited to higher energy levels, giving absorption spectra and inconsequence their compounds are often coloured e.g. non-stoichiometric NaCl is yellow, non-stoichiometricKCl is lilac.

F Centre

e–

Na+Cl–

F-centre in a Sodium chloride crystal

A+

Metal excess defects caused by interstitial cations.

A+

B–

A+

B–

A+

B– A+ B– A+

B–

A+

B–

A+

B–

A+

B–

A+

A+

B–

A+

B–

A+

B–

A+

B–

e–

Metal Deficiency Defect: This defect is caused bythe missing cation from its lattice site. The extra negativecharge may be balanced if the adjacent metal ion has higherpositive charge. This can be possible by compounds oftransition metals (variable valency). Crystals of FeO, FeSand NiO show this defect.

Properties of Solids

Electrical Properties: Solids can be broadly classified into three types, on the basis of electricalconductivity.

a) Metals (conductors)b) Insulatorsc) Semiconductors

Electrical conductivity of metals is very high and is of the order of 106 – 108 ohm–1 cm–1 whereas for insulators,it is of the order of 10–12 ohm–1 cm–1. Semiconductors have intermediate conductivity in the range of 102 – 10–9

ohm–1 cm–1. Electrical conductivity of solids may arise through the motion of electrons and holes (positive) orthrough the motion of ions. The conduction through electrons is called ntype conduction and through(positive) holes is called ptype conduction. Pure ionic solids where conduction can take place only throughmovement of ions are insulators. The presence of defects in the crystal structure increases their conductivity.

The conductivity of semiconductors and insulators is mainly due to the presence of interstitial electrons andpositive holes in the solids due to imperfections. The conductivity of semiconductors and insulators increaseswith increase in temperature while that of metals decrease.

Band Theory Consider a molecule with two atomic orbitals.The result must be that two molecular orbitals will be formedfrom these atomic orbitals: one bonding and one anti-bonding,separated by a certain energy

If this is expanded to a molecule with three atoms,assuming 1 atomic orbital for each, then the result mustbe that 3 molecular orbitals will be formed.

Now , let's take it to 10 atoms. As thenumber of molecular orbitals increases, theenergy difference between the lowest bondingand the highest anti-bondig increases, whilethe space between eachindividual orbital decreases

Consider a metal withan infinite number of atoms. This will form an

A+ A+ B-B-

A+

A+ A+

A+

A2+

B-

B-

B-

B-

B-

A+A+

B-

B-

B-

Metal deficiency due to cation vacancy

infinite number of molecular orbitals so close together they blur into one anotherforming a band.

Conductivity according to MOT is related to ease ofavailability of conduction [empty band ] band formovement of electrons from valance band1.valance band : highest occupied band and 2.Conduction band : lowest un occupied band

Energy gap of semiconducter

According to MOT thesubstance in which bandgap is intermediate ofConductors and insulatorand small as well, which make promotion of electron fromvalance band to conduction band possible

1. n-type Semiconductor 2 . p-type Semiconductor

n-type Semiconductor Doping with an element of extravalence electrons into a element. There is NOextra room for these electrons in the valence band;consequently, they are promoted into the conductionband, where they have access to many vacant orbitalswithin the energy band they occupy and serve aselectrical carrierse.g.Silicon (4v es-)doped withphosphorous(5 ves-)

In general 14 th group element with 15th group element

p-type Semiconductor : Doping with anelement of less electrons in order to create electronvacancies or positive holes in the system. Because thevalence band is incompletely filled, under the influenceof an applied field, electrons can move from occupiedmolecular orbitals to the few that are vacant, therebyallowing current to flow.

e.g.

Silicon (4 ves-)doped with aluminium (3 ves-)

In general 14th group element with 13th group element

Magnetic Properties

n= no of unpaired e-.

PARAMGNETISM

The materials which are weakly attracted by magneticfield are called paramagnetic materials. These materialshave permanent magnetic dipoles due to presence ofatoms, ions or molecules with unpaired electron. e.g. O2,Cu2+, Fe2+ etc. But these materials lose their magnetism in the absence ofmagnetic field.

Ferromagnetic Materials: The materials which show permanent magnetism even in the absence ofmagnetic field are called ferromagnetic materials. These materials are strongly attracted by the magnetic field.

e.g. Fe, Co, Ni and CrO2. Ferromagnetism arises due tospontaneous alignment of magnetic moments of ions oratoms in the same direction.

Alignment of magnetic moments in opposite directions in acompensatory manner and resulting in zero magneticmoment gives rise to antiferromagnetism

ferromagnetism ;

Anti ferromagnetism ;

Ferrimagnetism

for example, MnO, Mn2O3 and MnO2.to unequal number ofparallel and antiparallel magnetic dipoles give rise toferrimagnetism e.g. Fe3O4.

Ferromangetic and ferrimagnetic substances change into paramagnetic substances at higher temperature due torandomisation of spins. Fe3O4, is ferrimagnetic at room temperature and becomes paramagnetic at 850 K.

Property Definitions ExampleDiamagnetic Repelled by ex. M F.

because absence of unpaired electronN aCl

Paramagnetic W eakly attracted and show s m agnetismO nly in presence of ex. M F

O 2 ,Cu 2+

Ferromagnetic Strongly attracted and show s m agnetismeven in absence of ex.M F

Fe,Co,N i

Ferrimagnetic O bserved m agnetic m om ent is lesser thancalculated because unequal alignm ent

Fe3O 4

Antiferromagnetic

O bserved m agnetic m om ent is zerobecause of exactly equal no oppositealignm ent

M nO

Dielectric Properties: The electrons in insulators are closely bound to the individual atoms or ions andthus they do not generally migrate under the applied electric field. However, due to shift in charges, dipoles arecreated which results in polarisation. The alignment of these dipoles in different ways i.e. compensatory way(zero dipole) or noncompensatory way (net dipole) impart certain characteristic properties to solids.

If the dipoles align in such a way that there is net dipole moment in the crystals, these crystals are said to exhibitpiezoelectricity or piezoelectric effect i.e. when such crystals are subjected to pressure or mechanical stress,electricity is produced. Conversely, if an electric field is applied to such a crystal, the crystal gets deformed dueto generation of mechanical strain. This is called inverse piezoelectric effect.

Some crystals which on heating, acquire electric charges on opposite faces, are said to exhibit pyroelectriceffect.The solids, in which dipoles are spontaneously aligned in a particular direction, even in the absence of electricfield are called ferroelectric substances and the phenomenon is known as ferroelectricity. If the alternate dipolesare in opposite direction, then the net dipole moment will be zero and the crystal is called antiferroelectric

.

Ferroelectric solids – Barium titanate (BaTiO3), sodium potassium tartrate (Rochelle salt) and potassiumhydrogen phosphate (KH2PO4). Antiferroelectric – Lead Zirconate (PbZrO3).

LEVEL – I (CBSE LEVEL) Review Your Concept

1. An element crystallizes in a structure having a FCC unit cell of an edge 200 pm. Calculate its density if200g of this element contains 24 1023 atoms.

2. Copper crystal has a face centred cubic structure. Atomic radius of copper atom is 128 pm. What is thedensity of copper metal? Atomic mass of copper is 63.5.

3. An element occurs in BCC structure with a cell edge of 288 pm. The density of metal is7.2 g cm–3. How many atoms does 208g of the element contain?

4. The density of crystalline sodium chloride is 2.165 g cm–3. What is the edge length of the unit cell. Whatwould be the dimensions of cube containing one mole of NaCl?

5. The density of potassium bromide crystal is 2.75 g cm–3 and the length of an edge of a unit cell is 654pm. The unit cell of KBr is one of three types of cubic unit cells. How many formula units of KBr arethere in a unit cell? Does the unit cell have a NaCl or CsCl structure?

6. When heated above 916°C, iron changes its crystal structure from bodycentred cubic to cubic closedpacked structure. Assuming that the metallic radius of the atom does not change, calculate the ratio ofdensity of the BCC crystal to that of the CCP crystal.

7. The unit cell length of NaCl is observed to be 0.5627 nm by Xray diffraction studies; the measureddensity of NaCl is 2.164 g cm–3, Correlate the difference of observed and calculated density andcalculate % of missing Na+ and Cl– ions.

8. CsCl has cubic structure of ions in which Cs+ ion is present in the body centre of the cube. If density is3.99g cm–3.

a) Calculate the length of the edge of a unit cell.b) What is the distance between Cs+ and Cl– ions?c) What is the radius of Cs+ ion if the radius of Cl– ion is 180 pm?

9. An element having atomic mass 52, occurs in body centred cubic (BCC) structure with a cell edge of288 pm. The density of the element is 7.2 g cm–3. Evaluate Avagadro’s number?

10. The element having atomic mass 60 has face centred cubic unit cells. The edge length of the unit cell is400 pm. Find out density of the element.

LEVEL – II Brush Up Your Concepts

1. In a crystalline solid, having formula AB2O4, oxide ions are arranged in cubic close packed lattice whilecations A are present in tetrahedral voids and cations B are present in octahedral voids.i) What percentage of the tetrahedral voids is occupied by A?ii) What percentage of the octahedral voids is occupied by B?

2. Calculate the value of Avogadro number from the following data:Density of NaCl = 2.165 gm cm–3, distance between Na+ and Cl– ions in NaClcrystal = 281 pm

3. In a cubic close packed structure of mixed oxides, the lattice in made up of oxide ions, oneeighth oftetrahedral voids are occupied by divalent ions (X2+) while one half of octahedral voids are occupied bytrivalent ions (Y3+). What is the formula of the oxide?

4. Sodium crystallizes in b.c.c. lattice of side length 4.30 Å. How many atoms are present in a unit lattice?What is density of the metal? Atomic weight of Na = 23.

5. Sodium metal crystallises in body centred cubic lattice with cell edge, 4.29Å. What is the radius ofsodium atom?

6. Potassium crystallises in a ‘BCC’ lattice, (edge length, a = 5.20Å)a) What is the distance between nearest neighbours?b) How many neighbours does each K atom have?c) What is the density of crystalline K?

7. KBr crystallises in the FCC unit cell (NaCl type)a) How many K+ ions and how many Br– ions are in each unit cell?

b) What is the edgelength ifK

r = 1.33Å,Br

r = 1.95Å?c) What minimum value of r+/r– is needed to prevent anionanion contact in this structure?

8. Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit celldimensions are a = 6.8 Å, b = 4.4 Å and c = 7.2 Å. If the molar mass is 21.76. Calculate the density ofcrystal.

9. Compute the limiting radius ratio r+/r– for (a) a square planar crystal structure in which B– ions are at thecorner of the square and A+ ions are the centre, and (b) an equilateral triangular crystal structure with B–

at the apex and A+ at the centre.

10. A solid PQ has the NaCl structure. If the cation radius is 100 pm, calculate the maximum possible valueof the radius of the anion. Answer to Assignments (Subjective)

LEVEL – I1. 41.6 g cm–3 2. 8.9 g/cm3

3. 24.16 1023 atom 4. 3 cm5. 4, fcc 6. 0.9187. 0.016 g/cm3, 0.775% 8. a) 412 pm; b) 356.8 pm; c) 176.8 pm9. 6.04 1023 10. 6.2 g/cm3

LEVEL – II1. 12.5%, 50% 2. 6.02 1023

3. AB2O4 4. 2, 0.96 g –3cm5. 1.86Å 6. 4.05Å, 8, 0.925 g/cm3

7. 4, 6.56Å, 0.414 8. 0.6709 g cm–3

9. 0.414, 0.866 10. 241.5 pm

BY MUKESH SHARMA ,DPS JODHPUR