Solid State

SOLID STATE

By 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 and cannot 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 the characteristics 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 the entire 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 given

structure are more separated than along the face diagonal. This causes anisotropy

Types of crystalline solids

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

Unit cell

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)

Types of Lattices

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

Crystal System Bravais Lattices Parameters of Unit Cell Example

Intercepts Interfacial angle

1. Cubic Primitive, Face Centered, Body

Centered = 3 a = b = c === 90

o _Pb,Hg,Ag,Au

Diamond, NaCl, ZnS 2. Orthorhombic Primitive, Face Centered, Body

Centered, End Centered = 4 abc === 90

o _KNO

2, K2SO4

3. Tetragonal Primitive, Body Centered =2 a = bc === 90o _TiO 2,SnO2

4. Monoclinic Primitive, End Centered = 2 abc == 90o_,

 90o CaSO4,2H2O 5. Triclinic Primitive = 1 abc      900 _K 2Cr2O7, 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, CaCO} 3 Total = 14

Seven Crystal Systems and Fourteen Bravais Lattice

Simple TETRAGONAL Body Centred c a a

Simple End Centred

MONOCLINIC c a b b a a a

Simple Face Centred Body Centred

CUBIC

Simple End Centred

ORTHORHOMBIC

c

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 corners of the unit cell.

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

2 Face Centred: When atoms are present in all 8-corners and six face centres in a cubic unit cell then

this 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 cell

then 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 portion of 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 one eighth of each such point lies within the given unit cell.

ii) A point along an edge is shared by four unit cells and only onefourth of it lies within any one cell. iii) A facecentred point is shared by two unit cells and only one half of it is present in a given unit cell. iv) A bodycentred point lies entirely within the unit cell and contributes one complete point to the cell.

Close Packing in Crystals

RHOMBOHEDRAL a a a a a c b a b g TRICLINIC a 120o c a HEXAGONAL

In order to understand the packing of the constituent particles in a crystal, it is assumed that these particles are hard spheres of identical size. The packing of the crystals is done such that they occupy the maximum available space 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 a way that their centres are at the centres of an equilateral triangle. Each sphere is surrounded by six other similar spheres. This arrangement is called hexagonal close packing.

This arrangement can be extended in three dimensions by adjusting spheres on the top of hollows or voids of the 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 are marked as ‘a’ and ‘b’. The efficient way of placing the spheres in second layer is to place them in the ‘a’ voids of 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 are triangular while only ‘c’ voids of the second layer are triangular. The ‘d’ voids are combination of two triangular voids (one each of first and second layer) with the vertex of one triangle upwards and the vertex of other 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 threedimensional closest packing is obtained where the spheres in every third layer are vertically aligned to the first layer. This arrangement is called ABAB.,… pattern or hexagonal (HCP) close packing (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 cubic close packing (CCP). A A A A A A A A A A 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 A A A A A 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. It touches three spheres in the layer above and three spheres in the layer below. Thus a sphere is in direct contact with 12 other spheres. In other words, the coordination number of that sphere is 12. Coordination number is the number 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 closestpacked layers, and (b) rotation to show cubic symmetry

BY MUKESH SHARMA ,DPS JODHPUR FEATURES OF HEXAGONAL CLOSE PACKING

 this is an A-B-, A- B –type packing in which every 3rdlayer is similar to 1stlayer in pattern of packing

 it is resulted from packing of 3rdlayer point in tetrahedral void of 2ndlayer  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 1}st_{layer in pattern}

of packing

 it is resulted from packing of 3rd_{layer point in octahredral void of 2}nd_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 cubic Edge length = a Radius of sphere = r

As the spheres are touching each other, a = 2r No. of spheres per unit cell =1 8

8 =1

Volume of the sphere =_{4/3 r}3_{, Volume of the cube = a}3_=(2r)3 _8r3

fraction occupied (or packing fraction) = 3 3 4 r 3 8r  = 0.524% occupied = 52.4% b) In face centred cubic

PR = 4r

In right angled triangle PQR

PR = =  = 4r a 4 r

2 

Volume of the unit cell = a =

3 3 4 _r 32 _r 2 2       

No. of spheres in the unit cell = 8 1 6 1

8 2

   = 4 Volume of 4 spheres = ₄ 4 _r3 16 _r3

3 3

   

Fraction occupied (or packing fraction) =

3 3 16 r 3 32r 2  = 0.74 or % occupied = 74%

c) In body centred cubic: The atom at the body centre centre touches the spheres at the corners,

Body diagonal, PS = 4r

Face diagonal PR = _PR2 _QR2 _{= 2a}

and Body diagonal PS = _PR2_RS2 _{= 2a}2_a2 _{ 3a}

Now, 3a 4r a = 4 r

3 , Volume of unit cell = a

3₌ 64 r3 3 3 No. of spheres per unit cell =

1

8 1 1

8   

= 2 Volume of two spheres =

3 3

4 8

2 r

3 3

   

Fraction occupied (or packing fraction) =

3 3 8 r 3 64 r 3 3  = 0.68 or % occupied = 68% a r 2r r a a B C a 2 2 PQ QR _a2 _a2 _{ 2a 2a} 4r a a a P Q S R

Hexagonal Close Packing: Each corner atom would be common 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 = 12

6 (from 12 corners) + 2

2 (from 2 face centred + 3

1 (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 whose centres are A,B and D.

Hence FE = h 2 = 2 2 2r (2r) 3       

The height of unit cell (h) = 4r 2 3

The area of the base is equal to the area of six equilateral triangles, =₆ 3_(2r)2 4

 . The volume of the unit cell =

2 3 2 6 (2r) 4r 4 3   _. Therefore, PF = 3 2 4 6 πr 3 3 2 6 (2r) 4r 4 3    

0.74;VF0.26 BY MUKESH SHARMA ,DPS JODHPUR

Type of Lattice point Contribution to one unit cell

Corner of cube 1/8

Edge centre of cube 1/4

Facecenter of cube 1/2

Body Center of cube 1

Type of Lattice point Contribution to one unit cell

Corner of Hexagonal unit cell 1/6

Facecenter of 1/2

Body Center of cube 3

Edge centre of hexagonal [only voids] 1/3 C B a A _r D G E a A F h/2 3 / a E

a 2r



c

Type of UNIT CELL

Lattice

point Effective no. Atoms[Z] Coordination number b/w a andRelation r

Packing

effeicency Atoms incontact

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

corner Body

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

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

primitive 17 [1/6X12]+[1/2X2]+3=6 12 Shortesta=2r 74 hexagonalIn 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 3rdnearst 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 1st_and4th _{layer of FCC [corner to corner on body diogonal] √}

DistanceB/W1st_and2th_{layerofFCCanytwosuccessivelayer √}

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 atoms are at the cube centres and O atoms are at the edge centres. What is the formula of the compound?

Solution: Atoms of X are present at all the eight corners of the cube. Therefore, each atom of X at the

corner makes 1/8 contribution towards the unit cell. Number of atoms of X per unit cell = 8 1 1

8  

Y atom is present at the body centre, thus contribution of Y towards unit cell = 11 = 1

O 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 = 12 1

4  = 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 unit cells 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 = 8 1 8

 + 11 = 2 Number of atoms in 4.0g of potassium = 4 6.023 1023

39 

Number of unit cells in 4.0g of potassium = 4 6.023 1023

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 A while 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 sizes of the cation and anion, and their relative numbers. By simple calculations, we can work out as how many ions of a given size can be in contact with a smaller ion. Thus, we can predict the coordination number from the relative size of the ions.

Calculation of some limiting radius ratio values

a) Co_{ordination Number 3: The adjacent figure shows the}

smaller positive ion of radius r+ _{in contact with three larger} anion 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 E C D BD = BE cos30 r +_{+ r}– ₌ r r cos30 0.866     = 1.155r– r+ = 1.155r–_{– r+ = 0.155r}– r r   = 0.155

b) Co_{ordination Number 4: (tetrahedron)}

Angle ABC is the tetrahedral angle

 ABC = 109° 28’  ABD = 1 2(109° 28) = 54° 44 sin (ABD) = 0.8164 = AD r AB r r       r r 1 r 0.8164     _ = 1.225  r r   = 0.225 A B C D

c) Co_{ordination Number 6: (Octahedron): A cross}

section through an octahedral site is shown in the adjacent figure and the smaller positive ion touches six larger negative ions. (Only four negative ions are shown in the figure but there is one sphere above and the below the plane of paper).

In this figure

AB = r+_{+ r}–_{and BD = r}–

ABC = 45°cos(ABD) = 0.7071 = BD r AB r r      Solving, we get r r   = 0.414 A B C D

Limiting radius ratio r r         Coordination

number 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 NaCl

0.732 – 0.999 8 Body centred cubic CsCl

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

Solution: Edge length, a = 286 pm

For BCC, radius of atom, r = 3 a

4 = 43 286 = 1.732 286 4  = 123.8 pm IQ2:

Define coordination number, what is the coordination number of each sphere in

a) Cubic closepacked structures

b) Bodycentred closepacked structures

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

Case I: For cubic crystals of elements:

Let the edge of the unit cell = a pm No. of atoms present per unit cell = Z Atomic mass of the element = M

Volume of the unit cell = (a pm)3_{= a}3_pm3_{= a}3_10–30_cm3 Density of the unit cell,

= Mass of the unit cell Volume of the unit cell

a

=



 



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

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

Solution: For the facecentred cubic, Z = 4

= 3 3 30 0 Z M _{g / cm} a N 10    = 3 23 30 4 60 (400) (6.02 10 ) 10     = 6.23 g/cm 3 IQ3:

The unit cell in a crystal of diamond belongs to a cubic crystal system but doesn’t correspond to the Bravais

lattice. The volume of unit cell of diamond is 0.0454 nm3and the density of diamond is 3.52 g/cc. Find the

number of carbon atoms in a unit cell of diamond?

Classification of Ionic Structures

Simple ionic compounds are of the type AB or AB2 where A and B represent the positively and negatively charged ions respectively. (In any solid of the type AxBy, the ratio of coordination number of A to B would be y :x.

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

1. Rock salt (NaCl) type 2. Cesium chloride (CsCl) type 3. Zinc blende (ZnS) type

3 30 A Z M g./cm N 10    

1. Rock Salt (NaCl) type Structure: Cl– _is forming a FCC unit cell in which Na+ is in the octahedral voids. The coordination number of Na+ is 6 and that of Cl– would also be 6. Moreover, there are 4 Na+ ions and 4 Cl– _{ions per unit cell. The formula is Na}

4Cl4 i.e. NaCl. Other examples for this type of structure are all halides of alkali metals except CsCl and all oxides of alkaline earth metals except BeO.

Na+ Cl

-2. Zinc Blende Structure (ZnS): Sulphite ions are

face centred and zinc is present in alternate tetrahedral voids. Formula is Zn4S4, i.e. ZnS. Coordination number of Zn is 4 and that of sulphide is 4. Other substance that exists in this kind of a structure is BeO.

3. Cesium Chloride (CsCl) type structure:

CsCl has bodycentred cubic (bcc) arrangement. Each Cs+ion is surrounded by 8 Cl– ions and each Cl– ion is surrounded by 8 Cs+ _{ions i.e. this structure has 8:8} coordination. A unit cell of CsCl consists of only one unit of CsCl i.e. One Cs+ _{ion and one Cl}– _{ion. Few} examples of compounds having CsCl structure are CsBr, CsI, CsCN, TlCl, TlBr, TlI and TlCN.

Cs+ ion surrounded b by 8 Cl- ions

Structure of Ionic Compounds of the Type AB2: These are the ionic compounds having

cations and anions in the ratio 1:2. Most of these compounds have calciumfluorite (CaF2) type structure. These compounds have cubic close packing (CCP) arrangement in which Ca2+_{ions are present at the corners and the} centre of each face of the cube.

Each Ca2+ _{ion is surrounded by 8F}– _{ions i.e. it has a} coordination number of 8 whereas each F– _{ion is surrounded} by 4 Ca2+_{ions i.e. has a coordination number of 4. Thus this} structure has 8:4 coordination. Few examples of such compounds having CaF2 structure are BaF2, BaCl2, SrF2, SrCl2, CdF2, PbF2and ThO2.  2 Ca  F

In Na2O each oxide ion is coordinated to 8 Na+ ions and each Na+ ion to 4 oxide ions. Hence it has 4:8 coordination. This is called antifluorite 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 the sum 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 radius of 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 that with the removal of one (or more) electrons from the valence shell of the atom, the effective nuclear charge increases because now, it is acting on a smaller number of electrons, making the electron cloud pulled more inward towards the nucleus than before.

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

 2 Zn  2 S ions Zn four by lly tetrahedra surrounded ion S 2 2  

nuclear charge is now acting on a larger number of electrons. Consequently, the force of attraction on the electron 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 in the valence shell remains the same. In the same period, the radius of isoelectronic ions (having same number of electrons) 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 the maximum and minimum sizes of the anions filling the voids?

Solution: For closed packed AB type solid

r r   = 0.414 – 0.732 Minimum value of r–₌ r 75 0.732 0.732   pm = 102.5 pm Maximum value of r–₌ r 75 0.414 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 rocksalt structure.

Imperfection in Solids

An ionic crystal which has the same unit cell containing the same lattice points throughout the crystal is known as ideal crystal but crystals tend to have a perfectly ordered arrangement at only absolute zero. This arrangement corresponds to state of lowest energy. And as the temperature increases, the crystals start deviating from 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 a lattice 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 points are called vacancies or holes fig. (a). The number of missing positive and negative ions is the same so that the crystal remains neutral in all. Schottky defects are more common in ionic compounds with high coordination number, and where the sizes of positive and negative ions are almost equal for example, NaCl, KCl, CsCl and KBr.

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

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

In pure alkali metal halides, these defects are not very common because the ions cannot get into interstitial positions due to their 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 to form Schottky defects. When compounds such as NaCl, KCl, are heated with excess of their constituent metal vapours, or treated with high energy radiation, they become deficient in the negative ions and their formulae may be represented by AX1–, where is a small fraction. The crystal lattice has vacant anion sites which are

occupied by electrons. Anion sites occupied by electrons in this way are called F centres (F is an abbreviation Farbe, the German word for colour).

Interstitial ions and electrons: Metal excess defects also occur when an extra positive ion occupies an interstitial 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 much

more 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 an electric current. These free electrons may be excited to higher energy levels, giving absorption spectra and in consequence their compounds are often coloured e.g. non-stoichiometric NaCl is yellow, non-stoichiometric KCl 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 by the missing cation from its lattice site. The extra negative charge may be balanced if the adjacent metal ion has higher positive charge. This can be possible by compounds of transition metals (variable valency). Crystals of FeO, FeS and NiO show this defect.

Properties of Solids

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

a) Metals (conductors) b) Insulators

c) Semiconductors

Electrical conductivity of metals is very high and is of the order of 106_{– 10}8_ohm–1_cm–1_{whereas for insulators,} it is of the order of 10–12_ohm–1_cm–1_{. Semiconductors have intermediate conductivity in the range of 10}2_{– 10}–9 ohm–1_cm–1_{. Electrical conductivity of solids may arise through the motion of electrons and holes (positive) or} through the motion of ions. The conduction through electrons is called ntype conduction and through (positive) holes is called ptype conduction. Pure ionic solids where conduction can take place only through movement of ions are insulators. The presence of defects in the crystal structure increases their conductivity. The conductivity of semiconductors and insulators is mainly due to the presence of interstitial electrons and positive holes in the solids due to imperfections. The conductivity of semiconductors and insulators increases with 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 formed from 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 must be that 3 molecular orbitals will be formed.

 Now , let's take it to 10 atoms. As the number of molecular orbitals increases, the energy difference between the lowest bonding and the highest anti-bondig increases, while the space between each

individual orbital decreases  Consider a metal with an infinite number of atoms. This will form an

A+ _B- _A+ _B -A+ A+ _A+ A+ A2+ B -B -B -B -B -A+ A+ B -B -B

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

Conductivity according to MOT is related to ease of availability of conduction [empty band ] band for movement of electrons from valance band 1.valance band : highest occupied band and 2. Conduction band : lowest un occupied band

Energy gap of semiconducter

According to MOT the substance in which band gap is intermediate of Conductors and insulator

and small as well, which make promotion of electron from valance band to conduction band possible

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

n-type Semiconductor Doping with an element of extra valence electrons into a element. There is NO extra room for these electrons in the valence band; consequently, they are promoted into the conduction band, where they have access to many vacant orbitals within the energy band they occupy and serve as electrical carrierse.g.Silicon (4v es-)doped with phosphorous(5 ves-)

In general 14 th group element with 15th _{group element}

p-type Semiconductor : Doping with an element of less electrons in order to create electron vacancies or positive holes in the system. Because the valence band is incompletely filled, under the influence of an applied field, electrons can move from occupied molecular orbitals to the few that are vacant, thereby allowing current to flow.

e.g.

In general 14th group element with 13th _{group element}

Magnetic Properties

n

PARAMGNETISM

The materials which are weakly attracted by magnetic field are called paramagnetic materials. These materials have permanent magnetic dipoles due to presence of atoms, ions or molecules with unpaired electron. e.g. O2,

Cu2+_{, Fe}2+ _{etc. But these materials lose their magnetism in the absence of}

magnetic field.

Ferromagnetic Materials: The materials which show permanent magnetism even in the absence of

magnetic field are called ferromagnetic materials. These materials are strongly attracted by the magnetic field. e.g. Fe, Co, Ni and CrO2. Ferromagnetism arises due to spontaneous alignment of magnetic moments of ions or atoms in the same direction.

Alignment of magnetic moments in opposite directions in a compensatory manner and resulting in zero magnetic moment gives rise to antiferromagnetism

ferromagnetism

;

Anti ferromagnetism ;Ferrimagnetism

forexample, MnO, Mn2O3and MnO2.to unequal number of parallel and antiparallel magnetic dipoles give rise to ferrimagnetism e.g. Fe3O4.

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

Property Definitions Exam ple Diam agnetic Repelled by ex. M F.

becauseabsence of unpaired electron

NaCl Param agnetic Weakly attracted and shows m agnetism

Only in presence of ex. M F

O2,Cu2+

Ferrom agnetic Strongly attracted and shows m agnetism even in absence of ex.M F

Fe,Co,Ni

Ferrim agnetic Observed m agnetic m om ent is lesser than calculated because unequal alignm ent

Fe3O4

Antiferrom agn

etic Observed m agnetic m om ent is zerobecause of exactly equal no opposite alignm ent

M nO

Dielectric Properties: The electrons in insulators are closely bound to the individual atoms or ions and thus they do not generally migrate under the applied electric field. However, due to shift in charges, dipoles are created which results in polarisation. The alignment of these dipoles in different ways i.e. compensatory way (zero dipole) or noncompensatory 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 exhibit piezoelectricity 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 due to 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 pyroelectric effect.

The solids, in which dipoles are spontaneously aligned in a particular direction, even in the absence of electric field are called ferroelectric substances and the phenomenon is known as ferroelectricity. If the alternate dipoles are in opposite direction, then the net dipole moment will be zero and the crystal is called antiferroelectric .

Ferroelectric solids – Barium titanate (BaTiO3), sodium potassium tartrate (Rochelle salt) and potassium hydrogen phosphate (KH2PO4). Antiferroelectric – 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 if 200g 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 the density 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 is 7.2 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. What} would 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 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?

6. When heated above 916°C, iron changes its crystal structure from bodycentred cubic to cubic closed packed structure. Assuming that the metallic radius of the atom does not change, calculate the ratio of density 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 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.

8. CsCl has cubic structure of ions in which Cs+_{ion is present in the body centre of the cube. If density is} 3.99g 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 of 288 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 is 400 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 while cations 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 NaCl} crystal = 281 pm

3. In a cubic close packed structure of mixed oxides, the lattice in made up of oxide ions, oneeighth of tetrahedral voids are occupied by divalent ions (X2+_{) while one half of octahedral voids are occupied by} trivalent 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 of sodium 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)

b) What is the edgelength if r_K= 1.33Å, r_Br= 1.95Å?

c) What minimum value of r+_/r–_{is needed to prevent anionanion contact in this structure?}

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

9. Compute the limiting radius ratio r+/r–for (a) a square planar crystal structure in which B–ions are at the corner 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 value of the radius of the anion.

Answer to Assignments (Subjective)

LEVEL – I 1. 41.6 g cm–3 _{2. 8.9 g/cm}3 3. 24.161023atom 4. 3 cm 5. 4, fcc 6. 0.918 7. 0.016 g/cm3_{, 0.775% 8. a) 412 pm; b) 356.8 pm; c) 176.8 pm} 9. 6.041023 _{10. 6.2 g/cm}3 LEVEL – II 1. 12.5%, 50% 2. 6.021023 3. AB2O4 4. 2, 0.96 gcm–3 5. 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