Chapter 1 The Solid State by TEACHING CARE Online coaching and tuition classes

 

 

We know solids are the substances which have definite volume and definite shape. A solid is nearly incompressible state of matter. This is because the particles or units (atoms, molecules or ions) making up the solid are in close contact and are in fixed positions or sites. Now, let us study some characteristic properties of solids.

Solids can be distinguished from liquids and gases due to their characteristic properties. Some of these are as follows:

• Solids have definite volume, irrespective of the size of the
• Solids are rigid and have definite
• Solids are almost
• Many solids are crystalline in These crystals have definite pattern of angles and planes.
• The density of solids is generally greater than that of liquids and
• Solids diffuse very slowly as compared to liquids and
• Most solids melt on heating and become The temperature at which the solid melts and changes into liquid state under normal atmospheric pressure is called its normal melting point.
• Solids are not always crystalline in nature.
• Solids can be broadly classified into following two types :
• Crystalline solids/True solids (ii) Amorphous solids/Pseudo solids

(1)  Difference between crystalline and amorphous solids

 

Property	Crystalline solids	Amorphous solids
Shape Melting point Heat of fusion Compressibility   Cutting with a sharp edged tool Isotropy                  and Anisotropy Volume change   Symmetry Interfacial angles	They have long range order. They have definite melting point They have a definite heat of fusion They are rigid and incompressible   They are given cleavage i.e. they break into two pieces with plane surfaces They are anisotropic   There is a sudden change in volume when it melts. These possess symmetry These possess interfacial angles.	They have short range order. They do not have definite melting point They do not have definite heat of fusion These may not be compressed to any appreciable extent They are given irregular cleavage i.e. they break into two pieces with irregular surface They are isotropic There is no sudden change in volume on melting. These do not possess any symmetry. These do not possess interfacial angles.

Note : ® Isomorphism and polymorphism : Two subtances are said to be isomorphous if these possess similar

 

crystalline form and similar chemical composition e.g.,

Na2 SeO4

and

Na2 SO4 .

NaNO3

and

KNO₃ are not isomorphous

 

because they have similar formula but different crystalline forms. The existence of a substance in more than one crystalline form is known as polymorphism e.g., sulphur shows two polymorphic forms viz. rhomibic and monoclinic sulphur.

• Glass is a supercooled

 

 

• Classification of solids : Depending upon the nature of interparticle forces the solids are classified into four types :

 

Types of Solid	Constituents	Bonding             Examples	Physical Nature	M.P.	B.P.	Electrical Conductivity
Ionic	Positive and negative ions network	Coulombic        NaCl, KCl, CaO, MgO,  LiF,  ZnS, BaSO4                and K2SO4 etc.   Electron                SiO2 (Quartz), sharing                SiC,                             C (diamond), C(graphite) etc. (i)                                I2,S8, P4, CO2, Molecular         CH4, CCl4 etc. interactions (intermolec u-lar forces)    Starch, sucrose, (ii)                               water, dry ice or Hydrogen              drikold (solid bonding                                       CO2) etc.   Metallic                 Sodium  , Au, Cu, magnesium, metals and alloys London                  Noble gases dispersion force	Hard but brittle	High (≃1000K)	High (≃2000K)	Conductor (in molten state and in aqueous
	systematically					solution)
	arranged
Covalen	Atoms		Hard	Very high	Very high	Insulator       except
t	connected                                    in covalent bonds		Hard	(≃4000K)	(≃500K)	graphite
			Hard
Molecul ar	Polar or non- polar molecules		Soft	Low   (≃300K                            to 600K)	Low (≃ 450 to 800 K)	Insulator
			Soft	Low (≃400K)	Low	Insulator
					(≃373K to
					500K)
Metallic	Cations  in  a sea of electrons		Ductile malleable	High (≃800K to 1000 K)	High (≃1500K to 2000K)	Conductor
Atomic	Atoms		Soft	Very low	Very low	Poor thermal and
						electrical
						conductors

 

• Liquid Crystal : There are certain solids which when heated undergo two sharp phase transformations one after the Such solids first fuse sharply yielding turbid liquids and then further heating to a higher temperature these sharply change into clear liquids. The first temperature at which solids changes into turbid liquid is known as transition point and the second temperature at which turbid liquid changes into clear liquid is known as melting point. Such substances showing liquid crystal character are as follows :

p-chloesteryl benzoate, p – Azoxyamisole, Diethylbenzidine etc.

 

 

p- Chloesteryl benzoate

(Solid)

145^oC

p – Chloesteryl benzoate

(liquidcrystal )

178^oC

p – Chloesteryl benzoate

(Liquid)

 

 

A liquid crystal reflects only one colour, when light falls on it. If the temperature is changed it reflects different colour light. So, such liquid crystals can be used to detect even small temperature changes. The liquid crystals are of two types : (i) Nematic liquid crystals, (needle like), (ii) Smectic liquid crystals (soap like)

• Dispersion forces or London forces in solids : When the distribution of electrons around the nucleus is not symmetrical then there is formation of instantaneous electric Field produced due to this distorts

 

 

the electron distribution in the neighbouring atom or molecule so that it acquires a dipole moment itself. The two dipole will attract and this makes the basis of London forces or dispersion forces these forces are attractive in nature

 

and the interaction energy due to this is proportional to

æ 1 ö . Thus, these forces are important as short distances

ç 6 ÷

r

 

è     ø

(~-500 pm). This force also depends on the polarisability of the molecules.

4

• Amorphous Solids (Supercooled liquid) : Solids unlike crystalline solids, do not have an ordered arrangement of their constituent atoms or ions but have a disordered or random arrangement, are called amorphous solids. Ordinary glass (metal silicate), rubber and most of the plastics are the best examples of amorphous In fact, any material can be made amorphous or glassy either by rapidly cooling or freezing its

 

4

vapours for example,

SiO2

crystallises or quartz in which

SiO4-

tetrahedra are linked in a regular manner but on

 

melting and then rapid cooling, it gives glass in which

Properties of Amorphous solids

SiO4-

tetrahedron are randomly joined to each other.

 

• Lack of long range order/Existence of short range order : Amorphous solids do not have a long range order of their constituent atoms or However, they do have a short range order like that in the liquids.
• No sharp melting point/Melting over a
• Conversion into crystalline form on

Uses of Amorphous solids

• The most widely used amorphous solids are in the inorganic glasses which find application in construction, house ware, laboratory ware
• Rubber is amorphous solid, which is used in making tyres, shoe soles
• Amorphous silica has been found to be the best material for converting sunlight into electricity (in

photovoltaic cells).

“The branch of science that deals with the study of structure, geometry and properties of crystals is called crystallography”.

• Laws of crystallography : Crystallography is based on three fundamental Which are as follows
  • Law of constancy of interfacial angles : This law states that angle between adjacent corresponding faces of the crystal of a particular

substance is always constant inspite of different shapes and sizes. The size and shape of crystal depend upon the conditions of crystallisation. This law is also known as Steno’s Law.

• Law of rational indices : This law states that the intercepts of any face of a crystal along the crystallographic axes are either equal to unit intercepts (i.e., intercepts made by unit cell) a, b, c or some simple whole number multiples of them e.g., na, n‘ b, n”c, where n, n‘ and n” are simple whole numbers. The whole numbers n, n‘ and n” are called Weiss This law was given by Hally.
• Law of constancy of symmetry : According to this law, all crystals of a substance have the same elements of

 

 

• Designation of planes in crystals (Miller indices) : Planes in crystals are described by a set of integers (h, k and l) known as Miller Miller indices of a plane are the reciprocals of the fractional intercepts of that plane on the various crystallographic axes. For calculating Miller indices, a reference plane, known as parametral plane, is selected having intercepts a, b and c along x, y and z-axes, respectively. Then, the intercepts of the unknown plane are given with respect to a, b and c of the

parametral plane.

Thus, the Miller indices are :

h =                            a

intercept of the plane along x – axis

b

k = intercept of the plane along y – axis

l =                            c

intercept of the plane along z – axis

Consider a plane in which Weiss notation is given by

¥a : 2b : c . The Miller indices of this plane may be calculated as below.

 

• Reciprocals of the coefficients of Weiss indices = 1,

¥

1 , 1

2 1

 

• Multiplying by 2 in order to get whole numbers = 0,1, 2

Thus the Miller indices of the plane are 0, 1, and 2 and the plane is designated as the (012) plane, i.e.

k = 1, l = 2 .

 

h = 0 ,

 

The distance between the parallel planes in crystals are designated as interplanar spacing are given by the general formula,

dhkl . For different cubic lattices these

 

 

Where a is the length of cube side while h, k and l are the Miller indices of the plane.

Note : ® When a plane is parallel to an axis, its intercept with that axis is taken as infinite and the Miller will be zero.

• Negative signs in the Miller indices is indicated by placing a bar on the
• All parallel planes have same Miller
• The Miller indices are enclosed within i.e., brackets. Commas can be used for clarity.

 

Example 1:         Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c), (ii) ( ¥, 2b, c ) (a) 3, 2, 6 and 0, 1, 2                     (b) 4, 2, 6 and 0, 2, 1          (c) 6, 2, 3 and 0, 0, 1       (d) 7, 2, 3 and 1, 1, 1

Solution: (a)

 

(i) x	y	z	(ii)	x	y	z
2a	3b	c      Intercepts		¥	2b	c     Intercepts

 

 

 

 

2a       3b

 

 

a          b

1       1

c     Lattice parameters

c

2   3   1   ¥   2   1   3   2   6 Multiplying by LCM (6) 0   1   2 Multiplying by LCM (2)

1    Reciprocals

¥         2b

 

 

a          b

1       1

c     Lattice parameters

c

1    Reciprocals

 

 

 

Hence, the Miller indices are (3, 2, 6)                                   Hence, the Miller indices are (0, 1, 2).

Example 2.         Caculate the distance between 111 planes in a crystal of Ca. Repeat the calculation for the 222 planes. (a=0.556nm)

• 1 nm (b) 01.61 nm                        (c) 0.610 nm                        (d) None of the above

 

Solution:(b)      We have, d =              a            ; d

111 =

0.556

= 0.321nm

and d

222 =

0.556

= 0.161nm

 

 

The separation of the 111 planes is twice as great as that of 222 planes.

 

• Crystal : It is a homogeneous portion of a crystalline substance, composed of a regular pattern of structural units (ions, atoms or molecules) by plane surfaces making definite angles with each other giving a regular geometric
• Space lattice and Unit cell : A regular array of points (showing atoms/ions) in three dimensions is commonly called as a space lattice, or
• Each point in a space lattice represents an atom or a group of
• Each point in a space lattice has identical surroundings

A three dimensional group of lattice points which when repeated in space generates the crystal called unit cell. The unit cell is described by the lengths of its edges, a, b, c (which are related to the spacing between layers)

and the angles between the edges, a, b ,g .

 

• Symmetry in Crystal systems : Law of constancy of symmetry : According to this law, all crystals of a substance have the same elements of symmetry. A crystal possess following three types of symmetry :
  • Plane of symmetry : It is an imaginary plane which passes through the centre of a crystal can divides it into two equal portions which are exactly the mirror images of each

 

 

 

 

 

 

 

• Axis of symmetry : An axis of symmetry or axis of rotation is an imaginary line, passing through the crystal such that when the crystal is rotated about this line, it presents the same appearance more than once in one complete revolution i.e., in a rotation through 360°. Suppose, the same appearance of crystal is repeated, on rotating it through an angle of 360°/n, around an imaginary axis, is called an n-fold axis where, n is known as the

 

order of axis. By order is meant the value of n in

2p / n

so that rotation through

2p / n,

gives an equivalent

 

configuration. For example, If a cube is rotated about an axis passing perpendicularly through the centre so that the similar appearance occurs four times in one revolution, the axis is called a four – fold or a tetrad axis, [Fig (iii)]. If similar appearance occurs twice in one complete revolution i.e., after 180°, the axis is called two-fold axis of symmetry or diad axis [Fig (i)]. If the original appearance is repeated three times in one revolution i.e. rotation after 120°, the axis of symmetry is called three-fold axis of symmetry or triad axis [Fig (ii)]. Similarly, if the original appearance is repeated after an angle of 60° as in the case of a hexagonal crystal, the axis is called six-fold axis of symmetry or hexad axis [Fig (iv)].

 

 

• Centre of symmetry : It is an imaginary point in the crystal that any line drawn through it intersects the surface of the crystal at equal distance on either

Note : ® Only simple cubic system have one centre of symmetry. Other system do not have centre of symmetry.

• Element of symmetry : (i) The total number of planes, axes and centre of symmetries possessed by a crystal is termed as elements of

(ii) A cubic crystal possesses total 23 elements of symmetry.

 

 

(a) Plane of symmetry	( 3 + 6)	=	9
(b) Axes of symmetry	( 3 + 4 + 6)	=	13
(c) Centre of symmetry	(1)	=	1

Total symmetry = 23

• Formation of crystals : The crystals of the substance are obtained by cooling the liquid (or the melt) of the solution of that The size of the crystal depends upon the rate of cooling. If cooling is carried out slowly, crystals of large size are obtained because the particles (ions, atoms or molecules) get sufficient time to arrange themselves in proper positions.

 

(If loosing units dissolves as embryo and if gaining unit grow as a crystals).

• Crystal systems : Bravais (1848) showed from geometrical considerations that there can be only 14 different ways in which similar points can be arranged. Thus, there can be only 14 different space lattices. These 14 types of lattices are known as Bravais Lattices. But on the other hand Bravais showed that there are only seven types of crystal The seven crystal systems are :

 

(a) Cubic	(b) Tetragonal	(c) Orthorhombic	(d) Rhombohedral
(e) Hexagonal	(f) Monoclinic	(g) Triclinic

 

Bravais lattices corresponding to different crystal systems

 

Crystal system	Space lattice	Examples
Cubic	Simple  :  Lattice  points  at     Body  centered  :  Points     Face    centered      : the   eight   corners   of   the   at the eight corners and    Points  at  the  eight unit cells.                                                          at the body centred.               corners and at the six face centres.           Simple    :    Points    at    the      Body centered : Points at the eight corners and eight  corners  of  the  unit       at the body centre cell.       Simple:    Points      End  centered  :  Also      Body  centered      Face       centered:	Pb, Hg, Ag,
a = b = c ,		Au, Cu, ZnS ,
Here a, b and c are parameters (diamensions of a unit cell along three axes) size of crystals depend on parameters.		diamond, KCl,  NaCl, Cu2O,CaF2 and alums. etc.
a = b = g = 90o
ab and g are sizes of three                 angles between the axes.
Tetragonal		SnO2, TiO2,
a = b ¹ c ,		ZnO2, NiSO4
a = b = g = 90o		ZrSiO4 . PbWO4 ,
		white Sn etc.
Orthorhombic		KNO3 , K2SO4 ,

 

 

 

 

(Rhombic)

 a ¹ b ¹ c ,

at      the      eight corners of the

called side centered or   base        centered.

: Points at the eight          corners

Points      at      the eight  coreners

 PbCO₃, BaSO₄ ,

rhombic sulphur,

 

a = b = g = 90 ^o

unit cell.

Points at the eight corners and at two face centres opposite to each other.

and      at       the body centre

and at the six face centres.

 MgSO₄ . 7H₂O  etc.

 

 

 

 

 

 

 

 

 

 

Rhombohedral or Trigonal

 a = b = c ,

a = b = g ¹ 90^o

Simple : Points at the eight corners of the unit cell

 NaNO₃, CaSO₄ ,

calcite, quartz,

 As, Sb, Bi etc.

 

 

Hexagonal

 a = b ¹ c ,

a = b = 90^o

g = 120^o

Simple : Points at the twelve corners of the unit cell out lined  by  thick line.

or Points at the twelve corners of the hexagonal prism and at  the  centres of the two hexagonal faces.

 ZnO, PbS, CdS,

 HgS, graphite, ice,

 Mg, Zn, Cd etc.

 

 

 

 

 

 

 

 

 

 

Monoclinic

 a ¹ b ¹ c ,

a = g = 90^o, b ¹ 90^o

Simple : Points at the eight  corners of the unit cell

End centered : Point at the eight corners and at two face centres opposite to the each other.

 Na2SO4  .10H2O,

 Na2 B4 O7 .10H2O,

 CaSO₄ .2H₂O,

monoclinic sulphur etc.

 

 

 

 

Triclinic

 a ¹ b ¹ c ,

a ¹ b ¹ g ¹ 90^o

Simple : Points at the eight corners of the unit cell.

 CaSO₄ .5H₂O,

 K2Cr2O7 , H3 BO3

etc.

 

 

 

 

 

 

 

 

Note :           ® Out of seven crystal systems triclinic is the most unsymmetrical ( a ¹ b ¹ c,

a ¹ b ¹ g

¹ 90).

 

 

 

1

• Close packing in crystalline solids : In the formation of crystals, the constituent particles (atoms, ions or molecules) get closely packed together. The closely packed arrangement is that in which maximum available space is

occupied. This corresponds to a state of maximum density. The closer the packing, the greater is the stability of the packed system. It is of two types :

• Close packing in two dimensions : The two possible arrangement of close packing in two

1

• Square close packing : In which the spheres in the adjacent row lie just one over the other and show a horizontal as well as vertical alignment and form Each sphere in this arrangement is in contact with four spheres.
• Hexagonal close packing : In which the spheres in every second row are seated in the depression between the spheres of first row. The spheres in the third row are vertically aligned with spheres in first row. The similar pattern is noticed throughout the crystal structure. Each sphere in this arrangement is in contact with six other

Note : ® Hexagonal close packing is more dense than square close packing.

• In hexagonal close packing about 4% of available space is occupied by spheres. Whereas, square close packing occupies only 52.4% of the space by spheres.
• In square close packing the vacant spaces (voids) are between four touching spheres, whose centres lie at the corners of a square are called square voids. While in hexagonal close packing the vacant spaces (voids) are between three touching spheres, whose centres lie at the corners of an equilateral triangle are called triangular voids.
• Close packing in three dimensions : In order to develop three dimensional close packing, let us retain the hexagonal close packing in the first For close packing each

spheres in the second layer rests in the hollow at the centre of three

touching spheres in the layer as shown in figure. The spheres in the first layer are shown by solid lines while those in second layer are shown by broken lines. It may be noted that only half the triangular voids in the first layer are occupied by spheres in the second layer (i.e., either b or c). The unoccupied hollows or voids in the first layer are indicated by (c) in figure.

There are two alternative ways in which species in third layer can be arranged over the second layer,

• Hexagonal close packing : The third layer lies vertically above the first and the spheres in third layer rest in one set of hollows on the top of the second layer. This arrangement is called ABAB …. type and 74% of the available space is occupied by

 

 

 

• Cubic close packing : The third layer is different from the first and the spheres in the third layer lie on the other set of hollows marked ‘C’ in the first This arrangement is called ABCABC….. type and in this also 74% of the available space is occupied by spheres. The cubic close packing has face centred cubic (fcc) unit cell.

This arrangement is found in Be, Mg, Zn, Cd, Sc, Y, Ti, Zr.

This arrangement is found in Cu, Ag, Au, Ni, Pt, Pd, Co, Rh, Ca, Sr.

• Body centred cubic (bcc) : This arrangement of spheres (or atoms) is not exactly close packed. This structure can be obtained if spheres in the first layer

(A) of close packing are slightly opened up. As a result none of these spheres are in contact with each other. The second layer of spheres (B) can be placed on top of the first layer so that each sphere of the second layer is in contact with four spheres of the layer below it. Successive building of the third will be exactly like the first layer. If this pattern of building layers is repeated infinitely we get an

arrangement as shown in figure. This arrangement is found in Li, Na, K, Rb, Ba, Cs, V, Nb, Cr, Mo, Fe.

 

 

 

• Comparison of hcp, ccp and bcc

 

Property	Hexagonal close packed (hcp)	Cubic close packed (ccp)	Body centred cubic (bcc)
Arrangement of packing	Close packed	Close packed	Not close packed
Type of packing	AB AB AB A…..	ABC ABC A….	AB AB AB A……
Available                     space	74%	74%	68%
occupied
Coordination number	12	12	8
Malleability and ductility	Less malleable, hard and brittle	Malleable and ductile

• Interstitial sites in close packed structures : Even in the close packing of spheres, there is left some empty space between the This empty space in the crystal lattice is called site or void or hole. Voids are of following types

• Trigonal void : This site is formed when three spheres lie at the vertices of an equilateral Size of the trigonal site is given by the following relation.

where,   r = Radius of the spherical trigonal site

R = Radius of closely packed spheres

 

 

 

• Tetrahedral void : A tetrahedral void is developed when triangular voids (made by three spheres in one layer touching each other) have contact with one sphere either in the upper layer or in the lower This type of void is surrounded by four spheres and the centres of these spheres lie at the

apices of a regular tetrahedron, hence the name tetrahedral site for this void.

In a close packed structure, there are two tetrahedral voids associated with each sphere because every void has four spheres around it and there are eight voids around each sphere. So the number of tetrahedral voids is double the number of spheres in the crystal structure. The

maximum radius of the atoms which can fit in the tetrahedral voids relative to the radius of the sphere is calculated to

be 0.225: 1, i.e.,

 

r = 0.225 ,

R

where r is the radius of the tetrahedral void or atom occupying tetrahedral void and R is the radius of spheres forming tetrahedral void.

 

• Octahedral void : This type of void is surrounded by six closely packed spheres,

i.e. it is formed by six spheres. Out of six spheres, four are placed in the same plane touching each other, one sphere is placed from above and the other from below the plane of these spheres. These six spheres surrounding the octahedral void are present at the vertices of regular octahedron. Therefore, the number of octahedral voids is equal to the number of spheres. The ratio of the radius (r) of the atom or ion which can exactly fit in the octahedral void formed by spheres of radius R has been calculated to be 0.414, i.e.

 

 

 

• Cubic void : This type of void is formed between 8 closely packed spheres which occupy all the eight corner of cube e. this site is surrounded by eight spheres which touch each other. Here radius ratio is calculated to be 0.732, i.e.

Thus, the decreasing order of the size of the various voids is Cubic > Octahedral

> Tetrahedral > Trigonal

Important Tips

• At the limiting value of radius ratio r ⁺ / r ^– , the forces of attraction & repulsion are
• The most malleable metals (Cu, Ag, Au) have cubic close
• Cubic close packing has fcc (face centred cubic) unit cell
• Number of octahedral voids = Number of atoms present in the closed packed

• Number of tetrahedral voids = 2 × Number of octahedral voids = 2 × Number of

 

Simplest crystal system is to be studied in cubic system. Three types of cubic systems are following

• Simple cubic (sc) : Atoms are arranged only at the
• Body centred cubic (bcc) : Atoms are arranged at the corners and at the centre of the cube.
• Face centred cubic (fcc) : Atoms are arranged at the corners and at the centre of each
• Atomic radius : It is defined as the half of the distance between nearest neighbouring atoms in a It is expressed in terms of length of the edge (a) of the unit cell of the crystal.
  • Simple cubic structure (sc) : Radius of atom ‘r‘ = a

2

(ii)   Face centred cubic structure (fcc) : ‘r‘ =  a 

2

(iii)   Body centred cubic structure (bcc) : ‘r‘ =  3a

4

• Number of atoms per unit cell/Unit cell contents  : The total number of atoms contained in the unit cell for a simple cubic called the unit cell
  • Simple cubic structure (sc) : Each corner atom is shared by eight surrounding Therefore, it

 

contributes for

1 of an atom. \

8

Z = 8 ´ 1 = 1 atom per unit cell in crystalline solid.

8

 

 

 

 

• Face centered cubic structure (fcc) : The eight corners atoms contribute for

1 of an atom and thus

8

 

one atom per unit cell. Each of six face centred atoms is shared by two adjacent unit cells and therefore one face

 

centred atom contribute half of its share. \ Z = 6 ´ 1 = 3

2

atom per unit cell.

 

So, total Z = 3 + 1 = 4 atoms per unit cell.

• Body centered cubic structure (bcc) : Eight corner atoms contribute one atom per unit

 

 

Centre atom contribute one atom per unit cell. So, total 1 + 1 = 2 atoms per unit cells.

Z = 8 ´ 1 + 1 = 2

8

 

Note : ® Number of atoms in unit cell : It can be determined by the simplest relation = nc + nf + ni

 

 

Where

nc =

Number of atoms at the corners of the cube = 8

8     2     1

 

nf   =

n i =

Number of atoms at six faces of the cube = 6 Number of atoms inside the cube = 1

 

 

Cubic unit cell	nc	nf	ni	Total atom in per unit cell
Simple cubic (sc)	8	0	0	1
body centered cubic (bcc)	8	0	1	2
Face centered cubic (fcc)	8	6	0	4

• Co-ordination number (C.N.) : It is defined as the number of nearest neighbours or touching particles with other particle present in a crystal is called its co-ordination It depends upon structure of the crystal.
  • For simple cubic system N. = 6.
  • For body centred cubic system N. = 8
  • For face centred cubic system N. = 12.
• Density of the unit cell : It is defined as the ratio of mass per unit cell to the total volume of unit

Density of unit cell (r) = mass of unit cell ; r = Number of particles ´ mass of each particle or Z ´ M  

 

0

volume of unit cell

volume of the unit cell

a 3 ´ N

 

Where Z = Number of particles per unit cell, M = Atomic mass or molecular mass,

N 0 = Avogadro number

 

(6.023 ´ 10 23 mol -1 ) , a =

 

i.e.

Edge length of the unit cell= a  pm = a ´ 10 -10 cm ,

a 3 = volume of the unit cell

 

 

The density of the substance is same as the density of the unit cell.

• Packing fraction (P.F.) : It is defined as ratio of the volume of the unit cell that is occupied by spheres of the unit cell to the total volume of the unit

Let radius of the atom in the packing = r

Edge length of the cube = a

 

 

 

 

Volume of the cube V = a³

 

Volume of the atom (spherical) n =

4 pr ³ , then packing density = nZ

3                                     V

4 pr 3 Z

= 3      

a3

 

• Simple cubic unit cell : Let the radius of atom in packing is r. Atoms are present at the corner of the cube, each of the eight atom present at the eight corners shared amongst eight unit

 

Hence number of atoms per unit cell = 8 ´ 1

8

4 pr ³

= 1 , again r = a

2

 

\ P.F. = 3    = 0.52 ; % P.F. = 52%, then % of void = 100 – 52 = 48%

(2r)³

• Body centred cubic unit cell : Number of atoms per unit cell = 8 ´ 1 + 1 = 2 , r = 3a

8                  4

2 ´ 4 pr 3

P.F. =     3     = 0.68 ; % P.F. = 68%, then % of void = 100 – 68 = 32%

æ 4r ö3

 

• Face centred cubic unit cell : Number of atoms per unit cell = 4, r = 2a

4

4 ´ 4 pr 3

P.F. =      3     = 0.74 ; % P.F. = 74% , then % of void = 100 – 74=26%

æ 4r ö3

 

• Ionic radii : X-ray diffraction or electron diffraction techniques provides the necessary information regarding unit From the dimensions of the unit cell, it is possible to calculate ionic

radii.

rc + ra = a / 2

Let, cube of edge length ‘a‘ having cations and anions say NaCl structure. Then,

 

where rc

and ra

are radius of cation and anion.

 

 

 

Radius of Cl ^– =                              = a

3a

4

 

 

For body centred lattice say CsCl.

rc + ra   =    2

 

Radius ratio : Ionic compounds occur in crystalline forms. Ionic compounds are made of cations and anions. These ions are arranged in three dimensional array to form an aggregate of the type (A⁺B^–)_n . Since, the Coulombic forces are non-directional, hence the structures of such crystals are mainly governed by the ratio of the

 

radius of cation (r+ ) to that of anion (r– ). The ratio r+

to r–

(r+ / r– ) is called as radius ratio.

 

 

Radius ratio =	r+
	r–

 

The influence of radius ratio on co-ordination number may be explained as follows : Consider an ideal case of octahedral voids in close packing of anions with radius ratio 0.414 and co-ordination number six. An increase in size of cation increases the radius ratio from 0.414, then the anions move apart so as to accommodate the larger cation. As the radius ratio increases more and more beyond 0.732, the anions move further and further apart till a stage is obtained when more anions can be accommodated and this cation occupies a bigger void i.e., cubic void with co-ordination number eight.

When the radius ratio decreases from 0.414, the six anions would not be able to touch the smaller cation and in doing so, they overlap each other. This causes the cation to occupy a smaller void i.e., tetrahedral void leading to co-ordination number four

Limiting Radius ratios and Structure

 

Limiting radius ratio (r+)/(r–)	C.N.	Shape
< 0.155	2	Linear
0.155 – 0.225	3	Planar triangle
0.225 – 0.414	4	Tetrahedral
0.414 – 0.732	6	Octahedral
0.732 – 0.999 or 1	8	Body-centered cubic

 

 

 

 

 

Characteristics of Some Typical Crystal Structure

 

Crystal	Type of unit cell	Example		r+   r–	C.N.	Number of formula units of (AB, or AB2) per unit cell
CsBr	Body-centred	CsBr, TiCl		0.93	8 – 8	1
NaCl	Face-centred	AgCl, MgO		0.52	6 – 6	4
ZnS	Face-centred	ZnS		0.40	4 – 4	4
CaF2	Face-centred	CaF2,      SrF2,	CdF2,	0.73	8 – 4	4
		ThO2

Note : ® The ionic radius increases as we move from top to bottom in a group of periodic table for example :

Na ⁺ < K ⁺ < Rb⁺ < Cs⁺ and F ^– < Cl^– < Br^– < I ^–

 

• Along a period, usually iso-electronic ions are obtained g.

Na⁺, Mg²⁺, Al³⁺

(greater the nuclear charge, smaller the

 

size, Al³⁺ < Mg ²⁺ < Na⁺ )

 

Example : 3       A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB………………… Any packing

of spheres leaves out voids in the lattice. The percentage by volume of empty space of this is (a) 26%                                    (b) 21%                          (c) 18%                           (d) 16 %

Solution :(a)     The hexagonal base consists of six equilateral triangles, each with side 2r and altitude 2r sin 60°.

 

Hence, area of base = 6 é 1 (2r)(2r sin 60 ^o )ù = 6

3.r ²

 

êë 2                                             úû

The height of the hexagonal is twice the distance between closest packed layers.

The latter can be determined to a face centred cubic lattice with unit cell length a. In such a lattice, the distance

between closest packed layers is one third of the body diagonal, i.e. 3a , Hence

3

Height (h) = 2 é 3a ù = 2a

ë

û

êê 3 ú

Now, in the face centred lattice, atoms touch one another along the face diagonal,

 

Thus,

4r =

2 .a

 

 

With this, the height of hexagonal becomes :

Height (h) = 2 é 4r ù = é4 2 ù .r

 

2

ê       ú     ê

ë       û     êë

ú

3 úû

 

 

Volume of hexagonal unit is, V = (base area) ´ (height) = (6

3 r ² ) é 4  

ê

êë

ù

.rú = 24

úû

2.r ³

 

In one hexagonal unit cell, there are 6 atoms as described below :

• 3 atoms in the central layer which exclusively belong to the unit
• 1 atom from the centre of the There are two atoms of this type and each is shared between two hexagonal unit cells.
• 2 atoms from the There are 12 such atoms and each is shared amongst six hexagonal unit cells.

 

 

 

Now, the volume occupied by atoms = 6é 4 pr ³ ù

êë 3             úû

6æ 4 pr ³ ö

 

Fraction of volume occupied by atoms =

ç

  Volume occupied by atoms  =  è

Volume of hexagonal unit cell

÷

3

ø = p / 3

= 0.74.

 

Fraction of empty space = (1.00 – 0.74) = 0.26

Percentage of empty space = 26%

Example : 4       Silver metal crystallises in a cubic closest – packed arrangement with the edge of the unit cell having a length

a = 407 pm. . What is the radius of silver atom.

(a) 143.9 pm                        (b) 15.6 pm                          (c) 11.59 pm                        (d) 13.61 pm

Solution :(a)      AC ² + AB² = BC ²

 

here

AC = AB = a,

a ² + a ² = (4r)²

2a ² = 16r ²

a 2

BC = 4r

 

\             r ² =

8

 

\             r =    a    = 407

= 143.9 pm .

 

2         2

Example : 5       From the fact that the length of the side of a unit cell of lithium is 351 pm. Calculate its atomic radius. Lithium forms body centred cubic crystals.

(a) 152.69 pm                      (b) 62.71 pm                        (c) 151.98 pm                      (d) 54.61 pm

Solution : (c)     In (bcc) crystals, atoms touch each other along the cross diagonal.

 

Hence, Atomic radius (R) = a 3

4

= 351´    3 = 151.98 pm

4

 

Example : 6       Atomic radius of silver is 144.5 pm. The unit cell of silver is a face centred cube. Calculate the density of silver. (a) 10.50 g/cm³                             (b) 16.50 g/cm³                 (c) 12.30 g/cm³                  (d) 15.50 g/cm³

Solution :(a)     For (fcc) unit cell, atoms touch each other along the face diagonal.

 

Hence, Atomic radius (R) =

 

a =         = 4 ´ 144.5 pm = 408.70pm = 408.70 ´ 10^–10 cm

 

 

Density (D) = ZM ,

VN0

V = a ³

 

D =     ZM   ; where Z for (fcc) unit cell = 4 , Avagadro’s number

a ³ N 0

(N 0

) = 6.023 ´ 10 ²³ , Volume of cube

 

( V ) = (408.70 ´ 10 ^–10 )³ cm³ and M (Mol. wt.) of silver = 108,

 

D =                   4 ´ 108                          (408.70 ´ 10^–10 )³ ´ 6.023 ´ 10 ²³

= 10.50 g / cm³

 

 

 

Example : 7       Lithium borohydride (LiBH₄ ), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit

 

cell dimensions are : a = 6.81Å, b= 4.43Å, c=717Å. If the molar mass of density of the crystal is –

LiBH4

is 21.76

g mol ^-1 . The

 

(a) 0.668 g cm^-3

(b) 0.585g cm²

(c) 1.23 g cm^–3

(d) None

 

Solution : (a) We know that, r = ZM =

4 ´ (21.76 gmol ^–1 )

=                      -3

 

N 0 V

(6.023 ´ 10 ²³ mol ^–1 )(6.81 ´ 4.43 ´ 7.17 ´ 10^–24 cm³ )

0.668 g cm

 

Example : 8       A metallic elements exists as a cubic lattice. Each edge of the unit cell is 2.88Å. The density of the metal is

7.20 g cm^–3 . How many unit cells will be present in 100 gm of the metal.

 

(a) 5.82 ´ 10 ²³

(b) 6.33 ´ 10 ²³

(c) 7.49 ´ 10 ²⁴

(d) 6.9 ´ 10 ²⁴

 

Solution : (a) The volume of unit cell (V) = a³ = (2.88Å)³ = 23.9 ´ 10^–24 cm³

 

 

Volume of 100 g of the metal =

Mass Density

= 100

7.20

= 13.9cm²

 

 

Number of unit cells in this volume =

13.9cm³

23.9 ´ 10^–24 cm³

= 5.82 ´ 10 ²³

 

Example : 9       Silver crystallizes in a face centred cubic system, 0.408 nm along each edge. The density of silver is 10.6

g / cm³ and the atomic mass is 107.9 g / mol. Calculate Avogadro’s number. (a) 6.00 ´ 10 ²³ atom/mol                                                      (b) 9.31 ´ 10 ²³ atom/mol

(c) 6.23 ´ 10 ²³ atom/mol                                        (d) 9.61 ´ 10 ²³ atom/mol

 

Solution:(a)      The unit cell has a volume of (0.408 ´ 10^–9 m)³ = 6.79 ´ 10^–29 m³

volume of 1 mole of silver is,

per unit cell and contains four atoms. The

 

ë

û

107.9 g / molé(1 ´ 10-2 m)3 ù = 1.02 ´ 10^–5 m³ / mol ; where 107.9 g/mol is the molecular mass of the silver

ê      10.6g       ú

The number of unit cells per mol. is,

ç                  -29     3 ÷

1.02 ´ 10^–5 m³ / molæ          1unit cell      ö = 1.50 ´ 10 ²³ unit cells per mol.

6.79 ´ 10      m

è                               ø

æ 4atoms ö æ 1.50 ´ 10 ²³ unit cell ö                               23

 

and the number of atoms per mol. is,

ç unit cell ÷ ç

mol

÷ = 6.00 ´ 10

atom/mol.

 

è                ø è                                     ø

Example: 10      Fraction of total volume occupied by atoms in a simple cube is

 

 

(a) p

2

(d) p

8                                  6                                 6

 

Solution:(d)       In a simple cubic system, number of atoms a = 2r

4 pr ³      4 pr ³

\ Packing fraction = Volume occupied by one atom = 3              = 3       = p

Volume of unit cell                  a ³        (2r)³      6

 

 

 

 

Example: 11      A solid AB has the NaCl structure. If radius of cation the radius of the anion B^–

A⁺ is 120 pm, calculate the maximum possible value of

 

(a) 240 pm                            (b) 280 pm                            (c) 270 pm                            (d) 290 pm

Solution:(d)      We know that for the NaCl structure

 

 

radius of cation/radius of anion = 0.414;

rA+ rB–

B

= 0.414 ; r –

=    rA+

0.414

=   120 = 290pm

0.414

 

Example: 12      CsBr has a (bcc) arrangement and its unit cell edge length is 400 pm. Calculate the interionic                    distance     in

 

CsCl.

(a) 346.4 pm                        (b) 643 pm                            (c) 66.31 pm                        (d) 431.5 pm

Solution:(a)      The (bcc) structure of CsBr is given in figure

[CBSE 1993]

 

 

 

C

 

The body diagonal

AD = a

, where a is the length of edge of unit cell                  A

 

On the basis of figure

D

AD = 2(rCs + + rCl – )

 

 

 

a       = 2(r

• r ) or (r

+ r     ) = a 3 = 400 ´  3

= 200 ´ 1.732 = 346.4 pm

 

Cs+

Cl –

Cs +

Cl –                 2                 2

 

Ionic compounds consist of positive and negative ions arranged in a manner so as to acquire minimum potential energy (maximum stability). To achieve the maximum stability, ions in a crystal should be arranged in such a way that forces of attraction are maximum and forces of repulsion are minimum. Hence, for maximum stability the oppositely charged ions should be as close as possible to one another and similarly charged ions as far away as possible from one another. Among the two ions constituting the binary compounds, the larger ions (usually anions) form a close-packed arrangement (hcp or ccp) and the smaller ions (usually cations) occupy the interstitial voids. Thus in every ionic compound, positive ions are surrounded by negative ions and vice versa. Normally each ions is surrounded by the largest possible number of oppositely charged ions. This number of oppositely charged ions surrounding each ions is termed its coordination number.

Classification of ionic structures : In the following structures, a black circle would denote an anion and a

 

white circle would denote a cation. In any solid of the type Ax By

of B would be y : x .

the ratio of the coordination number of A to that

 

• Rock salt structure : The NaCl structure is composed of

Na +

and Cl – . The no. of

Na +

ions is equal to

 

that of

Cl – . The radii of

Na +

and

Cl –

are 95 pm and 181 pm respectively

rNa +

rCl –

= 95pm 181pm

= 0.524 . The radius

 

ratio of 0.524 for NaCl suggests an octahedral voids. Chloride is forming a fcc unit cell in which

Na ⁺

is in the

 

octahedral voids. The coordination number of

Na ⁺

is 6 and therefore that of Cl ^–

would also be 6. Moreover, there

 

are 4

Na ⁺

ions and 4

Cl ^–

ions per unit cell. The formula is

Na4 Cl4

i.e., NaCl. The other substances having this

 

kind of a structure are halides of all alkali metals except cesium, halides and oxides of all alkaline earth metals except berylium oxide.

 

 

 

• Zinc blende structure : Sulphide ions are face centred and zinc is present in alternate tetrahedral

 

Formula is

Zn₄ S₄ , i.e., ZnS. Coordination number of Zn is 4 and that of sulphide is also 4. Other substance that

 

exists in this kind of a structure is BeO.

 

The zine sulphide crystals are composed of

 

equal no. of

Zn+2

and

S 2-

ions. The radii of

 

two ions ( Zn+2 = 74 pm

and

S2-   = 184 pm ) led

 

to the radius ratio ( r + / r – ) as 0.40 which suggests a tetrahedral arrangement

 

rZn +2 rS 2-

= 74 pm = 0.40 184 pm

 

 

 

 

 

 

 

• Fluorite structure : Calcium ions are face centred and fluorite ions are present in all the tetrahedral

voids. There are four calcium ions and eight fluoride ions per unit cell. Therefore the formula is Ca4 F8 , (i.e. CaF₂ ).

The coordination number of fluoride ions is four (tetrahedral voids) and thus the coordination number of calcium ions is eight. Other substances which exist in this kind of structure are UO₂ and ThO₂ .

 

 

 

• Anti-fluorite structure : Oxide ions are face centred and lithium ions are present in all the tetrahedral There are four oxide ions and eight lithium ions per unit cell. As it can be seen, this unit cell is just the reverse of fluorite structure, in the sense that, the position of cations and anions is interchanged. Other substances which

 

exist in this kind of a structure are

Na2 O ,

K2 O

and

Rb2 O .

 

• Spinel and inverse spinel structure : Spinel is a mineral

(MgAl2 O4 ) . Generally they can be

 

represented as

M 2+ M 3+ O4

. Where

M 2+

is present in one-eighth of tetrahedral voids in a fcc lattice of oxide ions

 

2

and

M 3+

present in half of the octahedral voids.

M 2+

is usually Mg, Fe, Co, Ni, Zn and Mn,

M 3+

is generally Al,

 

Fe, Mn, Cr and Rh. e.g.,

ZnAl 2 O4 , Fe3 O4 , FeCr2 O4 etc.

 

• Cesium halide structure : Chloride ions are primitive cubic while the cesium ion occupies the centre of the unit cell. There is one chloride ion and one cesium ion per unit cell. Therefore the formula is CsCl. The coordination number of cesium is eight and that of chloride is ions is also eight. Other substances which exist in this kind of a structure are all halides of

 

The CsCl crystal is composed of equal no. of

Cs +

and

Cl –

ions. The radii of two ions ( Cs +   = 160pm

and

 

Cl – = 181pm ) led to radius ratio of rCs +

to rCl –

as 0.884

 

 

 

rCs +

rCl –

= 160pm = 0.884 181pm

 

Suggests a body centred cubic structure cubic structure having a cubic hole.

 

 

• Corundum structure : The general formula of compounds crystallizing in corundum structure is

Al2 O3 .

 

The closest packing is that of anions (oxide) in hexagonal primitive lattice and two-third of the octahedral voids are

 

filled with trivalent cations. e.g.,

Fe2 O3 ,

Al 2 O3

and Cr2 O3 .

 

• Pervoskite structure : The general formula is

ABO₃ . One of the cation is bivalent and the other is

 

tetravalent. e.g.,

CaTiO3 , BaTiO3 . The bivalent ions are present in primitive cubic lattice with oxide ions on the

 

centres of all six square faces. The tetravalent cation is in the centre of the unit cell occupying octahedral void.

 

 

 

Note : ® On applying high pressure, NaCl structure having 6:6 coordination number changes to CsCl structure having 8:8 coordination number similarly, CsCl having 8:8 coordination number on heating to 760 K changes to NaCl structure having 6:6 coordination number.

Depending upon the relative number of positive and negative ions present in ionic compounds, it is convenient to divide them into groups like AB, AB₂, AB₃, etc. Ionic compounds of the type AB and AB₂ are discussed below.

 

 

S. No.             Crystal Structure	Brief description	Examples	Co-ordination number	Number of formula units per unit cell
1.          Type AB	It has fcc arrangement in which	Halides of Li, Na, K, Rb,	Na+ = 6	4
Rock salt (NaCl) type	Cl – ions occupy the corners and face centres of a cube while	AgF,         AgBr,        NH4Cl, NH4Br, NH4I etc.	Cl – = 6
	Na +       ions  are   present   at   the
	body and edge of centres.
2.          Zinc     blende	It has ccp arrangement in which	CuCl, CuBr, CuI, AgI, BeS	Zn2+ = 4	4
(ZnS) type	S2– ions form fcc and each  Zn2+ ion is  surrounded  tetrahedrally		S 2– = 4
	by four S 2– ions and vice versa.
3.          Type AB2	It  has   arrangement   in   which	BaF2, BaCl2, SrF2	Ca 2+ = 8	4
Fluorite (CaF2) type	Ca2+  ions   form   fcc   with   each  Ca 2+   ions   surrounded   by    8 F –	SrCl2, CdF2, PbF2	F – = 4
	ions and each F –  ions by 4Ca2+
	ions.
4.          Antifluorite	Here negative ions form the ccp	Na2O	Na + = 4	4
type	arrangement       so      that       each positive ion is surrounded by 4		O 2– = 8
	negative ions and each negative
	ion by 8 positive ions
5.          Caesium	It has the bcc arrangement with	CsCl, CsBr, CsI, CsCN,	Cs+ = 8	1
chloride (CsCl) type	Cs+ at the body centre and  Cl – ions at the corners of a cube or	TlCl, TlBr, TlI and TlCN	Cl – = 8
	vice versa.

 

 

 

(iii) Crystal structure of some metals at room temperature and pressure :

 

X-ray diffraction and Bragg’s Equation : Crystal structure has been obtained by studying on the diffraction of X-rays by solids. A crystal, having constituents particles arranged in planes at very small distances in three dimension array, acts as diffraction grating for X– rays which have wavelengths of the same order as the spacing in crystal.

When a beam of X-rays passes through a crystalline solid, each atom in the beam scatters some of the radiations. If waves are on same phase means if their peak and trough coincides they add together to give a wave of greater amplitude. This enhancement of intensity is called constructive interference. If waves are out of phase, they cancel. This cancellation is called destructive interference.

Thus X– ray diffraction results from the scattering of X-rays by a regular arrangement of atoms or ions.

Bragg’s equation : Study of internal structure of crystal can be done with the help of X-rays. The distance of the constituent particles can be determined from diffraction value by Bragg’s equation,.

 

nl = 2d sinq

where, l = Wave length of X-rays, n = order of diffraction,

q = Angle of reflection, d = Distance between two parallel surfaces

 

The above equation is known as Bragg’s equation or Bragg’s law. The reflection corresponding to n = 1 (for a given family of planes) is called first order reflection; the reflection corresponding to n = 2 is the second order reflection and so on. Thus by measuring n (the order of reflection of the X-rays) and the incidence angle q, we can

 

know d/l.

d =     n     

 

l     2 sinq

From this, d can be calculated if l is known and vice versa. In X-ray reflections, n is generally set as equal to

1. Thus Bragg’s equation may alternatively be written as

l = 2 d sinq = 2 d_hkl sinq

Where d_hkl denotes the perpendicular distance between adjacent planes with the indices hkl.

 

 

 

 

Example : 15 The first order reflection (n = 1) from a crystal of the X-ray from a copper anode tube (l = 0.154nm) occurs at an angle of 16.3°. What is the distance between the set of planes causing the diffraction.

 

(a) 0.374nm

(b) 0.561nm

(c) 0.274nm

(d) 0.395 nm

 

 

Solution :(c)     From Bragg’s equation, nl = 2d sinq ;

d = n ´ l

= 1 ´ 0.154 = 0.154 nm = 0.274nm

 

2 sinq

2(sin 16.3)       2 ´ 0.281

 

Example : 16 The diffraction of barium with X-radiation of wavelength 2.29Å gives a first – order reflection at 30°. What is the distance between the diffracted planes.

(a) 3.29 Å                             (b) 4.39 Å                                (c) 2.29 Å                              (d) 6.29 Å

Solution :(c)     Using Bragg’s equation 2d sinq = nl

d = nl  , where d is the distance between two diffracted planes, q the angle to have maximum intensity of

2 sinq

diffracted X-ray beam, n the order of reflection and l is the wavelength

\ d = 1 ´ 2.29 Å = 2.29 Å                               æsin 30^o = 1 ö

2

è                       ø

2 ´ sin 30^o                                                  ç                       ÷

Example : 17   When an electron in an excited Mo atom falls from L to the K shell, an X-ray is emitted. These X-rays are

diffracted at angle of 7.75^o by planes with a separation of 2.64Å. What is the difference in energy between K–

shell and L-shell in Mo assuming a first-order diffraction. (sin 7.75^o = 0.1349)

 

 

 

Solution : (b)

(a) 36.88 ´10^–15 J

2d sinq = nl

(b) 27.88 ´ 10^–16 J

(c) 63.88 ´ 10^–17 J

(d) 64.88 ´ 10^–16 J

 

l = 2d sinq = 2 ´ 2.64 ´ 10^–10 ´ sin 7.75^o = 0.7123 ´ 10^–10 m

 

 

E = hc = 6.62 ´ 10^–34 ´ 3 ´ 10⁸

= 27.88 ´ 10^–16 J

 

l          0.7123 ´ 10^–10

 

Any deviation from the perfectly ordered arrangement constitutes a defect or imperfection. These defects sometimes called thermodynamic defects because the number of these defects depend on the temperature. Crystals may also possess additional defect due to the presence of impurities. Imperfection not only modify the properties of solids but also give rise to new properties. Any departure from perfectly ordered arrangement of atoms in crystals called imperfections or defects.

• Electronic imperfections : Generally, electrons are present in fully occupied lowest energy states. But at high temperatures, some of the electrons may occupy higher energy state depending upon the For example, in the crystals of pure Si or Ge some electrons are released thermally from the covalent bonds at temperature above 0 K. these electrons are free to move in the crystal and are responsible for electrical conductivity. This type of conduction is known as intrinsic conduction. The electron deficient bond formed by the release of an electron is called a hole. In the presence of electric field the positive holes move in a direction opposite to that of the electrons and conduct electricity. The electrons and holes in solids gives rise to electronic imperfections.

 

 

 

• Atomic imperfections/point defects : When deviations exist from the regular or periodic arrangement around an atom or a group of atoms in a crystalline substance, the defects are called point defects. Point defect in a crystal may be classified into following three types;

Point defects

• Stoichiometric defects (ii) Non- stoichiometric defects              (iii) Impurity defects

 

• Stoichiometric defects : The compounds in which the number of positive and negative ions are exactly in the ratios indicated by their chemical formulae are called stoichiometric The defects do not disturb the stoichiometry (the ratio of numbers of positive and negative ions) are called stoichiometric defects. These are of following types :
  • Schottky defects : This type of defect when equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained. This type of defect occurs in highly ionic compounds which have high co-ordination number and cations and anions of similar e.g., NaCl, KCl, CsCl and KBr etc.
  • Interstitial defects : This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the
  • Frenkel defects : This type of defect arises when an ion is missing from its lattice site and occupies an interstitial position. The crystal as a whole remains electrically neutral because the number of anions and cations remain same. Since cations are usually smaller than anions, they occupy interstitial sites. This type of defect occurs in the compounds which have low co-ordination number and cations and anions of different e.g., ZnS, AgCl and AgI etc. Frenkel defect are not found in pure alkali metal halides because the cations due to larger size cannot get into the interstitial sites. In AgBr both Schottky and Frenkel defects occurs simultaneously.

 

Consequences of Schottky and Frenkel defects : Presence of large number of Schottky defect lowers the density of the crystal. When Frenkel defect alone is present, there is no decrease in density. The closeness of the charge brought about by Frenkel defect tends to increase the dielectric constant of the crystal. Compounds having such defect conduct electricity to a small extent. When electric field is applied, an ion moves from its lattice site to occupy a hole, it creates a new hole. In this way, a hole moves from one end to the other. Thus, it conducts electricity across the crystal. Due to the presence of holes, stability (or the lattice energy) of the crystal decreases.

• Non-Stoichiometric defects : The defects which disturb the stoichiometry of the compounds are called non-stoichiometry defects. These defects are either due to the presence of excess metal ions or excess non- metal

 

 

 

• Metal excess defects due to anion vacancies : A compound may have excess metal anion if a negative ion is absent from its lattice site, leaving a ‘hole’, which is occupied by electron to maintain electrical This type of defects are found in crystals which are likely to possess Schottky defects. Anion vacancies in alkali metal halides are reduced by heating the alkali metal halides crystals in an atmosphere of alkali metal vapours. The ‘holes’ occupy by electrons are called F-centres (or colour centres).
• Metal excess defects due to interstitial cations : Another way in which metal excess defects may occur is, if an extra positive ion is present in an interstitial site. Electrical neutrality is maintained by the presence of an electron in the interstitial site. This type of defects are exhibit by the crystals which are likely to exhibit Frenkel defects g., when ZnO is heated, it loses oxygen reversibly. The excess is accommodated in interstitial sites, with electrons trapped in the neighborhood. The yellow colour and the electrical conductivity of the non- stoichiometric ZnO is due to these trapped electrons.

 

Consequences of Metal excess defects :

The crystals with metal excess defects are generally coloured due to the presence of free electrons in them.

The crystals with metal excess defects conduct electricity due to the presence of free electrons and are semiconductors. As the electric transport is mainly by “excess” electrons, these are called n-type (n for negative) semiconductor.

The crystals with metal excess defects are generally paramagnetic due to the presence of unpaired electrons at lattice sites.

Note : ® Colour Centres : Crystals of pure alkali metal halides such   as   NaCl,   KCl,   etc.   are   white. However, alkali metal halides becomes coloured on heating in excess of alkali metal vapour.   For example,   sodium chloride becomes yellow on heating in   presence   of   sodium   vapour.   These   colours   are   produced   due   to   the preferential absorption of some component of visible spectrum

due to some imperfections   called   colour   centres introduced   into the crystal.

When an alkali metal halide is heated in an atmosphere containing an excess of alkali metal vapour, the excess alkali metal atoms deposit on the crystal surface. Halide ions then diffuse to the surface where they combine with the metal atoms which have becomes ionised by loosing valence electrons. These electrons diffuse back into the crystal and occupy the vacant

 

 

 

sites created by the halide ions. Each electron is shared by all the alkali metal ions present around it and is thus a delocalized electrons. When the crystal is irradiated with white light, the trapped electron absorbs some component of white light for excitation from ground state to the excited state. This gives rise to colour. Such points are called F– centres. (German word Farbe which means colour) such excess ions are accompanied by positive ion vacancies. These vacancies serve to trap holes in the same way as the anion vacancies trapped electrons. The colour centres thus produced are called V-centres.

• Metal deficiency defect : These arise in two ways

By cation vacancy : in this a cation is missing from its lattice site. To maintain electrical neutrality, one of the nearest metal ion acquires two positive charge. This type of defect occurs in compounds where metal can exhibit variable valency. e.g., Transition metal compounds like NiO, FeO, FeS etc.

By having extra anion occupying interstitial site : In this, an extra anion is present in the interstitial position. The extra negative charge is balanced by one extra positive charge on the adjacent metal ion. Since anions are usually larger it could not occupy an interstitial site. Thus, this structure has only a theoretical possibility. No example is known so far.

Consequences of metal deficiency defects : Due to the movement of electron, an ion A⁺ changes to A⁺² ions. Thus, the movement of an electron from A⁺ ion is an apparent of positive hole and the substances are called p-type semiconductor

Impurity defect : These defects arise when foreign atoms are present at the lattice site (in place of host atoms) or at the vacant interstitial sites. In the former case, we get substitutional solid solutions while in the latter case, we get interstitial solid solution. The formation of the former depends upon the electronic structure of the impurity while that of the later on the size of the impurity.

 

Important Tips

 

• Berthallides is a name given to non-stoichiometric
• Solids containing F- centres are
• When NaCl is dopped with MgCl₂ the nature of defect produced is schottky
• AgBr has both Schottky & Frenkel

Some of the properties of solids which are useful in electronic and magnetic devices such as, transistor, computers, and telephones etc., are summarised below :

• Electrical properties : Solids are classified into following classes depending on the extent of conducting
  • Conductors : The solids which allow the electric current to pass through them are called conductors. These are further of two types; Metallic conductors and electrolytic conductors. In the metallic conductors the current is carries by the mobile electrons without any chemical change occurring in the In the electrolytic conductor like NaCl, KCl, etc., the current is carried only in molten state or in aqueous solution. This is because of

 

 

 

 

the movement of free ions. The electrical conductivity of these solids is high in the range Their conductance decrease with increase in temperature.

10⁴ – 10⁶ ohm^-1cm^-1 .

 

• Insulators : The solids which do not allow the current to pass through them are called insulators. g., rubber, wood and plastic etc. the electrical conductivity of these solids is very low i.e., 10 ^-12 – 10 ^-22 ohm^-1cm^-1 .
• Semiconductors : The solids whose electrical conductivity lies between those of conductors and insulators are called semiconductors. The conductivity of these solid is due to the presence of impurities. g. Silicon and Germanium. Their conductance increase with increase in temperature. The electrical conductivity of these solids is increased by adding impurity. This is called Doping. When silicon is doped with P (or As, group 15 elements), we get n-type semiconductor. This is because P has five valence electrons. It forms 4 covalent bonds with silicon and the fifth electron remains free and is loosely bound. This give rise to n-type semiconductor because current is carried by electrons when silicon is doped with Ga (or in In/Al, group 13 elements) we get p-type semiconductors.

Conductivity of the solids may be due to the movement of electrons, holes or ions.

Due to presence of vacancies and other defects, solids show slight conductivity which increases with temperature.

Metals show electronic conductivity.

The conductivity of semiconductors and insulators is mainly governed by impurities and defects. Metal oxides and sulphides have metallic to insulator behavior at different temperatures.

 

• Superconductivity : When any material loses its resistance for electric current, then it is called superconductor, Kammerlingh Onnes (1913) observed this phenomenon at 4K in mercury. The materials offering no resistance to the flow of current at very low temperature (2-5 K) are called superconducting

 

materials and phenomenon is called superconductivity. e.g.,

Nb₃ Ge alloy (Before 1986),

La1.25 Ba0.15 CuO4

 

(1986),

YBa 2 Cu3 O7

(1987) – super conductive at a temperature up to 92 K.

 

Applications

(a) Electronics,                                 (b) Building supermagnets,

(c) Aviation transportation,                    (d) Power transmission

 

 

 

“The temperature at which a material enters the superconducting state is called the superconducting transition

 

temperature,

(Tc ) ”. Superconductivity was also observed in lead (Pb) at 7.2 K and in tin (Sn) at 3.7K. The

 

phenomenon of superconductivity has also been observed in other materials such as polymers and organic crystals.

Examples are

(SN)_x, polythiazyl, the subscript x indicates a large number of variable size. (TMTSF)₂PF₆, where TMTSF is tetra methyl tetra selena fulvalene.

• Magnetic properties : Based on the behavior of substances when placed in the magnetic field, there are classified into five

 

Magnetic properties of solids

Properties                                 Description                              Alignment of Magnetic Dipoles

Examples               Applications

 

Diamagnetic           Feebly   repelled   by   the   magnetic

fields. Non-metallic elements (excepts O₂, S) inert gases and species with paired electrons are diamagnetic

Paramagnetic         Attracted by the magnetic field due

to    the     presence     of    permanent

All paired electrons             TiO₂,   V₂O₅,   NaCl,

C₆H₆ (benzene)

 

 

 

O2, Cu²⁺, Fe³⁺, TiO,

 Ti O , VO, VO ,

Insulator

 

 

 

 

Electronic appliances

 

magnetic           dipoles          (unpaired

2   3              2

 

electrons). In magnetic field, these tend to orient themselves  parallel to the direction of the field  and thus, produce magnetism in the substances.

Ferromagnetic     Permanent magnetism even in the

absence of magnetic field, Above a temperature            called     Curie temperature, there is no ferromagnetism.

At least one unpaired electron

 

 

 

Dipoles are aligned in the same direction

CuO

 

 

 

 

 

Fe, Ni, Co, CrO₂            CrO₂ is used

in audio and video tapes

 

Antiferromagn etic

This arises when the dipole alignment is zero due to equal and opposite alignment.

MnO,               MnO₂,                  – Mn₂O, FeO, Fe₂O₃;

NiO, Cr₂O₃, CoO, Co₃O₄,

 

Ferrimagnetic        This arises when there is net dipole

moment

Fe₃O₄, ferrites                            –

 

 

• Dielectric properties : When a non-conducting material is placed in an electrical field, the electrons and the nuclei in the atom or molecule of that material are pulled in the opposite directions, and negative and positive charges are separated and dipoles are generated, In an electric field :
• These dipoles may align themselves in the same direction, so that there is net dipole moment in the
• These dipoles may align themselves in such a manner that the net dipole moment in the crystal is Based on these facts, dielectric properties of crystals are summarised in table :

Dielectric properties of solids

 

 

 

 

Property                                              Description                                       Alignment of electric dipoles

Examples        Applications

 

Piezoelectricity When polar crystal is subjected to a mechanical stress, electricity is produced a case of piezoelectricity. Reversely if electric field is applied mechanical stress developed. Piezoelectric crystal acts as a mechanical electrical transducer.

• Piezoelectric crystals with permanent dipoles are said to have ferroelectricity
• Piezoelectric crystals with zero dipole are said to have antiferroelectricity

–                         Quartz,

 

 

Rochelle salt

BaTiO₃, KH₂PO₄,

 

PbZrO₃

Record players, capacitors, transistors, computer etc.

 

Pyroelectricity Small electric current is produced  due  to heating of some of polar crystals – a case of pyroelectricity

–                                                  Infrared

detectors

 

 

 

Important Tips

• Doping : Addition of small amount of foreign impurity in the host crystal is termed as It increases the electrical conductivity.
• Ferromagnetic property decreases from iron to nickel (Fe > Co > Ni) because of decrease in the number of unpaired
• Electrical conductivity of semiconductors and electrolytic conductors increases with increase in temperature, where as electrical conductivity of super conductors and metallic conductors decreases with increase in

 

These are the compounds with basic unit of (SiO₄)^4– anion in which each Si atom is linked directly to four oxygen atoms tetrahedrally. These tetrahedra link themselves by corners and never by edges. Which are of following types :

 

• Ortho silicates : In these discrete

SiO^4–

tetrahedra are present and there is no sharing of oxygen

 

4

atoms between adjacent tetrahedra e.g., Willamette Forestrite (Mg 2 SiO4 ) .

(Zn2 Si2 O4 ) , Phenacite

(Be ₂SiO₄ ) , Zircons

(ZrSiO4 )

and

 

• Pyrosilictes : In these silicates the two tetrahedral units share one oxygen atom (corner) between them

 

containing basic unit of (Si2 O7 )^6– anion e.g., Thortveitite (Sc 2 Si2 O7 ) and Hemimorphite

Zn3 Si2 O7 Zn(OH)2 H 2 O

 

• Cyclic or ring silicates : In these silicates the two tetrahedral unit share two oxygen atoms (two

 

corners) per tetrahedron to form a closed ring containing basic unit of Wollastonite (Ca3 Si3 O9 ) .

(SiO

2n–

)

3 n

e.g., Beryl

(Be3 Al 2 Si6 O18 ) and

 

• Chain silicates : The sharing of two oxygen atoms (two corners) per tetrahedron leads to the

 

formation of a long chain e.g., pyroxenes and Asbestos

CaMg 3 O(Si4 O11 ) and Spodumene

LiAl (Si2 O6 ).

 

 

 

• Sheet silicates : In these silicates sharing of three oxygen atoms (three corners) by each tetrahedron

5 n

unit results in an infinite two dimensional sheet of primary unit (Si₂O )^2n– . The sheets are held together by

electrostatic force of the cations that lie between them e.g., [Mg 3 (OH)2 (Si4 O10 )] and Kaolin, Al 2 (OH)4 (Si2 O5 ) .

• Three dimensional or frame work silicates : In these silicates all the four oxygen atoms (four

 

corners) of (SiO4 )^4–

tetrahedra are shared with other tetrahedra, resulting in a three dimensinal network with the

 

general formula (SiO2 )n

e.g., Zeolites, Quartz.

 

Important Tips

 

 

• Beckmann thermometer : Cannot be used to measure It is used only for the measurement of small differences in temperatures. It can and correctly upto 0.01^o
• Anisotropic behaviour of graphite : The thermal and electrical conductivities of graphite along the two perpendicular axis in the plane containing the hexagonal rings is 100 times more than at right angle to this
• Effect of pressure on melting point of ice : At high pressure, several modifications of ice are Ordinary ice is ice –I. The stable high pressure modifications of ice are designated as ice –II, ice – III, ice- V, ice – VI and ice – VII. When ice –I is compressed, its

melting point decreases, reaching – 22^o C at a pressure of about 2240 atm. A further increase in pressure transforms ice – I into ice – IIIs

whose melting point increases with pressure. Ice- VII, the extreme high-pressure modification, melts to form water at about 100°C and 20,000 atm pressure. The existence of ice-IV has not been confirmed.

• Isotropic : The substances which show same properties in all

• Anisotropic : Magnitude of some of the physical properties such as refractive index, coefficient of thermal expansion, electrical and thermal conductivities is different in different directions, with in the crystal