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Free Flip-Book Chemistry Class 12th by Study Innovations. 515 Pages

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Free Flip-Book Chemistry Class 12th by Study Innovations. 515 Pages

Free Flip-Book Chemistry Class 12th by Study Innovations. 515 Pages

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Solid state

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.

Characteristic Properties of Solids.

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

• Solids have definite volume, irrespective of the size of the container.
• Solids are rigid and have definite shape.
• Solids are almost incompressible.
• Many solids are crystalline in nature. These crystals have definite pattern of angles and planes.
• The density of solids is generally greater than that of liquids and gases.
• Solids diffuse very slowly as compared to liquids and gases.
• Most solids melt on heating and become liquids. 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 :
(i) Crystalline solids/True solids (ii) Amorphous solids/Pseudo solids
(1) Difference between crystalline and amorphous solids

Property Crystalline solids Amorphous solids
Shape They have long range order. They have short range order.
Melting point They have definite melting point They do not have definite melting point
Heat of fusion They have a definite heat of fusion They do not have definite heat of fusion
Compressibility They are rigid and incompressible These may not be compressed to any appreciable
Cutting with a sharp They are given cleavage i.e. they break They are given irregular cleavage i.e. they break
edged tool into two pieces with plane surfaces into two pieces with irregular surface
They are anisotropic They are isotropic
Isotropy and
Anisotropy There is a sudden change in volume when There is no sudden change in volume on melting.
it melts.
Volume change These possess symmetry These do not possess any symmetry.
These possess interfacial angles. These do not possess interfacial angles.
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., Na2SeO4 and Na2SO4 . 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 liquid.

(2) Classification of solids : Depending upon the nature of interparticle forces the solids are classified into
four types :

Types Constituents Bonding Examples Physical M.P. B.P. Electrical
of Solid Nature Conductivity
High High (≃2000K)
Ionic Positive and Coulombic NaCl, KCl, CaO, Hard but (≃1000K) Conductor (in
negative ions brittle Very high molten state and
Covalen network Electron MgO, LiF, ZnS, (≃500K) in aqueous
t systematically sharing solution)
arranged BaSO4 and Low (≃ 450 to
Molecul 800 K)
ar Atoms K2SO4 etc.
connected in Low
covalent bonds SiO2 (Quartz), Hard Very high (≃373K to Insulator except
Hard (≃4000K) 500K) graphite
SiC, C Hard
C(graphite) etc.

Polar or non- (i) I2,S8, P4, CO2, Low Insulator
polar Molecular CH4, CCl4 etc.
molecules interactions (≃300K to
(intermolec 600K)
u-lar forces)
Starch, sucrose, Soft Low (≃400K) Insulator
(ii) water, dry ice or
Hydrogen drikold (solid
bonding CO2) etc.

Metallic Cations in a Metallic Sodium , Au, Ductile High (≃800K High (≃1500K Conductor
Cu, magnesium,
sea of electrons metals and malleable to 1000 K) to 2000K)
Atomic Atoms London Soft Very low Very low Poor thermal and
dispersion Noble gases electrical
force conductors

(i) Liquid Crystal : There are certain solids which when heated undergo two sharp phase transformations
one after the other. 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 145oC p − Chloesteryl benzoate 178oC p − Chloesteryl benzoate
(liquidcrystal ) (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)

(ii) 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 pole. 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  1  . Thus, these forces are important as short distances
 r6 

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

(3) 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 solids. In fact, any material can be made amorphous or glassy either by rapidly cooling or freezing its
vapours for example, SiO2 crystallises or quartz in which SiO44− tetrahedra are linked in a regular manner but on

melting and then rapid cooling, it gives glass in which SiO44− tetrahedron are randomly joined to each other.

Properties of Amorphous solids

(i) Lack of long range order/Existence of short range order : Amorphous solids do not have a long range order
of their constituent atoms or ions. However, they do have a short range order like that in the liquids.

(ii) No sharp melting point/Melting over a range.

(iii) Conversion into crystalline form on heating.
Uses of Amorphous solids

(i) The most widely used amorphous solids are in the inorganic glasses which find application in construction,
house ware, laboratory ware etc.

(ii) Rubber is amorphous solid, which is used in making tyres, shoe soles etc.

(iii) 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”.

(1) Laws of crystallography : Crystallography is based on three

fundamental laws. Which are as follows

(i) Law of constancy of interfacial angles : This law states that angle Constancy of interfacial angles

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.

(ii) 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 indices. This law was given by Hally.

(iii) Law of constancy of symmetry : According to this law, all crystals of a substance have the same elements
of symmetry.

(2) Designation of planes in crystals (Miller indices) : Planes in crystals are described by a set of

integers (h, k and l) known as Miller indices. 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 z
parametral plane.

Thus, the Miller indices are :

h = intercept of the a along x - axis c
plane N

b b M y
k = intercept of the along y - axis aL

l = intercept of the c along z - axis A parametral plane (intercepts,
plane a, b, c along x, y and z axes)

Consider a plane in which Weiss notation is given by
∞a : 2b : c . The Miller indices of this plane may be calculated as below.

(i) Reciprocals of the coefficients of Weiss indices = 1 , 1 , 1
∞ 2 1

(ii) 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. h = 0 ,
k = 1, l = 2.

The distance between the parallel planes in crystals are designated as dhkl . For different cubic lattices these
interplanar spacing are given by the general formula,

d(hkl) = a
h2 + k2 + l 2

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 intercept.
All parallel planes have same Miller indices.
The Miller indices are enclosed within parenthesis. 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 c Lattice parameters ∞ 2b c Lattice parameters
a b c a b c

1 1 1 Reciprocals 1 1 1 Reciprocals
2 3 1 ∞ 2 1

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

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.
Solution:(b) (a=0.556nm)

(a) 016.1 nm (b) 01.61 nm (c) 0.610 nm (d) None of the above

We have, d = a ; d111 = 0.556 = 0.321nm and d222 = 0.556 = 0.161nm
h2 + k2 + 2 12 + 12 + 12 22 + 22 + 22

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

Study of Crystals.

(1) 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 form.

(2) Space lattice and Unit cell : A regular array of points (showing atoms/ions) in three dimensions is
commonly called as a space lattice, or lattice.

(i) Each point in a space lattice represents an atom or a group of atoms.
(ii) Each point in a space lattice has identical surroundings throughout.
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, α, β,γ .


αβ c Unit
γ Space lattice & unit cell

Unit cell

(3) 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 :

(i) 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 other.

(a) (b) (c)
Plane of symmetry
Rectangular plane Diagonal plane
of symmetry of symmetry

(ii) 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 2π / n so that rotation through 2π / 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)].

Fig. (i) Axis of two Fig. (ii) Axis of three Fig. (iii) Axis of four Fig. (iv) Axis of six
fold symmetry. fold symmetry. fold symmetry. fold symmetry.

(iii) Centre of symmetry : It is an imaginary point in the crystal that any Y Centre of
line drawn through it intersects the surface of the crystal at equal distance on Z symmetry
either side. of a cubic
Note :  Only simple cubic system have one centre of symmetry. Other X

system do not have centre of symmetry.

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

(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

(5) Formation of crystals : The crystals of the substance are obtained by cooling the liquid (or the melt) of
the solution of that substance. 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.

Atoms of molecules Dissolved → cluster dissolved → dissolved embryo → nucleus → crystal

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

(6) 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 systems. 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
Pb, Hg, Ag,
Cubic Simple : Lattice points at Body centered : Points Face centered : Au, Cu, ZnS ,
the eight corners of the at the eight corners and Points at the eight diamond, KCl,
a=b=c, unit cells. at the body centred. corners and at the NaCl, Cu2O,CaF2
six face centres. and alums. etc.
Here a, b and c are Simple : Points at the
parameters eight corners of the unit SnO2, TiO2,
(diamensions of a unit cell. ZnO2, NiSO4
cell along three axes) ZrSiO4 . PbWO4 ,
size of crystals depend white Sn etc.
on parameters.

α = β = γ = 90o

αβ and γ are sizes of

three angles

between the axes.

Tetragonal Body centered : Points at the eight corners and
at the body centre
a =b ≠ c,
α = β = γ = 90o

Orthorhombic Simple: Points End centered : Also Body centered Face centered: KNO3, K2SO4 ,

(Rhombic) at the eight called side centered : Points at the Points at the PbCO3, BaSO4 ,
corners of the or base centered. eight corners eight coreners rhombic sulphur,
a≠b≠c, unit cell. Points at the eight and at the and at the six MgSO4 . 7H2O etc.
α = β = γ = 90o corners and at two body centre face centres.
face centres opposite
to each other.

Rhombohedral Simple : Points at the eight corners of the unit cell NaNO3, CaSO4 ,
or Trigonal calcite, quartz,
As, Sb, Bi etc.
α = β = γ ≠ 90o Simple : Points at the twelve corners or Points at the twelve corners of the ZnO, PbS, CdS,
of the unit cell out lined by thick hexagonal prism and at the centres HgS, graphite, ice,
Hexagonal line. of the two hexagonal faces. Mg, Zn, Cd etc.

a =b ≠ c,
α = β = 90o
γ = 120o

Monoclinic Simple : Points at the eight corners End centered : Point at the eight Na2SO4 .10H2O,
of the unit cell corners and at two face centres Na2 B4 O7 .10 H 2O,
a≠b≠c, opposite to the each other. CaSO4 .2H2O,

α = γ = 90o, β ≠ 90o monoclinic
sulphur etc.

Triclinic Simple : Points at the eight corners of the unit cell. CaSO4 .5H2O,
K2Cr2O7, H3BO3
α ≠ β ≠ γ ≠ 90o etc.

Note :  Out of seven crystal systems triclinic is the most unsymmetrical ( a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90) .

Packing constituents in Crystals.

(1) Close packing in crystalline solids : In the formation of crystals, 11
the constituent particles (atoms, ions or molecules) get closely packed together. 42
The closely packed arrangement is that in which maximum available space is
occupied. This corresponds to a state of maximum density. The closer the 3
packing, the greater is the stability of the packed system. It is of two types :
Square close packing
(i) Close packing in two dimensions : The two possible arrangement
of close packing in two dimensions. 1 2
6 3
(a) 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 5 4
form square. Each sphere in this arrangement is in contact with four spheres.
Hexagonal close packing
(b) 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 spheres.

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

 In hexagonal close packing about 60.4% 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.

(ii) Close packing in three dimensions : In order to develop three dimensional close packing, let us retain

the hexagonal close packing in the first layer. For close packing each

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

touching spheres in the layer as shown in figure. The spheres in the first bbb
layer are shown by solid lines while those in second layer are shown by aaa a
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., c cc
a aa a a

either b or c). The unoccupied hollows or voids in the first layer are Close packing in three dimensions
indicated by (c) in figure.

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

(a) 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 spheres.

(b) 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 layer. 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.

A A B≡ C
(a) AB AB – type close packing (c) Hexagonal close packing (hcp) in three dimensions
(hexagonal close packing). (b) ABC ABC – type close
packing (cubic close

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


CC ≡



A Cubic close packing (ccp) ≡ face – centred cubic (fcc)

Cubic close packing (ccp) in three dimensions

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

(c) 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 AA AA AA
result none of these spheres are in contact with AA
each other. The second layer of spheres (B) can be B BB
placed on top of the first layer so that each sphere B
of the second layer is in contact with four spheres AA A A
of the layer below it. Successive building of the B B B A

third will be exactly like the first layer. If this pattern Body centred cubic (bcc) close packing in three dimensions
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.

(2) Comparison of hcp, ccp and bcc

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

Malleability and ductility

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

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

r = 0.155R

where, r = Radius of the spherical trigonal site Trigonal

R = Radius of closely packed spheres

Triogonal void

(ii) 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 layer. 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 Tetrahedral void
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., where r is the radius of the tetrahedral void or atom

r = 0.225 , occupying tetrahedral void and R is the radius of spheres
R forming tetrahedral void.

(a) Octahedral void : This type of void is surrounded by six closely packed spheres, Octahedral void
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.

r = 0.414

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

r = 0.732

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 equal.
 The most malleable metals (Cu, Ag, Au) have cubic close packing.
 Cubic close packing has fcc (face centred cubic) unit cell
 Number of octahedral voids = Number of atoms present in the closed packed arrangement.
 Number of tetrahedral voids = 2 × Number of octahedral voids = 2 × Number of atoms.

Mathematical analysis of Cubic system.

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 corners.

• 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 faces.

(1) Atomic radius : It is defined as the half of the distance between nearest neighbouring atoms in a crystal.
It is expressed in terms of length of the edge (a) of the unit cell of the crystal.

(i) Simple cubic structure (sc) : Radius of atom 'r' = a
(ii) Face centred cubic structure (fcc) : 'r' = a rr

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

(2) 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 content.

(i) Simple cubic structure (sc) : Each corner atom is shared by eight surrounding cubes. Therefore, it

contributes for 1 of an atom. ∴ Z = 8 × 1 = 1 atom per unit cell in crystalline solid.
8 8

(ii) Face centered cubic structure (fcc) : The eight corners atoms contribute for 1 of an atom and thus

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 atom per unit cell.

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

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

Centre atom contribute one atom per unit cell. So, total 1 + 1 = 2 atoms per unit cells. Z = 8 × 1 + 1 = 2

Note :  Number of atoms in unit cell : It can be determined by the simplest relation = nc + nf + ni
8 2 1

Where nc = Number of atoms at the corners of the cube = 8

nf = Number of atoms at six faces of the cube = 6

n i = Number of atoms inside the cube = 1

Cubic unit cell nc nf ni Total atom in per
unit cell
Simple cubic (sc) 80
body centered cubic (bcc) 80 01
Face centered cubic (fcc) 86


(3) 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 number. It depends upon structure of the crystal.

(i) For simple cubic system C.N. = 6.

(ii) For body centred cubic system C.N. = 8

(iii) For face centred cubic system C.N. = 12.

(4) Density of the unit cell : It is defined as the ratio of mass per unit cell to the total volume of unit cell.

Density of unit cell (ρ) = mass of unit cell ; ρ = Number of particles × mass of each particle or Z×M
volume of unit cell volume of the unit cell a3 × N0

Where Z = Number of particles per unit cell, M = Atomic mass or molecular mass, N0 = Avogadro number

(6.023 × 10 23 mol −1 ) , a = Edge length of the unit cell= a pm = a × 10−10 cm , a 3 = volume of the unit cell

i.e. ρ = a3 Z×M g / cm3
× N0 × 10−30

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

(5) 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 cell.

Let radius of the atom in the packing = r

Edge length of the cube = a

Volume of the cube V = a3

4 νZ 4 πr 3 Z
3 V 3
Volume of the atom (spherical) ν = πr 3 , then packing density = =


(i) 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 cells.

Hence number of atoms per unit cell = 8× 1 = 1, again r = a
8 2

4 πr 3
∴ P.F. = (2r)3 = 0.52 ; % P.F. = 52%, then % of void = 100 – 52 = 48%

(ii) Body centred cubic unit cell : Number of atoms per unit cell = 8× 1 +1 = 2, r = 3a
8 4

2 × 4 πr 3
P.F. = 3 = 0.68 ; % P.F. = 68%, then % of void = 100 – 68 = 32%
 3 

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

4 × 4 πr 3
P.F. =  3 = 0.74 ; % P.F. = 74% , then % of void = 100 – 74=26%
 2

Structure r related to a Volume of the atom (ν) Packing density
Simple cubic
r a 4  a  3 π 0.52
Face-centred cubic 2 3  2  6
= π =
Body-centred cubic
r= a 4 π  2 a  3 2π = 0.74
22 3 2 6

r= 3a 4 π  3a 3 3π = 0.68
4 3  4  8

(6) Ionic radii : X-ray diffraction or electron diffraction techniques provides the necessary information

regarding unit cell. From the dimensions of the unit cell, it is possible to calculate ionic

radii. Cl–

Let, cube of edge length 'a' having cations and anions say NaCl structure. a/2

Then, rc + ra = a / 2 90°
where rc and ra are radius of cation and anion. Cl–

Radii of chloride ion

Radius of Cl − = (a / 2)2 + (a / 2)2 = a
2 4

For body centred lattice say CsCl. rc + ra = 3a

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+

(b) r+/r– > 0.414 to 0.732 r+/r– < 0.414 (c)

r+/r– > 0.732 r+/r– = 0.414 Unstable
Coordination number Coordination number
increases from 6 to 8 Effect of radius ratio on co-ordination number decreases from 6 to 4

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

CaF2, SrF2,

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 e.g. Na+, Mg2+, Al3+ (greater the nuclear charge, smaller the

size, Al3+ < Mg2+ < Na+)

Example : 3 A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ............ Any packing
Solution :(a)
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 %

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

Hence, area of base = 6  1 (2r )(2r sin 60 o ) = 6 3.r 2
 2

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

Height (h) = 2  3a  2a
 =
 3  3

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  4r  = 4 2  . r
3  2   3 
  

Volume of hexagonal unit is, V = (base area) × (height) =(6 3 r 2 )  4 2 . r  = 24 2.r 3
 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 cell.

• 1 atom from the centre of the base. There are two atoms of this type and each is shared between two
hexagonal unit cells.

• 2 atoms from the corners. There are 12 such atoms and each is shared amongst six hexagonal unit cells.

Now, the volume occupied by atoms = 6 4 πr 3 
3 

6 4 πr 3 
 3 
Fraction of volume occupied by atoms = Volume occupied by atoms = = π / 3 2 = 0.74.
Volume of hexagonal unit cell 24 2.r 3

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 Cr
here AC = AB = a, BC = 4r

a 2 + a 2 = (4r)2

2a 2 = 16r 2 407 pm 2r

∴ r2 = a2 A 407 pm r
8 B

∴ r = a = 407 = 143.9 pm .
22 22

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) = a3 = 351× 3 = 151.98 pm
4 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.
Solution :(a)
(a) 10.50 g/cm3 (b) 16.50 g/cm3 (c) 12.30 g/cm3 (d) 15.50 g/cm3

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

Hence, Atomic radius (R) = a2

a = 4R = 4 × 144.5pm = 408.70pm = 408.70 × 10−10 cm

Density (D) = ZM , V = a3
VN 0

D= ZM ; where Z for (fcc) unit cell = 4 , Avagadro’s number (N 0 ) = 6.023 × 10 23 , Volume of cube
a3 N0

( V ) = (408.70 × 10−10 )3 cm3 and M (Mol. wt.) of silver = 108,

D = 4 × 108 = 10.50 g / cm3
(408.70 × 10−10 )3 × 6.023 × 10 23

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 LiBH4 is 21.76 g mol −1 . The
density of the crystal is –

(a) 0.668 g cm−3 (b) 0.585g cm2 (c) 1.23 g cm−3 (d) None

Solution : (a) We know that, ρ = ZM = 4 × (21.76 gmol −1 ) = 0.668 g cm−3
Example : 8 N0V (6.023 × 10 23 mol −1 )(6.81 × 4.43 × 7.17 × 10−24 cm3 )

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 × 1023 (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 = 100 = 13.9cm2
Density 7.20

Number of unit cells in this volume = 13.9cm3 = 5.82 × 10 23
23.9 × 10 −24 cm3

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 × 1023 atom/mol (b) 9.31 × 10 23 atom/mol

(c) 6.23 × 1023 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 per unit cell and contains four atoms. The
volume of 1 mole of silver is,

107.9 g / mol  (1 × 10 −2 m)3  = 1.02 × 10 −5 m3 / mol ; where 107.9 g/mol is the molecular mass of the silver
 10.6g 
 

The number of unit cells per mol. is,

1.02 × 10−5 m3 / mol 1unit cell  = 1.50 × 10 23 unit cells per mol.
 6.79 × 10 −29 m3 

and the number of atoms per mol. is,  4atoms   1.50 × 10 23 unit cell  = 6.00 × 10 23 atom/mol.
 unit cell  mol

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

(a) π (b) 3π (c) 2π (d) π
2 8 6 6

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

Volume occupied by one atom 4 πr 3 4 πr 3
Volume of unit cell 3 3
∴ Packing fraction = = a3 = (2r)3 = π

Example: 11 A solid AB has the NaCl structure. If radius of cation A+ is 120 pm, calculate the maximum possible value of
the radius of the anion B−

(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; r + = 0.414 ; rB− = rA+ = 120 = 290 pm
A 0.414 0.414


Example: 12 CsBr has a (bcc) arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in
[CBSE 1993]

(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 C

The body diagonal AD = a 3 , where a is the length of edge of unit cell A B
On the basis of figure
AD = 2(rCs + + r )
Cl −

a 3 = 2(r + + r − ) or (rCs + + rCl− ) = a3 = 400 × 3 = 200 × 1.732 = 346.4 pm
Cs Cl 2 2

Crystal structures and Method of determination.

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 the ratio of the coordination number of A to that

of B would be y : x .

(1) 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 + = 95 pm = 0.524 . The radius
rCl − 181pm

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.

= Cl–

Na+ ion surrounded Cl– ion surrounded
octahedrally octahedrally octahedrally by six Na+

Structure of NaCl (rock salt)

(2) Zinc blende structure : Sulphide ions are face centred and zinc is present in alternate tetrahedral voids.
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 S 2− = 184 pm ) led

to the radius ratio ( r + / r − ) as 0.40 which

suggests a tetrahedral arrangement

rZn +2 = 74 pm = 0.40
rS 2− 184 pm

=Zn2+ ion
= S2– ion

Structure of ZnS (Zinc blende)

(3) 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 .

= Ca2+

Structure of CaF2 (Fluorite)

(4) Anti-fluorite structure : Oxide ions are face centred and lithium ions are present in all the tetrahedral
voids. 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 Na2O , K2O and Rb2O .

(5) Spinel and inverse spinel structure : Spinel is a mineral (MgAl2O4 ) . 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

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., ZnAl2O4 , Fe3O4 , FeCr2O4 etc.

(6) 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 cesium.

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 + = 160 pm = 0.884
rCl − 181pm

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

Cs+ ion surrounded by 8 Cl– ions Cl– ion surround by 8 Cs+ ions

= Cs+
= Cl–

Structure of caesium chloride

(7) Corundum structure : The general formula of compounds crystallizing in corundum structure is Al2O3 .
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., Fe2O3 , Al2O3 and Cr2O3 .

(8) 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.

NaCl Pressure CsCl
6 : 6 Co − orditnation number Temp. 8 : 8 Co − orditnation 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 Brief description Examples Co-ordination Number of
1. Structure number formula
2. units per
3. Type AB It has fcc arrangement in which Halides of Li, Na, K, Rb, Na+ = 6 unit cell
4. Rock salt AgF, AgBr, NH4Cl, Cl− = 6 4
5. (NaCl) type Cl− ions occupy the corners and NH4Br, NH4I etc.
face centres of a cube while Zn2+ = 4 4
Zinc blende CuCl, CuBr, CuI, AgI, BeS S 2− = 4
(ZnS) type Na + ions are present at the 4
body and edge of centres. BaF2, BaCl2, SrF2 Ca 2+ = 8
Type AB2 It has ccp arrangement in which SrCl2, CdF2, PbF2 F− =4 4
Fluorite S2− ions form fcc and each Zn2+
(CaF2) type ion is surrounded tetrahedrally Na2O Na + = 4 1
by four S 2− ions and vice versa. O 2− = 8
Antifluorite It has arrangement in which CsCl, CsBr, CsI, CsCN,
type Ca2+ ions form fcc with each TlCl, TlBr, TlI and TlCN Cs+ = 8
Ca 2+ ions surrounded by 8F − Cl − = 8
chloride ions and each F − ions by 4Ca2+
(CsCl) type ions.
Here negative ions form the ccp
arrangement so that each
positive ion is surrounded by 4
negative ions and each negative
ion by 8 positive ions
It has the bcc arrangement with

Cs+ at the body centre and Cl−
ions at the corners of a cube or
vice versa.

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

Li Be

Na Mg Al Body centred
Face centred
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Simple cubic
Zr Nb Mo Tc Ru Rh Pd Ag Cd Hexagonal close
Rbv Sr Y packed

Cs Ba

Experimental method of determining Crystal structure.

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,.

nλ = 2d sinθ where, λ = Wave length of X-rays, n = order of diffraction,

θ = 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 θ, we can

know d/λ. dn
λ = 2 sinθ

From this, d can be calculated if λ 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

λ = 2 d sinθ = 2 dhkl sinθ

Where dhkl 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 (λ = 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, nλ = 2d sinθ ; d= n×λ = 1 × 0.154 = 0.154 nm = 0.274nm
2 sinθ 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 sinθ = nλ

d= nλ , where d is the distance between two diffracted planes, θ the angle to have maximum intensity of
2 sinθ

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

∴ d= 1 × 2.29 Å = 2.29 Å  sin 30o = 1 
2 × sin 30o  2 

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 7.75o 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 7.75o = 0.1349)

(a) 36.88 ×10−15 J (b) 27.88 × 10−16 J (c) 63.88 × 10−17 J (d) 64.88 × 10−16 J

Solution : (b) 2d sinθ = nλ

λ = 2d sinθ = 2 × 2.64 × 10−10 × sin 7.75o = 0.7123 × 10−10 m

E = hc = 6.62 × 10−34 × 3 × 108 = 27.88 × 10−16 J
λ 0.7123 × 10−10

Defects or Imperfections in Solids.

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.

(1) 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 temperature. 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.

(2) 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

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

(i) 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 compounds. 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 :

(a) 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 sizes. e.g., NaCl, KCl, CsCl and KBr etc.

(b) Interstitial defects : This type of defect is caused due to the presence of ions in the normally vacant
interstitial sites in the crystals.

(c) 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 sizes. 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.

A+ B– A+ B– A+ B– A+ B– A+ B– A+ B–
B– A+ B– A+ A+
B+ A– A+
B– A+ B– A+

A+ B– A+ B– B– A+ B– B– A+ B–
Ideal Crystal Schottky defect Frenkel defect

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.

(ii) 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 ions.

(a) 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
neutrality. 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).

(b) 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 e.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.

A+ B– A+ B– A+ B– A+ B–
B– A+ B– A+
B– A+ B– A+ A+ e– A+ B–
Metal excess defect due to anion vacancy
A+ B– A+ B–
Metal excess defect due to extra cation

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)

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 A+ B– A+ B–

the crystal. Cation vacancy

When an alkali metal halide is heated in an atmosphere B– B– A+
containing an excess of alkali metal vapour, the excess alkali Metal having higher

metal atoms deposit on the crystal surface. Halide ions then A+ B– A+2 B– charge

diffuse to the surface where they combine with the metal atoms

which have becomes ionised by loosing valence electrons. These B– A+ B– A+

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.

(c) 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 compounds.
 Solids containing F- centres are paramagnetic.
 When NaCl is dopped with MgCl2 the nature of defect produced is schottky defect.
 AgBr has both Schottky & Frenkel defect.

Properties of Solids .

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

(1) Electrical properties : Solids are classified into following classes depending on the extent of conducting

(i) 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 matter. 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 104 − 106 ohm−1cm−1 .
Their conductance decrease with increase in temperature.

(ii) Insulators : The solids which do not allow the current to pass through them are called insulators. e.g.,
rubber, wood and plastic etc. the electrical conductivity of these solids is very low i.e., 10−12 − 10−22 ohm−1cm−1 .

(iii) 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. e.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

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

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.

Insulator like Conductivity Metal like
FeO, Fe2O3 Insulator – to –metal TiO
MnO, MnO2 Ti 2 O 3 CrO2
Cr2 O3 V2O3 ReO3

(2) 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), YBa2 Cu3O7 (1987) – super conductive at a temperature up to 92 K.


(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.

(3) Magnetic properties : Based on the behavior of substances when placed in the magnetic field, there are
classified into five classes.

Magnetic properties of solids

Properties Description Alignment of Magnetic Examples Applications
Diamagnetic Dipoles Insulator
Feebly repelled by the magnetic TiO2, V2O5, NaCl,
Paramagnetic fields. Non-metallic elements All paired electrons C6H6 (benzene) Electronic
(excepts O2, S) inert gases and appliances
species with paired electrons are At least one unpaired O2, Cu2+, Fe3+, TiO,
diamagnetic electron Ti2O3, VO, VO2 ,
Attracted by the magnetic field due CuO
to the presence of permanent
magnetic dipoles (unpaired
electrons). In magnetic field, these
tend to orient themselves parallel
to the direction of the field and
thus, produce magnetism in the

Ferromagnetic Permanent magnetism even in the Dipoles are aligned in Fe, Ni, Co, CrO2 CrO2 is used
the same direction in audio and
absence of magnetic field, Above a
video tapes
temperature called Curie

temperature, there is no


Antiferromagn This arises when the dipole MnO, MnO2, –
etic alignment is zero due to equal and –
opposite alignment. Mn2O, FeO, Fe2O3;

NiO, Cr2O3, CoO,


Ferrimagnetic This arises when there is net dipole Fe3O4, ferrites

(4) 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 :

(i) These dipoles may align themselves in the same direction, so that there is net dipole moment in the crystal.

(ii) These dipoles may align themselves in such a manner that the net dipole moment in the crystal is zero.

Based on these facts, dielectric properties of crystals are summarised in table :

Dielectric properties of solids

Property Description Alignment of electric Examples Applications
Piezoelectricity dipoles
When polar crystal is subjected to a – Quartz, Record
Pyroelectricity mechanical stress, electricity is produced a players,
case of piezoelectricity. Reversely if electric – Rochelle capacitors,
field is applied mechanical stress salt transistors,
developed. Piezoelectric crystal acts as a BaTiO3, computer etc.
mechanical electrical transducer. KH2PO4,

• Piezoelectric crystals with permanent PbZrO3
dipoles are said to have ferroelectricity
• Piezoelectric crystals with zero dipole detectors
are said to have antiferroelectricity

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

Important Tips

 Doping : Addition of small amount of foreign impurity in the host crystal is termed as doping. It increases the electrical conductivity.
 Ferromagnetic property decreases from iron to nickel (Fe > Co > Ni) because of decrease in the number of unpaired electrons.

 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 temperature.


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 :

(1) Ortho silicates : In these discrete SiO44− tetrahedra are present and there is no sharing of oxygen
atoms between adjacent tetrahedra e.g., Willamette (Zn2Si2O4 ) , Phenacite (Be2SiO4 ) , Zircons (ZrSiO4 ) and
Forestrite (Mg 2SiO4 ) .

(2) Pyrosilictes : In these silicates the two tetrahedral units share one oxygen atom (corner) between them
containing basic unit of (Si2O7 )6− anion e.g., Thortveitite (Sc 2Si2O7 ) and Hemimorphite Zn3Si2O7 Zn(OH)2 H 2O

(3) 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 (SiO3 )n2n− e.g., Beryl (Be3 Al2Si6O18 ) and
Wollastonite (Ca3Si3O9 ) .

(4) 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 CaMg3O(Si4 O11 ) and Spodumene LiAl (Si2O6 ).

(5) Sheet silicates : In these silicates sharing of three oxygen atoms (three corners) by each tetrahedron
unit results in an infinite two dimensional sheet of primary unit (Si2O5 )n2n− . The sheets are held together by
electrostatic force of the cations that lie between them e.g., [Mg3 (OH)2(Si4 O10 )] and Kaolin, Al2(OH)4 (Si2O5 ) .

(6) 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 temperature. It is used only for the measurement of small differences in
temperatures. It can and correctly upto 0.01o

 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 plane.

 Effect of pressure on melting point of ice : At high pressure, several modifications of ice are formed. 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 − 22o 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 directions.

 Anisotropic : Magnitude of some of the physical properties such as refractive index, coefficient of thermal expansion, electrical and

thermal conductivities etc. is different in different directions, with in the crystal


A solution is a homogenous mixture of two or more substances, the composition of which may vary within
limits. “A solution is a special kind of mixture in which substances are intermixed so intimately that they can not be
observed as separate components”. The dispersed phase or the substance which is to be dissolved is called solute,
while the dispersion medium in which the solute is dispersed to get a homogenous mixture is called the solvent. A
solution is termed as binary, ternary and quartenary if it consists of two, three and four components respectively.


Solubility of as substance in the solvent is the measure of the capacity of the latter to dissolve the former at a
given temperature and pressure. “Solubility of a substance may be defined as the amount of solute dissolved in
100gms of a solvent to form a saturated solution at a given temperature”. A saturated solution is a solution which
contains at a given temperature as much solute as it can hold in presence of dissolveding solvent. Any solution may
contain less solute than would be necessary to to saturate it. Such a solution is known as unsaturated solution.
When the solution contains more solute than would be necessary to saturate it then it is termed as supersaturated
solution. It is metastable.

Kinds of solutions.

All the three states of matter (gas , liquid or solid) may behave either as solvent or solute. Depending on the
state of solute or solvent, mainly there may be following nine types of binary solutions.

Solvent Solute Example

Gas Gas Mixture of gases, air.

Gas Liquid Water vapours in air, mist.

Gas Solid Sublimation of a solid into a gas, smoke.

Liquid Gas CO2 gas dissolved in water (aerated drinks).
Liquid Liquid Mixture of miscible liquids, e.g., alcohol in water.

Liquid Solid Salt in water, sugar in water.
Solid Gas
Solid Liquid Adsorption of gases over metals; hydrogen over palladium.
Solid Solid
Mercury in zinc, mercury in gold, CuSO4.5H2O.
Homogeneous mixture of two or more metals (alloys), e.g., copper in gold, zinc in

Among these solutions the most significant type of solutions are those which are in liquid phase and may be
categorised as : (i) Solid in liquid solutions, (ii) Liquid in liquid solutions and (iii) Gas in liquid solutions.

(i) Solid in liquid solutions : For the solid in liquid solutions the solid is referred to as solute. The amount
of solute that is dissolved in 100g of a solvent, to form a saturated solution at a particular temperature is called

solubility. The solubility of a solid solute in liquid depends upon
(a) Nature of solute and solvent
(b) Temperature : Usually the solubility of the solute increases with increase in temperature (e.g.,

KI, KNO3 , NH4 Br ) but in some cases increase in solubility is negligible (e.g. NaCl ) and in cases of some salts
(e.g., NaOH, Na2SO4 , CeSO4 ) solubility decreases with increase in temperature.

Cause of miscibility of solids in liquids : The basic cause of solubility of solid solute in liquid solvent can be
summed up in one line, i.e., “Similia similibus solvanter” meaning like dissolves like which implies that polar
solvents dissolve polar solutes and non-polar solvents dissolve non-polar solutes.

For ionic solutes, the solubility, in general, is related to the magnitude of hydration (or salvation) energy
and lattice energy. In general if the magnitude of hydration energy is greater than lattice energy, the solute is
soluble otherwise it is insoluble.

(ii) Liquid in liquid solutions : When two liquids are mixed, the mixture may be of the following types :

(a) The two components may be almost immiscible : In this case, one of the liquids is polar while the other is
of non-polar nature. For example, benzene and water.

(b) The miscibility of the component may be partial : If the intermolecular attraction of one liquid is different
from intramolecular attraction of the other, there may be a partial miscibility of the two liquids. For example, ether
and water.

(c) The two components may be completely miscible : In this case, the two liquids are of the same nature, i.e.,
they are either polar (like alcohol and water) or non-polar (like benzene and hexane).

Cause of Miscibility of Liquids

(a) Chemically alike liquids dissolve in one another more freely as compared to others, for example, alkanes
are miscible in all proportions with one another. Alkanes are however, not miscible with water because they cannot
form H-bonds with water molecules.

(b) Dipole-Dipole interactions also play an important role in forming liquid solutions.

(c) Molecular sizes of liquids which are mutually soluble, are also approximately same.

(iii) Gas in Liquid solutions : Most of the gases are soluble in water to some extent. The solubility of gas in
water generally depends upon the following factors

(a) Nature of the gas : In general the gases which are easily liquefiable are more soluble in water.

(b) Temperature : The dissolution of gas in water is exothermic process. Hence, the solubility of gas decreases
with rise in temperature.

(c) Pressure : Effect of pressure on the solubility of gas is explained by “Henry’s law”. According to Henry’s law “m
Solubility ∝ Pressure; S = KP where K is Henry’s constant.
Higher the value of K at given pressure, the lower is the solubility of the gas in the liquid. Value of K increases
with increase in temperature while solubility of gas decreases. It is due to this reason that aquatic species are more
comfortable in cold water rather than hot water.
Applications of Henry's law : Henry's law finds several applications in industry and explains some biological
phenomena. Notable among these are :
(i) To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
(ii) To minimise the painful effects accompanying the decompression of deep sea divers, oxygen diluted with
less soluble helium gas is used as breathing gas.
(iii) In lungs, where oxygen is present in air with high partial pressure of oxygen to form oxohaemoglobin. In
tissues where partial pressure of oxygen is low. Oxohaemoglobin releases oxygen for utilization in cellular activities.

Methods of expressing concentration of solution.

Concentration of solution is the amount of solute dissolved in a known amount of the solvent or solution. The
concentration of solution can be expressed in various ways as discussed below.

(1) Percentage : It refers to the amount of the solute per 100 parts of the solution. It can also be called as
parts per hundred (pph). It can be expressed by any of following four methods :

(i) Weight to weight percent (% w/w) = Wt. of solute × 100
Wt. of solution

e.g., 10% Na2CO3 solution w/w means 10g of Na2CO3 is dissolved in 100g of the solution. (It means
10g Na2CO3 is dissolved in 90g of H 2O )

(ii) Weight to volume percent (% w/v) = Wt. of solute × 100
Volume of solution

e.g., 10% Na2CO3 (w/v) means 10g Na2CO3 is dissolved in 100 cc of solution.

(iii) Volume to volume percent (% v/v) = Vol. of solute × 100
Vol. of solution

e.g., 10% ethanol (v/v) means 10 cc of ethanol dissolved in 100 cc of solution.

(iv) Volume to weight percent (% v/w) = Vol. of solute × 100
Wt. of solution

e.g., 10% ethanol (v/w) means 10 cc of ethanol dissolved in 100g of solution.

(2) Parts per million (ppm) and parts per billion (ppb) : When a solute is present in trace quantities, it is
convenient to express the concentration in parts per million and parts per billion. It is the number of parts of solute
per million (106 ) or per billion (109 ) parts of the solution. It is independent of the temperature.

ppm = mass of solute component × 106 ; ppb = mass of solute component × 109
Total mass of solution Total mass of solution

Example:1 15 g of methyl alcohol is dissolved in 35 g of water. The weight percentage of methyl alcohol in solution is
Solution : (a)
(a) 30% (b) 50% (c) 70% (d) 75%

 Weight percentage = Weight of solute × 100
Weight of solution

Total weight of solution = (15 + 35) g = 50 g

Weight percentage of methyl alcohol = Weight of methyl alcohol × 100 = 15 × 100 = 30%
Weight of solution 50

Example: 2 Sea water contains 5.8 × 10 −3 g of dissolved oxygen per kilogram. The concentration of oxygen in parts per
million is

(a) 5.8 ppm (b) 58.5ppm (c) 0.58ppm (d) 0.05ppm

Solution: (a) Part per million = Mass of solute × 106 = 5.8 × 10−3 g × 106 = 5.8 ppm
Mass of solution 103 g

Example: 3 A 500gm toothpaste sample has 0.2g fluoride concentration. The concentration of fluoride ions in terms of

ppm level is [AIIMS 1994]

(a) 250 ppm (b) 200 ppm (c) 400 ppm (d) 1000 ppm

Solution: (c) ppm of F− ions = Mass of solute × 10 6 = 0.2 × 106 = 400 ppm
Mass of solution 500

(3) Strength : The strength of solution is defined as the amount of solute in grams present in one litre (or
dm3 ) of the solution. It is expressed in g/litre or (g / dm3 ).

Strength = Mass of solute in grams
Volume of solution in litres

(4) Normality (N) : It is defined as the number of gram equivalents (equivalent weight in grams) of a solute
present per litre of the solution. Unit of normality is gram equivalents litre–1. Normality changes with temperature
since it involves volume. When a solution is diluted x times, its normality also decreases by x times. Solutions in
term of normality generally expressed as,

N = Normal solution; 5N = Penta normal, 10N = Deca normal; N / 2 = semi normal
N / 10 = Deci normal; N / 5 = Penti normal

N / 100 or 0.01 N = centinormal, N / 1000 or 0.001= millinormal

Mathematically normality can be calculated by following formulas,

(i) Normality (N) = Number of g.eq. of solute = Weight of solute in g. solution (l)
Volume of solution (l) g. eq. weight of solute × Volume of

(ii) N = Wt. of solute per litre of solution ,
g eq. wt. of solute

(iii) N = Wt. of solute × Vol. of 1000
g.eq. wt. of solute solution in ml

(iv) N = Percent of solute × 10 ,
g eq. wt. of solute

(v) N = Strength in g l -1 of solution
g eq. wt. of solute

(vi) N = Wt% × density × 10
Eq. wt.

(vii) If volume V1 and normality N1 is so changed that new normality and volume N 2 and V2 then,
N1V1 = N 2V2 (Normality equation)

(viii) When two solutions of the same solute are mixed then normality of mixture (N) is

N = N 1 V1 + N 2V2
V1 + V2

(ix) Vol. of water to be added i.e., (V2 − V1) to get a solution of normality N 2 from V1 ml of normality N1

V2 − V1 =  N1 − N2  V1

(x) If Wg of an acid is completely neutralised by V ml of base of normality N

Wt. of acid = VN ; Similarly, Wt. of base = Vol. of acid × N of acid
eq. wt. of acid 1000 g eq. wt. of base 1000

(xi) When Va ml of acid of normality Na is mixed with Vb ml of base of normality Nb

(a) If Va Na = Vb Nb (Solution neutral)

(b) If Va Na > Vb Nb (Solution is acidic)

(c) If Vb Nb > Va Na (Solution is basic)

(xii) Normality of the acidic mixture = Va N a + Vb Nb
(Va + Vb )

(xiii) Normality of the basic mixture = Vb Nb + Va N a
(Va + Vb )

(xiv) N = No. of meq * of solute (* 1 equivalent = 1000 milliequivalent or meq.)
Vol. of solution in ml

Example: 4 Normality of a solution containing 9.8 g of H 2SO4 in 250 cm3 of the solution is
Solution: (a)
[MP PMT 1995, 2003; CMC Vellore 1991; JIPMER 1991]
Example: 5
Solution: (b) (a) 0.8 N (b) 1N (c) 0.08 N (d) 1.8 N
Example: 6
Solution: (a) Eq. wt. of H 2 SO4 = Mol. mass of H 2SO4 = 98 = 49
Example: 7 Basicity of H 2SO4 2

∴ Number of g equivalent of H 2SO4 = Weight in g = 9.8 = 0.2
Eq. mass 49

250 cm3 of solution contain H 2SO4 = 0.2g equivalent

∴ 1000 cm3 of the solution contain H 2SO4 = 0.2 × 1000g equivalent = 0.8g equivalent

Hence normality of the solution = 0.8 N

Amount of NaOH present in 200 ml of 0.5N solution is [AIIMS 1992]

(a) 40 g (b) 4 g (c) 0.4 g (d) 4.4 g

Wt. of solute = N × V × g eq. wt. = 0.5 × 200 × 40 = 4g
1000 1000

50 ml of 10N H 2SO4 , 25ml of 12N HCl and 40 ml of 5N HNO3 were mixed together and the volume of
the mixture was made 1000ml by adding water. The normality of the resulting solution will be

(a) 1N (b) 2N (c) 3N [MP PMT 1998, 2002, Kerala CET 2003]

(d) 4 N

N1V1 + N 2V2 + N 3 V3 = N 4 V4
50 × 10 + 25 × 12 + 40 × 5 = N 4 × 1000 or N 4 = 1N

100ml of 0.3N HCl is mixed with 200ml of 0.6N H 2SO4 . The final normality of the resulting solution will
be [DPMT 1994]

(a) 0.1 N (b) 0.2 N (c) 0.3N (d) 0.5N

Soltuion: (d) 100ml of 0.3N HCl contains HCl = 0.03g eq.,

200ml of 0.6N H 2SO4 contains H 2SO4 = 0.6 × 200 = 0.12g eq.

Total g eq. = 0.15 , Total volume = 300ml Finally normality = 0.15 × 1000 = 0.5

Alternatively N1V1 + N 2V2 = N 3V3 i.e., 0.3 × 100 + 0.6 × 200 = N 3 × 300 or 0.3 + 1.2 = 3N 3

or N 3 = 1.5 / 3 = 0.5

Example: 8 An aqueous solution of 6.3g oxalic acid dihydrate is made up to 250ml . The volume of 0.1N NaOH

required to completely neutralize 10ml of this solution is [IIT 2001; CPMT 1986]

(a) 40 ml (b) 20 ml (c)10 ml (d) 4 ml

Solution: (a) Normality of oxalic acid solution = 6.3 × 1000 = 0.4
63 250
Example: 9
Solution: (b) N1V1 = N 2 V2

0.1 × V1 = 0.4 × 10 or V1 = 40 ml

10.6 g of Na2CO3 was exactly neutralised by 100 ml of H 2SO4 solution. Its normality is

(a) 1 N (b) 2 N (c) 1.5 N (d) 0.5 N

Weight of base (w) = 10.6g ; g eq. wt. of base = 53; Vol. of acid (V) = 100 ml ; Normality of acid (N) = ?

g w wt. = V× N ; 10.6 = 100 × N ; N = 1000 × 10.6 = 2
eq. 1000 53 1000 100 × 53

(4) Molarity (M) : Molarity of a solution is the number of moles of the solute per litre of solution (or number
of millimoles per ml. of solution). Unit of molarity is mol/litre or mol/dm3 For example, a molar (1M) solution of

sugar means a solution containing 1 mole of sugar (i.e., 342 g or 6.02 × 1023 molecules of it) per litre of the
solution. Solutions in term of molarity generally expressed as,

1M = Molar solution, 2M = Molarity is two, M or 0.5 M = Semimolar solution,

M or 0.1 M = Decimolar solution, M or 0.01 M = Centimolar solution
10 100

M or 0.001 M = Millimolar solution

• Molarity is most common way of representing the concentration of solution.

• Molarity is depend on temperature as, Molarity ∝ 1

• When a solution is diluted (x times), its molarity also decreases (by x times)
Mathematically molarity can be calculated by following formulas,

(i) M = No. of moles of solute (n) ,
Vol. of solution in litres

(ii) M = Wt. of solute (in gm) per litre of solution
Mol. wt. of solute

(iii) M = Wt. of solute (in gm) × Vol. of 1000 in ml.
Mol. wt. of solute solution

(iv) M = No. of millimoles of solute
Vol. of solution in ml

(v) M = Percent of solute × 10
Mol. wt. of solute

(vi) M = Strength in gl -1of solution
Mol. wt. of solute

(vii) M = 10 × Sp. gr. of the solution × Wt. % of the solute
Mol. wt. of the solute

(viii) If molarity and volume of solution are changed from M1, V1 to M 2 , V2 . Then,

M1V1 = M 2V2 (Molarity equation)

(ix) In balanced chemical equation, if n1 moles of reactant one react with n2 moles of reactant two. Then,

M 1 V1 = M 2V2
n1 n2

(x) If two solutions of the same solute are mixed then molarity (M) of resulting solution.

M = M 1 V1 + M 2V2
(V1 + V2 )

(xi) Volume of water added to get a solution of molarity M 2 from V1 ml of molarity M1 is

V2 − V1 =  M1 − M2  V1

Relation between molarity and normality

Normality of solution = molarity × Molecular mass
Equivalent mass

Normality × equivalent mass = molarity × molecular mass

For an acid, Molecular mass = basicity
Equivalent mass

So, Normality of acid = molarity × basicity.

For a base, Molecular mass = Acidity
Equivalent mass

So, Normality of base = Molarity × Acidity.

Example:10 The molarity of pure water (d = 1 g / l) is [KCET 1993; CMC 1991, CPMT 1974, 88,90]
Solution: (c)
(a) 555 M (b) 5.55 M (c) 55.5 M (d) None

Consider 1000ml of water

Mass of 1000ml of water = 1000 ×1 = 1000 g

Number of moles of water = 1000 = 55.5

Molarity = No. of moles of water = 55.5 = 55.5M
Volume in litre 1

Example:11 The number of iodine atoms present in 1cm3 of its 0.1M solution is [Pb. CET 1990]
Solution: (d)
(a) 6.02 × 10 23 (b) 6.02 × 10 22 (c) 6.02 × 1019 (d) 1.204 × 10 20

1 cc of 0.1M I2 solution = 0.1 = 10−4 mol = 10−4 × 6.02 × 10 23 molecules

= 6.02 × 1019 molecules = 2 × 6.02 × 1019 atoms = 1.204 × 10 20

Example:12 Equal volumes of 0.1M AgNO3 and 0.2M NaCl are mixed. The concentration of NO3− ions in the mixture

will be [MP PMT 1996; CPMT 1991; Pb. PMT 1994; BHU 1994]

(a) 0.1M (b) 0.05M (c) 0.2 M (d) 0.15M

Solution: (b) AgNO3 + NaCl → AgCl + NaNO3
0.1M 0.2 M
Solution: (c) 0.1M AgNO3 reacts with 0.1M NaCl to produce

0.1M AgCl and 0.1M NaNO3

∴ NO3− = 0.1M = 0.05M

[ when equal volumes are mixed dilution occurs]

The molarity of H 2SO4 solution that has a density of 1.84g / cc at 35o C and contains 98% by weight is

[CPMT 1983, 2000; CBSE 1996, 2000; AIIMS 2001]

(a) 4.18M (b) 8.14M (c) 18.4M (d) 18M

Molarity Wt. of solute 1000 98 1000 18.4 M
= Mol. wt. × Vol. of solution (in ml.) = 98 × 54.34 =

 Vol. of solution = mass = 100 = 54.34 ml 
density 1.84

Example:14 Amount of oxalic acid ((COOH)2.2H 2O) in grams that is required to obtain 250ml of a semi-molar solution
Solution: (c) is [CBSE 1994]

Example:15 (a) 17.25 g (b) 17.00 g (c) 15.75 g (d) 15.00 g

Molecular mass of oxalic acid = 126

1000ml of 1M oxalic acid require oxalic acid = 126g

∴ 250 ml of 1M oxalic acid will require oxalic acid = 126 × 250 = 31.5g

Hence 250ml of M oxalic acid will require oxalic acid = 31.5 × 1 = 15.75 g
2 2

∴ Mass of oxalic acid required = 15.75g .

Volume of 10M HCl should be diluted with water to prepare 2.00L of 5M HCl is [AIEEE 2003]

(a) 2L (b) 1.5L (c) 1.00L (d) 0.5L

Solution: (c) In dilution, the following equation is applicable :

M 1 V1 = M 2V2

10M HCl 5M HCl

10 × V1 = 5 × 2.00

V1 = 5 × 2.00 L = 1.00L

Example:16 The volume of 95% H 2SO4 (density = 1.85g cm−3 ) needed to prepare 100cm3 of 15% solution of H 2SO4

(density = 1.10g cm−3 ) will be [CPMT 1983]

(a) 5 cc (b) 7.5 cc (c) 9.4 cc (d) 12.4cc

Solution: (c) Molarity of 95% H 2SO4 = 95 × 100 1 × 1000 = 17.93M
98 / 1.85
Solution: (c) Molarity of 15% H 2SO4 = 15 × 100 1 × 1000 = 1.68 M
98 / 1.10

M1V1 = M 2 V2

(95% H 2SO4 ) (15% H 2SO4 )

17.93 × V1 = 1.68 × 100 or V1 = 9.4cm3

The molarity of a 0.2N Na2CO3 solution will be [MP PMT 1987]

(a) 0.05M (b) 0.2M (c) 0.1M (d) 0.4M

N = M × molecular mass (M 2 )
Equivalent mass (E)

So, M = N× Equivalent mass (E)
Molecular mass (M 2 )

Equivalent mass of salt = Molecular mass
Total positive valency

Equivalent mass of Na 2 CO3 = M2

M = 0.2 × M2 / 2 ; M = 0.2 = 0.1M
M2 2

Example:18 If 20ml of 0.4 N NaOH solution completely neutralises 40ml of dibasic acid. The molarity of acid solution is

[EAMCET 1992; DPMT 1994; JIPMER 1994]

(a) 0.1M (b) 0.2M (b) 0.3M (d) 0.4M

Solution: (a) We know N1V1 = N 2V2

0.4 × 20 = N 2 × 40 ; N 2 = 0.2N

Molarity = Normality of acid ; Here acid solution is dibasic, so M = 0.2 = 0.1M
basicity 2

(5) Molality (m) : It is the number of moles or gram molecules of the solute per 1000 g of the solvent. Unit of

molality is mol / kg . For example, a 0.2 molal (0.2m) solution of glucose means a solution obtained by dissolving

0.2 mole of glucose in 1000gm of water. Molality (m) does not depend on temperature since it involves

measurement of weight of liquids. Molal solutions are less concentrated than molar solution.

Mathematically molality can be calculated by following formulas,

(i) m = Number of moles of the solute × 1000 = Strength per 1000 grams of solvent
Weight of the solvent in grams Molecular mass of solute

(ii) m = No. of gm moles of solute
Wt. of solvent in kg

(iii) m = Wt. of solute × 1000 in g
Mol. wt. of solute Wt. of solvent

(iv) m = No. of millimoles of solute
Wt. of solvent in g

(v) m = 10 × solubility
Mol. wt. of solute

(vi) m = 1000 × wt. % of solute (x)
(100 − x) × mol. wt. of solute

(vii) m = (1000 × sp. 1000 × Molarity wt. of solute)
gravity) − (Molarity × Mol.

Relation between molarity (M) and molality (m)

Molality (m) = Molarity mass
Molarity × molecular
Density −

Molarity (M) = Molality × density
Molality × molecular mass

Example:19 H 2SO4 solution whose specific gravity is 1.98g ml −1 and H 2SO4 by volume is 95%. The molality of the
Solution: (c) solution will be

Example:20 (a) 7.412 (b) 8.412 (c) 9.412 (d) 10.412
Solution: (b)
H 2SO4 is 95% by volume

Wt. of H 2SO4 = 95g

Vol. of solution = 100ml

∴ moles of H 2SO4 95 and weight of solution = 100 × 1.98 = 198g
= 98

Weight of water = 198 − 95 = 103 g

Molality = 95 × 1000 = 9.412
98 × 103

Hence molality of H 2SO4 solution is 9.412

The density of H 2SO4 solution is 1.84 gm ml −1 . In 1 litre solution H 2SO4 is 93% by volume then, the

molality of solution is [UPSEAT 2000]

(a) 9.42 (b) 10.42 (c) 11.42 (d) 12.42

Given H 2SO4 is 93% by volume

Wt. of H 2SO4 = 93 g

Volume of solution =100ml  Density = mass ∴ mass = d × volume

∴ weight of solution = 100 × 1.84 g = 184 g

wt. of water = 184 − 93 = 91 g

Molality = Moles in kg = 93 × 1000 = 10.42
wt. of water 98 × 91

(6) Formality (F) : Formality of a solution may be defined as the number of gram formula masses of the

ionic solute dissolved per litre of the solution. It is represented by F . Commonly, the term formality is used to

express the concentration of the ionic solids which do not exist as molecules but exist as network of ions. A solution

containing one gram formula mass of solute per litre of the solution has formality equal to one and is called formal

solution. It may be mentioned here that the formality of a solution changes with change in temperature.

Formality (F)= Number of gram formula masses of solute = Mass of ionic solute (g)
Volume of solution in litres mass of solute) × (Volume
(gm. formula of solution (l))

Thus, F = WB (g) or WB (g) × 1000
GFM × V(l) GFM × V(ml)

(7) Mole fraction (X) : Mole fraction may be defined as the ratio of number of moles of one component to
the total number of moles of all the components (solvent and solute) present in the solution. It is denoted by the
letter X . It may be noted that the mole fraction is independent of the temperature. Mole fraction is dimensionless.
Let us suppose that a solution contains the components A and B and suppose that WA g of A and WB g of B are
present in it.

Number of moles of A is given by, nA = WA and the number of moles of B is given by, nB = WB

where M A and M B are molecular masses of A and B respectively.

Total number of moles of A and B = nA + nB

Mole fraction of A, XA = nA ; Mole fraction of B, XB = nB
nA + nB nA + nB

The sum of mole fractions of all the components in the solution is always one.

XA + XB = nA + nB =1.
nA + nB nA + nB

Thus, if we know the mole fraction of one component of a binary solution, the mole fraction of the other can

be calculated.

Relation between molality of solution (m) and mole fraction of the solute (XA).

XA = m
55.5 + m

Example: 21 A solution contains 16gm of methanol and 10gm of water, mole fraction of methanol is

(a) 0.90 (b) 0.090 (c) 0.1 [BHU 1981, 87; EAMCET 2003]

(d) 1.9

Solution: (b) Mass of methanol = 16g , Mol. mass of CH3OH = 32

∴ No. of moles of methanol = 16 = 0.5 moles

No. of moles of water = 90 = 5 moles

∴ Mole fraction of methanol = 0.5 = 0.090
5 + 0.5

Example: 22 A solution has 25% of water, 25% ethanol and 50% acetic acid by mass. The mole fraction of each

component will be [EAMCET 1993]

(a) 0.50, 0.3, 0.19 (b) 0.19, 0.3, 0.50 (c) 0.3, 0.19, 0.5 (d) 0.50, 0.19, 0.3

Solution: (d) Since 18g of water = 1mole

25g of water = 25 = 1.38 mole

Similarly, 46g of ethanol = 1 mole

25g of ethanol = 25 = 0.55 moles

Again, 60g of acetic acid = 1 mole

50g of acetic acid = 50 = 0.83 mole

∴ Mole fraction of water = 1.38 = 0.50
1.38 + 0.55 + 0.83

Similarly, Mole fraction of ethanol = 1.38 + 0.55 = 0.19
0.55 + 0.83

Mole fraction of acetic acid = 1.38 + 0.83 + 0.83 = 0.3

(8) Mass fraction : Mass fraction of a component in a solution is the mass of that component divided by the
total mass of the solution. For a solution containing wA gm of A and wB gm of B

Mass fraction of A = wA ; Mass fraction of B = wB
wA + wB wA + wB

Note :  It may be noted that molality, mole fraction, mass fraction etc. are preferred to molarity, normality,

etc. because the former involve the weights of the solute and solvent where as later involve volumes of solutions.
Temperature has no effect on weights but it has significant effect on volumes.

(9) Demal unit (D) : The concentrations are also expressed in “Demal unit”. One demal unit represents one
mole of solute present in one litre of solution at 0o C .

Colligative properties.

Certain properties of dilute solutions containing non-volatile solute do not depend upon the nature of the
solute dissolved but depend only upon the concentration i.e., the number of particles of the solute present in the
solution. Such properties are called colligative properties. The four well known examples of the colligative properties

(1) Lowering of vapour pressure of the solvent.
(2) Osmotic pressure of the solution.

(3) Elevation in boiling point of the solvent.

(4) Depression in freezing point of the solvent.

Since colligative properties depend upon the number of solute particles present in the solution, the simple
case will be that when the solute is a non-electrolyte. In case the solute is an electrolyte, it may split to a number of
ions each of which acts as a particle and thus will affect the value of the colligative property.

Each colligative property is exactly related to any other and thus if one property is measured, the other can be
calculated. The study of colligative properties is very useful in the calculation of molecular weights of the solutes.

Lowering of vapour pressure.

The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given
temperature is called vapour pressure of the liquid. The vapour pressure of a liquid depends on

(1) Nature of liquid : Liquids, which have weak intermolecular forces, are volatile and have greater vapour
pressure. For example, dimethyl ether has greater vapour pressure than ethyl alcohol.

(2) Temperature : Vapour pressure increases with increase in temperature. This is due to the reason that with
increase in temperature more molecules of the liquid can go into vapour phase.

(3) Purity of liquid : Pure liquid always has a vapour pressure greater than its solution.

Raoult’s law : When a non-volatile substance is dissolved in a liquid, the vapour pressure of the liquid

(solvent) is lowered. According to Raoult’s law (1887), at any given temperature the partial vapour pressure (pA) of
any component of a solution is equal to its mole fraction (XA) multiplied by the vapour pressure of this component in

the pure state (p 0 ) . That is, pA = p 0 × XA

The vapour pressure of the solution (Ptotal ) is the sum of the parital pressures of the components, i.e., for the
solution of two volatile liquids with vapour pressures pA and pB .

Ptotal = pA + pB = ( p 0 × X A ) + ( p 0 × XB )

Alternatively, Raoult’s law may be stated as “the relative lowering of vapour pressure of a solution containing
a non-volatile solute is equal to the mole fraction of the solute in the solution.”

Relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour
pressure of the pure solvent. It is determined by Ostwald-Walker method.

Mole fraction of the solute is defined as the ratio of the number of moles of solute to the total numbr of moles
in solution.

Thus according to Raoult’s law,

p0 − p n w
p0 n+ N
= = w m W
m M

where, p = Vapour pressure of the solution; p0 = Vapour pressure of the pure solvent

n = Number of moles of the solute; N = Number of moles of the solvent

w and m = weight and mol. wt. of solute; W and M = weight and mol. wt. of the solvent.
Limitations of Raoult’s law
• Raoult’s law is applicable only to very dilute solutions.
• Raoult’s law is applicable to solutions containing non-volatile solute only.
• Raoult’s law is not applicable to solutes which dissociate or associate in the particular solution.

Example: 23 34.2 g of canesugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be
Solution: (a)
Example: 24 (a) 0.0099 (b) 1.1597 (c) 0.840 (d) 0.9901
Solution: (d)
Example: 25 PA0 − PA = WB / WB / MA / MA = 34.2 / 342 = 0.1 = 0.0099
PA0 MB + WA 34.2 / 342 + 180 / 18 10.1
Solution: (a)
Example:26 Lowering in vapour pressure is the highest for [Roorkee 1989; BHU 1997]
Solution: (c)
(a) 0.2m urea (b) 0.1 m glucose (c) 0.1m MgSO4 (d) 0.1m BaCl 2
PA0 − PA = Molality × (1 − αx + xα + γx)
Solution: (a) PA0

The value of PA0 − PA is maximum for BaCl 2 .

Vapour pressure of CCl4 at 25o C is 143 mm Hg 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in

100 ml of CCl4 . Find the vapour pressure of the solution. (Density of CCl4 = 1.58 g / cm3 ) [CBSE PMT 1996]

(a) 141.93 mm (b) 94.39 mm (c) 199.34 mm (d) 143.99 mm

PA0 − PA = nB ; 143 − Ps 0.5 / 65 or Ps = 141.93 mm
PA0 nA 143 = 158 / 154

The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of

toluene in vapour pressure in contact with equimolar solution of benzene and toluene is [Pb. CET 1988]

(a) 0.50 (b) 0.6 (c) 0.27 (d) 0.73

For equimolar solutions, X B = XT = 0.5

PB = X B × PB0 = 0.5 × 160 = 80 mm

PT = X T × PT0 = 0.5 × 60 = 30 mm

PTotal = 80 + 30 = 110 mm

Mole fraction of toluene in vapour phase = 30 = 0.27

The vapour pressure of a solvent decreases by 10 mm of mercury when a non-volatile solute was added to the

solvent. The mole fraction of the solute in the solution is 0.2. What should be if the decrease in vapour

pressure is to be 20 mm of mercury then the mole fraction of the solvent is [CBSE PMT 1998]

(a) 0.8 (b) 0.6 (c) 0.4 (d) None

∆P / P 0 = X 2

Hence ∆P / P0 = X2 / X 1 i.e. 10 / 20 = 0.2 / X B or XB = 0.4

Mole fraction of solvent = 1 − 0.4 = 0.6

Example:28 The vapour pressure of a solvent A is 0.80 atm. When a non-volatile substance B is added to this solvent its
Solution: (a)
vapour pressure drops to 0.6 atm. the mole fraction of B in the solution is [MP PMT 2000]

(a) 0.25 (b) 0.50 (c) 0.75 (d) 0.90

∆P / P 0 = X B or X B = 0.2 / 0.8 = 0.25

deal and Non-Ideal solution.

(1) Ideal solution : An ideal solution may be defined as the solution which obeys Raoult’s law over the
entire range of concentration and temperature and during the formation of which no change in enthalpy and no
change in volume takes place. So for ideal solutions the conditions are,

(i) It should obey Raoult’s law, i.e., PA = PA0 X A and PB = PB0 X B .
(ii) ∆Hmixing = 0

(iii) ∆Vmixing = 0

The solutions in which solvent-solvent and solute-solute interactions are almost of the same type as solvent-
solute interactions, behave nearly as ideal solutions.

This type of solutions are possible if molecules of solute and solvent are almost of same size and have identical
polarity. For example, solutions of following pairs almost behave as ideal solutions,

n-Heptane and n-hexane.; Chlorobenzene and bromobenzene.

Benzene and toluene; Ethyl bromide and ethyl iodide.

Ethylene bromide and ethylene chloride; Carbon tetrachloride and silicon tetrachloride.

For such solutions the vapour pressure of the solution is always intermediate between the vapour pressures of
pure components A and B, i.e., PA0 and PB0 .

(2) Non-Ideal solution : The solutions which do not obey Raoult’s law and are accompanied by change in
enthalpy and change in volume during their formation are called non-ideal solutions. so, for non-ideal solutions the
conditions are :

(i) It does not obey Raoult's law. PA ≠ PA0 X A ; PB ≠ PB0 X B P=PA+PB
(ii) ∆Hmixing ≠ 0
Vapour Pressure
(iii) ∆Vmixing ≠ 0 PB

Either component of non-ideal binary solution do not follow Raoult's law. PA

The non-ideal solutions are further divided into two types :

(a) Solutions showing positive deviations. (b) Solutions showing negative XA = 1 Mole Fraction XA = 0
deviations. XB = 0 XB = 1

(a) Solutions showing positive deviation : In this type of deviations, Partial and total vapour pressure curves
the partial vapour pressure of each component (say A and B) of solution is for solutions that show positive
deviations from Raoult’s law

greater than the vapour pressure as expected according to Raoult’s law. This type of deviations are shown by the

solutions in which solvent-solvent and solute-solute interactions are stronger than solvent-solute interactions.

For the non-ideal solutions exhibiting positive deviation.

PA > PA0 X A , PB > PB0 X B ; ∆H mixing = + ve ; ∆Vmixing = + ve

e.g. of solutions showing positive deviations

(CH3 )2 CO + CS2 ; (CH3 )2 CO + C2 H5OH
C6 H6 + (CH3 )2 CO ; CCl4 + C6 H6
CCl4 + CHCl3 ; CCl4 + C6 H5CH3
H 2O + CH3OH ; H 2O + C2 H5OH
CH3CHO + CS2 ; CHCl3 + C2 H5OH

(b) Solutions showing negative deviation : In this type of deviations the partial vapour pressure of each
component of solution is less than the vapour pressure as expected according to Raoult’s law. This type of
deviations are shown by the solutions in which solvent-solvent and solute-solute interactions are weaker than
solvent-solute interactions.

For non-ideal solution showing negative deviation. Vapour Pressure P=PA+PB
PA < PA0 X A , PB < PB0 X B ; ∆H mixing = − ve ; ∆Vmixing = − ve
e.g. of solutions showing negative deviations PB
CH 3COOH + C5 H5 N (pyridine)
CHCl3 + (CH 3 )2 CO ; CHCl3 + C6 H6 PA
CHCl3 + (C2 H5 )2 O ; H 2O + HCl
H 2O + HNO3 ; (CH 3 )2 CO + C6 H5 NH 2 XA = 1 Mole Fraction XA = 0
XB = 0 XB = 1

A vapour pressure curve showing
negative deviation (solid lines) from
ideal behaviour (dotted lines)

Differences between ideal and non-ideal solutions

Ideal solutions Solutions with positive deviations Solutions with negative deviations
A……B interactions are similar to A……B interactions are greater than
A……A and B……B interactions A……B interactions are smaller than A……A and B……B interactions
A……A and B……B interactions
PA = PA0 X A; PB = PB0 XB PA < PA0 X A ; PB < PB0 X B
PA > PA0 X A ; PB > PB0 X B
∆H sol. = 0 ∆H sol. < 0
∆H sol. > 0
∆Vmix = 0 ∆Vmix < 0
Do not form azeotrope ∆Vmix > 0 point Exhibit maximum boiling azeotropy

Exhibit minimum boiling

Azeotropic mixture.

Azeotropes are defined as the mixtures of liquids which boil at constant temperature like a pure liquid and
possess same composition of components in liquid as well as in vapour phase. Azeotropes are also called constant
boiling mixtures because whole of the azeotropes changes into vapour state at constant temperature and their
components can not be separated by fractional distillation. Azeotropes are of two types as described below :

(1) Minimum boiling azeotrope : For the solutions with positive deviation there is an intermediate

composition for which the vapour pressure of the solution is maximum and hence, boiling point is minimum. At this

composition the solution distills at constant temperature without change in composition. This type of solutions are
called minimum boiling azeotrope. e.g;

Components Mass % of B Boiling points (K)
AB 95.57 A Azeotrope
71.69 373 351.3 351.1
H2O C2 H5OH 373 350.72
67 334 370 332.3
H2O C2 H5CH 2OH 6.8 329.25 312.2
CHCl 3 C2 H5OH
(CH 3 )2 CO CS2

(2) Maximum boiling azeotrope : For the solutions with negative deviations there is an intermediate
composition for which the vapour pressure of the solution is minimum and hence, boiling point is maximum. At this
composition the solution distill`s at constant temperature without the change in composition. This type of solutions
are called maximum boiling azeotrope. e.g

Components Mass % of B Boiling points (K)

AB A B Azeotrope

H 2O HCl 20.3 373 188 383

H2O HNO3 58.0 373 359 393.5

H2O HClO4 71.6 373 383 476

Osmosis and Osmotic pressure of the solution.

The flow of solvent from pure solvent or from solution of lower concentration into solution of higher
concentration through a semi-permeable membrane is called Osmosis. Osmosis may be divided in following types,

(1) Exo-Osmosis : The outward osmotic flow of water from a cell containing an aqueous solution through a
semi-permeable membrane is called as Exo-osmosis. For example, egg (after removing hard shell) placed in conc.
NaCl solutions, will shrink due to exo-osmosis.

(2) Endo-osmosis : The inward flow of water into the cell containing an aqueous solution through a semi-
permeable membrane is called as endo-osmosis. e.g., an egg placed in water swells up due to endo-osmosis.

(3) Reverse osmosis : If a pressure higher than osmotic pressure is applied on the solution, the solvent will
flow from the solution into the pure solvent through the semi-permeable membrane. Since here the flow of solvent
is in the reverse direction to that observed in the usual osmosis, the process is called reverse osmosis.

Differences between osmosis and diffusion

Osmosis Diffusion
In osmosis movement of molecules takes place through In diffusion there is no role of semi-permeable
a semi-permeable membrane. membrane.
It involves movement of only solvent molecules from
one side to the other. It involves passage of solvent as well as solute
Osmosis is limited to solutions only. molecules from one region to the other.

Osmosis can be stopped or reversed by applying Diffusion can take place in liquids, gases and
additional pressure on the solution side. solutions.

Diffusion can neither be stopped nor reversed

Osmotic pressure (π) : Osmotic pressure may be defined in following ways,

The osmotic pressure of a solution at a particular temperature may be defined as the excess hydrostatic

pressure that builds up when the solution is separated from the solvent by a semi-permeable membrane. It is

denoted by π. Or

Osmotic pressure may be defined as the excess pressure which must be applied to a solution in order to
prevent flow of solvent into the solution through the semi-permeable membrane.


Osmotic pressure is the excess pressure which must be applied to a given solution in order to increase its
vapour pressure until it becomes equal to that of the solution.


Osmotic pressure is the negative pressure which must be applied to (i.e. the pressure which must be
withdrawn from) the pure solvent in order to decrease its vapour pressure until it becomes equal to that of the

Measurements of osmotic pressure : Following methods are used for the measurement of osmotic
pressure : (i) Pfeffer’s method, (ii) Morse and Frazer’s method, (iii) Berkeley and Hartley’s method, (iv)
Townsend’s negative pressure method, (v) De Vries plasmolytic method.

Determination of molecular mass of non-volatile solute from osmotic pressure (π) :

According to Van’t Hoff equation,

π = nB RT ; MB = WB RT where, WB = known mass of solute in gm
V πV

This method is especially suitable for the determination of molecular masses of macromolecules such as
proteins and polymers. This is due to the reason that for these substances the values of other colligative properties
such as elevation in boiling point or depression in freezing point are too small to be measured. On the other hand,
osmotic pressure of such substances are measurable. Conditions for getting accurate value of molecular mass are,

(i) The solute must be non-volatile.

(ii) The solution must be dilute.

(iii) The solute should not undergo dissociation or association in the solution.

Van’t Hoff’s solution equation :

The osmotic pressure is a colligative property. For a given solvent the osmotic pressure depends only upon
the molar concentration of solute but does not depend upon its nature. Osmotic pressure is related to the number of
moles of the solute by the following relation :

πV = nRT or π = n RT ⇒ π = CRT ; here, C= concentration of solution in moles per litre

R = gas constant; T = temperature; n = number of moles of solute; V = volume of solution

Above eq. is called Van’t Hoff’s solution equation

Relation of osmotic pressure with different colligative properties : Osmotic pressure is related to
relative lowering of vapour pressure, elevation of boiling point and depression of freezing point according to the
following relations.

(1) π =  PAo − PA  × dRT (2) π = ∆Tb dRT (3) π = ∆Tf dRT
PAo MB × 1000 × Kb × 1000 × K f

In the above relations, π = Osmotic pressure; d = Density of solution at temperature T; R = Universal gas
constant; M B = Mol. Mass of solute; Kb = Molal elevation constant of solvent; K f = Molal depression constant of


Isotonic, Hypertonic and Hypotonic solutions

Isotonic or iso-osmotic solutions : Two solutions of different substances having same osmotic pressure at
same temperature are known as isotonic solutions.

For isotonic solutions, π 1 = π 2 Primary Condition …..(i)

Also, C1 = C2

or n1 = n2 Secondary Conditions
V1 V2

or w1 = w2 …..(ii)
m1 V1 m2 V2

Eq. (ii) holds good only for those solutes which neither possess the tendency to get associate nor dissociate in

solution, e.g.,

(a) Urea and glucose are isotonic then, π 1 = π 2 and C1 = C2

(b) Urea and NaCl are isotonic then, π 1 =π2 but C1 ≠ C2


(c) Urea and Benzoic acid are isotonic then, π 1 = π 2 but C1 ≠ C2

Hypertonic and Hypotonic Solution : The solution which has more osmotic pressure than the other
solution is called as hypertonic solution and the solution which has lesser osmotic pressure than the other is called
as hypotonic solution.

The flow of solvent is always from lower osmotic pressure to higher osmotic pressure i.e. from hypotonic to
hypertonic solution.

Example: 29 Osmotic pressure is 0.0821 atm at a temperature of 300 K. find concentration in mole/litre [Roorkee 1990]

Solution: (c) (a) 0.033 (b) 0.066 (c) 0.33 × 10 −2 (d) 3
Example: 30
C = P = 0.0821 = 1 = 0.33 × 10 −2 mole / litre
RT 0.0821 × 300 300

The osmotic pressure of 5% (mass-volume) solution of cane sugar at 150o C (mol. mass of sugar = 342) is

[BHU 1995]

(a) 4 atm (b) 5.07 atm (c) 3.55 atm (d) 2.45 atm

Solution: (b) C = 5 × 1 × 1000 = 50 M ; P = 50 × 0.082 × 423 = 5.07 atm
342 100 342 342

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