class 12 cbse

1. THE SOLID STATE
UNIT -1
CLASS 12

3. liquids and gases are called fluids because of their ability to flow. The
fluidity in both of these states is due to the fact that the molecules are
free to move about.
On the contrary, the constituent particles in solids have fixed positions and can only
oscillate about their mean positions. This explains the rigidity in solids.
These properties depend upon the nature of constituent particles and the binding
forces operating between them.

4. In this Unit, we shall discuss different possible arrangements of
particles resulting in several types of structures and explore why
different arrangements of structural units lend different properties to
solids. We will also learn how these properties get modified due to the
structural imperfections
Objectives
Todescribe general characteristics of solid state;
Todistinguish between amorphous and crystalline solids;
Toclassify crystalline solids on the basis of the nature of binding forces;
Todefine crystal lattice and unit cell;
Toexplain close packing of particles;
Todescribe different types of voids and close packed structures;
Tocalculate the packing efficiency of different types of cubic unit cells
To describe the imperfections in solids and their effect on properties;
Tocorrelate the electrical and magnetic properties of solids and
their structure.

5. General Characteristics of Solid State
(i) They have definite mass, volume and shape.
(ii) Intermolecular distances are short.
(iii) Intermolecular forces are strong.
(iv) Their constituent particles (atoms, molecules
or ions) have fixed positions and can only oscillate
about their mean positions.
(v) They are incompressible and rigid

6. CLASSIFICATION OF SOLID STATE
Solids can be classified as crystalline or amorphous on the
basis of the nature of order present in the arrangement of their
constituent particles.
Crystalline solid
They have definite and regular geometry due to definite and
orderly arrangement of atoms, ions or molecules in three-
dimensional space.
They Melt at a sharp and characteristic temperature
Crystalline solids are anisotropic.
These are considered as true solids
Example: NaCl, KCl, Sugar, Quartz, etc

7. Amorphous solids
They do not have any pattern of arrangement of atoms, ions
or molecules and, thus do not have any definite geometrical
shape
Amorphous solids do not have sharp melting point
Amorphous solids are isotropic
These are considered pseudo solids or supercooled liquid.
Example :Plastic, Glass, Rubber etc

8. Crystalline solids are anisotropic in nature,
that is, some of their physical properties like
electrical resistance or refractive index show
different values when measured along
different directions in the same crystals.
ANISOTROPY
Amorphous solids are isotropic in nature. Their properties such as
mechanical strength, refractive index and electrical conductivity, etc.,are
same in all directions.
ISOTROPY

9. Pseudo solids are solids which are considered
to be solid though they resemble liquid in many
respects . They flow very slowly at room
temperature and are considered as supercooled
liquids. Amorphous solids are considered to
be pseudo solid. For e.g. Glass.
Pseudo solids
Glass objects from ancient civilisations are found to become
milky in appearance because of some crystallisation.

10. Classification of Crystalline solid
Crystalline solids can be classified in various ways.
on the basis of nature of intermolecular forces or bonds that hold the
constituent particles together. These are — (i) Van der waals forces;
(ii) Ionic bonds; (iii) Covalent bonds; and (iv) Metallic bonds. On this
basis,
crystalline solids are classified into four categories viz., molecular, ionic,
metallic and covalent solids.
Molecular Solids
(i) Non polar Molecular Solids: They comprise either atoms, for example,
argon and helium or the molecules formed by non polar covalent
bonds, for example, H2, Cl2 and I2. In these solids, the atoms or
molecules are held by weak dispersion forces or London forces
These solids are soft and non-conductors of electricity. They have low melting
points and are usually in liquid or gaseous state at room temperature and pressure

11. (ii) Polar Molecular Solids: The molecules of
substances like HCl, SO2,
etc. are formed by polar covalent bonds. The molecules
in such solids are held together by relatively stronger
dipole-dipole interactions. These solids are soft and
non-conductors of electricity.
Their melting points are higher than those of non polar
molecular solids yet most of these are gases or liquids
under room temperature and pressure. Solid SO2 and
solid NH3 are some examples of such solids

12. (iii) Hydrogen Bonded Molecular Solids: The molecules of such
solidscontain polar covalent bonds between H and F, O or N
atoms.
Strong hydrogen bonding binds molecules of such solids like
H2O(ice). They are non-conductors of electricity. Generally they
are volatile liquids or soft solids under room temperature and
pressure.

14. CRYSTAL LATTICE AND UNIT CELL
The characteristic feature of crystalline solid is regular and repeating arrangement
of constituent particles in space. The regular three dimensional arrangement of
constituent particle in a crystal is known as crystal lattice.
The smallest part of the crystal lattice is known as unit cell. Look at the role of
unit cell in crystal lattice.

15. A unit cell is characterised by 6 parameters:
•3- edges a, b and c (refer the diagram given above). The edges may or may
not be perpendicular to each other.
•3 angles α (between b and c) ,β (between a and c) and γ (between a and b).
Classification of unit cell:
i) Primitive unit cells: The constituent particles are only present at corners of unit
cell.
ii) Centred unit cells: The constituent particles occupy other positions also beside the
corners.

16. Centred unit cell can be further classified into 3-types:
a) Body- centred unit cells: Other than corner it contains one constituent particle at the
centre of the body of unit cell.
b) Face-centred unit cells: Other than corner it contains one constituent particle at the
centre of each face of unit cell.

17. c) End- centred unit cells: Other than corner it contains one constituent particle
at the centre of any two opposite faces of unit cell.

18. On the basis of 6 parameters (3 edges and 3 angles) of unit cell mentioned
above, total seven types of primitive unit cells are possible. The primitive unit
cells show variations in form of centred unit cells. The total number of possible
unit cells (primitive and centred) in 3-dimensional lattice is 14. This is
called Bravais Lattice. The detail of the lattice structure in tabular form is given
below.

22. CLOSED PACKED STRUCTURE
As we the constituent particles in solid are in form of sphere. The spheres in solid are
arranged in different way to leave minimum vacant space. These arrangements of spheres in
different layers form the closed packed structure of solid. The crystals are formed in closed
packed structures. Similarly spheres in solid are also arranged in 3-dimension to form close-
packed structure

23. .
a) Close-packing in one-dimension :
•Spheres are arranged in a row with touching each other as shown below :

25. In this type of arrangement spheres of
both the layers are perfectly aligned
horizontally and vertically. This type of
arrangement is known as AAA type
closed packing of sphere in 3-dimension.
The possible smallest geometrical 3-
dimensional shape would be cube. Thus
this type of arrangement generate simple
cubic lattice with primitive type of unit
cell.

26. (i) Hexagonal closed packing in three dimension:
•Placing second layer over the first layer-
- Spheres in the second layer are fit in the depression of first layer.
Let the number of close packed spheres be N, then:
The number of octahedral voids generated = N
The number of tetrahedral voids generated = 2N
•Placing the third layer above second layer –
- Covering tetrahedral voids : If all the tetrahedral voids of second layer is covered by third
layer, then third layer would be exactly aligned with the first layer as shown below :

29. This forms the ABCABC... type of arrangement of sphere and is known as Cubic closed
packing (ccp) or face centred cubic (fcc) structure.
Coordination number: The number of nearest neighbour touching a particle in closed packed
structure is known as the coordination number of constituent particles.
In both type of crystal lattice (hcp, ccp or fcc) the coordination number for the constituent
particle is 12.

30. Q-1)How many lattice points are there in one unit cell of following lattices? (i) body centred cubic
cell, and (ii) a face centred cubic cell?
(i) For body centred unit cell:
8 lattice points at corner + 1 lattice point at centre of the body = 9 lattice points
(ii) For face centred unit cell:
8 lattice points at corner + 6 lattice point at face-centre of each face = 14 lattice points
Q.2 A cubic solid is made of two elements X and Y. Atoms Y are at the corners of the cube and
X at the body centre. What is the formula of the compound? [CBSE 2006]
Number of X atom per unit cell = 1
Number of Y atom per unit cell = 1/8 x 8 = 1
Formula of compound = XY
Q.3 A cubic solid is made of two elements X and Y. Atoms Y(anions) are at the corners of the cube
and X(cations) at present at face-centre of the cubic lattice. What is the formula of the compound?
Number of X atom per unit cell = 1/2 x 6 = 3
Number of Y atom per unit cell = 1/8 x 8 = 1
Formula of compound = XY3

31. Intext Questions:
Q.1 A compound is formed by two element A and B. Element B
occupy all the ccp position and element A occupy all the
octahedral voids. Find the formula of compound.
A.1 In ccp number of atoms per unit cell = 4
Number of octahedral voids = same as number of atoms present at ccp = 4
Number of atoms of element B (at ccp) = 4
Number of atoms of element A (octahedral voids) = 4
Ratio of element A : B = 4 : 4 = 1 : 1
Formula of compound = AB

32. Q.2 A compound is formed by two elements A and B. The element B
form ccp and element A occupy 2/3 of tetrahedral voids. What is the
formula of compound?
A.2 Let the number of atoms of element B form ccp: N
The number of tetrahedral voids would be: 2N
Since 2/3 of tetrahedral voids are occupied therefore number of atom of
element A would be: 2N x 2/3 = 4/3 N
Ratio of element A: B = 4/3: 1 or
Simple whole number ratio A: B = 4: 3
Formula of compound: A4B3.

33. Q.3 An oxide of aluminium is formed where oxide ions occupy
all the hcp positions and aluminium ion occupy 2/3 of
octahedral voids. What is the formula of compound?
A.3 Let the number of atoms of oxide form hcp: N
The number of aluminium ions in octahedral voids would be: N
Since 2/3 of octahedral voids are occupied therefore number of
aluminium ion would be: N x 2/3 = 2/3 N
Ratio of element Aluminium ion: oxide ion = 2/3N: 1N or
Simple whole number ratio Aluminium ion: oxide ion = 2: 3
Formula of compound: Al2O3.

34. Question : A compound forms hexagonal close-packed structure.
What is the total number of voids in 0.5 mol of it? How many of these
are tetrahedral voids?
Answer: Number of particles in 1 mol of compound = 6.022 X 1023
Number of particles in 0.5 mol = 0.5 X6.022 X 1023 = 3.011 X 1023
Number of octahedral voids = number of atoms or particles.
Number of tetrahedral voids = 2 X number of particles
Therefore number of octahedral in the given compound = 3.011 X 1023
Number of tetrahedral voids = 2 X 3.011 X 1023 = 6.022 X 1023
The total number of voids = 3.011 X 1023 + 6.022 X 1023
= 9.033 X 1023
Number of tetrahedral voids = 6.022 X 1023

35. PACKING EFFICIENCY:
Packing efficiency is defined as percentage of total space filled by the constituent particles in crystal.
Packing efficiency in hcp and ccp structures:
Let us consider the unit cell of ccp in which sphere ABCDEFGH are occupied at corner. And sphere
a,b,c,d,e,f are placed at centre of face.

39. 7. The compound CuCl [Formula mass = 99 g mol-1] has FCC structure like ZnS. Its
density is 3.4 g cm-3. What is the length of the edge of unit cell? [NA = 6.02 X 1023 mol-1]
Ans.

40. 8. An element with density 10 g cm-3forms a cubic cell with edge length of 3 X 10-8 cm. What is
the nature of cubic unit cell if the atomic mass of the element is 81 g mol?
Answer:

41. An element with molar mass 27 g mol-1 forms a cubic unit cell with edge length 4.05 X 10-8cm.
If its density is 2.7 g cm-3, what is the nature of the cubic unit cell?

42. An element X (molar mass = 60 g mol-1) has a density of 6.23 g cm-3. Identify
the type of cubic unit cell, if the edge length of the unit cell is 4 X 10-8cm.

43. An element with density 11.2 g cm-3 forms an fee lattice with edge length of 4 X 10-8cm.
Calculate the atomic mass of the element. (Given: NA = 6.022 X 1023 mol-1)

44. 23. An element with density 2.8 g cm-3 forms a fee unit cell with edge length 4 X 10-8 cm.
Calculate the molar mass of the element. (Given: N. = 6.022 X 1023 mol-1).

45. 40. Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm-3,
calculate atomic radius of niobium using its atomic mass 93 u.