INTRODUCTION

A solid is defined as that form of matter which possesses rigidity and hence possesses a definite shape and a definite volume. Unlike gases and liquids in which the molecules are free to move about and hence constitute fluid state, in a solids the constituent particles are not free to move but oscillate about their fixed positions.

CLASSIFICATION OF SOLIDS

Solids are broadly classified into two types crystalline solids and amorphous solids.

A crystalline solid is a substance whose constituent particles possess regular orderly arrangement e.g. Sodium chloride, sucrose, diamond etc.

An amorphous solid is a substance whose constituent particles do not possess a regular orderly arrangement e.g. glass, plastics, rubber, starch, and proteins. Though amorphous solids do not possess long range regularity, in some cases they may possess small regions of orderly arrangement. These crystalline parts of an otherwise amorphous solid are known as crystallites.

An amorphous solid does not posses a sharp melting point. It undergoes liquefication over a broad range of temperature. The amorphous solid do not posses any characteristic heat of fusion. When an amorphous solid is cut with the help of sharp edged knife it results in an irregular cut.

Amorphous substances are also, sometimes, referred to as super cooled liquids because they posses disorderly arrangement like liquids. In fact many amorphous solids such as glass are capable flowing. Careful examination of the window panes of very old houses reveals that the panes are thicker at the bottom than at the top because the glass has flown under constant influence of gravity.

DISTINCTION BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS

USES OF AMORPHOUS SOLIDS

Amorphous solids such as glass and plastics are very important materials and are widely used in construction, house ware, laboratory ware etc. Amorphous silica is likely to be the best material for converting sunlight into electricity (photovoltaic). Another well known amorphous solid is rubber which is used in making tyres shoes soles etc.

SPACE LATTICE OR CRYSTAL LATTICE

All crystals consists of regularly repeating array of atoms, molecules or ions which are the structural units (or basic units). It is much more convenient to represent each unit of pattern by a point, called lattice point, rather than drawing the entire unit of pattern. This results in a three dimensional orderly arrangement of points called a space lattice or a crystal lattice

Thus, a space lattice may be defined as a regular three dimensional arrangement of identical points in space or it can be defined as an array of points showing how molecules, atoms or ions are arranged at different sites in three dimensional space.

It must be noted that

(a) Each lattice point has the same environment as that of any other point in the lattice

(b) The constituent particles have always to be represented by a lattice point, irrespective of whether it contains a single atom or more than one atoms

UNIT CELL

A unit cell is the smallest repeating unit in space lattice which when repeated over and over again results in a crystal of the given substance. Unit cell may be also defined as a three dimensional group of lattice points that generates the whole lattice on repetition.

BRAVAIS LATTICES

The French crystallographer August Bravais in 1848 showed from geometrical consideration that there can be only 14 different ways in which similar points can be arranged in a three dimensional space. Thus the total no. of space lattices belonging to all the seven basic crystal system but together is only 14.

Types of unit cells

1. Simple unit cell having lattice points only at the corners is called simple, primitive or basic unit cell. A crystal lattice having primitive unit cell is called simple crystal lattice

2. Face centred cubic lattice (fcc) – A unit cell in which the lattice point is at the centre of each face as well as at the corner.

3. Body centred cubic lattice (bcc) A unit cell having a lattice point at the centre of the body as well as at the corners.

Another type of unit cell, called end – centred unit cell is possible for orthorhombic and monoclinic crystal types.

In an end centred there are lattice points in the face centres of only one set of faces in addition to the lattice pints at the corners of the unit cell

The various types of unit cells possible for different crystal classes (in all seven) are given below in tabular form

The Bravais space lattices associated with various crystal system are show in fig below

CALCULATION OF NUMBER OF PARTICLES PER UNIT CELL

The no of atom in a unit cell can be calculated by keeping in view following points

1. An atom at the corners is shared by eight unit cells. Hence the contribution of an atom at the corner to a particular cell $=\frac{1}{8}$.

2. An atom at the face is shared by two unit cells. Hence the contribution of an atom at the face to a particular cell = $\frac{1}{2}$

3. An atom at the edge centre is shared by four unit cells in the lattice and hence contributes only $\frac{1}{4}$ to a particular unit cell.

4. An atom at the body centre of a unit cell belongs entirely to it, so its contribution = 1

The number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence it helps to predict the formula of the compound

SIMPLE CUBIC LATTICE

There are eight atoms at the corners. Each corner atom makes 1/8 contribution to the unit cell.

$\therefore$ No. of atoms present in the unit cell = $\frac{1}{8}\times 8=1$

BODY CENTRED CUBIC (BCC)

BCC has 8 atoms at the corners and one atom, within the body. Each corner atom makes 1/8 contribution and the contribution of atom within the body = 1

$\therefore$ No of atoms present in bcc = $\frac{1}{8}\times 8$(at corner) + 1(at the body centre)

= 1+1 =2

FACE CENTRED CUBIC (FCC)

Fcc has 8 atoms at the corners and 6 atoms on the faces (one on each face)

Contribution by atoms at the corners = $\frac{1}{8}\times 8=1$

Contribution by atom on the face = $\frac{1}{2}\times 6=3$

$\therefore$ Number of atoms present in fcc unit cell = 1+3 = 4

Illustration 1. A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at face centres. Derive the formula of the compound.

Solution: As 'A' atom are present at the 8 corners of the cube therefore no of atoms of A in the unit cell = $\frac{1}{8}\times 8=1$

As B atoms present at the face centres of the cube, therefore no of atoms of B in the unit cell = $\frac{1}{2}\times 6=3$

Hence the formula of compound is AB₃.

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

Solution: In bcc unit cell there are 8 atoms at the corners of the cube and one atom at the body centre

$\therefore$ No of atoms per unit = $8\times \frac{1}{8}+1=2$

No of atoms is 4.0 g of potassium = $\frac{4}{39}\times 6.023\times {10}^{23}$

$\therefore$ No of unit cells in 4.0 g potassium = $\frac{4}{39}\times \frac{6.023\times {10}^{23}}{2}$= 3.09´10²²

Illustration 3. A compound made up of A and B atoms have a crystalline structure, in which A forms Hexagonal close packed structure and B occupies 2/3 of octahedral holes. What will be the simplest molecular formula?

Solution: Effective number of A atoms forming HCP = 6

Effective number of octahedral holes in HCP = 6

So molecular formula will

$A_6{B}_{6\times 2/3}=A_6{B}_4$= A₃B₂

CALCULATION INVOLVING UNIT CELL DIMENSIONS

From the unit cell dimensions, it is possible to calculate the volume of the unit cell. Knowing the density of the metal. We can calculate the mass of the atoms in the unit cell. The determination of the mass of a single atom gives an accurate determination of Avogadro constant.

Suppose edge of unit cell of a cubic crystal determined by X – Ray diffraction is a, d is density of the solid substance and M is the molar mass, then in case of cubic crystal

Volume of a unit cell = a3

Mass of the unit cell = no. of atoms in the unit cell x mass of each atom = Z x m

Here Z = no. of atoms present in one unit cell

m = mass of a single atom

Mass of an atom present in the unit cell = $\frac{m}{N_A}$

$\therefore$ Density d = $\frac{massofunitcell}{volumeofunitcell}=\frac{Z.m}{a^3}$

d = $\frac{Z.M}{a^3\times N_A}$

Note:

Density of the unit cell is same as the density of the substance

Illustration 4. An element having atomic mass 60 has face centred cubic unit cell. The edge length of the unit cell is 400 pm. Find out the density of the element?

Solution: Unit cell edge length = 400 pm

= 400x10-10 cm

Volume of unit cell = (400x10-10)3 = 64x10-24 cm3

Mass of the unit cell = No. of atoms in the unit cell x mass of each atom

No. of atoms in fcc unit cell = $8\times \frac{1}{8}+6\times \frac{1}{2}=4$

$\therefore$ Mass of unit cell = $\frac{4\times 60}{6.023\times {10}^{23}}$

Density of unit cell = $\frac{massofunitcell}{volumeofunitcell}=\frac{4\times 60}{6.023\times {10}^{23}\times 64\times {10}^{-24}}=6.2g/c{m}^3$

Illustration 5. An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element?

Solution: Volume of unit cell = (288x10-10)3 cm3 = 2.39x10-23cm3

Volume of 208 g of the element = $\frac{mass}{volume}=\frac{208}{7.2}=28.88c{m}^3$

No of unit cells in this volume = $\frac{28.88}{2.39\times {10}^{-23}}=12.08\times {10}^{23}$

Since each bcc unit cell contains 2 atoms

$\therefore$ no of atom in 208 g = 2x12.08x1023 = 24.16x1023 atom