1. The Solid State

The vast majority of solid substances like high temperature superconductors, biocompatible plastics, silicon chips, etc. are destined to play an ever expanding role in future development of science.

We are mostly surrounded by solids and we use them more often than liquids and gases. For different applications we need solids with widely different properties. These properties depend upon the nature of constituent particles and the binding forces operating between them. Therefore, study of the structure of solids is important. The correlation between structure and properties helps in discovering new solid materials with desired properties like high temperature superconductors, magnetic materials, biodegradable polymers for packaging, biocompliant solids for surgical implants, etc.

From our earlier studies, we know that 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. In crystalline solids, the constituent particles are arranged in regular patterns.

In this Unit, we shall discuss different possible arrangements of particles resulting in several types of structures. The correlation between the nature of interactions within the constituent particles and several properties of solids will also be explored. How these properties get modified due to the structural imperfections or by the presence of impurities in minute amounts would also be discussed.

1.1 General Characteristics of Solid State

In Class XI you have learnt that matter can exist in three states namely, solid, liquid and gas. Under a given set of conditions of temperature and pressure, which of these would be the most stable state of a given substance depends upon the net effect of two opposing factors. Intermolecular forces tend to keep the molecules (or atoms or ions) closer, whereas thermal energy tends to keep them apart by making them move faster. At sufficiently low temperature, the thermal energy is low and intermolecular forces bring them so close that they cling to one another and occupy fixed positions. These can still oscillate about their mean positions and the substance exists in solid state. The following are the characteristic properties of the 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.

1.2 Amorphous and crystalline solids

Solids can be classified as crystalline or amorphous on the basis of the nature of order present in the arrangement of their constituent particles. A crystalline solid usually consists of a large number of small crystals, each of them having a definite characteristic geometrical shape. In a crystal, the arrangement of constituent particles (atoms, molecules or ions) is ordered. It has long range order which means that there is a regular pattern of arrangement of particles which repeats itself periodically over the entire crystal. Sodium chloride and quartz are typical examples of crystalline solids. An amorphous solid (Greek amorphos = no form) consists of particles of irregular shape. The arrangement of constituent particles (atoms, molecules or ions) in such a solid has only short range order. In such an arrangement, a regular and periodically repeating pattern is observed over short distances only. Such portions are scattered and in between the arrangement is disordered. The structures of quartz (crystalline) and quartz glass (amorphous) are shown in Fig. 1.1 (a) and (b) respectively. While the two structures are almost identical, yet in the case of amorphous quartz glass there is no long range order. The structure of amorphous solids is similar to that of liquids. Glass, rubber and plastics are typical examples of amorphous solids. Due to the differences in the arrangement of the constituent particles, the two types of solids differ glass in their properties.

Crystalline solids have a sharp melting point. On the other hand, amorphous solids soften over a range of temperature and can be moulded and blown into various shapes. On heating they become crystalline at some temperature. Some glass objects from ancient civilisations are found to become milky in appearance because of some crystallisation. Like liquids, amorphous solids have a tendency to flow, though very slowly. Therefore, sometimes these are called pseudo solids or super cooled liquids. Glass panes fixed to windows or doors of old buildings are invariably found to be slightly thicker at the bottom than at the top. This is because the glass flows down very slowly and makes the bottom portion slightly thicker.

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. This arises from different arrangement of particles in different directions. This is illustrated in Fig. 1.2. Since the arrangement of particles is different along different directions, the value of same physicalproperty is found to be different along each direction.
Amorphous solids on the other hand are isotropic in nature. It is because there is no long range order in them and arrangement is irregular along all the directions. Therefore, value of any physical property would be same along any direction. These differences are summarised in Table 1.1.

Table 1.1: Distinction between Crystalline and Amorphous Solids
Property Crystalline solids Amorphous solids
Shape Definite characteristics geometrical shape Irregular shape
Melting Point Melt at a sharp and characteristic temperature Gradually soften over a range of temperature
Cleavage property When cut with a sharp edged tool, they split into two pieces and the newly generated surfaces are plain and smooth When cut with a sharp edged tool , they cut into two pieces with irregular surfaces
Heat of fusion They have a definite and characterstics heat of fusion They do not have definite heat of fusion
Anisotrophy Nature order in of constituent particles Anisotrophy in nature true solids long range order Isotropic in nature pseudo solids or super cooled liquids only short range orders.

Amorphous solids are useful materials. Glass, rubber and plastics find many applications in our daily lives. Amorphous silicon is one of the best photovoltaic material available for conversion of sunlight into electricity.

Intext Questions

1.1 Why are solids rigid?
1.2 Why do solids have a definite volume?
1.3 Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
1.4 Why is glass considered a super cooled liquid?

1.3 Classification of crystalline solids

In Section 1.2, we have learnt about amorphous substances and that they have only short range order. However, most of the solid substances are crystalline in nature. For example, all the metallic elements like iron, copper and silver; non – metallic elements like sulphur, phosphorus and iodine and compounds like sodium chloride, zinc sulphide and naphthalene form crystalline solids.
Crystalline solids can be classified on the basis of nature of intermolecular forces operating in them into four categories viz., molecular, ionic, metallic and covalent solids. Let us now learn about these categories.

1.3.1 Molecular Solids

Molecules are the constituent particles of molecular solids. These are further sub divided into the following categories:
(i) Non polar Molecular Solids: They comprise of 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 about which you have learnt in Class XI. 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.
(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.
(iii) Hydrogen Bonded Molecular Solids: The molecules of such solids contain 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.

1.3.2 Ionic Solids

Ions are the constituent particles of ionic solids. Such solids are formed by the three dimensional arrangements of cations and anions bound by strong coulombic (electrostatic) forces. These solids are hard and brittle in nature. They have high melting and boiling points. Since the ions are not free to move about, they are electrical insulators in the solid state. However, in the molten state or when dissolved in water, the ions become free to move about and they conduct electricity.

1.3.3 Metallic Solids

Metals are orderly collection of positive ions surrounded by and held together by a sea of free electrons. These electrons are mobile and are evenly spread out throughout the crystal. Each metal atom contributes one or more electrons towards this sea of mobile electrons. These free and mobile electrons are responsible for high electrical and thermal conductivity of metals. When an electric field is applied, these electrons flow through the network of positive ions. Similarly, when heat is supplied to one portion of a metal, the thermal energy is uniformly spread throughout by free electrons. Another important characteristic of metals is their lustre and colour in certain cases. This is also due to the presence of free electrons in them. Metals are highly malleable and ductile.

1.3.4 Covalent or Network Solids

A wide variety of crystalline solids of non-metals result from the formation of covalent bonds between adjacent atoms throughout the crystal. They are also called giant molecules. Covalent bonds are strong and directional in nature, therefore atoms are held very strongly at their positions. Such solids are very hard and brittle. They have extremely high melting points and may even decompose before melting. They are insulators and do not conduct electricity. Diamond (Fig. 1.3) and silicon carbide are typical examples of such solids. Graphite is soft and a conductor of electricity. Its exceptional properties are due to its typical structure (Fig. 1.4). Carbon atoms are arranged in different layers and each atom is covalently bonded to three of its neighbouring atoms in the same layer. The fourth valence electron of each atom is present between different layers and is free to move about. These free electrons make graphite a good conductor of electricity. Different layers can slide one over the other. This makes graphite a soft solid and a good solid lubricant.

Intext questions
1.6 Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
1.7 Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
1.8 Ionic solids conduct electricity in molten state but not in solid state. Explain.
1.9 What type of solids are electrical conductors, malleable and ductile?

1.4 Crystals Lattice and unit cells

The main characteristic of crystalline solids is a regular and repeating pattern of constituent particles. If the three dimensional arrangement of constituent particles in a crystal is represented diagrammatically, in which each particle is depicted as a point, the arrangement is called crystal lattice. Thus, a regular three dimensional arrangement is called crystal lattice. Thus, a regular three dimensional arrangement of points in space is called a crystal lattice.
A portion of a crystal lattice is shown in fig. 1.5.

There are only 14 possible three dimensional lattices. These are called Bravais Lattices (after the French mathematician who first described them). The following are the characteristics of a crystal lattice:
(a) Each point in a lattice is called lattice point or lattice site.
(b) Each point in a crystal lattice represents one constituent particle which may be an atom, a molecule (group of atoms) or an ion.
(c) Lattice points are joined by straight lines to bring out the geometry of the lattice.
Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
A unit cell is characterised by:
(i) its dimensions along the three edges, a, b and c. These edges may or may not be mutually perpendicular.
(ii) angles between the edges, α (between b and c) β (between a and c) and γ (between a and b). Thus, a unit cell is characterised by six parameters, a, b c, α, β and γ. These parameters of a typical unit cell are shown in Fig. 1.6.

1.4.1 Primitive and Centred Unit Cells

Unit cells can be broadly divided into two categories, primitive and centred unit cells.
a) Primitive Unit Cells
When constituent particles are present only on the corner positions of a unit cell, it is called as primitive unit cell.
(b) Centred Unit Cells
When a unit cell contains one or more constituent particles present at positions other than corners in addition to those at corners, it is called a centred unit cell. Centred unit cells are of three types:
(i) Body-Centred Unit Cells: Such a unit cell contains one constituent particle (atom, molecule or ion) at its body-centre besides the ones that are at its corners.
(ii) Face-Centred Unit Cells: Such a unit cell contains one constituent particle present at the centre of each face, besides the ones that are at its corners.
(iii) End-Centred Unit Cells: In such a unit cell, one constituent particle is present at the centre of any two opposite faces besides the ones present at its corners.

In all, there are seven types of primitive unit cells (Fig. 1.7).

Their characteristics along with the centred unit cells they can form have been listed in Table 1.3.

Table 1.3: Seven Primitive Unit Cells and their Possible Variations as Centred Unit Cells
Crystal System Possible variation Axile distance or edge lengths Axile angle Examples
Cubic Primitive body
Body-centered
Face-Centered
a=b=c α = β = γ = 90° NaCl, Zinc blende, Cu
Tetragonal Primitive,
Body-centered
a=b≠c α = β = γ = 90° White tin, SnO2,TiO2, CaSO4
Orthorhombic Primitive,
Body-centered,
Face-centered,
End-centered
a≠b≠c α = β = γ = 90° Rhombic sulphur,
KNO3, BaSO4
Hexagonal Primitive a=b≠c α = β = 90°
γ = 120°
Graphite, Zno,Cds
Rhombohedral or trigonal Primitive a=b=c α = β = γ ≠ 90° Calcite (CaCO3), HgS(cinnabar)
Monoclinic Primitive,
End-centered
a≠b≠c α = γ =90°
β ≠ 120°
Monoclinic sulphur,Na2SO4.10H2O
triclinic Primitive a ≠ b ≠ c α ≠ β ≠ γ ≠ 90° K2Cr2O7, CuSo4.5H2O,
H3BO3


1.5 Number of Atoms in a unit Cell

We know that any crystal lattice is made up of a very large number of unit cells and every lattice point is occupied by one constituent particle (atom, molecule or ion). Let us now work out what portion of each particle belongs to a particular unit cell.
We shall consider three types of cubic unit cells and for simplicity assume that the constituent particle is an atom.

1.5.1 primitive Cubic Unit cell

Primitive cubic unit cell has atoms only at its corner. Each atom at a corner is shared between eight adjacent unit cells as shown in Fig. 1.8, four unit cells in the same layer and four unit cells of the upper (or lower) layer. Therefore, only 1/8th of an atom (or molecule or ion) actually belongs to a particular unit cell. In Fig. 1.9, a primitive cubic unit cell has been depicted in three different ways. Each small sphere in Fig. 1.9 (a) represents only the centre of the particle occupying that position and not its actual size. Such structures are called open structures. The arrangement of particles is easier to follow in open structures. Fig. 1.9 (b) depicts space-filling representation of the unit cell with actual particle size and Fig. 1.9 (c) shows the actual portions of different atoms present in a cubic unit cell.

In all, since each cubic unit cell has 8 atoms on its corners, the total number of atoms in one unit cell is 8 ×1/8 = 1 atom.

1.5.2 Body-Centered Cubic Unit Cell

A body-centred cubic (bcc) unit cell has an atom at each of its corners and also one atom at its body centre. Fig. 1.10 depicts (a) open structure (b) space filling model and (c) the unit cell with portions of atoms actually belonging to it. It can be seen that the atom at the body centre wholly belongs to the unit cell in which it is present. Thus in a body-centered cubic (bcc) unit cell:

(i) 8 corners 1/8 per corner atom = 8 × 1/8=1 atom

(ii) 1 body centre atom = 1 1= 1 atom
∴ Total number of atoms per unit cell = 2 atoms

1.5.3 face-Centered Cubic Unit Cell

A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube. It can be seen in Fig. 1.11 that each atom located at the face-centre is shared between two adjacent unit cells and only 1/2 of each atom belongs to a unit cell. Fig. 1.12 depicts (a) open structure (b) space-filling model and (c) the unit cell with portions of atoms actually belonging to it. Thus, in a face-centred cubic (fcc) unit cell:

(i) 8 corners atoms 1/8 atom per unit cell= 8 × 1/8 = 1 atom
(ii) 6 face – centered atoms 1/2 atom per unit cell = 6 × 1/2 = 3 atoms

∴ Total number of atoms per unit cell = 4 atoms

Intext Questions

1.10 Give the significance of a ‘lattice point’.
1.11 Name the parameters that characterise a unit cell.
1.12 Distinguish between
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
1.13 Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

1.6 Closed Packed Structures

In solids, the constituent particles are close-packed, leaving the minimum vacant space. Let us consider the constituent particles as identical hard spheres and build up the three dimensional structure in three steps.
(a) Close Packing in One Dimension
There is only one way of arranging spheres in a one dimensional close packed structure, that is to arrange them in a row and touching each other (Fig. 1.13).

In this arrangement, each sphere is in contact with two of its neighbours. The number of nearest neighbours of a particle is called its coordination number. Thus, in one dimensional close packed one dimension arrangement, the coordination number is 2.

(b) Close Packing in Two Dimensions

Two dimensional close packed structure can be generated by stacking (placing) the rows of close packed spheres. This can be done in two different ways.

(i) The second row may be placed in contact with the first one such that the spheres of the second row are exactly above those of the first row. The spheres of the two rows are aligned horizontally as well as vertically. If we call the first row as ‘A’ type row, the second row being exactly the same as the first one, is also of ‘A’ type. Similarly, we may place more rows to obtain AAA type of arrangement as shown in Fig. 1.14 (a).

In this arrangement, each sphere is in contact with four of its neighbours. Thus, the two dimensional coordination number is 4. Also, if the centres of these 4 immediate neighbouring spheres are joined, a square is formed. Hence this packing is called square close packing in two dimensions.

(ii) The second row may be placed above the first one in a staggered manner such that its spheres fit in the depressions of the first row. If the arrangement of spheres in the first row is called ‘A’ type, the one in the second row is different and may be called ‘B’ type. When the third row is placed adjacent to the second in staggered manner, its spheres are aligned with those of the first layer. Hence this layer is also of ‘A’ type. The spheres of similarly placed fourth row will be aligned with those of the second row (‘B’ type). Hence this arrangement is of ABAB type. In this arrangement there is less free space and this packing is more efficient than the square close packing. Each sphere is in contact with six of its neighbours and the two dimensional coordination number is 6. The centres of these six spheres are at the corners of a regular hexagon (Fig. 1.14b) hence this packing is called two dimensional hexagonal close- packing. It can be seen in Figure 1.14 (b) that in this layer there are some voids (empty spaces). These are triangular in shape. The triangular voids are of two different types. In one row, the apex of the triangles are pointing upwards and in the next layer downwards.

(c) Close Packing in Three Dimensions

All real structures are three dimensional structures. They can be obtained by stacking two dimensional layers one above the other. In the last Section, we discussed close packing in two dimensions which can be of two types; square close-packed and hexagonal close-packed. Let us see what types of three dimensional close packing can be obtained from these.

(i) Three dimensional close packing from two dimensional square close-packed layers: While placing the second square close-packed layer above the first we follow the same rule that was followed when one row was placed adjacent to the other. The second layer is placed over the first layer such that the spheres of the upper layer are exactly above those of the first layer. In this arrangement spheres of both the layers are perfectly aligned horizontally as well as vertically as shown in Fig. 1.15. Similarly, we may place more layers one above the other. If the arrangement of spheres in the first layer is called ‘A’ type, all the layers have the same arrangement. Thus this lattice has AAA…. type pattern. The lattice thus generated is the simple cubic lattice, and its unit cell is the primitive cubic unit cell (See Fig. 1.9).

(ii) Three dimensional close packing from two dimensional hexagonal close packed layers: Three dimensional close packed structure can be generated by placing layers one over the other.

(a) Placing second layer over the first layer
Let us take a two dimensional hexagonal close packed layer ‘A’ and place a similar layer above it such that the spheres of the second layer are placed in the depressions of the first layer. Since the spheres of the two layers are aligned differently, let us call the second layer as B. It can be observed from Fig. 1.16 that not all the triangular voids of the first layer are covered by the spheres of the second layer. This gives rise to different arrangements. Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called

tetrahedral voids because a tetrahedron is formed when the centres of these four spheres are joined. They have been marked as ‘T’ in Fig. 1.16. One such void has been shown separately in Fig. 1.17.

At other places, the triangular voids in the second layer are above the triangular voids in the first layer, and the triangular shapes of these do not overlap. One of them has the apex of the triangle pointing upwards and the other downwards. These voids have been marked as ‘O’ in Fig. 1.16. Such voids are surrounded by six spheres and are called octahedral voids. One such void has been shown separately in Fig. 1.17. The number of these two types of voids depend upon the number of close packed spheres.
Let the number of close packed spheres be N, then:
The number of octahedral voids generated = N
The number of tetrahedral voids generated = 2N

(b) Placing third layer over the second layer
When third layer is placed over the second, there are two possibilities.

(i) Covering Tetrahedral Voids: Tetrahedral voids of the second layer may be covered by the spheres of the third layer. In this case, the spheres of the third layer are exactly aligned with those of the first layer. Thus, the pattern of spheres is repeated in alternate layers. This pattern is often written as ABAB ……. pattern. This structure is called hexagonal close packed (hcp) structure (Fig. 1.18). This sort of arrangement of atoms is found in many metals like magnesium and zinc.

(ii) Covering Octahedral Voids: The third layer may be placed above the second layer in a manner such that its spheres cover the octahedral voids. When placed in this manner, the spheres of the third layer are not aligned with those of either the first or the second layer. This arrangement is called “C’ type. Only when fourth layer is placed, its spheres are aligned with those of the first layer as shown in Figs. 1.18 and 1.19. This pattern of layers is often written as ABCABC ……….. This structure is called cubic close packed (ccp) or face-centred cubic (fcc) structure. Metals such as copper and silver crystallise in this structure.

Both these types of close packing are highly efficient and 74% space in the crystal is filled. In either of them, each sphere is in contact with twelve spheres. Thus, the coordination number is 12 in either of these two structures.

1.6.1 Formula of a compound and number of Voids Filled

Earlier in the section, we have learnt that when particles are close- packed resulting in either ccp or hcp structure, two types of voids are generated. While the number of octahedral voids present in a lattice is equal to the number of close packed particles, the number of tetrahedral voids generated is twice this number. In ionic solids, the bigger ions (usually anions) form the close packed structure and the smaller ions (usually cations) occupy the voids. If the latter ion is small enough then tetrahedral voids are occupied, if bigger, then octahedral voids. Not all octahedral or tetrahedral voids are occupied. In a given compound, the fraction of octahedral or tetrahedral voids that are occupied, depends upon the chemical formula of the compound, as can be seen from the following examples.

Example 1.1 A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound?

Solution The ccp lattice is formed by the element Y. The number of octahedral voids generated would be equal to the number of atoms of Y present in it. Since all the octahedral voids are occupied by the atoms of X, their number would also be equal to that of the element Y. Thus, the atoms of elements X and Y are present in equal numbers or 1:1 ratio. Therefore, the formula of the compound is XY.

Example 1.2 Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B?
solution
The number of tetrahedral voids formed is equal to twice the number of atoms of element B and only 2/3rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3):1 or 4:3 and the formula of the compound is A4B3.

Locating Tetrahedral and Octahedral Voids

We know that close packed structures have both tetrahedral and octahedral voids. Let us take ccp (or fcc) structure and locate these voids in it.
(a) Locating Tetrahedral Voids
Let us consider a unit cell of ccp or fcc lattice [Fig. 1(a)]. The unit cell is divided into eight small cubes.

Each small cube has atoms at alternate corners [Fig. 1(a)]. In all, each small cube has 4 atoms. When joined to each other, they make a regular tetrahedron. Thus, there is one tetrahedral void in each small cube and eight tetrahedral voids in total. Each of the eight small cubes have one void in one unit cell of ccp structure. We know that ccp structure has 4 atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.

(b) Locating Octahedral Voids
Let us again consider a unit cell of ccp or fcc lattice [Fig. 2(a)]. The body centre of the cube, C is not occupied but it is surrounded by six atoms on face centres. If these face centres are joined, an octahedron is generated. Thus, this unit cell has one octahedral void at the body centre of the cube. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. [Fig. 2(b)]. It is surrounded by six atoms, three belonging to the same unit cell (2 on the corners and 1 on face centre) and three belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 th of each void belongs to a particular unit cell.

Thus in cubic close packed structure:
Octahedral void at the body-centre of the cube = 1
12 octahedral voids located at each edge and shared between four unit cells
= 12 × 1/4 =3

∴ Total number of octahedral voids = 4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to this number.

1.7 Packing Efficiency

In whatever way the constituent particles (atoms, molecules or ions) are packed, there is always some free space in the form of voids. Packing efficiency is the percentage of total space filled by the particles. Let us calculate the packing efficiency in different types of structures.

1.7.1 Packing Efficiency in hcp and ccp

Both types of close packing (hcp and ccp) are equally efficient. Let us calculate the efficiency of packing in ccp structure. In Fig. 1.20 let the unit cell edge length be ‘a’ and face diagonal AC = b.

In Δ ABC
AC2=b2=BC2+AB2
=a2+a2=2a2 or
b=√2a
If r is the radius of the sphere, we find
b=4r=√2a
or a = 4r/√2=2√2r

( we can also write, r = a/2√2)

We know, that each unit cell in ccp structure, has effectively 4 spheres. Total volume of four spheres is equal to 4 × ( 4 /3 ) πr 3 and volume of the cube is a or (2√2r)3.

Therefore,
packing efficiency= (Volume occupied by four sphere in the unit cell × 100/Total volume of the unit cell)%
=4×(4/3)πr3× 100/(2√2r)%
=((16/3)πr3×100/16√r3)%=74%

1.7.2 Efficiency of packing in Body-Centred Cubic

From Fig. 1.21, it is clear that the atom at the centre will be in touch with the other two atoms diagonally arranged.

In Δ EFD,

b2 = a2 + a2 = 2a2
b = √2a
Now in Δ AFD

c2 = a2 + b2 = a2 + 2a2 = 3a2
c =√3a
The length of the body diagonal c is equal to 4r, where r is the radius of the sphere (atom), as all the three spheres along the diagonal touch each other.
Therefore, √3a = 4r
a=4r/√3
Also we can write, r = (√3/4)a
In this type of struture, total number of atoms is 2 and their volume is 2 × (4/3)πr3
Volume of the cube, a3 will be equal to (4r/√3)3 or a3 = (4r/√3)3
Therefore,
Packing efficiency = (Volume occupied by two spheres in the unit cell × 100/Total Volume of the unit cell )%
=(2×(4/3)πr3×100/[(4/√3)r]3)%
=(8/3)πr3×100/64/√3)r3)% = 68%

1.7.3 Packing Efficiency in Simple cubic Lattice

In a simple cubic lattice the atoms are located only on the corners of the cube. The particles touch each other along the edge (Fig. 1.22). Thus, the edge length or side of the cube ‘a’, and the radius of each particle, r are related as a = 2r

The volume of the cubic unit cell = a3 = (2r)3 = 8r3 Since a simple cubic unit cell contains only 1 atom The volume of the occupied space = (4/3)πr3
∴ Packing efficiency
=(Volume of one atom/volume of cubic unit cell)× 100%
=(4/3πr3)/8r3× 100 =π/6×100
=52.36% = 52.4%
Thus, we may conclude that ccp and hcp structures have maximum packing efficiency.

1.8 Calculation Involving Unit cell

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 method of determination of Avogadro constant. Suppose, edge length of a unit cell of a cubic crystal determined by X-ray diffraction is a, d the density of the solid substance and M the molar mass. In case of cubic crystal:
Volume of a unit cell = a3
Mass of the unit cell = number of atoms in unit cell × mass of each atom = z × m
(Here z is the number of atoms present in one unit cell and m is the mass of a single atom)
Mass of an atom present in the unit cell:
m =M/NA (M is molar mass)

Therefore, density of the unit cell
=mass of unit cell/ volume of unit cell
=z.m/a3 = z.M/a3.N3 or d= zM/a3NA

Remember, the density of the unit cell is the same as the density of the substance. The density of the solid can always be determined by other methods. Out of the five parameters (d, z M, a and NA), if any four are known, we can determine the fifth.

Example 1.3

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 the unit cell = (288 pm)3
= (288×10-12 m) = (288×10-10cm)3
= 2.39×10-23 cm3
Volume of 208 g of the element
=mass/density=208 g/7.2gcm−3 = 28.88 cm 3
Number of unit cells in this volume
=28.88cm 3/2.39 × 10−23 cm 3 / unit cell
= 12.08×1023 unit cells

Since each bcc cubic unit cell contains 2 atoms, therefore, the total number of atoms in 208 g = 2 (atoms/unit cell) × 12.08 × 1023 unit cells
= 24.16×1023 atoms

Example 1.4

X-ray diffraction studies show that copper crystallises in an fcc unit cell with cell edge of 3.608×10-8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm3, calculate the atomic mass of copper.
Solution
In case of fcc lattice, number of atoms per unit cell, z = 4 atoms
Therefore, M = dNA a3/Z

=8.92 g cm3 × 6.022 × 1023 atoms mol−1 × (3.608 × 10 −8 cm)3/4 atoms
= 63.1 g/mol
Atomic mass of copper = 63.1u

Example 1.5

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).
solution
Since the lattice is ccp, the number of silver atoms per unit cell = z = 4
Molar mass of silver = 107.9 g mol = 107.9×10 kg mol-1
Edge length of unit cell = a = 408.6 pm = 408.6×10–12 m

Density, d = z.M/a3.NA

=(4 × 107.9 × 10−3 kg mol −1)/(408.6 × 10-12m)3 (6.022 × 1023 mol-1)=10.5×103kg m-3

= 10.5 g cm-3

Intext Question

1.14 What is the two dimensional coordination number of a molecule in square close-packed layer?
1.15 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?
1.16 A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
1.17 Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?
1.18 An element with molar mass 2.7×10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×103 kg m-3, what is the nature of the cubic unit cell?

1.9 Imperfections in Solids

Although crystalline solids have short range as well as long range order in the arrangement of their constituent particles, yet crystals are not perfect. Usually a solid consists of an aggregate of large number of small crystals. These small crystals have defects in them. This happens when crystallisation process occurs at fast or moderate rate. Single crystals are formed when the process of crystallisation occurs at extremely slow rate. Even these crystals are not free of defects. The defects are basically irregularities in the arrangement of constituent particles. Broadly speaking, the defects are of two types, namely, point defects and line defects. Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance, whereas the line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points. These irregularities are called crystal defects. We shall confine our discussion to point defects only.

1.9.1 Types of Point defects

Point Defects can be classified into three types : (i) stoichiometric defects (ii) impurity defects and (iii) non-stoichiometric defects.
(a) Stoichiometric Defects
These are the point defects that do not disturb the stoichiometry of the solid. They are also called intrinsic or thermodynamic defects. Basically these are of two types, vacancy defects and interstitial defects.
(i) Vacancy Defect: When some of the lattice sites are vacant, the crystal is said to have vacancy defect (Fig. 1.23). This results in decrease in density of the substance. This defect can also develop when a substance is heated.

(ii) Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have interstitial defect (Fig. 1.24). This defect increases the density of the substance.

Vacancy and interstitial defects as explained above can be shown by non-ionic solids. Ionic solids must always maintain electrical neutrality. Rather than simple vacancy or interstitial defects, they show these defects as Frenkel and Schottky defects.

(iii) Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site (Fig. 1.25). It creates a vacancy defect at its original site and an interstitial defect at its new location.
Frenkel defect is also called dislocation defect. It does not change the density of the solid. Frenkel defect is shown by ionic substance in which there is a large difference in the size of ions, for example, ZnS, AgCl, AgBr and AgI due to small size of Zn2+ and Ag+ ions.

(iv) Schottky Defect: It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal (Fig. 1.26).

Like simple vacancy defect, Schottky defect also decreases the density of the substance. Number of such defects in ionic solids is quite significant. For example, in NaCl there are approximately 106 Schottky pairs per cm3 at room temperature. In 1 cm3 there are about 1022 ions. Thus, there is one Schottky defect per 1016ions. Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For example, NaCl, KCl, CsCl and AgBr. It may be noted that AgBr shows both, Frenkel as well as Schottky defects.

(b) Impurity Defects

If molten NaCl containing a little amount of SrCl2 is crystallised, some of the some of the sites of Na+ ions are occupied by Sr2+ (Fig.1.27). Each Sr2+ replaces two Na+ ions. It occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions. Another similar example is the solid solution of CdCl2 and AgCl.

(c) Non-Stoichiometric Defects

The defects discussed so far do not disturb the stoichiometry of the crystalline substance. However, a large number of non-stoichiometric inorganic solids are known which contain the constituent elements in non-stoichiometric ratio due to defects in their crystal structures. These defects are of two types: (i) metal excess defect and (ii) metal deficiency defect.
(i) Metal Excess Defect
• Metal excess defect due to anionic vacancies: Alkali halides like NaCl and KCl show this type of defect. When crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by loss of electron by sodium atoms to form Na+ ions. The released electrons diffuse into the crystal and occupy anionic sites (Fig. 1.28). As a result the crystal now has an excess of sodium. The anionic sites occupied by unpaired electrons are called F-centres (from the German word Farbenzenter for colour centre). They impart yellow colour to the crystals of NaCl. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac).

• Metal excess defect due to the presence of extra cations at interstitial sites: Zinc oxide is white in colour at room temperature. On heating it loses oxygen and turns yellow.

Now there is excess of zinc in the crystal and its formula becomes Zn1+xO. The excess Zn 2+ ions move to interstitial sites and the electrons to neighbouring interstitial sites.

(ii) Metal Deficiency Defect
There are many solids which are difficult to prepare in the stoichiometric composition and contain less amount of the metal as compared to the stoichiometric proportion. A typical example of this type is FeO which is mostly found with a composition of Fe0.95O. It may actually range from Fe0.93O to Fe0.96O. In crystals of FeO some Fe2+ cations are missing and the loss of positive charge is made up by the presence of required number of Fe3+ ions.

1.10 Electrical Properties

Solids exhibit an amazing range of electrical conductivities, extending over 27 orders of magnitude ranging from 10-20 to 107 ohm-1 m-1. Solids can be classified into three types on the basis of their conductivities.
(i) Conductors: The solids with conductivities ranging between 104 to 107 ohm–1m–1 are called conductors. Metals have conductivities in the order of 107 ohm–1m–1 are good conductors.
(ii) Insulators : These are the solids with very low conductivities ranging between 10-20 to 10-10ohm-1m-1 .
(iii) Semiconductors : These are the solids with conductivities in the intermediate range from 10-6 to 104 ohm-1 m-1 .

1.10.1 Conduction Electricity in Metals

A conductor may conduct electricity through movement of electrons or of ions. Metallic conductors belong to the former category and electrolytes to the latter.
Metals conduct electricity in solid as well as molten state. The conductivity of metals depend upon the number of valence electrons available per atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other as to form a band. If this band is partially filled or it overlaps with a higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal shows conductivity(Fig. 1.29 a).
If the gap between filled valence band and the next higher unoccupied band (conduction band) is large, electrons cannot jump to it and such a substance has very small conductivity and it behaves as an insulator (Fig. 1.29 b).

1.10.2 Conduction of Electricity in Semi-conductors

In case of semiconductors, the gap between the valence band and conduction band is small (Fig. 1.29c). Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semiconductors increases with rise in temperature, since more electrons can jump to the conduction band. Substances like silicon and germanium show this type of behaviour and are called intrinsic semiconductors.

The conductivity of these intrinsic semiconductors is too low to be of practical use. Their conductivity is increased by adding an appropriate amount of suitable impurity. This process is calleddoping. Doping can be done with an impurity which is electron rich or electron deficient as compared to the intrinsic semiconductor silicon or germanium. Such impurities introduce electronic defects in them.

(a) Electron – rich impurities
Silicon and germanium belong to group 14 of the periodic table and have four valence electrons each. In their crystals each atom forms four covalent bonds with its neighbours (Fig. 1.30 a). When doped with a group 15 element like P or As, which contains five valence electrons, they occupy some of the lattice sites in silicon or germanium crystal (Fig. 1.30 b). Four out of five electrons are used in the formation of four covalent bonds with the four neighbouring silicon atoms. The fifth electron is extra and becomes delocalised. These delocalised electrons increase the conductivity of doped silicon (or germanium). Here the increase in conductivity is due to the negatively charged electron, hence silicon doped with electron-rich impurity is called n-type semiconductor.

(b) Electron – deficit impurities
Silicon or germanium can also be doped with a group 13 element like B, Al or Ga which contains only three valence electrons. The place where the fourth valence electron is missing is called electron hole or electron vacancy (Fig. 1.30 c). An electron from a neighbouring atom can come and fill the electron hole, but in doing so it would leave an electron hole at its original position. If it happens, it would appear as if the electron hole has moved in the direction opposite to that of the electron that filled it. Under the influence of electric field, electrons would move towards the positively charged plate through electronic holes, but it would appear as if electron holes are positively charged and are moving towards negatively charged plate. This type of semi conductors are called p-type semiconductors.

Applications of n-type and p-type semiconductors

Various combinations of n-type and p-type semiconductors are used for making electronic components. Diode is a combination of n-type and p-type semiconductors and is used as a rectifier. Transistors are made by sandwiching a layer of one type of semiconductor between two layers of the other type of semiconductor. npn and pnp type of transistors are used to detect or amplify radio or audio signals. The solar cell is an efficient photo-diode used for conversion of light energy into electrical energy.
Germanium and silicon are group 14 elements and therefore, have a characteristic valence of four and form four bonds as in diamond. A large variety of solid state materials have been prepared by combination of groups 13 and 15 or 12 and 16 to simulate average valence of four as in Ge or Si. Typical compounds of groups 13 – 15 are InSb, AlP and GaAs. Gallium arsenide (GaAs) semiconductors have very fast response and have revolutionised the design of semiconductor devices. ZnS, CdS, CdSe and HgTe are examples of groups 12 – 16 compounds. In these compounds, the bonds are not perfectly covalent and the ionic character depends on the electronegativities of the two elements.
It is interesting to learn that transition metal oxides show marked differences in electrical properties. TiO, CrO2 and ReO3 behave like metals. Rhenium oxide, ReO3 is like metallic copper in its conductivity and appearance. Certain other oxides like VO, VO2, VO3 and TiO3 show metallic or insulating properties depending on temperature.

1.11 Magnetic Properties

Every substance has some magnetic properties associated with it. The origin of these properties lies in the electrons. Each electron in an atom behaves like a tiny magnet. Its magnetic moment originates from two types of motions (i) its orbital motion around the nucleus and (ii) its spin motion around the nucleus and (ii) its spin around its own axis (Fig. 1.31). Electron being a charged particle and undergoing these motions can be considered as a small loop of current which possesses a magnetic moment. Thus, each electron has a permanent spin and an orbital magnetic moment associated with it. Magnitude of this magnetic moment is very small and is measured in the unit called Bohr magneton, μB. It is equal to 9.27 × 10-24 A m2 .

On the basis of their magnetic properties, substances can be classified into five categories: (i) paramagnetic (ii) diamagnetic (iii) ferromagnetic (iv) antiferromagnetic and (v) ferrimagnetic.

(i) Paramagnetism: Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired electrons which are attracted by the magnetic field. O2, Cu2+, Fe3+, Cr3+ are some examples of such substances.

(ii) Diamagnetism: Diamagnetic substances are weakly repelled by a magnetic field. H2O, NaCl and C6H6 are some examples of such substances. They are weakly magnetised in a magnetic field in opposite direction. Diamagnetism is shown by those substances in which all the electrons are paired and there are no unpaired electrons. Pairing of electrons cancels their magnetic moments and they lose their magnetic character.

(iii) Ferromagnetism: A few substances like iron, cobalt, nickel, gadolinium and CrO2 are attracted very strongly by a magnetic field. Such substances are called ferromagnetic substances. Besides strong attractions, these substances can be permanently magnetised. In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance the domains are randomly oriented and their magnetic moments get cancelled. When the substance is placed in a magnetic field all the domains get oriented in the direction of the magnetic field (Fig. 1.32 a) and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.

(iv) Antiferromagnetism: Substances like MnO showing anti- ferromagnetism have domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each other’s magnetic moment (Fig. 1.32 b).

(v) Ferrimagnetism: Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers (Fig. 1.32 c). They are weakly attracted by magnetic field as compared to ferromagnetic substances. Fe3O4 (magnetite) and ferrites like MgFe2O4 and ZnFe2O4 are examples of such substances. These substances also lose ferrimagnetism on heating and become paramagnetic.

Intext Questions

1.19 What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
1.20 What type of stoichiometric defect is shown by:
(i) ZnS
(ii) AgBr
1.21 Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
1.22 Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
1.23 A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
1.24 What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

Summary

Solids have definite mass, volume and shape. This is due to the fixed position of their constituent particles, short distances and strong interactions between them.
In amorphous solids, the arrangement of constituent particles has only short range order and consequently they behave like super cooled liquids, do not have sharp melting points and are isotropic in nature. In crystalline solids there is long range order in the arrangement of their constituent particles. They have sharp melting points, are anisotropic in nature and their particles have characteristic shapes. Properties of crystalline solids depend upon the nature of interactions between their constituent particles. On this basis, they can be divided into four categories, namely: molecular, ionic, metallic and covalent solids. They differ widely in their properties.
The constituent particles in crystalline solids are arranged in a regular pattern which extends throughout the crystal. This arrangement is often depicted in the form of a three dimensional array of points which is called crystal lattice. Each lattice point gives the location of one particle in space. In all, fourteen different types of lattices are possible which are called Bravais lattices. Each lattice can be generated by repeating its small characteristic portion called unit cell. A unit cell is characterised by its edge lengths and three angles between these edges. Unit cells can be either primitive which have particles only at their corner positions or centred. The centred unit cells have additional particles at their body centre (body- centred), at the centre of each face (face-centred) or at the centre of two opposite faces (end-centred). There are seven types of primitive unit cells. Taking centred unit cells also into account, there are fourteen types of unit cells in all, which result in fourteen Bravais lattices.
Close-packing of particles result in two highly efficient lattices, hexagonal close-packed (hcp) and cubic close-packed (ccp). The latter is also called face- centred cubic (fcc) lattice. In both of these packings 74% space is filled. The remaining space is present in the form of two types of voids-octahedral voids and tetrahedral voids. Other types of packing are not close-packings and have less efficient packing of particles. While in body-centred cubic lattice (bcc) 68% space is filled, in simple cubic lattice only 52.4 % space is filled.
Solids are not perfect in structure. There are different types of imperfections or defects in them. Point defects and line defects are common types of defects. Point defects are of three types – stoichiometric defects, impurity defects and non-stoichiometric defects. Vacancy defects and interstitial defects are the two basic types of stoichiometric point defects. In ionic solids, these defects are present as Frenkel and Schottky defects. Impurity defects are caused by the presence of an impurity in the crystal. In ionic solids, when the ionic impurity has a different valence than the main compound, some vacancies are created. Non- stoichiometric defects are of metal excess type and metal deficient type. Sometimes calculated amounts of impurities are introduced by doping in semiconductors that change their electrical properties. Such materials are widely used in electronics industry. Solids show many types of magnetic properties like paramagnetism, diamagnetism, ferromagnetism, antiferromagnetism and ferrimagnetism. These properties are used in audio, video and other recording devices. All these properties can be correlated with their electronic configurations or structures.

Exercises
1.1 Define the term ‘amorphous’. Give a few examples of amorphous solids.

1.2 What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

1.3 Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide (P4O10)
(ii) Ammonium phosphate (NH4)3PO4
(iii) SiC
(iv) I2
(v) P4
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si

1.4
(i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?

1.5 How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

1.6′Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

1.7 How will you distinguish between the following pairs of terms:
(i) Hexagonal close-packing and cubic close-packing?
(ii) Crystal lattice and unit cell ?
(iii) Tetrahedral void and octahedral void ?

1.8 How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred

1.9 Explain
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.

1.10 Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).

1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm and density is 10.5 g cm–3, calculate the atomic mass of silver.

1.12 A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

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

1.14 If the radius of the octahedral void is r and radius of the atoms in close- packing is R, derive relation between r and R.

1.15 Copper crystallises into a fcc lattice with edge length 3.61 × 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3.

1.16 Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

1.17 What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

1.18 Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

1.19 Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

1.20 Classify each of the following as being either a p-type or a n-type semiconductor:
(i) Ge doped with In
(ii) Si doped with B.

I. Multiple Choice Questions (Type-I)

1. Which of the following conditions favours the existence of a substance in the solid state?

(i) High temperature
(ii) Low temperature
(iii) High thermal energy
(iv) Weak cohesive forces

2. Which of the following is not a characteristic of a crystalline solid?

(i) Definite and characteristic heat of fusion.
(ii) Isotropic nature.
(iii) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal.
(iv) A true solid

3. Which of the following is an amorphous solid?

(i) Graphite (C)
(ii) Quartz glass (SiO2)
(iii) Chrome alum
(iv) Silicon carbide (SiC)

4. Which of the following arrangements shows schematic alignment of magnetic moments of antiferromagnetic substances?

5. Which of the following is true about the value of refractive index of quartz glass?

(i) Same in all directions
(ii) Different in different directions
(iii) Cannot be measured
(iv) Always zero

6. Which of the following statement is not true about amorphous solids?

(i) On heating they may become crystalline at certain temperature.
(ii) They may become crystalline on keeping for long time.
(iii) Amorphous solids can be moulded by heating.
(iv) They are anisotropic in nature.

7. The sharp melting point of crystalline solids is due to ___________.

(i) a regular arrangement of constituent particles observed over a short distance in the crystal lattice.
(ii) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
(iii) same arrangement of constituent particles in different directions.
(iv) different arrangement of constituent particles in different directions.

8. Iodine molecules are held in the crystals lattice by ____________.

(i) london forces
(ii) dipole-dipole interactions
(iii) covalent bonds
(iv) coulombic forces

9. Which of the following is a network solid?

(i) SO2 (Solid)
(ii) I2
(iii) Diamond
(iv) H2O (Ice)

10. Which of the following solids is not an electrical conductor?

(A) Mg(s)
(B) TiO (s)
(C) I2 (s)
(D) H2O (s)

(i) (A) only
(ii) (B) Only
(iii) (C) and (D)
(iv) (B), (C) and (D)

11. Which of the following is not the characteristic of ionic solids?

(i) Very low value of electrical conductivity in the molten state.
(ii) Brittle nature.
(iii) Very strong forces of interactions.
(iv) Anisotropic nature.

12. Graphite is a good conductor of electricity due to the presence of __________.

(i) lone pair of electrons
(ii) free valence electrons
(iii) cations
(iv) anions

13. Which of the following oxides behaves as conductor or insulator depending upon temperature?

(i) TiO
(ii) SiO2
(iii) TiO3
(iv) MgO

14. Which of the following oxides shows electrical properties like metals?

(i) SiO2
(ii) MgO
(iii) SO2(s)
(iv) CrO2

15. The lattice site in a pure crystal cannot be occupied by _________.

(i) molecule
(ii) ion
(iii) electron
(iv) atom

16. Graphite cannot be classified as __________.

(i) conducting solid
(ii) network solid
(iii) covalent solid
(iv) ionic solid

17. Cations are present in the interstitial sites in __________.

(i) Frenkel defect
(ii) Schottky defect
(iii) Vacancy defect
(iv) Metal deficiency defect

18. Schottky defect is observed in crystals when __________.

(i) some cations move from their lattice site to interstitial sites.
(ii) equal number of cations and anions are missing from the lattice.
(iii) some lattice sites are occupied by electrons.
(iv) some impurity is present in the lattice.

19. Which of the following is true about the charge acquired by p-type semiconductors?

(i) positive
(ii) neutral
(iii) negative
(iv) depends on concentration of p impurity

20. To get a n-type semiconductor from silicon, it should be doped with a substance with valence__________.

(i) 2
(ii) 1
(iii) 3
(iv) 5

21. The total number of tetrahedral voids in the face centred unit cell is __________.

(i) 6
(ii) 8
(iii) 10
(iv) 12

22. Which of the following point defects are shown by AgBr(s) crystals?

(A) Schottky defect (B) Frenkel defect
(C) Metal excess defect (D) Metal deficiency defect

(i) (A) and (B)
(ii) (C) and (D)
(iii) (A) and (C)
(iv) (B) and (D)

23. In which pair most efficient packing is present?

(i) hcp and bcc
(ii) hcp and ccp
(iii) bcc and ccp
(iv) bcc and simple cubic cell

24. The percentage of empty space in a body centred cubic arrangement is ________.

(i) 74
(ii) 68
(iii) 32
(iv) 26

25. Which of the following statement is not true about the hexagonal close packing?

(i) The coordination number is 12.
(ii) It has 74% packing efficiency.
(iii) Tetrahedral voids of the second layer are covered by the spheres of the third layer.
(iv) In this arrangement spheres of the fourth layer are exactly aligned withthose of the first layer.

26. In which of the following structures coordination number for cations and anions in the packed structure will be same?

(i) Cl ion form fcc lattice and Na+ ions occupy all octahedral voids of the unit cell.
(ii) Ca2+ ions form fcc lattice and F ions occupy all the eight tetrahedral voids of the unit cell.
(iii) O2– ions form fcc lattice and Na+ ions occupy all the eight tetrahedral voids of the unit cell.
(iv) S2– ions form fcc lattice and Zn2+ ions go into alternate tetrahedral voids of the unit cell.

27. What is the coordination number in a square close packed structure in two dimensions?

(i) 2
(ii) 3
(iii) 4
(iv) 6

28. Which kind of defects are introduced by doping?

(i) Dislocation defect
(ii) Schottky defect
(iii) Frenkel defects
(iv) Electronic defects

29. Silicon doped with electron-rich impurity forms ________.

(i) p-type semiconductor
(ii) n-type semiconductor
(iii) intrinsic semiconductor
(iv) insulator

30. Which of the following statements is not true?

(i) Paramagnetic substances are weakly attracted by magnetic field.
(ii) Ferromagnetic substances cannot be magnetised permanently.
(iii) The domains in antiferromagnetic substances are oppositely oriented with respect to each other.
(iv) Pairing of electrons cancels their magnetic moment in the diamagnetic substances.

31. Which of the following is not true about the ionic solids?

(i) Bigger ions form the close packed structure.
(ii) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size.
(iii) Occupation of all the voids is not necessary.
(iv) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.

32. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field becuase ________.

(i) all the domains get oriented in the direction of magnetic field.
(ii) all the domains get oriented in the direction opposite to the direction of magnetic field.
(iii) domains get oriented randomly.
(iv) domains are not affected by magnetic field.

33. The correct order of the packing efficiency in different types of unit cells is ________.

(i) fcc < bcc < simple cubic (ii) fcc > bcc > simple cubic
(iii) fcc < bcc > simple cubic
(iv) bcc < fcc > simple cubic

34. Which of the following defects is also known as dislocation defect?

(i) Frenkel defect
(ii) Schottky defect
(iii) Non-stoichiometric defect
(iv) Simple interstitial defect

35. In the cubic close packing, the unit cell has ________.

(i) 4 tetrahedral voids each of which is shared by four adjacent unit cells.
(ii) 4 tetrahedral voids within the unit cell.
(iii) 8 tetrahedral voids each of the which is shared by four adjacent unit cells.
(iv) 8 tetrahedral voids within the unit cells.

36. The edge lengths of the unit cells in terms of the radius of spheres constituting fcc,bcc and simple cubic unit cell are respectively________.

37. Which of the following represents correct order of conductivity in solids?

(i) κmetals >> κinsulators< κsemiconductors
(ii) κmetals<< κinsulators < κsemiconductors
(iii) κmetals κsemiconductors> κinsulators = zero
(iv) κmetals< κsemiconductors > κinsulators ≠ zero

II. Multiple Choice Questions (Type-II)

Note : In the following questions two or more options may be correct.

38. Which of the following is not true about the voids formed in 3 dimensional hexagonal close packed structure?

(i) A tetrahedral void is formed when a sphere of the second layer is present above triangular void in the first layer.
(ii) All the triangular voids are not covered by the spheres of the second layer.
(iii) Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular shapes of these voids do not overlap.
(iv) Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer.

39. The value of magnetic moment is zero in the case of antiferromagnetic substances because the domains ________.

(i) get oriented in the direction of the applied magnetic field.
(ii) get oriented opposite to the direction of the applied magnetic field.
(iii) are oppositely oriented with respect to each other without the application of magnetic field.
(iv) cancel out each other’s magnetic moment.

40. Which of the following statements are not true?

(i) Vacancy defect results in a decrease in the density of the substance.
(ii) Interstitial defects results in an increase in the density of the substance.
(iii) Impurity defect has no effect on the density of the substance.
(iv) Frankel defect results in an increase in the density of the substance.

41. Which of the following statements are true about metals?

(i) Valence band overlaps with conduction band.
(ii) The gap between valence band and conduction band is negligible.
(iii) The gap between valence band and conduction band cannot be determined.
(iv) Valence band may remain partially filled.

42. Under the influence of electric field, which of the following statements is true about the movement of electrons and holes in a p-type semi conductor?

(i) Electron will move towards the positvely charged plate through electron holes.
(ii) Holes will appear to be moving towards the negatively charged plate.
(iii) Both electrons and holes appear to move towards the positively charged plate.
(iv) Movement of electrons is not related to the movement of holes.

43. Which of the following statements are true about semiconductors?

(i) Silicon doped with electron rich impurity is a p-type semiconductor.
(ii) Silicon doped with an electron rich impurity is an n-type semiconductor.
(iii) Delocalised electrons increase the conductivity of doped silicon.
(iv) An electron vacancy increases the conductivity of n-type semiconductor.

44. An excess of potassium ions makes KCl crystals appear violet or lilac in colour since ________.

(i) some of the anionic sites are occupied by an unpaired electron.
(ii) some of the anionic sites are occupied by a pair of electrons.
(iii) there are vacancies at some anionic sites.
(iv) F-centres are created which impart colour to the crystals.

45. The number of tetrahedral voids per unit cell in NaCl crystal is ________.

(i) 4
(ii) 8
(iii) twice the number of octahedral voids.
(iv) four times the number of octahedral voids.

46. Amorphous solid can also be called ________.

(i) pseudo solids
(ii) true solids
(iii) super cooled liquids
(iv) super cooled solids

47. A perfect crystal of silicon (Fig. 1.1) is doped with some elements as given in the options. Which of these options show n-type semiconductors?

48. Which of the following statements are correct?

(i) Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic.
(ii) Ferrimagnetic substances do not lose ferrimagnetism on heating and remain ferrimagnetic.
(iii) Antiferromagnetic substances have domain structures similar to ferromagnetic substances and their magnetic moments are not cancelled by each other.
(iv) In ferromagnetic substances all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field.

49. Which of the following features are not shown by quartz glass?

(i) This is a crystalline solid.
(ii) Refractive index is same in all the directions.
(iii) This has definite heat of fusion.
(iv) This is also called super cooled liquid.

50. Which of the following cannot be regarded as molecular solid?

(i) SiC (Silicon carbide)
(ii) AlN
(iii) Diamond
(iv) I2

51. In which of the following arrangements octahedral voids are formed?

(i) hcp
(ii) bcc
(iii) simple cubic
(iv) fcc

52. Frenkel defect is also known as ________.

(i) stoichiometric defect
(ii) dislocation defect
(iii) impurity defect
(iv) non-stoichometric defect

53. Which of the following defects decrease the density?

(i) Interstitial defect
(ii) Vacancy defect
(iii) Frankel defect
(iv) Schottky defect

III. Short Answer Type

54. Why are liquids and gases categorised as fluids?
55. Why are solids incompressible?
56. Inspite of long range order in the arrangement of particles why are the crystals usually not perfect?
57. Why does table salt, NaCl, some times appear yellow in colour?
58. Why is FeO (s) not formed in stoichiometric composition?
59. Why does white ZnO (s) becomes yellow upon heating?
60. Why does the electrical conductivity of semiconductors increase with rise in temperature?
61. Explain why does conductivity of germanium crystals increase on doping with galium.
62. In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
63. Under which situations can an amorphous substance change to crystalline form?

IV. Matching Type

Note : In the following questions match the items given in Column I with the items given in Column II. In some questions more than one item of Column I and Column II may match.

64. Match the defects given in Column I with the statements in given Column II.

Column I Column II
(i) Simple vacancy defect (a) shown by non-ionic solids and increases density of the solid.
(ii) Simple interstitial defect (b) shown by ionic solids and decreases density of the solid.
(iii) Frenkel defect (c) shown by non ionic solids and density of the solid decreases
(iv) Schottky defect d) shown by ionic solids and density of the solid remains the same.

65. Match the type of unit cell given in Column I with the features given in Column II.

Column I Column II
(i) Primitive cubic unit cell (a) Each of the three perpendicular edges compulsorily have the different edge length i.e; a≠b≠c.
(ii) Body centred cubic unit cell (b) Number of atoms per unit cell is one.
(iii) Face centred cubic unit cell (c) Each of the three perpendicular edges compulsorily have the same edge length i.e; a = b = c
(iv) End centred orthorhombic (d) In addition to the contribution from unit cell the corner atoms the number of atoms present in a unit cell is one.
(e) In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three.

66. Match the types of defect given in Column I with the statement given in Column II.

Column I Column II
(i) Impurity defect (a) NaCl with anionic sites called F-centres
(ii) Metal excess defect (b) FeO with Fe3+
(iii) Metal deficiency defect (c) NaCl with Sr2+ and some cationic sites vacant

67 Match the items given in Column I with the items given in Column II.

Column I Column II
(i) Mg in solid state (a) p-Type semiconductor
(ii) MgCl2 in molten state (b) n-Type semiconductor
(iii) Silicon with phosphorus (c) Electrolytic conductors
(iv) Germanium with boron (d) Electronic conductors

68. Match the type of packing given in Column I with the items given in Column II.

Column I Column II
(i) Square close packing in two dimensions (a) Triangular voids
(ii) Hexagonal close packing in two dimensions (b) Pattern of spheres is repeated in every fourth layer
(iii) Hexagonal close packing in three dimensions (c) Coordination number 4
(iv) Cubic close packing in three dimensions (d) Pattern of sphere is repeated in alternate layers

V. Assertion and Reason Type

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(i) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.

69. Assertion : The total number of atoms present in a simple cubic unit cell is one.
Reason : Simple cubic unit cell has atoms at its corners, each of which is shared between eight adjacent unit cells.

70. Assertion : Graphite is a good conductor of electricity however diamond belongs to the category of insulators.
Reason : Graphite is soft in nature on the other hand diamond is very hard and brittle.

71. Assertion : Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four.
Reason : Besides the body centre there is one octahedral void present at the centre of each of the six faces of the unit cell and each of which is shared between two adjacent unit cells.

72. Assertion : The packing efficiency is maximum for the fcc structure.
Reason : The cordination number is 12 in fcc structures.

73. Assertion : Semiconductors are solids with conductivities in the intermediate range from 10–6–104 ohm–1m–1.
Reason : Intermediate conductivity in semiconductor is due to partially filled valence band.

VI. Long Answer Type

74. With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
75. Show that in a cubic close packed structure, eight tetrahedral voids are present per unit cell.
76. How does the doping increase the conductivity of semiconductors?
77. A sample of ferrous oxide has actual formula Fe0.93O1.00. In this sample what fraction of metal ions are Fe2+ ions? What type of nonstoichiometric defect is present in this sample?

ANSWERS

I. Multiple Choice Questions (Type-I)

1. (ii)      2. (ii)      3. (ii)      4. (iv)      5. (i)      6. (iv)      7. (ii)      8. (i)      9. (iii)
10. (iii) 11. (i)      12. (ii)      13. (iii)      14. (iv)      15. (iii)      16. (iv)      17. (i)      18. (ii) 19. (ii) 20. (iv)      21. (ii)      22. (i)      23. (ii)      24. (iii)      25. (iv)      26. (i)      27. (iii) 28. (iv) 29. (ii)      30. (ii)      31. (iv)      32. (i)      33. (ii)      34. (i)      35. (iv)      36. (i)   37. (i)

II. Multiple Choice Questions (Type-II)

38. (iii), (iv)      39. (iii), (iv)      40. (iii), (iv)      41. (i), (ii), (iv)      42. (i), (ii)      43. (ii), (iii)      44. (i), (iv)      45. (ii), (iii)      46. (i), (iii)      47. (i), (iii)      48. (i), (iv)      49. (i), (iii)      50. (i), (ii), (iii)      51. (i), (iv)      52. (i), (ii)      53. (ii), (iv)

III. Short Answer Type

54. The liquids and gases have a property to flow i.e. the molecules can move past and tumble over one another freely. Hence, they have been categorised as fluids.

55. The distance between the constituent particles (atoms, ions, molecules etc.) is very less in solids. On bringing them still closer repulsion will start between electron clouds of these particles. Hence, they cannot be brought further close to each other.

56. Crystals have long range repeated pattern of arrangement of constitutent particles but in the process of crystallisation some deviations from the ideal arrangement (i.e. defects) may be introduced, therefore, crystals are usually not perfect.

57. Yellow colour in sodium chloride is due to metal excess defect due to which unpaired electrons occupy anionic sites. These sites are called F-centres. These electrons absorb energy from the visible region for the excitation which makes crystal appear yellow.

58. In the crystals of FeO, some of the Fe2+ cations are replaced by Fe3+ ions. Three Fe2+ ions are replaced by two Fe3+ ions to make up for the loss of positive charge. Eventually there would be less amount of metal as compared to stoichiometric proportion.

59. On heating ZnO loses oxygen according to the following reaction.

Zn2+ ions and electrons move to interstitial sites and F-centres are created which impart yellow colour to ZnO(s).

60. The gap between conduction band and valence band is small in semiconductors (Fig. 1.1), therefore, electrons from the valence band can jump to the conduction band on increasing temperature. Thus they become more conducting as the temperature increases.

61. On doping germanium with galium some of the positions of lattice of germanium are occupied by galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearby germanium atom is not satisfied. The place remains vacant. This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the gap, thereby creating a hole in its original position. Under theinfluence of electric field electrons move towards positively charged plates through these holes and conduct electricity. The holes appear to move towards negatively charged plates.

62. M2N3

63. See page no. 3 of NCERT textbook for Class XII.

IV. Matching Type

64. (i) → (c) (ii) → (a) (iii) → (d) (iv) → (b)
65. (i) → (b), (c) (ii) → (c), (d) (iii) → (c), (e) (iv) → (a), (d)
66. (i) → (c) (ii) → (a) (iii) → (b)
67. (i) → (d) (ii) → (c) (iii) → (b) (iv) → (a)
68. (i) → (c) (ii) → (a) (iii) → (d) (iv) → (b)

V. Assertion and Reason Type

69. (i) 70. (ii) 71. (iii) 72. (ii) 73. (iii)

VI. Long Answer Type

74. [Hint : Draw structure and discuss]
75. [Hint :D raw structure and discuss]
76. See page no. 26 of NCERT textbook for Class XII.
77. Let the formula of sample be

(Fe2+)x(Fe3+)yO.

On looking at the given formula of the compound

x + y = 0.93 …………………………………. (1)

Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen. Therefore,

2x + 3y = 2 …………………………………… (2)
⇒ x + 3/2 y = 1 ……………………………….. (3)

On subtracting equation (1) from equation (3) we have

3/2 y – y = 1 – 0.93
⇒ 1/2 y = 0.07
⇒ y = 0.14

On putting the value of y in equation (1) we get,

x + 0.14 = 0.93
⇒ x = 0.93 – 0.14
x = 0.79

Fraction of Fe2+ ions present in the sample = 0.79/0.93 = 0.81

Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.

20 thoughts on “1. The Solid State

  1. Hmm it appears like your site ate my first
    comment (it was extremely long) so I guess I’ll just sum it up what I had written and say, I’m thoroughly enjoying your blog.

    I too am an aspiring blog blogger but I’m still new to the whole thing. Do you have any suggestions for novice blog writers? I’d definitely appreciate it.

  2. Old edition NCERT book solid state QNo !.20( ii), question was B doped with Si regarding semiconductor. Reference chemistry books have given answer n type semiconductor. I could not make me understand.. some school printed that question and the teachers followed that answer. I had some contradictions regarding the question. But I get on line NCERT book question has been changed,
    1.20 (ii) Si doped with B. I am very happy to see that the confusing question has been amended. So accordingly answer comes p type semiconductor. Many students and teachers will be definitely benefitted confirm me

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