The Solid State – Definition, Characteristics, Dielectric Properties of Solids

The Solid State

You know from your previous classes that all three states Of matter are made of postludes, wSScfc osar be the molecules or ions. However, the arrangement of particles differs in each <m fe a gas, this precedes ece apart whereas in a liquid they are closer.

But in both gases and liquids, the oatstoenipszixSes-haver? arrangement and they are in motion. In contrast, in a solid, the particles are tightly packed and; are net free he move due to the strong forces of interaction, between them—they only oscillate about their fared pcasSkaas.

Therefore, solids are rigid and have a definite volume. As you know, liquids too have a despite voters. But they do not have a fixed shape, which solids do.

Crystalline And Amorphous Solids

Based on the order of arrangement of particles, solids can be categorised as crystalline or amorphous. Cryste-fre solids are those in which the particles (atoms, molecules or ions) are arranged in a pattern, that repeats itself over a long range.

Examples are sugar, ice, salt and metals. The particles of amorphous solids, on the other hand, are arranged in a pattern that repeats itself over a very short range. Examples are wax, butter, skin powder, glass and plastics.

Solid state chemistry class 12 notes

Let us now look at a few properties of crystalline and amorphous solids. The differences in their presences arise due to the difference in the arrangement of their constituent particles.

Crystalline solids are anisotropic, meaning that their physical properties, such as electrical resistance, thermal expansion and refractive index, differ when measured along different directions in the same crystal This is because the arrangement of particles differs in different directions.

However, amorphous solids are isotropic, meaning that their physical properties are the same when measured along am. directions. lids are because their constituent particles are more or less randomly arranged.

A crystalline solid has a sharp melting point. In contrast, an amorphous solid liquefies over a range of temperatures. For example, glass and plastics soften over a range of temperatures and can be moulded or Hewn into various shapes, which they retain on cooling. Like liquids, amorphous solids tend to Sow, although very slowly.

Therefore, they are called pseudo solids or supercooled liquids. Glass panes of old buildings are thicker at the bottom than at the top because of this tendency of amorphous solids to Bow.

When you cut a crystalline solid with a sharp object, it breaks into pieces which have smooth and regular surfaces. In other words, you can cut a crystalline solid. If you were to cut an amorphous solid with a sharp 1 object, it would break into pieces with irregular surfaces.

Another important difference between crystalline and amorphous solids is that the former have definite heats of fusion whereas the latter do not.

Types Of Crystalline Solids

Crystalline solids are grouped into different types depending on the nature of the constituent particles and the bond formed between them. Such solids can be classified as ionic, covalent, metallic and molecular.

Ionic solids:

They are composed of anions and cations that are held together by electrostatic forces. For example, common salt, NaCl contains NaT and Cl- ions. Some other examples of ionic solids are CsCl, CaF₂ and ZnS.

Definition and characteristics of solids in chemistry

Ionic solids have high melting points; they are hard and brittle, and conduct electricity when molten or in solution. Many compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.

Covalent solids:

They are made up of atoms of the same or different elements held together by a network of covalent bonds.

Diamond is the most common example of a covalent solid. Silicon and silicon dioxide are also covalent solids.

These solids are very hard, and strong and have high melting points due to the presence of strong covalent bonds.

Diamond melts above 3500. The oxide of silicon, SiO₂ (commonly called silica), exists in several forms with different crystal structures. Such different forms of the same compound are called polymorphs and the phenomenon, of polymorphism.

SiO₂ does not contain discrete SiO₂ molecules. The large size of Si, together with the decreased tendency of Si to form multiple bonds, leads to the formation of Si—O single bonds in Si04 tetrahedral units. The tetrahedra are linked together in a three-dimensional network by the sharing of each O between two Si.

In a single crystal of quartz, SiO₂, all SiO₄ units are cross-linked into a single unit resulting in a giant molecule or covalent crystal. The strong covalent bonds account for the hardness, very high melting point, and electrical and thermal insulating properties of silica (the localisation of electrons in the covalent bonds prevents their moving freely under an applied electric field).

Metallic solids:

They comprise a network of positive ions whose positions are fixed. The valence electrons are free to move. The force of attraction between the valence electrons and the positive ions is called a metallic bond. Iron, copper and aluminium are examples of metallic solids.

Most metallic solids are very hard and strong, and all of them exhibit high thermal and electrical conductivity; they are malleable and ductile. The melting points of metallic solids vary depending on the strength of metallic bonding. Alkali metals melt below 473 K, most transition metals melt at temperatures above 1273 K. Mercury is a liquid at room temperature.

Molecular solids:

They consist of molecules held together by intermolecular forces. They are classified as polar or nonpolar on the basis of tire type of tire intermolecular force. The physical properties of molecular solids depend on the strength of the intermolecular forces.

In polar molecular solids, the intermolecular forces are dipole-dipole forces. Examples are solid NH₃ and solid HCl. They are soft and do not conduct electricity. Although gases or liquids at room temperature, they melt at higher temperatures than nonpolar molecular solids do. Some polar molecular solids are hydrogen-bonded. Examples are ice and solid ammonia.

Nonpolar molecular solids are bound by van der Waals forces (or dispersion forces), which are very weak. They may comprise small symmetric molecules such as H₂, N₂, O and F₂. Because of the weak intermolecular forces, their melting points are very low (below -73 K). They are soft and do not conduct electricity.

Crystal And Space Lattice

Now let us study the structure of crystalline solids, in which the particles making up the crystal have a basic, periodic arrangement.

A crystal may be defined as a three-dimensional pattern in which a structural motif is repeated in such a way that the environment of every motif is the same throughout the crystal. In other words, the crystal appears exactly the same at one point as it does at a series of other equivalent points. The motif or the repeat unit may be an atom, a molecule or a group of such entities.

To understand the arrangement of atoms/molecules in a crystal, first let us consider a set of parallel motifs separated by a distance say a (the structural motif is represented by a point)

This is a one-dimensional pattern. Now, if we consider the placement of such motifs at a distance b in the second direction, i.e., the y-direction, we can build a two-dimensional pattern by simply arranging the motifs in such a way that in the x-direction, they are separated by a distance a and in the y-direction, by a distance.

If the z-direction is also considered, we get a three-dimensional pattern. If we arrange the points in space so as to represent the actual disposition of the motifs in the crystal, we get what is called a space lattice.

A space lattice is thus a pattern formed by points representing the structural disposition of particles—atoms, of these. Each particle occupies a latte point in the array, Vie npm? the lattice of a crystal into identical parallelepiped by joining the lattice paints with straight Each such pM&fktepip&h with h ’Sat Bask repealing mat of the arrangement of atoms, form or molecule called a unit edh of a nail -cdi in three dimensions generates the entire pattern of a crystal.

Shows the possible unit cells- in a particular two-dimensional array, This arrangement can be extended in a parallel way to a lattice. The choice of a unit cell for a particular crystal depends on various factors such as symmetry and volume since there can be more than one way of placing atoms in a volume. At the present level of learning, we do not go into further details.

Primitive Unit Cell

A space lattice can be divided into unit cells in different ways. The lattice points In a unit cell could be placed only si the comers, or at the comers and inside the body of the cell. The former kind of unit cell Is called a primitive ve trust .cell (Hgnre 1.6).

Wherever the arrangement of particles an a crystal, in each case the lattice contains a volume represented by a unit cell which is regularly repeated throughout the crystal.

In order to describe the geometry of a unit cell for a frre^dimensional lattice, and we must specify the lengths F and c along the three axes, x, y and z, and also the three angles a,p and yin a three-dimensional parallelepiped.

There are 7 types of primitive unit cells and based on these we have seven crystal systems. The simplest unit cell is a cubic unit cell, where all sides are equal and all angles are 901

Cubic Unit Cell

There are three types of cubic unit cells—primitive, body-centred and face-centred. In the primitive unit cell (or a simple cubic unit cell), the atoms are located at the corners of the cell. In the body-centred unit cell, apart from the comers, the atoms are placed at the centre of the cell also. In the face-centred cubic (fee) structure, the atoms are placed at the centre of each face apart from the corners of the cube.

The number of atoms per unit cell varies according to the type of the cell. Assume a three-dimensional arrangement of cubes in space. Imagine four cubes arranged in such a manner that their faces form a square and four cubes arranged similarly above them.

Dielectric properties of solids explained

Consider an atom that is at the centre of this arrangement. It is in fact an atom in the comer of a cube. It is being shared by eight cubes. Only 1/8 of the atom belongs to one cube. The same is the case with the other 7 atoms situated at the comers.

Hence, the number of atoms in a primitive cubic unit cell is \(8 \times \frac{1}{8}=1\).

In the case of the body-centred cubic (bcc) structure, the central atom belongs only to one cube. One atom from the comers and one at the centre of the body amount to two atoms per body-centred cubic unit cell. In the face-centred cubic (fee) arrangement, there are four atoms per unit cell. One atom is obtained from the comers as in the case of a primitive cubic unit cell.

Each atom at the face is shared by two cubes; there are six faces and hence \(6 \times \frac{1}{2}=3\) atoms are obtained from the faces.

We have seen that three types of cubic unit cells are possible. Regular stacking of each one of these unit cells results in a particular type of space lattice. Taking into account symmetry considerations (which we will not go into here), it is seen that for the tetragonal crystal system, there are two types of unit cells possible—primitive and body-centred.

In the other crystal systems, one or more than one type of unit cell is possible, amounting to a total of 14 types of unit cells. In 1848, Auguste Bravais showed that the arrangement of these 14 types results in 14 different lattices. They are therefore known as Bravais lattices.

Close Packing In Metallic Crystals

Particles of the same type, like atoms in a metal or anions of an ionic solid, can pack closely resulting in a stable structure of crystals. The particles of a crystal pack in a manner that gives the maximum possible density. Let us consider the various possibilities of this arrangement in a crystal, where the constituent particles are assumed to be spheres and of the same size.

The solid state and its importance in material science

First of all, let us consider the formation of one edge of a crystal. The closest packing in a row of spheres will be one where the spheres are simply touching each other.

In this type of arrangement, each sphere is in contact with two of its neighbours. The coordination number of a particle is the number of its nearest neighbours. In Fluence, in a one-dimensional arrangement of spheres, the coordination number is two.

Now, another row can be aligned below this in any one of the following two ways in order to obtain one face or plane of the crystal. One possibility is that the second row is placed in such a way that each sphere lies exactly above the sphere in the first row; this gives rise to an arrangement of spheres in a square. Hence this arrangement is called square packing.

The coordination number in square close packing in two dimensions is four since each sphere is in contact with the four nearest neighbours. In another type of arrangement, the spheres of the second row can lie in the depressions between the spheres in the first row. The same thing can be repeated for the third row to obtain a hexagonal packing arrangement where one sphere is surrounded by six other spheres in a plane.

The coordination number in the hexagonal close packing in two dimensions is six. Among these arrangements, we find that hexagonal packing results in a close-packed structure leaving less space between the particles of the crystal. If such planes of spheres are stacked one over the other we obtain the complete three-dimensional structure of the crystal.

Let us see how such a structure can be formed. Consider the hexagonal packing of spheres to form the first layer. In this case, a sphere can be surrounded at the maximum by six other spheres of the same size.

Then each of these spheres can again be surrounded by six such other spheres. This arrangement called closest packing can extend indefinitely in a single layer, say layer A.

Now a second layer, B, can be formed above the first layer A still maintaining close packing. If you look at, you will find that there are two types of holes, one type which lies on the same straight horizontal line and the other type which is diagonally opposite to the first ones.

It is not possible to keep spheres of the second layer together on both types of holes. The spheres of the second layer are kept on top of the holes in the first layer so that each particle in layer B is in contact with three particles in layer A.

The third layer can now be stacked above layer B in two ways. In the first arrangement, the spheres are placed exactly as in layer A, so that the third layer resembles layer A and the fourth layer resembles layer B. Thus, the stacking continues as ABABAB …

What is solid state in chemistry – definition and examples

This arrangement is called the hexagonal closest packing (hep). Examples include Cd, Co, Li, Mg, Na and Zn from the metallic solids. A noncubic unit cell is formed from this arrangement.

In the second type of arrangement, which is called cubic closest packing, the third layer is formed by placing the spheres in the voids or interstices between tire spheres in layer B such that the newly formed layer C does not resemble either layer A or layer B.

Then, the fourth layer is formed by keeping spheres in the voids of layer C and exactly above the spheres in layer A. This is followed by a pattern which is the same as in layer B and then by one which is the same as in layer C.

This type of packing is therefore of the ABCABC … type and is called cubic closest packing or face-centred cubic arrangement (fee) (the smallest unit cell that can describe this arrangement is the face-centred cubic). Examples include Ag, Al, Ca, Cu, Ni and Pb.

Coordination number:

The coordination number of an atom or ion in a crystal is the number of neighbouring atoms or ions it touches. In both the hexagonal closest packed and cubic closest packed structures each particle touches 12 nearest neighbours. This can be understood if we consider three layers at a time and consider a sphere at the centre of a layer.

In the middle layer, there are six nearest neighbours for both hep and cup structures. In the layers above and below, three spheres from each layer touch this central sphere. Thus, in all, there are 12 spheres in close proximity, touching the central sphere. Hence, the coordination number is 12 in both cases.

Packing efficiency in CCP and hep arrangements:

In the help arrangement, there are six atoms belonging to a hexagonal prism, Let us see how. Assume a three-dimensional arrangement of such prisms. There are three spheres in layer B belonging exclusively to this prism. The central sphere at the top of the prism is shared by two prisms (one prism placed exactly above the one shown in the figure). Hence, only half of this sphere belongs to this prism.

There is another sphere of the same kind at the bottom; it also contributes only 1/2 to this prism. Then, the third type of sphere we must consider is placed at the corners. Each sphere in a corner is shared by 6 prisms; 3 in the same plane and 3 above it. There are 12 (6 on top, 6 at bottom) such spheres.

Hence, the spheres from corners contribute to the extent of \(12 \times \frac{1}{6}=2\) for the prism under consideration.

Therefore, the total number of spheres per prism = \(3+\frac{1}{2}+\frac{1}{2}+2=6\)

As you know, the unit cell in a ccp structure is face-centred cubic. You also know that there are four atoms per fee unit cell.

Even when the spheres are closely packed, there is a gap between them due to their shape. Packing efficiency or packing fraction is the fraction of the total space of a unit cell which is occupied by the spheres. To determine packing efficiency, we must first find the total volume of the cell and then the total volume of the spheres. The volume of spheres divided by the volume of the unit cell gives the packing fraction. For a cup structure, the unit cell is the volume of a cube where 7 is the edge length of the cube.

= a³

As there are four atoms per unit cell, the volume occupied by 4 atoms

⇒ \(4 \times \frac{4}{3} \pi r^3\)

where r is the radius of each sphere. Now, we have to relate r and a. As the spheres are in close contact along the face diagonal in a ccp structure, the face diagonal = 4r = √2 a

⇒ \(r=\frac{\sqrt{2}}{4} a\)

Substituting the expression for r from Equation (3) in Equation (2), we get

volume occupied by atoms ⇒ \(4 \times \frac{4}{3} \times \frac{22}{7} \times\left(\frac{\sqrt{2}}{4} a\right)=0.74 a^3\)

Hence, packing fraction ⇒ \(0.74 \frac{a^3}{a^3}=0.74 \text { or } 74 \%\)

Only 74% of the space in a cube is occupied by spheres/atoms in a ccp structure.

Even in a hep structure, the packing efficiency is 74%.

Other structures:

Most metals, other than those cited as forming a hep or a fee structure, crystallise in a bcc structure, where there is one particle in the centre of a cube and one in each corner of the cube. Those particles which are in the same plane do not touch each other.

Here the central particle touches four particles in the upper layer and four in the lower layer, resulting in a coordination number of 8. The crystal structure consists of such repeating units. The particles are not closely packed. Examples of elements with a bcc structure are Cr, Ba, Mo, W and Fe.

There is yet another arrangement possible though again it is not close-packed. It is called the simple cubic structure and is exhibited by polonium.

Here each sphere in a plane touches four of its nearest neighbours and the next layer is stacked in such a manner that each sphere in the second layer is exactly above a sphere in the first layer. This results in an arrangement where the spheres occupy all the comers of a cube and none is at the centre.

The coordination number can be determined by considering three layers at a time. In the middle layer, the central sphere is surrounded by four other spheres and by one sphere each in the upper and lower layers. Thus, the total number of nearest neighbours is 6. In the simple cubic structure, the coordination number of each atom is thus 6.

Packing efficiency In other cubic structures:

In a simple cubic unit cell, the atoms touch each other along the side of the tire cube. If the radius of the atom is jR, the volume of the cell is (2R)³.

The volume of each atom is = \(\frac{4}{3} \pi R^3\)

Therefore, packing efficiency = \(=\frac{\frac{4}{3} \pi R^3}{(2 R)^3}=\frac{\pi}{6}=0.52\)

Only 52% of the total space of the cube is occupied.

In a body-centred cubic unit cell, the atoms on the main body diagonal, and not at the sides, touch each other.

As can be seen, the length of the main body diagonal in terms of the radius of the sphere is 4R. Let us now find the side of the cube.

Let the side of the cube be s. Hence, AB =BC = s.

Consider A ABC. AB and BC are sides of the cube and AC is the face diagonal. ∠B = 90º.

We know that,

⇒ \((A B)^2+(B C)^2=(A-C)^2\)

or, \(s^2+s^2=(A C)^2\)

or, \((A C)^2=2 s^2\)

or, \(A C=\sqrt{2} s,\)

Since BD is also a face diagonal, AC = BD = √2s.

Now, consider A ABD. Again ZB = 90º

⇒ \((A B)^2+(B D)^2=(A D)^2\)

or, \(s^2+(\sqrt{2} s)^2=(A D)^2\)

or, \((A D)^2=s^2+2 s^2=3 s^2 .\)

But AD = 4R.

Hence, \(16 R^2=3 s^2 \text { or } s^2=\frac{16}{3} R^2\)

or, \(s=\frac{4}{\sqrt{3}} R\)

Hence, one side of the cube is \(\left(\frac{4}{\sqrt{3}}\right)^R\)

Volume of the cube = \(\left(\frac{4}{\sqrt{3}} R\right)^3=\frac{64 R^3}{3 \sqrt{3}}\)

Volume of one atom = \(\frac{4}{3} \pi R^3\)

Packing efficiency \(\frac{2 \times \frac{4}{3} \pi R^3}{\frac{64 R^3}{3 \sqrt{3}}}=0.68=68 \%.\)

Voids Or Holes In Close-Packed Structures

In the closest packed arrangement of spheres, the depressions formed by placing the spheres are called voids or holes. We find two types of holes or voids, the octahedral hole and the tetrahedral hole.

An octahedral hole is created when six spheres are in contact while a tetrahedral hole is one formed by four spheres in contact, When the spheres of layer B are placed in such a manner that they occupy the holes in the first layer A, there is a void below every sphere of layer B and above every sphere of layer A.

There are three spheres forming a triangle in one layer and one sphere in another layer, resulting in a total of four spheres around this void. Hence, it is called a tetrahedral void. When the triangles formed in the two layers are placed with their vertices in opposite directions, an octahedral void is obtained. This is obtained when the arrangement in the two layers A and B is such that there is a void in layer B above a void in layer A.

Both the hep and ccp structures have tetrahedral as well as octahedral holes. There are two tetrahedral holes for each atom in both structures. And for each atom, there is only one octahedral hole in the two structures. Shows the octahedral and tetrahedral holes in a ccp unit cell. Such units when repeated give rise to the three-dimensional structure of the crystal.

Location of voids:

Tetrahedral void Consider the fee unit cell. There are spheres at the corners of the cell and also at the centre of each face. Imagine this cube to be divided into eight smaller cubes of equal dimensions.

A close look at each of these smaller cubes shows that all the corners do not contain spheres; in fact, oriy alternate corners do Neither the resulting faces nor the centre contain any sphere.

Out of the eight corners of the smaller cubes, only four are occupied. On joining these four comers containing spheres a regular tetrahedron is obtained It is a Vacant space and the spheres surrounding it form a tetrahedron.

Hence, the void is called a tetrahedral void or tetrahedral hole Since one unit cell can be divided into eight cubes, each of which contains a tetrahedral hole, there are eight such holes per fee unit cell. We know that there are four atoms per fee unit cell and so the number of tetrahedral holes per atom in a fee unit cell is 8 holes/4 atoms = 2 holes per atom.

Octahedral void Again consider a face-centred cubic unit cell. It does not contain a sphere at the centre However, there are six faces, each of which contains a sphere. If the spheres at these faces are connected an octahedron is formed. Hence, we have an octahedral hole at the centre of a fee unit cell. There is an octahedral hole at the edge of the cube also.

In order to locate the hole, consider four cubes of fee unit cells arranged in the form of a rectangular parallelepiped. At the centre of this structure, there is an edge which is shared by all four cubes The spheres at the end of this edge, and those at the faces surrounding this edge, can be connected to form an octahedron. Since there is no sphere at the centre of this octahedron, it is an octahedral hole or octahedral void.

This hole, as can be, is shared by four fee unit cells. So only one-fourth of it belongs to a unit cell. There are twelve edges in a fee unit cell and each of these contains an octahedral hole.

So there are in all \(12 \times \frac{1}{4}=3\) holes per unit cell which are present at the edges.

Apart from this, there is one octahedral hole at the 4 centre which belongs completely to one fee unit cell. In all, there are 3 + 1 = 4 octahedral holes per fee unit cell. Since there are four atoms in a ccp structure, the number of octahedral holes per atom is one.

Ionic Compounds

By now you know that in a crystalline solid, the atoms, ions or molecules are arranged in a definite repeating pattern. The particles behave as though they are spheres and arrange themselves as closely as possible.

The type of packing depends upon of the interaction between the different kinds of spheres. This type of packing of the particles is the basis of the definite structure of crystalline solids.

In the case of ionic compounds, crystals consist of two different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure: therefore depends on the geometry of the ions.

Any ion formed of a single atom behaves as a charged sphere and is surrounded by oppositely charged ions in all directions. Anions tend to be surrounded by cations and vice versa.

The force of attraction between the ions is the same in all directions. Stable ionic crystals result if each ion is surrounded by as many ions as possible of opposite charge and the oppositely charged ions (cations and anions) are in contact with each other. The final structure and the extent of packing is decided by the relative sizes of anions and cations and their relative numbers.

In simple ionic structures, anions, which are generally large, are arranged in a close-packed manner, touching each other with the smallest possible space between them. And cations are found to occupy the voids or holes (the space between the anions).

General properties of solids – rigidity, shape, and volume

The holes can be either octahedral or tetrahedral. If the anions that surround the hole are arranged in a tetrahedral manner then the hole is called a tetrahedral hole and the anions are hence four in number. If the anions that surround the hole are arranged in an octahedral manner, an octahedral hole is obtained and we have six anions.

If the cations are small, they tend to occupy tetrahedral holes and those that are relatively bigger occupy octahedral holes. Sometimes the cations may be too large to be able to occupy either of these holes. In such a case, the anions assume a simple cubic structure occupying all the comers of a cube and the cations occupy the cubic holes thus formed, resulting in a body-centred cubic structure.

Let us consider the case where the anions occupy the comers of a cube and the cations occupy all the tetrahedral holes. As the number of one type of atoms/ions per unit cell in a ccp structure is four, the number of anions per unit cell is four. As the tetrahedral holes are eight in number, the number of cations per unit cell is eight.

Hence, the ratio of cations to anions is 8: 4, i.e., 2:1. The formula of the compound is then M₂X, where M is the cation and X is the anion. Similarly, if the cations occupy octahedral holes, the ratio is 1:1 and the formula of the compound will be MX.

Structures Of Ionic Compounds

The relative number of cations and anions varies in different ionic compounds. For the sake of convenience, ionic compounds are divided into four types—MX, MX₂, MX₃ and M₂X.

Ionic compounds of the formula MX generally possess any one of the following structures NaCl or rocksalt structure, caesium chloride or CsCl structure and zinc sulphide, ZnS structure. All these are cubic structures.

Shows the NaCl structure, where both Na+ and Cl^– ions form a fee (face-centred cubic) structure. You can imagine two separate fee lattices formed by Na⁺ and Cl^– ions, which then interpenetrate each other. The coordination number of the cations as well as the anions is 6.

In the structure, the Cs⁺ ion is bigger than the Na⁺ ion. Therefore, the C^– ions occupy the corners of the cube and the Cs⁺ ion is positioned at the centre, resulting in a body-centred cubic structure.

The ionic radius of Cs⁺ is 169 pm and that of Cl^– is 181 pm. When we place Cl^– ions at the comers, each of them touching the adjacent one, a cubic hole is formed and Cs⁺ can occupy this hole, resulting in a bcc structure. The CsCI unit cell can also be drawn with Cs⁺ at the comers and CP at the centre. The bee structure is a result of two interpenetrating simple cubic structures. As against the 6-6 coordination in NaCl, in the CsCI structure, there is an 8-8 coordination.

In the zinc sulphide (ZnS) structure, the large sulphide ions form a face-centred cubic structure that is almost close-packed. The zinc ions occupy alternate tetrahedral holes. Out of the eight tetrahedral sites available, only four sites are occupied.

Only half the tetrahedral sites are occupied as the compound has 1; 1 stoichiometry and for each sulphide ion there are 2 tetrahedral holes.

Both zinc and sulphur have a coordination number of four. ZnS is also known to form a hexagonal structure called the wurtzite structure. Here the sulphide ions form an hep array and zinc ions occupy half of the tetrahedral holes.

Compounds of formula MX₂, which have a 1: 2 stoichiometry (i.e., two anions for every cation), crystallise in the fluorite (CaF₂) structure and the tetragonal rutile (TiO₂) structure. The Ca²⁺ ions in fluorite are in a face-centred cubic arrangement and the fluoride (F^–) ions occupy the tetrahedral holes.

There are, thus, four cations per unit cell and eight anions (equal to the number of tetrahedral holes) resulting in an 8-4 coordination or 1-2 valence type of compound. The coordination of the cation is eight and that of the anion is four.

In the rutile (TiOz) structure, each Ti⁴⁺ ion is attached to 6O^2- ions and each O^2- ion is in turn attached to three Ti⁴⁺ ions. Hence, Ti⁴⁺ ions have a coordination number of 6 and the O^2- ions have a coordination number of 3.

Radius Ratio

The structure of an ionic compound depends largely on the relative sizes of the cation(s) and anion(s) constituting it. Let us see how the sizes of the ions determine the coordination number. Since cations are smaller than anions, let us consider the size requirements for a cation to fit into the interstitial space between anions.

In Δ ABC, cos BCA = cos (30°) = \(\frac{\sqrt{3}}{2}\)

But \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{r^{-}}{r^{+}+r^{-}},\)

where r+ and r- are the radii of the cation and anion respectively.

From Equations 1 and 2, we get

⇒ \(\frac{r^{-}}{r^{+}+r^{-}}=\frac{\sqrt{3}}{2}\)

or, \(\frac{r^{+}+r^{-}}{r^{-}}=\frac{2}{\sqrt{3}}\)

⇒ \(\frac{r^{+}}{r^{-}}+1=\frac{2}{\sqrt{3}}\)

⇒ \(\frac{r^{+}}{r^{-}}=\frac{2}{\sqrt{3}}-1=0.155\)

The radii must satisfy the condition

⇒ \(r^{+}=0.155 r^{-}\)

Based on such calculations, Pauling laid down some rules as criteria for close packing in ionic crystals. These are known as radius ratio rules.

The radius ratio of any ionic compound is given by p = \(=\frac{r_s}{r_1}\), the ratio of the radius h of the smaller ion to the radius of the larger ion. In general, since cations are smaller than anions, the radius ratio is \(\frac{r^{+}}{r^{-}}\).

The cations and anions get as close to each other as possible as determined by their ratio. The larger the radius ratio, the larger is the coordination number of the cation. When the radius ratio is larger, the size of the cation is large and more anions can surround the cation.

Hence, the coordination number is also larger. According to Pauling’s radius ratio rules, if p is between 0.225 and 0.414, then a 4-4 coordination (zinc blende or wurtzite, tetrahedral coordination) is possible, where a cation touches 4 anions], if it is between 1,414 and 0.732 then the compound adepts a 6-6 coordination structure (NaCl, octahedral coordination) and If coordination structure is formed (CsCb cubic holes occupied),

Ionic Radii

X-ray studies of ionic crystals give us information about the dimensions of unit cells from which a of ionic radii can be obtained, assuming that the ions are spheres. Table 1,4 gives some of the Ionic radius values,

Note: pm denotes picometre, which is 1 x 10 ¯¹² metre.

Example 1. On the basis of data given in Tables 1.3 and 1.4, predict the structures of NaCl, KCl and
Solution:

NaCl: From Table 1.4, r_Na+ = 102 pm and r_cl- = 181 pm.

∴ \(\frac{r^{+}}{r^{-}}=\frac{102}{181}=0.564\)

As you can see in Table 1.3, the structure is octahedral. Therefore, there is 6-6 coordination, KCl:  rK = r+ = 138 pm, ra. = r¯ =181 pm,

∴ \(\frac{r^{+}}{r^{-}}=\frac{102}{181}=0.564\)

Li₂S: From Table 1.4, r+ =59 pm and r- =184 pm

∴ \(\frac{r^{+}}{r^{-}}=0.32\) the structure is tetrahedral.

Example 2. The edge length of the unit cell of NaCl is 5,66 A., Assuming close contact of ts a’ and ^– ions, calculate the ionic radius of the Cl Ion given that the ionic radius of the NaT ion is 10² pm.
Solution:

Given

The edge length of the unit cell of NaCl is 5,66 A., Assuming close contact of ts a’ and ^– ions

Along the edge, two Na+ ions and one Cl¯ ion are in close contact. While the two Na+ ions are centred at the corners, each one contributing only a length equal to its radius to the edge length, the Cl ion is in the middle and 2r¯ is its contribution to edge length. If the edge length is a, then

⇒ \(a=2 r^{+}+2 r^{-}\)

= 5.66Å

= 566 pm

r⁺ =102 pm  (given)

∴ \(2 r^{-}=566-2 r^{+}\)

2r¯ = 566 – (2×102)

⇒ \(r^{-}=\frac{(566-204)}{2}=\frac{362}{2}\)

r¯ = 181 pm.

The ionic radius of the Cl” ion is 181 pm.

Calculations Involving Unit Cell Dimensions

Knowledge of the edge length of the unit cell helps us to calculate the density of the crystals and also the mass of atoms in a unit cell. Let us see how density can be calculated.

If a is the edge length of a unit cell of the crystal, the volume is given by the cube of the edge length,

i.e., the volume of the unit cell = a³.

If the mass and volume are known, the density of a unit cell can be determined, which is the same as the density of the substance.

Mass of unit cell = sum of masses of all atoms present per unit cell since in a crystal of a single element all the atoms are of the same type.

Mass of unit cell = number of atoms per unit cell x mass of a single atom = \(n \times \frac{M}{N_{\mathrm{A}}}\)

where M is the atomic mass of the element and NA is the Avogadro constant.

Now,

density of until cell, d= \(n \times \frac{M}{\left(a^3 N_{\mathrm{A}}\right)}\)

If M is in grams and a is in centimetres, then the unit of density is g/cm³.

In the case of molecules, n is the number of molecules per unit cell and M is the molar mass.

Example 1. The density of sodium chloride (NaCl) is 2,163 g cm¯³. Find the edge length of a unit cell of NaCl.
Solution:

Given

The density of sodium chloride (NaCl) is 2,163 g cm¯³.

The molar mass of NaCl = 23 + 35.5 = 58.5 g mol¯¹

Molar volume of NaCl = \(\frac{58.5}{2.163}=27 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\)

Volume of a molecule of NaCl = \(\frac{27}{6.023 \times 10^{23}}=4.48 \times 10^{-23} \mathrm{~cm}^3\) (∵ one mole contains 6.0^-23 x 10^-23 number of molecules).

One unit cell of a fee structure contains 4 molecules.

Hence, volume of unit cell = 4 x 4.48 x 10^-23 = 17.9 x 10^-23 cm³.

The volume of the unit cell in terms of edge length a = a3.

⇒ \(a^3=17.9 \times 10^{-23}\)

∴ \(a=\sqrt[3]{17.9 \times 10^{-23}}=\left(17.9 \times 10^{-23}\right)^{1 / 3} \mathrm{~cm}\)

Example 2. Compound CsCl crystallises in a bcc lattice with a unit cell edge length of 405 pm. Calculate the radius of Ca+ if that of Cr^– is 181 pm.
Solution:

Given

Compound CsCl crystallises in a bcc lattice with a unit cell edge length of 405 pm.

In a body-centred cubic structure, the ions are in close contact along the main body diagonal. In the figure given below the dotted line BC represents the main body diagonal.

Consider the triangle ADB.

If the edge length is a, by Pythagoras’ theorem, a² + a² = square of face diagonal = AB².

∴ AB = √2a.

The face diagonal, one edge and the main body diagonal form a right-angled triangle with the main diagonal as a hypotenuse Face diagonal (AG4B).

Hence, a² + {AB)² -(BC)²

⇒ a² +2a² – (BC)²

⇒ BC = √3a.

But along the main diagonal, we have two anions and one cation in contact; its length is \(2 r^{+}+2 r^{-}\)

⇒ \(\sqrt{3} a=2 r^{+}+2 r^{-}\)

⇒ \(\sqrt{3} \times 405 \mathrm{pm}=2 \times r^{+}+2 \times 181\)

⇒ \(\frac{(\sqrt{3} \times 405)-(2 \times 181)}{2}=r^{+}\)

r⁺=169.7 pm

The ionic radius of the Cs+ ion ≈ 170 pm.

Example 3. NaCl forms a fee structure. If the ionic radii of Na⁺ and Cl^– are 10² pm and 180 pm respectively, find the edge length and hence the volume of a unit cell. Also, calculate the density (the atomic masses of Na and Cl are 23 and 35.5 respectively).
Solution:

Given

NaCl forms a fee structure. If the ionic radii of Na⁺ and Cl^– are 10² pm and 180 pm respectively

Since the ions are closely packed and touching each other, edge length

a = 2r⁺ +2r^– =2×102 + 2×180 = 204+ 360

a =564 pm = 564 x 10^-10 cm.

Volume of unit cell = \(a^3=(564)^3 \times\left(10^{-10}\right)^3=1.79 \times 10^{-28} \mathrm{~cm}^3\)

Molar mass of NaCl = 23 + 35.5 = 58.5 g.

Mass of one molecule of NaCl = \(\frac{58.5}{N_{\mathrm{A}}}\)

There are 4 molecules of NaCl per unit cell.

Hence total mass of unit cell = \(\frac{4 \times 58.5}{6.023 \times 10^{23}} \mathrm{~g}=3.885 \times 10^{-22} \mathrm{~g}\)

Density = \(\frac{\text { mass }}{\text { volume }}=2.16 \mathrm{~g} \mathrm{~cm}^{-3}\)

Example 4. Europium (Eu) is the block element present in the lanthanide series with atomic mass 152. It crystallises in a body-centred cubic structure. Its density is 5.26 g cm¯³. Calculate the radius of an Eu atom.
Solution:

Given

Europium (Eu) is the block element present in the lanthanide series with atomic mass 152. It crystallises in a body-centred cubic structure. Its density is 5.26 g cm¯³.

The body diagonal of the cube = √3 a.

There are three atoms of Eu along this diagonal (one full and 2 half)

Hence, 4K = √3a

⇒ \(R=\frac{\sqrt{3} a}{4}\) (R = radius of atom).

There are 2 atoms in a unit cell of a bcc structure.

∴ \(\rho=\frac{2 \times M}{a^3 \times N_A}\)

⇒ \(5.26=\frac{2 \times 152}{a^3 \times 6.023 \times 10^{23}},\)

⇒ \(a=\sqrt[3]{\frac{2 \times 152}{5.26 \times 6.023 \times 10^{23}}} \mathrm{~cm}=4.58 \times 10^{-8} \mathrm{~cm}\)

= 485 pm.

∴ \(R=\frac{\sqrt{3} \times 458}{4}=198.3 \mathrm{pm}\)

This problem can be tackled in another way.

Alternative method:

Molar volume (volume occupied by 1 mole) = \(\frac{1}{5.26} \times 152=28.9 \mathrm{~cm}^3\)

Volume actually occupied by the Eu atoms without the empty space between them = 28.9 x 0.68 = 19.7 cm³ mol¯¹,

where 0.68 is the packing efficiency.

Volume of one atom = \(\frac{19.7}{6.023 \times 10^{23}}=3.26 \times 10^{-23} \mathrm{~cm}^3=\frac{4}{3} \pi R^3\)

Radius \(\left(\frac{3 V}{4 \pi}\right)^{1 / 3}=1.98 \times 10^{-8} \mathrm{~cm}=198 \mathrm{pm}\)

Example 5. KCI forms a fee structure. Its density is 1.984 g cm³. If the edge length of its unit cell is 629 pm, find the molar mass of KCI.
Solution:

Given

KCI forms a fee structure. Its density is 1.984 g cm³. If the edge length of its unit cell is 629 pm

Edge length = 629 pm = 629 x 10^-12 m = 629 x 10^-10cm.

Density = \(\frac{\text { mass }}{\text { volume }}\)

⇒ \(1.984=\frac{n \times M}{a^3 N_{\mathrm{A}}}\)

⇒ \(1.984=\frac{4 \times M}{\left(629 \times 10^{-10}\right)^3 \times 6.023 \times 10^{23}}\)

∴ M = 74.3.

the molar mass of KCI = 74.3 g mol-1

Example 6. The compound CuCl has a structure similar to that of ZnS. The edge length of the unit cell is 500 pm. Calculate the density (atomic masses: Cu = 63, Cl = 35.5).
Solution:

Given

The compound CuCl has a structure similar to that of ZnS. The edge length of the unit cell is 500 pm.

As you already know p = \(\frac{n M}{a^3 N_{\mathrm{A}}}\)

where p is the density, n is the number of atoms per unit cell, M is the molar mass, a is the edge length and NA is the Avogadro constant.

Therefore, p = \(\frac{4 \times 98.5}{125 \times 10^{-24} \times 6.02 \times 10^{23}}=5.22 \mathrm{~g} \mathrm{~cm}^{-3}\)

(In case of an fcc lattice, n=4).

Example 7. Chromium has a monatomic body-centred cubic structure. The edge length of its unit cell is 300 pm. What is i density (atomic mass of Cr = 52)?
Solution:

Given

Chromium has a monatomic body-centred cubic structure. The edge length of its unit cell is 300 pm.

a = 300 pm = 300 x 10^-10 cm.

⇒ \(\rho=\frac{n M}{a^3 N_A}=\frac{2 \times 52}{\left(300 \times 10^{-10}\right)^3 \times 6.023 \times 10^{23}}=6.39 \mathrm{~g} \mathrm{~cm}^{-3}\)

Example 8. An element has a ccp lattice with a cell edge length of 500 pm. The density of the element is 10 g/cm³. How many atoms are present in 270 g of the element?
Solution:

Given

An element has a ccp lattice with a cell edge length of 500 pm. The density of the element is 10 g/cm³.

Volume of one unit cell = \(=(300 \mathrm{pm})^3=\left(300 \times 10^{-10} \mathrm{~cm}\right)^3\)

⇒ \(=27 \times 10^6 \times 10^{-30} \mathrm{~cm}^3=27 \times 10^{-24} \mathrm{~cm}^3\)

⇒ \(2.7 \times 10^{-23} \mathrm{~cm}^3\)

Volume of 270 g of element = \(\frac{\text { mass }}{\text { density }}=\frac{270 \mathrm{~g}}{10 \mathrm{~g} / \mathrm{cm}^3}=27 \mathrm{~cm}^3 \text {. }\)

Number of unit cells in this volume = \(\frac{27}{2.7 \times 10^{-23}}=10^{24}\)

Since there are 4 atoms in one unit cell of a ccp structure, in 10²⁴ unit cells there will be 4 x 10²⁴ atoms.

Example 9. An element crystallises in a cubic lattice and is found to have a density of 705 g/cm³. The edge length of a unit cell is 250 pm and there are 4 x 10²⁴  atmos in 234 g of the element. Find the number of atoms in a unit cell.
Solution:

Given

An element crystallises in a cubic lattice and is found to have a density of 705 g/cm³. The edge length of a unit cell is 250 pm and there are 4 x 10²⁴  atmos in 234 g of the element

Proceeding as in the previous example, the volume of one unit cell=\(\left(250 \times 10^{-10} \mathrm{~cm}\right)^3=15625 \times 10^{-2} \mathrm{~cm}^3.\)

Number of unit cell in this volume = \(\frac{\text { mass }}{\text { density }}=\frac{234}{7.5}=31.2 \mathrm{~cm}^3\)

Given that the number of atoms in 234 g of the elements is 4 x 10²⁴, the number of atoms in one unit cell = \(\frac{4 \times 10^{24}}{2 \times 10^{24}}=2\).

Hence, there are 2 atmos per unit cell.

Imperfections In Solids

We have studied different types of lattices and you may think that all crystalline solids are perfect in that all unit cells consist of a perfect arrangement of atoms/ions/molecules and the unit cells line up sequentially to form a three-dimensional space lattice with no distortion.

However, only a few crystals have a complete and perfect arrangement of their entities. There may be fewer or more entities placed randomly in a unit cell.

Most crystals suffer from such types of imperfections, which are called defects. Crystal defects occur in points, along lines or along planes. Based on the number of dimensions to which they extend they may be classified as point defects (0 dimension, 0 D defect) involving one or two lattice sites, line defects (1 D defect, involving a row of a lattice), plane defects (2 D defect, when a plane of sites is imperfect) or bulk or volume defects extending in all the three dimensions.

These defects have an important effect on the mechanical, electrical, magnetic and optical properties of solids. Here we will confine ourselves to point defects. Point defects arise due to imperfections at one or more points in a lattice.

Point Defect

A point defect in a crystal may result from any of the following:

1. The creation of the vacancy
2. The presence of an atom/ion of the parent compound in the interstice
3. Substitution of an atom/ion by another atom/ion (the presence of an impurity)
4. The presence of interstitial impurities

The first two are called stoichiometric defects since the}- do not disturb the stoichiometry of the compound, the number of positive and negative ions are in the same ratio as indicated by their chemical formula.

The third one is classified as an impurity defect while the fourth one results in a nonstoichiometric defect due to deviation from the ideal stoichiometric composition of the compound.

Stoichiometric defect:

1. Vacancy defect f creation of vacancy) An atom may be missing from the place where it ought to be present, resulting in the creation of vacancy. Such a defect is called a vacancy defect- The density of the substance decreases due to this defect

2. Interstitial defect When an atom (or ion or molecule) occupies an interstitial site (space between atom/ Ions in the crystal), It results in an interstitial defect in the crystal. There is an increase in the density of the crystal in this case.

In the case of ionic solids, one or a combination of both of the above two defects may be observed. The imperfections in such solids are mainly classified as Schottky defects and Frenkel defects.

Schottky defect In an ionic solid, when a vacancy at an anionic site is compensated for by a vacancy at a cationic site, thus maintaining electrical neutrality, the defect is termed a Schottky defect.

The greater the number of Schottky pairs, the lower the density of the solid. This is the principal defect in alkali halides. Such defects are responsible for the optical and electrical properties of Nad 1 in KT pairs are vacant in Nad at room temperature.

Frenkel defect When an ion occupies an interstitial site (in between the ions) causing a vacancy at the original place, the defect is called a Frenkel defect A vacancy defect occurs at the original site resulting in an interstitial defect at the new site.

Here the density and the electrical neutrality of the solid are the same as that of the normal solid. Such defects occur when the anion and cation differ greatly in size so that the smaller ion can fit into the interstitial sites. A Frenkel defect in AgCl (which also has a Nad structure).

CaF₂ also has predominantly Frenkel defects but here it is the anion which occupies the interstitial site. SrF₂, PbF₂, ThO₂, UO₂ and ZrO, are other compounds shouting Frenkel defects similar to those in CaF¯.

Impurity defects (presence of substitutional impurity):

In ionic compounds, a cation may be replaced by another cation of similar size. Hus is what is called a substitutional impurity. For example, during the precipitation of BaSO₄ from a solution, if some Sr²⁺ ions are also present, then the Sr^2- ions occupy the lattice of BaS04 crystals. Similarly, some of the sites of Na⁺ ions in NaCl may be occupied by Sr²⁺ ions if molten Nad is crystallised in the presence of SrCl₂.

Nonstoichiometric defects:

When there is a change in the stoichiometry of the solid due to imperfections, the defect is called a nonstoichiometric defect There are two types of such defects—the metal-excess defect and the metal-deficiency defect.

Metal-excess defect This defect may arise due to the creation of anionic vacancies or dun-to (ho presence of extra cations at the interstitial sites, An example of the former type Is shown when Mt in cut rifle of Na vapour, This results in the formation of a colour centre, which imparls yellow Colour to Mat I crystals, 1 his is discussed in the section on colour centres.

The second type of metal-excess defect is observed in nQ, Here (the extra %w’ cations occupy the jMfefsHlidi sites and electrons occupy other interstitial sites to maintain electrical neutrality.

This defers is observed in chrysalis exhibiting Frenkel defect, When nO is healed, it loses oxygen and electrons to form it 31 and turns yellow in colour. These excess cations are trapped in interstitial sites with the electrons occupying the nearly interstitial sites.

Effects Of Imperfections In Solids

ionic conductivity in solids:

The most important aspect of point defects is that they make 11 possible for atoms or ions to move through the structure. If there were no defects it would be difficult to imagine the movement of ions in the la I life, Two mechanisms are possible—the vacancy mechanism (which may lie described as the movement of Him vacancy rather than the ion) and the interstitial mechanism, These form what is called Die hopping model.

Colour centres:

Sodium chloride is colourless when pure at room temperature, It is interesting to note that when it is healed in sodium vapour (for that matter any alkali halide in the alkali metal vapour) it allows a greenish-yellow colour.

In this process, NnCl becomes slightly nonstoichiometric to give \(\mathrm{Na}_{1+\delta} \mathrm{Cl} \text { with } \delta \ll 1\) What happens Is that,

When alkali metal halides are heated in Hie alkali metal vapour, the anions migrate to the surface of the crystal to combine with Hie neutral metal atoms. The metal atoms ionise by losing an electron and then combine with the anions to form the salt, As a result, anion vacancies are produced and the electrons released by the metal occupy the anionic vacancies.

The anionic sites occupied by electrons are called N’enlies, (F stands for Farbenzentre, which means colour centre in Herman.) A series of energy levels are available for Hie alec! runs in the colour centres and the energy required to transfer from one level to another falls hi the visible range of electromagnetism radiation.

Crystalline vs amorphous solids – differences and examples

When visible light falls on the crystal, the excitation of the electrons imparts a characteristic colour (yellow in ease of NaCl) to the crystal. Similarly, KCl heated in potassium vapour exhibits a violet colour and LiCl heated in Li vapour shows a pink colour.

Electrical Properties Of Solids

Solids can be conductors, semiconductors or insulators depending upon whether and to what extent they conduct electricity. Electrical conductivity is determined by the ease of movement of electrons past the atoms under the influence of an electric field.

It is the reciprocal of electrical resistivity and is measured in Sm^– (siemens per metre). Metals are good conductors of electricity with conductivity in the order of 106 to 108 Sm~ The conductivity in metals depends on the number of electrons available for conduction.

Semiconductors have conductivity in the range of 1Q^– to 10′ Sm^-1 while insulators have a conductivity of 10^-13 Sm^-1. A semiconductor is a substance with a conductivity which can be varied by several orders of magnitude by altering its chemical composition or by increasing electrical potential.

One important characteristic of metals is their ability to conduct electricity which increases as the temperature decreases. In contrast, insulators and semiconductors show a decrease in v conductivity as the temperature is decreased.

Ionic compounds are ordinarily nonconductors in the solid state at room temperature. However, there is a significant increase in conductivity on increasing the temperature or if an impurity is present.

Sometimes, a nonstoichiometric defect can make an insulator a semiconductor. For example, NiO is known to have variable stoichiometry. When prepared at a low temperature (1100 K), it is an insulator, pale green and its stoichiometry is 1: 1.

If nickel is oxidised in the presence of an excess of oxygen at 1500 K, the stoichiometry of the compound changes and is obtained. In this process, oxygen molecules are absorbed onto the surface, dissociate and undergo a redox reaction with some Ni2+ ions to form Ni3+ and O2- ions.

To restore electrical neutrality, some Ni2+ ions diffuse out to the surface and leave cation vacancies inside tire crystals. The colour of the compound so formed is black and it is a moderately good electrical conductor.

NiO can be represented as \(\mathrm{Ni}_{1-3 x}^{2+} \mathrm{Ni}_{2 x}^{3+} \mathrm{v}_x \mathrm{O}\), where v denotes vacancies and x is the number of vacancies. Accordingly, the number of Ni²⁺ and Ni³⁺ ions are indicated in NiO. In this case, there can be movement of electrons from Ni²⁺ to Ni³⁺; the Ni³⁺ ion is thus converted to Ni²⁺.

This leads to the formation of Ni³⁺ at another ‘ point. This ion (Ni³⁺) can again be converted into Ni²⁺ by the movement of an electron. This phenomenon involving the movement of electrons is responsible for the conduction of electricity. NiO is also called a hopping semiconductor.

The tire difference between the conductance of a conductor, that of a semiconductor and that of an insulator is explained based on the band theory of solids. According to the molecular orbital theory (MO theory), when atomic orbitals (AOs) on two atoms combine or mix, they form sets of higher-energy (called antibonding) and lower-energy (called bonding) molecular orbitals (MOs).

The total number of MOs obtained is equal to the number of AOs that combine. As the number of atoms in a molecule increases, so does that of MOs. With the rise in the number of MOs, their energy differences become small and we obtain a continuous band of MOs or energy levels.

Consequently, the MO theory for metals is called band theory. Let us take the example of sodium metal. The valence orbital in sodium is 3s. In diatomic sodium, the 3s orbitals of the two Na atoms combine to form one bonding and one antibonding molecular orbital.

When it is triatomic, three 3s orbitals will combine and when it is Na, n orbitals will combine. The difference in energy between successive MOs in a Na molecule decreases with the increase in the number of sodium atoms forming a continuous band.

The lower band is made of bonding MOs while the upper one is made of antibonding MOs. The rules for filling the MOs are the same as for AOs, with each MO accommodating 2 electrons and the lower-energy MO being filled first followed by the higher-energy MOs. Hence, the lower band is filled with electrons but the upper band is empty. This may be seen from the arrangement of electrons.

In the presence of electrical potential, the electrons can be shifted from one set of energy levels to the other or more precisely from the lower band to the upper band. If the upper band is filled, no such shifting can

In the case of magnesium (electronic configuration [Ar] 3s²) the 3s and 3p orbitals form bands, which overlap in energy, and the resulting composite band is only partially filled. Magnesium is thus a conductor of electricity.

As the valence shells of many metal atoms are only partly filled, all MOs will not be filled; some of the higher MOs will therefore be only partly filled or even empty. The filled band is generally referred to as the tire valence band, and the higher energy band is called the conduction band, In hi case of metals, the two bands overlap and easy electron transfer from the valence to the conduction band can occur, facilitating the conduction of electricity.

Metals are thus good conductors of electricity. However, the electrical conductivity of a metal decreases with increasing temperature. This is because the particles undergo increased vibrational motion about their lattice sites and this vibration disrupts the flow of electrons through the crystal.

In the case of semiconductors, the gap between the valence and tire conduction bands is small. Thus, some thermally excited electrons can move into their conduction band and tire substances can conduct electricity. An insulator, however, has a filled valence band and an appreciable gap between tire valence and the conduction bands

Semiconductors

Silicon and germanium are the most common examples of semiconductors; the width of the forbidden band (band gap which contains no allowed energy states) in them is very small and hence the electrons can easily move from the valence to the conduction band.

These are called intrinsic semiconductors. Their conductivity increases with an increase in temperature because electrons gain thermal energy with the rise in temperature and jump from the valence band to the conduction hand. However, their conductivity is too low for practical purposes.

If small amounts of impurities are incorporated into the lattice of these elements their conductivity increases. Crystal defects arising due to the incorporation of impurities can facilitate the movement of electrons through the solid under the influence of an electric field.

The impurity is called a dopant and the process is called doping- Such materials whose conductivity is controlled by the addition of dopants are called extrinsic semiconductors. Doping alters the number of charge carriers and increases conductivity to a large degree.

Two types of conduction mechanisms may be observed in semiconductors. Any electron that is promoted to the conduction band is a negative charge carrier and moves towards the positive electrode under an applied potential.

The valence electron levels that are left behind in the valence band may be regarded as positive holes. Positive holes move when electrons enter them leaving their own positions. Effectively, therefore, positive holes move in a direction opposite to that of electrons.

If doping increases tire conduction-band electron population, an n-type (negative charge carrier) semiconductor is produced. If doping removes electrons from the valence band, the semiconductor is of the p-type (positive charge carrier). These defects increase with temperature rise and hence the conductivity also increases with temperature.

In order to understand tire concepts better, let us take the example of Si, which is a Group 14 element. Alloys formed from the elements of Group 14 (say Si) and Group 15 are n-type semiconductors. Each atom of the element from Group 15 adds one extra valence electron to the alloy.

Let us see how this happens. Pure, crystalline silicon has four valence electrons at room temperature, forming four covalent bonds with the adjacent silicon atoms in a tetrahedral arrangement. If this crystal is now doped with a Group 15 element, say P, the added P atoms occupy normal Si positions in the structure.

But each P atom has five valence electrons. Of the five valence electrons, four are used to form covalent bonds (as if the dopant were a silicon atom) and one is extra not needed for bonding. According to the band theory, this extra electron occupies a discrete level which is slightly below the bottom of the conduction band.

This level acts as a donor level, meaning that the electrons here have sufficient thermal energy to move into the conduction band, where they are free to move. Since the number of electrons in the conduction band of silicon doped with phosphorus is greater than that in pure silicon, the conductivity of silicon doped in this manner is greater than that of pure silicon.

This type of semiconductor in which atoms with more valence electrons occupy sites in a crystal structure, adding extra electrons (negatively charged particles) to the structure, is called an n-type semiconductor. Germanium doped with arsenic also produces an n-type semiconductor.

If a p-type semiconductor has to be formed from Si, elements of Group 13 (for example Ga) are added to a silicon crystal. (We say that Si is doped with Ga.) Group 13 elements have one valence electron less than the elements of Group 14. Thus, in the case of Si-doped with Ga, the dopant atom has only three valence electrons when it is expected to form four covalent bonds.

Thus, one of the Ga-Si bonds must be deficient by one electron and as per the band theory, the energy level associated with this Ga-Si bond does not form part of the valence band but forms a discrete level just above the valence band. This level is known as the acceptor level since it is capable of accepting an electron.

The gap between the acceptor level and the top of the valence band is small. Consequently, electrons from the valence band have sufficient thermal energy to move to the acceptor level. This leaves behind positive holes in the valence band. These holes can move about in the crystal, contributing to electrical conduction. Shows a comparison of the bands in the three types of semiconductors.

Applications of semiconductors:

Semiconductors find use in transistors, silicon chips, photocells, etc. Doped semiconductors are essential components in the modem solid-state electronic devices, found in radios, televisions, calculators and computers.

The simplest example is the pn junction. In this, silicon is doped in such a way that half is n-type and the other half is p-type. The p-type region has positive charge carriers (holes) while the n-type region has negative charge carriers (electrons).

The electrons in this case can flow spontaneously from the n-type to the p-type region across the junction (band gap) combining with the holes in the p-type region to form negative ions. As electrons diffuse, they leave positively charged ions (donors) in the n region.

Similarly, holes near the pn interface begin to diffuse into the n-type region leaving localised ions (acceptors) with a negative charge. The regions near the pn interface lose their neutrality and become charged forming the space charge region, or depletion layer. The space charge then acts as a barrier to further electron flow.

Types of solids – molecular, ionic, covalent, and metallic

If an external potential difference is applied to the sample such that the p-type end is positive and the n-type end is negative, current flows through the circuit. Electrons enter the sample from the right-hand electrode, flow through the conduction band of the n-type region, drop into the valence band of the p-type region at the pn junction and then flow through the valence band via the positive holes to leave at the left-hand electrode. A continuous current cannot flow in the opposite direction. The pn junction is a rectifier in the sense that

Gallium arsenide, GaAs, is a semiconductor used in solar cells, LEOs, etc, On applying an external difference, the electrons flow through the pn junction arid and finally fall ink) the holes of the p-type context, During this process they emit energy which is in the form of light (in case of silicon semiconductors tuggy fa released in the form of heat).

Thus, these act as light-emitting diodes (LEDs) in electronic displays, The red light in red laser pointers, bar code readers and CD players is all due to the presence of this semi/x/ruictetor, Partially sub stituting gallium with aluminium or arsenic with phosphorus changes the band gap energy faulting in diodes exhibiting various colours such as yellow, orange and green.

The general formula for such semiconductors is Transistors are typically single crystals of silicon doped to give three zones, They may contain pop or npn junctions and act as voltage amplifiers and oscillators in radios, televisions, hi-fi circuits and in amputees.

There are also controlled-valency semiconductors (in which the valency of the combining atoms is modified m per the need) which find application as thermisters—temperature-sensitive sensors. An example is Li_0.05 Ni_0.05O, It is a hopping semiconductor and its conductivity depends on

Some semiconductors are photoconductive, Le, their conductivity increases greatly on irradiation with light, Amorphous selenium is an example and forms an essential component of the photocopying process.

Superconductivity

In 1911 Kamerlingh Onnes discovered that mercury offers no resistance to the flow of electric current at the very low temperature of 4 K, This phenomenon of offering no resistance to the flow of electricity is called superconductivity. Most metals exhibit superconductivity between 2 K and 5 K,

Since 5 K is a temperature at which one cannot work normally, attempts were made to discover materials offering superconductivity at higher temperatures. Examples of such materials are Tl₂Ca₂ha₂Cu₃O₁₀; (125 K); Bi₂Ca₂Sr₂Cu₃O₁₀ (105 K) and YBa₂Cu₂O₇ (90 K). (The values in brackets indicate the temperatures at which time substances become superconducting,)

Magnetic Properties Of Solids

Before studying the magnetic properties of solids Let us see how a substance behaves under the influence of a magnetic field. A magnetic field produces lines of force that penetrate the medium in which it is applied.

The density of these lines of force is called magnetic flux density, B. The magnetic field H and the magnetic flux density are related as B~p0H in a vacuum, where p() is the permeability of free space.

If a magnetic material is placed in the field, it can change the magnetic flux density. The field of the sample in the applied field is known as its magnetisation, M., Now, the magnetic flux density is given by

⇒ \(B=\mu_0 H\)

There is another parameter, magnetic susceptibility, y (% ~ Magnetisation is usually discussed in terms of χ (chi).

Solids can be broadly classified into two types—(1) those materials which when placed in a magnetic field are drawn away from it and (2) those materials which are drawn towards a magnetic field.

The first type of materials are called diamagnetic materials, In such materials, all the electrons are paired and the magnetic moment of one of a pair is compensated by the equal and opposite moment of the other. When such a material is kept in a magnetic field, the number of lines of force emanating from the magnet is reduced, and the magnetic flux density is reduced.

They are weakly repelled by a magnetic field. Examples of diamagnetic substances are H₂O and C₆H₆, ‘Flic magnetism in the second type of materials is a clue to the presence of Urns, terms or molecules containing unpaired electrons. Magnetic behaviour is mainly exhibited by compounds of transition metals and lanthanides/ many of which possess unpaired d and f electrons respectively.

The unpaired electrons may be oriented M random on the different atoms, when the material is called paramagnetic f, both substances have permanent dipoles and are weakly attracted to a magnetic field. Examples are O₂ and CuO,

Substances which are strongly attracted to a magnetic field are said to be ferromagnetic. They can be permanently magnetised. Examples are iron, cobalt and nickel.

The metal ions of such substances, In the solid state, cluster together in small areas called domains. Each domain behaves like a small magnet. When the sample is demagnetised, the domains are oriented at random and their magnetic moments get cancelled. In the presence of a magnetic field, the domains are permanently aligned in one direction, that of the field, and the sample becomes permanently magnetised.

The orientation of domains in opposite directions leads to antiferromagnetism. The magnetic moments cancel each other out. Examples of antiferromagnetic substances are MnO and NiO, When the magnetic dipoles of the domains are aligned in parallel and antiparallel directions in unequal numbers, a net magnetic moment results and the substance is ferrimagnetic. Examples are ferrites (AFe₂O), where A is a divalent metal) and Fe₃O₄. Ferrimagnetic materials are not so strongly attracted to a magnetic field as ferromagnetic substances.

The magnetic properties are the outcome of the magnetic moments associated with the electrons. The magnetic moment of an unpaired electron arises from two causes, one due to orbital motion around the nucleus and the other due to the spin of the electron. The spin component is of more importance. If the electron is visualised as a bundle of negative charge spinning on its axis, the magnitude of the resulting spin moment pB is 1.73 Bohr magnetons. The Bohr magneton is the fundamental unit of magnetic moment.

It is defined as \(1 \mathrm{BM}=\frac{e h}{(4 \pi m c)}\), where e is the electronic charge, h is the Planck constant, m is the electronic mass and c is the velocity of light. It is equal to 9.27 x 10^-24 A m²,

The spin magnetic moment of an electron is ±ps. The orbital magnetic moment is given by the product of the magnetic quantum number and the Bohr magneton.

The effect of temperature on magnetic behaviour

The temperature dependence of the magnetic susceptibility of a paramagnetic substance is given by Curie’s law.

⇒ \(\chi=\frac{\dot{C}}{T},\)

where C is a constant called Curie constant and T is the temperature in kelvin. Curie’s law takes different forms in the case of ferromagnetic and antiferromagnetic substances as the variation of χ is observed to be different

Two temperatures called the Curie temperature and the N6el temperature, are significant. Ferromagnetic materials change to paramagnetic substances above a temperature called the Curie temperature (Tc). Materials which are antiferromagnetic at low temperatures become paramagnetic above a certain temperature called the Neel temperature (TN).

Metals and alloys:

Iron, cobalt and nickel are ferromagnetic while chromium and manganese are antiferromagnetic at low temperatures. The oxides MnO, FeO, CoO and NiO are all antiferromagnetic at low temperatures and change to paramagnetic above the Neel temperature.

Spinels owe their name to the mineral spinel, MgAl₂O₄.

They have the general formula AB₂O₄, A being a divalent ion and B being a trivalent ion. The commercially important spinels known as ferrites are of the type MFe204; where M is a divalent ion such as Fe²⁺, Ni²⁺, Cu²⁺ and Mg²⁺. They are all either antiferromagnetic or ferrimagnetic. Some garnets [M₃M₂ (SiO₄)₃; where M₁₁ = Ca²⁺, Mg²⁺ or Fe³⁺ and Mm = Al³⁺, Ca³⁺ or Fe³⁺ ] are important ferrimagnetic materials. For example, pyrope (Mg₃Al₂Si₃O₁₂) and andradite (Cr₃Fe₂Si₃O₁₂) are ferrimagnetic.

A major application of ferro- and ferrimagnetic materials is in transformers and motor cores. Magnetic bubble memory devices for storing information use thin films of garnets deposited on nonmagnetic substances.

The Solid State Multiple Choice Questions

Question 1. Bonding in diamonds is

1. Covalent
2. Ionic
3. Dipole
4. Metallic

Answer: 1. Covalent

Question 2. During the formation of a solid,

1. Some Energy Is Lost
2. Some Energy Is Gained
3. Energy Remains Constant
4. Some Energy May Be Gained Or Lost Depending On The System

Answer: 1. Some Energy Is Lost

Question 3. Tetrahedral bonding is characteristic of

1. Ionic Bonds
2. Molecular Bonds
3. Metallic Bonds
4. Covalent Bonds

Answer: 4. Covalent Bonds

Question 4. Ionic solids have

1. A Low Melting Point
2. A High Melting Point
3. A Moderate Melting Point
4. None Of These

Answer: 3. A Moderate Melting Point

Question 5. The bond between ice molecules is

1. Ionic
2. Covalent
3. A Hydrogen
4. Metallic

Answer: 3. A Hydrogen

Question 6. Among the following, the strongest bond is the

1. Ionic Bond
2. Hydrogen Bond
3. Metallic Bond
4. Covalent Bond

Answer: 1. Ionic Bond

Properties of crystalline solids – symmetry, lattice structure

Question 7. Ionic bonds are mainly formed in

1. Inorganic Compounds
2. Metals
3. Organic Compounds
4. None Of These

Answer: 1. Inorganic Compounds

Question 8. Molecular solids have

1. Very Low Melting Points
2. Fairly Low Melting Points
3. Very High Melting Points
4. None Of These

Answer: 3. Very High Melting Points

Question 9. Metallic solids are generally

1. Hard And Brittle
2. Soft And Plastically Deformable
3. Malleable And Ductile
4. None Of These

Answer: 2. Soft And Plastically Deformable

Question 10. Among the following, the element which has a covalently bonded crystal structure is

1. Al
2. Pb
3. Ge
4. Bi

Answer: 3. Ge

Question 11. Materials having different properties along different directions are called

1. Isotropic
2. Anisotropic
3. Amorphous
4. None Of These

Answer: 2. Anisotropic

Question 12. The tiny fundamental block which, when repeated in space indefinitely, generates a crystal is called

1. A Primitive Cell
2. A Lattice
3. A Unit Cell
4. None Of These

Answer: 3. A Unit Cell

Question 13. How many basic crystal systems are possible?

1. Four
2. Five
3. Six
4. Seven

Answer: 4. Seven

Question 14. How many types of Bravais lattices are possible in crystals?

1. 7
2. 14
3. 8
4. 5

Answer: 2. 14

Question 15. The number of lattice points in a primitive cell is

1. 4
2. 2
3. 8
4. 1

Answer: 4. 1

Question 16. A unit cell with crystallographic dimensions abc, a \(a \neq b \neq c, \alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}\), B 90° corresponds to

1. Calcite
2. Graphite
3. Rhombic Sulphur
4. Monoclinic Sulphur

Answer: 4. Monoclinic Sulphur

Question 17. Which of the following structures does a crystal, with a b c and interfacial angles a = B = y = 90°, have?

1. Cubic
2. Tetragonal
3. Trigonal
4. Orthorhombic

Answer: 4. Orthorhombic

Question 18. Cubic close packing of equal-sized spheres is described by

1. AB AB AB AB…
2. AB AC AC AB…
3. ABC ACB ABC ACB…
4. ABC ABC ABC…

Answer: 4. ABC ABC ABC…

Question 19. Which of the following metals has an FCC structure?

1. Al
2. Cu
3. Pb
4. All Of These

Answer: 4. All Of These

Question 20. The percentage of total volume occupied by the particles in a bcc structure is

1. 32
2. 68
3. 50
4. 74

Answer: 2. 68

Question 21. The FCC structure is often called

1. Cubic Close Packed
2. Hexagonal close-packed
3. The Graphite Structure
4. The diamond structure

Answer: 1. Cubic Close Packed

Question 22. The correct order of sizes of voids (holes) is

1. Trigonal > Tetrahedral > Octahedral
2. Hexagonal Close Packed
3. Octahedral > Tetrahedral > Trigonal
4. The Diamond Structure

Answer: 1. Trigonal > Tetrahedral > Octahedral

Question 23. The tetrahedral void has a coordination number of

1. Two
2. Three
3. Four
4. Eight

Answer: 3. Four

Question 24. The octahedral void has a coordination number of

1. Two
2. Six
3. Eight
4. Four

Answer: 2. Six

Question 25. In close-packed (hcp, ccp) arrangements of lattices comprising n atoms of a kind, the number of tetrahedral and octahedral voids present respectively are

1. 2n and n
2. n and 2n
3. n and n
4. 2n and 2n

Answer: 1. 2n and n

Question 26. The coordination number of the simple cubic structure is

1. 6
2. 8
3. 12
4. 4

Answer: 1. 6

Question 27. The effective number of atoms belonging to the unit cell of a simple cubic structure is

1. 8
2. 1
3. 4
4. 6

Answer: 2. 1

Question 28. The atomic packing efficiency of a simple cubic structure is

1. 0.68
2. 0.74
3. 1.00
4. 0.52

Answer: 4. 0.52

Question 29. The only element with a simple cubic structure is

1. Silver
2. Polonium
3. Zinc
4. Iron

Answer: 2. Polonium

Question 30. In the crystal structure of NaCl, the arrangement of Clions is

1. Fcc
2. Bcc
3. Both Fcc And Bcc
4. None Of These

Answer: 1. Fcc

Question 31. Interstitial impurities are a

1. Surface Defect
2. Point Defect
3. Line Defect
4. Volume Defect

Answer: 2. Point Defect

Question 32. In ionic crystals, an ion displaced from a regular site to an interstitial site results in what is called a/an

1. Electronic Defect
2. Schottky Defect
3. Frenkel Defect
4. None Of These

Answer: 3. Frenkel Defect

Question 33. In ionic crystals, the missing of a cation-anion pair results in a/an

1. Electronic Defect
2. Schottky Defect
3. Frenkel Defect
4. None Of These

Answer: 2. Schottky Defect

Question 34. In a solid-crystal lattice, a cation leaves its original site and moves to an interstitial position. The lattice defect is called a/an

1. Interstitial Defect
2. Vacancy Defect
3. Frenkel Defect
4. Schottky Defect

Answer: 3. Frenkel Defect

Characteristics of amorphous solids and their behavior

Question 35. An ionic solid with the Schottky defect contained in its structure

1. An Equal Number Of Cation And Anion Vacancies
2. Anion Vacancies And Interstitial Anions
3. Cation Vacancies
4. Cation Vacancies And Interstitial Cations

Answer: 1. An Equal Number Of Cation And Anion Vacancies

Question 36. For \(\mathrm{NaCl}, r_{\mathrm{Na}^{\prime}} / r_{\mathrm{Cl}}=0.325\) The ratio of the coordination numbers of the ions is

1. 6:6
2. 4:4
3. 8:4
4. 6:12

Answer: 1. 6:6

Question 37. Which of these is found in AgCI?

1. Frenkel defect involving cations
2. Frenkel defect involving anions
3. Schottky defect
4. Interstitial defect

Answer: 1. Frenkel defect involving cations

Question 38. The point defect found in Call Is called

1. Frenkel Defect Involving Cations
2. Schottky Defect
3. Frenkel Defect Involving Anions
4. Edge Dislocation Defect

Answer: 3. Frenkel Defect Involving Anions

Question 39. To prepare an n-type semiconductor, the element to be added to Si is

1. Germanium
2. Arsenic
3. Aluminum
4. Indium

Answer: 2. Arsenic

Question 40. To prepare a p-type semiconductor, the element to be added to Ge is

1. Silicon
2. Arsenic
3. Aluminum
4. Antimony

Answer: 3. Aluminum

Question 41. Which type of atoms are added to Ge to prepare an n-type semiconductor?

1. Trivalent
2. Pentavalent
3. Tetravalent
4. Divalent

Answer: 2. Pentavalent

Question 42. Which type of atoms are added to Si to prepare a p-type semiconductor?

1. Trivalent
2. Pentavalent
3. Tetravalent
4. None of these

Answer: 1. Trivalent

Question 43. An example of a superconductor Is

1. Cu
2. Si
3. Ge
4. Hg

Answer: 4. Hg

Question 44. A superconductor exhibits…… resistance.

1. Small
2. Large
3. Zero
4. Infinite

Answer: 3. Zero

Question 45. One Bohr magneton equals

1. 9.27 × 10^-24 Am²
2. 9.1 x 10^-31 Am²
3. 9.27 x 10^-16 Am²
4. 9.1 x 10^-24 Am²

Answer: 1. 9.27 × 10^-24 Am²

Question 46. The temperature at which an antiferromagnetic substance changes to a paramagnetic substance is called the

1. Curie-Weiss temperature
2. Curie temperature
3. Debye temperature
4. Néel temperature

Answer: 4. Néel temperature

Question 47. The transition from the ferromagnetic to the paramagnetic state takes place at the

1. Curie temperature
2. Curie-Weiss temperature
3. Néel temperature
4. Debye temperature

Answer: 1. Curie temperature