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Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
Physics Chemistry Biology

Class 12th (Chemistry) Chapters
1. The Solid State 2. Solutions 3. Electrochemistry
4. Chemical Kinetics 5. Surface Chemistry 6. General Principles And Processes Of Isolation Of Elements
7. The P-Block Elements 8. The D-And F-Block Elements 9. Coordination Compounds
10. Haloalkanes And Haloarenes 11. Alcohols, Phenols And Ethers 12. Aldehydes, Ketones And Carboxylic Acids
13. Amines 14. Biomolecules 15. Polymers
16. Chemistry In Everyday Life



Unit 1 The Solid State



General Characteristics Of Solid State

Matter exists in three states: solid, liquid, and gas. The state of a substance at a given temperature and pressure is determined by a balance between two opposing factors:

At sufficiently low temperatures, thermal energy is minimal, allowing strong intermolecular forces to hold particles in fixed positions. Although fixed, these particles can still oscillate around their mean positions, defining the solid state.

Characteristic properties of solids:


Amorphous And Crystalline Solids

Solids are classified based on the arrangement pattern of their constituent particles:

Two-dimensional structure showing the ordered arrangement in crystalline quartz and the disordered arrangement in amorphous quartz glass

Differences in particle arrangement lead to distinct properties:

Summary of distinctions:

Property Crystalline solids Amorphous solids
ShapeDefinite characteristic geometrical shapeIrregular shape
Melting pointMelt at a sharp and characteristic temperatureGradually soften over a range of temperature
Cleavage propertySplit into two pieces with plain, smooth surfaces when cutCut into two pieces with irregular surfaces when cut
Heat of fusionDefinite and characteristic enthalpy of fusionDo not have definite enthalpy of fusion
AnisotropyAnisotropic in nature (properties vary with direction)Isotropic in nature (properties are same in all directions)
NatureTrue solidsPseudo solids or supercooled liquids
Order in arrangementLong range orderOnly short range order

Some solids appear amorphous but are actually polycrystalline, consisting of many randomly oriented tiny crystals. Metals are often polycrystalline, appearing isotropic overall despite individual anisotropic crystals.

Amorphous solids like glass, rubber, and plastics are widely used. Amorphous silicon is important in solar cells.



Classification Of Crystalline Solids

Crystalline solids can be classified based on the nature of the intermolecular forces or bonds holding their constituent particles together. This classification divides them into four categories:


Molecular Solids

Constituent particles are molecules. These are further sub-divided based on the intermolecular forces:

  1. Nonpolar Molecular Solids: Constituent particles are either atoms (like Ar, He) or nonpolar molecules ($\textsf{H}_2$, $\textsf{Cl}_2$, $\textsf{I}_2$, $\textsf{CO}_2$). Held together by weak dispersion forces (London forces). These are soft, non-conductors of electricity, have low melting points, and are usually gases or liquids at room temperature and pressure.
  2. Polar Molecular Solids: Constituent particles are molecules formed by polar covalent bonds ($\textsf{HCl}$, $\textsf{SO}_2$, $\textsf{NH}_3$). Held together by stronger dipole-dipole interactions. These are soft, non-conductors of electricity, have higher melting points than nonpolar molecular solids, but are often gases or liquids at room temperature and pressure.
  3. Hydrogen Bonded Molecular Solids: Molecules contain polar covalent bonds between $\textsf{H}$ and highly electronegative atoms like $\textsf{F}$, $\textsf{O}$, or $\textsf{N}$. Held together by strong hydrogen bonds. Example: Ice ($\textsf{H}_2\textsf{O}$). These are non-conductors of electricity. Generally volatile liquids or soft solids at room temperature and pressure.

Ionic Solids

Constituent particles are ions (cations and anions). Formed by three-dimensional arrangement of ions held together by strong coulombic (electrostatic) forces. Examples: $\textsf{NaCl}$, $\textsf{MgO}$, $\textsf{ZnS}$, $\textsf{CaF}_2$. These solids are hard and brittle. They have high melting and boiling points. In the solid state, ions are fixed, so they are electrical insulators. In the molten state or when dissolved in water, ions are free to move and conduct electricity.


Metallic Solids

Constituent particles are positive metal ions (cations) surrounded by and held together by a 'sea' of mobile or delocalised electrons (metallic bonding). Each metal atom contributes valence electrons to this mobile pool, which are spread throughout the crystal. Examples: $\textsf{Fe}$, $\textsf{Cu}$, $\textsf{Ag}$, $\textsf{Mg}$. These solids are typically hard (though variable), malleable (can be hammered into sheets), and ductile (can be drawn into wires). Mobile electrons are responsible for their high electrical and thermal conductivity in both solid and molten states, as well as their characteristic lustre and color.


Covalent Or Network Solids

Constituent particles are atoms held together by a network of continuous covalent bonds extending throughout the crystal. These are also called giant molecules. Covalent bonds are strong and directional, holding atoms in fixed positions. Examples: Diamond ($\textsf{C}$), silicon carbide ($\textsf{SiC}$), quartz ($\textsf{SiO}_2$), aluminum nitride ($\textsf{AlN}$). These are very hard and brittle, have extremely high melting points (often decompose before melting), and are electrical insulators.

An exception is Graphite ($\textsf{C}$). It is a covalent solid but is soft and conducts electricity. Its structure consists of layers of carbon atoms, where each carbon is covalently bonded to three neighbors in the same layer. The fourth valence electron is delocalised between layers, allowing it to conduct electricity. Weak Van der Waals forces between layers allow them to slide, making graphite soft and a good lubricant.

Network structure of diamond showing continuous covalent bonding
Layered structure of graphite

Summary of properties of different crystalline solids:

Type of Solid Constituent Particles Bonding/Attractive Forces Examples Physical Nature Electrical Conductivity Melting Point
Molecular solidsMoleculesDispersion or London forcesAr, CCl4, H2, I2, CO2SoftInsulatorVery low
MoleculesDipole-dipole interactionsHCl, SO2SoftInsulatorLow
MoleculesHydrogen bondingH2O (ice)HardInsulatorLow
Ionic solidsIonsCoulombic or electrostaticNaCl, MgO, ZnS, CaF2Hard but brittleInsulators in solid state but conductors in molten state and in aqueous solutionsHigh
Metallic solidsPositive ions in a sea of delocalised electronsMetallic bondingFe, Cu, Ag, MgHard but malleable and ductileConductors in solid state as well as in molten stateFairly high
Covalent or network solidsAtomsCovalent bondingSiO2 (quartz), SiC, C (diamond), AlNVery hardInsulatorsVery high
AtomsC (graphite)SoftConductor (exception)Very high

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 Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?

Answer:

1.1 Solids are rigid because their constituent particles (atoms, molecules or ions) are held in fixed positions by strong intermolecular forces and can only oscillate about their mean positions. This restricts their movement, making the solid rigid.

1.2 Solids have a definite volume because their constituent particles are held closely together by strong intermolecular forces and occupy fixed positions. The distances between particles are short and relatively constant, resulting in a fixed volume.

1.3 Amorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.

Crystalline solids: Naphthalene, benzoic acid, potassium nitrate, copper.

1.4 A solid whose refractive index is the same along all directions is isotropic in nature. This indicates that the solid is likely an amorphous solid. Amorphous solids have only short-range order and disordered arrangement over long distances, making their properties the same in all directions.

Amorphous solids do not show cleavage property. Crystalline solids with their long-range order and definite planes show cleavage property when cut with a sharp tool, splitting into pieces with smooth surfaces. Amorphous solids cut into pieces with irregular surfaces.


Intext Questions

1.5 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.

1.6 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?

1.7 Ionic solids conduct electricity in molten state but not in solid state. Explain.

1.8 What type of solids are electrical conductors, malleable and ductile?

Answer:

1.5 Classification based on intermolecular forces:

  • Ionic solids: Potassium sulphate ($\textsf{K}_2\textsf{SO}_4$), Zinc sulphide ($\textsf{ZnS}$)
  • Metallic solids: Tin ($\textsf{Sn}$), Rubidium ($\textsf{Rb}$)
  • Molecular solids:
    • Nonpolar: Benzene ($\textsf{C}_6\textsf{H}_6$), Argon ($\textsf{Ar}$)
    • Polar: Urea ($\textsf{NH}_2\textsf{CONH}_2$, due to dipole-dipole interactions), Ammonia ($\textsf{NH}_3$)
    • Hydrogen bonded: Water ($\textsf{H}_2\textsf{O}$, in solid state as ice)
  • Covalent or Network solids: Graphite ($\textsf{C}$), Silicon carbide ($\textsf{SiC}$)

1.6 Solid A is a very hard electrical insulator in both solid and molten states and has an extremely high melting point. This description matches the properties of a covalent or network solid (like diamond or silicon carbide).

1.7 Ionic solids are made up of positive and negative ions held by strong electrostatic forces. In the solid state, these ions are fixed in their lattice positions and are not free to move. Therefore, they do not conduct electricity. In the molten state or when dissolved in water, the ionic bonds are broken, and the ions become mobile (free to move). The movement of these charged ions constitutes electric current, so ionic solids conduct electricity in molten state or in aqueous solution.

1.8 Solids that are electrical conductors, malleable, and ductile are metallic solids. These properties are due to the presence of mobile, delocalised electrons in the metallic lattice, which can move freely under an electric field and also allow the metal to be deformed without breaking.



Crystal Lattices And Unit Cells

In crystalline solids, constituent particles are arranged in a highly ordered, repeating pattern. This pattern can be visualised using a space lattice or crystal lattice, which is a three-dimensional arrangement of points representing the locations of the constituent particles (motifs).

Diagram showing a motif, a two-dimensional space lattice, and a crystal structure formed by placing the motif on the lattice points

Characteristics of a crystal lattice:

A crystal lattice can be completely described by its smallest repeating unit, called a unit cell. The entire crystal structure is generated by repeating the unit cell through translational displacements in three dimensions.

Diagram showing how a hypothetical two-dimensional crystal structure is generated by repeating its unit cell

In three dimensions, a unit cell is characterised by six parameters:

Illustration of parameters of a unit cell (edge lengths a, b, c and angles alpha, beta, gamma)

Primitive And Centred Unit Cells

Unit cells are broadly categorised into two types:

  1. Primitive Unit Cells: Constituent particles are located only at the corner positions of the unit cell.
  2. Centred Unit Cells: Contain one or more constituent particles at positions other than the corners, in addition to those at the corners. There are three types of centred unit cells:
    • Body-Centred Unit Cells (bcc): Have one particle at the body centre in addition to the corner particles.
    • Face-Centred Unit Cells (fcc): Have one particle at the center of each face in addition to the corner particles.
    • End-Centred Unit Cells: Have one particle at the center of any two opposite faces in addition to the corner particles.

Based on the possible variations in edge lengths and angles, there are seven basic types of primitive unit cells, known as the seven crystal systems. A French mathematician, Bravais, showed that considering the possible centred unit cells within these crystal systems, there are only 14 unique three-dimensional lattices, called Bravais lattices.

Illustrations of the seven crystal systems: Cubic, Tetragonal, Orthorhombic, Monoclinic, Triclinic, Hexagonal, Rhombohedral

Characteristics of the seven primitive unit cells and their possible centred variations:

Crystal system Possible variations Axial distances or edge lengths Axial angles Examples
CubicPrimitive, Body-centred, Face-centreda = b = c$\alpha = \beta = \gamma = 90^\circ$NaCl, Zinc blende, Cu
TetragonalPrimitive, Body-centreda = b $\neq$ c$\alpha = \beta = \gamma = 90^\circ$White tin, SnO2, TiO2, CaSO4
OrthorhombicPrimitive, Body-centred, Face-centred, End-centreda $\neq$ b $\neq$ c$\alpha = \beta = \gamma = 90^\circ$Rhombic sulphur, KNO3, BaSO4
HexagonalPrimitivea = b $\neq$ c$\alpha = \beta = 90^\circ$, $\gamma = 120^\circ$Graphite, ZnO, CdS
Rhombohedral or TrigonalPrimitivea = b = c$\alpha = \beta = \gamma \neq 90^\circ$Calcite (CaCO3), HgS (cinnabar)
MonoclinicPrimitive, End-centreda $\neq$ b $\neq$ c$\alpha = \gamma = 90^\circ$, $\beta \neq 90^\circ$Monoclinic sulphur, Na2SO4.10H2O
TriclinicPrimitivea $\neq$ b $\neq$ c$\alpha \neq \beta \neq \gamma \neq 90^\circ$K2Cr2O7, CuSO4. 5H2O, H3BO3
Illustrations of unit cells for the 14 Bravais lattices within the seven crystal systems

Intext Questions

1.9 Give the significance of a ‘lattice point’.

1.10 Name the parameters that characterise a unit cell.

1.11 Distinguish between

(i) Hexagonal and monoclinic unit cells

(ii) Face-centred and end-centred unit cells.

1.12 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.

Answer:

1.9 The significance of a lattice point is that it represents the position of a constituent particle (an atom, a molecule, or an ion) in the crystal lattice. It's a point in space that indicates the location of the repeating structural unit.

1.10 A unit cell is characterised by six parameters:

  • The dimensions of its edges along the three axes: a, b, and c.
  • The angles between the edges: $\alpha$ (angle between edges b and c), $\beta$ (angle between edges a and c), and $\gamma$ (angle between edges a and b).

1.11 Distinctions:

(i) Hexagonal and monoclinic unit cells:

Feature Hexagonal Unit Cell Monoclinic Unit Cell
Axial lengthsa = b $\neq$ ca $\neq$ b $\neq$ c
Axial angles$\alpha = \beta = 90^\circ$, $\gamma = 120^\circ$$\alpha = \gamma = 90^\circ$, $\beta \neq 90^\circ$
Primitive unit cellsOnly Primitive typePrimitive and End-centred types

(ii) Face-centred and end-centred unit cells:

Feature Face-Centred Unit Cell (fcc) End-Centred Unit Cell
Particle locationParticles at all corners AND at the center of EACH of the six faces.Particles at all corners AND at the center of ANY TWO opposite faces.
Number of effective particles contributed by face centers6 face centers $\times$ 1/2 per face = 3 particles2 face centers $\times$ 1/2 per face = 1 particle

1.12 Portion of an atom in a cubic unit cell that is part of its neighboring unit cell:

(i) Atom located at a corner: An atom at a corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore, the portion of an atom located at a corner that belongs to its neighboring unit cell is $1/8$th. Each unit cell gets $1/8$th from each of its 8 corners.

Diagram showing a corner atom shared by 8 adjacent cubic unit cells

(ii) Atom located at the body-centre: An atom at the body centre of a cubic unit cell is located entirely within that unit cell and is not shared with any other unit cell. Therefore, the portion of an atom located at the body-centre that is part of its neighboring unit cell is 0. The entire atom belongs to the unit cell it is in.



Number Of Atoms In A Unit Cell

In a crystalline solid, constituent particles are located at the lattice points of the unit cell. We need to calculate the effective number of particles belonging to a single unit cell based on their positions (corners, faces, body centre).

Let's consider cubic unit cells (primitive, body-centred, face-centred), assuming identical atoms as constituent particles.


Primitive Cubic Unit Cell

A primitive cubic unit cell has atoms only at its eight corners. As seen in Fig. 1.12, each corner atom is shared by eight adjacent cubic unit cells (four in the same layer, four in the layer above/below). Thus, only $1/8$th of a corner atom belongs to any one unit cell.

Diagram illustrating how a corner atom in a simple cubic unit cell is shared by 8 adjacent unit cells

Effective number of atoms per primitive cubic unit cell:

Total number of atoms per primitive cubic unit cell = 1 atom.

Representations of a primitive cubic unit cell: open structure, space-filling model, and unit cell with actual portions of atoms

Body-Centred Cubic Unit Cell

A body-centred cubic (bcc) unit cell has atoms at its eight corners and one atom at the body centre. The atom at the body centre is entirely within the unit cell and is not shared with any other unit cell.

Representations of a body-centred cubic unit cell: open structure, space-filling model, and unit cell with actual portions of atoms

Effective number of atoms per body-centred cubic unit cell:

Total number of atoms per body-centred cubic unit cell = $1 + 1$ = 2 atoms.


Face-Centred Cubic Unit Cell

A face-centred cubic (fcc) unit cell has atoms at its eight corners and one atom at the center of each of its six faces. Each atom at a face centre is shared between two adjacent unit cells.

Diagram illustrating how a face-centered atom in a cubic unit cell is shared by 2 adjacent unit cells

Thus, $1/2$ of a face-centred atom belongs to any one unit cell.

Representations of a face-centred cubic unit cell: open structure, space-filling model, and unit cell with actual portions of atoms

Effective number of atoms per face-centred cubic unit cell:

Total number of atoms per face-centred cubic unit cell = $1 + 3$ = 4 atoms.



Close Packed Structures

Constituent particles in solids are often packed as closely as possible to minimise vacant space. This close packing can be built up by stacking layers of identical hard spheres.


Close Packing In One Dimension

Spheres are arranged in a single row, touching each other.

Close packing of spheres in one dimension (single row)

Each sphere is in contact with two neighbors. The coordination number (number of nearest neighbors) is 2.


Close Packing In Two Dimensions

Formed by stacking rows of one-dimensionally close-packed spheres. Two ways of stacking:

  1. Square Close Packing (AAA type): Second row spheres are placed directly above the first row, aligned horizontally and vertically (A type over A type).
    Two-dimensional square close packing (AAA type) showing spheres aligned vertically and horizontally
    Each sphere contacts four neighbors. Two-dimensional coordination number is 4. Joining centers of neighbors forms a square.
  2. Hexagonal Close Packing (ABAB type): Second row spheres are placed in the depressions of the first row (staggered). Third row aligned with first (A), fourth with second (B), etc. (ABAB type).
    Two-dimensional hexagonal close packing (ABAB type) showing staggered arrangement
    This arrangement has less empty space and is more efficient. Each sphere contacts six neighbors. Two-dimensional coordination number is 6. Joining centers of neighbors forms a hexagon. Triangular voids (empty spaces) are present between the spheres. These voids have two different orientations (apex pointing up or down).

Close Packing In Three Dimensions

Three-dimensional structures are built by stacking two-dimensional layers. Two types of 2D layers are possible: square close-packed and hexagonal close-packed.

  1. From 2D Square Close-Packed Layers (AAA... type): Stacking square close-packed layers directly on top of each other (A type over A type over A type...).
    Three-dimensional structure formed by stacking 2D square close-packed layers directly on top of each other (AAA type), resulting in simple cubic lattice
    This forms a simple cubic lattice, with a primitive cubic unit cell.
  2. From 2D Hexagonal Close-Packed Layers: More efficient 3D packing is obtained by stacking hexagonal close-packed layers.
    1. Placing the second layer (B) over the first layer (A): Spheres of the second layer are placed in the depressions of the first. This generates two types of voids (empty spaces):
      • Tetrahedral voids (T): Formed wherever a sphere of the second layer is above a void of the first layer (or vice versa). Surrounded by four spheres, forming a tetrahedron when centers are joined.
      • Octahedral voids (O): Formed where triangular voids in the first and second layers overlap (but their apexes point in opposite directions). Surrounded by six spheres, forming an octahedron when centers are joined.
      Stack of two layers of hexagonal close-packed spheres showing tetrahedral (T) and octahedral (O) voids
      Illustrations showing the geometry of tetrahedral and octahedral voids
      If N is the number of close-packed spheres, then the number of octahedral voids is N, and the number of tetrahedral voids is 2N.
    2. Placing the third layer over the second layer: Two possibilities:
      • Covering Tetrahedral Voids (ABAB... type): Spheres of the third layer cover the tetrahedral voids of the second layer and are aligned with the first layer. This results in a repeating pattern ABAB... This structure is called hexagonal close packed (hcp) structure. Metals like Magnesium ($\textsf{Mg}$) and Zinc ($\textsf{Zn}$) often crystallise in hcp.
      • Illustrations of hexagonal close-packed (hcp) structure showing ABAB stacking of layers
      • Covering Octahedral Voids (ABCABC... type): Spheres of the third layer cover the octahedral voids of the second layer. The third layer is not aligned with the first or second layer. The fourth layer is aligned with the first. This results in a repeating pattern ABCABC... This structure is called cubic close packed (ccp) structure or face-centred cubic (fcc) structure. Metals like Copper ($\textsf{Cu}$) and Silver ($\textsf{Ag}$) crystallise in ccp/fcc.
      • Illustrations of cubic close-packed (ccp) structure showing ABCABC stacking of layers, resulting in a face-centred cubic unit cell
    Both hcp and ccp/fcc structures are highly efficient, filling 74% of the total space. In both structures, each sphere has 12 nearest neighbors, so the coordination number is 12.

Formula Of A Compound And Number Of Voids Filled

In ionic solids, larger ions (usually anions) often form a close-packed structure (ccp or hcp), while smaller ions (usually cations) occupy some of the tetrahedral or octahedral voids. The fraction of voids occupied depends on the compound's stoichiometry.

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?

Answer:

In a ccp lattice, the effective number of atoms Y per unit cell is 4. The number of octahedral voids generated in a ccp lattice is equal to the number of atoms in the lattice, which is 4.

Since atoms of element X (cations) occupy all the octahedral voids, the number of atoms X per unit cell is also 4.

Thus, the ratio of atoms X : Y is $4:4$ or $1:1$.

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?

Answer:

In an hcp lattice, let the number of atoms B per unit cell be N (though the effective number in an hcp unit cell is 6, we can use N for the ratio calculation, considering a portion of the lattice). The number of tetrahedral voids generated is twice the number of atoms in the lattice, so number of tetrahedral voids = 2N.

Atoms of element A occupy $2/3$rd of the tetrahedral voids.

Number of atoms A = $(2/3) \times \text{Number of tetrahedral voids} = (2/3) \times 2\text{N} = 4\text{N}/3$.

Ratio of atoms A : B = $(4\text{N}/3) : \text{N}$. Dividing both sides by N, we get $4/3 : 1$. Multiplying by 3 to get whole numbers, the ratio is $4:3$.

The formula of the compound is A$_4$B$_3$.


Locating Tetrahedral And Octahedral Voids

Let's locate these voids within a cubic close-packed (ccp) or face-centred cubic (fcc) unit cell.

Effective number of atoms in a ccp unit cell is 4.

(a) Locating Tetrahedral Voids:

This confirms that the number of tetrahedral voids (8) is twice the effective number of atoms (4) in a ccp unit cell (2N).

Diagram showing the location of eight tetrahedral voids within a ccp unit cell and the geometry of a single tetrahedral void

(b) Locating Octahedral Voids:

Total number of octahedral voids in a ccp unit cell:

Total number of octahedral voids per ccp unit cell = $1 + 3$ = 4.

Diagram showing the location of octahedral voids at the body centre and edge centres of a ccp unit cell

This confirms that the number of octahedral voids (4) is equal to the effective number of atoms (4) in a ccp unit cell (N).

Intext Questions

1.13 What is the two dimensional coordination number of a molecule in square close-packed layer?

1.14 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?

1.15 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?

Answer:

1.13 The two-dimensional coordination number of a molecule in a square close-packed layer (AAA type) is 4, as each sphere is in contact with four neighboring spheres.

1.14 The compound forms a hexagonal close-packed (hcp) structure. In any close-packed structure (hcp or ccp), let the number of particles be N.

Number of octahedral voids = N

Number of tetrahedral voids = 2N

Total number of voids = N + 2N = 3N

The number of particles in 0.5 mol is N = 0.5 mol $\times$ Avogadro's number ($\textsf{N}_\textsf{A}$).

$\textsf{N}_\textsf{A} = 6.022 \times 10^{23} \text{ mol}^{-1}$.

Number of particles in 0.5 mol = $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$.

Total number of voids = $3 \times \text{N} = 3 \times (3.011 \times 10^{23}) = 9.033 \times 10^{23}$ voids.

Number of tetrahedral voids = $2 \times \text{N} = 2 \times (3.011 \times 10^{23}) = 6.022 \times 10^{23}$ tetrahedral voids.

1.15 Element N forms ccp lattice. In a ccp unit cell, the effective number of atoms N is 4. The number of tetrahedral voids in a ccp lattice is twice the number of atoms, so number of tetrahedral voids = $2 \times 4 = 8$.

Atoms of element M occupy $1/3$rd of the tetrahedral voids.

Number of atoms M = $(1/3) \times \text{Number of tetrahedral voids} = (1/3) \times 8 = 8/3$.

Ratio of atoms M : N = $(8/3) : 4$. To simplify to whole numbers, multiply by 3: $8 : 12$. Divide by 4: $2:3$.

The formula of the compound is M$_2$N$_3$.



Packing Efficiency

Packing efficiency is the percentage of the total space in a crystal that is filled by the constituent particles. The remaining space is in the form of voids. Packing efficiency is calculated as:

$\textsf{Packing efficiency} = \frac{\textsf{Volume occupied by particles in the unit cell}}{\textsf{Total volume of the unit cell}} \times 100\%$

Let's calculate the packing efficiency for different cubic structures, assuming particles are identical hard spheres.


Packing Efficiency In Hcp And Ccp Structures

Both hcp and ccp (fcc) structures have the highest packing efficiency among common crystal lattices. Let's calculate the efficiency for the ccp structure.

In a ccp structure, the unit cell is a face-centred cubic (fcc) cell. Atoms are at corners and face centers. Particles touch along the face diagonal.

Let the edge length of the unit cell be 'a' and the radius of each sphere be 'r'. The face diagonal (b) is related to the edge length by the Pythagorean theorem in a face-diagonal triangle (e.g., ABC in Fig 1.24): $b^2 = a^2 + a^2 = 2a^2$, so $b = a\sqrt{2}$.

Diagram of a cubic close-packed (fcc) unit cell showing face diagonal and its relation to edge length and sphere radius

Along the face diagonal, the spheres at the corners and the face center touch each other. The length of the face diagonal (b) is equal to $4r$ (r from one corner atom, 2r from the face-centered atom, r from the other corner atom): $b = r + 2r + r = 4r$.

So, $4r = a\sqrt{2}$, which gives the relationship between edge length and radius: $a = \frac{4r}{\sqrt{2}} = 2r\sqrt{2}$.

In a ccp/fcc unit cell, the effective number of atoms is $z = 4$.

Volume occupied by spheres in the unit cell = $z \times \text{Volume of one sphere} = 4 \times (\frac{4}{3}\pi r^3) = \frac{16}{3}\pi r^3$.

Volume of the unit cell = $a^3 = (2r\sqrt{2})^3 = (2\sqrt{2})^3 r^3 = 8 \times 2\sqrt{2} r^3 = 16\sqrt{2} r^3$.

Packing efficiency = $\frac{\textsf{Volume occupied by spheres}}{\textsf{Volume of unit cell}} \times 100\% = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2} r^3} \times 100\%$

Packing efficiency = $\frac{\frac{1}{3}\pi}{\sqrt{2}} \times 100\% = \frac{\pi}{3\sqrt{2}} \times 100\% \approx \frac{3.14159}{3 \times 1.41421} \times 100\% \approx \frac{3.14159}{4.24264} \times 100\% \approx 0.7404 \times 100\% \approx 74\%$.

Both hcp and ccp/fcc structures have a packing efficiency of 74%.


Efficiency Of Packing In Body-Centred Cubic Structures

In a body-centred cubic (bcc) structure, atoms are at corners and the body centre. Particles touch each other along the body diagonal.

Let the edge length be 'a' and radius 'r'. Consider a triangle along a face diagonal (e.g., EFD in Fig 1.25) with face diagonal $b = a\sqrt{2}$. Now consider a triangle along the body diagonal (e.g., AFD) with edge 'a', face diagonal 'b', and body diagonal 'c'. $c^2 = a^2 + b^2 = a^2 + (a\sqrt{2})^2 = a^2 + 2a^2 = 3a^2$, so $c = a\sqrt{3}$.

Diagram of a body-centred cubic (bcc) unit cell showing body diagonal and its relation to edge length and sphere radius

Along the body diagonal, the corner atom, the body-centered atom, and the opposite corner atom touch each other. The length of the body diagonal (c) is equal to $4r$: $c = r + 2r + r = 4r$.

So, $4r = a\sqrt{3}$, which gives the relationship between edge length and radius: $a = \frac{4r}{\sqrt{3}}$.

In a bcc unit cell, the effective number of atoms is $z = 2$.

Volume occupied by spheres in the unit cell = $z \times \text{Volume of one sphere} = 2 \times (\frac{4}{3}\pi r^3) = \frac{8}{3}\pi r^3$.

Volume of the unit cell = $a^3 = (\frac{4r}{\sqrt{3}})^3 = \frac{64r^3}{3\sqrt{3}}$.

Packing efficiency = $\frac{\textsf{Volume occupied by spheres}}{\textsf{Volume of unit cell}} \times 100\% = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100\%$

Packing efficiency = $\frac{8\pi/3}{64/(3\sqrt{3})} \times 100\% = \frac{8\pi}{64/\sqrt{3}} \times 100\% = \frac{\pi\sqrt{3}}{8} \times 100\% \approx \frac{3.14159 \times 1.73205}{8} \times 100\% \approx \frac{5.4413}{8} \times 100\% \approx 0.6802 \times 100\% \approx 68\%$.


Packing Efficiency In Simple Cubic Lattice

In a simple cubic lattice, atoms are only at the corners. Particles touch each other along the edges of the cube.

Diagram of a simple cubic unit cell showing atoms touching along the edge

Let the edge length be 'a' and radius 'r'. Since particles touch along the edge, the edge length is equal to twice the radius: $a = 2r$.

In a simple cubic unit cell, the effective number of atoms is $z = 1$.

Volume occupied by spheres in the unit cell = $z \times \text{Volume of one sphere} = 1 \times (\frac{4}{3}\pi r^3) = \frac{4}{3}\pi r^3$.

Volume of the unit cell = $a^3 = (2r)^3 = 8r^3$.

Packing efficiency = $\frac{\textsf{Volume occupied by spheres}}{\textsf{Volume of unit cell}} \times 100\% = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100\%$

Packing efficiency = $\frac{4\pi/3}{8} \times 100\% = \frac{\pi}{6} \times 100\% \approx \frac{3.14159}{6} \times 100\% \approx 0.5236 \times 100\% \approx 52.4\%$.

Conclusion: Close-packed structures (hcp and ccp) have the highest packing efficiency (74%), followed by bcc (68%), and simple cubic (52.4%) has the lowest packing efficiency among these cubic structures.



Calculations Involving Unit Cell Dimensions

The dimensions of a unit cell can be used to calculate properties like density or atomic/molar mass if other parameters are known. The density of a unit cell is the same as the density of the substance it represents.

Density (d) of a unit cell is given by:

$\textsf{d} = \frac{\textsf{Mass of the unit cell}}{\textsf{Volume of the unit cell}}$

Mass of the unit cell = Number of atoms in the unit cell (z) $\times$ Mass of a single atom (m).

The mass of a single atom (m) can be calculated from the molar mass (M) and Avogadro's number ($\textsf{N}_\textsf{A}$): $\textsf{m} = \frac{\textsf{M}}{\textsf{N}_\textsf{A}}$.

Volume of a cubic unit cell with edge length 'a' is $\textsf{a}^3$.

So, for a cubic crystal, the density formula is:

$\textsf{d} = \frac{\textsf{z} \times \textsf{m}}{\textsf{a}^3} = \frac{\textsf{z} \times \frac{\textsf{M}}{\textsf{N}_\textsf{A}}}{\textsf{a}^3} = \frac{\textsf{zM}}{\textsf{a}^3 \textsf{N}_\textsf{A}}$

This equation relates density (d), number of atoms per unit cell (z), molar mass (M), edge length (a), and Avogadro's number ($\textsf{N}_\textsf{A}$). If any four of these are known, the fifth can be calculated.

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/cm$^3$. How many atoms are present in 208 g of the element?

Answer:

For a bcc structure, the number of atoms per unit cell, z = 2.

Cell edge length, a = 288 pm $= 288 \times 10^{-12} \text{ m} = 288 \times 10^{-10} \text{ cm}$.

Density, d = 7.2 g/cm$^3$.

Volume of the unit cell = a$^3 = (288 \times 10^{-10} \text{ cm})^3 = (288)^3 \times 10^{-30} \text{ cm}^3 = 23887872 \times 10^{-30} \text{ cm}^3 \approx 2.39 \times 10^{-23} \text{ cm}^3$.

Volume of 208 g of the element = $\frac{\text{mass}}{\text{density}} = \frac{208 \text{ g}}{7.2 \text{ g/cm}^3} \approx 28.89 \text{ cm}^3$.

Number of unit cells in 208 g = $\frac{\text{Total volume}}{\text{Volume of one unit cell}} = \frac{28.89 \text{ cm}^3}{2.39 \times 10^{-23} \text{ cm}^3/\text{unit cell}} \approx 12.09 \times 10^{23}$ unit cells.

Since each bcc unit cell contains 2 atoms, the total number of atoms in 208 g = Number of unit cells $\times$ Atoms per unit cell

= $(12.09 \times 10^{23} \text{ unit cells}) \times (2 \text{ atoms/unit cell}) \approx 24.18 \times 10^{23} \text{ atoms}$.


Example 1.4 X-ray diffraction studies show that copper crystallises in an fcc unit cell with cell edge of 3.608$\times$10$^{-8}$ cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm$^3$, calculate the atomic mass of copper.

Answer:

For an fcc unit cell, the number of atoms per unit cell, z = 4.

Cell edge length, a = $3.608 \times 10^{-8}$ cm.

Density, d = 8.92 g/cm$^3$.

Avogadro's number, $\textsf{N}_\textsf{A} = 6.022 \times 10^{23} \text{ mol}^{-1}$.

We use the formula: $\textsf{d} = \frac{\textsf{zM}}{\textsf{a}^3 \textsf{N}_\textsf{A}}$. We need to calculate the molar mass M.

$\textsf{M} = \frac{\textsf{d} \times \textsf{a}^3 \times \textsf{N}_\textsf{A}}{\textsf{z}}$

$\textsf{M} = \frac{(8.92 \text{ g/cm}^3) \times (3.608 \times 10^{-8} \text{ cm})^3 \times (6.022 \times 10^{23} \text{ mol}^{-1})}{4}$

$\textsf{M} = \frac{8.92 \times (3.608)^3 \times 10^{-24} \times 6.022 \times 10^{23}}{4} \text{ g/mol}$

$\textsf{M} = \frac{8.92 \times 47.04 \times 10^{-1} \times 6.022}{4} \text{ g/mol}$

$\textsf{M} = \frac{8.92 \times 4.704 \times 6.022}{4} \text{ g/mol}$

$\textsf{M} = \frac{252.8}{4} \text{ g/mol} \approx 63.2 \text{ g/mol}$.

The atomic mass of copper is approximately 63.2 u.


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).

Answer:

Silver forms a ccp lattice, which is equivalent to an fcc unit cell. So, the number of atoms per unit cell, z = 4.

Atomic mass = 107.9 u. Molar mass, M = 107.9 g/mol $= 107.9 \times 10^{-3}$ kg/mol.

Edge length, a = 408.6 pm $= 408.6 \times 10^{-12}$ m $= 408.6 \times 10^{-10}$ cm.

Avogadro's number, $\textsf{N}_\textsf{A} = 6.022 \times 10^{23} \text{ mol}^{-1}$.

We need to calculate the density, d.

$\textsf{d} = \frac{\textsf{zM}}{\textsf{a}^3 \textsf{N}_\textsf{A}}$

Using units of g and cm:

$\textsf{d} = \frac{(4 \text{ atoms}) \times (107.9 \text{ g/mol})}{(408.6 \times 10^{-10} \text{ cm})^3 \times (6.022 \times 10^{23} \text{ mol}^{-1})}$

$\textsf{d} = \frac{4 \times 107.9}{(408.6)^3 \times 10^{-30} \times 6.022 \times 10^{23}} \text{ g/cm}^3$}

$\textsf{d} = \frac{431.6}{(408.6)^3 \times 6.022 \times 10^{-7}} \text{ g/cm}^3$}

$(408.6)^3 \approx 68350000 \approx 6.835 \times 10^7$

$\textsf{d} = \frac{431.6}{6.835 \times 10^7 \times 6.022 \times 10^{-7}} \text{ g/cm}^3$}

$\textsf{d} = \frac{431.6}{6.835 \times 6.022} \text{ g/cm}^3$}

$\textsf{d} = \frac{431.6}{41.16} \text{ g/cm}^3} \approx 10.48 \text{ g/cm}^3$.

Using units of kg and m:

$\textsf{d} = \frac{(4) \times (107.9 \times 10^{-3} \text{ kg/mol})}{(408.6 \times 10^{-12} \text{ m})^3 \times (6.022 \times 10^{23} \text{ mol}^{-1})}$

$\textsf{d} = \frac{431.6 \times 10^{-3}}{(408.6)^3 \times 10^{-36} \times 6.022 \times 10^{23}} \text{ kg/m}^3$}

$\textsf{d} = \frac{431.6 \times 10^{-3}}{6.835 \times 10^7 \times 10^{-36} \times 6.022 \times 10^{23}} \text{ kg/m}^3$}

$\textsf{d} = \frac{431.6 \times 10^{-3}}{6.835 \times 6.022 \times 10^{-6}} \text{ kg/m}^3$}

$\textsf{d} = \frac{431.6 \times 10^{3}}{41.16} \text{ kg/m}^3} \approx 10.48 \times 10^3 \text{ kg/m}^3 = 10500 \text{ kg/m}^3$.


Intext Questions

1.16 Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?

1.17 An element with molar mass 2.7$\times$10$^{-2}$ kg mol$^{-1}$ forms a cubic unit cell with edge length 405 pm. If its density is 2.7$\times$10$^3$ kg m$^{-3}$, what is the nature of the cubic unit cell?

Answer:

1.16 Comparing the packing efficiencies calculated:

  • Simple cubic (i): 52.4%
  • Body-centred cubic (ii): 68%
  • Hexagonal close-packed (hcp) lattice (iii): 74% (ccp/fcc also 74%)

The lattice with the highest packing efficiency is the hexagonal close-packed (hcp) lattice, which is also equivalent to face-centred cubic (ccp).

1.17 Molar mass, M = $2.7 \times 10^{-2}$ kg/mol.

Edge length, a = 405 pm $= 405 \times 10^{-12}$ m.

Density, d = $2.7 \times 10^3$ kg/m$^3$.

Avogadro's number, $\textsf{N}_\textsf{A} = 6.022 \times 10^{23} \text{ mol}^{-1}$.

We use the formula: $\textsf{d} = \frac{\textsf{zM}}{\textsf{a}^3 \textsf{N}_\textsf{A}}$. We need to find the value of z (number of atoms per unit cell) to determine the nature of the cubic unit cell.

$\textsf{z} = \frac{\textsf{d} \times \textsf{a}^3 \times \textsf{N}_\textsf{A}}{\textsf{M}}$

$\textsf{z} = \frac{(2.7 \times 10^3 \text{ kg/m}^3) \times (405 \times 10^{-12} \text{ m})^3 \times (6.022 \times 10^{23} \text{ mol}^{-1})}{2.7 \times 10^{-2} \text{ kg/mol}}$

$\textsf{z} = \frac{2.7 \times 10^3 \times (405)^3 \times 10^{-36} \times 6.022 \times 10^{23}}{2.7 \times 10^{-2}}$

$\textsf{z} = \frac{10^3 \times (405)^3 \times 10^{-36} \times 6.022 \times 10^{23}}{10^{-2}}$

$\textsf{z} = 10^3 \times (4.05 \times 10^2)^3 \times 10^{-36} \times 6.022 \times 10^{23} \times 10^2$

$\textsf{z} = 10^3 \times (4.05)^3 \times 10^6 \times 10^{-36} \times 6.022 \times 10^{23} \times 10^2$

$\textsf{z} = (4.05)^3 \times 6.022 \times 10^{(3+6-36+23+2)} = (4.05)^3 \times 6.022 \times 10^{-2}$}

$(405)^3 \approx 66430000 \approx 6.643 \times 10^7$

$\textsf{z} = \frac{2.7 \times 10^3 \times 6.643 \times 10^7 \times 10^{-36} \times 6.022 \times 10^{23}}{2.7 \times 10^{-2}}$

$\textsf{z} = \frac{10^3 \times 6.643 \times 10^7 \times 10^{-36} \times 6.022 \times 10^{23}}{10^{-2}}$

$\textsf{z} = 6.643 \times 6.022 \times 10^{(3+7-36+23-(-2))} = 39.99 \times 10^{-1}$}

Using original edge length: $(405 \times 10^{-12})^3 = (4.05 \times 10^{-10})^3 = (4.05)^3 \times 10^{-30} \approx 66.43 \times 10^{-30}$.

$\textsf{z} = \frac{2.7 \times 10^3 \times 66.43 \times 10^{-30} \times 6.022 \times 10^{23}}{2.7 \times 10^{-2}}$

$\textsf{z} = \frac{10^3 \times 66.43 \times 10^{-30} \times 6.022 \times 10^{23}}{10^{-2}}$

$\textsf{z} = 66.43 \times 6.022 \times 10^{(3-30+23+2)} = 399.9 \times 10^{-2} \approx 3.99 \approx 4$.

Since the number of atoms per unit cell (z) is approximately 4, the nature of the cubic unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).



Imperfections In Solids

Even though crystalline solids have long-range order, they are not perfectly ordered. Real crystals contain irregularities in the arrangement of constituent particles, called crystal defects or imperfections. Defects are more common when crystallisation occurs rapidly or at moderate rates. Even slow crystallisation yields crystals with some defects.

Defects are classified into two main types:

We will focus on point defects.


Types Of Point Defects

Point defects are categorised into three types:

  1. Stoichiometric defects: These point defects do not change the overall stoichiometry of the solid. They are also called intrinsic or thermodynamic defects.
  2. Impurity defects: Arise due to the presence of foreign atoms or ions in the crystal lattice.
  3. Non-stoichiometric defects: Disturb the stoichiometry of the solid, resulting in a non-stoichiometric composition.

(a) Stoichiometric Defects

These are defects where the ratio of positive and negative ions (or atoms) remains the same as in the ideal stoichiometric formula. Two basic types:

Vacancy and interstitial defects, as described above, are typically observed in non-ionic solids. In ionic solids, defects must maintain electrical neutrality. Instead of simple vacancy or interstitial defects, ionic solids show these defects in slightly different forms: Frenkel and Schottky defects.

(b) Impurity Defects

Impurity defects arise when foreign atoms or ions are present in the crystal lattice. In ionic solids, if the impurity ions have a different charge (valence) than the host ions, vacancies may be created to maintain electrical neutrality.

Example: Crystallising molten $\textsf{NaCl}$ containing a small amount of $\textsf{SrCl}_2$. $\textsf{Sr}^{2+}$ ions (from $\textsf{SrCl}_2$) have a +2 charge, while $\textsf{Na}^{+}$ ions have a +1 charge. To maintain neutrality, one $\textsf{Sr}^{2+}$ ion replaces two $\textsf{Na}^{+}$ ions. One $\textsf{Na}^{+}$ site is occupied by $\textsf{Sr}^{2+}$, and the other $\textsf{Na}^{+}$ site remains vacant. This creates cation vacancies equal in number to the $\textsf{Sr}^{2+}$ ions introduced.

Diagram showing cation vacancy created in NaCl crystal by doping with SrCl2

Similar impurity defects are observed in solid solutions of $\textsf{CdCl}_2$ and $\textsf{AgCl}$ (where $\textsf{Cd}^{2+}$ replaces $\textsf{Ag}^{+}$ ions, creating $\textsf{Ag}^{+}$ vacancies).

(c) Non-Stoichiometric Defects

These defects disturb the ideal stoichiometric ratio of elements in a crystal. They result in solids with a non-stoichiometric composition (e.g., $\textsf{Fe}_{0.95}\textsf{O}$). Two types:

Intext Questions

1.18 What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

1.19 What type of stoichiometric defect is shown by: (i) ZnS (ii) AgBr

1.20 Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

1.21 Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.

Answer:

1.18 When a solid (specifically a non-ionic solid) is heated, a vacancy defect can arise. Increased thermal energy causes some particles to move out of their lattice sites, leaving them vacant. This physical property affected is the density of the solid. The density decreases because the number of particles per unit volume decreases while the volume remains essentially unchanged.

1.19 Stoichiometric defects shown by:

(i) $\textsf{ZnS}$ shows Frenkel defect. $\textsf{Zn}^{2+}$ ions are much smaller than $\textsf{S}^{2-}$ ions, allowing the smaller $\textsf{Zn}^{2+}$ cations to be dislocated from their lattice sites to interstitial sites.

(ii) $\textsf{AgBr}$ is interesting as it shows both Frenkel and Schottky defects. Frenkel defect occurs due to the small size of $\textsf{Ag}^{+}$ ions. Schottky defect occurs because the $\textsf{Ag}^{+}$ and $\textsf{Br}^{-}$ ions are of relatively similar size.

1.20 When an ionic solid containing cations of a certain valence is doped with a cation of higher valence, cation vacancies are created to maintain electrical neutrality. For example, consider doping molten $\textsf{NaCl}$ (where cations are $\textsf{Na}^{+}$, valence +1) with a small amount of $\textsf{SrCl}_2$ (where cations are $\textsf{Sr}^{2+}$, valence +2). When crystallisation occurs, some $\textsf{Na}^{+}$ lattice sites are occupied by $\textsf{Sr}^{2+}$ ions. Each $\textsf{Sr}^{2+}$ ion carries a +2 charge, replacing two $\textsf{Na}^{+}$ ions, each carrying a +1 charge. To maintain the overall positive charge equal to the negative charge from anions, if one $\textsf{Sr}^{2+}$ occupies one $\textsf{Na}^{+}$ site, another $\textsf{Na}^{+}$ site must remain vacant. So, for every $\textsf{Sr}^{2+}$ ion introduced, one cation vacancy is created. The number of cation vacancies is equal to the number of $\textsf{Sr}^{2+}$ ions present as impurity.

1.21 Ionic solids with anionic vacancies due to metal excess defects develop colour because electrons occupy these anionic vacancies (forming F-centres). These electrons can absorb energy from visible light and get excited to higher energy levels. When they return to their ground state, they emit light of specific wavelengths, which corresponds to the color observed.

Example: When crystals of $\textsf{NaCl}$ are heated in an atmosphere of sodium vapor, $\textsf{Na}^{+}$ ions occupy surface sites, and $\textsf{Cl}^{-}$ ions diffuse to the surface and combine with $\textsf{Na}$ atoms ($\textsf{Na} + \textsf{Cl}^{-} \to \textsf{NaCl} + \textsf{e}^{-}$). The released electrons diffuse into the crystal and occupy the anionic vacancies left by the $\textsf{Cl}^{-}$ ions. These occupied sites are F-centres, and they cause the $\textsf{NaCl}$ crystals to appear yellow.



Electrical Properties

Solids exhibit a vast range of electrical conductivities. Based on conductivity, solids are classified into three types:

  1. Conductors: High conductivity ($10^4$ to $10^7 \text{ ohm}^{-1}\text{m}^{-1}$). Metals are good conductors ($\sim 10^7 \text{ ohm}^{-1}\text{m}^{-1}$).
  2. Insulators: Very low conductivity ($10^{-20}$ to $10^{-10} \text{ ohm}^{-1}\text{m}^{-1}$).
  3. Semiconductors: Intermediate conductivity ($10^{-6}$ to $10^4 \text{ ohm}^{-1}\text{m}^{-1}$).
  4. Electrical conductivity in solids can be due to the movement of electrons (metallic conductors) or ions (electrolytes).


Conduction Of Electricity In Metals

Metals conduct electricity in both solid and molten states. Their conductivity is explained by band theory, based on the formation of molecular orbitals from atomic orbitals of metal atoms. These molecular orbitals are very close in energy, forming continuous energy bands.

In metals, the valence band is either partially filled or overlaps with the conduction band. Electrons can easily move into the empty energy levels within the partially filled band or readily jump to the overlapping conduction band when an electric field is applied. This free movement of electrons results in high conductivity.

Band diagram for a conductor showing a partially filled band or overlapping valence and conduction bands

In insulators, there is a large energy gap (called the forbidden gap) between the filled valence band and the unoccupied conduction band. Electrons require significant energy to jump across this gap, so conductivity is very low.

Band diagram for an insulator showing a large forbidden gap between valence and conduction bands

Conduction Of Electricity In Semiconductors

In semiconductors, the energy gap between the valence band and the conduction band is smaller than in insulators. At room temperature, some electrons gain enough thermal energy to jump from the valence band to the conduction band, allowing for some conductivity. Conductivity increases with increasing temperature as more electrons can jump to the conduction band.

Band diagram for a semiconductor showing a small forbidden gap between valence and conduction bands

Pure semiconductors like silicon ($\textsf{Si}$) and germanium ($\textsf{Ge}$) are called intrinsic semiconductors. Their conductivity is too low for practical use.

The conductivity of intrinsic semiconductors can be significantly increased by adding a suitable impurity, a process called doping. Doping introduces electronic defects and creates extrinsic semiconductors.

Types of doping based on impurity valence:

  1. Doping with electron-rich impurities (Group 15 elements): Elements from Group 14 ($\textsf{Si}$, $\textsf{Ge}$) have 4 valence electrons. When doped with a Group 15 element (P, As) which has 5 valence electrons, the impurity atom replaces a host atom in the crystal lattice. Four of the impurity's valence electrons form covalent bonds with neighbors, but the fifth electron is extra and becomes delocalised. These free electrons increase conductivity. Since conductivity is due to negatively charged electrons, the semiconductor is called an n-type semiconductor ('n' for negative).
  2. Diagram showing doping of a Group 14 semiconductor (Silicon) with a Group 15 impurity (Phosphorus), creating an extra electron
  3. Doping with electron-deficit impurities (Group 13 elements): When doped with a Group 13 element (B, Al, Ga) which has only 3 valence electrons, the impurity atom replaces a host atom. The three valence electrons form covalent bonds with neighbors, but there is a missing electron in the fourth bond, creating an electron hole or electron vacancy. An electron from a neighboring atom can fill this hole, leaving a hole at its original position. Under an electric field, electrons move towards the positive terminal by filling holes, making it appear as if the positive holes are moving towards the negative terminal. Since conductivity is due to these apparent positive holes, the semiconductor is called a p-type semiconductor ('p' for positive).
  4. Diagram showing doping of a Group 14 semiconductor (Silicon) with a Group 13 impurity (Boron), creating an electron hole

Applications of n-type and p-type semiconductors:

Other semiconductors are formed by combining elements from different groups to achieve an average valence of four, similar to Group 14 elements. Examples: Group 13-15 compounds (InSb, AlP, GaAs - Gallium arsenide is fast response), Group 12-16 compounds (ZnS, CdS, CdSe, HgTe - ionic character varies).

Transition metal oxides also show varying electrical properties (metallic or insulating) depending on the material and temperature (e.g., $\textsf{TiO}$, $\textsf{CrO}_2$, $\textsf{ReO}_3$ are metallic; $\textsf{VO}$, $\textsf{VO}_2$, $\textsf{VO}_3$, $\textsf{TiO}_3$ show temperature-dependent behavior).



Magnetic Properties

Every substance has magnetic properties originating from the electrons within its atoms. Each electron acts as a tiny magnet due to its motion:

Diagram showing the magnetic moments associated with electron orbital motion and spin motion

Each electron has a permanent spin and orbital magnetic moment. The magnitude is measured in Bohr magneton ($\mu_\textsf{B}$), equal to $9.27 \times 10^{-24} \text{ A m}^2$. Magnetic properties of substances arise from the net magnetic moment of the atoms, which depends on the number of unpaired electrons and how their magnetic moments are oriented.

Based on magnetic behavior in an external magnetic field, substances are classified into five types:


Paramagnetism

Paramagnetic substances are weakly attracted by a magnetic field. They become weakly magnetised in the same direction as the applied field but lose magnetism when the field is removed. This property is due to the presence of one or more unpaired electrons, whose magnetic moments align with the external field. Examples: $\textsf{O}_2$, $\textsf{Cu}^{2+}$, $\textsf{Fe}^{3+}$, $\textsf{Cr}^{3+}$.


Diamagnetism

Diamagnetic substances are weakly repelled by a magnetic field. They become weakly magnetised in the opposite direction to the applied field. This property is shown by substances where all electrons are paired. Paired electrons have opposite spins, and their magnetic moments cancel each other out, resulting in no net magnetic moment. Examples: $\textsf{H}_2\textsf{O}$, $\textsf{NaCl}$, $\textsf{C}_6\textsf{H}_6$, $\textsf{Ti}^{4+}$ (e.g., in $\textsf{TiO}_2$).


Ferromagnetism

Ferromagnetic substances are very strongly attracted by a magnetic field. They can become permanently magnetised even after the external field is removed. In solid state, metal ions of ferromagnetic substances are grouped into small regions called domains, each acting as a tiny magnet with aligned magnetic moments. In an unmagnetised state, these domains are randomly oriented, canceling out their magnetic moments. When a magnetic field is applied, domains align in the direction of the field, producing a strong net magnetic moment. This alignment persists after the field is removed. Examples: $\textsf{Fe}$, $\textsf{Co}$, $\textsf{Ni}$, $\textsf{Gd}$, $\textsf{CrO}_2$. Ferromagnetism is the basis for permanent magnets.

Schematic alignment of magnetic moments in ferromagnetic materials in the presence of an external magnetic field

Antiferromagnetism

Antiferromagnetic substances have a domain structure similar to ferromagnetic substances, but their domains are aligned in opposite directions in equal numbers. This results in the cancellation of their magnetic moments, giving a net magnetic moment of zero. These substances do not show magnetism in an external field. Example: $\textsf{MnO}$.

Schematic alignment of magnetic moments in antiferromagnetic materials, showing opposite and equal alignment

Ferrimagnetism

Ferrimagnetic substances have domains with magnetic moments aligned in parallel and anti-parallel directions in unequal numbers. The anti-parallel moments are not equal and do not completely cancel, resulting in a net magnetic moment, but it is smaller than that of ferromagnetic substances. They are weakly attracted by a magnetic field compared to ferromagnetic substances. Examples: $\textsf{Fe}_3\textsf{O}_4$ (magnetite), ferrites like $\textsf{MgFe}_2\textsf{O}_4$, $\textsf{ZnFe}_2\textsf{O}_4$. Ferrimagnetic substances become paramagnetic on heating as the domain alignment is lost above a certain temperature.

Schematic alignment of magnetic moments in ferrimagnetic materials, showing parallel and anti-parallel alignment in unequal numbers

Intext Questions

1.22 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?

1.23 What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

Answer:

1.22 A group 14 element (like Silicon or Germanium) has 4 valence electrons. To make an n-type semiconductor, we need to introduce electron-rich impurities, meaning the impurity atom should have more valence electrons than the group 14 element. Therefore, the impurity should belong to Group 15 of the periodic table (like Phosphorus or Arsenic), which has 5 valence electrons. The extra electron from the group 15 impurity contributes to conductivity.

1.23 Ferromagnetic substances would make better permanent magnets. Justification: Both ferromagnetic and ferrimagnetic substances show net magnetism in an external field. However, in ferromagnetic substances, the magnetic moments of the domains are all aligned in the same direction in the presence of a field, resulting in a much stronger magnetic effect. Crucially, this ordering and strong magnetic moment persist even after the external magnetic field is removed, allowing them to become permanently magnetised. Ferrimagnetic substances have anti-parallel alignment in unequal numbers, leading to a smaller net magnetic moment, and their magnetism is generally weaker and may not persist as strongly upon removal of the field compared to ferromagnetics.



Intext Questions



Question 1.1. Why are solids rigid?

Answer:

Question 1.2. Why do solids have a definite volume?

Answer:

Question 1.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

Answer:

Question 1.4. Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?

Answer:

Question 1.5. 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.

Answer:

Question 1.6. 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?

Answer:

Question 1.7. Ionic solids conduct electricity in molten state but not in solid state. Explain.

Answer:

Question 1.8. What type of solids are electrical conductors, malleable and ductile?

Answer:

Question 1.9. Give the significance of a ‘lattice point’.

Answer:

Question 1.10. Name the parameters that characterise a unit cell.

Answer:

Question 1.11. Distinguish between

(i) Hexagonal and monoclinic unit cells

(ii) Face-centred and end-centred unit cells.

Answer:

Question 1.12. Explain how much portion of an atom located at (i) corner and (ii) bodycentre of a cubic unit cell is part of its neighbouring unit cell.

Answer:

Question 1.13. What is the two dimensional coordination number of a molecule in square close-packed layer?

Answer:

Question 1.14. A compound forms hexagonal close-packed structure. What is the total number of voids in $0.5 \text{ mol}$ of it? How many of these are tetrahedral voids?

Answer:

Question 1.15. A compound is formed by two elements $M$ and $N$. The element $N$ forms ccp and atoms of $M$ occupy $1/3 \text{rd}$ of tetrahedral voids. What is the formula of the compound?

Answer:

Question 1.16. Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?

Answer:

Question 1.17. An element with molar mass $2.7 \times 10^{-2} \text{ kg mol}^{-1}$ forms a cubic unit cell with edge length $405 \text{ pm}$. If its density is $2.7 \times 10^3 \text{ kg m}^{-3}$, what is the nature of the cubic unit cell?

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Question 1.18. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

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Question 1.19. What type of stoichiometric defect is shown by:

(i) $ZnS$

(ii) $AgBr$

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Question 1.20. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

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Question 1.21. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.

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Question 1.22. 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?

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Question 1.23. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

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Exercises



Question 1.1. Define the term 'amorphous'. Give a few examples of amorphous solids.

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

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Question 1.3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

(i) Tetra phosphorus decoxide ($P_4O_{10}$)

(ii) Ammonium phosphate ($(NH_4)_3PO_4$)

(iii) $SiC$

(iv) $I_2$

(v) $P_4$

(vi) Plastic

(vii) Graphite

(viii) Brass

(ix) $Rb$

(x) $LiBr$

(xi) $Si$

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Question 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?

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Question 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.

The formula to determine the atomic mass ($M$) of an unknown metal using its density ($d$) and unit cell dimensions is given by:

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

where:

$d$ is the density of the metal

$a$ is the edge length of the unit cell

$N_A$ is Avogadro's number ($6.022 \times 10^{23} \text{ mol}^{-1}$)

$z$ is the number of atoms per unit cell (for simple cubic $z=1$, for bcc $z=2$, for fcc $z=4$)

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Question 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?

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Question 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?

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Question 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

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Question 1.9. Explain

(i) The basis of similarities and differences between metallic and ionic crystals.

(ii) Ionic solids are hard and brittle.

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Question 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).

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Question 1.11. Silver crystallises in fcc lattice. If edge length of the cell is $4.07 \times 10^{-8} \text{ cm}$ and density is $10.5 \text{ g cm}^{-3}$, calculate the atomic mass of silver.

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Question 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 $Q$?

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Question 1.13. Niobium crystallises in body-centred cubic structure. If density is $8.55 \text{ g cm}^{-3}$, calculate atomic radius of niobium using its atomic mass $93 \text{ u}$.

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Question 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 $R$.

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Question 1.15. Copper crystallises into a fcc lattice with edge length $3.61 \times 10^{-8} \text{ cm}$. Show that the calculated density is in agreement with its measured value of $8.92 \text{ g cm}^{-3}$.

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Question 1.16. Analysis shows that nickel oxide has the formula $Ni_{0.98}O_{1.00}$. What fractions of nickel exist as $Ni^{2+}$ and $Ni^{3+}$ ions?

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

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Question 1.18. Non-stoichiometric cuprous oxide, $Cu_2O$ 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?

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Question 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.

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Question 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 $B$.

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Question 1.21. Gold (atomic radius = $0.144 \text{ nm}$) crystallises in a face-centred unit cell. What is the length of a side of the cell?

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Question 1.22. In terms of band theory, what is the difference

(i) between a conductor and an insulator

(ii) between a conductor and a semiconductor?

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Question 1.23. Explain the following terms with suitable examples:

(i) Schottky defect

(ii) Frenkel defect

(iii) Interstitials

(iv) F-centres.

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Question 1.24. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is $125 \text{ pm}$.

(i) What is the length of the side of the unit cell?

(ii) How many unit cells are there in $1.00 \text{ cm}^3$ of aluminium?

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Question 1.25. If $NaCl$ is doped with $10^{-3} \text{ mol %}$ of $SrCl_2$, what is the concentration of cation vacancies?

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Question 1.26. Explain the following with suitable examples:

(i) Ferromagnetism

(ii) Paramagnetism

(iii) Ferrimagnetism

(iv) Antiferromagnetism

(v) $12-16$ and $13-15$ group compounds.

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