Solid State JEE Notes

A solid state is a state of matter besides a liquid and gaseous state. In general, solids are substances that are rigid and have a fixed shape and size. The intermolecular forces are very strong, and the empty space between the atoms or molecules is very small. The study of solids is called solid-state chemistry.

Table of Contents

• Cause of Existence of Solid State
• Types of Solids
• Crystal Lattices and Unit Cells
• Packing Fraction
• Coordination Number
• Radius Ratio
• Types of Voids
• Types of Ionic Structures
• Crystal Defects
• Solved Examples
• Practice Problems
• Frequently Asked Questions

Cause of Existence of Solid State

The following are the reasons for the existence of the solid state:

• Intermolecular Forces
• Thermal Energy

Intermolecular Forces

The force of attraction between the particles in a solid is called intermolecular force. This force holds all the particles together and makes the existence of solid states possible. The intermolecular force depends on the following interactions:

1. Dipole-dipole interactions: It is the attractive force between polar molecules.
2. Ion-dipole interactions: This type of interaction is between ions and polar molecules.
3. Ion-induced dipole interactions: A nonpolar molecule, upon obtaining a charge, behaves as induced dipoles.
4. Dipole-induced dipole interactions: A non-polar molecule is polarised by an ion placed near it.
5. Dispersion forces or London forces: This kind of force arises due to the movement of electrons, thus creating a temporary positive and negative charged region.

Thermal Energy

The energy between the particles in a solid is called thermal energy. It is also called kinetic energy, and it is due to the movement of the particles inside the solid lattice. When the temperature is increased, there will be an increase in the thermal energy.

Characteristics of solids

• High density
• Rigidity
• Low compressibility
• Definite shape and volume

Types of Solids

Solids are classified on the basis of the following parameters:

1. Based on the different properties
2. Based on the bonding present in the building blocks

Based on the different properties, solids are generally classified into the following categories:

• Crystalline Solids
• Amorphous Solids

Crystalline Solids

Crystalline solids are those whose atoms, molecules or ions have an ordered arrangement extending over a long range. For example, Rock salt, NaCl.

Amorphous Solids

Amorphous solids are those whose constituent particles are randomly arranged and have an ordered long-range structure. For example, rubber, glass, etc.

Based on the bonding present in the building blocks

Application of Solid State

• Salt is an example of crystalline solids. It is used in cooking
• Glass is an example of an amorphous solid. It is widely used in window panes etc.
• Gels and polymers are used in the proper working of electronic and mechanical equipment
• It is used in the medical industry

Crystal Lattices and Unit Cells

The crystal lattice is the symmetrical three-dimensional structural arrangement of atoms, ions or molecules inside crystalline solids as points.

Characteristics of Crystal Lattice

• In the crystal lattice, each atom, molecule or ion is represented by a single point. These points are called the lattice points or lattice sites
• Straight lines are used to join the lattice points
• When the lattice points are joined with the straight lines, we get a 3D view of the structure. This 3D view is called the crystal lattice or the Bravais lattice

Unit Cell

The unit cell is the simplest repeating unit in the crystal lattice. A unit cell can be either primitive cubic (simple cubic), body-centred cubic (BCC), face-centred cubic (FCC) unit cell or end-centred unit cell.

Primitive or simple cubic unit Cell

In the primitive cubic cell, the atoms are present only at the corners.

Body-centred unit cell

When in addition to the particles present at the corner, one particle is also present at the centre of the unit cell. It is called the body-centred unit cell.

Face-centred unit cell

When the particles are present at the corners and at the centre of each cell, it is called the face-centred unit cell.

End-centred unit cell

The particles are present at the centres of opposite faces in addition to the particles at the corners.

Density of a unit cell

The density of a unit cell is given by the ratio of mass and volume of the unit cell. The mass of a unit cell is equal to the product of the number of atoms in a unit cell and the mass of each atom in a unit cell.

Mass of unit cell = number of atoms in unit cell × mass of each atom = z × m

Where, z = number of atoms in the unit cell

m = Mass of each atom

The mass of the atom is given in terms of the Avogadro number and molar mass = M/N_A

Here, M = molar mass

N_A is Avogadro’s number

The volume of the unit, V = a³

The density of the unit cell = mass of unit cell/volume of the unit cell.

Packing Fraction

Packing faction or packing efficiency is the ratio of the volume of the unit cell that is occupied by spheres of the unit cell to the volume of the unit cell.

Packing cell = [Volume occupied by all the spheres in the unit cell/Total volume of the unit cell] x 100

Packing Efficiency in Face-centred Cubic (Hexagonal close packing and cubic close packing)

Hexagonal close packing (hcp) and cubic close packing (ccp) have the same packing efficiency. Let “r” be the radius of the sphere and “a” be the edge length. The face diagonal “b” can be calculated with the help of the Pythagoras theorem.

The volume of the four spheres = 4 x (4/3 πr²)

In fcc, the face-centred spheres are in touch with the corner spheres. Therefore, the face diagonal AC is equal to four times the radius of the sphere.

AC = 4r

From the right-angled triangle ACD

4r = √2a

Or a = (4/√2) r

a = 2√2 r

The volume of the cube = (2√2 r)³

Perfect efficiency = [Volume occupied by four spheres in the unit cell/Total volume of the unit cell] x 100

= [4 x (4/3) πr³ x 100]/[16√2r³]

= 74 %

Packing Efficiency in a Body-centred Cubic Structure

In a body-centred cubic unit cell, one atom is located at the body centre, and the remaining atoms are at the corners of the cube. Let us take a unit cell of edge length “a”. The length of the body diagonal “c” can be calculated with the help of the Pythagoras theorem.

c² = a² + b²

The length of the face diagonal, b = √2 a

⇒ c² = 3a²

⇒ c = √3 a

The radius of the sphere, r = length of the body diagonal/ 4

⇒ r = c/4 = (√3/4)a

⇒ a = (4/√3) r

In body-centred cubic structures, each unit cell has two atoms.

Perfect efficiency = [Volume occupied by four spheres in the unit cell/ Total volume of the unit cell] x 100

= [2 x (4/3) πr³ x 100]/[4/√3) r]³

= 68 %

Packing Efficiency in Simple Cubic Unit Cell

In a simple cubic unit cell, atoms are located at the corners of the cube. Let us take a unit cell of edge length “a”. Radius of an atom can be given as,

r = a/2

⇒ a = 2r

In simple cubic structures, each unit cell has only one atom.

Perfect efficiency = [Volume occupied by four spheres in the unit cell/ Total volume of the unit cell] x 100

= [(4/3) πr³ x 100]/[2r]³

= 52.4 %

Types of Packing

Closed packing of atoms: This is the most efficient way of packing. 74% of available space is occupied by spheres, and the coordination number is 12.

(i) Hexagonal closed packing A-B-A-B): Each layer has a hexagonal arrangement of touching spheres, and the third layer is similar to the first layer.

(ii) Cubic closed packing (A-B-C-A-B-C): AB layers are similar to the HCP arrangement, but the third layer is offset from both A and B layers. The fourth layer is exactly on top of the layer.

Coordination Number

Coordination number, also known as ligancy, is the number of atoms, ions, or molecules that a central atom or ion carries in a complex or coordination compound or in a crystal as its closest neighbours.

Radius Ratio

The radius ratio refers to the ratio of the smaller ionic radius (cation) to the larger ionic radius (anion).

Significance of Radius Ratio

• The structure of the ionic solids can be predicted using the radius ratio.
• Stoichiometry and the size of the ions determine the structure of the ionic compound.
• In the crystals, the cations are surrounded by the largest possible number of anions around them.
• If the radius ratio is high, then the coordination number of the cations and anions is also high.
• If the size of the cation is very small and the size of the anions is very large, then the value of the radius radio will be small. The packing between the anions, in this case, will be very close to each other, and the system becomes unstable. Therefore, the structure will change to some suitable stable arrangement.
• The radius ratio at which the anions will just touch each other and the central cation is called the critical radius ratio.

Coordination Numbers and Radius Ratio

If the radius ratio is exactly equal to 0.414, then a cation would fit exactly into the octahedral void. Similarly, if the radius ratio is exactly equal to 0.225, then the cation would fit exactly into a tetrahedral void.

The radius ratio will be exactly equal to 0.414 if the cation fits exactly into an octahedral void of close-pack anions and has a coordination number of 6. When the radius ratio is greater than this, the anions will move away from each other to accommodate more cations. This arrangement is relatively unstable. If the radius ratio is further increased, the anions will move further away, and the cations will move to an octahedral void with a coordination number 8. This happens when the radius ratio is greater than 0.732.

If the radius ratio is below 0.414, the six anions will not be able to touch the cation. This is an unstable state, so the cations will move to the smaller tetrahedral void with coordination number 4.

Types of Voids

Voids in solid states mean the vacant space between the constituent particles in a closed-packed structure.

1) Tetrahedral voids: In a cubic closed-packed structure, spheres of the second layer lie above the triangular voids of the first layer. Each sphere touches the three spheres of the first layer. By joining the centre of these four spheres, it forms a tetrahedron, and the empty space left over by joining the centre of these spheres forms a tetrahedral void.

In a closed-packed structure, the number of tetrahedral voids is two times the number of spheres.

Let the number of spheres be n.

Then, the number of tetrahedral voids will be 2n.

2) Octahedral voids: Adjacent to tetrahedral voids, you can find octahedral voids. When the triangular voids of the first layer coincide with the triangular voids of the layer above or below it, we get a void that is formed by enclosing six spheres. This vacant space formed by combining the triangular voids of the first layer and that of the second layer is called octahedral voids.

If the number of spheres in a close-packed structure is n, then the number of octahedral voids will be n.

Types of Ionic Structures

Ionic compounds are formed by the simultaneous arrangement of the cations and anions in the unit cell. The larger of the two species will occupy the major positions, and the smaller one will occupy the voids. This is decided on the basis of the radius ratio.

Rock Salt Structure (NaCl): Cl^– will occupy the corners and the face centres, and Na⁺ will occupy the octahedral voids in FCC crystal. The coordination number of Na⁺ and Cl^– is 6.

Zinc Blende (Sphalerite) Structure (ZnS): S^2- ions will occupy the main positions, and Zn⁺² are present in the alternate tetrahedral voids in FCC crystal. The coordination number of S^2- and Zn⁺² is 4.

Fluorite Structure (CaF₂): Ca⁺² ions will occupy the main positions, and F^– ions will occupy the tetrahedral voids in the FCC structure. The coordination number of Ca⁺² is 8, and that of F^– is 4.

Anti-fluorite Structure: Common in alkali oxides like Na₂O, Li₂O etc., O^2- ions will occupy the FCC, and the Li⁺ ions will occupy the tetrahedral voids. The coordination number of Li+ is 4, and that of O^2- is 8.

Cesium Halide Structure(CsCl): Cl^– will be at the corners and Cs^– in the centre.

Corundum Structure(Al₂O₃): O^2- forming hcp and Al^3- filling ⅔ octahedral voids.

Rutile Structure (TiO₂): O^2- forming hcp, while Ti^4- ions occupy half of the octahedral voids.

Perovskite Structure (CaTiO₃): Ca²⁺ in the corner of the cube, O^2- at the face centre and Ti⁴⁺ at the centre of the cube.

Crystal Defects

When the temperature is increased, defects are created, which are classified into three types.

• Stoichiometric defects
• Non-stoichiometric defects
• Impurity defects

Stoichiometric Defects

The defect in which the stoichiometry of the compound remains the same, as shown by their chemical formula.

(i) Schottky Defect

This defect is caused by the ions missing from lattice points in pairs. It is usually seen in ionic solids with high coordination numbers like NaCl, KCl, and KBr.

(ii) Frenkel Defect

It is caused due to the creation of lattice vacancy as a result of misplaced ions in interstitial sites. It is common in solids with low coordination numbers like ZnS, and AgBr.

Non-stoichiometric Defects

This defect is shown by the d-block elements of the periodic table. These defects are classified as metal excess defects and metal deficiency defects.

(i) Metal excess defect due to anionic vacancy

A negative ion is replaced by an electron

(ii) Metal excess defect due to the presence of interstitial cation

The extra metal ion is present in the lattice, and electrons are also present in the interstitial site.

(iii) Metal deficiency defect

In this type of defect, some of the cations are missing from the lattice site, and to maintain its electrical neutrality, the remaining cations increase their valency.

Impurity Defects

In the molten state, NaCl contains a small amount of SrCl₂ in the form of impurity in the crystal. Due to the presence of SrCl_2, some sites of Na⁺ ions are occupied by Sr²⁺. To maintain the neutrality of the crystal, one Sr²⁺ replaces two ions of Na^+, and the remaining site of Na⁺ will be vacant. For example, CdCl₂ and AgCl.

Points to Remember

• Solids are substances that are rigid and have a fixed shape and size.
• Solids have high density, rigidity, low compressibility and a definite shape and volume.
• Solids are classified based on the different properties and on the bonding present in the building blocks.
• Crystalline solids are those whose atoms, molecules or ions have an ordered arrangement extending over a long range.
• Amorphous solids are those whose constituent particles are randomly arranged and have an ordered long-range structure.
• Based on the bonding present in the building blocks, the solids are classified into ionic, molecular, covalent and metallic.
• The crystal lattice is the symmetrical three-dimensional structural arrangement of atoms, ions or molecules inside crystalline solids as points.
• Unit Cell is the simplest repeating unit in the crystal lattice.
• Packing faction or packing efficiency is the ratio of the volume of the unit cell that is occupied by spheres of the unit cell to the volume of the unit cell.
• The coordination number is the number of atoms, ions, or molecules that a central atom or ion carries in a complex or coordination compound or in a crystal as its closest neighbours.
• Voids in solid states mean the vacant space between the constituent particles in a closed-packed structure.
• The radius ratio refers to the ratio of the smaller ionic radius (cation) to the larger ionic radius (anion).
• When the temperature is increased, stoichiometric defects, non-stoichiometric defects or impurity defects are created.

Solved Examples

1) A solid compound contains X, Y and Z atoms in a cubic lattice with X atoms occupying the corners. Y atoms in the body-centred positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound?

1) XY₂Z₃

2) XYZ₃

3) X₂Y₂Z₃

4) X₈YZ₆

5) XYZ

Answer: 2) XYZ₃

Solution:

At the corner, the number of atoms X in the unit cell = (⅛) x 8 = 1

At the body centres positions, the number of atoms Y in the unit cell = 1

At the centres of faces of the unit cell, the number of atoms of Z in the unit cell = (½) x 6 = 3

Therefore, the empirical formula of the compound = XYZ_3.

2) Structure of a mixed oxide is cubic close-packed (c.c.p). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A, and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is :

1) ABO₂

2) A₂BO₂

3) A₂B3O₄

4) AB₂O₂

Answer: 4) AB₂O₂

Solution:

We know, in cubic close-packed,

If the number of atoms = n

The number of octahedral voids = n

The number of tetrahedral voids = 2n

Number of A⁺² = 2n/4

Number of B⁺ = n

Number of O^-2 = n

Ratio A⁺²: B⁺:O^2-

n/2 : n: n

½ : 1: 1

1: 2: 2

3) Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of the lithium will be:

1) 151.8 pm

2) 75.5 pm

3) 300.5 pm

4) 240.8 pm

Answer: 1) 151.8 pm

Solution:

Since lithium metal crystallises in a body-centred cubic crystal, atomic radius, r = (√3/4) x a

Here, a is the edge length

r = (√3/4) x 351

= 151.8 pm

4) Silver (atomic weight = 108 g mol^-1 ) has a density of 10.5 g cm^-3. The number of silver atoms on a surface of an area of 10^-12 m² can be expressed in scientific notation as y × 10^x. The value of x is:

1) 3

2) 5

3) 7

4) 9

Answer: 3) 7

Solution:

The density of the crystal = 10.5 g cm^-3

Silver crystallises in FCC unit cell with 4 atoms per unit cell

ρ = (4 x 108)/ (6.023 x 10²³ x a³)

10.5 = (4 x 108)/ (6.023 x 10²³ x a³)

⇒ a³ = 6.83 x 10^-23 cm³

a = 4 x 10^-8 cm

= 4 x 10^-10 m

The surface area of the unit cell

⇒ a² = 1.6 x 10^-19 m²

Number of unit cells on 10^-12 m² surface

= 10^-12/ (1.6 x 10^-19)

= 6.25 x 10⁶

There are two atoms on one face of the unit cell

Number of atoms on 10^-12 m² surface = 2 x number of unit cell = 1.25 x 10⁷

⇒ x = 7

5) The three states of matter are solid, liquid and gas. Which of the following statements is true about them?

1) Gases and liquids have viscosity as a common property

2) The molecules in all three states possess random translational motion

3) Gases cannot be converted into solids without passing through the liquid phase

4) Solids and liquids have pressure as a common property

Answer: 1) Gases and liquids have viscosity as a common property

Solution:

Both gases and liquids possess viscosity.

6) Total volume of atoms present in a face-centred cubic unit cell of metal is (r = atomic radius)

1) 20/3 πr³

2) 24/3 πr³

3) 12/3 πr³

4) 16/3 πr³

Answer: 4) 16/3 πr³

Solution:

There are 4 atoms per unit cell in the fcc structure.

The volume of an atom (V) = (4/3)πr³

In fcc,

The total volume of atoms = 4 x V

= 4 x (4/3)πr³

= (16/3)πr³

Practice Problems

1) A closed flask contains water in all its three states solid, liquid and vapour at OºC. In this situation, the average kinetic energy of water molecules will be:

1) The greatest in all the three states

2) The greatest in the vapour state

3) The greatest in the liquid state

4) The greatest in the solid-state

2) The pure crystalline substance, on being heated gradually, first forms a turbid liquid at constant temperature and still, at a higher temperature, turbidity completely disappears. The behaviour is a characteristic of substance forming.

1) Allotropic crystals

2) Liquid crystals

3) Isometric crystals

4) Isomorphous crystals

3) In the laboratory, sodium chloride is made by burning sodium in the atmosphere of chlorine. The salt obtained is yellow in colour. The cause of the yellow colour is:

1) Presence of Na⁺ ions in the crystal lattice

2) Presence of Cl^– ions in the crystal lattice

3) Presence of electrons in the crystal lattice

4) Presence of face-centred cubic crystal lattice

4) A solid has a structure in which W atoms are located at the corners of a cubic lattice, O atoms at the centre of edges and Na atoms at the centre of the cube. The formula for the compound is:

1) NaWO₂

2) NaWO₃

3) Na2WO₃

4) NaWO₄

5) X- rays of wavelength equal to 0.134 nm give the first-order diffraction from the surface of a crystal when the value of θ is 10.5º. The distance between the planes in the crystal parallel to the surface is (sin 10.5º = 0.1822):

1) 0.368 nm

2) 0.184 nm

3) 0.726 nm

4) 0.92 nm

6) In a face-centred cubic arrangement of A and B atoms in which A atoms are at the corners of the unit cell, and B atoms at the face centres, one of the A atoms is missing from one corner in the unit cell. The simplest formula of the compound is:

1) A₇B₃

2) AB₃

3) A₇B₂₄

4) A_7/8B₃

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Solid State Important JEE Questions and Answers

Solids and Unit Cells – JEE Concepts

Solid State – Contribution of Particles and Cubic Lattice

Solid State – Voids in FCC and HCP

Solid State – Top 10 Most Important JEE Main Questions