NCERT Solutions for Class 12 Chemistry Chapter 1 – The Solid State (Textbook Questions Solved)

Below are the detailed and accurate NCERT textbook solutions for Class 12 Chemistry Chapter 1 – The Solid State. These answers help you prepare for CBSE board exams with a clear understanding of concepts and step-by-step explanations.

NCERT Exercise Questions with Answers

Q1.1. Why are solids rigid?
Ans: In solids, the constituent particles are held in fixed positions due to strong intermolecular forces. They can only oscillate about their mean positions, which gives solids their rigidity.

Q1.2. Why do solids have a definite volume?
Ans: Solids have strong intermolecular forces and minimal intermolecular space. This structural rigidity resists compression, giving solids a fixed and definite volume.

Q1.3. Classify the following as amorphous or crystalline solids:
Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fibreglass, copper
Ans:

• Crystalline solids: Benzoic acid, potassium nitrate, copper

• Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibreglass

Q1.4. Why is glass considered a supercooled liquid?
Ans: Glass behaves like a supercooled liquid because it lacks a sharp melting point and flows very slowly over time. This is evident from its slightly thicker base in old windowpanes.

Q1.5. A solid has the same refractive index in all directions. What type of solid is it? Would it show cleavage?
Ans: It is an amorphous solid (isotropic in nature). Amorphous solids do not exhibit clean cleavage; they break into irregular pieces.

Q1.6. Classify the following solids by intermolecular forces:
Sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubidium, argon, silicon carbide
Ans:

• Ionic: Sodium sulphate, zinc sulphide

• Metallic: Copper, rubidium

• Molecular: Benzene, urea, ammonia, water, argon

• Covalent (network): Diamond, silicon carbide

Q1.7. A solid is very hard, an electrical insulator in both solid and molten states, and has a high melting point. What type is it?
Ans: It is a covalent or network solid.

Q1.8. Why do ionic solids conduct electricity in the molten state but not in the solid state?
Ans: In the solid state, ions are fixed and cannot move. In the molten state, ions become mobile and can conduct electricity.

Q1.9. What type of solids are good conductors and malleable?
Ans: Metallic solids are electrical conductors, malleable, and ductile.

Q1.10. What is the significance of a lattice point?
Ans: A lattice point represents the position of an atom, ion, or molecule in the crystal lattice. It defines the structure and shape of the crystal.

Q1.11. Parameters that characterize a unit cell:
Ans:

• Edge lengths: a, b, c

• Angles between edges: α (between b & c), β (between a & c), γ (between a & b)

Q1.12. Distinguish between:
(i) Hexagonal vs Monoclinic unit cells
(ii) Face-centered vs End-centered unit cells
Ans:
(i) Hexagonal: a = b ≠ c; α = β = 90°, γ = 120°
Monoclinic: a ≠ b ≠ c; α = γ = 90°, β ≠ 90°
(ii) Face-centered: particles at all corners and face centers
End-centered: particles at all corners and centers of two opposite faces

Q1.13. Atom contributions in a cubic unit cell:
(i) Corner atom: 1/8
(ii) Body-centered atom: 1 (not shared)

Q1.14. 2D coordination number in a square close-packed layer:
Ans: 4

Q1.15. Total number of voids in 0.5 mol of hcp structure and number of tetrahedral voids:
Ans:
Atoms = 3.011 × 10²³
Octahedral voids = 3.011 × 10²³
Tetrahedral voids = 6.022 × 10²³
Total = 9.033 × 10²³

Q1.16. Compound formed by M and N (M occupies 1/3 tetrahedral voids in ccp):
Ans:
Ratio M:N = 2:3
Formula = M₂N₃

Q1.17. Lattice with highest packing efficiency:
Ans: Hexagonal close-packed (hcp) = 74%

Q1.18. Identify the unit cell from density and edge length data:
Ans: The unit cell is face-centered cubic (fcc) or cubic close-packed (ccp).

Q1.19. What defect arises when a solid is heated?
Ans: Vacancy defect, which decreases the density of the solid.

Q1.20. Stoichiometric defects in (i) ZnS (ii) AgBr:
Ans:
(i) Frenkel defect
(ii) Both Frenkel and Schottky defects

Q1.21. How vacancies form when doping with a higher-valent cation:
Ans: To maintain electrical neutrality, two lower-valent cations are replaced by one higher-valent cation, creating cation vacancies.

Q1.22. Why do anionic vacancies cause color in some solids?
Ans: In NaCl, Cl⁻ ions leave vacancies that are occupied by electrons (F-centres). These absorb visible light and impart yellow color.

Q1.23. Group to use for n-type doping of a Group 14 element:
Ans: Group 15

Q1.24. Which makes better permanent magnets – ferromagnetic or ferrimagnetic substances?
Ans: Ferromagnetic substances, because their magnetic domains remain aligned after removing the external field.

Class 12 Chemistry – Chapter 1: The Solid State (NCERT Solutions)

This section provides detailed, easy-to-understand solutions to all in-text and exercise questions from NCERT Class 12 Chemistry Chapter 1: The Solid State. The answers are structured to support conceptual clarity, exam preparation, and SEO-focused educational content.

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

Answer: Amorphous solids are substances in which the constituent particles are not arranged in a regular, repeating pattern. These solids lack long-range order, and their physical properties are isotropic, meaning they are the same in all directions.

Examples:

• Glass

• Plastic

• Rubber

• Asphalt

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

Answer: Quartz is a crystalline solid where silicon and oxygen atoms are arranged in a regular 3D structure. In contrast, glass is an amorphous solid without a well-defined arrangement.

Quartz can be converted into glass by melting it and then cooling it rapidly. This rapid cooling prevents the atoms from forming a crystalline structure, resulting in amorphous glass.

Classify the following solids as: (i) Ionic (ii) Molecular (iii) Covalent (iv) Metallic — P₄, plastic, graphite, brass, Rb, LiBr, Si, SiO₂, diamond, copper.

Answer:

• Ionic: LiBr, Rb

• Molecular: P₄, plastic

• Covalent: Graphite, Si, SiO₂, diamond

• Metallic: Brass, copper

What types of solids are electrical conductors, malleable and ductile?

Answer: Metallic solids are electrical conductors, malleable, and ductile. This is due to the presence of free electrons (also called a sea of electrons) that move throughout the structure, allowing electrical conduction and flexibility.

Give the significance of a ‘lattice point’.

Answer: A lattice point refers to a position in a crystal lattice that represents the location of an atom, ion, or molecule. Each point defines the spatial arrangement of the particles in the crystalline solid.

Name the parameters that characterise a unit cell.

Answer: A unit cell is characterized by the following six parameters:

1. Edge lengths: a, b, and c

2. Angles between the edges: α (between b & c), β (between a & c), γ (between a & b)

Name the three types of cubic unit cells.

Answer:

1. Simple Cubic Unit Cell (SC)

2. Body-Centered Cubic Unit Cell (BCC)

3. Face-Centered Cubic Unit Cell (FCC)

Explain how much portion of an atom located at (i) corner and (ii) body centre of a cubic unit cell is shared by adjacent unit cells.

Answer:

• (i) Corner: Each corner atom is shared among 8 unit cells, so only 1/8 of the atom belongs to one unit cell.

• (ii) Body centre: The body-centred atom is not shared with any other unit cell, so it belongs entirely or 1 to the unit cell.

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

Answer: In a square close-packed layer, each particle is in contact with 4 neighbouring particles. Therefore, the 2D coordination number is 4.

A compound forms hexagonal close-packed structure. What is the total number of atoms in 1 unit cell of this structure?

Answer: In a hexagonal close-packed (hcp) structure:

• 12 corner atoms × 1/6 = 2

• 2 face-centre atoms × 1/2 = 1

• 3 atoms completely inside = 3

Total atoms = 2 + 1 + 3 = 6 atoms

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

Answer:

• (i) FCC: 8 corner atoms × 1/8 + 6 face atoms × 1/2 = 1 + 3 = 4 atoms

• (ii) BCC: 8 corner atoms × 1/8 + 1 body atom = 1 + 1 = 2 atoms

• (iii) SC: 8 corner atoms × 1/8 = 1 atom

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

Answer: (i) Both metallic and ionic crystals have strong electrostatic forces:

• Similarities: High melting points, crystalline structures

• Differences: Metallic solids have mobile electrons (conduct electricity), ionic solids have immobile ions and conduct only in molten or aqueous state.

(ii) Ionic solids are hard due to strong ionic bonds. They are brittle because shifting ions of like charges can come close and repel, breaking the structure.

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

Answer: When a higher valency cation replaces a lower valency cation in an ionic solid (e.g., Al³⁺ replacing Na⁺), to maintain electrical neutrality, vacancies are created. For example, two Na⁺ ions are replaced by one Al³⁺ ion, and one Na⁺ site becomes vacant.

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

Answer: In the solid state, ions are fixed and cannot move. In the molten state, ions become mobile and can carry electric current. Hence, conductivity occurs only in the molten or aqueous state.

What type of defect can arise when a solid is heated? Which physical property is affected by it and how?

Answer: Vacancy defect arises when a solid is heated. Some atoms leave their lattice sites, creating vacancies. This increases entropy and lowers density because mass remains the same but volume occupied effectively increases.

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

Answer:

• (i) ZnS: Frenkel defect – Zn²⁺ ions occupy interstitial sites.

• (ii) AgBr: Shows both Frenkel and Schottky defects.

Explain how electrical neutrality is maintained in solids with Schottky and Frenkel defects.

Answer: In Schottky defect, equal numbers of cations and anions are missing. In Frenkel defect, cations leave their lattice sites but stay within the crystal (interstitial site). In both cases, the number of positive and negative charges remains balanced, preserving electrical neutrality.

Describe the two main types of semiconductors and contrast their conduction mechanisms.

Answer:

• n-type semiconductor: Doping with a group 15 element adds extra electrons (negative charge carriers).

• p-type semiconductor: Doping with a group 13 element creates holes (positive charge carriers).

Conduction:

• n-type: movement of electrons

• p-type: movement of holes

Non-stoichiometric cuprous oxide, Cu₂O can be prepared in the 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?

Answer: In Cu₂O with less copper, some Cu⁺ ions are missing. To maintain neutrality, some Cu⁺ ions get oxidized to Cu²⁺, creating holes (positive charge carriers). This makes it a p-type semiconductor.

Ferrimagnetism is observed only in solids but not in liquids and gases. Why?

Answer: Ferrimagnetism arises from the alignment of magnetic moments in a crystal lattice. This ordered arrangement is possible only in solids, where particles have fixed positions—not in liquids or gases where particles move randomly.

What is meant by the term ‘coordination number’?

Answer: The coordination number is the number of nearest neighbour atoms surrounding a particle in a crystal structure. For example, in FCC and HCP structures, the coordination number is 12.

What is the coordination number of atoms: (i) in a cubic close-packed structure? (ii) in a body-centred cubic structure?

Answer:

• (i) Cubic close-packed (FCC): 12

• (ii) Body-centred cubic (BCC): 8

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

Answer: Use the formula: Where:

• : atomic mass

• : density

• : edge length (in cm)

• : Avogadro’s number (6.022×10²³)

• : number of atoms per unit cell

A cubic solid is made of two elements, P and Q. Atoms of Q are at corners, and P at face centres. What is the formula of the compound?

Answer:

• Q atoms at 8 corners × 1/8 = 1

• P atoms at 6 faces × 1/2 = 3

Formula = P₃Q

An element has a body-centred cubic structure with a cell edge of 288 pm. The density is 7.2 g/cm³. Calculate atomic mass.

Answer: Given:

• a = 288 pm = 2.88 × 10⁻⁸ cm

• Z = 2 (BCC)

• d = 7.2 g/cm³

•  

Gold (atomic radius = 144 pm) crystallizes in a face-centred unit cell. What is the length of a side of the unit cell?

Answer: In FCC:

So, the unit cell edge length = 407 pm