CBSE Class 12 Chemistry Chapter 1: The Solid State - NCERT Solutions
This comprehensive set of NCERT Solutions for CBSE Class 12 Chemistry, Chapter 1: The Solid State, provides detailed explanations for all exercises. It covers fundamental concepts such as the definition and examples of amorphous solids, the distinction between amorphous solids like glass and crystalline solids like quartz, and the classification of various substances into ionic, metallic, molecular, network (covalent), and amorphous types. The solutions also explain the significance of coordination number in different crystal structures (cubic close-packed and body-centered cubic) and demonstrate how to determine the atomic mass of an unknown metal using its density and unit cell dimensions. These solutions are designed to help students grasp the core principles of solid-state chemistry, clarify doubts, and prepare effectively for their board examinations by offering step-by-step problem-solving guidance.
Quick info
| Board | CBSE |
|---|---|
| Class | Class 12 |
| Subject | Chemiry |
| Session | 2026 |
| Language | English |
| Type | NCERT Solutions |
| Chapter | Chapter 1: The Solid State - NCERT Exercises Solutions |
Chapter summary
This chapter's NCERT Solutions for Class 12 Chemistry focus on the fundamental properties of solids. It covers the definition and examples of amorphous solids, differentiating them from crystalline solids like quartz. Students will learn to classify various compounds based on their bonding and structure (ionic, metallic, molecular, covalent/network, amorphous). Key concepts like coordination number in different packing structures (CCP and BCC) and the relationship between unit cell properties (density, edge length) and atomic mass are explained. The exercises provide a solid foundation for understanding the solid state.
Learning outcomes
- Define amorphous solids and provide examples.
- Differentiate between amorphous solids (like glass) and crystalline solids (like quartz).
- Classify solids into ionic, metallic, molecular, network (covalent), and amorphous types.
- Explain the concept of coordination number in crystal lattices.
- Determine the coordination number for cubic close-packed and body-centered cubic structures.
- Calculate the atomic mass of a metal using unit cell properties and density.
Topics covered
Paper topics
- Amorphous Solids
- Crystalline Solids
- Examples of Amorphous Solids
- Quartz vs. Glass
- Classification of Solids
- Ionic Solids
- Metallic Solids
- Molecular Solids
- Network (Covalent) Solids
- Coordination Number
- Cubic Close-Packed Structure
- Body-Centred Cubic Structure
- Unit Cell Properties
- Density of Unit Cell
- Atomic Mass Calculation
Important topics
- Classification of Solids
- Amorphous vs. Crystalline Solids
- Coordination Number in CCP and BCC
- Calculating Atomic Mass from Unit Cell Data
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Questions and Solutions
Question 1.1
Amorphous solids are characterized by the absence of a long-range, ordered arrangement of their constituent particles. These particles can be atoms, ions, or molecules, and they are arranged in a disordered or irregular manner. Due to this irregular structure, amorphous solids exhibit the following properties:
- They are isotropic, meaning their physical properties (like electrical resistance or refractive index) are the same in all directions.
- They do not have a sharp melting point; instead, they soften gradually over a range of temperatures.
- They are sometimes referred to as 'pseudo solids' or 'super cooled liquids' because they exhibit some liquid-like properties.
- They do not possess a definite heat of fusion.
- When cut with a sharp tool, they tend to break into pieces with irregular surfaces.
Common examples of amorphous solids include:
- Glass
- Rubber
- Plastic
- Certain polymers
Question 1.2
The fundamental difference between glass and quartz lies in the arrangement of their constituent particles:
- Glass: It is an amorphous solid. Its constituent particles (silicon and oxygen atoms in a disordered network) exhibit only short-range order. This means there is some degree of order over short distances, but it does not extend throughout the entire solid.
- Quartz: It is a crystalline solid. Its constituent particles are arranged in a highly ordered, repeating three-dimensional pattern, exhibiting both short-range and long-range order.
Conditions for converting quartz into glass:
Quartz can be converted into glass by a process involving heating and rapid cooling. When quartz is heated to a sufficiently high temperature (above its melting point) and then cooled very quickly, the ordered crystalline structure of quartz does not have enough time to reform. Instead, the particles get 'frozen' in a disordered arrangement, resulting in the formation of glass.
Question 1.3
- Tetra phosphorus decoxide (P4O10)
- Ammonium phosphate ((NH4)3PO4)
- SiC
- I2
- P4
- Plastic
- Graphite
- Brass
- Rb
- LiBr
- Si
The classification of the given solids based on their bonding and structure is as follows:
- Ionic Solids: These solids are formed by the electrostatic attraction between oppositely charged ions. They typically have high melting points and are hard but brittle.
- (ii) Ammonium phosphate ((NH4)3PO4) - Contains ammonium cations (NH4+) and phosphate anions (PO43-).
- (x) LiBr - Contains lithium cations (Li+) and bromide anions (Br-).
- Metallic Solids: These solids consist of metal atoms held together by metallic bonds, characterized by a 'sea' of delocalized electrons. They are good conductors of heat and electricity.
- (viii) Brass - An alloy, which is a mixture of metals (copper and zinc), exhibiting metallic properties.
- (ix) Rb (Rubidium) - An alkali metal, forming a metallic lattice.
- Molecular Solids: In these solids, molecules are held together by weaker intermolecular forces (like van der Waals forces or dipole-dipole interactions). They are generally soft and have low melting points.
- (i) Tetra phosphorus decoxide (P4O10) - Consists of discrete P4O10 molecules.
- (iv) I2 (Iodine) - Consists of discrete I2 molecules.
- (v) P4 (White Phosphorus) - Consists of discrete P4 molecules.
- Covalent Network Solids: These solids have atoms joined in a continuous network by covalent bonds. They are typically very hard and have very high melting points.
- (iii) SiC (Silicon Carbide) - A network of silicon and carbon atoms joined by covalent bonds.
- (vii) Graphite - A form of carbon with a layered structure where atoms within layers are covalently bonded.
- (xi) Si (Silicon) - A metalloid forming a network covalent structure similar to diamond.
- Amorphous Solids: These solids lack a regular, long-range internal structure.
- (vi) Plastic - A general term for amorphous polymeric materials.
Question 1.4
To determine the atomic mass of an unknown metal, we can utilize the relationship between its density, unit cell dimensions, and the number of atoms within the unit cell. The fundamental formula connecting these quantities is derived from the definition of density:
Density (d) =
\frac{\text{Mass of the unit cell}}{\text{Volume of the unit cell}}
Let:
- 'a' be the edge length of the cubic unit cell.
- 'd' be the density of the metal.
- 'm' be the atomic mass of the metal (which we want to find).
- 'z' be the number of atoms per unit cell. (This value depends on the type of unit cell: z=1 for simple cubic, z=2 for body-centred cubic (BCC), and z=4 for face-centred cubic (FCC)).
The mass of the unit cell can be expressed as the product of the number of atoms in the unit cell and the mass of a single atom (which is related to the atomic mass). The volume of a cubic unit cell is given by the cube of its edge length.
Mass of the unit cell = z × m
Volume of the unit cell = a3
Substituting these into the density formula, we get: d = \frac{z \times m}{a^3}
To find the atomic mass 'm', we can rearrange this formula: m = \frac{d \times a^3}{z}
Therefore, by measuring or knowing the density (d) of the metal, the edge length (a) of its unit cell (often determined using X-ray diffraction), and identifying the type of unit cell to know 'z', we can calculate the atomic mass (m) of the unknown metal.
Question 1.5
Coordination Number: The coordination number of an atom in a crystal lattice is defined as the number of nearest neighbouring atoms (or ions or molecules) that are in direct contact with it. It essentially indicates how many atoms are directly surrounding a central atom.
Coordination Numbers in Specific Structures:
- (a) Cubic Close-Packed (CCP) Structure: In a CCP structure (which is identical to the face-centred cubic, FCC, lattice), each atom is surrounded by 12 nearest neighbours. Imagine an atom in the middle layer; it touches 4 atoms in its own layer, 4 atoms in the layer above, and 4 atoms in the layer below. Thus, the coordination number is 12.
- (b) Body-Centred Cubic (BCC) Structure: In a BCC structure, the central atom is located at the body centre of the cube, and it touches the 8 atoms located at the corners of the cube. The atoms at the corners do not touch each other along the edges or faces. Therefore, the coordination number of an atom in a BCC structure is 8.
Common mistakes
- Confusing amorphous and crystalline solids.
- Incorrectly classifying substances based on bonding and structure.
- Misidentifying the coordination number for different crystal structures.
- Errors in applying the formula relating density, unit cell dimensions, and atomic mass.
Revision tips
- Review the definitions and examples of amorphous and crystalline solids thoroughly.
- Create a table to categorize different types of solids with their properties and examples.
- Memorize the coordination numbers for common crystal structures (CCP and BCC).
- Practice the formula for calculating atomic mass from unit cell data, ensuring correct unit conversions.
Practice MCQs
Q1. Which of the following is an example of an amorphous solid?
Explanation: Glass is an amorphous solid because its constituent particles lack long-range order, unlike crystalline solids such as quartz, sodium chloride, and diamond.
Q2. What is the coordination number of atoms in a body-centred cubic (BCC) structure?
Explanation: In a body-centred cubic (BCC) structure, each atom is surrounded by 8 nearest neighbours, hence its coordination number is 8.
Q3. Which type of solid is Graphite?
Explanation: Graphite is classified as a network covalent solid due to its structure where carbon atoms are covalently bonded in a lattice.
Q4. The formula relating density (d), atomic mass (m), number of atoms per unit cell (z), and edge length (a) of a cubic unit cell is:
Explanation: The density of a unit cell is calculated as the mass of the unit cell divided by its volume. Mass of unit cel* m, and Volum, leading to * m / .
Q5. Which of the following is NOT a characteristic of amorphous solids?
Explanation: Amorphous solids do not have a definite heat of fusion; they melt over a range of temperatures. They are isotropic and possess only short-range order.
Frequently asked questions
What is the main difference between amorphous solids and crystalline solids?
Amorphous solids have constituent particles arranged randomly with only short-range order, while crystalline solids have a regular, repeating arrangement of particles with long-range order.
How can quartz be converted into glass?
Quartz can be converted into glass by heating it to a high temperature and then cooling it rapidly. This process disrupts the long-range order of the crystalline structure.
What does coordination number signify in a crystal lattice?
The coordination number represents the number of nearest neighbouring particles (atoms, ions, or molecules) that surround a particular particle in the crystal lattice.
What are the coordination numbers for cubic close-packed (CCP) and body-centred cubic (BCC) structures?
The coordination number for a cubic close-packed (CCP) structure is 12, and for a body-centred cubic (BCC) structure, it is 8.
How are the properties of solids like glass and quartz related to their structure?
Glass, being amorphous, has irregular particle arrangement leading to properties like melting over a range of temperatures and isotropic behaviour. Quartz, a crystalline solid, has a regular structure resulting in sharp melting points and anisotropic properties.
Can we determine the atomic mass of an unknown metal using its unit cell properties?
Yes, if the density and the dimensions of the unit cell (edge length) are known, along with the type of unit cell (to determine 'z', the number of atoms per unit cell), the atomic mass can be calculated using the formula d = z * m / a^3.
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