Page No 30: - Chapter 1 The Solid State Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 1.1:

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

Answer:

Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.

Question 1.2:

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

Answer:

The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders.

Quartz can be converted into glass by heating and then cooling it rapidly.

Question 1.3:

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

(i) Tetra phosphorus decoxide (P₄O₁₀) (vii) Graphite

(ii) Ammonium phosphate (NH₄)₃PO₄ (viii) Brass

(iii) SiC (ix) Rb

(iv) I₂ (x) LiBr

(v) P₄ (xi) Si

Answer:

Ionic → (ii) Ammonium phosphate (NH₄)₃PO₄, (x) LiBr

Metallic → (viii) Brass, (ix) Rb

Molecular → (i) Tetra phosphorus decoxide (P₄O₁₀), (iv) I₂, (v) P₄.

Covalent (network) → (iii) SiC, (vii) Graphite, (xi) Si

Amorphous → (vi) Plastic

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?

Answer:

(i) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number.

(ii) The coordination number of atoms

(a) in a cubic close-packed structure is 12, and

(b) in a body-centred cubic structure is 8

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.

Answer:

By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.

Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the mass of one atom of the metal and ‘z’ be the number of atoms in the unit cell.

Now, density of the unit cell 

[Since mass of the unit cell = Number of atoms in the unit cell × mass of one atom]

[Volume of the unit cell = (Edge length of the cubic unit cell)³]

From equation (i), we have:

Now, mass of one atom of metal (m) 

Therefore, 

If the edge lengths are different (say a, b and c), then equation (ii) becomes:

M = d(abc)NAz     (iv)

From equations (iii) and (iv), we can determine the atomic mass of the unknown metal.

Question 1.6:

‘Stability of a crystal is reflected in the magnitude of its melting point’. 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?

Answer:

Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.

The melting points of the given substances are:

Solid water → 273 K

Ethyl alcohol → 158.8 K

Diethyl ether → 156.85 K

Methane → 89.34 K

Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest.