NCERT Solutions Class 12 Chemistry Chapter 1 The Solid State

NCERT Class 12 Chemistry Solutions The Solid State - NCERT Solutions prepared for CBSE students by the best teachers in Delhi. 

Class XII  Chemistry

Chapter 1 – The Solid State

Question 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 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 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 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 5:

'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.

Question 6:

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?

Answer

(i). A 2-D hexagonal close-packing contains two types of triangular voids (a and b). Let us call this 2-D structure as layer A. Now, particles are kept in the voids present in layer A (it can be easily observed from figures 2 and 3 that only one of the voids will be occupied in the process, i.e., either a or b). Let us call the particles or spheres present in the voids of layer A as layer B. Now, two types of voids are present in layer B (c and d). Unlike the voids present in layer A, the two types of voids present in layer B are not similar. Void c is surrounded by 4 spheres and is called the tetrahedral void. Void d is surrounded by 6 spheres and is called the octahedral void.

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