The Solid State

Chemistry

NCERT

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

Solution :

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.

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

Solution :

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.

3   Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.$\\$ (i) Tetra phosphorus decoxide $(P_4 0_{10 })$$\\$ (ii) Ammonium phosphate $(NH_4 )_3 PO_4$$\\$ (iii) SiC$\\$ (iv)$I_2$$\\$ (v)$ P_4$$\\$ (vi) Plastic$\\$ (vii) Graphite$\\$ (viii) Brass$\\$ (ix) Rb$\\$ (x) LiBr$\\$ (xi) Si

Solution :

Ionic $\to$ (ii) Ammonium phosphate $(NH_4 )_3 PO_4 , $(x)LiBr$\\$ Metallic $\to $ (viii) Brass, (ix) Rb$\\$ Molecular $\to $ (i) Tetra phosphoursdecoxide($P_4 O_{10} ),$(iv)$ l_2$ (v)$ P_4$$\\$ Covalent (network) $\to $ (iii) SiC, (vii) Graphite, (xi) Si $\\$ Amorphous $\to$ (vi) Plastic

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?

Solution :

(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 $8$

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.

Solution :

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 atomic mass of the metal and $‘z’$ be the number of atoms in the unit cell. $\\$ Now,density of the unit cell=$\dfrac{\text{Mass of the unit cell}}{\text{Volume of the unit cell}}$$\\$ $\implies d=\dfrac{zm}{a^3}(i)$$\\$ [Since mass of the unit cell = Number of atoms in the unit cell x Atomic mass]$\\$ (Volume of the unit cell = (Edged length of the cubic unit cell)$ ^3$ From equation (i) we have:$\\$ $M=\dfrac{da^3}{z}(ii)$$\\$ Now, mass of the metal (m) =$\dfrac{\text{Atomic Mass(M)}}{\text{Avogadro's number(N_A)}}$$\\$ Therefore,$M=\dfrac{da^3N_A}{z}(iii)$$\\$ Therefore,If the edge lengths are different (say a, b and c), then equation (ii) becomes:$\\$ $M=\dfrac{d(abc)N_A}{z}(iv)$$\\$ From equation (iii) and (iv), we can determine the atomic mass of the unknown metal.