Question

Give reasons for the following :
(i) Oxygen is a gas but sulphur is a solid.
(ii) $O_3$ acts as a powerful oxidising agent.
(iii) $Bi{H}_3$ is the strongest reducing agent amongst all the hydrides of Group 15 elements.

Answer 1

(i) Oxygen due to its small size and high electronegativity forms $p\pi -p\pi$ double bond with other oxygen atom (O = O) to form $O_2$ molecules. The intermolecular forces of attraction between $O_2$ molecules are weak van der Waals forces and hence oxygen is a diatomic gas at room temperature. But due to its larger size sulphur does not form $p\pi -p\pi$ double bonds and hence does not form diatomic molecules. It exists as octaatomic molecules $\left(S_8\right)$ having puckered ring structure. The forces of attraction holding $S_8$ molecules together are much stronger. Hence sulphur is a solid.

(ii) $O_3$ is a thermodynamically unstable compound. On heating it readily decomposes to give dioxygen and nascent oxygen.
$O_3\stackrel{\Delta }{\to }{O}_2+O$ (nascent oxygen)
Since nascent oxygen is very reactive, therefore, $O_3$ acts as a powerful oxidising agent.

(iii) Among the hydrides of group 15 elements, $Bi{H}_3$ is least stable due to largest size of Bi resulting in a very weak Bi - H bond. Therefore, it has maximum tendency to liberate hydrogen. Hence it is the strongest reducing agent among the hydrides of group 15 elements.