(i) C is a reducing agent while $\mathrm{O}_2$ is an oxidising agent. If excess of carbon is burnt in a limited supply of $\mathrm{O}_2$, CO is formed in which oxidation state of C is +2 but when $\mathrm{O}_2$ is in excess CO formed gets oxidised to $\mathrm{CO}_2$ in which oxidation state of C is + 4. (a) $\underset{\text{(Excess)}}{2\mathrm{C}_{(s)}}+\mathrm{O}_{2(g)}$ → $\overset{+2}{2\mathrm{CO}_{(g)}}$ ; (b) $\mathrm{C}_{(s)} + \underset{\text{(Excess)}}{\mathrm{O}_{2(g)}}$ → $\overset{+4}{\mathrm{CO}_{2(g)}}$ (ii) $\mathrm{P}_4$ is a reducing agent while $\mathrm{Cl}_2$ is an oxidising agent. When excess of $\mathrm{P}_4$ is used, $\mathrm{PCl}_3$ is formed in which the oxidation state of P is +3. When excess of $\mathrm{Cl}_2$ is used, the initially formed $\mathrm{PCl}_3$ reacts further to form $\mathrm{PCl}_5$ in which the oxidation state of P is +5. (a) $\underset{\text{(Excess)}}{\mathrm{P}_{4(s)}} + 6\mathrm{Cl}_{2(g)}$ → $\overset{+3}{4\mathrm{PCl}_3}$ (b) $\mathrm{P}_{4(s)} + \underset{\text{(Excess)}}{10\mathrm{Cl}_{2(g)}}$ → $\overset{+5}{44\mathrm{PCl}_5}$ (iii) Na is a reducing agent and $\mathrm{O}_2$ is an oxidising agent. When excess of Na is used, sodium oxide is formed in which the oxidation state of O is –2. If, excess of $\mathrm{O}_2$ is used, $\mathrm{Na_2O_2}$ is formed in which the oxidation state of O is –1. (a) $\underset{\text{(Excess)}}{4\mathrm{Na}_{(s)}} + \mathrm{O}_{2(g)}$ → $\overset{-2}{2\mathrm{Na}_2\mathrm{O}_{(s)}}$ (b) $2\mathrm{Na}_{(s)} + \underset{\text{(Excess)}}{\mathrm{O}_{2(g)}}$ → $\mathrm{Na}_2\mathrm{O}_{\overset{-1}{2(s)}}$