Explanation:
Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions,
$\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{2O}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-2}{\text{CO}}_2+\stackrel{+1-2}{\text{2H}}_2\text{O}$
$\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{4Cl}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-1}{\text{CCl}}_4+\stackrel{+1-1}{\text{4HCl}}\text{}$
$2\stackrel{0}{\text{F}}_2+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{-1}{\text{2F}^-+}\stackrel{+2-1}{\text{OF}_2}+\stackrel{+1-2}{\text{H}_2\text{O}}$
$\stackrel{+4-2}{\text{2NO}_2}+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+3-2}{\text{NO}^-_2}+\stackrel{+5-2}{\text{NO}^-_3}+\stackrel{+1-2}{\text{H}_2\text{O}}$
Thus, in reaction (d), N is both oxidized as well as reduced since the O.N. of N increases from +4 in NO2 to +5 in NO–3 and decreases from +4 in NO2 to +3 in NO–2.
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| List $-I$ | List $-II$ |
| $(A)$ Phenelzine | $(p)$ Pyrimidine |
| $(B)$ Chloroxylenol | $(q)$ Furan |
| $(C)$ Uracil | $(r)$ Hydrazine |
| $(D)$ Ranitidine | $(s)$ Phenol |