Which of the following is disproportionation reaction?

Reactions in which the same substance is oxidized as well as reduced are called disproportionation reaction. Writing the oxidation number of each element above its symbol in the given reactions.

$\overset{+ 4}{C}\overset{+ 1}{{H}_{4}} + 2{{O}_{2}}^{0} \to \overset{+ 4}{C}\overset{- 2}{{O}_{2}} + 2\overset{+ 1}{{H}_{2}}\overset{- 2}{O}$

$\overset{- 4}{C}\overset{+ 1}{{H}_{4}} + 4{{Cl}_{2}}^{0} \to \overset{+ 4}{C}C{l}_{4} +4\overset{+ 1}{H} \overset{- 1}{C} l$

$2{F}_{2}^{0} + 2\overset{- 2}{O}\overset{+ 1}{H^{-}} \to 2\overset{- 1}{F} + \overset{+ 2}{O}{F}_{2}^{- 1} + \overset{- 1}{{H}_{2}}\overset{- 2}{O}$

$2\overset{+ 4}{N}\overset{- 2}{{O}_{2}} + 2\overset{- 2}{O}\overset{+ 1}{H^{-}} \to \overset{+ 3}{N}\overset{- 2}{{{O}_{2}}^{-}} + \overset{+ 5}{N}\overset{- 2}{{{O}_{3}}^{-}} + \overset{+ 1}{{H}_{2}}\overset{- 2}{O}$

Thus, in the reaction of $N{O}_{2}$ and $OH^{-}, N$ is both oxidized as well as reduced. Hence, the oxidation number increases from $+ 4$ in $N{O}_{2}$ to $+ 5$ in $N{O}_{3}^{-}$ and decreases from $+ 4$ in $N{O}_{2}$ to $+ 3$ in $N{O}_{2}^{-} .$