Identify the disproportionation reaction

1.

2.

3.

4.

Hint: Same substance is oxidised as well as reduced in a compound
Reactions in which the same substance is oxidised as well as reduced are called disproportionation reactions.
Writing the oxidation number of each element above its symbol in the given reactions
(1) $\stackrel{-4}{\mathrm{C}}{\stackrel{+1}{\mathrm{H}}}_{4}+2{\stackrel{0}{\mathrm{O}}}_{2}\to \stackrel{+4}{\mathrm{C}}{\stackrel{-2}{\mathrm{O}}}_{2}+2{\stackrel{+1}{\mathrm{H}}}_{2}\stackrel{-2}{\mathrm{O}}$
(2) $\stackrel{-4}{\mathrm{C}}{\stackrel{+1}{\mathrm{H}}}_{4}+4{\stackrel{0}{\mathrm{Cl}}}_{2}\to \stackrel{+4}{\mathrm{C}}{\stackrel{-1}{\mathrm{Cl}}}_{4}+4\stackrel{+1}{\mathrm{H}}\stackrel{-1}{\mathrm{Cl}}$
(3) $2{\stackrel{0}{\mathrm{F}}}_{2}+2\stackrel{-2}{\mathrm{O}}\stackrel{+1}{{\mathrm{H}}^{-}}\to 2\stackrel{-1}{{\mathrm{F}}^{-}}+\stackrel{+2}{\mathrm{O}}{\stackrel{-1}{\mathrm{F}}}_{2}+{\stackrel{+1}{\mathrm{H}}}_{2}\stackrel{-2}{\mathrm{O}}$
(4) $2\stackrel{+4}{\mathrm{N}}{\stackrel{-2}{\mathrm{O}}}_{2}+2\stackrel{-2}{\mathrm{O}}\stackrel{+1}{{\mathrm{H}}^{-}}\to \stackrel{+3}{\mathrm{N}}\stackrel{-2}{{\mathrm{O}}_{2}^{-}}+\stackrel{+5}{\mathrm{N}}\stackrel{-2}{{\mathrm{O}}_{3}^{-}}+{\stackrel{+1}{\mathrm{H}}}_{2}\stackrel{-2}{\mathrm{O}}$
Thus, in reaction (4), N is both oxidised as well as reduced since the O.N. of N increases from +4 in NO2 to +5 in ${\mathrm{NO}}_{2}^{-}$ and decreases +4 in NO3 to +3 in ${\mathrm{NO}}_{3}^{-}$