(N/A) The oxidation numbers of carbon in $(CN)_{2}$,$CN^{-}$,and $CNO^{-}$ are $+3$,$+2$,and $+4$ respectively.
$1$. In $(CN)_{2}$,let the oxidation number of $C$ be $x$. Since $N$ is $-3$,$2(x - 3) = 0$,so $x = +3$.
$2$. In $CN^{-}$,the oxidation number of $C$ is $+2$ (since $x + (-3) = -1$).
$3$. In $CNO^{-}$,the oxidation number of $C$ is $+4$ (since $x - 3 - 2 = -1$).
The reaction is:
$(C^{+3}N)_{2}(g) + 2OH^{-}(aq) \rightarrow C^{+2}N^{-}(aq) + C^{+4}NO^{-}(aq) + H_{2}O(l)$
It can be observed that the same element (carbon) is being reduced (from $+3$ to $+2$) and oxidized (from $+3$ to $+4$) simultaneously. Reactions in which the same substance is both reduced and oxidized are known as disproportionation reactions. Thus,the alkaline decomposition of cyanogen is an example of a disproportionation reaction.