Ncert Exercises & Solutions

The oxidation numbers of carbon in ${(CN)}^{2}$ , ${CN}^{-} and {CNO}^{-}$ are +3, +2 and +4 respectively.

These are obtained as shown below:

Let the oxidation number of C be x.

${(CN)}_{2}$
$2 (x - 3) = 0$
$: x = 3$

$C N^{-} x - 3 = - 1$
$: x = 2$

$C N O^{-}$

$x - 3 - 2 = - 1$
$: x = 4$

The oxidation number of carbon in the various species is:

${(\overset{+ 1}{C} N)}_{2 (g)} + 2 O {H^{-}}_{(a q)} \to \overset{+ 2}{C} {N^{-}}_{(aq)} + \overset{+ 4}{C} {N^{-}}_{(aq)} + H_{2} O_{(l)}$

It can be easily observed that the same compound is being reduced and oxidised
simultaneously in the given equation. Reactions in which the same compound is reduced
and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline
decomposition of cyanogen is an example of a disproportionation reaction.

8.21 : The ${Mn}^{3 +}$ ion is unstable in solution and undergoes disproportionation to give ${Mn}^{2 +}, {MnO}_{2} and H^{+}$ ion. Write a balanced ionic equation for the reaction.

NEETprep Answer:

The oxidation numbers of carbon in ${(CN)}^{2}$ , ${CN}^{-} and {CNO}^{-}$ are +3, +2 and +4 respectively.

These are obtained as shown below:

Let the oxidation number of C be x.

${(CN)}_{2}$
$2 (x - 3) = 0$
$: x = 3$

$C N^{-} x - 3 = - 1$
$: x = 2$

$C N O^{-}$

$x - 3 - 2 = - 1$
$: x = 4$

The oxidation number of carbon in the various species is:

${(\overset{+ 1}{C} N)}_{2 (g)} + 2 O {H^{-}}_{(a q)} \to \overset{+ 2}{C} {N^{-}}_{(aq)} + \overset{+ 4}{C} {N^{-}}_{(aq)} + H_{2} O_{(l)}$

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