An example of a disproportionation reaction is:

JEE MAIN 2019
$2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$
$2 N a B r + C l_{2} \to 2 N a C l + B r_{2}$
$2 C u B r \to C u B r_{2} + C u$
$2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 I_{2} + 8 H_{2} O$

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Solution

The correct option is C $2 C u B r \to C u B r_{2} + C u$
$(A) 2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$

-In this reaction, manganese has $+ 7$ oxidation state in $K M n O_{4}$ and $+ 6$ and $+ 4$ oxidation states in $K_{2} M n O_{4}$ and $M n O_{2}$ respectively. This indicates manganese is only getting reduced. And $O$ is getting oxidised from $- 2$ to $0$ . So, this reaction is not a disproportionation reaction.

$(B) 2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 I_{2} + 8 H_{2} O$
-In this reaction, manganese has $+ 7$ oxidation state in $M n O_{4}^{-}$ and $+ 2$ oxidation state in the product side.
And $I$ is getting oxidised from $- 1$ to $0$ .
So, this reaction is not a disproportionation reaction.

$(C) 2 C u B r \to C u B r_{2} + C u$
-In this reaction, copper is $+ 1$ in $C u B r$ and $+ 2$ oxidation state in $C u B r_{2}$ and zero oxidation state elemental form. This implies, In this reaction copper is getting both oxidized as well as reduced. Therefore, this reaction is an example of a disproportionation reaction.

$(D) 2 N a B r + C l_{2} \to 2 N a C l + B r_{2}$
In this reaction, $B r$ is getting oxidised from $- 1$ to $0$ and $C l$ is getting reduced from $0$ to $- 1$ . Hence, this is not a disproportionation reaction.

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