Answer: B (\(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}\)). In a disproportionation reaction the same element is both oxidised and reduced.
Chlorine starts at 0 in \(\text{Cl}_2\). In \(\text{NaCl}\) it is \(-1\) (reduced); in \(\text{NaOCl}\) it is \(+1\) (oxidised). The same element, chlorine, is both oxidised and reduced.
In A copper is oxidised while nitrogen is reduced; in C iron is reduced while iodine is oxidised; in D nitrogen is reduced while oxygen is oxidised. These involve different elements, so only B is disproportionation. Answer B.
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