Which of the following reactions are disproportionation reactions? (a) $2\mathrm{Cu}{}^{+}\to \mathrm{Cu}{}^{2+}+\mathrm{Cu}$ (b) $3\mathrm{MnO}{}_4{}^{2-}+4\mathrm{H}{}^{+}\to 2\mathrm{MnO}{}_4{}^{-}+\mathrm{MnO}{}_2+2\mathrm{H}{}_2\mathrm{O}$ (c) $2\mathrm{KMnO}{}_4\to \mathrm{K}{}_2\mathrm{MnO}{}_4+\mathrm{MnO}{}_2+\mathrm{O}{}_2$ (d) $2\mathrm{MnO}{}_4{}^{-}+3\mathrm{Mn}{}^{2+}+2\mathrm{H}{}_2\mathrm{O}\to 5\mathrm{MnO}{}_2+4\mathrm{H}{}^{+}$ Select the correct option from the following:

Solution & Explanation

\textbf{Answer:} (a) (a) and (b) only. \textbf{Solution:} In a disproportionation reaction the SAME element in a single oxidation state is simultaneously oxidised and reduced. (a) $2\mathrm{Cu}{}^{+}\to \mathrm{Cu}{}^{2+}+\mathrm{Cu}$: Cu($+1$) goes to Cu($+2$) and Cu(0) — disproportionation. ✓ (b) $3\mathrm{MnO}{}_4{}^{2-}\to 2\mathrm{MnO}{}_4{}^{-}+\mathrm{MnO}{}_2$: Mn($+6$) goes to Mn($+7$) and Mn($+4$) — disproportionation. ✓ (c) $2\mathrm{KMnO}{}_4\to \mathrm{K}{}_2\mathrm{MnO}{}_4+\mathrm{MnO}{}_2+\mathrm{O}{}_2$: here Mn($+7$) is reduced to $+6$ and $+4$, but oxygen ($-2$) is oxidised to $\mathrm{O}{}_2$ (0). Two different elements change, so it is not a disproportionation of a single element. ✗ (d) $2\mathrm{MnO}{}_4{}^{-}+3\mathrm{Mn}{}^{2+}\to 5\mathrm{MnO}{}_2$: Mn($+7$) is reduced to $+4$ and Mn($+2$) is oxidised to $+4$. This is a comproportionation (two different oxidation states converging), not a disproportionation. ✗ Therefore only (a) and (b) are disproportionation reactions — option (a).