Which one of the following is an example of disproportionation reaction?

We need to identify which reaction is a disproportionation reaction.

What is a disproportionation reaction?

A disproportionation reaction is one in which the same element in a single oxidation state is simultaneously oxidized and reduced.

$$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$$

Mn in $$MnO_4^{2-}$$: oxidation state = +6

Mn in $$MnO_4^-$$: oxidation state = +7 (oxidized)

Mn in $$MnO_2$$: oxidation state = +4 (reduced)

Here, Mn(+6) is both oxidized to Mn(+7) and reduced to Mn(+4). This is a disproportionation reaction.

$$MnO_4^- + 4H^+ + 4e^- \to MnO_2 + 2H_2O$$

This is a half-reaction (reduction only), not a disproportionation.

$$10I^- + 2MnO_4^- + 16H^+ \to 2Mn^{2+} + 8H_2O + 6I_2$$

$$I^-$$ is oxidized to $$I_2$$, and $$Mn^{+7}$$ is reduced to $$Mn^{+2}$$. Different elements are oxidized and reduced — this is a simple redox reaction.

$$8MnO_4^- + 3S_2O_3^{2-} + H_2O \to 8MnO_2 + 6SO_4^{2-} + 2OH^-$$

$$Mn^{+7}$$ is reduced to $$Mn^{+4}$$, and $$S^{+2}$$ is oxidized to $$S^{+6}$$. Different elements are oxidized and reduced — not a disproportionation.

The correct answer is Option A: $$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$$.