1. Concept of disproportionation

A disproportionation reaction is one in which the same species undergoes both:

• oxidation, and
• reduction.

So, the same element in the same oxidation state should form two products with higher and lower oxidation states.

2. Check option A

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

Here, reactant is $MnO_4^{2-}$.

Let oxidation state of Mn in $MnO_4^{2-}$ be $x$:

$x + 4(-2) = -2$ $x - 8 = -2$ $x = +6$

Now in products:

• In $MnO_4^-$: $x + 4(-2) = -1 \Rightarrow x = +7$
• In $MnO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$

So Mn goes from $+6$ to:

• $+7$ (oxidation)
• $+4$ (reduction)

Thus, the same species $MnO_4^{2-}$ is both oxidized and reduced.

✅ Option A is a disproportionation reaction.

3. Check option B

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

Mn changes from:

• in $MnO_4^-$: $+7$
• in $MnO_2$: $+4$

This is only reduction, not disproportionation.

❌ Option B is not disproportionation.

4. Check option C

$10I^- + 2MnO_4^- + 16H^+ \to 2Mn^{2+} + 8H_2O + 5I_2$

Here:

• $I^-$ is oxidized to $I_2$
• $MnO_4^-$ is reduced to $Mn^{2+}$

Two different species are involved in oxidation and reduction.

❌ Option C is not disproportionation.

5. Check option D

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

Here:

• $MnO_4^-$ is reduced to $MnO_2$
• $S_2O_3^{2-}$ is oxidized to $SO_4^{2-}$

Again, oxidation and reduction occur for different species.

❌ Option D is not disproportionation.

6. Final conclusion

The disproportionation reaction is:

$\boxed{A}$