$MnO_4^{2-}$ undergoes a disproportionation reaction in an acidic medium,but $MnO_4^{-}$ does not. Give a reason.

In $MnO_4^{2-}$,the oxidation number of $Mn$ is $+6$. Since it is not in its maximum oxidation state,it can be oxidized to $+7$ and reduced to lower oxidation states like $+4$ (in $MnO_2$). This simultaneous oxidation and reduction is called a disproportionation reaction.
The reaction is: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^{-} + MnO_2 + 2H_2O$.
In $MnO_4^{-}$,the oxidation number of $Mn$ is $+7$,which is the maximum possible oxidation state for $Mn$ (as it has $d^5s^2$ configuration). Since it cannot be further oxidized,$MnO_4^{-}$ does not undergo a disproportionation reaction.