$$\text{MnO}_4^{2-}$$, in acidic medium, disproportionates to :

We need to find the disproportionation products of $$MnO_4^{2-}$$ in acidic medium and recall that disproportionation is a reaction where the same element is simultaneously oxidized and reduced; here Mn is in the +6 oxidation state in $$MnO_4^{2-}$$.

In acidic medium the reaction is given by: $$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$

On verifying oxidation states, Mn in $$MnO_4^{2-}$$ is +6, in $$MnO_4^-$$ it is +7 (oxidation) and in $$MnO_2$$ it is +4 (reduction), which confirms disproportionation since Mn(+6) is oxidized to Mn(+7) and reduced to Mn(+4).

The correct answer is Option (1): $$MnO_4^-$$ and $$MnO_2$$.