MnO_ 4 ^ 2- , in acidic medium, disproportionates to :

\mathrm{MnO}_{4}{ }^{2-}

, in acidic medium, disproportionates to :

Options

A $\mathrm{MnO}_{4}{ }^{-}$ and $\mathrm{MnO}_{2}$
B $\mathrm{Mn}_{2} \mathrm{O}_{7}$ and $\mathrm{MnO}_{2}$
C $\mathrm{Mn}_{2} \mathrm{O}_{7}$ and MnO
D $\mathrm{MnO}_{4}{ }^{-}$ and MnO

Correct answer

A. $\mathrm{MnO}_{4}{ }^{-}$ and $\mathrm{MnO}_{2}$

Step-by-step solution

MnO_4^{2-}

(manganate) has Mn in +6 oxidation state.
In acidic medium, it undergoes disproportionation where Mn is simultaneously oxidized and reduced:
Oxidation:

+6 \rightarrow +7

(forms

MnO_4^-

, permanganate)
Reduction:

+6 \rightarrow +4

(forms

MnO_2

)
Reaction:

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

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