${MnO}_{4}^{2 -}$ (1 mole) in neutral aqueous medium disproportionates to give

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1/3 mole of ${Mn}_{2} O_{7}^{}$ and 1/3 mole of ${MnO}_{2}^{}$

2/3 mole of ${Mn}_{2} O_{7}^{}$ and 1/3 mole of ${MnO}_{2}^{}$

2/3 mole of ${MnO}_{4}^{-}$ and 1/3 mole of ${MnO}_{2}^{}$

1/3 mole of ${MnO}_{4}^{-}$ and 2/3 mole of ${MnO}_{2}^{}$

answer is A.

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Detailed Solution

$3 K_{2} {MnO}_{4} + 2 H_{2} O \to 2 {KMnO}_{4} + {MnO}_{2} + 4 KOH$
${MnO}_{4}^{2 -}$ in neutral aqueous medium is disproportionate to $\frac{2}{3}$ mole of ${MnO}_{4}^{-}$ and $\frac{1}{3}$ mole of ${MnO}_{2}^{}$

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