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

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$\frac{1}{3} mole of {MnO}_{4}^{-} and \frac{2}{3} mole of {MnO}_{2} .$

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

$\frac{1}{3} mole of {Mn}_{2} O_{7} and \frac{1}{3} mole of {MnO}_{2}$

$\frac{2}{3} mole of {Mn}_{2} O_{7} and \frac{1}{3} mole of {MnO}_{2}$

answer is A.

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Detailed Solution

${3 MnO}_{4}^{2-} + {4 H}^{+} \to 2 {MnO}_{4}^{-} + {MnO}_{2} + 2 H_{2} O$

3 moles ${MnO}_{4}^{2-} \to 2 {moles MnO}_{4}^{-} + 1 {mole MnO}_{2}$

1 mole ${MnO}_{4}^{2-} \to \frac{2}{3} {mole MnO}_{4}^{-} + \frac{1}{3} {mole MnO}_{2}$

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