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In a reaction
${AO}_{2}^{-} + {MnO}_{4}^{-} + H^{+} \to A^{+ n} + {Mn}^{+ 2}$
1 mole of ${AO}_{2}^{- 1}$ is oxidized to $A^{+ n}$ in presence of 0.4 mole acidified potassium permanganate. Then "n" value is

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answer is D.

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Detailed Solution

5 moles of ${AO}_{2}^{- 1}$ reacts with (n-3) moles of ${MnO}_{4}^{-}$
5(0.4) = n – 3
n = + 5

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