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The oxidizing power of ${[C r O_{4}]}^{2 -}, {[M n O_{4}]}^{2 -}$ and ${[F e O_{4}]}^{2 -}$ follows the order.

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${[C r O_{4}]}^{2 -} < {[M n O_{4}]}^{2 -} < {[F e O_{4}]}^{2 -}$

${[M n O_{4}]}^{2 -} < {[F e O_{4}]}^{2 -} < {[C r O_{4}]}^{2 -}$

${[F e O_{4}]}^{2 -} < {[M n O_{4}]}^{2 -} < {[C r O_{4}]}^{2 -}$

${[C r O_{4}]}^{2 -} < {[F e O_{4}]}^{2 -} < {[M n O_{4}]}^{2 -}$

answer is A.

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Detailed Solution

Oxidation state of Fe in ${[F e O_{4}]}^{2 -}$ is +6, Cr in ${[C r O_{4}]}^{2 -}$ is +6 and oxidation state of Mn in ${[M n O_{4}]}^{2 -}$ is +6.

Ferrate is very unstable, hence it is strong oxidizing agent.

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