The correct descending order of oxidising power of the following is:

C r_{2} O_{7}^{2 -} > M n O_{4}^{-} > V O_{2}^{+}

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M n O_{4}^{-} > C r_{2} O_{7}^{2 -} > V O_{2}^{+}

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V O_{2}^{+} > M n O_{4}^{-} > C r_{2} O_{7}^{2 -}

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M n O_{4}^{-} > V O_{2}^{+} > C r_{2} O_{7}^{2 -}

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C r_{3} O_{7}^{2 -} > V O_{2}^{+} > M n O_{4}^{-}

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Solution

The correct option is A $M n O_{4}^{-} > C r_{2} O_{7}^{2 -} > V O_{2}^{+}$
The ions in which the central metal atom is present in the highest oxidation state will have the highest oxidizing power. In $V O_{2}^{+}$ , vanadium is present in the +5 oxidation state, while in $C r_{2} O_{7}^{2 -}$ ion, $C r$ is present in the +6 oxidation state. Similarly in the $M n O_{4}^{-}$ , $M n$ is present in the +7 oxidation state.

Oxidizing power depends upon the stability of the reduction product obtained. Since lower oxides of $M n$ are more stable than that of $C r$ and $V$ .

So the order of increasing stability of the reduced species is $M n^{2 +} > C r^{3 +} > V^{3 +}$

The correct descending order of oxidizing power is:
$M n O_{4}^{-} > C r_{2} O_{7}^{2 -} > V O_{2}^{+}$

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Similar questions

V O_{2}^{+} < C r_{2} O_{7}^{2 -} < M n O_{4}^{-}

V O_{2}^{+}, M n O_{4}^{-}, C r_{2} O_{7}^{2 -}

V O_{2}^{+}

C r_{2} O_{7}^{2 -}

M n O_{4}^{2 -}

C r_{2} O_{7}^{2 -} + H^{+} + I^{-} \to C r^{3 +} + I_{2} + H_{2} O

C r_{2} O_{7}^{2 -} + F e_{2}^{+} + H^{+} \to C r^{3 +} + F e^{3 +} + H_{2} O

M n O^{4 -} + S O_{3}^{2 -} + H^{+} \to M n^{2 +} + S O_{4}^{2 -} + H_{2} O

M n O^{4 -} + H^{+} + B r^{-} \to M n^{2 +} + B r_{2} + H_{2} O

Complete and balance the following reactions:

(i) ${C r_{2} O_{7}}^{2 -} + H^{+} + I^{-} \to$

(ii) ${M n O_{4}}^{-} + N O_{2} + H^{+} \to$

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