Choose correct order of oxidizing power:-

VO₂⁺ < Cr₂O₇^2- < MnO4^-

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Cr₂O₇^2-2⁺ < MnO₄^-

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MnO₄^- < Cr₂O₇^2-+

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VO₂⁺ < MnO₄^- < Cr₂O₇^2-

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Solution

The correct option is A VO₂⁺ < Cr₂O₇^2- < MnO4^-
This is due to the increasing stability of the lower species to which they are reduced.
$M n O_{4}^{-}$ $(M n^{7 +})$ is reduced to $(M n^{2 +})$ which is having half-filled stable $d^{5}$ configuration.

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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 -}

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

V O_{2}^{+}

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

M n O_{4}^{2 -}

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