Arrange the following in the decreasing order of oxidising power
$A = M n {O_{4}}^{-}$
$B = C r_{2} {O_{7}}^{2 -}$
$C = V {O_{2}}^{+}$

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$B > A > C$

$C > A > B$

$C > B > A$

$A > B > C$

answer is A.

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Detailed Solution

The ions in which the central metal atom is present in the highest oxidation state will have the highest oxidizing power. In VO₂+, vanadium is present in the +5 oxidation state, while in Cr₂O₇²⁻ion, Cr is present in the +6 oxidation state. Similarly in the MnO₄⁻, Mn is present in the +7 oxidation state.

Oxidizing power depends upon the stability of the reduction product obtained. Since lower oxides of Mn are more stable than that of Cr and V.

So the order of increasing stability of the reduced species is Mn²⁺>Cr³⁺>V³⁺

The correct descending order of oxidizing power is:
MnO₄⁻>Cr₂O₇²⁻>VO₂⁺

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