The correct order of oxidizing power of the g | vedclass

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Question

(A) In $VO_2^{+}$,the oxidation state of $V$ is $+5$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. In $MnO_4^{-}$,the oxidation state of $Mn

Vedclass · Accepted Answer

$VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$

Vedclass · Answer

(A) In $VO_2^{+}$,the oxidation state of $V$ is $+5$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. In $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$. Oxidizing power is related to the tendency of the central metal atom to undergo reduction,which is generally higher for transition metals in higher oxidation states. Comparing the oxidation states: $Mn (+7) > Cr (+6) > V (+5)$. Therefore,the correct order of oxidizing power is $VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$.

The correct order of oxidizing power of the given ions is

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