Question

Choose correct order of oxidizing power:-

• A. $V{O}_2^{+}$ < $C{r}_2{O}_7^{2-}$ < $Mn{O}_4^{-}$
• B. $C{r}_2{O}_7^{2-}$ <$V{O}_2^{+}$ < $Mn{O}_4^{-}$
• C. $Mn{O}_4^{-}$ < $C{r}_2{O}_7^{2-}$ < $V{O}_2^{+}$
• D. $V{O}_2^{+}$ < $Mn{O}_4^{-}$ < $C{r}_2{O}_7^{2-}$

Answer 1

The correct option is A $V{O}_2^{+}$ < $C{r}_2{O}_7^{2-}$ < $Mn{O}_4^{-}$

This is due to the increasing stability of the lower species to which they are reduced.
$Mn{O}_4^{-}$ $\left(M{n}^{7+}\right)$ is reduced to $\left(M{n}^{2+}\right)$ which is having half-filled stable $d^5$ configuration.