Question

$Mn{O}_4^{2-}$ undergoes disproportionation reaction in acidic medium but $Mn{O}_4^{-}$ does not. Give reason.

Answer 1

Disproportionation reaction : Reactions in which the same species gets oxidised and reduced simultaneously.

The highest oxidation state of $Mn=+7$
Oxidation state of $Mn$ in $Mn{O}_4^{2-}$
$x+4\times (-2)=-2$
$x-8=-2$
$x=+6$
Thus, Mn is in $+6$ oxidation state in $Mn{O}_4^{2-}$.

Oxidation state of $Mn$ in $Mn{O}_4^{-}$
$x+4\times (-2)=-1$
$x-3=-1$
$x=+7$
Thus, $Mn$ is in $+7$ oxidation state.

In $Mn{O}_4^{-},Mn$ is in highest oxidation state $(+7)$, and thus cannot oxidise further. It can only get reduced, while in $Mn{O}_4^2-$,

$Mn$ is in $+6$ oxidation state due to which it can undergo reduction and oxidation both.
Hence, $Mn{O}_4^{2-}$ undergoes disproportionation reaction in acidic medium.

$3Mn{O}_4^{2-}+4H^{+}\to 2Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$

Oxidation state:
Reactant Product
$Mn=+6H=+1Mn=+7Mn=+4H=+1$
$O=-2O=-2O=-2O=-2$

Here, $Mn$ is getting oxidised from $+6$ oxidation state in $Mn{O}_4^{2-}$ to $+7$ oxidation state in $Mn{O}_4^{-}$ in oxidation state.

It is also getting reduced from $+6$ oxidation state in $Mn{O}_4^{2-}$ to $+4$ oxidation state in $Mn{O}_2$.