The correct match between Medium of reaction and Equivalent weight of ${\mathrm{KMnO}}_4$ is:

Given data: Medium of reaction and equivalent weight of ${KMnO}_4$ Formula/concept used: Calculation of equivalent weight Detailed explanation: In all the below reactions, $Mn$ in ${MnO}_4{ }^{-}$is in $+7$ oxidation state.

$\begin{array}{l}5e^{-}+{MnO}_4{}^{-}\to {Mn}^{+2}\text{ (in acidic medium) }\\ 1e^{-}+{MnO}_4{}^{-}\to {MnO}_4{}^{2-}\text{ (in basic medium) }\\ 3e^{-}+{MnO}_4{}^{-}\to {MnO}_2\text{ (in neutral medium) }\end{array}$

In acidic medium,

Change in oxidation state = (7-2) = 5

Or no. of electrons participated in the reaction $=5$ Equivalent weight of ${KMnO}_4=158/5=31.6$ In basic medium, Change in oxidation state $=(7-6)=1$ Or no. of electrons participated in the reaction $=1$ Equivalent weight of ${KMnO}_4=158/1=158$. In neutral medium, (or weakly alkaline) Change in oxidation state $=(7-4)=3$ Or no. of electrons participated in the reaction $=3$ Equivalent weight of ${KMnO}_4=158/3=52.67$