The chemical reaction ${MnO}_{\mathit{4}}^{\mathit{-}}+e^{\mathit{-}}\rightleftharpoons Mn{O}_{\mathit{4}}^{\mathit{2}\mathit{-}}$ takes place in.

In an acidic medium, there is a gain of 5 electrons from Mn

${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

In a neutral medium, there is a gain of 3 electrons from Mn

${\mathrm{MnO}}_4^{-}+2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{MnO}}_2+4{\mathrm{OH}}^{-}$

But in the case of the basic medium, there is a gain of only one electron while conversion.

${\mathrm{MnO}}_4^{-}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{MnO}}_4^{2-}$

Hence, the correct answer is option (A) A basic medium.