Equivalent weight of $Mn{O}_4^{⊝}$ in acidic, neutral and basic media are in ratio of :

The correct option is C $3:5:15$

Equation for reduction of ${Mn{O}_4}^{-}$ in acidic medium:

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

Here, the oxidation number is changing from +7 to +2, the difference is 5.

$\therefore Ew=\frac{Mw}{5}$

In neutral medium:

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

Here, the oxidation number is changing from +7 to +4, the difference is 3.

$\therefore Ew=\frac{Mw}{3}$

In basic medium:

${Mn{O}_4}^{-}+e^{-}\to {Mn{O}_4}^{2-}$

Here, the oxidation number is changing from +7 to +6, the difference is 1.

$\therefore Ew=\frac{Mw}{1}$

Ratio is $\frac{1}{5}:\frac{1}{3}:\frac{1}{1}=3:5:15$