The equivalent weight of ${\mathrm{MnSO}}_4$ is half its molecular weight when it is converted to

Solution:

$\text{Equivalent\hspace{0.17em}\hspace{0.17em}}\mathrm{weight}=\frac{\mathrm{molecular}\mathrm{weight}}{\mathrm{valency}\mathrm{factor}}$

If valency factor is 2, then equivalent weight will be equal to half its molecular weight.

In ${\mathrm{MnSO}}_4,$ the oxidation state of Mn is +II

In ${\mathrm{Mn}}_2{\mathrm{O}}_3$, the oxidation state of Mn is +III

In ${\mathrm{MnO}}_2$, the oxidation state of Mn is +IV

In ${\mathrm{MnO}}_4^{-}$, the oxidation state of Mn is +VII

In ${\mathrm{MnO}}_4^{2-}$, the oxidation state of Mn is +VI

Thus, when ${\mathrm{MnSO}}_4$ is converted into ${\mathrm{MnO}}_2$, then the valency factor is 2, and the equivalent weight of ${\mathrm{MnSO}}_4$ will be half of its molecular weight.