Question

The equivalent weight of $MnS{O}_4$ is half its molecular weight when it is converted to:

• A. ${Mn}_2{O}_3$
• B. $Mn{O}_2$
• C. $Mn{O}_4^{\ominus }$
• D. $Mn{O}_4^{2-}$

Answer 1

The correct option is B $Mn{O}_2$

$\stackrel{+2}{Mn}S{O}_4^{2-}⟶\stackrel{+4}{Mn}{O}_2+2e^{-}$

Here, change in oxidation of $Mn=2$.

So, $n-$factor $=2$.

Equivalent weight $=\frac{M}{2}$

Therefore, the equivalent weight of ${MnSO}_4$ is half of its molecular weight when it is converted to ${MnO}_2$