Question

The equivalent weight of ${MnSO}_4$ is said to be half its molecular weight, when it is converted to:

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

Answer 1

The correct option is B ${MnO}_2$

Equivalent mass is half of molecular mass. Therefore, valency factor of $MnS{O}_4=2$.

Now, oxidation state of Mn in $MnS{O}_4=+2$

Oxidation in the resultant compound should be 0 or +4.

In $Mn{O}_2$, the oxidation state of Mn is +4.

Hence, option B satisfies the condition.