Question

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

• A. ${Mn}_2{O}_3$
• B. $Mn{O}_2$
• C. ${MnO}_4^{-}$
• D. none of these

Answer 1

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

For equivalent weight of $MnS{O}_4$ to be half of its molecular weight, change in oxidation state must be equal to $+2$. It is possible only when oxidation state of $Mn$ in product is $+4$. Since oxidation state of $Mn$ in $MnS{O}_4$ is $+2$

In $Mn{O}_2$ Oxidation state of $Mn=+4$

Change in Oxidation state of $Mn=+4-(+2)=+2$