Question

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

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

Answer 1

Answer: (B)

Equivalent weight in redox system is defined as :
$E=\frac{\text{Molar mass}}{\text{n-factor}}$
Here $n$-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, $n$-factoris $2$ because equivalent weight is half of molecular weight. Also,
$n$-factor $MnS{O}_4\to \frac{1}{2}M{n}_2{O}_{3}1(+2\to +3)$
$MnS{O}_4\to Mn{O}_{2}2(+2\to +4)$
$MnS{O}_4\to Mn{O}_4^{-}5(+2\to +7)$
$MnS{O}_4\to Mn{O}_4^2-4(+2\to +6)$
Therefore , MnS$O_4$ converts to $Mn{O}_2$