Some Basic Concepts of Chemistry 2 Question 23
23. The equivalent weight of $\mathrm{MnSO}_{4}$ is half of its molecular weight, when it converts to
$(1988,1 \mathrm{M})$
(a) $\mathrm{Mn}{2} \mathrm{O}{3}$
(b) $\mathrm{MnO}_{2}$
(c) $\mathrm{MnO}_{4}^{-}$
(d) $\mathrm{MnO}_{4}^{2-}$
Objective Question II (More than one correct option)
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Solution:
- Equivalent weight in redox system is defined as :
$$ E=\frac{\text { Molar mass }}{n \text {-factor }} $$
Here $n$-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, $n$-factor is 2 because equivalent weight is half of molecular weight. Also,
$$ \begin{aligned} & \boldsymbol{n} \text {-factor } \mathrm{MnSO}{4} \longrightarrow \frac{1}{2} \mathrm{Mn}{2} \mathrm{O}{3} \quad 1(+2 \longrightarrow+3) \ & \mathrm{MnSO}{4} \longrightarrow \mathrm{MnO}{2} \quad 2(+2 \longrightarrow+4) \ & \mathrm{MnSO}{4} \longrightarrow \mathrm{MnO}{4}^{-} \quad 5(+2 \longrightarrow+7) \ & \mathrm{MnSO}{4} \longrightarrow \mathrm{MnO}_{4}^{2-} \quad 4(+2 \longrightarrow+6) \end{aligned} $$
Therefore, $\mathrm{MnSO}{4}$ converts to $\mathrm{MnO}{2}$.
