Question

The equivalent weight of \(\mathrm{MnSO}_{4}\) is half its molecular weight when it is converted to
(a) \(\mathrm{Mn}_{2} \mathrm{O}_{3}\)
(b) \(\mathrm{MnO}_{2}\)
(c) \(\mathrm{MnO}_{4}^{-}\)
(d) \(\mathrm{MnO}^{2-}{ }_{4}\)

Answer 1

Correct option (b) \(\mathrm{MnO}_{2}\)
Explanation:
Equivalent weight = molecular weight/valency factor
If valency factor is 2, then equivalent weight will be equal to its molecular weight.
In \(\mathrm{MnSO}_{4}\), the oxidation state of \(\mathrm{Mn}\) is +ll
In \(\mathrm{Mn}_{2} \mathrm{O}_{3}\), the oxidation state of \(\mathrm{Mn}\) is \(+I I I\)
In \(\mathrm{MnO}_{2}\), the oxidation state of \(\mathrm{Mn}\) is + IV
In \(\mathrm{MnO}_{4}^{-}\), the oxidation state of \(\mathrm{Mn}\) is + VII
In \(\mathrm{MnO}^{2-} 4\), the oxidation state of \(\mathrm{Mn}\) is \(+\mathrm{VI}\)
Thus, when \(\mathrm{MnSO}_{4}\) is converted into \(\mathrm{MnO}_{2}\), then the valency factor is 2 , and the equivalent weight of \(\mathrm{MnSO}_{4}\) will be half of its molecular weight.

Answer 2

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Answer 3

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