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Q)

The equivalent weight of $ MnSO_4$ is half its molecular weight when it is converted to

$(a)\;Mn_2O_3 \\ (b)\;MnO_2 \\(c)\;MnO_4 \\(d)\;MnO_4^2$

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A)
Equivalent weight $=\large\frac{Molecular\; weight}{Change\;in\;O.N\;of\; Mn}$
$\qquad= \large\frac{Molecular\;weight}{2}$
Thus change in oxidation state of $Mn$ must be 2.
Oxidation state of $Mn$ in $MnSO_4=+2$
So, the oxidation state of $Mn$ in the new compound should be $+4$
The oxidation states of $Mn$ in $MnO_3,MnO_2.MnO_4^{-}$ and $MnO_4^{2-}$.
Compounds are $+3,+4,+7$ and respectively
Hence b is the correct answer.
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