The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted to

Options

A. (a)`Mn_(2)O_(3)`
B. (b)`MnO_(2)`
C. (c )`MnO_(4)^(-)`
D. (d)`MnO_(4)^(2-)`

Answer

B

Explanation

Equivalent `"weight"=("molecular weight")/("valency factor")`
If valency is 2, then equivalent weight will be equal to its molecular weight.
In `MnSO_(4)`, the oxidation state of Mn is +II
In `Mn_(2)O_(3)`, the oxidation state of Mn is +III
In `MnO_(2)`, the oxidation state of Mn is +IV
In `MNO_(4)`, the oxidation state of Mn is +VII
In `Mn_(4)^(2-)`, the oxidation state of Mn is +VI
Thus, when `MnSO_(4)` is converted into `MnO_(2)`, then the valency factor is 2, and the equivalent weight of `MnSO_(4)` will be half of its molecular weight.