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

Options

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

Answer

B

Explanation

Eq. mass = `("Molecular weight")/("Change in oxidation no of Mn")=("Mol.wt.")/(4-2)=("Mol. Wt.")/(2)`
(O.N. of Mn in `MnSO_(4)=+2`,O.N. of Mn in `MnO_(2)=+4`).