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

Options

A. `MnO_(4)^(2-)`
B. `Mn_(O_(3))`
C. `MnO_(2)`
D. `MnO_(1)^(-)`

Answer

3

Explanation

`overset(+2)(M)nSO_(4) rarr overset(+6)(M)nO_(4)^(2-)`
`overset(+2)(M)nSO_(4) rarr overset(+3)(M)n_(2)O_(3)`
`overset(+2)(M)nSO_(4) rarr overset(+4)(M)nO_(2)`
`overset(+2)(M)nSO_(4) rarr overset(+6)(M)nO_(4)^(-)`
Only in `3^(rd)` option, change in oxidation number `(ON)` of `Mn` per formula unit of `MnSO_(4)=2`. Thus,
Equivalent mass of `MnSO_(4)=("Formula mass")/2`