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

Options

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

Answer

B

Explanation

Equivalent weight in redox system is defined as :
`E = ("Molar mass")/("n-factor")`
Here n-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, n-factor is because equivalent weight is half of molecular weight. Also,
`{:("n-factor",MnSO_(4),rarr,(1)/(2)Mn_(2)O_(3),1(+2 rarr +3)),(,MnSO_(4),rarr,MnO_(2),2(+2 rarr +4)),(,MnSO_(4),rarr ,MnO_(4)^(-),5(+2 rarr +7)),(,MnSO_(4),rarr,MnO_(4)^(2-),4(+2 rarr +6)):}`
Therefore, `MnSO_(4)`converts to `MnO_(2)`.