`KMnO_(4)` (purple) is reduced to `K_(2)MnO_(4)` (green) by `SO_(3)^(2-)` in basic medium. `1` mole of `KMnO_(4)` is reduced by

Options

A. `1` mole of `SO_(3)^(2-)`
B. `2` mole of `SO_(3)^(2-)`
C. `1.5` mole of `SO_(3)^(2-)`
D. `0.5` mole of `SO_(3)^(2-)`

Answer

D

Explanation

Multiply by change in oxidation number
`2MnO_(4)^(-)+SO_(3)^(2-) rarr SO_(4)^(2-)+2MnO_(4)^(2-)`
Thus, `2MnO_(4)^(-)=0.5 SO_(3)^(2-)`
`1 MnO_(4)^(2-)=0.5 SO_(3)^(2-)`