Acidified `KMnO_(4)` can be decolourised by:

Options

A. `H_(2)O_(2)//H^(+)`
B. `SO_(2)//H^(+)`
C. `C_(2)O^(2)_(4)//H^(+)`
D. All of these

Answer

B

Explanation

`MnO_(4)^(-)+SO_(2)overset(H)rarr "Colourless" Mn^(2+)`
So it cam be bleached by `SO_(2)//H^(+)`.