When sulphur dioxide is passed in an acidified `K_(2)Cr_(2)O_(7)` solution, the oxidation state of sulphur is changed from:

Options

A. `+4` to `+6`
B. `+6` to `+4`
C. `+4` to 0
D. `+4` to `+2`

Explanation

`ul(K_(2) Cr_(2)O_(7) +4H(2) SO_(4)rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 7H_(2)O+ 3(O))`
`SO_(2) + (O) + H_(2) O rarr H_(2) SO_(4) ] xx3`
`K_(2)Cr_(2)O_(7) + H_(2)SO_(4) + 3SO_(2) rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + H_(2)O`
In `SO_(2)` , ox. state of `S= + 4 `. In `SO_(4)^(2-)`,ox.state of `S= +6`