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When SO(2) is passed through an acidifie...

When `SO_(2)` is passed through an acidified `K_(2)Kr_(2)O_(7)` solution, the oxidation state of sulphur changes from

A

`+4` to `+6`

B

`+6` to `+4`

C

`+4` to 0

D

`+4` to `+2`

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(2)Cr_(2)O_(7)` acts as a strong oxidising agent. It oxidises other compounds and gets itself reduced. In `SO_(2)` sulphure is in +4 state so it gets oxidised to `+6` state which is maximum group oxidation state.
`K_(2)Cr_(2)O_(7)+ 4H_(2)SO_(4)toK_(2)SO_(4)+ Cr_(2) (SO_(4))_(3))+7H_(2)O+35`
`[SO_(2)+ [O]+H_(2)O to H_(2)SO_(4)]xx3`
`K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+3SO_(2)toK_(2)SO_(4)+Cr_(2) (SO_(4))_(3)+H_(2)O`
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