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`

Answer

A

Explanation

`K_(2)Cr_(2)O_(7)` acts as a strong oxidising agent . It oxidises other compounds and gets itself reduced . In `SO_(2)`, S 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+3S[SO_(2)+[O]+H_(2)OtoH_(2)SO_(4)]xx3)/(K_(2)Cr_(2)+[O]+H_(2)SO_(4)+2SO_(2)toK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O)`