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

Options

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

Answer

C

Explanation

When `SO_(2)` passes through acidified `K_(2)Cr_(2)O_(7)` solution the following reactions take place `K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)rarrK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+4H_(2)O+3[O]`
`[SO_(2)+[O]+H_(2)OrarrH_(2)SO_(4)]xx3`
`K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+3SO_(2)rarrK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O`
In ` SO_(2)`,oxidation state of S is `+4 but "in" SO_(4)^(2)`,oxidation state changes to `+6`.