Which of the following reactions represents disproportionation ?

Options

A. `CrO_(5) rarr Cr^(3+)+O_(2)`
B. `IO_(3)^(-)+I^(-)+H^(+)rarr I_(2)`
C. `CrO_(2)Cl_(2)+NaOHrarr Na_(2)CrO_(4)+NaCl+H_(2)O`
D. `Na_(2)S_(2)O_(3)+H_(2)SO_(2)rarr Na_(2)SO_(4)+SO_(2)+S_(8)+H_(2)O`

Answer

4

Explanation

`Na_(2)S_(2)O_(3)+H_(2)SO_(4)rarr Na_(2)SO_(4)+SO_(2)+S_(8)+H_(2)O`
Oxidation No. of `S` in `Na_(2)S_(2)O_(3)=+2`
Oxidation No. of `S` in `H_(2)SO_(4)=+6`
Oxidation No. of `S` in `SO_(4)=+4`
Oxidation No. of `S` in `Na_(2)SO_(4)=+6`
Oxidation No. of `S` in `S_(8)=0`
So, Sulphur can oxidise from `+2` to `+4` as well as reduce to `0(S_(8))`.