The oxidation states of sulphur in the anions `SO_(3)^(-2),S_(2)O_(4)^(2-)andS_(2)O_(6)^(2-)` follow the order :

Answer

B, C, D

Explanation

For neutral molecule, sum of oxidation number of all the atoms is equal to zero and for a charged species equals to the charge on the ion.
In `H_(2)S`, the oxidation number of `H` is `+1`,
`H_(2)S=2+x-0`
`x= -2`
Oxidation number of `S= -2`
In `H_(2)SO_(4)`, the oxidation numoer of `H` is `+1`, for `O` it is `H_(2)SO_(4)= 2+x-8= 0`
`x=+6`
Oxidation number of `S` is `+6`.
In `S_(2)O_(4)^(2-)`, oxidation number of `O` is `-2`,
`S_(2)O_(4)^(2-): , 2x-8= -2`
`x=3`
`:.` oxidation numbers of `S` is `+3`
In `S_(2)O_(8)^(2-)`, we have one peroxide bond, therefore, for two oxygens atom, oxidation number is `-1` and for the order six oxygen atoms, the oxidation number is `-2`.
or oxidation number of `S=6`.
In `HSO_(3)^(ө)`, oxidation number for `H` is `+1` and oxidation number of `O` is `-2`
`1+`(oxidation number of S) `+3(-2)= -1` or oxidation number of `S= +4`
It may be noted that sulphur is in different oxidation number states in different compounds. The is true in the next example also in which `Cl` is in different oxidation state.