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

Options

A. ` S_2 O_6^(2-) lt S_2 O_4^(2-) lt SO_3^(2-)`
B. ` S_2O_4^(2-) lt SO_3^(2-) lt S_2O_6^(2-)`
C. `SO_3^(2-) lt S_(2)O_(6)^(2-) lt SO_(3)^(2-)`
D. ` S_2O_4^(2-) lt S_2O_6^(2-) lt SO_3^(2-)`

Answer

B

Explanation

We can work out the oxidation number of S by using the fact that in polyatomic ions, the algebraic sum of all the oxidation numbers must equal the charge on the ion.
Let the oxidation number of S be x.
(a) `SO_3^(2-) rArr + 3 (-2) =- 2`
`x- 6=-2`
` x= +4`
(b) `S_2O_4^(2-) rArr 2x +4 (-2) =-2`
` 2 x - 8 =- 2`
` x= +3`
(c) ` S_2O_6^(2-) rArr 2x + 6(-2) =-2`
` 2x- 12 =-2`
`x= +5`
Thus, the order of oxidation number is
`overset(+3)(S_(2))O_(4)^(2-) lt overset(+4)(S)O_(3)^(2-) lt overset(+5)(S_(2))O_(6)^(2-)`