Question

The oxidation state of sulphur in the anions $S_2{O}_4^{2-}$, $S_2{O}_4^{2-}$ and $S_2{O}_6^{2-}$ follows the order:

• A. $S_2{O}_6^{2-}$-<$S_2{O}_4^{2-}$<$S{O}_3^{2-}$
• B. $S_2{O}_4^{2-}$<$S{O}_3^{2-}$<$S_2{O}_6^{2-}$
• C. $S{O}_3^{2-}$<$S_2{O}_4^{2-}$<$S_2{O}_6^{2-}$
• D. $S_2{O}_4^{2-}$<$S_2{O}_6^{2-}$-<$S{O}_3^{2-}$

Answer 1

The correct option is B $S_2{O}_4^{2-}$<$S{O}_3^{2-}$<$S_2{O}_6^{2-}$

Oxidation state of $S_2{O}_4^{2-}$

$2\left(x\right)+4(-2)=-2$

$2x=8-2$

$2x=6$

$x=3$

Oxidation state of $S{O}_3^{2-}$

$x+3(-2)=-2$

$x=6-2$

$x=4$

Oxidation state of $S_2{O}_6^{2-}$

$2\left(x\right)+6(-2)=-2$

$2x=12-2$

$2x=10$

$x=5$

So the oxidation state of sulphur in the anions $S_2{O}_4^{2-}$, $S_2{O}_4^{2-}$ and $S_2{O}_6^{2-}$ follows the order.$S_2{O}_4^{2-}$<$S{O}_3^{2-}$<$S_2{O}_6^{2-}$.

Hence option B is correct.