Question

The oxidation state of sulphur in the anions ${SO}_3^{2-},{SO}_4^{2-},S_2{O}_4^{2-},S_2{O}_6^{2-}$ is in the order :

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

Answer 1

The correct option is C ${SO}_4^{2-}>S_2{O}_6^{2-}>{SO}_3^{2-}>S_2{O}_4^{2-}$

Consider the oxidation state of sulphur as $x$. We know that the charge on oxygen atom is $-2$. Hence the oxidation states of sulphur in the given four anions are calculated as follows:

$S{O}_3^{2-}\to x+(-6)=-2\Rightarrow x=+4$
$S{O}_4^{2-}\to x+(-8)=-2\Rightarrow x=+6$
$S_2{O}_4^{2-}\to 2x+(-8)=-2\Rightarrow x=+3$
$S_2{O}_6^{2-}\to 2x+(-12)=-2\Rightarrow x=+5$
Hence $S{O}_4^{2-}>S_2{O}_6^{2-}>S{O}_3^{2-}>S_2{O}_4^{2-}$