Question

The oxidation states of sulphur in the anions, $S{O}_3^{2-}$, $S_2{O}_4^{2-}$ and $S_2{O}_6^{2-}$ follow the order :

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

Answer 1

Answer: (A)

$S_2{O}_4^{2-}$<$S{O}_3^{2-}$<$S_2{O}_6^{2-}$

Sulphur in $S{O}_3^{-2},S_2{O}_4^{-2},$ and $S_2{O}_6^{-2}$ has $+4$, $+3$ and $+5$ oxidation state respectively.
Hence, the order is $S_2{O}_4^{2-}$<$S{O}_3^{2-}$<$S_2{O}_6^{2-}$.