Redox Reactions - Result Question 7
####7. 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 [2003]
(a) $S_2 O_6{ }^{2-}<S_2 O_4{ }^{2-}<SO_3{ }^{2-}$
(b) $S_2 O_4{ }^{2-}<SO_3{ }^{2-}<S_2 O_6{ }^{2-}$
(c) $SO_3{ }^{2-}<S_2 O_4{ }^{2-}<S_2 O_6{ }^{2-}$
(d) $S_2 O_4{ }^{2-}<S_2 O_6{ }^{2-}<SO_3{ }^{2-}$
Show Answer
Solution:
- (b) $SO_3^{2-} \to S$ is in +4 oxidation state $S_2 O_4^{2-} \to S$ is in +3 oxidation state $S_2 O_6^{2-} \to S$ is in +5 oxidation state
The structure of $S_2 O_4{ }^{2-}$ and $S_2 O_6{ }^{2-}$ are symmetrical. Thus, both sulphur atoms are in same oxidation state. This is not the case with $S_2 O_3{ }^{2-}$ or $S_4 O_6{ }^{2-}$ ions.
