SATHEE: Redox Reactions - Result Question 7

Redox Reactions - Result Question 7

####7. 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 [2003]

(a) $S_{2} O_{6}^{2 -} < S_{2} O_{4}^{2 -} < S O_{3}^{2 -}$

(b) $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$

(d) $S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -} < S O_{3}^{2 -}$

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Solution:

(b) $S O_{3}^{2 -} \to S$ 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.

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Redox Reactions - Result Question 7

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