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Q.
The oxidation states of S-atoms in $S _{4} O _{6}^{2-}$ form left to right respectively are:
Redox Reactions
Solution:
The tetrathionate anion, $S _{4} O _{6}^{2-}$ is a sulphur oxoanion derived from the compound tetrathionic acid, $H _{4} S _{4} O _{6}$ anion $S _{4} O _{6}^{2-}$ can be viewed as the product formed from reaction between lewis base, $S _{2}{ }^{2-}$ and lewis acid, $SO _{3}$
$S _{2}^{2-}+2 SO _{3} \rightarrow S _{4} O _{6}^{2-}$
Two of the sulphur atoms are present in the oxidation state of zero and two are in oxidation state $+5 .$
The structure of $S _{4} O _{6}^{2-}$
Since; each of the terminal sulphur atom is connected to two oxygen atom by a double bond and to one oxygen atom by a single bond, the oxidation state of these terminal sulphur atoms is $+5 .$ Since; two central sulphur atoms are linked to each other by a single bond, and the sulphur is further attached to similar species on either side, the electron pair forming S-S bond remains in the centre and hence; each of the two central sulphur atoms has on oxidation state of zero. Thus; the average oxidation state of the four sulphur atom is
