Question

The oxidation states of S-atoms in $S_4{O}_6^{2-}$ form left to right respectively are:

• A. + 6, 0, 0, + 6
• B. + 3, 1, + 1, + 3
• C. + 5, 0, 0, +5
• D. +4, +1, + 1, +4

Answer 1

Answer: (C)

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, $S{O}_3$

$S_2^{2-}+2S{O}_3\to S_4{O}_6^{2-}$

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

$(2\times 5+2\times 0)/4=2.5$