In the thiosulphate ion $\mathrm{S}_2\mathrm{O}_3^{2-}$, what is the AVERAGE oxidation number of sulphur?
A$0$
B$+2$
C$+4$
D$+6$
Answer & Solution
Correct answer: B. $+2$
1. The species $\mathrm{S}_2\mathrm{O}_3^{2-}$ carries an overall charge of $-2$.
2. Therefore the algebraic sum of oxidation numbers of all atoms equals $-2$.
3. Let the average oxidation number of S be $x$. With 2 S atoms and 3 O atoms ($-2$ each):
$2x + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2$.
4. This is an AVERAGE — the two S atoms in thiosulphate are actually inequivalent: one is central and S(VI)-like ($+5$), the other terminal and S(-II)-like ($-1$). The average happens to be $+2$.
5. Your syllabus ("Paradox of Fractional Oxidation Number", §7.3) explicitly notes that average values like this are useful for stoichiometry even when individual atoms are in different states.
_Source: syllabus Chemistry Part 2, Ch 7, §7.3 + "The Paradox of Fractional Oxidation Number" box, p. 10._
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