What is the AVERAGE oxidation number of sulphur in the tetrathionate ion $\mathrm{S}_4\mathrm{O}_6^{2-}$?

1. The species $\mathrm{S}_4\mathrm{O}_6^{2-}$ has overall charge $-2$, so the sum of oxidation numbers of all atoms equals $-2$. 2. Let the average oxidation number of S be $x$. There are 4 S atoms and 6 O atoms (each O is $-2$): $4x + 6(-2) = -2$. 3. Simplify: $4x - 12 = -2 \Rightarrow 4x = +10 \Rightarrow x = +2.5$. 4. The fractional value is an AVERAGE — in the actual tetrathionate structure $\mathrm{S–S(O_3)–(O_3)S–S}$, two terminal S atoms are $+5$-like and two central S atoms are $0$, averaging to $+2.5$. 5. Option A is the average for $\mathrm{S}_2\mathrm{O}_3^{2-}$ (thiosulphate). Options C and D are common arithmetic slips (subtracting instead of adding, or wrong charge). _Source: NCERT Class 11 Chemistry Part 2, Ch 7, §7.3 + "Paradox of Fractional Oxidation Number" sidebar, p. 10._