In polythionic acid, H2SxO6 (x = 3 to 5) the oxidation state(s) of sulphur is/are:

Step 1 — Structure of polythionic acids $\mathrm{H_2S_xO_6}$:

Polythionic acids have the general structure: $\mathrm{HO_3S - (S)_{x-2} - SO_3H}$

The two terminal S atoms are bonded to oxygen (like in $\mathrm{SO_3^{2-}}$), while the middle $(x-2)$ S atoms form a chain with no oxygen.

Step 2 — Oxidation states:

• Terminal S atoms (bonded to O): Each terminal S is in a $-\mathrm{SO_3H}$ group. In $\mathrm{SO_3H}$: $x + 3(-2) + 1 = 0$ (considering the S-S bond contributes 0) → effectively $x = +5$

• Middle S atoms (in S–S chain, no oxygen): These S atoms have oxidation state = 0 (similar to elemental sulfur)

Step 3 — Conclusion: Polythionic acids contain S in two oxidation states: 0 (middle chain S) and +5 (terminal S).

Answer: Option (3) — 0 and +5 only