In polythionic acid, H$$_2$$S$$_x$$O$$_6$$ ($$x$$ = 3 to 5) the oxidation state(s) of sulphur is/are:

We start with the general molecular formula of polythionic acid: $$\mathrm{H_2S_xO_6},$$ where $$x = 3, 4, 5.$$

The experimental structures (deduced from salts, X-ray data and stepwise oxidation studies) show a straight chain of sulphur atoms. The two end sulphur atoms are each bonded to three oxygens and one sulphur, forming $$\text{-SO}_3\text{H}$$ groups. The remaining $$x-2$$ sulphur atoms are present only in the S-S chain and are not bonded to oxygen.

Whenever a sulphur atom is bonded to three oxygens in the $$\text{-SO}_3\text{H}$$ environment (exactly like in sulphurous acid fragments), its oxidation number is taken as an unknown $$+a.$$ Each “internal” sulphur that is attached only to neighbouring sulphur atoms carries no hetero-atoms, so its oxidation number is zero.

Hence, in $$\mathrm{H_2S_xO_6}$$ we have:

• 2 hydrogen atoms, each with oxidation number $$+1.$$

• 6 oxygen atoms, each with oxidation number $$-2.$$

• 2 terminal sulphur atoms with oxidation number $$+a$$ (to be found).

• $$(x-2)$$ internal sulphur atoms with oxidation number $$0.$$

Because the molecule is electrically neutral, the algebraic sum of all oxidation numbers must be zero. Writing this balance, we get:

$$2(+1) \;+\; 2(+a) \;+\; (x-2)(0) \;+\; 6(-2) \;=\; 0.$$

Simplifying step by step:

$$2 + 2a + 0 - 12 = 0,$$

$$2a - 10 = 0,$$

$$2a = 10,$$

$$a = +5.$$

Thus each terminal sulphur is in the $$+5$$ oxidation state, while every internal sulphur is in the $$0$$ oxidation state.

Therefore, the possible oxidation states of sulphur found in any polythionic acid $$\mathrm{H_2S_xO_6}$$ are $$0$$ and $$+5$$ only.

Hence, the correct answer is Option C.