Which of the following oxoacids of sulphur contains "S" in two different oxidation states?

We need to find which oxoacid of sulphur has sulphur in two different oxidation states.

In H$$_2$$S$$_2$$O$$_3$$ (thiosulphuric acid), the structure is derived from H$$_2$$SO$$_4$$ by replacing one oxygen with a sulphur atom. The two sulphur atoms are not equivalent — one is in the central position bonded to oxygen atoms, and the other is the terminal sulphur replacing oxygen. For the central sulphur, it is bonded to 3 oxygen atoms and 1 sulphur atom, giving it an oxidation state of +5. For the terminal sulphur, it replaces an oxygen atom (O$$^{2-}$$), so its oxidation state is −1. Verification: (+1)$$\times$$2 + (+5) + (−1) + (−2)$$\times$$3 = 2 + 5 − 1 − 6 = 0 $$\checkmark$$. Sulphur exists in two different oxidation states (+5 and −1).

In H$$_2$$S$$_2$$O$$_6$$ (dithionic acid), the structure is HO$$_3$$S-SO$$_3$$H and both sulphur atoms are in identical environments. Using the formula: 2(+1) + 2x + 6(−2) = 0 $$\Rightarrow$$ 2 + 2x − 12 = 0 $$\Rightarrow$$ x = +5, so both sulphur atoms have oxidation state +5 and only one oxidation state.

In H$$_2$$S$$_2$$O$$_7$$ (pyrosulphuric/disulphuric acid), the structure is HO-SO$$_2$$-O-SO$$_2$$-OH and both sulphur atoms are in identical environments. Using the formula: 2(+1) + 2x + 7(−2) = 0 $$\Rightarrow$$ 2 + 2x − 14 = 0 $$\Rightarrow$$ x = +6, so both sulphur atoms have oxidation state +6 and only one oxidation state.

In H$$_2$$S$$_2$$O$$_8$$ (peroxodisulphuric acid/Marshall's acid), the structure is HO-SO$$_2$$-O-O-SO$$_2$$-OH and it contains a peroxide (O-O) linkage. Accounting for the peroxide oxygen atoms (each −1): 2(+1) + 2x + 6(−2) + 2(−1) = 0 $$\Rightarrow$$ 2 + 2x − 12 − 2 = 0 $$\Rightarrow$$ x = +6, so both sulphur atoms have oxidation state +6 and only one oxidation state.

Only H$$_2$$S$$_2$$O$$_3$$ has sulphur in two different oxidation states (+5 and −1). Hence, the correct answer is Option A.