1. Write the general formula

Polythionic acids have formula: $H_2S_xO_6 \quad (x=3\text{ to }5)$

We need the oxidation states of sulphur atoms.

2. Assign known oxidation states

• Hydrogen: $+1$ each
• Oxygen: $-2$ each

Let the sum of oxidation states of all sulphur atoms be $S_{\text{total}}$.

Since the molecule is neutral, $2(+1) + S_{\text{total}} + 6(-2)=0$ $2 + S_{\text{total}} -12 =0$ $S_{\text{total}}=+10$

So, in every polythionic acid $H_2S_xO_6$, the total oxidation number of all sulphur atoms together is $+10$.

3. Use the structure of polythionic acid

Polythionic acids have the chain structure: $HO_3S-S-S-\cdots-S-SO_3H$

That means:

• The two terminal sulphur atoms are bonded to oxygen atoms as $-SO_3H$ groups.
• The inner sulphur atoms are only bonded to sulphur atoms.

4. Find oxidation state of terminal sulphur atoms

For each terminal sulphur in $-SO_3H$:

• three $S-O$ bonds make sulphur positive
• one $S-S$ bond contributes $0$
• one $S-OH$ is also counted through oxygen

This corresponds to oxidation state: $+5$

So, the two terminal sulphur atoms are each at $+5$.

Their total contribution is: $2\times(+5)=+10$

5. Find oxidation state of inner sulphur atoms

Since total oxidation state of all sulphurs is $+10$, the remaining inner sulphur atoms must together contribute: $+10-(+10)=0$

Each inner sulphur is bonded only to sulphur atoms. Since bonds between identical atoms contribute zero oxidation number, $\text{oxidation state of each inner sulphur} = 0$

6. Conclusion

Thus, sulphur shows oxidation states: $0 \text{ and } +5$

So the correct option is: D: $0$ and $+5$ only.

7. Comparison with stored correct answer

Stored correct answer = D

This matches our derived answer.