1. We need to identify the product formed when thiosulphate ion, $\mathrm{S_2O_3^{2-}}$, is oxidised by permanganate, $\mathrm{MnO_4^-}$, in neutral or alkaline medium.

2. In neutral/alkaline solution, permanganate is a strong oxidising agent and commonly gets reduced as: $\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$ This means it can oxidise thiosulphate significantly.

3. Let us examine the oxidation state change in thiosulphate:

• In $\mathrm{S_2O_3^{2-}}$, the two sulphur atoms are not equivalent.
• On strong oxidation, thiosulphate is converted to sulphate, $\mathrm{SO_4^{2-}}$.

A standard oxidation half-reaction for thiosulphate to sulphate in alkaline/neutral conditions is: $\mathrm{S_2O_3^{2-} + 10OH^- \rightarrow 2SO_4^{2-} + 5H_2O + 8e^-}$ This shows that the oxidation product is clearly $\mathrm{SO_4^{2-}}$.

4. Now check the options:

• A: $\mathrm{S_2O_7^{2-}}$ — pyrosulphate; not the usual oxidation product here.
• B: $\mathrm{S_2O_8^{2-}}$ — peroxodisulphate; formed by electrochemical oxidation, not by this reaction.
• C: $\mathrm{SO_3^{2-}}$ — sulphite is a lower oxidation state product, so not an oxidation product of thiosulphate under strong oxidising conditions.
• D: $\mathrm{SO_4^{2-}}$ — correct.

5. Therefore, in neutral or alkaline solution, $\mathrm{MnO_4^-}$ oxidises thiosulphate to: $\boxed{\mathrm{SO_4^{2-}}}$