In neutral or alkaline solution, $$MnO_4^-$$ oxidises thiosulphate to

We need to determine the product when $$MnO_4^-$$ oxidises thiosulphate ($$S_2O_3^{2-}$$) in neutral or alkaline solution.

In neutral or alkaline medium, permanganate is reduced to $$MnO_2$$:

$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \quad \cdots (i)$$

Mn changes from $$+7$$ to $$+4$$, gaining 3 electrons.

In $$S_2O_3^{2-}$$, the average oxidation state of S is:

$$2x + 3(-2) = -2 \implies x = +2$$

The average oxidation state of sulphur is $$+2$$.

With a strong oxidizing agent like $$MnO_4^-$$ in alkaline medium, thiosulphate undergoes complete oxidation to sulphate ($$SO_4^{2-}$$), where S is in the $$+6$$ state:

$$S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-} + 10H^+ + 8e^-$$

Converting to basic medium by adding $$10OH^-$$ to both sides:

$$S_2O_3^{2-} + 5H_2O + 10OH^- \rightarrow 2SO_4^{2-} + 10H_2O + 8e^-$$

Simplifying:

$$S_2O_3^{2-} + 10OH^- \rightarrow 2SO_4^{2-} + 5H_2O + 8e^- \quad \cdots (ii)$$

Each sulphur atom changes from $$+2$$ to $$+6$$, losing 4 electrons per atom (8 electrons total for 2 atoms).

Multiply equation (i) by 8 and equation (ii) by 3 to equalize electrons (24 electrons):

$$8MnO_4^- + 16H_2O + 24e^- \rightarrow 8MnO_2 + 32OH^-$$

$$3S_2O_3^{2-} + 30OH^- \rightarrow 6SO_4^{2-} + 15H_2O + 24e^-$$

Adding both half-reactions:

$$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$$

Conclusion:

In neutral or alkaline solution, $$MnO_4^-$$ completely oxidises thiosulphate to $$SO_4^{2-}$$.

Hence, the correct answer is Option D.