What are the respective oxidation states of sulphur atoms numbered 1 to 4 in tetrathionate ion shown below?

The oxidation states of sulfur atoms in $S_4O_6^{2-}$ are determined from the structure $^-O\text{--}S^1(=O)_2\text{--}S^2\text{--}S^3\text{--}S^4(=O)_2\text{--}O^-$ by applying electronegativity-based bond contributions.
Each S–O bond contributes +1 to sulfur where oxygen is more electronegative; a double bond contributes twice this value. S–S bonds contribute zero.

For $S^1$: contributions from one S–O single bond (+1), two S=O double bonds ($2 \times +2$), and one S–S bond (0), so oxidation state $= 1 + 4 + 0 = +5$.

By symmetry, $S^4$ is identical to $S^1$ and thus also has oxidation state +5.

For $S^2$: two S–S bonds, both contributing 0, so oxidation state $= 0$.

By symmetry, $S^3$ is identical to $S^2$ and also has oxidation state 0.

The sum confirms the overall charge: $+5 + 0 + 0 + 5 + 6 \times (-2) = +10 - 12 = -2$, as required.

The oxidation states are +5, 0, 0, +5, so the correct multiple-choice answer is $\boxed{B}$.