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

Step-by-step solution

The oxidation states of sulfur atoms in S₄O₆²⁻ are determined from the structure ^-O -- S^1(=O)₂ -- S^2 -- S^3 -- S^4(=O)₂ -- 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 +2 ), and one S–S bond (0), so oxidation state = 1 + 4 + 0 = +5 . By symmetry, S^4 is identical to S^1 and