Thiosulphate reacts differently with iodine and bromine in the reactions given below:
2 S₂ O₃²⁻+I₂ → S₄ O₆²⁻+2 I⁻ ; S₂ O₃²⁻+4 Br₂+5 H₂ O → 2 SO₄²⁻+8 Br⁻+10 H⁺
Which of the following statement justifies the above dual behaviour of thiosulphate?

Solution

Chemistry ma’am · 3 steps

Track sulphur's oxidation state: bromine pushes it all the way to +6, iodine only to +2.5.

1. Start point. In S₂O₃²⁻, two S atoms share a total of +4, so each sulphur averages +2. Both halogens start at 0 and end as -1 halide ions, so both are reduced.

   Oxidant	Product of S₂O₃²⁻	S oxidation state	Electrons lost per S₂O₃²⁻
   I₂	S₄O₆²⁻	+2 → +2.5	1
   Br₂	SO₄²⁻	+2 → +6	8

2. Compare the oxidants. Thiosulphate is oxidised in both reactions, but bromine takes 8 electrons from each S₂O₃²⁻ while iodine takes only 1. Pulling sulphur up to its highest state, +6, needs the stronger oxidant.
3. So. The dual behaviour comes from bromine being a stronger oxidant than iodine, which is option (c). This fits E°(Br₂/Br⁻) = 1.09 V, well above E°(I₂/I⁻) = 0.54 V.

The idea to keep. When one reductant gives different products with different oxidants, the stronger oxidant drives it to the higher oxidation state. Halogen oxidising power falls down the group: F₂ > Cl₂ > Br₂ > I₂.

Answer from the NTA final answer key. Solution by PrepWiser.