Consider the reaction:
`2S_(2)O_(3)^(2-)(aq)+I_(2)(s) rarr S_(4)O_(6)^(2-)(aq) + 2I^(Θ)(aq)`
`2S_(2)O_(3)^(2-)(aq) + 2Br_(2)(l) + 5H_(2)O(l) rarr 2SO_(4)^(2-)(aq) + 4Br^(Θ)(aq)+10H^(o+)(aq)`
Why does the same reductant, thiosulphate, react differently with iodine and bromine?

Explanation

The O.N. of in `S_2O_3^(2-)` is +2, in `S_4O_6^(2-)` it is +2.5 (average) and in `SO_4^(2-)` it is +6. Since `Br_2` is a stronger oxidising agent than , `I_2` , it oxidises S of `S_2O_3^(2-)` to `SO_4^(2-)` (with higher O.N. of +6) .However , `I_2` being a weaker oxidising agnet oxidises S of `S_2O_3^(2-)` into a lower O.S. of +2.5 in `S_4O_6^(2-)` . Therefore , `S_2O_3^(2-)` reacts differently with `Br_2 and I_2`.