Consider the reactions:
$2S_2O_3^{2-}{(aq)} + I_{2(s)} \to S_4O_6^{2-}{(aq)} + 2I^{-}{(aq)}$
$S_2O_3^{2-}{(aq)} + 2Br_{2(l)} + 5H_2O_{(l)} \to 2SO_4^{2-}{(aq)} + 4Br^{-}{(aq)} + 10H^{+}{(aq)}$
Why does the same reductant,thiosulphate,react differently with iodine and bromine?

(N/A) The average oxidation number $(O.N.)$ of $S$ in $S_2O_3^{2-}$ is $+2$.
$Br_2$ is a stronger oxidizing agent than $I_2$.
$Br_2$ oxidizes the sulfur in thiosulfate to the $+6$ oxidation state (in $SO_4^{2-}$).
$I_2$ is a weaker oxidizing agent and can only oxidize the sulfur to an average oxidation state of $+2.5$ (in $S_4O_6^{2-}$).
Therefore,the extent of oxidation depends on the strength of the oxidizing agent,causing the thiosulfate to react differently.