Thiosulphate reacts differently with iodine and bromine in the reactions given below: $$2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$$, $$S_2O_3^{2-} + 5Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 4Br^- + 10H^+$$. Which of the following statement justifies the above dual behaviour of thiosulphate?

Explain why thiosulphate reacts differently with iodine and bromine.

With $$I_2$$, thiosulphate reacts according to $$2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$$, in which sulfur is oxidized from +2 to +2.5, indicating mild oxidation; with $$Br_2$$, the reaction $$S_2O_3^{2-} + 5Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 4Br^- + 10H^+$$ shows sulfur oxidized from +2 to +6, indicating strong oxidation.

Bromine, being a stronger oxidizing agent than iodine, can oxidize thiosulphate all the way to sulfate ($$SO_4^{2-}$$, S in the +6 state), while iodine, being milder, oxidizes it only partially to tetrathionate ($$S_4O_6^{2-}$$).

The correct answer is Option (1): Bromine is a stronger oxidant than iodine.