1. Analyze the reaction with iodine

The given reaction is $2\,\mathrm{S}_2\mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4\mathrm{O}_6^{2-}+2\,\mathrm{I}^-$

Here, iodine changes as: $\mathrm{I}_2 \rightarrow 2\,\mathrm{I}^-$ So iodine is reduced, hence it acts as an oxidizing agent.

Now thiosulphate changes from $\mathrm{S}_2\mathrm{O}_3^{2-}$ to tetrathionate $\mathrm{S}_4\mathrm{O}_6^{2-}$, which means thiosulphate is oxidized.

2. Analyze the reaction with bromine

The given reaction is $\mathrm{S}_2\mathrm{O}_3^{2-}+5\,\mathrm{Br}_2+5\,\mathrm{H}_2\mathrm{O} \rightarrow 2\,\mathrm{SO}_4^{2-}+4\,\mathrm{Br}^-+10\,\mathrm{H}^+$

Here bromine changes as: $\mathrm{Br}_2 \rightarrow \mathrm{Br}^-$ So bromine is also reduced, hence bromine also acts as an oxidizing agent.

Thiosulphate is oxidized much further here, all the way to sulfate: $\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}$

3. Why is the behaviour different?

With iodine, thiosulphate is oxidized only to tetrathionate.

With bromine, thiosulphate is oxidized more strongly to sulfate.

This shows that bromine is a stronger oxidizing agent than iodine.

Indeed, standard reduction potentials support this: $E^\circ(\mathrm{Br}_2/\mathrm{Br}^-) > E^\circ(\mathrm{I}_2/\mathrm{I}^-)$ So bromine can oxidize thiosulphate to a higher oxidation state product.

4. Check the options

• A: False. Thiosulphate is oxidized in both reactions, not reduced by iodine.
• B: False. Bromine is not weaker; it is stronger.
• C: True. Bromine is a stronger oxidant than iodine.
• D: False. Both bromine and iodine undergo reduction, not oxidation.

5. Final answer

The correct option is: $\boxed{\text{C}}$