QJEEM24Ch841f: JEE Main 2024 Chemistry - Class 11 JEEM24Ch841f Solution: Thiosulphate reacts differently with iodine and bromine in the…

In the given reactions:

$\begin{aligned}& 2 \mathrm{S}_2\mathrm{O}_3^{2-} + \mathrm{I}_2 \rightarrow \mathrm{S}_4\mathrm{O}_6^{2-} + 2 \mathrm{I}^{-} \\\\& \mathrm{S}_2\mathrm{O}_3^{2-} + 5 \mathrm{Br}_2 + 5 \mathrm{H}_2\mathrm{O} \rightarrow 2 \mathrm{SO}_4^{2-} + 4 \mathrm{Br}^{-} + 10 \mathrm{H}^{+}\end{aligned}$

In the first reaction with iodine ( $I_2$ ), thiosulfate ( $S_2O_3^{2-}$ ) acts as a reducing agent, converting $I_2$ to $I^{-}$ , while it is itself oxidized to $S_4O_6^{2-}$ . This indicates that iodine is acting as an oxidant (oxidizing agent), getting reduced in the process.

In the second reaction with bromine ( $Br_2$ ), thiosulfate is oxidized more extensively, being converted all the way to sulfate ( $SO_4^{2-}$ ), with hydrogen ions ( $H^+$ ) and bromide ions ( $Br^-$ ) also produced. This shows that bromine acts as a stronger oxidant compared to iodine, as it promotes a complete oxidation of $S_2O_3^{2-}$ to $SO_4^{2-}$ .

Given these observations, the correct option that justifies the dual behavior of thiosulfate reacting with iodine and bromine differently is:

Option C: Bromine is a stronger oxidant than iodine

This statement correctly explains why thiosulfate is oxidized to $S_4O_6^{2-}$ in the presence of iodine, and to $SO_4^{2-}$ in the presence of bromine. Bromine's stronger oxidizing ability leads to a more complete oxidation of thiosulfate compared to the oxidation by iodine.