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.