Reaction of Aqueous Sodium Thiosulfate with Chlorine

When an aqueous solution of sodium thiosulfate ($\mathrm{Na_2S_2O_3}$) reacts with chlorine gas ($\mathrm{Cl_2}$), a strong redox reaction takes place. Chlorine is a much stronger oxidizing agent than iodine; hence, while iodine oxidizes thiosulfate mildly to tetrathionate ($\mathrm{S_4O_6^{2-}}$), chlorine oxidizes the sulfur in thiosulfate all the way to its highest oxidation state, sulfate ($\mathrm{SO_4^{2-}}$ / $\mathrm{HSO_4^-}$).

The balanced chemical reaction is:
$\mathrm{Na_2S_2O_3} + 4\mathrm{Cl_2} + 5\mathrm{H_2O} \longrightarrow 2\mathrm{NaHSO_4} + 8\mathrm{HCl}$

In ionic form:
$\mathrm{S_2O_3^{2-}} + 4\mathrm{Cl_2} + 5\mathrm{H_2O} \longrightarrow 2\mathrm{SO_4^{2-}} + 8\mathrm{Cl^-} + 10\mathrm{H^+}$

Because of this reaction, sodium thiosulfate acts as an "antichlor" to remove excess chlorine in bleaching processes.

Therefore, the product formed from the given options is $\mathrm{NaHSO_4}$.

The correct option is (B).