Step-by-step Solution:

1. Identify Reactants and Reaction Type: The reaction is between aqueous sodium thiosulfate ($Na_2S_2O_3$) and chlorine ($Cl_2$). Sodium thiosulfate acts as a reducing agent, while chlorine is a strong oxidizing agent. This indicates a redox reaction will occur.

2. Determine the Oxidation and Reduction Half-Reactions:

   • Oxidation of Thiosulfate: In the thiosulfate ion ($S_2O_3^{2-}$), the average oxidation state of sulfur is +2. Chlorine is a strong oxidizing agent, so it will oxidize the sulfur to its highest stable oxidation state, which is +6 in the sulfate ion ($SO_4^{2-}$). Tetrathionate ($S_4O_6^{2-}$) is formed only with mild oxidizing agents like iodine. The unbalanced oxidation half-reaction is: $S_2O_3^{2-} \rightarrow SO_4^{2-}$ To balance this half-reaction in an acidic medium (as we'll see, the reaction produces acid):

     • Balance sulfur atoms: $S_2O_3^{2-} \rightarrow 2SO_4^{2-}$
     • Balance oxygen atoms by adding water ($H_2O$): $S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-}$
     • Balance hydrogen atoms by adding protons ($H^+$): $S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-} + 10H^+$
     • Balance the charge by adding electrons ($e^-$): The charge on the left is -2. The charge on the right is $2(-2) + 10(+1) = +6$. To balance, we add 8 electrons to the right side. $S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-} + 10H^+ + 8e^-$

   • Reduction of Chlorine: Chlorine gas ($Cl_2$), with an oxidation state of 0, is reduced to chloride ions ($Cl^-$), with an oxidation state of -1. The balanced reduction half-reaction is: $Cl_2 + 2e^- \rightarrow 2Cl^-$

3. Combine the Half-Reactions: To get the overall ionic equation, we need to balance the electrons. We multiply the reduction half-reaction by 4 so that the number of electrons (8e-) is the same in both half-reactions.

   • Oxidation: $S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-} + 10H^+ + 8e^-$
   • Reduction: $4Cl_2 + 8e^- \rightarrow 8Cl^-$ Adding them together and canceling the electrons gives the overall ionic equation: $S_2O_3^{2-} + 4Cl_2 + 5H_2O \rightarrow 2SO_4^{2-} + 8Cl^- + 10H^+$

4. Determine the Final Products in Molecular Form: The initial reactant was sodium thiosulfate ($Na_2S_2O_3$), so we have $Na^+$ as spectator ions. The products in the aqueous solution are sulfate ions ($SO_4^{2-}$), chloride ions ($Cl^-$), hydrogen ions ($H^+$), and the spectator sodium ions ($Na^+$). The reaction produces a large amount of $H^+$ ions, making the solution strongly acidic. In a strongly acidic solution, sulfate ions ($SO_4^{2-}$) react with $H^+$ to form bisulfate ions ($HSO_4^-$). $SO_4^{2-} + H^+ \rightleftharpoons HSO_4^-$ So, the main sulfur-containing species will be $HSO_4^-$. The products can be represented as sodium bisulfate ($NaHSO_4$) and hydrochloric acid ($HCl$). Let's write the full balanced molecular equation: $Na_2S_2O_3 + 4Cl_2 + 5H_2O \rightarrow 2NaHSO_4 + 8HCl$ The products are sodium bisulfate and hydrochloric acid.

5. Evaluate the Options:

   • A: $Na_2S_4O_6$ (Sodium tetrathionate) is formed with mild oxidizing agents like $I_2$, not strong ones like $Cl_2$. So, A is incorrect.
   • B: $NaHSO_4$ (Sodium bisulfate) is one of the main products of the reaction, representing the oxidized form of sulfur. So, B is correct.
   • C: $NaCl$ (Sodium chloride) can be considered to be formed from the $Na^+$ and $Cl^-$ ions present in the solution. However, $NaHSO_4$ represents the fate of the primary reactant ($S_2O_3^{2-}$), making it the more significant and specific product of the transformation.
   • D: $NaOH$ (Sodium hydroxide) is a base. The reaction produces acid, so this option is incorrect.

Therefore, the most appropriate answer is NaHSO${}_4$.