A solution of `I_(2)` in aqueous Kl on reaction with an aqueous solution of `Na_(2)S_(2)O_(3)` gets decolourised. The reaction taking place here is

Options

A. `Na_(2)S_(2)O_(3)+H_(2)O+I_(2)rarrNa_(2)S_(2)O_(4)+2HI`
B. `2Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2NaI`
C. `Na_(2)S_(2)O_(3)+2H_(2)O+2I_(2)rarrNa_(2)S_(2)O_(5)+4HI`
D. `Na_(2)S_(2)O_(3)+2H_(2)O+2I_(2)rarrNa_(2)S_(4)O_(8)+4HI`

Answer

To solve the problem, we need to understand the reaction that occurs when a solution of iodine (I₂) in aqueous potassium iodide (KI) reacts with sodium thiosulfate (Na₂S₂O₃).

### Step-by-Step Solution:

1. **Identify the Reactants:**
- We have iodine (I₂) dissolved in potassium iodide (KI).
- We are reacting this with sodium thiosulfate (Na₂S₂O₃).

2. **Understand the Nature of the Reaction:**
- Iodine (I₂) is a brownish color in solution. The reaction with sodium thiosulfate will lead to a decolorization of this solution.
- Sodium thiosulfate acts as a reducing agent in this reaction.

3. **Write the Reaction Equation:**
- The balanced chemical reaction can be written as:
\[ I_2 + 2 Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2 NaI \]
- Here, iodine (I₂) is reduced to iodide ions (NaI), and sodium thiosulfate is oxidized to tetrathionate (Na₂S₄O₆).

4. **Explain the Decolorization:**
- The decolorization occurs because the brown color of iodine disappears as it is converted into colorless iodide ions (NaI).
- The formation of tetrathionate (Na₂S₄O₆) is responsible for the change in color.

5. **Determine the Correct Option:**
- Based on the reaction, we can conclude that the correct option is the one that represents the formation of Na₂S₄O₆ and NaI from the reaction of I₂ with Na₂S₂O₃.

### Final Answer:
The reaction taking place here is:
\[ I_2 + 2 Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2 NaI \]
Thus, the correct option is **B**.