When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2) iodine is liberated which can be titrated against a standard solution of sodium thiosulphate, this is a quantitative method for estimating `O_(3)` gas. when liberated `I_(2)` and sodium thiosulphate will react, then product is/are:

Options

A. `S_(4)O_(6)^(2-)`
B. `SO_(4)^(2-)`
C. `S_(2)O_(4)^(2-)`
D. `S^(0)`

Explanation

To determine the products formed when iodine (I₂) reacts with sodium thiosulfate (Na₂S₂O₃), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When ozone (O₃) reacts with potassium iodide (KI) in a buffered solution, iodine (I₂) is liberated. This liberated iodine can then react with sodium thiosulfate.

2. **Writing the Reaction**:
- The reaction between iodine and sodium thiosulfate can be represented as:
\[ I_2 + 2 Na_2S_2O_3 \rightarrow 2 NaI + Na_2S_4O_6 \]
- In this reaction, iodine (I₂) is reduced to iodide ions (I⁻), and sodium thiosulfate (Na₂S₂O₃) is oxidized to form sodium tetrathionate (Na₂S₄O₆).

3. **Identifying the Products**:
- From the balanced reaction, we can see that the products formed are sodium iodide (NaI) and sodium tetrathionate (Na₂S₄O₆).

4. **Final Answer**:
- Therefore, the products of the reaction between liberated iodine and sodium thiosulfate are:
- Sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI).

### Conclusion:
The primary product of interest in this context is the sodium tetrathionate ion (S₄O₆²⁻), which corresponds to option 1 in the given choices.