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`

Explanation

To solve the question regarding the reaction of ozone (O₃) with potassium iodide (KI) and the subsequent reaction with sodium thiosulfate (Na₂S₂O₃), we will break down the steps involved in the reactions and identify the products formed.

### Step-by-Step Solution:

1. **Identify the Initial Reaction**:
Ozone (O₃) reacts with potassium iodide (KI) in a buffered solution. The reaction can be represented as:
\[ O_3 + 2KI \rightarrow I_2 + 2KOH + O_2 \]
Here, ozone oxidizes iodide ions (I⁻) to iodine (I₂), while itself being reduced.

2. **Understand the Products of the First Reaction**:
From the above reaction, we see that the products formed are:
- Iodine (I₂)
- Potassium hydroxide (KOH)
- Oxygen (O₂)

3. **Titration of Iodine**:
The liberated iodine (I₂) is then titrated against a standard solution of sodium thiosulfate (Na₂S₂O₃). The reaction between iodine and sodium thiosulfate can be represented as:
\[ I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI \]
In this reaction, iodine is reduced, and the products formed are:
- Sodium tetrathionate (Na₂S₄O₆)
- Sodium iodide (NaI)

4. **Final Products**:
The final products from the reaction of iodine with sodium thiosulfate are:
- Sodium tetrathionate (Na₂S₄O₆)
- Sodium iodide (NaI)

### Conclusion:
The products formed when iodine (I₂) reacts with sodium thiosulfate (Na₂S₂O₃) are sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI).