In this reaction `2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)SO_(4)O_(6)+2NaI_(2)` ,`NaI_(2)`acts as:

Options

A. Reducing agent
B. Oxidising agent
C. Oxidising agent as well as reducing agent
D. None of these

Answer

To determine the role of NaI₂ in the reaction \(2Na_{2}S_{2}O_{3} + I_{2} \rightarrow Na_{2}SO_{4} + 2NaI\), we need to analyze the oxidation states of the elements involved in the reaction. Here’s a step-by-step solution:

### Step 1: Identify the oxidation states of the elements in the reactants and products.

1. **Sodium (Na)**: In both \(Na_{2}S_{2}O_{3}\) and \(NaI\), sodium has an oxidation state of +1.
2. **Sulfur (S)**: In \(Na_{2}S_{2}O_{3}\), we can calculate the oxidation state of sulfur:
- Let the oxidation state of sulfur be \(x\).
- The formula for \(Na_{2}S_{2}O_{3}\) is:
\[ 2(+1) + 2x + 3(-2) = 0 \]
\[ 2 + 2x - 6 = 0 \implies 2x - 4 = 0 \implies x = +2 \]
- Therefore, the oxidation state of sulfur in \(Na_{2}S_{2}O_{3}\) is +2.

3. **Iodine (I)**: In \(I_{2}\), the oxidation state is 0. In \(NaI\), iodine has an oxidation state of -1.

### Step 2: Determine the changes in oxidation states.

- **Sulfur**: Changes from +2 in \(Na_{2}S_{2}O_{3}\) to +6 in \(Na_{2}SO_{4}\). This indicates that sulfur is **oxidized**.
- **Iodine**: Changes from 0 in \(I_{2}\) to -1 in \(NaI\). This indicates that iodine is **reduced**.

### Step 3: Identify the oxidizing and reducing agents.

- An **oxidizing agent** is a substance that gains electrons (is reduced) and causes another substance to be oxidized.
- A **reducing agent** is a substance that loses electrons (is oxidized) and causes another substance to be reduced.

From our analysis:
- Since iodine goes from 0 to -1, it is reduced. Therefore, \(I_{2}\) acts as the **oxidizing agent**.
- Since sulfur goes from +2 to +6, it is oxidized. Therefore, \(Na_{2}S_{2}O_{3}\) acts as the **reducing agent**.

### Conclusion

In the reaction \(2Na_{2}S_{2}O_{3} + I_{2} \rightarrow Na_{2}SO_{4} + 2NaI\), \(NaI\) (which contains iodine in the -1 oxidation state) acts as a product of the reduction of iodine, thus it is not an oxidizing agent or reducing agent in this context. The correct answer is that \(NaI\) does not act as an oxidizing or reducing agent; it is simply a product formed from the reaction.

### Final Answer

**NaI acts as none of these (option D)**.