Maximum number of moles of electrons taken up by one mole of `NO_(3)^(-)` when it is reduced to :

Options

A. `NH_3`
B. `NH_2OH`
C. `NO`
D. `NO_2`

Explanation

To determine the maximum number of moles of electrons taken up by one mole of \( \text{NO}_3^- \) when it is reduced, we need to analyze the oxidation states of nitrogen in various possible reduction products.

### Step-by-Step Solution:

1. **Identify the oxidation state of nitrogen in \( \text{NO}_3^- \)**:
- Let the oxidation state of nitrogen be \( x \).
- The oxidation states of oxygen is \(-2\).
- The equation based on the charge of the ion is:
\[ x + 3(-2) = -1 \]
- Simplifying this gives:
\[ x - 6 = -1 \implies x = +5 \]
- Thus, the oxidation state of nitrogen in \( \text{NO}_3^- \) is +5.

2. **Consider possible reduction products**:
- We will consider the following compounds as potential reduction products:
- \( \text{NH}_3 \) (Ammonia)
- \( \text{NH}_2\text{OH} \) (Hydroxylamine)
- \( \text{NO} \) (Nitric oxide)
- \( \text{NO}_2 \) (Nitrogen dioxide)

3. **Calculate the oxidation state of nitrogen in each product**:
- **For \( \text{NH}_3 \)**:
\[ x + 3(+1) = 0 \implies x = -3 \]
- **For \( \text{NH}_2\text{OH} \)**:
\[ x + 2(+1) + (-2) = 0 \implies x = -1 \]
- **For \( \text{NO} \)**:
\[ x + 1(-2) = 0 \implies x = +2 \]
- **For \( \text{NO}_2 \)**:
\[ x + 2(-2) = 0 \implies x = +4 \]

4. **Calculate the change in oxidation state (Δ) for each product**:
- **For \( \text{NH}_3 \)**:
\[ \Delta = 5 - (-3) = 8 \]
- **For \( \text{NH}_2\text{OH} \)**:
\[ \Delta = 5 - (-1) = 6 \]
- **For \( \text{NO} \)**:
\[ \Delta = 5 - 2 = 3 \]
- **For \( \text{NO}_2 \)**:
\[ \Delta = 5 - 4 = 1 \]

5. **Determine the maximum number of moles of electrons**:
- The maximum change in oxidation state occurs when \( \text{NO}_3^- \) is reduced to \( \text{NH}_3 \), which corresponds to a change of 8.
- Therefore, the maximum number of moles of electrons taken up by one mole of \( \text{NO}_3^- \) is **8 moles**.

### Final Answer:
The maximum number of moles of electrons taken up by one mole of \( \text{NO}_3^- \) when it is reduced to \( \text{NH}_3 \) is **8 moles**.