The oxidation number of nitrogen atoms in `NH_(4)NO_(3)` are:

Options

A. `+3,+3`
B. `+3,-3`
C. `-3,-5`
D. `-5,+3`

Answer

To determine the oxidation numbers of nitrogen atoms in the compound \( NH_4NO_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of the Compound**:
The compound \( NH_4NO_3 \) consists of two nitrogen atoms: one in the ammonium ion (\( NH_4^+ \)) and one in the nitrate ion (\( NO_3^- \)).

2. **Assign Oxidation Number for Hydrogen**:
The oxidation number of hydrogen is always +1. Since there are 4 hydrogen atoms in \( NH_4^+ \):
\[ \text{Oxidation number of H} = +1 \times 4 = +4 \]

3. **Set Up the Equation for Ammonium Ion**:
Let the oxidation number of nitrogen in \( NH_4^+ \) be \( x \). The total charge of the ammonium ion is +1. Therefore, we can set up the equation:
\[ x + 4(+1) = +1 \]
Simplifying this gives:
\[ x + 4 = 1 \implies x = 1 - 4 \implies x = -3 \]
So, the oxidation number of nitrogen in \( NH_4^+ \) is -3.

4. **Assign Oxidation Number for Oxygen**:
The oxidation number of oxygen is typically -2. In the nitrate ion \( NO_3^- \), there are 3 oxygen atoms:
\[ \text{Oxidation number of O} = -2 \times 3 = -6 \]

5. **Set Up the Equation for Nitrate Ion**:
Let the oxidation number of nitrogen in \( NO_3^- \) be \( y \). The total charge of the nitrate ion is -1. Therefore, we can set up the equation:
\[ y + (-6) = -1 \]
Simplifying this gives:
\[ y - 6 = -1 \implies y = -1 + 6 \implies y = +5 \]
So, the oxidation number of nitrogen in \( NO_3^- \) is +5.

6. **Conclusion**:
The oxidation numbers of nitrogen in \( NH_4NO_3 \) are:
- Nitrogen in \( NH_4^+ \): -3
- Nitrogen in \( NO_3^- \): +5

### Final Answer:
The oxidation numbers of nitrogen in \( NH_4NO_3 \) are -3 and +5.

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