`{:("Column-I",underset(("Oxidation sate of N"))(Column-II")),((a)NO,(p)+5),((b) NO2, (q) + 3),((c) NO_(2)^(–) ,(r) + 4),((d) NO_(3)^(–) ,(s) + 2):}`

Answer

To determine the oxidation states of nitrogen in the given compounds, we will analyze each compound one by one.

### Step-by-Step Solution:

1. **For NO (Nitric Oxide)**:
- The oxidation state of oxygen (O) is typically -2.
- Let the oxidation state of nitrogen (N) be \( x \).
- The overall charge of the molecule is neutral (0).
- Therefore, we can set up the equation:
\[ x + (-2) = 0 \]
- Solving for \( x \):
\[ x = +2 \]
- **Oxidation state of N in NO is +2**.

2. **For NO2 (Nitrogen Dioxide)**:
- The oxidation state of oxygen (O) is -2.
- Let the oxidation state of nitrogen (N) be \( x \).
- The overall charge of the molecule is neutral (0).
- Therefore, we can set up the equation:
\[ x + 2(-2) = 0 \]
- Solving for \( x \):
\[ x - 4 = 0 \implies x = +4 \]
- **Oxidation state of N in NO2 is +4**.

3. **For NO2⁻ (Nitrite Ion)**:
- The oxidation state of oxygen (O) is -2.
- Let the oxidation state of nitrogen (N) be \( x \).
- The overall charge of the ion is -1.
- Therefore, we can set up the equation:
\[ x + 2(-2) = -1 \]
- Solving for \( x \):
\[ x - 4 = -1 \implies x = +3 \]
- **Oxidation state of N in NO2⁻ is +3**.

4. **For NO3⁻ (Nitrate Ion)**:
- The oxidation state of oxygen (O) is -2.
- Let the oxidation state of nitrogen (N) be \( x \).
- The overall charge of the ion is -1.
- Therefore, we can set up the equation:
\[ x + 3(-2) = -1 \]
- Solving for \( x \):
\[ x - 6 = -1 \implies x = +5 \]
- **Oxidation state of N in NO3⁻ is +5**.

### Summary of Results:
- (a) NO → +2
- (b) NO2 → +4
- (c) NO2⁻ → +3
- (d) NO3⁻ → +5

### Matching with Column-II:
- (a) NO → (s) +2
- (b) NO2 → (r) +4
- (c) NO2⁻ → (q) +3
- (d) NO3⁻ → (p) +5

### Final Answer:
- (a) NO → (s) +2
- (b) NO2 → (r) +4
- (c) NO2⁻ → (q) +3
- (d) NO3⁻ → (p) +5