Assertion:- `N_(2)^(+)` is more stable than `N_(2)^(-)`
Reason:- `N_(2)^(+)` has less electrons in antibonding orbital .

Options

A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
C. If Assertion is True, but the Reason is False.
D. If both Assertion & Reason are false

Answer

To solve the question regarding the stability of \( N_2^+ \) and \( N_2^- \), we will analyze the molecular electronic configurations of both species and evaluate their bond orders and the presence of electrons in antibonding orbitals.

### Step-by-Step Solution:

1. **Identify the Total Number of Electrons:**
- For \( N_2^+ \): Nitrogen has 7 electrons, so \( N_2 \) has \( 2 \times 7 = 14 \) electrons. Since \( N_2^+ \) has lost one electron, it has \( 14 - 1 = 13 \) electrons.
- For \( N_2^- \): \( N_2 \) has 14 electrons, and since \( N_2^- \) has gained one electron, it has \( 14 + 1 = 15 \) electrons.

2. **Write the Molecular Electronic Configuration:**
- For \( N_2^+ \) (13 electrons):
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \]
- For \( N_2^- \) (15 electrons):
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^*^1 \]

3. **Determine the Bond Order:**
- Bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]
- For \( N_2^+ \):
- Bonding electrons = 10 (from \( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} \))
- Antibonding electrons = 3 (from \( \sigma_{1s}^*, \sigma_{2s}^* \))
- Bond Order = \( \frac{10 - 3}{2} = 3.5 \)
- For \( N_2^- \):
- Bonding electrons = 10
- Antibonding electrons = 4 (from \( \sigma_{1s}^*, \sigma_{2s}^*, \pi_{2p_x}^*, \sigma_{2p_z} \))
- Bond Order = \( \frac{10 - 4}{2} = 3 \)

4. **Evaluate Stability Based on Antibonding Electrons:**
- \( N_2^+ \) has fewer antibonding electrons (3) compared to \( N_2^- \) (4).
- More antibonding electrons generally indicate less stability.

5. **Conclusion:**
- Since \( N_2^+ \) has fewer electrons in antibonding orbitals, it is more stable than \( N_2^- \).
- Therefore, the assertion that \( N_2^+ \) is more stable than \( N_2^- \) is true, and the reason that \( N_2^+ \) has less electrons in antibonding orbitals is also true.

### Final Answer:
Both the assertion and reason are correct, and the reason correctly explains the assertion.