Bond order of `N_2^(+) and N_2^-` are same. Which relation is correct for `N_2^(+) and N_2^(-)` ?

Options

A. Bond energy of `N_2^(+)` = bond energy of `N_2^(-)`
B. Bond energy of `N_2^(+)gt` bond energy of `N_2^(-)`
C. Bond energy of `N_2^(+)lt` bond energy of `N_2^(-)`
D. Bond energy of `N_2^(+)ge` bond energy of `N_2^(-)`

Answer

To solve the question regarding the bond order and the relationship between \( N_2^+ \) and \( N_2^- \), we will follow these steps:

### Step 1: Determine the Bond Order
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{(N_b - N_a)}{2} \]
where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of antibonding electrons.

### Step 2: Write the Molecular Orbital Configuration
1. For \( N_2^+ \):
- The electronic configuration is:
\[ (1s)^2 (1s^*)^2 (2s)^2 (2s^*)^2 (2p_x)^2 (2p_y)^2 (2p_z)^1 \]
- Bonding electrons \( N_b = 10 \) (from \( 1s, 2s, 2p_x, 2p_y \))
- Antibonding electrons \( N_a = 4 \) (from \( 1s^*, 2s^*, 2p_z^* \))
- Bond Order \( = \frac{10 - 4}{2} = 3 \)

2. For \( N_2^- \):
- The electronic configuration is:
\[ (1s)^2 (1s^*)^2 (2s)^2 (2s^*)^2 (2p_x)^2 (2p_y)^2 (2p_z)^2 (2p_x^*)^1 (2p_y^*)^1 \]
- Bonding electrons \( N_b = 10 \)
- Antibonding electrons \( N_a = 5 \)
- Bond Order \( = \frac{10 - 5}{2} = 2.5 \)

### Step 3: Compare Bond Energies
Since both \( N_2^+ \) and \( N_2^- \) have the same bond order of 2.5, we need to analyze their bond energies.

- **Stability and Bond Energy**:
- \( N_2^+ \) has fewer antibonding electrons compared to \( N_2^- \). More antibonding electrons in \( N_2^- \) lead to greater instability, which decreases bond strength.
- Therefore, the bond energy of \( N_2^- \) will be less than that of \( N_2^+ \).

### Conclusion
The correct relation is:
\[ \text{Bond Energy of } N_2^- < \text{Bond Energy of } N_2^+ \]