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 determine the correct relation between \( N_2^+ \) and \( N_2^- \), we will analyze their bond orders and bond energies using Molecular Orbital Theory (MOT).

### Step 1: Determine the Electron Configuration for \( N_2^+ \)
- **Total Electrons**: \( N_2 \) has 14 electrons (7 from each nitrogen atom). For \( N_2^+ \), we remove one electron, giving us 13 electrons.
- **Molecular Orbital Filling**: The order of filling for \( N_2 \) is:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^* \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^* \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)

For \( N_2^+ \) (13 electrons), the filling will be:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^* \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^* \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^1 \)

### Step 2: Calculate Bond Order for \( N_2^+ \)
- **Bonding Electrons**: 10 (from \( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} \))
- **Antibonding Electrons**: 3 (from \( \sigma_{1s}^*, \sigma_{2s}^*, \sigma_{2p_z}^* \))
- **Bond Order Formula**:
\[ \text{Bond Order} = \frac{\text{Bonding Electrons} - \text{Antibonding Electrons}}{2} \]
\[ \text{Bond Order for } N_2^+ = \frac{10 - 3}{2} = \frac{7}{2} = 3.5 \]

### Step 3: Determine the Electron Configuration for \( N_2^- \)
- **Total Electrons**: For \( N_2^- \), we add one electron to \( N_2 \), giving us 15 electrons.
- **Molecular Orbital Filling**: The filling will be:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^* \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^* \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^* \) or \( \pi_{2p_y}^* \) (the additional electron goes into one of the antibonding orbitals)

### Step 4: Calculate Bond Order for \( N_2^- \)
- **Bonding Electrons**: 10 (same as before)
- **Antibonding Electrons**: 4 (3 from \( \sigma_{1s}^*, \sigma_{2s}^*, \sigma_{2p_z}^* \) and 1 from \( \pi_{2p_x}^* \) or \( \pi_{2p_y}^* \))
- **Bond Order Formula**:
\[ \text{Bond Order for } N_2^- = \frac{10 - 4}{2} = \frac{6}{2} = 3 \]

### Step 5: Compare Bond Energies
- Since \( N_2^+ \) has a higher bond order (3.5) compared to \( N_2^- \) (3), it implies that \( N_2^+ \) is more stable and has a greater bond energy.
- Therefore, the relation is:
\[ \text{Bond Energy of } N_2^+ > \text{Bond Energy of } N_2^- \]

### Conclusion
The correct relation for \( N_2^+ \) and \( N_2^- \) is:
\[ \text{Bond Energy of } N_2^+ > \text{Bond Energy of } N_2^- \]