Explain why `N_(2)` has a greater bond dissociation energy than `N_(2)^(+)` while `O_(2)` has lesser bond dissociation energy than `O_(2)^(+).`

Answer

To explain why \( N_2 \) has a greater bond dissociation energy than \( N_2^+ \), while \( O_2 \) has a lesser bond dissociation energy than \( O_2^+ \), we will analyze the bond order of each molecule and its correlation with bond dissociation energy.

### Step-by-Step Solution:

1. **Understanding Bond Dissociation Energy**:
- Bond dissociation energy is the energy required to break a bond in a molecule. It is directly related to the bond order of the molecule.

2. **Calculate Bond Order for \( N_2 \)**:
- The molecular orbital configuration for \( N_2 \) is \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \).
- The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]
- For \( N_2 \):
\[ \text{Bond Order} = \frac{10 - 2}{2} = 3 \]

3. **Calculate Bond Order for \( N_2^+ \)**:
- For \( N_2^+ \), one electron is removed from the molecular orbital configuration, typically from the highest energy level, which is \( \pi_{2p_x} \) or \( \pi_{2p_y} \).
- The new configuration for \( N_2^+ \) becomes:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \]
- Therefore, the bond order for \( N_2^+ \) is:
\[ \text{Bond Order} = \frac{9 - 2}{2} = 2.5 \]

4. **Comparison of Bond Dissociation Energies**:
- Since \( N_2 \) has a bond order of 3 and \( N_2^+ \) has a bond order of 2.5, \( N_2 \) has a greater bond dissociation energy than \( N_2^+ \) because higher bond order correlates with higher bond dissociation energy.

5. **Calculate Bond Order for \( O_2 \)**:
- The molecular orbital configuration for \( O_2 \) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \]
- The bond order for \( O_2 \) is:
\[ \text{Bond Order} = \frac{10 - 6}{2} = 2 \]

6. **Calculate Bond Order for \( O_2^+ \)**:
- For \( O_2^+ \), one electron is removed from the \( \pi_{2p_y} \) orbital:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^0 \]
- The bond order for \( O_2^+ \) is:
\[ \text{Bond Order} = \frac{9 - 6}{2} = 1.5 \]

7. **Comparison of Bond Dissociation Energies for \( O_2 \) and \( O_2^+ \)**:
- Since \( O_2 \) has a bond order of 2 and \( O_2^+ \) has a bond order of 2.5, \( O_2 \) has a lesser bond dissociation energy than \( O_2^+ \) because lower bond order correlates with lower bond dissociation energy.

### Conclusion:
- \( N_2 \) has a greater bond dissociation energy than \( N_2^+ \) due to its higher bond order (3 vs. 2.5).
- \( O_2 \) has a lesser bond dissociation energy than \( O_2^+ \) because of its lower bond order (2 vs. 2.5).