How will bond order in `N_(2) and O_(2)` be influenced in the following ionization process ?
(a) ` N_(2) to N_(2)^(+) + e^(-) (b) O_(2) to O_(2)^(+) + e^(-)`

Explanation

To analyze how the bond order in \(N_2\) and \(O_2\) is influenced by the ionization processes \(N_2 \to N_2^+ + e^-\) and \(O_2 \to O_2^+ + e^-\), we will follow these steps:

### Step 1: Understand the Molecular Orbital (MO) Configuration
1. **For \(N_2\)**:
- Total electrons in \(N_2\) = 14 (7 from each nitrogen atom).
- The MO configuration is:
- \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Bonding electrons = 10, Anti-bonding electrons = 4.
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