During oxidation , in which of the following bond order increase ?

Options

A. `N_(2)`
B. `CO`
C. `O_(2)^(-)`
D. Both (2) & (3)

Answer

To determine in which of the following molecules the bond order increases during oxidation, we will analyze the bond order of N2, CO, and O2- before and after oxidation.

### Step-by-Step Solution:

1. **Understanding Oxidation**:
- Oxidation refers to the loss of electrons. When a molecule is oxidized, it loses one or more electrons, which can affect its bond order.

2. **Bond Order Calculation**:
- The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \times (\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}) \]

3. **Analyzing N2**:
- **Molecular Orbital Configuration of N2**:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \]
- **Bonding Electrons**: 10 (2 from each sigma and pi bond)
- **Antibonding Electrons**: 4 (2 from sigma* and 2 from pi*)
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{1}{2} \times (10 - 4) = \frac{6}{2} = 3 \]
- **Upon Oxidation (N2 to N2+)**:
- One electron is lost from the bonding orbitals.
- New Bonding Electrons: 9
- Bond Order Calculation:
\[ \text{Bond Order for N2+} = \frac{1}{2} \times (9 - 4) = \frac{5}{2} = 2.5 \]
- **Conclusion**: Bond order decreases from 3 to 2.5.

4. **Analyzing CO**:
- **Molecular Orbital Configuration of CO**:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]
- **Bonding Electrons**: 10
- **Antibonding Electrons**: 4
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{1}{2} \times (10 - 4) = 3 \]
- **Upon Oxidation (CO to CO+)**:
- One electron is lost from the antibonding orbitals.
- New Antibonding Electrons: 3
- Bond Order Calculation:
\[ \text{Bond Order for CO+} = \frac{1}{2} \times (10 - 3) = \frac{7}{2} = 3.5 \]
- **Conclusion**: Bond order increases from 3 to 3.5.

5. **Analyzing O2-**:
- **Molecular Orbital Configuration of O2-**:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^2 \]
- **Bonding Electrons**: 10
- **Antibonding Electrons**: 6
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{1}{2} \times (10 - 6) = 2 \]
- **Upon Oxidation (O2- to O2)**:
- One electron is lost from the antibonding orbitals.
- New Antibonding Electrons: 5
- Bond Order Calculation:
\[ \text{Bond Order for O2} = \frac{1}{2} \times (10 - 5) = \frac{5}{2} = 2.5 \]
- **Conclusion**: Bond order increases from 1.5 to 2.

### Final Conclusion:
- **N2**: Bond order decreases from 3 to 2.5 (does not increase).
- **CO**: Bond order increases from 3 to 3.5 (increases).
- **O2-**: Bond order increases from 1.5 to 2 (increases).

Thus, the correct answer is **both CO and O2-** experience an increase in bond order during oxidation.