To solve the question about which pair is expected to have the same bond order, we need to calculate the bond order for the given pairs of molecules using the Molecular Orbital Theory (MOT). The bond order is calculated using the formula:
\[
\text{Bond Order} = \frac{1}{2} \times (n_B - n_A)
\]
where \( n_B \) is the number of electrons in bonding molecular orbitals and \( n_A \) is the number of electrons in anti-bonding molecular orbitals.
### Step 1: Calculate the bond order for O2
1. **Total Electrons**: O2 has 16 electrons (8 from each oxygen atom).
2. **Molecular Orbital Configuration**: According to the provided information, the configuration for O2 is:
- \( \sigma_{1s}^2 \)
- \( \sigma^{*}_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^{*}_{2s}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \), \( \pi_{2p_y}^2 \)
- \( \pi^{*}_{2p_x}^1 \), \( \pi^{*}_{2p_y}^1 \)
3. **Count Electrons**:
- **Bonding Electrons (n_B)**: 10 (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- **Anti-bonding Electrons (n_A)**: 6 (from \( \sigma^{*}_{1s}, \sigma^{*}_{2s}, \pi^{*}_{2p_x}, \pi^{*}_{2p_y} \))
4. **Bond Order Calculation**:
\[
\text{Bond Order for O2} = \frac{1}{2} \times (10 - 6) = \frac{4}{2} = 2
\]
### Step 2: Calculate the bond order for N2
1. **Total Electrons**: N2 has 14 electrons (7 from each nitrogen atom).
2. **Molecular Orbital Configuration**: The configuration for N2 is:
- \( \sigma_{1s}^2 \)
- \( \sigma^{*}_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^{*}_{2s}^2 \)
- \( \pi_{2p_x}^2 \), \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
3. **Count Electrons**:
- **Bonding Electrons (n_B)**: 10 (from \( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} \))
- **Anti-bonding Electrons (n_A)**: 4 (from \( \sigma^{*}_{1s}, \sigma^{*}_{2s} \))
4. **Bond Order Calculation**:
\[
\text{Bond Order for N2} = \frac{1}{2} \times (10 - 4) = \frac{6}{2} = 3
\]
### Step 3: Calculate the bond order for O2⁺ and N2⁻
1. **O2⁺**: This has 15 electrons (removing one from the anti-bonding orbital).
- **Configuration**: Same as O2, but one less electron in the anti-bonding orbital.
- **Bonding Electrons (n_B)**: 10
- **Anti-bonding Electrons (n_A)**: 5
\[
\text{Bond Order for O2⁺} = \frac{1}{2} \times (10 - 5) = \frac{5}{2} = 2.5
\]
2. **N2⁻**: This has 15 electrons (adding one to the anti-bonding orbital).
- **Configuration**: Same as N2, but one more electron in the anti-bonding orbital.
- **Bonding Electrons (n_B)**: 10
- **Anti-bonding Electrons (n_A)**: 5
\[
\text{Bond Order for N2⁻} = \frac{1}{2} \times (10 - 5) = \frac{5}{2} = 2.5
\]
### Conclusion
- The bond order for O2 is 2, for N2 is 3, for O2⁺ is 2.5, and for N2⁻ is 2.5.
- Therefore, the pair that has the same bond order is **O2⁺ and N2⁻**, which both have a bond order of 2.5.
### Answer
The correct option is **B: O2⁺ and N2⁻**.
