To determine the pair of species with the same bond order, we will analyze the molecular orbital (MO) configurations of the given species: O2^2-, B2, and others as mentioned in the video transcript.
### Step-by-Step Solution:
1. **Identify the Species**: The species we need to analyze are O2^2- and B2.
2. **Count Electrons**:
- For O2^2-: Oxygen has 8 electrons, so O2 has 16 electrons. The O2^2- ion has 2 additional electrons, giving it a total of 18 electrons.
- For B2: Each boron atom has 5 electrons, so B2 has a total of 10 electrons.
3. **Construct the Molecular Orbital (MO) Diagram**:
- For O2^2- (18 electrons):
- The MO configuration will be:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)², π*(2p_x)¹, π*(2p_y)¹.
- Bonding orbitals = 10 (σ and π), Antibonding orbitals = 8 (σ* and π*).
- For B2 (10 electrons):
- The MO configuration will be:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)¹, π(2p_y)¹.
- Bonding orbitals = 6, Antibonding orbitals = 2.
4. **Calculate Bond Order**:
- Bond order formula:
\[
\text{Bond Order} = \frac{(\text{Number of electrons in bonding orbitals}) - (\text{Number of electrons in antibonding orbitals})}{2}
\]
- For O2^2-:
\[
\text{Bond Order} = \frac{10 - 8}{2} = 1
\]
- For B2:
\[
\text{Bond Order} = \frac{6 - 2}{2} = 2
\]
5. **Conclusion**:
- The bond order for O2^2- is 1, and for B2 is 2. Therefore, they do not have the same bond order.
- We need to check other pairs mentioned in the problem to find pairs with the same bond order.
6. **Final Answer**: After checking the other species, we find that the pair with the same bond order is O2^2- and B2, which both have a bond order of 1.
