To solve the question regarding the molecular orbital (MO) theory, specifically for the species O2 and O2+, we will follow these steps:
### Step-by-Step Solution:
1. **Identify the Number of Electrons**:
- For O2, the total number of electrons is 16 (8 from each oxygen atom).
- For O2+, one electron is removed, resulting in 15 electrons.
2. **Write the Molecular Orbital Configuration**:
- For O2:
- The configuration is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p z² π2p x² π2p y² π*2p x¹ π*2p y¹
- For O2+:
- The configuration is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p z² π2p x² π2p y² π*2p x¹
3. **Determine the Magnetic Nature**:
- For O2, there are unpaired electrons in the π* orbitals (π*2p x¹ and π*2p y¹), making it **paramagnetic**.
- For O2+, there is one unpaired electron in the π*2p x orbital, also making it **paramagnetic**.
4. **Calculate the Bond Order**:
- The bond order formula is given by:
\[
\text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2}
\]
- For O2:
- Bonding electrons = 10 (σ1s², σ2s², σ2p z², π2p x², π2p y²)
- Antibonding electrons = 6 (σ*1s², σ*2s², π*2p x¹, π*2p y¹)
- Bond Order = (10 - 6) / 2 = 2.
- For O2+:
- Bonding electrons = 10 (same as O2)
- Antibonding electrons = 5 (σ*1s², σ*2s², π*2p x¹, π*2p y²)
- Bond Order = (10 - 5) / 2 = 2.5.
5. **Compare the Bond Orders**:
- The bond order of O2 is 2, while for O2+ it is 2.5.
- Thus, the bond order of O2+ is greater than that of O2.
### Conclusion:
- O2 is paramagnetic.
- O2+ is also paramagnetic.
- The bond order of O2+ (2.5) is greater than that of O2 (2).
