Draw molecular orbital diagrams for O2-, O22-, and O2. Which has the highest bond order? Which would be paramagnetic, and which would be diamagnetic? Can you draw good dot structures that correspond to each of these ions or molecules?
The Correct Answer and Explanation is:
To compare O₂, O₂⁻, and O₂²⁻, we will:
- Draw molecular orbital (MO) diagrams
- Determine bond orders
- Classify each as paramagnetic or diamagnetic
- Provide Lewis (dot) structures
1. Molecular Orbital Diagrams and Bond Order
For oxygen and other elements with atomic numbers ≥ 8, the molecular orbital energy ordering is:
σ(1s), σ*(1s), σ(2s), σ*(2s), σ(2p_z), π(2p_x) = π(2p_y), π(2p_x) = π(2p_y)**, σ(2p_z)*
Oxygen has 8 electrons, so O₂ has 16 total valence electrons (8 each).
We fill MOs in the order above.
➤ O₂ (16 e⁻ total):
- Bonding orbitals: σ(2s), σ(2p_z), π(2p_x), π(2p_y) = 8 electrons
- Antibonding orbitals: σ*(2s), π*(2p_x), π*(2p_y) = 4 electrons
- Bond order = (bonding – antibonding) / 2 = (10 – 6) / 2 = 2
- Paramagnetic (2 unpaired electrons in π* orbitals)
➤ O₂⁻ (17 e⁻):
- Add 1 e⁻ to π*(2p_x) or π*(2p_y):
- Bond order = (10 – 7) / 2 = 1.5
- Paramagnetic (1 unpaired electron)
➤ O₂²⁻ (18 e⁻):
- Add another e⁻ to complete both π* orbitals
- Bond order = (10 – 8) / 2 = 1
- Diamagnetic (all electrons paired)
2. Lewis Dot Structures
- O₂ → O=O with one unpaired electron on each O (diradical); hard to represent well with dots
- O₂⁻ → O–O⁻ with a single bond and a lone electron (radical ion)
- O₂²⁻ → O–O²⁻ with a single bond, all electrons paired
3. Summary
| Molecule/Ion | Bond Order | Magnetic Property | Dot Structure Summary |
|---|---|---|---|
| O₂ | 2 | Paramagnetic | O=O with unpaired electrons |
| O₂⁻ | 1.5 | Paramagnetic | O–O⁻ with one unpaired e⁻ |
| O₂²⁻ | 1 | Diamagnetic | O–O²⁻, all electrons paired |
Conclusion
- Highest bond order: O₂ (2)
- Paramagnetic: O₂ and O₂⁻
- Diamagnetic: O₂²⁻
MO theory explains both bond strength and magnetism, which dot structures alone can’t fully convey—especially for O₂.
