Molecular Orbital Theory
Lewis structures and VSEPR tell you a molecule's shape, but they cannot explain why oxygen is magnetic or why He2 does not exist. Molecular orbital (MO) theory can. It treats electrons as belonging to the whole molecule, not to one atom or one bond.
What Is a Molecular Orbital?
When two atoms bond, their atomic orbitals combine to form molecular orbitals that spread over both nuclei. Two atomic orbitals combine to give two molecular orbitals:
- Bonding orbital (σ, π): lower in energy than the atomic orbitals. Electron density builds up between the nuclei, holding them together.
- Antibonding orbital (σ*, π*): higher in energy than the atomic orbitals. It has a node between the nuclei, so electron density there is zero — electrons in this orbital weaken the bond.
The same filling rules apply as for atoms: orbitals fill lowest-energy first (Aufbau), one electron per orbital before pairing (Hund's rule), and at most two electrons per orbital with opposite spins (Pauli exclusion).
MO Energy-Level Diagrams for Second-Period Diatomics
For a second-period diatomic molecule, the valence molecular orbitals in order of increasing energy are built from the 2s and 2p atomic orbitals: σ2s, σ*2s, then the 2p-derived set, then σ*2p. But the order of the 2p-derived orbitals flips partway across the period:
- B2, C2, N2: the π2p orbitals sit below σ2p. Order: σ2s < σ*2s < π2p < σ2p < π*2p < σ*2p.
- O2, F2: the σ2p orbital sits below the π2p orbitals. Order: σ2s < σ*2s < σ2p < π2p < π*2p < σ*2p.
Why the flip? In B2, C2, and N2, the 2s and 2p atomic orbitals are close in energy, so they mix (s–p mixing). This mixing pushes the σ2p orbital up in energy, above the π2p pair. In O2 and F2, the 2s orbital is much lower in energy than the 2p, so the mixing is negligible and σ2p stays below π2p.
Bond Order
Bond order measures how strong a bond is — it is the molecular-orbital version of counting single, double, and triple bonds:
$$\text{Bond order} = \frac{(\text{bonding electrons}) - (\text{antibonding electrons})}{2}$$A higher bond order means a shorter, stronger bond. A bond order of zero means no bond at all — the molecule does not exist.
Why He2 Does Not Exist — and Why B2 Is Magnetic
He2 would have 4 electrons: (σ1s)2(σ*1s)2. Bonding electrons = 2, antibonding = 2: $$\text{Bond order} = \frac{2 - 2}{2} = 0$$ The antibonding electrons cancel the bonding ones exactly, so there is no net bond. Two helium atoms simply drift apart.
B2 is the surprise case. It has 6 valence electrons: (σ2s)2(σ*2s)2(π2p)2. Its bond order is $(4 - 2)/2 = 1$, a single bond — but the last two electrons sit one each in the two π2p orbitals, so B2 has 2 unpaired electrons and is paramagnetic, even though its Lewis structure shows all electrons paired. Only MO theory predicts this.
Practice Problems
- Write the MO electron configuration of Li2, calculate its bond order, and state whether it is paramagnetic or diamagnetic.
- C2 has 8 valence electrons. Calculate its bond order and state whether it is paramagnetic or diamagnetic.
- O2+ has one fewer electron than O2. Calculate its bond order and predict how many unpaired electrons it has.
- Use bond order to explain why Be2 does not exist as a stable molecule.
Answers: 1. (σ2s)2, bond order 1, diamagnetic 2. Bond order 2, diamagnetic (all electrons paired) 3. Bond order 2.5, 1 unpaired electron (paramagnetic) 4. (σ2s)2(σ*2s)2 gives bond order (2 − 2)/2 = 0, so no bond