Main ideas of molecular orbital theory
Mulliken and Hund developed MO theory. Its main points:
- Electrons in a molecule are in molecular orbitals (MOs) that spread over the whole molecule, not over one atom.
- Only atomic orbitals of similar energy and the right symmetry combine.
- Number of MOs formed = number of atomic orbitals combined. Half are bonding, half antibonding.
- A bonding MO has lower energy and more stability than the atomic orbitals; an antibonding MO has higher energy.
- MOs fill like atomic orbitals: lowest energy first (Aufbau), at most two electrons with opposite spins (Pauli), and one in each equal orbital before pairing (Hund).
LCAO: making molecular orbitals
LCAO means Linear Combination of Atomic Orbitals. For atoms A and B:
σ = ψA + ψB (waves add, electron density builds up between nuclei → bonding MO)
σ* = ψA − ψB (waves cancel between nuclei, leaving a node → antibonding MO)
The bonding MO is lowered in energy by about as much as the antibonding MO is raised. So if both are full, there is no net gain.
Conditions for combining
- Similar energy (1s combines with 1s, not with 2p).
- Same symmetry about the molecular axis (2pz with 2pz; 2px cannot combine with 2pz).
- Good overlap.
Types of MOs
σ MOs are symmetric around the bond axis (from s orbitals or head-on 2pz). π MOs come from side-on 2px or 2py; they have electron density above and below the axis. π2px and π2py have the same energy.
Energy order and bond order
For Li₂, Be₂, B₂, C₂, N₂ (2s–2p mixing pushes σ2p up):
σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz
For O₂, F₂, Ne₂:
σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz
Bond order = ½ (Nb − Na), Nb = electrons in bonding MOs, Na = electrons in antibonding MOs.
- Bond order > 0: the molecule is stable. Bond order 0: it does not exist.
- Higher bond order → higher bond enthalpy, shorter bond length.
- Any unpaired electron → paramagnetic; all paired → diamagnetic.
Bonding in homonuclear diatomic molecules
| Molecule | Electrons | Bond order | Magnetism |
|---|---|---|---|
| H₂ | 2 | 1 | diamagnetic |
| He₂ | 4 | 0 (does not exist) | — |
| Li₂ | 6 | 1 | diamagnetic |
| Be₂ | 8 | 0 (not stable) | — |
| B₂ | 10 | 1 | paramagnetic (2 unpaired in π) |
| C₂ | 12 | 2 | diamagnetic |
| N₂ | 14 | 3 | diamagnetic |
| O₂ | 16 | 2 | paramagnetic (2 unpaired in π*) |
| F₂ | 18 | 1 | diamagnetic |
| Ne₂ | 20 | 0 (does not exist) | — |
C₂ is unusual: its bond order 2 comes from two π bonds and no σ2p bond.
Oxygen ions: O₂⁺ (BO 2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1). So the bond length goes O₂⁺ < O₂ < O₂⁻ < O₂²⁻.
Try it: fill the ladder yourself
Draw the MO ladder for O₂ on paper. Use small arrows for electrons. Fill 12 valence electrons: two per level, one in each π* before pairing. Count bonding and antibonding electrons and find the bond order. Now predict O₂⁻ (add one arrow). Check both in the free-play step.
Key formulas and definitions
- σ (bonding) = ψA + ψB; σ* (antibonding) = ψA − ψB
- Bond order = ½ (Nb − Na)
- Number of MOs = number of atomic orbitals combined
- Unpaired electrons > 0 ⇒ paramagnetic; all paired ⇒ diamagnetic
- Bond order ↑ ⇒ stability ↑, bond enthalpy ↑, bond length ↓
Worked examples
1. Find the bond order of H₂ and H₂⁺.
Step 1: H₂: σ1s² → Nb = 2, Na = 0 → ½(2 − 0) = 1. Step 2: H₂⁺: σ1s¹ → ½(1 − 0) = 0.5. Answer: H₂ 1, H₂⁺ 0.5. Both exist; H₂ is more stable.
2. Show that He₂ does not exist, but He₂⁺ can.
Step 1: He₂: σ1s² σ*1s² → ½(2 − 2) = 0 → no bond. Step 2: He₂⁺: σ1s² σ*1s¹ → ½(2 − 1) = 0.5. Answer: He₂ has bond order 0 and does not exist; He₂⁺ has 0.5 and can exist.
3. Write the MO configuration of N₂ and find its bond order.
Step 1: 14 electrons: KK σ2s² σ*2s² (π2px² = π2py²) σ2pz². Step 2: Valence Nb = 2 + 4 + 2 = 8, Na = 2. Step 3: Bond order = ½(8 − 2) = 3. Answer: 3 (a triple bond), diamagnetic.
4. Write the MO configuration of O₂ and explain its magnetism.
Step 1: 16 electrons: KK σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px¹ = π*2py¹). Step 2: Nb = 8, Na = 4 → bond order = 2. Step 3: The two π* electrons are in separate orbitals (Hund) → 2 unpaired. Answer: bond order 2, paramagnetic.
5. Compare the bond orders of O₂⁺, O₂⁻ and O₂²⁻.
Step 1: O₂⁺: remove one π* electron → Nb 8, Na 3 → 2.5. Step 2: O₂⁻: add one π* → Na 5 → 1.5. Step 3: O₂²⁻: Na 6 → 1. Answer: O₂⁺ 2.5 > O₂ 2 > O₂⁻ 1.5 > O₂²⁻ 1; bond length goes the other way.
6. Is B₂ paramagnetic or diamagnetic?
Step 1: 10 electrons: KK σ2s² σ*2s² (π2px¹ = π2py¹). (For B₂, π2p is below σ2p.) Step 2: The last two electrons go one each into the two π orbitals. Step 3: Bond order = ½(4 − 2) = 1. Answer: paramagnetic (2 unpaired), bond order 1.
7. Which is more stable: N₂ or N₂⁺? And O₂ or O₂⁺?
Step 1: N₂⁺ loses a bonding σ2p electron → BO = ½(7 − 2) = 2.5 < 3. So N₂ is more stable. Step 2: O₂⁺ loses an antibonding π* electron → BO = ½(8 − 3) = 2.5 > 2. So O₂⁺ is more stable. Answer: N₂ > N₂⁺, but O₂⁺ > O₂.
Common mistakes
- Using the O₂ energy order (σ2p below π2p) for B₂, C₂ and N₂. For those, π2p comes first.
- Pairing electrons in π* before each π* orbital has one. Hund's rule gives O₂ two unpaired electrons.
- Counting only valence electrons for Nb and forgetting Na from σ*2s. Always count both.
- Thinking an antibonding orbital is "empty space". It is a real orbital; electrons in it weaken the bond.