Hybridization & molecular orbital theory
The idea
Hybridization and molecular orbital (MO) theory are two complementary answers to a question Lewis structures dodge: what are the actual orbitals doing in a bond? Hybridization is the local fix — mixing one atom's s and p orbitals into sp, sp² or sp³ sets that point along the observed VSEPR geometry, with leftover p orbitals forming π bonds. It explains methane's four identical bonds and the rigidity of a C=C double bond cleanly and quickly.
MO theory is the global model: atomic orbitals across the whole molecule combine into delocalized bonding and antibonding orbitals, filled bottom-up like atomic levels. Its scorecard is bond order = (bonding electrons − antibonding electrons)/2, and its signature triumph is O₂ — Lewis predicts all electrons paired, yet liquid oxygen clings to a magnet. MO filling puts two unpaired electrons in degenerate π* orbitals, predicting both the double-bond strength and the paramagnetism in one stroke.
Resist the misconception that hybridization is a physical event atoms undergo before bonding. Both schemes are models — mathematical re-descriptions of the same electrons — and you choose whichever lens fits the question: hybridization for geometry and σ/π counting, MO for magnetism, bond order, and delocalization.
Worked example
Use molecular orbital theory to determine the bond order of O₂ and predict whether the molecule is paramagnetic or diamagnetic.
- Count the valence electrons that enter the MO diagram: each oxygen brings 6, so 12 electrons fill the 2s- and 2p-derived molecular orbitals.
- Fill in energy order for O₂ (where σ2p lies below π2p): σ2s² σ*2s² σ2p² π2p⁴ π*2p². The first 10 electrons fill through π2p; the last 2 must enter the degenerate π* pair.
- Apply Hund's rule to the two π* orbitals: the electrons occupy separate orbitals with parallel spins rather than pairing, leaving 2 unpaired electrons.
- Compute bond order: bonding electrons number 8 (σ2s, σ2p, two π2p), antibonding number 4 (σ*2s, two π*2p singles), so bond order = (8 − 4)/2 = 2 — a double bond, matching the measured bond length and strength.
- Interpret against experiment: unpaired electrons make O₂ paramagnetic, exactly what the magnet test on liquid oxygen shows and what no correct Lewis structure can reproduce.
Answer. O₂ has bond order 2 and two unpaired π* electrons, so it is paramagnetic — a prediction unique to MO theory.
Check your understanding
- Why does filling antibonding orbitals weaken a bond, and what does a bond order of zero predict about a molecule like He₂?
- How would the bond order and magnetism change if you removed one electron from O₂ to form O₂⁺, and what does that imply about its bond length?
- When would you reach for hybridization instead of MO theory, and what kinds of questions force you the other way?
- How does the sp² + leftover-p picture of ethene explain why rotation about a C=C bond is restricted while rotation about C–C is free?
Build the foundations first
Hybridization & molecular orbital theory builds on these concepts. If any feel shaky, start there.