Hybridization
High-Yield Summary
- Hybridization is a mathematical model (rooted in LCAO) that reconciles ground-state electron configuration with the bonding geometry actually observed in molecules — it's not a physical process.
- sp³ (e.g. methane): 1s + 3p → 4 degenerate orbitals, tetrahedral, 109.5°, 25% s / 75% p character, 0 unhybridized p orbitals (single bonds only).
- sp² (e.g. ethene): 1s + 2p → 3 degenerate orbitals, trigonal planar, 120°, 33% s / 66% p character, 1 unhybridized p orbital (enables a double bond).
- sp (e.g. ethyne): 1s + 1p → 2 degenerate orbitals, linear, 180°, 50% s / 50% p character, 2 unhybridized p orbitals (enables a triple bond).
- The number of unhybridized p orbitals always equals the number of pi bonds the atom forms; higher %s character → shorter, stronger bonds.
Hybridization States
| Hybridization | Geometry / angle / %s-%p / unhybridized p |
|---|---|
| sp³ | Tetrahedral, 109.5°, 25% s / 75% p, 0 unhybridized p |
| sp² | Trigonal planar, 120°, 33% s / 66% p, 1 unhybridized p |
| sp | Linear, 180°, 50% s / 50% p, 2 unhybridized p |
Key Terms
- Hybridization
- Mathematical model (LCAO) generating hybrid orbitals from weighted sums of atomic orbitals to explain observed bonding geometry.
- Degenerate orbitals
- Orbitals of equal energy — hybrid orbitals within a given hybridization state are always degenerate.
- Resonance
- Occurs when two+ valid Lewis structures (same atom arrangement, different electron placement) describe a molecule; true structure is a hybrid of all contributing forms, increasing stability via delocalization.
- Resonance hybrid
- The actual, delocalized electronic structure of a molecule — not a mixture the molecule switches between, but its single true state.
Ranking Resonance Structures by Stability
- 1Satisfy the octet rule — structures where all atoms have a complete octet (exceptions like H) are generally more stable.
- 2Minimize formal charges — best structures spread electrons to produce the fewest overall formal charges.
- 3Place negative formal charges on more electronegative atoms (positive charges on less electronegative atoms).
- 4If structures satisfy all rules equally (e.g. NO₃⁻'s three forms), they contribute equally to the resonance hybrid.
Common MCAT Trap
- More unhybridized p orbitals = more pi bonds, not more sigma bonds — sp always still forms sigma bonds to its two substituents plus 2 pi bonds from the 2 leftover p orbitals.
- Resonance structures are not different molecules in equilibrium — the molecule has one true structure, the resonance hybrid, at all times.
- Signs of resonance to spot fast: allylic lone pairs, allylic carbocations, lone pairs adjacent to carbocations, pi bonds between atoms of differing electronegativity, conjugated rings (e.g. benzene).
Quick Recall
What hybridization does a carbon with a double bond adopt, and how many unhybridized p orbitals does it have?
Why does sp hybridization produce the strongest, shortest C–C bond among sp³/sp²/sp?
List the formal-charge rules for ranking resonance structures, in order.
What bond angle and geometry does sp² hybridization produce?