Hybrid Orbitals from Domain Count

The observed geometry comes first: methane has four equivalent C–H bonds arranged tetrahedrally. In a localized valence-bond description, four directional carbon orbitals can be constructed by combining one 2s and three 2p orbital functions. Those four sp³ hybrids point toward the four hydrogen atoms and overlap with their 1s orbitals.

Hybridization is a mathematical way of describing bonding, not an experimentally observed sequence in which an isolated atom physically mixes orbitals before bonding. It is also not the only valid bonding model. The simple free-atom configuration alone does not fully describe a molecule, whose electrons and orbital energies change upon bond formation.

The number of orbitals is conserved: three atomic orbitals produce three hybrids, not two or four. The superscript in sp² counts the p orbitals used, not the number of π bonds. A lone pair can occupy one hybrid direction, so NH₃ and H₂O both have a four-domain sp³ description in this elementary model even though their molecular shapes differ.

An sp² atom keeps one unhybridized p orbital perpendicular to the plane, and an sp atom keeps two. Those leftover p orbitals are exactly what form pi bonds, which is why sp² carbon appears in double bonds and sp carbon in triple bonds.

For ethene H₂C=CH₂, each carbon has three σ-bond directions and is described as sp². Its remaining p orbital overlaps side-on with the other carbon's p orbital to make one π component. For ethyne HC≡CH, each carbon has two σ directions and is sp; two unhybridized p orbitals on each C form two perpendicular π components. The multiple bond supplies one σ direction, not a separate domain for every bond line.

Start from a valid Lewis structure. At the atom of interest, count σ connections and lone-pair directions. Match two, three or four domains to sp, sp² or sp³ for the ordinary second-period centers considered here. Then check how many p orbitals are left for any π bonding. Carbon in H–C≡N has two directions and two leftover p orbitals; oxygen in H₂O has two bond directions plus two lone pairs and no leftover p in the simple sp³ model.

The ideal angle belongs to the hybrid framework, not an exact measured angle for every molecule. Heavy-atom hydrides and delocalized lone pairs can require a less literal model. Do not extend this table into a claim that SF₆ must use literal d-orbital hybrids, or label a conjugated amide nitrogen sp³ solely because one Lewis contributor shows three bonds and a lone pair; its lone pair participates in a planar delocalized system.