Molecular Geometry & Bonding
what the shape decides
Concept

Hybrid orbital — where it appears

A combination of the s and p functions on one centre, chosen to point along the directions a molecule's bonds take. It is a change of basis rather than a physical state: the same wavefunction is described equally well by the unmixed orbitals, so nothing measurable depends on the choice.

Named by 5 essays across 2 fields — each of them below, with the objects they name alongside it.

One double bond, two descriptions. A carbon–carbon double bond drawn twice in the plane perpendicular to the molecule: as a σ orbital along the axis with a π orbital above and below it, and as two equivalent bent bonds tilted 50.8 degrees either side of the axis. The two descriptions are related by a rotation and have the same density everywhere.

Two bent bonds, or a σ and a π

A carbon–carbon double bond is drawn two ways and the two look like rival claims about what is there. They are one occupied space in two bases, related by a rotation of exactly forty-five degrees: the density is identical to the last bit a double holds, the bent pair are sp⁵ hybrids at 50.77° to the axis, and their charge sits 0.235 Å off the plane of the molecule where neither canonical orbital's does.

bonding · Hybrids The inequality, decided twice. The quantity the localisation criterion compares — the centroid separation over twice the off-axis dipole — computed with hydrogenic radial functions and with Slater ones at the same effective charges. Below one the localised description is a pair of bent components; above it, σ and π. The hydrogenic answer is 1.8090 and the Slater answer is 0.3937, and they are on opposite sides.

The node that decided a picture

A hydrogenic calculation cannot say whether formaldehyde's localised description is two bent components or σ and π, because the integral that decides it came out at 0.00718 where it should be the largest in the molecule. Replacing the hydrogenic radial functions with nodeless Slater ones moves that overlap by a factor of ninety-seven, leaves the π overlap identical to the last bit, and flips the answer: the description is bent.

bonding · Hybrids What a rotation group buys over a single angle. The Boys functional at the canonical orbitals, at the best mixing of the two bonds with the lone pair left alone, and at the maximum over the whole three-dimensional rotation group. The first step is 11.70 per cent and the second, which no two-orbital treatment can reach, is a further 2.50.

An interior maximum a third orbital allows

Two orbitals related by a mirror plane give a Boys functional with no linear term, so it has no interior maximum and no angle is ever searched for. Add the oxygen lone pair and the search over a three-dimensional rotation group finds one — eight and a half degrees of lone pair mixed into two bent bonds, worth a further two and a half per cent that no pair of orbitals could reach.

bonding · Hybrids The criterion is the reciprocal of the p amplitude. The two lone pairs' mixing criterion — twice the dipole between them over the separation of their centroids — against the p fraction of the outward hybrid. The curve is 1/√(p fraction), written down rather than fitted; the marks are the numerical integrals. They agree to 15 parts in a million at every hybrid, and the criterion never falls to one, so the mixed description wins at every s character there is.

The five figures were an identity

Two lone pairs satisfy the two-orbital mixing criterion by a margin of 1.732128, against √3 = 1.732051 — agreement to five figures on a number assembled from three integrals over a numerical grid. It is exact. The criterion is the reciprocal of the p amplitude of the outward hybrid, and everything else in the molecule cancels out of it.

bonding · Hybrids The zero at π/R follows parity, not bonding. The bonding and antibonding combinations of three atomic functions on nitrogen, each along the bond and each normalised to its own largest value, against momentum in units of π/R. For 2s the bonding combination is zero at π/R and the antibonding one is not. For 2p along the bond it is the other way round: the σ bond carries a sine, is zero at the origin and near its largest at π/R, and the antibonding combination carries the cosine. For 2p across the bond the π bond carries the cosine again. The factor is a cosine exactly when the orbital's inversion parity matches the atomic function's.

The zero is a parity, not a bond

A hydrogen-like σ bond has a momentum profile along its axis that vanishes at π/R, and it is natural to read that zero as a bond's signature. Built from 2p functions pointing along the axis, the σ bond carries a sine instead and sits at 83 per cent of its peak there. Which factor an orbital carries is decided by whether its inversion parity matches its atom's, and bonding has nothing to do with it.

orbitals · Orbital

Named alongside it

The objects these essays reach for when they reach for this one.

LocalisationClosed formModel limitUnitary transformationBent bondDipole momentEffective nuclear chargeLone pairLöwdin orthogonalisationMolecular orbitalOne-electron modelsOrthogonality

All concepts