Molecular Geometry & Bonding
what the shape decides
Concept

Orthogonality — where it appears

The vanishing of the overlap between two functions, which makes their coefficients into independent shares. Where it fails, squaring a coefficient stops being a weight and every scheme for repairing that is a convention.

Named by 14 essays across 4 fields — each of them below, with the objects they name alongside it.

1s with 2px at 2.8 bohr. The two orbitals in the plane containing both nuclei, with the regions where their product is positive and negative shown faintly. The overlap integral is the signed volume of that product, and where symmetry makes the two regions mirror images it comes out exactly zero. Contours drawn: 1s at 50% of its density, |ψ| = 1.48e-1; 2px at 50% of its density, |ψ| = 3.16e-2.

Exactly zero

Where symmetry forbids an interaction the overlap is not small. It is zero — and computing it and finding arithmetic noise is a different kind of statement from computing it and finding a small number.

bonding · Overlap sp3 hybrids. The directions the hybrids point, with the angle between them computed from the coefficients rather than quoted. The set is an orthogonal transformation of the atomic orbitals, so it describes the same space in different coordinates.

Hybrids are a basis

An sp³ hybrid set is an orthogonal matrix applied to the atomic orbitals. Rotating a basis changes no observable, so asking whether the electrons are really in hybrids is asking which coordinate system nature prefers.

bonding · Hybrids The s character a bond angle requires. The fraction of the one s orbital each of two equivalent hybrids must carry in order to meet at a given angle, from pure p at 90 degrees to half s at 180. The marks are measured bond angles — H₂O, NH₃, H₂S, PH₃ — and the column beside them is the budget each implies: what one bond hybrid takes, and what is left for the lone pairs to share. The relation is orthogonality, not a model of bonding, and nothing here is fitted.

The angle does not fix the hybridisation

Two equivalent hybrids are orthogonal only at one relation between their s character and the angle between them. Run it backwards on measured bond angles and water's bonds come out sp³·⁹⁹, hydrogen sulfide's sp²⁷, and phosphine's lone pair takes eighty-four per cent of the one s orbital.

shape · Hybrids Four bonds, or one a₁ and three t₂. The same four occupied orbitals written in two bases. On the left each orbital sits on one bond; on the right one is shared over all four hydrogens and three follow the Cartesian directions. The transformation between them is orthogonal, so the density is unchanged.

The localisation transformation, demonstrated

Four equivalent bonds or one a₁ and three t₂ — two descriptions of methane's bonding electrons that disagree about everything except the electron density, which they agree about to the last bit a double can hold.

bonding · Hybrids Two orbitals, 4 electrons, S = 0 and S = 0.25. Two interacting orbitals with 4 electrons in them, drawn twice: once with the overlap set to zero and once with it kept at 0.25. Dropping the overlap makes the two shifts equal, which is the picture usually taught; keeping it makes the upper level rise by more than the lower falls, which is why four electrons in two orbitals is a repulsion.

The antibonding level goes up more

The two-level diagram every course draws is symmetric, and the symmetry is an artefact of setting the overlap to zero. Keep it, and the upper level rises further than the lower one falls — which is why helium has no molecule and why closed shells push each other apart.

bonding · Overlap How far two hybrids are from orthogonal. The overlap between two equivalent s–p hybrids of a stated label, against the angle between them. Each curve crosses zero at exactly one angle — sp3 at 109.47°, sp2 at 120.00°, sp at 180.00° — and a molecule whose measured angle is not that angle has hybrids that overlap. The largest here is cyclopropane at 0.63.

Hybrids that were never orthogonal

Two sp³ hybrids are orthogonal at 109.47° and nowhere else. Water's are drawn at 104.5° and overlap by 0.062; cyclopropane's are drawn at 60° and overlap by 0.625, which is not a small correction to a basis but a description that has stopped being one.

shape · Hybrids Why a character table has the rows it has. All 20 groups this site defines, with the two counts that fix the shape of every character table: the number of representations equals the number of classes, and the sum of the squares of their dimensions equals the order of the group. Neither column pair differs anywhere in the table, and there is no room for another row in any of them.

Why a character table stops where it stops

A character table is square, and both of its dimensions are forced. The number of representations equals the number of classes, the squares of their dimensions sum to the order of the group, and between them there is no room for another row.

symmetry · Representation The wrong shape, fitted as well as it can be. The exact hydrogen 1s orbital and the best sums of one, two, three and six Gaussians, each with its exponents optimised for the energy. Three of them already reproduce the exact function to 99.94 per cent by overlap, which is why the method works at all — and the two places it goes wrong, at the nucleus and far out, are exactly where the other faces of this figure look.

A Gaussian is the wrong shape

Sixty years of molecular calculation are built on functions that get the two ends of an orbital wrong. A Gaussian has no cusp at the nucleus and dies too fast far away, and no number of them fixes either — while three of them already reproduce hydrogen's 1s to better than 99.9 per cent by overlap, and that is why the method works.

orbitals · Basis Two answers from one projector: E1g. The E1g projection operator of benzene, applied to the pz function on one atom and then to the one on its neighbour. Both results belong to the same two-dimensional representation and span the same subspace; neither is more correct than the other; and they are different pictures, overlapping by 0.500. The circle areas are the coefficients and the two colours are their signs.

The projector is unique, the basis is not

A reduction says how many times each representation appears. It cannot say which combinations of orbitals they are — that takes a projection operator, and for a degenerate representation the operator returns a different pair of orbitals depending on which function it is handed first. Both pairs are correct, they span the same space, and every energy computed from either is identical.

symmetry · Representation The cost of a ratio decided somewhere else. Four bases containing exactly the same six primitives, differing only in how many of the linear coefficients the calculation may choose. The horizontal axis is the effective nuclear charge, which is this one-electron problem's only knob for a different environment; the contraction was fitted at one. At 1.238 — the exponent H₂⁺ chooses when a bond forms — the fully contracted basis is 0.0283 hartree above what the same six functions could give, and one freed coefficient removes most of it.

A contraction is a decision made once

Every published basis set freezes its primitive functions into fixed combinations, on an isolated atom, before any molecule is in sight. Freeing one coefficient recovers three quarters of what that costs — and freeing it at the other end of the basis recovers one per cent.

orbitals · Basis The ionic weight against how much ionic structure is in the wavefunction. The percentage each convention calls ionic, at a fixed structure overlap, as the amount of ionic structure in the wavefunction is raised from none to the molecular orbital value. They meet at both ends of the sweep and disagree everywhere between. Every curve is a weight and every set sums to one.

A weight that depends on how it is weighed

The ionic character of a two-electron bond is quoted as a percentage. For one wavefunction at hydrogen's bond length, three conventions in the literature give 18.73, 34.74 and 5.88 per cent — a factor of six — and on a wavefunction with no ionic structure in it at all, one of them still reports a quarter.

bonding · Models Three bent bonds, at a hundred and one degrees to each other. A carbon–carbon triple bond in its localised description: three equivalent bent bonds, spaced by 101.54 degrees, each tilted 63.43 degrees off the axis and each carrying 0.17 of an s orbital — an sp⁵ hybrid. Its charge sits 0.32 ångström off the axis, where every canonical orbital's sits on it. The mixing that makes the three equivalent is a rotation in the three-dimensional occupied space, so the density is untouched.

Three bent bonds, and the same hybrid

A triple bond localises into three bent bonds at 101.537 degrees to one another, each an sp⁵ hybrid with exactly one sixth s character. A double bond's two bent components are sp⁵ hybrids at 101.537 degrees. The two are the same hybrid, built out of different frameworks and in different numbers, and the reason is that a third of a half is a half of a third.

shape · 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

Named alongside it

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

BasisOverlap integralUnitary transformationLocalisationModel limitConventionHybridisationLone pairLöwdin orthogonalisations charactersp³ hybridsBond angle

All concepts