Molecular Geometry & Bonding
what the shape decides
Concept

Basis — where it appears

A set of functions a wavefunction is expanded in, chosen for convenience rather than found in the molecule. Every observable is unchanged by a change of basis, so arguing about which basis is real is arguing about a coordinate system.

Named by 37 essays across 8 fields — each of them below, with the objects they name alongside it.

methane — Td. The molecule with the point group found from its coordinates rather than looked up: every candidate operation was applied and kept when it permuted the atoms among themselves.

Hybridisation does not explain

Methane's photoelectron spectrum has two bands, not one. Four equivalent sp³ bonding orbitals cannot produce that, and the resolution is that hybridisation was never a claim about what a measurement would find.

wrong · Hybrids H₂O: 3 distinct modes. The displacement of every atom in 3 normal modes of H₂O, drawn from the eigenvectors of the mass-weighted Hessian. Under each is the internal coordinate with the largest share of the motion and how large that share is; where no coordinate holds nine tenths, the mode is not a motion of one bond or one angle and is labelled so.

Normal modes are not bond stretches

Water has two stretching frequencies and two O–H bonds, and it is almost irresistible to pair them off. Computed, each mode is exactly half in one bond and half in the other, and neither frequency belongs to a bond at all.

spectra · Normal mode benzene — D6h. The molecule with the point group found from its coordinates rather than looked up: every candidate operation was applied and kept when it permuted the atoms among themselves.

Delocalisation

Benzene does not alternate between two structures. It has one structure, and the two Kekulé forms are basis functions in a description of it — which is a different and much less exciting claim than the one usually made.

beyond · Delocalisation Orbitals at the 90 per cent contour. Several orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density. Contours drawn: 1s at 90% of its density, |ψ| = 3.94e-2; 2s at 90% of its density, |ψ| = 7.40e-3; 2pz at 90% of its density, |ψ| = 9.48e-3.

Orbitals are not where the electron is

A many-electron atom has no exact orbitals at all. The orbital picture is a basis for an approximation — an extremely good one — and treating it as a description of reality is the source of most of the confusion in this subject.

wrong · Approximation 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 Td character table. The irreducible representations of the molecule's point group. The class headings carry the number of operations found by generating the group from the coordinates, and those counts were produced before this table was opened.

Character tables and reduction

A molecule's point group is a list of matrices, not a label. Generating them, sorting them into classes, and dividing by the group order turns any set of orbitals into a statement about how many energies there can be.

symmetry · Representation Splitting goes with overlap. For each pair, the atomic levels on the outside and the combinations they form in the middle, with the splitting drawn in proportion to the computed overlap. A pair that symmetry forbids does not split at all, because its overlap is exactly zero.

Molecular orbital and valence bond

Two frameworks, taught as rivals, describing the same molecules. One starts from delocalised orbitals and localises; the other starts from localised bonds and delocalises. Pushed far enough they meet.

bonding · Models Orbitals at the 90 per cent contour. Several orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density. Contours drawn: 2pz at 90% of its density, |ψ| = 9.48e-3; 2px at 90% of its density, |ψ| = 9.49e-3; 2py at 90% of its density, |ψ| = 9.49e-3.

Complex harmonics against real ones

The p orbitals every chemist draws are not eigenfunctions of anything. They are real combinations of the complex solutions, chosen because they point along axes — and the choice is invisible until a magnetic field makes it matter.

orbitals · Orbital The Td character table. The irreducible representations of the molecule's point group. The class headings carry the number of operations found by generating the group from the coordinates, and those counts were produced before this table was opened.

Degeneracy is a group theorem

How many orbitals can share an energy is decided before any energy is computed. The dimensions of a group's irreducible representations are the only degeneracies it permits, and a molecule with no representation larger than one cannot have a degenerate level at all.

symmetry · Representation 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 methane: 2 valence bands. The measured valence photoelectron bands of methane, each labelled with the symmetry species of the orbital it comes from, and beside them the species the valence basis spans — the central atom's s and p functions and one s on each ligand, reduced in the molecule's own group. A band carrying a species the reduction does not produce would stop this figure being drawn.

What a photoelectron spectrum measures

The bands of a photoelectron spectrum are routinely read off as orbital energies. They are ionisation energies, which is a different quantity — and the identification rests on two errors of about an electronvolt each that happen to have opposite signs.

wrong · Photoelectron Orbitals at the 90 per cent contour. Several orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density. Contours drawn: 2px at 90% of its density, |ψ| = 9.49e-3; 2py at 90% of its density, |ψ| = 9.49e-3; 2pz at 90% of its density, |ψ| = 9.48e-3.

A filled shell has no shape

Sum the angular densities of a complete p shell and the answer is 3/4π in every direction, to sixteen decimal places. A filled d shell gives 5/4π. The lobes are in the decomposition and not in the density, and nothing that measures a closed-shell atom can see them.

orbitals · Contour 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 water: 4 valence bands. The measured valence photoelectron bands of water, each labelled with the symmetry species of the orbital it comes from, and beside them the species the valence basis spans — the central atom's s and p functions and one s on each ligand, reduced in the molecule's own group. A band carrying a species the reduction does not produce would stop this figure being drawn.

Water's lone pairs are not a pair

Every course draws two equivalent lone pairs on water, pointing away from the hydrogens like a pair of ears. Its photoelectron spectrum shows the two bands they would produce at 12.6 and 14.7 electronvolts, two point one apart, in different symmetry species.

bonding · Photoelectron The exponent the molecule chooses, and what it buys. The 1s exponent that minimises the energy of a one-electron diatomic, against the separation of the nuclei, with the binding curves at that exponent and at the free atom's. Held at ζ = 1 the bond comes out at 2.49 bohr and binds 0.0648 hartree; with the exponent free it comes out at 2.00 bohr at ζ = 1.238 and binds 0.0865. The exact answer for this molecule is 2.00 bohr and 0.1026.

The atom does not bring its own orbital

Build a one-electron diatomic from two hydrogen 1s functions and it comes out 25 per cent too long and 37 per cent too weakly bound. Let the molecule choose how large those functions are and the bond length is right to three figures, at an exponent of 1.238 — the orbital contracts by a quarter when the bond forms.

orbitals · Orbital 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 valence-bond and molecular-orbital directions in one plane. The valence-bond function and the molecular-orbital function as two directions, at the angle their overlap requires — 45.0°, since they overlap by 0.7071. The exact ground state lies in the plane they span at every repulsion, to twelve decimal places, and swings from one to the other as the repulsion grows without ever arriving. Neither picture is a special case of the other and the answer is not either of them.

Two pictures, one plane

Molecular orbital theory and valence bond theory are taught as rival descriptions of a two-electron bond. In a model small enough to solve exactly they are two vectors in a two-dimensional space, the exact answer lies in the plane they span at every repulsion, and it is neither of them at any repulsion but two.

bonding · Models Where the electron is, and how fast it is going. The radial distribution in position on the left and in momentum on the right, for the same orbitals. The two run opposite ways: the 1s is the most compact in space and the widest in momentum, and every excited orbital that spreads out in one narrows in the other. Both are normalised, both are the same function, and neither is more fundamental than the other — the transform loses nothing and adds nothing.

The orbital in momentum space

Every orbital has a second picture as complete as the first and almost never drawn. Nothing is added by taking it — it is the same function in the other variable — but the uncertainty product falls out of it, and the functions quantum chemistry is built from turn out to be the only ones that attain the bound.

orbitals · Orbital The Compton profile, exact and fitted. The momentum density integrated over the two perpendicular directions, for the exact 1s and for three fitted bases. The exact curve is 8/3π(1 + q²)³ in closed form; the fitted ones are sums of Gaussians and are cusped differently at the origin, which is the position-space cusp showing up as a shape in momentum.

The measurement a basis was not fitted to

Six Gaussians reproduce hydrogen's energy to eleven parts in a hundred thousand and its Compton profile to three parts in a thousand — twenty-five times worse, on a quantity an X-ray scattering experiment measures directly. The gap between the two errors widens as the basis is improved, because the energy is the one property a variational fit is best at.

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 Two errors, opposite signs, four orders of magnitude apart. A finite basis makes H₂⁺'s binding too large by letting each atom borrow the other's functions, and too small by describing the molecule incompletely. Both are computed here against the exact binding of 0.102634 hartree. The second is thousands of times the first at every basis size, and it is the first that counterpoise removes — so the corrected number is further from the true one than the uncorrected at every row of this table.

The basis the other atom lent

Two atoms in a molecule are described in each other's functions and the separated atoms are not, so the molecule is treated better than the pieces and the binding comes out too large. That is the basis set superposition error, it is removed by a standard correction, and for H₂⁺ in four Gaussians a centre it is six tenths of a microhartree against an incompleteness error of twelve millihartree — a factor of eighteen thousand the other way.

orbitals · Basis The two halves change places. What fraction of the counterpoise correction belongs to the lighter of two unlike atoms, against their separation. At a bonding distance it is 1.26 per cent — essentially the whole correction is the heavier atom's — and by 9.0 bohr it is 67. The two change places at 6.29 bohr. A symmetric pair's share is exactly a half everywhere, which is what makes one frozen number a complete description there and nowhere else.

A correction that is two functions

A symmetric pair's counterpoise correction splits exactly in half, at every separation, to the last digit — which is why one frozen number describes it. Give the two atoms different charges and the split runs from 0.03 per cent to 67, changing places at 6.29 bohr: the correction a single number was standing in for is two functions of different shapes.

orbitals · Basis The pair's regime is not a property of the pair. How much the pair's bonding responds to its own overlap — the ratio of what it is bonded by at an overlap of 0.4 to what it is bonded by at 0.1 — with and without a third orbital coupled to both. Alone it is 11.83, which is the regime in which bonding tracks overlap. With a third orbital present it falls to 1.46, 0.92, 0.74 — and two of those are below one, meaning a fourfold increase in the overlap between the two atoms buys them less bonding rather than more. The coupling comes from the overlap by the Wolfsberg–Helmholz rule with K = 1.75, which is fitted rather than derived. Every stabilisation here inherits that; the shape of the curve against separation does not, because K is a constant.

A regime that belongs to the neighbours

Two orbitals at the same energy are bonded in proportion to their overlap and two far apart are barely bonded at all — two regimes, and the natural question is whether the regime is a property of the pair. It is not. Put a third orbital beside them and the pair's response to its own overlap falls from twelvefold to less than one: more overlap buys less bonding.

wrong · Overlap One half is a curve and the other is not. The two halves of a counterpoise correction for an unequal pair, against separation, on a logarithmic axis. The heavier centre's falls smoothly over two decades; the lighter centre's scatters over more than one decade between neighbouring points. It is not a rough function — it is a difference of two energies of order a hartree whose difference is a millionth, and the solver does not have seven figures to spare.

The half that cannot be computed

How many points does each half of a counterpoise correction need to interpolate? The natural expectation is two different numbers. The answer is that the question is not yet askable: the lighter centre's half is a difference of two energies agreeing to six figures, its second differences are seven per cent of its own value, and no interpolation of it means anything. The third thing worth checking — the symmetric-pair check — works perfectly.

orbitals · Basis The trimer's correction, and the sum of its pairs. The counterpoise correction of a three-fragment system computed directly — each fragment's energy alone less its energy in the whole trimer's basis — against the sum of the three pairwise corrections, on a logarithmic axis. The sum is the larger everywhere the difference is above the solver's noise: 30.4 per cent at 1.6 bohr and nothing by six.

The assembly that counts one share twice

A counterpoise correction is divided unequally between its two centres, and the first place that matters is a three-fragment system, where the pairwise corrections are added up and the assembly must double-count one share and undercount another. It does: the heavy centre is over-corrected by seventeen per cent and the light ones under-corrected by two and a half, and the two do not cancel.

orbitals · Basis 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 Three cages, three answers, and none of them reassuring. For each cage: how many descriptions the search finds, what share the commonest takes, how far the whole set spreads in the functional, whether the commonest is the best, and the verdict. A search that always agrees with itself is agreeing about a choice that does not matter; a search whose descriptions genuinely differ does not return the best one.

Fifty descriptions of one molecule

A search whose largest basin takes ninety-eight per cent of its starts will report one description however long it is run, and nobody runs four thousand starts when the first fifty agree. Whether the rare ones are worse descriptions or merely rarer is one number per description, already computed and never looked at. On two of three cages they are not worse — they are the same answer, to parts per million.

beyond · Multicentre One geometry, four electron counts, four answers. The bond order between two orbitals with exactly no overlap and no resonance integral, as the third orbital's energy is swept, at every count the trio can hold. With none it is identically zero. With two it is positive and rises past one. With four it is negative and reaches -0.954. With six it is a horizontal line — the third orbital's energy stops mattering entirely.

A filled shell is not an empty statement

A bond order of −0.954 between two orbitals with no overlap and no resonance integral invites the prediction that at six electrons — every level occupied, the sum over a complete set — it would be exactly zero. It is exactly one seventh, and the reason is that a complete set in a non-orthogonal basis sums to the inverse of the overlap matrix, which has entries where the overlap has none.

wrong · Overlap The two overlaps, squared, at the measured bond lengths. For each chromium(III) donor: the σ overlap squared, a metal 3d(z²) against the donor's p(z), and the π overlap squared, a 3d(xz) against its p(x). Everything is computed — the radial functions from Slater's rules, the separation from the measured bond length, the integral by quadrature. Chloride's π overlap is 3.5 times fluoride's, which is the opposite of what the overlap argument required of it.

The overlap the model is not proportional to

Without a computed π overlap, the natural argument reasons about one instead: the denominator over-predicts the halide trend, so the overlap must shrink down the group to cancel part of it. Computed, it grows — 3.5 times from fluoride to chloride. And the fitted parameter changes sign across the series, which no ratio of squared overlaps can do.

applied · Ligand field Two charges, and three ligands no charge reaches. The metal effective charge each ligand would need on its own for the model's ratio to equal the fitted one. The band is the range Slater's rules allow chromium. Two ligands have an answer, both far outside it and 2.30 apart from each other. Chloride needs more than the overlap rule can be trusted to compute. Ammonia's fitted parameter is exactly zero and cyanide's is negative, and a quotient of squares is neither.

A contraction that cannot reach three of them

The angular overlap model's own derivation gives a π/σ ratio that disagrees with the fitted parameters by up to sixfold, and the metal's contraction is the obvious candidate to account for it. The whole range Slater's rules allow moves the ratio by a factor of two. Two ligands need charges far outside it, one needs a charge past where the overlap rule can be trusted at all, and two are unreachable at any charge because a quotient of squared overlaps cannot be zero or negative.

applied · Ligand field A thirty-four per cent variation that is entirely the truncation. The field at which the one avoided crossing sits, against the tilt, computed in the five functions a field in the xz plane couples and in the whole nine-function shell. The truncated answer runs from 1.6435e-5 to 2.2014e-5 — a factor of 1.34. The whole shell's is 2.0876e-5 at every direction, and equals the truncated answer at zero tilt, where the truncation is exact because the field is along z and the excluded functions genuinely do not couple.

The variation was the basis

Solved in the five functions a field in one plane couples, the tilted Stark problem has one avoided crossing at every tilt and a crossing field that moves by a third across ninety degrees. Solved in the whole nine-function shell the field does not move at all — the same number at every direction, to eleven decimal places — and the thirty-four per cent was the truncation.

symmetry · Representation One spectrum along three directions, and three different sets of estimates. The two-state estimate gap ÷ 2d for every pair of the shell's functions the field couples, with the field along z, along x and at the tilt the defect sweep used. Along z there are three distinct estimates, along x three different ones and at the tilt eight, none equal to any of the axial three. The exact spectrum is identical along all three directions, and its two minima are drawn as vertical lines: the one between coupled levels at 0.0196 and the tangency of uncoupled levels at 0.0400. An estimate is a property of the axes the functions were written along, and a feature is not.

Two levels cannot make a minimum

The one minimum between coupled levels in a Stark shell sits at 0.0196 of the s–p gap, and the nearest two-state estimate at 0.0192 — two per cent away, which reads as the estimates having been aimed at the right feature all along. They were not. Two coupled levels only ever separate, so no estimate can be where its own pair is closest. The minimum belongs to a third level, exists only while the d level sits within a quarter of the s–p gap, and meets the estimate by crossing it.

symmetry · Representation Every window sits above the value it was meant to reach. For each of the five chromium(III) complexes, the whole range of π/σ ratios the model can produce as the ligand's donor atom is taken through every oxidation state it has — from its bare nucleus to its closed-shell anion — drawn as a bar, with the fitted parameter marked beneath it. The three ligands whose fitted parameter is positive have bars that begin above it and never come down. The other two have fitted parameters of zero and of a negative number, which a quotient of squared overlaps cannot be at any charge.

The correction that moves three of them backwards

Every ligand radial function in the angular overlap sweeps was a neutral atom's, while three of the five donors carry a formal charge. Giving each one the charge it actually has moves three of the five computed ratios — and moves all three away from the fitted parameter, none towards it. The whole window each donor's own oxidation states allow sits above the value it was meant to reach.

applied · Ligand field

Named alongside it

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

Model limitOverlap integralApproximationOne-electron modelsConventionDegeneracyMolecular orbitalOrthogonalityConvergenceLocalisationReference stateWavefunction

All concepts