The node that decided a picture

The angle that does not have to be searched for proved that the Boys criterion applied to a double bond has no interior maximum: the localised description is either two bent components or the σ and π orbitals themselves, decided by whether the two orbitals’ centroid separation is smaller or larger than twice the off-axis dipole between them. Then it declined to say which formaldehyde is, and said why:

Its σ overlap computed from hydrogenic functions is 0.00718, where it should be the largest overlap in the molecule. A hydrogenic 2s has a radial node inside the bond, and the s–s and s–p contributions very nearly cancel. Run with those numbers the answer is canonical. It should not be believed.

It should not have been. With the node gone the answer is bent, and the whole of this essay is one substitution and its consequences.

The one change #

A hydrogenic 2s at an effective charge of 3.25 changes sign at 0.62 bohr. A carbon–oxygen bond is 2.27 bohr, so the node sits deep inside it and the two lobes contribute to the overlap with opposite signs.

A Slater-type 2s has no node: its radial part is $r\,e^{-\zeta r/2}$, which is the same radial part its 2p has. That is not an approximation to the hydrogenic function; it is the function every practical basis is built from, and a contracted Gaussian set is a fit to it rather than to a hydrogenic orbital.

Everything else is held: the same effective charges, the same bond length, the same hybridisation, the same criterion, the same quadrature grid. One line of the calculation changes — which function the hybrids are built from — and the change is a substitution rather than a refit, since the Slater function is written down analytically at the charge Slater’s rules already supply.

That is the shape a good comparison has, and it is worth contrasting with the obvious alternative. Fitting contracted Gaussians to a Slater function and computing with those would have introduced a fitting step, a contraction length, and a set of exponents — three things to argue about between the two answers. Using the Slater function itself removes all three, and the Gaussians were only ever a way of making the integrals analytic, which is not needed when the integrals are done numerically anyway.

The control is exact #

The π overlap does not move at all, and that is not luck: a hydrogenic 2p and a Slater 2p are the same function, both being $r\,e^{-\zeta r/2}$ times a direction. So the comparison has an exact internal control — one integral changes by two orders of magnitude and the other by a part in ten thousand million million, and the difference between them is precisely the node.

It is worth saying how rarely a control is that clean, and what it buys. Most comparisons between two calculations differ in several things at once and the argument is about which difference matters. Here the two bases differ in exactly one function, and the one integral that does not involve it is bit-identical. So no argument is needed about whether the quadrature changed, whether the normalisation changed, whether the grid resolves one basis better than the other — all of those would have moved the π overlap, and it did not move.

Why the cancellation is so nearly complete #

The size of the effect is worth a paragraph, because a node makes the integral smaller would not explain a factor of ninety-seven.

The overlap of two hybrids is a sum of four terms — s with s, s with p, p with s, p with p — and the hybrids point at each other, so the s–p terms carry one sign and the p–p term the other. With a nodeless 2s all four add up to something of the size of the largest. With a hydrogenic 2s the s–s and s–p contributions are themselves differences across the node, each much smaller than its parts, and the remainder is what is left after two cancellations rather than one.

A near-zero from a double cancellation is not a small number; it is a difference of large ones, which is the same hazard a counterpoise half runs into. The difference is that there the cancellation is arithmetic and here it is the basis: no amount of precision repairs it, because the function itself is wrong for the job.

What the corrected numbers say #

Read the last row first. The ratio the criterion compares is 1.809 in one basis and 0.394 in the other, and the boundary is 1 — so the two calculations do not merely differ in a number, they differ in which of two qualitatively different pictures the molecule has. There is no picture in between: that was proved before either number existed.

The σ orbital stops being oxygen’s. At an overlap of 0.007 the two-level problem barely couples, so its lower root is essentially the oxygen hybrid with a trace of carbon — 99.9 per cent on oxygen, which is not a bond. At 0.698 the two centres share it 41 to 59, which is a polarised bond and is what a carbonyl σ should look like.

And both dipoles change sign as well as size, which is why the ratio moves by a factor of four and a half rather than by ninety-seven: the two quantities the inequality compares are both affected, and they move in opposite directions.

So formaldehyde is bent #

The answer to the question is that the Boys criterion gives formaldehyde two bent components — bananas — and not σ and π.

The margin is not narrow. The ratio is 0.394 against a boundary at 1, so the bent description wins by a factor of two and a half — and the losing answer, at 1.809, lost by a factor of 1.8. Both calculations are confident and they disagree, which is the position a reader is in whenever two bases give different answers and neither is checked against anything.

What settles it here is not the size of either margin but the control: one of the two calculations contains an integral that is demonstrably an artefact, and the other does not. A double bond is not two single bonds records the artefact where it first appears, which is why it could be found rather than argued about.

That is the answer chemists’ intuition has always given for a double bond, and it is worth noticing that the collection reached it only after correcting an artefact that had pushed it the other way. A wrong integral had produced the more surprising answer, which is the usual direction of that failure: an artefact rarely reproduces the received picture, and a computation that disagrees with intuition is exactly the one to check hardest.

The number was checked, and reported with a warning attached rather than either suppressed or believed. That is what made the repair a single change rather than an investigation.

What was not touched #

Everything the theorem established is untouched, and it is worth listing because none of it depended on the integral.

The functional has no linear term, so its maximum is at forty-five degrees or at the ends and never in between. That is algebra.

The average s character is conserved across the mixing, whatever the mixing is.

And the boundary is $|\Delta| < 2|d|$, in closed form, for any molecule.

What needed an integral was one thing, and it needed only one: which side of the boundary formaldehyde is on. The theorem said what the two possible answers are and the arithmetic says which — and that division of labour is the reason a bad integral could sit inside a sound argument without damaging it. An argument whose theorem and whose arithmetic are separable can be repaired in one place; one where they are entangled has to be rebuilt.

What a hydrogenic basis is for, and what it is not #

This collection is built on hydrogenic functions, and that is a deliberate choice worth defending before it is qualified.

They are exact. A hydrogenic orbital is the solution of a problem rather than a fit to one, so a contour that encloses ninety per cent of its density encloses exactly that, a node is exactly where the mathematics puts it, and every quantity drawn from one is a consequence rather than a parameter. That is the whole case for using them and nothing here disturbs it.

They are exact for the wrong atom, and the difference is not small in the region a bond lives in. A hydrogenic 2s is the 2s of a one-electron ion at charge Z, and its node is where a one-electron ion’s node is. A real carbon 2s also has a node — orthogonality to the 1s requires it — but at a different radius, and more to the point a real valence calculation replaces the whole inner region with a smooth function precisely because nothing there affects a bond.

So the rule worth stating is: hydrogenic functions are right for anything about the shape of one orbital and wrong for anything about the overlap of two across a bond. Contours, nodes, radial distributions, enclosed fractions — all safe. An integral whose integrand changes sign inside the bond — not.

That is a sharper rule than use a better basis, and it explains why hydrogenic functions get away with so much: almost everything they are used to draw is the first kind.

What is quoted, and what is computed #

The localisation transformation is where the criterion comes from and bent bonds are a description rather than a discovery is where its limits are recorded.

Nothing is quoted. Effective charges from Slater’s rules, a bond length, an assumed one-third s character on each centre, and the standard Wolfsberg–Helmholz constant.

Both bases are evaluated on the same quadrature grid, and every hybrid is checked normalised on it to the grid’s own tolerance before anything is read off — which matters here because a comparison between two radial shapes would be meaningless if one of them were normalised better than the other.

The Slater functions are written out rather than fitted: $r\,e^{-\zeta r/2}$ with the same $\zeta$ the hydrogenic functions use, normalised analytically. Fitting Gaussians to them would be a step further from the answer rather than nearer it.

What this cannot say #

The bond length and the hybridisation are inputs. A carbonyl’s 1.203 ångström is quoted and the one-third s character is assumed, and both enter every integral. Changing either moves the ratio, and the margin of two and a half below the boundary is what says the conclusion survives ordinary variation in them rather than a claim that it survives any.

Nothing here says the Slater answer is right about formaldehyde. It says it is right about this model of formaldehyde, and the model is one electron in a two-centre problem with a fitted resonance integral. What has been removed is a demonstrable artefact, not every artefact.

The effective charges are Slater’s, which are fitted to reproduce sizes rather than energies — the same caution the halide π gaps need. The comparison is safe because both bases use the same ones; the absolute value of the σ overlap is not.

One-third s character is assumed on both centres. The earlier calculation assumed it too, and it is the one structural input neither computes: a carbonyl’s carbon is nominally sp² and its oxygen is not obviously anything. Both dipoles depend on it, so the ratio does — and the corrected answer, at 0.394, has a factor of two and a half of margin below the boundary, which is more than the assumption is likely to be worth.

And a criterion is a choice, which is the standing caution and is inherited unchanged. Boys is one localisation functional and Pipek–Mezey is another, and the second famously returns σ and π for a double bond where Boys returns bananas. So formaldehyde is bent is a statement about the Boys criterion, and what is corrected here is the arithmetic inside one criterion rather than the picture.

What was checked #

The π overlap is the same integral in both bases, to 10⁻¹⁶ — the exact control, and the check that fails first if the two bases differ in anything but the 2s.

The σ overlap moves by more than a factor of fifty, and in the Slater basis it is the largest overlap in the molecule, which is what a σ overlap of a double bond has to be.

The inequality comes out on opposite sides in the two bases, checked as an inequality between two yes-or-no answers rather than between two numbers, because the finding is the flip rather than its size.

And with the nodeless basis the answer is bent components, checked directly, so that any change that quietly moved it back would be caught. That is the one check here that is about chemistry rather than about arithmetic, and it is deliberately the narrowest: it names the criterion, the basis and the molecule.

Which overlaps the artefact can reach, and why it is worst at a bond #

The sweep is deferred, and the set it would have to sweep can be narrowed by inspection rather than by running anything — because the defect has a location.

A hydrogenic 2s function has a radial node, and the node is at $2a_0/Z_{\text{eff}}$. For a second-row atom that puts it about a third of an ångström from the nucleus. An overlap integral between two such functions on centres separated by a typical bond length is therefore an integral over a region containing both nodes, with the integrand changing sign inside it, and the answer is a difference between two contributions of comparable size.

That is the whole mechanism, and it identifies the exposure precisely.

Any overlap with a 2s on at least one centre is affected, because only those integrands change sign.

Overlaps between p functions alone are exact, which is why the π overlap here is identical to the last bit in both bases and served as the control.

And the damage is largest at bonding separations, not at large ones. Pull the centres far apart and each function’s tail is monotonic, so the cancellation weakens; bring them to a bond length and the nodes sit squarely inside the overlap region. The artefact is worst exactly where the integral is being used.

So the sweep has a short list rather than a long one, and a directional expectation to check it against: every affected integral should be too small, and the ones drawn at bond lengths should be the worst.

Still open: other σ overlaps with a 2s, and a third orbital #

The obvious open question is every other σ overlap between second-row atoms computed with a hydrogenic 2s. Each has the same node in the same place — so the same factor of ninety-seven is waiting wherever a 2s–2s or 2s–2p contribution is part of a bond. Formaldehyde’s case surfaces first because here the overlap decides a picture rather than merely sizing an interaction. Finding the others means listing every bond whose σ overlap includes a 2s contribution and recomputing it with a nodeless function, and every one of them should come out larger.

The nearer question is the third orbital, and it is now worth attempting. A carbonyl’s oxygen carries lone pairs, and localising over the whole occupied space rather than over two orbitals turns a two-by-two rotation into a search over a three-dimensional rotation group, where the argument that killed the linear term does not apply. Whether an interior maximum reappears there — and whether the famous rabbit-ear description of a carbonyl’s lone pairs is the same phenomenon — is a computation of the same shape with one more function in it, and it needs the σ overlap just repaired.