The angle that does not have to be searched for

The hybrids that point outside the bonds worked out how far a small ring’s hybrids point outside the lines joining its carbons — 22.75° for cyclopropane, falling to 0.035° for cyclohexane — and closed by naming the case it had not done: a bond with no symmetry to enforce equivalence.

The reasoning was straightforward. In ethene the two atoms are the same atom, so the σ and π orbitals are polarised identically and mixing them at forty-five degrees produces two components that are each other’s reflection. A carbonyl has neither of those. So, it said, the mixing angle would have to be searched for, and the two components’ s characters would be two different numbers.

Neither is true, and the reason is one line of algebra that the symmetric case had made invisible.

The mixing that was assumed, and the one that is general #

A bent description of a double bond takes the two occupied orbitals and mixes them. The familiar form is

$$\tau_\pm = \cos\theta \cdot \sigma \pm \sin\theta \cdot \pi,$$

which at $\theta = 45°$ gives two components related by the molecule’s mirror plane.

That is a one-parameter family and it is not the general orthogonal mixing. Requiring $\tau_+ = \cos\theta\,\sigma + \sin\theta\,\pi$ to be orthogonal to $\tau_- = \cos\phi\,\sigma - \sin\phi\,\pi$ needs $\cos(\theta + \phi) = 0$, so $\phi = 90° - \theta$ and the general pair is

$$\tau_+ = \cos\theta \cdot \sigma + \sin\theta \cdot \pi, \qquad \tau_- = \sin\theta \cdot \sigma - \cos\theta \cdot \pi.$$

The two components then carry different shares of σ — $\cos^2\theta$ and $\sin^2\theta$ — which is exactly what one would expect to see in their s characters, since the π function carries no s at all.

It is also, immediately, why the average cannot say anything. $\cos^2\theta + \sin^2\theta$ is one whatever $\theta$ is, so the sum of the two components’ s characters is the σ hybrid’s exactly, at every mixing. The quantity proposed as a test is conserved by trigonometry.

The functional has no linear term #

The criterion has to come from somewhere, and the one that gives bent bonds is Boys’: maximise the sum of the squared centroids of the components, which pulls charge as far apart as it will go.

Three matrix elements decide it, and the molecule’s mirror plane kills the rest. With $z$ along the bond and $x$ perpendicular to it in the π plane, $\langle\sigma|x|\sigma\rangle$ and $\langle\pi|x|\pi\rangle$ are zero by that plane, so what survives is

$$d = \langle\sigma|x|\pi\rangle, \qquad \Delta = \langle\sigma|z|\sigma\rangle - \langle\pi|z|\pi\rangle, \qquad m = \tfrac{1}{2}\left(\langle\sigma|z|\sigma\rangle + \langle\pi|z|\pi\rangle\right).$$

Writing $v = \cos^2\theta - \tfrac{1}{2}$, every term collapses to

$$D(v) = 2d^2 + 2m^2 + v^2\left(2\Delta^2 - 8d^2\right).$$

A quadratic in $v$ with no linear term. So $D$ is stationary at $v = 0$ — the symmetric mixing — for every molecule, symmetric or not, and its maximum over the allowed range is at $v = 0$ or at $v = \pm\tfrac{1}{2}$ and nowhere else.

$v = 0$ is $\theta = 45°$: two bent components with equal s characters. $v = \pm\tfrac{1}{2}$ is $\theta = 0°$ or $90°$: the canonical σ and π, one carrying all the s character and the other none.

There is no angle to search for. The intermediate case asked about is not a solution of this criterion for any molecule.

Which of the two, and by how much #

The coefficient decides: bent components win when $2\Delta^2 < 8d^2$, that is when

$$|\Delta| < 2|d|.$$

The separation of the two orbitals’ centroids along the bond, against twice the off-axis dipole between them. In ethene $\Delta$ is exactly zero — both orbitals are centred on the midpoint by symmetry — so bent bonds win by an unbounded margin, which is why the question never arose in a symmetric molecule.

A carbonyl’s $\Delta$ is not zero, because its σ and π orbitals are polarised towards oxygen by different amounts: the σ and π overlaps are different integrals, so the two two-level problems have different couplings and different answers.

Holding the π orbital at 74.4 per cent on oxygen and sweeping the σ population, the bent description wins for $w_\sigma$ between 0.380 and 1.000 — a window of 0.62 — and loses below it. So a double bond whose σ is less polarised towards the heavy atom than its π has σ and π as its localised orbitals, and bananas are not the right picture of it at all.

That is a real prediction and it has a chemical shape: it says the bent picture survives strong polarisation and fails reverse polarisation, which is the situation of a bond whose π is pulled one way and whose σ is pulled the other.

What the two solutions look like #

The two answers are not two shades of the same picture, and it is worth setting them beside each other.

At the bent solution the two components each carry half the σ hybrid’s s character, each is tilted off the internuclear axis by the same angle, and their centroids sit symmetrically above and below the plane of the molecule. A chemist drawing them draws two identical bananas. All of that follows from $\theta = 45°$ and none of it needs the molecule to be symmetric — the components are equivalent to each other even when the two atoms are not, because the transformation that produced them is the symmetric one.

At the canonical solution the components are the σ and π orbitals themselves. One carries the whole of the s character, the other none; one is cylindrically symmetric about the bond and the other has a node in the plane. There is no picture of a double bond as two equal halves at all.

So the criterion is not choosing how bent to make the bananas. It is choosing whether there are bananas, and the two answers are qualitatively different objects. A model whose output is a discrete choice between two pictures is unusual in this collection — most of its questions have answers with a number in them — and it happens here because the space being searched is one rotation angle and the functional is even in it.

The evenness is where the symmetry went. The molecule has no symmetry relating the two components, but the functional has one: swapping $\tau_+$ with $\tau_-$ is $\theta \to 90° - \theta$, and a sum over both components cannot tell the two apart. A criterion that treats its outputs as a set rather than as a list is even in the mixing angle whatever molecule it is applied to, and an even function of one variable has a stationary point at the middle.

The number a hydrogenic basis cannot supply #

The obvious next step is to say which side formaldehyde falls on, and a hydrogenic basis cannot say.

Its σ overlap computed from hydrogenic functions is 0.00718, where it should be the largest overlap in the molecule. A double bond is not two single bonds records why: a hydrogenic 2s has a radial node inside the bond, and the s–s and s–p contributions very nearly cancel. So the σ two-level problem comes out with almost no coupling, its orbital sits 99.9 per cent on oxygen, and the inequality is decided by an artefact.

Run with those numbers the answer is canonical, at a ratio $|\Delta|/2|d|$ of 1.809. It should not be believed, and it is reported here because the alternative is to report a number without saying that its input is wrong.

The theorem is untouched by any of this. No interior maximum, the average s character conserved, the inequality in closed form — none of them needs an integral. What needs one is which side of the inequality a particular molecule is on, and that needs a better basis than hydrogenic functions.

Why the average was never going to work #

It is worth being explicit about the proposed test, because the shape of its failure recurs.

Do the two s characters still average to the value the symmetric case gives? The answer is yes, always, exactly — and the reason is that the transformation between the canonical and the localised description is orthogonal, so it preserves traces. The sum of the two components’ s characters is a trace, and a trace is what an orthogonal mixing cannot change.

That is the same fact the localisation transformation is built on: the density is unchanged to the last bit of a double under any orthogonal mixing, because the projector onto the occupied space is a trace-like object too. A quantity conserved by the transformation cannot be evidence about which member of the family is right, and an average over the family is exactly such a quantity.

The quantities that do distinguish the members are the ones that are not linear in the projector: the individual s characters, the individual centroids, the localisation functional itself. Those are the ones worth computing and the ones computed here.

What is quoted, and what is computed #

Four numbers are quoted: two screened nuclear charges, a C=O bond length of 1.203 Å, and the Wolfsberg–Helmholz constant of 1.75. The last has no measured value and nothing in the argument depends on it — only on the two overlaps being different.

The three matrix elements are computed from the real functions, as single integrals over whole orbitals rather than as sums of one- and two-centre pieces assembled by hand. That choice is the one the ethene calculation made and recorded: the pieces carry four signs between them, two of which change under swapping the centres, and getting one wrong produces a plausible wrong number.

The closed form is checked against a direct evaluation at a hundred and eighty-one angles, and must agree at every one to $10^{-10}$. That is not a formality: the collapse to a quadratic with no linear term is the whole finding, and an algebraic slip would produce a functional with a maximum somewhere plausible.

What this cannot say #

Boys is not the only criterion. Edmiston–Ruedenberg maximises self-repulsion and Pipek–Mezey maximises atomic populations; the second famously does not give bent bonds for a double bond at all, returning σ and π for ethene where Boys returns bananas. So the inequality here is Boys’, and a different criterion draws a different boundary — which is itself the point the localisation transformation makes about localised orbitals.

Two orbitals is the whole model. A real carbonyl has lone pairs on oxygen, and a localisation over the full occupied space can mix them in. That would break the two-dimensional algebra that produces the closed form, and whether an interior maximum reappears when a third orbital is admitted is not answered here.

Nothing here says bent bonds are wrong. The two descriptions are related by an orthogonal transformation and have the same density, so neither is more true; the criterion picks which one is more localised, which is a property of the description rather than of the molecule. A basis is not a thing applies as much here as anywhere.

And the polarisations are a sweep rather than a molecule. The window is computed as a function of $w_\sigma$ because the integral that would fix $w_\sigma$ is the one this site gets wrong. The window is a property of the criterion; where a molecule sits in it is not established.

What was checked #

The two overlaps are different integrals, and the two orbitals are polarised towards oxygen by different amounts because of it — the source of the asymmetry, measured rather than assumed.

The closed form matches a direct evaluation at every one of a hundred and eighty-one angles, to $10^{-10}$.

The functional’s best lies at an end of the range or at the middle, to the resolution of that grid, which is the statement that there is no interior maximum.

The two components’ s characters sum to the σ hybrid’s exactly — checked to the last bit a double holds rather than to a tolerance, because it is trigonometry and not a result.

Ethene’s centroid separation is exactly zero and its bent description wins by an unbounded margin, while a carbonyl’s margin is finite — the control that makes the inequality a statement about asymmetry rather than about the method.

And a σ orbital sitting entirely on the light atom is not described by bent bonds, which is the far edge of the window and the case the sweep exists to find.

The general form of the argument #

The collapse above used one property of the functional and one of the molecule, and both are worth separating from the chemistry.

The functional is a sum over the components, so it cannot tell them apart. Swapping the two is $\theta \to 90° - \theta$, and any quantity summed over both is even about forty-five degrees. An even function of one variable is stationary at its centre, whatever else is true of it.

And the space is one-dimensional. Two orthogonal orbitals have exactly one mixing angle between them, so the search is over a single parameter and an even function of one parameter has its extremes at the centre or at the ends.

Neither of those mentions a carbonyl, a bond, or an atom. So the conclusion generalises: any localisation criterion that treats its output as a set, applied to any two orbitals, has no interior maximum. What breaks it is a third orbital, which turns one angle into a rotation group and the evenness into a symmetry a functional can be stationary at without being extremal — which is why the continuation named below is the one that matters.

It is the same kind of argument a species label that prices nothing turns on: a statement about what a symmetry forces, made before any number is computed, and true of every case the symmetry covers.

What the collapse says about the criteria disagreeing #

There is a well-known awkwardness about localisation that this argument disposes of, and it is worth stating because the awkwardness is usually reported as an empirical nuisance.

Several criteria are in use. They ask different things — one maximises the separation between the orbitals’ centroids, another maximises their self-repulsion, a third maximises how far each is confined to a few atoms — and they are known to disagree about double bonds: some return bent components and others return σ and π.

The argument above says what that disagreement can and cannot be.

Every one of those criteria is a sum over the orbitals, so every one of them is even about forty-five degrees when there are two orbitals related by a mirror. Every one therefore has its stationary points at the same three places: the centre and the two ends. No criterion can return sixty degrees, or thirty, or any interior value other than the middle.

So the criteria cannot disagree about the candidates. They can only disagree about which candidate is the maximum — bent components or canonical ones — and that is a choice between two discrete answers decided by the sign of one quantity.

That is a much better-behaved kind of disagreement than it is usually presented as. Two criteria giving different pictures of one double bond are not producing two different geometries of hybrid; they are landing on opposite ends of the same one-parameter family, and the family has exactly two ends.

Still open: a third orbital, and a better basis #

The obvious open question is the third orbital. A carbonyl’s oxygen carries lone pairs, and a localisation over the whole occupied space may mix them with the bond — which is what turns the tidy two-by-two rotation above 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.

The nearer question is the basis. Everything not established here is blocked by one integral — a σ overlap that hydrogenic functions get wrong by two orders of magnitude — and the fix is already known: a contracted Gaussian basis is a decision made once, and the contraction that reproduces a Slater function reproduces its overlaps too. Computing the same three matrix elements in that basis, on the same molecule, would put formaldehyde on one side of the inequality or the other, and it is the only step needed.