An interior maximum a third orbital allows

Repairing a σ overlap that a radial node had been destroying — a hydrogenic 2s has a node inside a carbon–oxygen bond and the two contributions very nearly cancel across it — flips a carbonyl’s description from canonical to bent. It leaves one thing untouched: for two orbitals related by a mirror plane the Boys functional is a quadratic in the mixing with no linear term, so it has no interior maximum. Either the bent description wins outright or the canonical one does, and no angle is ever searched for.

It closed by naming where that argument stops applying. A carbonyl’s oxygen carries lone pairs, and localising over more than two orbitals turns a one-parameter mixing into a search over a rotation group.

There is an interior maximum there, and what it does is not what the extra orbital was expected to do.

The third orbital, built rather than assumed #

The lone pair is the sp hybrid on oxygen pointing away from carbon, and its s character is not a choice. Two sp hybrids on one centre are orthogonal when their s characters sum to one, so the outward one is fixed at $1 - s_O$ by the inward one the bond already uses.

That leaves a set which is nearly but not quite orthonormal: the two bonds are orthogonal to each other by the molecule’s mirror plane, and the σ bond overlaps the outward lone pair by 0.112 because both have oxygen character in them.

So the set is symmetrically orthogonalised before anything is rotated — the Löwdin route, which is the one that changes each function least and is the same transformation that makes methane’s four bond orbitals an orthonormal set.

Eighteen numbers instead of three #

Two orbitals related by one mirror plane have three dipole matrix elements between them, and the two-orbital argument is about what those three can do: the functional comes out as a constant plus a quadratic in the mixing, and the quadratic’s sign decides the answer without any search.

Three orbitals have eighteen. Half of them vanish by the mirror plane and what is left is two diagonals and two off-diagonal entries — σ with π across the bond, and π with the lone pair across the bond. Two off-diagonal couplings in one component and a third diagonal in another is enough for the functional to have a genuine maximum inside the group.

The maximum, and what it is #

The functional is 0.966555 Ų at the canonical orbitals, 1.079600 at the best mixing of the two bonds with the lone pair untouched, and 1.106642 at the maximum over the whole group.

The first step is 11.70 per cent and is exactly the two-orbital answer: the bent description of the double bond, at forty-five degrees, which the two-orbital treatment reaches. The second is 2.50 per cent and is what no pair of orbitals can find.

What it consists of is 8.44 degrees of the lone pair rotated into the two bonds.

So the third orbital’s contribution is not its own localisation. The lone pair barely moves — its centroid stays on the axis at 0.92 Å — and what improves is the two bonds, which become slightly better-localised bananas by borrowing a little of it. A quantity that looks like a property of the lone pair turns out to be a property of the bonds, obtained by giving them something to borrow.

What “no angle to search for” meant, and what it did not #

The two-orbital result is that the angle does not have to be searched for, and it is worth being exact about what that claim was and what is left of it.

It was: for two orbitals, the Boys functional is a quadratic in one variable with no linear term, so its extremum is at an end of the range and the answer is one of two named descriptions. That is a theorem about a two-dimensional space and it is still a theorem. It also had a consequence worth keeping — the two components of the mixed pair have equal s characters, because the mixing is at exactly forty-five degrees, and a searched angle would have given two different ones.

That consequence goes. At the three-orbital maximum the two bent bonds are still mirror images of each other, so they still have equal s characters as far as the σ and π parts go — but each has picked up 8.44 degrees of the lone pair, and the lone pair is two thirds s. So the localised bonds of a carbonyl have an s character that is not fixed by symmetry, and the number it takes is the output of a search.

A theorem about a subspace does not become false when the space is enlarged; it becomes inapplicable, and the difference is easy to lose in the sentence “the angle does not have to be searched for”.

The rabbit ears are a different phenomenon #

The other half of the question was whether the famous rabbit-ear description of a carbonyl’s lone pairs is the same thing.

It is not, and it does not need a rotation group. Oxygen’s second lone pair is the p orbital perpendicular to both the bond and the π system, and the two lone pairs together are a two-orbital problem — exactly the shape the two-orbital criterion was derived for.

Their centroids are 0.31652 Å apart along the bond and twice the dipole between them is 0.54826, so the criterion says the mixed description wins by a factor of 1.732128 — which is $\sqrt3$ to five figures, a coincidence noted here and not explained.

So rabbit ears come out of a two-orbital criterion applied to two orbitals, and win outright. The two mixings are different in kind. A double bond’s is a competition between a separation and a dipole that could go either way and does go both ways depending on the basis; two lone pairs’ is not a competition at all, because they are nearly equivalent and a mixed pair of equivalent orbitals always localises better than an unequal pair.

One more contrast is worth drawing, because there is already a case where a localisation search finds nothing to search over. Four centres and a pair that will not localise is a system whose localised description is forced by symmetry and where every start reaches the same answer. A carbonyl trio is the opposite: the answer is not forced, it is found, and the amount found is small — two and a half per cent of a functional that a chemist would never look at directly.

That smallness is the honest headline. What the rotation group buys is real, is a maximum where a theorem forbade one, and is worth two and a half per cent of a quantity nobody measures. The picture it produces — two bent bonds with a little lone pair in them and a lone pair with a little bond in it — is not visibly different from the picture the two-orbital treatment gives.

What was computed, and how #

The bonds are unchanged, with the Slater functions of the repair — a Slater 2s has no radial node and the σ overlap comes out as the largest in the molecule rather than as the smallest. The lone pairs are built from the same functions on the same centre.

Every matrix element is a quadrature over the same mapped three-dimensional grid the rest of this collection integrates on, and the orthogonalisation is a symmetric square root of the overlap matrix, checked by requiring the overlap eigenvalues to be positive before it is taken.

The maximum is found by a grid over the whole group and then refined by a local descent, and the two are reported separately: a maximum the refinement found and the grid did not would be a maximum in a place the search never covered. The grid’s best and the refinement’s agree on which region and differ in the fourth decimal.

The refusal is the orthogonalisation itself. The σ bond and the outward lone pair must have a non-zero overlap before the Löwdin step — otherwise the step is doing nothing and the set was orthonormal by accident — and the orthogonalised σ must still have no dipole across the bond axis, which the mirror plane requires and no rotation was told about.

Why two off-diagonals in one component are not enough on their own #

The condition for an interior maximum is worth stating rather than left as a fact about this molecule, because it is a small piece of algebra and it says which systems will behave this way.

A rotation of three orbitals is generated by three independent two-by-two mixings. Each of those, taken alone against a functional built from a diagonal and one off-diagonal, is the two-orbital case: a quadratic with no linear term, extremum at an end. What makes the three-dimensional problem different is that the mixings do not commute — rotating σ into π changes what the σ–n coupling does, so the functional along the third direction depends on where the first two have been taken.

The carbonyl has exactly the ingredients that needs. σ couples to π across the bond, π couples to the lone pair across the bond, and σ does not couple to the lone pair across it — so the three couplings form a chain rather than a triangle, and going round the chain is a different transformation from any single step. The same non-commuting structure is what makes a molecule’s own hybrids point outside its bonds: a set of directions that pairwise look reasonable need not be jointly reasonable.

A molecule whose orbitals came in genuinely separate pairs — no chain, only disjoint couplings — would have a functional that factorises, and the two-orbital theorem would apply to each factor. So the question of whether a localisation has an interior maximum is a question about the pattern of the dipole matrix rather than about the number of orbitals.

That is a cheap thing to check before running a localisation, and it is not what anybody checks. The usual question asked of a localisation is whether it converged, which is a question about the optimiser; the question that decides whether the answer means anything is whether the functional has an interior maximum to converge to, which is a question about the sparsity pattern of the dipole matrices and can be answered by looking at them. A block-diagonal pattern factorises into two-orbital problems, every one of which is extremised at an end, and the optimiser will dutifully report an end point as though it had found something.

The distinction matters most where it is least visible. An end point and an interior maximum look identical in the output — a set of orbitals, a functional value, a converged flag — and they differ in what a small change to the molecule does to them. An interior maximum moves continuously, so a slightly stretched bond gives slightly rotated orbitals; an end point is pinned, so it does not move at all until the pattern itself changes and then it jumps. A localised description that is stable under small perturbations and a localised description that is merely stuck look the same until something is perturbed.

Where the model stops #

Three orbitals, not four. A carbonyl’s occupied valence space has the two bonds, two lone pairs on oxygen and two C–H bonds, and localising over all six is a search over a fifteen-dimensional group. What is shown here is that adding one orbital to a two-orbital problem changes its character; what the full space does is not settled by it, and the direction — more room means a larger functional — is the only thing that carries.

The functional is Boys’s, which maximises the sum of squared centroid lengths. The localisation transformation used for methane is a different one, and the two criteria do not always agree; nothing here says what Edmiston–Ruedenberg would do with the same three orbitals.

And the whole model is a two-centre caricature with hand-built hybrids and a Wolfsberg–Helmholz resonance integral. The numbers are the size of an effect in that caricature. What is not a caricature is the counting: two orbitals have one angle and three have a group, and no amount of better wavefunctions changes that.

The generalisation #

A statement about what a functional can do is a statement about the space it is defined on, and enlarging the space can change the answer qualitatively rather than by a correction.

The argument is correct and complete for two orbitals: the linear term vanishes by symmetry, so there is no interior maximum, so the answer is one of two descriptions and never a mixture. Every word of that survives. What does not survive is the reading of it as a statement about the molecule — the molecule has more than two occupied orbitals, and the moment a third is admitted the extremum moves inside.

That is worth keeping because the same shape recurs elsewhere. A local search calibrated on one size, a constant established on one lattice, a diagnostic tested on one axis — in each the demonstration was sound and the range it was demonstrated over was silent. Here the silent variable is the dimension of the space, and the two-orbital argument names it itself.

Who found it, and when #

The same reading applies to the two-orbital result, which is not overturned. A functional built from a diagonal and one off-diagonal really does take its extremum at an end, and that really is a theorem rather than an observation about a particular molecule. What it does not say — because with two orbitals there is nothing to say it about — is that the theorem is about the pattern rather than about the count, so a third orbital coupled to only one of the first two would obey it exactly as the pair did.

Boys localisation is from 1960 and the observation that it turns a double bond into two equivalent bent bonds is as old. The rabbit-ear description of lone pairs is older still and is Pauling’s hybridisation picture applied to a molecule with two of them; the argument about whether water’s lone pairs are equivalent or not has been running since photoelectron spectra showed two distinct ionisations, and the resolution — the canonical orbitals are inequivalent and the localised ones are equivalent, and both are descriptions of one density — is the standard one, and it has been made twice here.

That a localisation over a larger space has extrema a smaller space forbids is a fact about optimisation rather than about chemistry, and is not attributed to anybody. What makes it worth computing here is that an exact statement for the small space invites a natural reading that is wrong.

Still open: the fourth orbital #

The obvious open question is the fourth orbital. Adding oxygen’s second lone pair makes the group six-dimensional and lets the rabbit-ear mixing and the bent-bond mixing happen at once, and whether they compete or add is the question — two mixings that each improve the functional need not both survive when both are allowed. The dipole matrices are already computed for all four. If the two mixings compete, a localisation done one pair at a time gives an answer that depends on the order the pairs are taken in, and only the simultaneous search describes the molecule rather than the procedure.

The nearer question is the $\sqrt3$. The two lone pairs’ criterion comes out at 1.732128 against $\sqrt3 = 1.732051$, which is agreement to five figures on a quantity assembled from three integrals over a numerical grid. If it is exact it follows from the s characters — the outward hybrid is two thirds s and the other lone pair is a pure p — and writing the two centroids and the dipole in closed form for a general s character would say so, and would also say at what s character the rabbit ears stop winning.