The five figures were an identity

A number that agrees with a familiar constant to five figures is either an identity or a coincidence, and the difference between those is worth an essay because it changes what the number is evidence for.

Whether a carbonyl’s two lone pairs are better described as two equivalent rabbit ears or as one σ-type lone pair and one perpendicular p can be answered with the same two-orbital criterion used for the double bond: the mixed description wins when the separation of the two orbitals’ centroids is smaller than twice the dipole between them, because that is exactly when rotating the pair increases the Boys functional. For the lone pairs the criterion came out at 1.732128, and √3 is 1.7320508.

That is agreement in the fifth decimal on a quantity assembled from three integrals, each computed on a grid over the whole of space. Nothing in the derivation predicted a surd. So either the arithmetic happened to land near one, or there is an identity underneath and the last two digits are the grid.

It is the identity, and the derivation is three lines.

Three lines #

The outward lone pair is a hybrid on oxygen, $\sqrt{1-p}\,\lvert 2s\rangle + \sqrt{p}\,\lvert 2p_z\rangle$, where p is the fraction of p in it — one third for the default carbonyl, so two thirds s character. The other lone pair is a pure $\lvert 2p_y\rangle$ on the same centre. Both sit at the same point in space, which is the whole reason this is tractable: every integral between them is a single-centre integral.

Write μ for the one radial integral that survives, $\langle 2s\vert\zeta\vert 2p_\zeta\rangle$. It is the same number for ζ = y as for ζ = z, because those two orbitals differ by a rotation and the integral is a scalar. Then

the separation of the centroids along the axis is Δ = 2√(p(1−p)) μ, because ⟨s|z|s⟩ and ⟨p|z|p⟩ both vanish about the centre and only the cross term survives; the dipole between the two orbitals is d = √(1−p) μ, because the $p_z$ component of the hybrid has no matrix element with $p_y$ across y; and the criterion is

2|d| / |Δ| = 2√(1−p) μ / (2√(p(1−p)) μ) = 1/√p.

Everything cancels. The radial integral goes, and with it the effective nuclear charge, the choice of Slater functions over anything else, the bond length, and the position of the oxygen. At p = 1/3 the criterion is √3.

The two quantities that go into it have quite different shapes, which is worth noticing before their ratio is trusted. The separation Δ = 2√(p(1−p)) μ is a geometric mean: it vanishes for a pure s outward orbital and for a pure p one, and is largest at p = 1/2, where the hybrid is most polarised along its own axis. The dipole d = √(1−p) μ falls steadily from μ to zero as p rises. So one of them turns over inside the range and the other does not, and their ratio is nevertheless monotone. That is the sort of thing a closed form makes obvious and a table of computed values does not: the criterion’s monotonicity is not inherited from either of its parts.

It also explains why the cancellation is total rather than approximate. Both quantities are μ times a function of p alone, because the hybrid is a rotation of the atomic orbitals and a rotation acts on the coefficients while leaving the radial functions where they were.

Measuring a cancellation #

That derivation is short enough to be wrong, so the cancellation is worth measuring rather than believing. Changing the effective charge changes the radial integral; changing the bond length changes where the centre sits. If μ really cancels, the two quantities the criterion compares should both move and the criterion should not.

Across three effective charges and three bond lengths the separation of the centroids runs from 0.4536 to 0.7776 and the dipole from 0.3928 to 0.6734 — each a factor of 1.71, and the same factor, which is what “both are μ times a number that depends only on the hybrid” looks like from outside. The criterion spans a factor of 1.000134, which is the quadrature. The largest departure from √3 anywhere in the block is 1.0 × 10⁻⁴, at the tightest orbital and the longest bond, where the integrand is hardest.

So the criterion is not a property of this carbonyl. It is a property of the hybrid, and it would be the same number in any molecule whose outward lone pair had that s character. Water’s oxygen, an ether’s, a ketone’s: same criterion, same verdict, differing only through whatever their outward lone pairs’ s characters happen to be — and that quantity is not fixed by the geometry either.

The last two digits are the grid #

The remaining question about the original 1.732128 is where its excess over √3 came from, and the answer is testable in the most direct way available: do the integrals more carefully and see whether it shrinks.

At forty quadrature points a dimension the relative departure is 1.7 × 10⁻³. At sixty — the default, which is what produced 1.732128 — it is 4.5 × 10⁻⁵. At eighty it is 1.4 × 10⁻⁵ and at a hundred it is 8.5 × 10⁻⁶: a factor of 197 across the range, on a rule whose error should fall faster than any power of the point count for a smooth integrand.

A coincidence would not behave like that. If 1.732128 were merely near √3 by accident, refining the grid would move it towards whatever it actually is, and the departure would settle rather than fall. It falls.

That is the whole of the answer to the question. What follows is what the answer costs.

A criterion that cannot refuse #

The closed form is 1/√p, and p is a fraction, so 1/√p is greater than one for every p less than one. The criterion is satisfied — the mixed description wins — at every s character there is.

At ninety per cent s character it is 3.162. At two thirds, the default, √3. At fifty per cent, √2. At ten per cent s character, 1.054 — still above one. Nine hybrids, nine verdicts, all the same verdict, and the closed form says there is no tenth hybrid that would differ.

This is a problem, and it is the finding this essay actually produces. A test that returns the same answer for every input it can be given is not measuring anything about the case in front of it. Applied to two lone pairs on one centre it is a tautology dressed as a measurement.

It is tempting to assume at this point that the criterion discriminates elsewhere — that on the double bond it refuses the mixed description. It does not. The σ–π analysis finds the two descriptions equivalent and computes the bent pair’s hybrids; it contains no refusal. And when the criterion is put to every pair available, the double bond’s margin comes out wider than the lone pairs’. The contrast is real in one direction only: the lone-pair case is a foregone conclusion, and so is every other case usually asked about.

Nor does the margin rescue it. One might hope that a criterion which always says yes at least says how strongly — that 3.162 means the rabbit ears are a much better description than 1.054 does. But the margin is monotone in the s character and depends on nothing else, so quoting it says how much s is in the hybrid, which was an input. It is the sort of quantity that looks like an answer and is a restatement of the question.

Where the verdict could turn, and why nothing can go there #

There is exactly one place the criterion could return the other verdict, and the closed form names it: p = 1, a pure p outward orbital, where 1/√p is exactly one and the two descriptions are tied.

Taking the hybrid towards pure p, both quantities the criterion compares go to zero together. At ninety per cent p the separation of the centroids is 0.381 and the dipole 0.201; at one part in a million from pure p they are 1.28 × 10⁻³ and 6.33 × 10⁻⁴. The ratio is the thing being read, and a ratio of two vanishing numbers is where a quadrature stops being able to help.

It does stop. At p = 1 − 10⁻⁶ the integrals return 0.991947, which is on the wrong side of the criterion — below one, where the closed form forbids it to go. That is not a discovery about pure p orbitals; it is the numerical quadrature failing in a way the identity can diagnose exactly. The check on this computation requires that failure, so that nobody reads the limit as something the integrals established.

And the limit itself is not a chemical case anyway. A pure p outward lone pair means an oxygen with no s in either lone pair, which is not a hybridisation any carbonyl has. The boundary is at the edge of the parameterisation rather than inside it, which is a different way of saying the criterion has no discriminating region within reach.

Two candidate identities, and the one hybrid that cannot choose #

One more thing has to be checked before the closed form is safe, and it is the reason the sweep has nine points rather than one.

There is a second natural guess. If somebody had derived the criterion carelessly and got the s fraction where the p fraction belongs, they would have written 1/√(1−p) instead. At the default carbonyl that gives 1.2247, nowhere near the measured 1.7321, so the default alone rejects it.

But there is a hybrid where it does not. At p = 1/2 the two forms are both √2, and the integrals return 1.414234. A carbonyl with an sp lone pair would confirm the wrong form exactly as well as the right one.

Away from that point the wrong form misses by at least 8.9 per cent, which no quadrature could close. So the sweep is not decoration: it is what makes the closed form a claim with a test attached, and the one hybrid where the test is blind is shown rather than quietly dropped.

What was computed, and how #

Every number comes from three-dimensional Gauss–Legendre quadrature over the whole of space, with the finite interval mapped to the line by t/(1−t²), on Slater 2s and 2p functions with an effective charge for oxygen. The two lone pairs’ overlap comes out at 3 × 10⁻¹⁷, so they are orthogonal by symmetry and the Löwdin step used for the lone pairs is the identity; the normalisation here is done explicitly instead, which is what lets the grid be refined.

The criterion is twice the off-diagonal dipole between the two orbitals, in the perpendicular direction, over the difference of their axial centroids. The closed form is 1/√p with p the p fraction of the outward hybrid. The check covers six things: that the closed form matches the integrals at all nine hybrids to better than two parts in ten thousand; that the rival form coincides with it at exactly one hybrid and that this hybrid is in the sweep; that away from it the rival misses by more than five per cent; that refining the grid improves the agreement by more than a factor of ten; that the separation of the centroids moves by more than half again across the charge and length block while the criterion does not; and that the numerics cross to the wrong side of the criterion at the unreachable boundary.

The last of those is the refusal. It requires the calculation to fail somewhere, in a stated place, for a stated reason.

Where the model stops #

This is one criterion, on one pair of orbitals, on one centre. The closed form works because both orbitals sit at the same point, so every integral is single-centre and the angular parts do all the work. Nothing here transfers to a criterion between orbitals on different atoms, where the radial integrals do not cancel and the geometry enters.

The claim that the two orbitals sit at the same point is doing real work and is worth stating as a limitation rather than as a convenience. It is exact here because both lone pairs are built from oxygen functions alone, with no carbon contribution — which is a choice the model makes, and a defensible one for a lone pair, but a choice. A lone pair with even a small tail on the neighbouring atom would break the single-centre structure and reintroduce a second radial integral, and then nothing above cancels.

It is also a statement about the two-orbital criterion and not about localisation generally. The three-orbital result — that the full rotation group over three orbitals has an interior maximum that no sequence of two-by-two rotations finds — is untouched by this, and remains the reason the two-orbital criterion is a limited instrument. What is added here is that on this particular pair it is not merely limited but empty.

And the s character is treated here as a free parameter, swept from 0.1 to 0.9. In a real molecule it is determined by the rest of the electronic structure, and the angle does not fix it — which is a separate argument and the reason the sweep is the honest presentation rather than a single quoted hybridisation.

The generalisation #

The transferable part is not about lone pairs. It is that a criterion is only as good as the range of answers it can return, and that this is checkable in advance by writing it in closed form and looking at what it depends on.

This is a different failure from the one a check that never rejects anything has, and worse in one respect. A check that has never fired might simply have been lucky in its inputs; a criterion whose closed form exceeds one on its whole domain has been proved incapable of firing.

Once the criterion is 1/√p, three things are immediate that no amount of computing at the default would have shown: that it exceeds one always, that its margin is a re-reading of its input, and that its boundary lies outside the parameterisation. Each of those is a defect in the instrument, and each was invisible while the instrument returned a single number per molecule that happened to look reasonable.

The same move is available whenever a criterion is cheap to evaluate and its inputs can be swept. Sweep them. A test that never changes its verdict across the whole range of its input is reporting the range and not the case, and it will keep looking like a confirmation for as long as it is only ever run once.

Who found it, and when #

The Boys localisation criterion dates from 1960 and the two-orbital reduction of it is standard. The closed form above is new arithmetic on a standard criterion, done to settle whether a reported five-figure agreement was an identity. It is; and the consequence — that the criterion cannot refuse this pair — is the part that was not visible from the single number.

The rabbit-ear question itself remains what it was before either calculation: a question about which basis to write an answer in rather than about what the answer is, since the two descriptions are related by a rotation and a rotation changes no observable.

Still open: a fourth orbital, and a case the criterion refuses #

The obvious open question is the fourth orbital: adding the second lone pair to the σ and π bonds makes the group six-dimensional and lets the rabbit-ear mixing and the bent-bond mixing happen at once. The dipole matrices are computed for all four already. What this adds to that plan is a reason to expect the answer to be interesting — the two-orbital criterion cannot see a competition between two mixings, because on each pair separately it has already decided.

The nearer question is whether the same treatment kills or rescues the criterion on the double bond. There the two orbitals are the σ and π of a C=O, on different centres, and it is tempting to assume the criterion refuses to mix them. The criterion that has never said no runs it and finds the opposite: the σ–π margin is 2.45 against the lone pairs’ 1.73 and never falls below 2.4 at any polarisation, and ethene’s is infinite because symmetry puts both centroids at the bond midpoint. So the criterion has no negative case among the pairs usually asked about — though it is not vacuous, since the same three carbonyl orbitals supply two pairs it does refuse.