A verdict that flips and changes nothing

Every margin the two-orbital mixing criterion had produced sat a long way from its boundary. The criterion that has never said no listed them — 0.111, 0.739, 1.732, 2.455 and infinity — and made the sharpest point available about them: nothing lies between 0.74 and 1.73, so a test whose only job is to say which side of one a pair falls on had never been tried anywhere near one. A threshold that no case approaches is neither supported nor refuted by any of its cases.

That essay named where a near case might come from. The carbonyl’s polarisation does not bring one; it moves the double bond’s margin between 2.4 and 39 and never lower. But the same construction is built from two numbers that describe the heteroatom — its Slater exponent and its electronegativity — and those can be changed instead. Replace the oxygen with carbon, nitrogen or fluorine and keep everything else, and the molecule runs through an isoelectronic series: the carbanion H₂C=CH⁻, the imine H₂C=NH, formaldehyde H₂C=O and the cation H₂C=F⁺. Twelve valence electrons each, a σ bond, a π bond and an outward lone pair on the heteroatom, in every case. The question is whether any pair’s margin crosses one along the way, and if so, what changes when it does.

One pair crosses. Nothing changes.

The heteroatom as two numbers #

The construction is the one the earlier carbonyl essays used. Carbon and the heteroatom each carry Slater-type 2s and 2p functions; the σ bond is a two-level combination of an sp hybrid on each atom and the π bond a combination of the two p functions perpendicular to the plane; the lone pair is the heteroatom’s outward hybrid. The three are Löwdin-orthogonalised and every dipole matrix element between them is a numerical integral on one grid.

The heteroatom enters in two places. Its Slater exponent sets how compact its functions are — 3.25 for carbon, 3.90 for nitrogen, 4.55 for oxygen, 5.20 for fluorine, Slater’s own rule for a 2p electron. Its Mulliken electronegativity, the mean of the first ionisation energy and the electron affinity, sets its diagonal energy in the two-level problems: 6.26 eV for carbon, 7.23 for nitrogen, 7.54 for oxygen and 10.41 for fluorine. Nitrogen’s comes out below the line from carbon to oxygen because its electron affinity is slightly negative — the anion is not bound — and that is used as tabulated rather than smoothed. Electronegativity is not one quantity is the reason for choosing one scale and saying which; Mulliken’s is the one already in the construction.

Everything else is held fixed. The bond length is formaldehyde’s 1.203 Å for every member, so that only the atom changes; the s fraction of the heteroatom’s σ hybrid is a third. Between members, both numbers are interpolated linearly, which turns the four molecules into a path with a continuous parameter along it. No atom sits between oxygen and fluorine, and nothing below claims one does. The path exists so that a crossing can be located, not so that it can be assigned to an element.

The pair that crosses is the one nobody proposes #

The σ–π margin falls along the whole series. At the carbanion it is about 244, because a carbon–carbon σ and π have almost coincident centroids — the same symmetry that makes ethene’s margin infinite, broken only by the outward lone pair’s hybrid, which sits on one carbon and not the other. It drops to 3.90 at nitrogen, 2.455 at oxygen, and 1.179 at fluorine. It approaches the boundary and stays above it at every member.

The π–n margin rises: 0.470 at carbon, 0.642 at nitrogen, 0.739 at oxygen and 1.350 at fluorine. It rises at every step of the path and it crosses one.

So the first pair whose verdict a physical change flips is the π bond and the lone pair. It is one of the two pairs the earlier audit found nobody puts to the criterion — a bond and a lone pair are not rival descriptions of one region, so no one proposes rotating them into each other. The pair that was offered as the instrument’s control, the evidence that it could say no, is the pair that stops saying no.

The reason is in the two quantities the margin is a ratio of. The off-diagonal dipole between π and n barely changes along the series, 0.25 Å at carbon and 0.23 at fluorine. What changes is their separation. As the heteroatom becomes more electronegative the π bond is drawn towards it — the π orbital’s share on the heteroatom goes from a half at carbon to 0.74 at oxygen and higher still at fluorine — and its centroid moves along the bond towards the lone pair’s. The separation falls from 1.06 Å to 0.34, and somewhere in that fall it passes twice the dipole.

A third of the way from oxygen to fluorine #

The crossing was located by regula falsi on the path parameter, with the margin evaluated afresh at each trial point. It takes eight evaluations to bring the margin within a millionth of one, at an exponent of 4.772 and an electronegativity of 8.52 eV, 34% of the way from oxygen to fluorine.

It is tempting to look for an element at 8.52 eV. Chlorine’s Mulliken electronegativity is close to it, 8.3 eV, but chlorine’s valence shell is the third and its exponent is not on this line, so the coincidence is a coincidence. The honest statement is narrower: in a family of second-row heteroatoms described by Slater exponent and Mulliken electronegativity, the π–lone-pair verdict is refused at oxygen and endorsed at fluorine, and the change happens nearer the oxygen end.

The location depends on the two things held fixed, and they were released one at a time.

Over bond lengths from 1.15 to 1.334 Å the oxygen member stays between 0.717 and 0.815 and the fluorine member between 1.303 and 1.458; the crossing lies between them at every length. Along the heteroatom’s s fraction it holds from 0.15 to 0.5 and fails at the ends. A strongly s-rich fluorine hybrid, at 0.7, pushes the lone pair’s centroid back towards the atom and the fluorine member is refused too, at 0.941. A strongly p-rich oxygen, at 0.15, brings the oxygen member to 0.941, within six per cent of the line.

So this is the near-boundary case that was missing, and it is near the boundary in the sense that matters. A heteroatom’s electronegativity moves the verdict across it, and so does its hybridisation, by comparable amounts. For the first time a reasonable change to the molecule changes the answer.

Two yeses that cannot both be taken #

At fluorine the criterion says yes twice: σ–π at 1.179 and π–n at 1.350. Both pairs contain π, and π can be rotated fully into only one partner. The two-orbital criterion has no way to express that, because each verdict is computed for a pair with the third orbital absent.

The functional itself can express it. Every rotation of the three is the same occupied space in another basis, and the Boys functional — the sum of the squared centroids, which the localisation essays maximise — was evaluated at four points: the canonical orbitals, the best rotation of σ–π alone, the best rotation of π–n alone, and the maximum over the whole rotation group of the three.

The canonical orbitals give 1.1235. Rotating σ–π alone to forty-five degrees gives 1.1459. Rotating π–n alone gives 1.1702 — more. On the pair-by-pair reckoning, the π–n mixing is the better of the two endorsed rotations, by 0.024.

The maximum over all three orbitals is 1.1801, and it is not the π–n mixing. It is two bent bonds, each sharing π almost exactly equally — 0.496 and 0.504 — with the third localised orbital 95.2% the canonical lone pair. The orbitals the larger pair gain proposes are not where the functional peaks. They are not a peak of the three-orbital functional at all: from the π–n optimum, one further rotation of σ against either partner climbs by 0.00087. A point from which a single rotation climbs is not a maximum, so the configuration the stronger yes endorses is a point the localisation passes through rather than one it could stop at.

That is the content of an interior maximum a third orbital allows seen from the other side. There, a third orbital made a maximum available that no two-orbital rotation could reach. Here, a third orbital removes one that a two-orbital rotation appeared to offer.

The same orbitals on both sides of the line #

If crossing the boundary mattered to the molecule’s localised description, the description should look different at fluorine from how it looks at oxygen. It does not.

At carbon the lone pair is kept 98.5% itself, at nitrogen 98.0%, at oxygen 97.8% and at fluorine 95.2%, and at every member the other two localised orbitals are bent bonds sharing π equally. The lone pair drifts a little further from purity as the heteroatom becomes more electronegative, and it drifts smoothly — there is no break at the crossing, and nothing in the localised set that corresponds to a verdict changing.

The verdict that flipped is a statement about what the functional would do if π and n were the only two orbitals in the molecule. They are not, and in the three-orbital problem π’s mixing partner is σ at every member of the series, refused π–n or endorsed π–n alike.

No order wins #

A pairwise localisation — the way the Boys functional is usually maximised in practice — applies one two-orbital rotation at a time and sweeps over the pairs until nothing improves. If two endorsed mixings genuinely competed, the outcome could depend on which pair went first: start with π–n and the sweep might settle on a lone-pair-rich description; start with σ–π and it might settle on bent bonds. A description that depends on the order of an algorithm describes the algorithm.

Three orders were run at H₂C=F⁺ until the functional changed by less than 10⁻¹² in a sweep, one of them starting with π–n. All three reach 1.180063, the same value as the search over the whole rotation group, and all three end with the lone pair kept and π shared between two bent bonds. The run that starts by taking the π–n mixing simply climbs away from it.

There is one sign that fluorine is closer to a contest than the other members, and it is in the number of sweeps. At carbon, nitrogen and oxygen the sweeps converge in four to eight passes. At fluorine they take 83 to 139. The functional is flat across the region between the two mixings — the π–n optimum is only 0.0099 below the maximum and the climb out of it starts at under a thousandth — and a pairwise method crosses a flat region slowly. The two mixings are nearly balanced at fluorine. Nearly balanced is not tied, and the pairwise answer and the global one agree.

A refusal that was never a verdict #

Building the series meant computing all three pairs at every member, including the σ–lone-pair pair, and that pair turned out to have been misread.

The criterion’s either-or — the canonical pair or a forty-five-degree mixing, never anything between — comes from a property of the pairs it was derived for. Two pictures, one plane and the essays after it wrote the functional for σ and π, where the mirror plane of the molecule puts the off-diagonal dipole across the bond while the separation runs along it. With the two perpendicular, the functional is a quadratic in the mixing with no linear term, and its maximum must sit at one end or the other.

σ and the outward lone pair both lie on the bond axis. Their off-diagonal dipole lies along it too, parallel to the separation, and then the functional is not a quadratic without a linear term. Writing the pair’s contribution as twice the squared length of (Δ/2)·cos 2θ + d·sin 2θ, with both vectors along one line, the maximum is where tan 2θ equals 2|d|/|Δ| — the margin itself.

For formaldehyde’s σ–n pair that is half the arctangent of 0.111: 3.17 degrees. A direct scan of the functional puts its maximum at the same angle to three decimal places, and the same holds at every member of the series, from 3.05 degrees at carbon to 5.21 at fluorine. The canonical pair is not a maximum, and neither is the forty-five-degree mixing; the best rotation is a small one.

So the margin of 0.111 never measured a refusal. The criterion was applied to a pair for which it does not decide anything, and the number it returned is the tangent of an angle rather than a verdict. The earlier audit counted σ–n as one of two refusals; it is a case outside the test, and that essay’s figures and its count are corrected to say so. The audit’s conclusion survives on the other refusal, π–n at 0.739, whose dipole lies across the bond and for which the either-or does hold. The criterion can say no, but on one case rather than two.

What the picture leaves out #

The heteroatom is two numbers. A real fluorine, nitrogen or carbanion carbon differs from oxygen in a great deal more than its exponent and its electronegativity — in its charge, in the number and kind of its other lone pairs, in the bond length, in the H–C–H angle — and the construction holds all of that fixed. H₂C=F⁺ is a cation and H₂C=CH⁻ an anion; the model sees their charges only through the two numbers the series moves.

The trio is three orbitals, not twelve electrons. Each real molecule has a second lone pair — two on fluorine, one more on oxygen — and C–H bonds, and the localised description of the whole molecule could differ from the localised description of three of its orbitals. The four-orbital question, with both lone pairs in, remains open, and it is where a genuine contest between mixings would have most room.

And the Boys functional is one localisation criterion among several. Edmiston–Ruedenberg and Pipek–Mezey weigh orbitals differently and do not always agree with Boys about bent bonds. The finding is that this criterion’s own pairwise verdict does not predict this criterion’s own maximum, which is a statement about the instrument, not about which description a molecule prefers.

A test passed at the only place it could fail #

The criterion had never been tried near its boundary, and the obvious worry was that near the boundary it would turn out to be fragile — that a small change in the molecule would flip its answer back and forth. That happens: a third of the way from oxygen to fluorine, or at a slightly more p-rich oxygen, the π–n verdict changes sign.

The less obvious result is that the flip is inconsequential. The thing the criterion was meant to decide — whether two orbitals in the molecule’s localised description are canonical or mixed — does not change when the verdict does, because the molecule has a third orbital that takes π first. A two-orbital verdict is a correct statement about a two-orbital problem, and the one place in the series where it could have told the three-orbital problem something, it does not.

That reframes the earlier finding about the cases usually asked about. There, every pair proposed as rival descriptions cleared the boundary by a wide margin, and the worry was that the instrument was being pointed only where it had to agree. Here a case is found that genuinely sits on the boundary, and the instrument’s verdict there is the least informative thing on the page. A criterion is informative near its boundary only if crossing the boundary means something downstream, and for a pair that shares an orbital with a stronger partner, it does not.

Who measured what #

The Boys functional is Foster and Boys’s, from 1960, and the pairwise Jacobi sweep is Edmiston and Ruedenberg’s way of maximising such functionals. Slater’s rules for the exponents date from 1930 and Mulliken’s electronegativity from 1934; the ionisation energies and electron affinities behind the four values are the tabulated atomic ones. The isoelectronic series, the crossing, the three-orbital maxima across it, and the observation that the σ–lone-pair pair lies outside the two-orbital criterion’s scope are new here.

How the numbers were obtained #

Each member is the carbonyl trio rebuilt with the heteroatom’s Slater exponent and Mulliken electronegativity, at 1.203 Å and an s fraction of a third. Along the path, seven points were evaluated at intervals of half a member. At each, the three orbitals were Löwdin-orthogonalised and the nine dipole matrix elements integrated, and for each pair the separation along the bond, the off-diagonal dipole split into components along and across the bond, the margin and the maximising angle were recorded.

The crossing was found by regula falsi in the Illinois form, which halves the retained end’s value when it is kept twice so that one end cannot stall.

At each member the pair optima were found by a quarter-degree scan refined by golden-section search. The three-orbital maximum was found by a grid over the rotation group refined by coordinate steps, and confirmed by Jacobi sweeps in three orders to a change below 10⁻¹² a sweep.

Several things were required to hold, and each could have failed. The π–n margin had to rise at every step of the path. The crossing point had to be a root of the margin minus one to 10⁻⁴. σ–π had to stay above one at every member. At fluorine both shared pairs had to be endorsed and π–n’s pair gain had to be the larger. At every member the maximum had to keep the lone pair at least 94% itself and share π equally between the other two. All three orders had to reach the group maximum to 2 × 10⁻⁴, and the π–n optimum had to fail to be a maximum. For σ–n, the scanned angle had to match the closed form and lie strictly inside the interval, while π–n’s sat at an end. And the oxygen member had to be refused and the fluorine member endorsed at every bond length tried.

Still open: four orbitals, and a pair that shares nothing #

The four-orbital problem is the first question, and this series sharpens it. With both lone pairs present, fluorine’s lone pairs are two of three and the π bond has three partners rather than two. The near-balance found here — 83 to 139 sweeps where other members take eight — suggests the contest is close enough that a fourth orbital could tip it, and a heteroatom series is the natural way to look for where.

The other question is whether any pair crosses the boundary without sharing an orbital with a stronger partner. A crossing on such a pair would be a verdict whose flip changes the description, and it would be the first case in which the two-orbital criterion’s boundary means something for a molecule. The lone pairs themselves are one candidate, since rabbit-ear mixing involves no bond; their margin was shown to be fixed by the s fraction alone, and an s fraction near one is where it would approach the line.