Three-centre bonding, computed

Sulfur hexafluoride has six bonds to an atom whose valence shell holds eight electrons. The textbook resolution — borrow two 3d orbitals — is wrong, and hypervalency without d orbitals sets out why: the 3d orbitals lie about ten electron volts too high and are far too diffuse to contribute more than a few per cent.

The account that works is three orbitals in a line with four electrons in them, and it is small enough to solve completely.

The matrix #

Three orbitals, two bonds. Put the ligand orbitals at positions 0 and 2 and the central one at 1, and the matrix is

$$\begin{pmatrix} h & k & 0 \\ k & 0 & k \\ 0 & k & h \end{pmatrix},$$

where $h$ is how much lower a ligand orbital sits than the central one, in units of $\beta$, and $k$ is the strength of each bond. Fluorine’s values here are $h = 3.0$ and $k = 0.7$, quoted from the standard heteroatom tables — fitted numbers, with the caveat that applies throughout Hückel with a heteroatom.

There is no bond between the two ligands, so the corner entries are zero. That single zero is where most of the interesting behaviour comes from.

Diagonalising gives levels at

$$3.297, \qquad 3.000, \qquad -0.297$$

with the four electrons filling the lower two.

The orbital with nothing in the middle #

The middle eigenvector is the one worth looking at. Its coefficients are

$$(-0.707, \qquad 0.000, \qquad +0.707)$$

— equal and opposite on the two ligands, and exactly zero on the central atom.

That zero is not a numerical coincidence and does not depend on $h$ or $k$ at all. The arrangement has a mirror plane through the central atom, so every eigenvector must be symmetric or antisymmetric under reflection in it. This one is antisymmetric, and an antisymmetric function vanishes wherever the mirror plane passes — which is exactly where the central atom sits.

So one of the two occupied orbitals has no amplitude on the central atom whatever, and its two electrons live entirely on the ligands.

That is the resolution of the octet question. The central atom’s orbital appears in the filled bonding orbital with a coefficient of 0.288, and again in the empty antibonding one. Two electrons in the bonding orbital, weighted by that coefficient, put a small fraction of a pair on the centre. Nothing needs an expanded shell, because nothing was expanded.

Where the charge goes #

The charges come out at

$$1.917, \qquad 0.165, \qquad 1.917,$$

so each ligand carries nearly two electrons and the central atom carries a sixth of one. In an ordinary two-centre bond each atom would carry one.

That lopsidedness is a prediction rather than an embarrassment, and it accounts for the pattern that motivated the model in the first place: hypervalent compounds have electronegative ligands. Sulfur hexafluoride, phosphorus pentachloride, the xenon fluorides, the interhalogens — fluorine, oxygen, chlorine, and essentially nothing else. If the arrangement puts negative charge on the ligands, only ligands that will accept negative charge can support it.

A three-centre four-electron bond to methyl groups would put a formal negative charge on carbon, and hexamethylsulfur does not exist. Nor does hexahydridosulfur. The list of what is missing is as good a test of the model as the list of what exists, and the d-orbital account predicts neither list, because a d orbital is as available to a methyl group as to a fluorine.

Two electrons, and the other family #

The same three-by-three matrix with two electrons instead of four is the three-centre two-electron bond, and it is the bonding in diborane’s bridges and in every boron cluster.

There only the bonding orbital is filled. The non-bonding orbital is empty, so the ligands carry no excess and the arrangement needs no charge separation — which is why it works with hydrogen bridges, where the four-electron version would not.

Two electrons, three atoms, one bonding orbital. Here the name “electron-deficient” is fair in the plainest sense: there are fewer electron pairs than there are lines a chemist would want to draw. What the arithmetic says is that the lines were never the right thing to count.

Diborane is the case everyone meets first. Six atoms contribute twelve valence electrons; the four terminal B–H bonds take eight; and the four that remain hold the two bridges together, one pair per bridge, each pair spread over a boron, a hydrogen and the other boron. Drawn conventionally the molecule appears to need sixteen electrons and has twelve, and the discrepancy is not a discrepancy at all — it is a drawing convention meeting a molecule that does not obey it.

The two families are the same eigenvalue problem with different fillings, and between them they cover the two large classes of compound that the octet rule declares impossible — one by having too many electrons for the central atom and one by having too few.

What the parameters do and do not decide #

It is worth separating, as it always is here, the part of the answer that is the fit from the part that is the structure.

Vary $h$ from 2 to 4 — a range covering every plausible ligand — and three things happen. The levels slide, because $h$ is where the ligand orbitals sit. The charge on the ligands rises, because a lower ligand orbital holds its electrons more tightly. And the bonding orbital’s central coefficient falls, because a larger energy gap means less mixing.

Three things do not happen. The non-bonding orbital’s central coefficient does not move off zero. The ordering — bonding, non-bonding, antibonding — does not change. And the filling does not change, because four electrons in three orbitals is a count rather than a computation.

So every qualitative statement in this essay survives the parameters and every number in it does not. The octet is not exceeded at any $h$; the ligands carry the charge at any $h$; the amount they carry is a fitted quantity.

That split is the same one what a lone pair is worth finds in VSEPR and Hückel with a heteroatom finds in pyridine, and it is becoming the characteristic result of this collection: the useful predictions are the ones symmetry and counting make, and the parameters decorate them.

The molecule the arithmetic is usually applied to has six bonds and four valence orbitals, and the three-centre construction is how the count is made to work without invoking a d orbital. What this essay computes is the three-centre unit itself; how many of them a hypervalent molecule needs is a counting question answered elsewhere.

What was computed, and how #

The matrix is written out rather than built by substitution, because the electron count does not follow the usual substitution rule: the central atom contributes an empty orbital and each ligand contributes a lone pair, so it is four electrons in three orbitals however the parameters are set. That count is the name of the thing, and deriving it from a rule designed for carbon chains would have been a way of getting it wrong quietly.

The levels and coefficients come from a Jacobi diagonalisation, checked against the residual $Av = xv$, the orthonormality of the vectors, and the two trace relations — generalised here to be read off the matrix, so that they still say something when the diagonal is occupied and the bonds are not all one.

The pairing test runs in the refusing direction. This system has heteroatoms, so its levels must not be symmetric about $\alpha$, and they are not: 3.297, 3.000, −0.297 is symmetric about nothing. A version of the arithmetic that had dropped the diagonal entries would produce a symmetric spectrum, pass every other check, and be describing the allyl radical instead of a bond to fluorine.

The bond orders and charges are the usual sums over occupied coefficients, so the 0.389 and the 1.917 are not a separate method — they are bond order from the eigenvectors applied to a three-atom system.

Where the model stops #

This is a σ framework treated with π methods. The matrix is the one Hückel theory uses for conjugated systems, applied here to orbitals pointing along a line rather than perpendicular to a plane. The arithmetic transfers exactly; the justification for neglecting everything else is weaker, because a σ system has more orbitals close in energy than a π system does.

One axis at a time. Describing sulfur hexafluoride as three of these arrangements at right angles treats them as independent, and they are not. The real molecule’s ligand combinations span $a_{1g} \oplus e_g \oplus t_{1u}$ in Oh, and only the $t_{1u}$ set corresponds to what has been solved here — a decomposition that follows from character tables and reduction.

The parameters are fitted, and the charges move with them. What does not move is the zero coefficient and the direction the charge goes, both of which follow from the mirror plane.

Nothing here is an energy in physical units, so no comparison with a two-centre bond’s strength is available. “About four fifths of a single bond” is a statement about a computed bond order, which is a different quantity from a bond enthalpy and should not be read as one.

The geometry is assumed and not derived. That the three atoms are collinear, and that sulfur’s six ligands sit at the vertices of an octahedron, are inputs here. They are also what a repulsion argument predicts independently — six equivalent domains minimise at the octahedron, as VSEPR, computed shows — so the two accounts agree, but the agreement is a coincidence of two models rather than a derivation within one.

And the ligand orbital is treated as a single function. A fluorine brings three 2p orbitals, one pointing at sulfur and two perpendicular to the bond; only the first is in this matrix. The perpendicular pair carries the lone pairs and is the reason fluorine is a poor π donor, and none of that is represented.

Why the picture is so hard to draw #

There is a practical reason the d-orbital story survived as long as it did, and it is worth naming because it is about pictures rather than about chemistry.

A three-centre bond has no line to draw. The conventional structural formula gives one line per shared pair between two atoms, and the whole content of this arrangement is that a pair is shared between three. Chemists write sulfur hexafluoride with six lines because there is no other notation, and six lines around sulfur is twelve electrons, and twelve electrons needs an explanation — so an explanation was supplied.

The hybridisation label followed the drawing rather than the other way round. Six lines means six orbitals; carbon’s four are $sp^3$; so sulfur’s six must be $d^2sp^3$, and two d orbitals were conscripted to make the count work. Every step of that reasoning is about the notation.

The arithmetic has no such difficulty. It never counts lines; it counts electrons in orbitals, and the orbitals are whatever the eigenvectors say. That is the same lesson hybridisation does not explain draws from methane’s photoelectron spectrum: a labelling scheme built to justify a drawing is not a description of anything, and the moment a measurement is brought to bear it comes apart.

The generalisation #

The three-centre case is where the two-centre habit visibly stops being general, and the instructive part is that nothing in the arithmetic changes when it does.

The same eigensolver, the same filling rule, the same bond orders from the same sums over coefficients. What changes is the size of the matrix, and with it how many atoms an orbital is spread over. A two-centre bond is the case where the matrix happens to be two by two, and there is no sense in which the arithmetic treats it as normal.

Run the construction with more atoms and it becomes the band limit; run it on a ring and it becomes aromaticity; run it on a polyhedron and it becomes the electron-counting rules for clusters. Where two-centre bonding stops makes that continuity the argument rather than an aside.

The habit worth keeping is to ask, of any bonding picture, how many centres its orbitals are spread over — and to notice that the honest answer is rarely two, and that two is a drawing convention rather than a result.

The bond order is a half, and a diffraction measurement says so #

The model gives every bond in a three-centre four-electron arrangement a bond order of one half — two bonding electrons shared over two bonds — and that is a structural prediction rather than a bookkeeping one. A half-order bond should be markedly longer than a whole one between the same two atoms, and the comparison is available because both species exist.

Take iodine. The bond in the diatomic is 2.666 ångström and is a straightforward single bond. The triiodide ion is the three-centre case, and its two bonds measure about 2.90 — longer by nearly a quarter of an ångström, in a system with the same two atoms and no other difference.

The size is right as well as the sign. Pauling’s empirical relation between bond order and length, $r(n) = r(1) - 0.60\log_{10} n$, predicts that halving the order lengthens the bond by 0.18 ångström. Applied to iodine that gives 2.85 against a measured 2.90, and applied to the analogous dichloroiodate ion — 2.32 in iodine monochloride, 2.55 in the ion — it gives 2.50 against 2.55. Both are five picometres out, in the same direction, from a relation with no hypervalency in it at all.

The asymmetric cases are the better evidence, and they are the ones a d-orbital account has no room for. Triiodide is often not symmetric in the solid state: depending on which cation it is crystallised with, the two bonds can come out at something like 2.83 and 3.04 rather than equal. A description in which the central atom uses a hybrid to make two ordinary bonds has no reason to permit that. A three-centre arrangement does — the two bonds share one bonding pair, so lengthening one shortens the other at almost no cost, and the total is nearly fixed while the split is soft.

That softness is the signature. A pair of independent bonds resists being made unequal; a pair sharing one bonding orbital barely notices, and the arrangement in the crystal is then decided by whatever the surroundings prefer rather than by the anion itself.

So the arithmetic of the three-level problem produces two checkable structural consequences and both hold: bonds longer than single ones by the amount a bond order of a half implies, and an unusual indifference to being made unequal. Neither is a statement about energies, which is what the model cannot supply — both are statements about lengths, which is what a diffraction experiment returns.

Who found it, and when #

Rundle and Pimentel proposed the three-centre four-electron description independently in 1951, for the trihalide ions and for the xenon fluorides, and it was the correct account from the day it appeared. The d-orbital story it should have replaced had been standard since the 1930s.

Longuet-Higgins’s work on the boranes in the late 1940s established the three-centre two-electron bridge, and Lipscomb’s structural work through the 1950s turned it into a systematic scheme for the whole cluster family — the work recognised by his 1976 Nobel Prize. Wade’s rules, which count skeletal electron pairs and predict cluster shapes, followed in 1971 and are the same accounting carried to polyhedra.

The persistence of the d-orbital account in teaching is the part worth recording. The alternative had been available since 1951, calculations settled the question in the 1980s, and hybridisation schemes labelled $d^2sp^3$ are still printed. Collecting that kind of gap is what the field this belongs to exists for.

What the model cannot supply #

The octet is not exceeded even where it appears to be, and a three-atom bond is ordinary Hückel arithmetic with one more row in the matrix. What it cannot supply is the strength of any of these bonds, which is what a model working in units of an unfitted parameter can never give.