Hypervalency is about the ligands

Hypervalency without d orbitals removes the standard explanation of sulfur hexafluoride — that sulfur uses its 3d orbitals — on the ground that those orbitals are far too high in energy to contribute meaningfully. Three-centre bonding, computed then supplies the replacement: three orbitals in a line with four electrons in them give a bonding level, a level with exactly zero amplitude on the centre, and an empty antibonding one, so twelve electrons can be accommodated around an atom with only s and p available.

Both of those read the argument as a permission: the arrangement is possible, therefore no d orbitals are needed.

Read forwards it is a requirement, and this essay is about what it requires.

Where the charge goes, and why it is not adjustable #

Take the three-orbital system with all three atoms alike. The two occupied levels are:

• a bonding combination over all three centres, $(1, \sqrt2, 1)/2$;
• a non-bonding one, $(1, 0, -1)/\sqrt2$, with exactly no amplitude in the middle.

The second is fixed by symmetry rather than by any parameter: the middle atom sits on a mirror plane, the level is antisymmetric under it, and an antisymmetric function has a node where the plane is.

So of the four electrons, two are shared over all three centres and two sit entirely on the ends. Counting them up: each end holds $1.5$ electrons and brought one, and the middle holds $1.0$ and brought two.

$$q_{\text{end}} = -0.5, \qquad q_{\text{middle}} = +1.0.$$

With no electronegativity difference at all. The charge separation is not a consequence of fluorine being electronegative; it is a consequence of the level structure, and fluorine’s electronegativity only makes it larger.

And a more electronegative ligand takes more #

Give the terminal atoms a Coulomb integral of $\alpha + h\beta$, which is this site’s usual parameter for an electronegative atom, and sweep $h$:

Monotone in $h$, and never below half whatever the parameters are.

That is the requirement, and it is a chemically substantial one. To hold this arrangement, the central atom must give up half an electron per axis at the very least and more in practice; and the ligands must be willing to accept it.

Which is why the hypervalent species are the ones they are. SF₆, PF₅, ClF₃, XeF₂, IF₇, and the oxides and the chlorides. There is no SH₆, no PH₅, and no CH₆ — a pattern where two-centre bonding stops sets up as the general case rather than the exception — hydrogen has nowhere to put the charge, and carbon is not electronegative enough to take it from anything worth taking it from.

This is a prediction of the model rather than a rationalisation after the fact: the model says the arrangement costs a specific amount of charge transfer, and the compounds that exist are the ones where that transfer is affordable.

It also sorts the borderline cases in the right order. Xenon difluoride exists and xenon dichloride barely does; iodine forms IF₇ and no iodine heptahydride; phosphorus forms PF₅ and PCl₅ and, in the gas phase, no PH₅ at all. In each pair the compound that exists is the one whose ligand is further to the right of the periodic table — which is exactly the ordering the table above produces, since a larger $h$ is a ligand more willing to take the charge the arrangement forces on it.

The bond order, which is not what the counting says #

There is a second number here and it corrects a slogan.

The usual gloss on the three-centre four-electron bond is that its two bonding electrons are shared over two links, so each link is “half a bond”. Compute the bond order from the eigenvectors — the sum over occupied levels of the product of the two coefficients, which is what bond order from the eigenvectors establishes — and it comes out at

$$p = \frac{1}{\sqrt2} = 0.7071$$

for each link, with the two together at $1.414$.

Not a half, and not a whole. The counting argument divides two electrons between two links and gets a half; the computation says the bonding combination is not spread evenly — the middle atom carries half of it — so the products of coefficients across each link are larger than the naive share. The non-bonding level contributes exactly nothing, since one of its two coefficients is zero at every link.

Both numbers are worth having and the counting one is what is usually quoted. What they agree about is the direction: two lines drawn on paper are worth less than two bonds, and a structure of sulfur hexafluoride with six lines in it is overstating the bonding by about a third.

It is worth saying what the counting and the computation disagree about elsewhere, because the gap is general. Benzene’s π bond orders are all exactly two thirds against a naive share of two electrons over each of six links, and that is a system where the two accounts agree unusually well. The three-centre case is the same discrepancy with the agreement removed.

Three systems, one atom #

Sulfur hexafluoride’s twelve bonding electrons are three of these systems on three orthogonal axes, and the charges add.

The central sulfur ends up at about $+3$ in this model with $h = 0$ and at $+4.7$ with $h = 2$, which is a fluorine-like value. Every fluorine carries between $-0.5$ and $-0.8$.

Those are large charges and they are the model’s, not measurements. What can be said is the direction and the ordering: the model requires a substantial positive charge on the central atom, real determinations of the charge distribution in SF₆ do give the sulfur a large positive charge, and the compounds that exist are those with ligands able to absorb it.

And the symmetry half of the argument is exact rather than modelled. Six bonds and four orbitals reduces the six fluorine σ functions in the octahedral group and gets $a_{1g} \oplus e_g \oplus t_{1u}$; sulfur’s 3s supplies the $a_{1g}$ and its 3p the $t_{1u}$, and the $e_g$ pair has nothing on sulfur to combine with. So two of the six ligand combinations are strictly non-bonding, which is the octahedral version of the “no amplitude in the middle” level above, and it is a group-theoretical statement rather than a parametrised one.

The refusal: two electrons instead of four #

The sign of everything above is a fact about the filling, and the check is to change it.

Put two electrons into the same three orbitals rather than four. Now only the bonding level is occupied, the non-bonding one is empty, and the density is $(0.5, 1.0, 0.5)$: the middle atom gains and the ends lose.

That is the three-centre two-electron bond — diborane’s B–H–B bridge, which what one pair can hold together computes the bond order of — and its charge pattern is the opposite of the four-electron case’s. Same three orbitals, same symmetry, opposite sign of charge separation, and the only difference is two electrons.

So the charge separation in the four-electron case is not a general property of three-centre bonding. It comes specifically from occupying a level that has no amplitude in the middle, and a check that reported the same sign for both fillings would be measuring nothing.

What the octet rule was actually about #

The word hypervalent carries an assumption worth separating out, because the argument above dissolves it.

An octet is not a limit on how many atoms can be attached; it is a limit on how many bonding orbitals a central atom’s own valence shell can supply. With one s and three p functions there are four, and four bonding orbitals hold eight electrons. That is the whole of the rule.

Sulfur hexafluoride does not break it. Its sulfur supplies four bonding orbitals — the $a_{1g}$ from 3s and the three $t_{1u}$ from 3p — holding eight electrons, exactly as the rule says. The other four electrons are in the $e_g$ pair, which is on the fluorines and has no sulfur amplitude at all.

So the molecule has twelve electrons in the bonding region and eight of them around the sulfur, which is what the rule constrains. The apparent violation is an artefact of counting an electron pair as “belonging to” a bond that is drawn as a line between two atoms, and the three-centre picture is a way of not doing that.

That reading also explains why the rule is stated so confidently for the second row and breaks down below it. It is not that a third-row atom acquires new orbitals; it is that a larger, more polarisable central atom can carry the positive charge the arrangement requires, and a second-row atom of the same electronegativity cannot. Which is the same conclusion as the section above, arrived at from the counting side.

The word itself, and why it is being retired #

Hypervalent is falling out of use in the literature, and the reasons are the ones above rather than a fashion.

The word presupposes that a valence has been exceeded, and the count it refers to — eight electrons — turns out to be a count of what the central atom’s own orbitals can hold in bonding combinations, which sulfur hexafluoride does not exceed. So the term names a violation of a rule that is not violated.

What replaces it is a description rather than a label: electron-rich three-centre bonding, or simply the molecular-orbital diagram. Both say what the arrangement is, and neither implies that anything has broken.

The change is not merely terminological, because the old word carried a wrong explanation with it. If a molecule is hypervalent, something must have been added to let it exceed the limit, and the something offered was d orbitals — an explanation that survived for fifty years, that hypervalency without d orbitals shows is quantitatively wrong, and that is still in current textbooks.

What can be done instead is compute what the arrangement costs, in a quantity that separates the compounds that exist from the ones that do not. That is a better kind of account than a permission: it explains SF₆ and it explains the absence of SH₆ with the same arithmetic, which the d-orbital story never could — sulfur’s 3d orbitals are exactly as available in one as in the other.

Where the model stops #

Hückel, so no repulsion — the model Hückel theory and what it gets right sets out, with the same caution about what a one-electron picture can be asked for. The charges computed here are what a one-electron model with the stated parameters gives. A real molecule’s electrons repel, which opposes charge separation, so the true charges are smaller than these — and the requirement is what the argument turns on, not the size.

The parameter $h$ is not measured. It stands for electronegativity and is calibrated by convention rather than by any experiment on these molecules.

No σ framework, no geometry. The three orbitals are a line with equal spacings; nothing here computes a bond length or asks why the axes are orthogonal. The arrangement of the six ligands is the repulsion result VSEPR, computed produces, and this argument takes it as given rather than deriving it.

And the three axes are treated as independent. In sulfur hexafluoride they are not quite: the three three-centre systems share the same central atom and therefore compete for its charge, so the charges do not simply add. The octahedral reduction in the previous section is the way of doing the same calculation without that approximation, and it agrees about the count of non-bonding electrons — four in the eg pair, which is four of the twelve, matching two per axis over three axes only after the shared bookkeeping is untangled.

Charges are not observable. There is no operator whose expectation value is “the charge on an atom” — every population analysis is a convention for dividing a continuous density among nuclei, and different conventions give different numbers for the same wavefunction. The differences between the rows of the table above are more reliable than any row of it, which is the same caution the dipole is not a sum of bonds applies to a partial charge.

One further consequence follows from the charge requirement and is easy to check against the chemistry. If the arrangement forces charge onto the ligands, then a hypervalent centre should be electrophilic — the positive charge is real and it is exposed.

That is what these molecules do. Phosphorus pentafluoride and sulfur tetrafluoride are attacked at the central atom by anything nucleophilic; xenon difluoride is a fluorinating agent because its xenon is positive enough to give fluoride away. The exception proves the point: sulfur hexafluoride is famously inert, and the reason usually given is steric — six fluorines leave no room to approach the sulfur — which is a statement about access rather than about charge, and is consistent with the sulfur being as positive as the model says.

The compounds at the boundary, which fail one step at a time #

If the arrangement is a demand on the ligands, then walking down a group should not simply switch the compounds off. It should make them progressively harder to have, and the phosphorus pentahalides walk down that boundary one halogen at a time.

Phosphorus pentafluoride is a stable gas, trigonal bipyramidal, and molecular as a solid, a liquid and a gas alike. Fluorine takes the charge the arrangement demands without complaint.

Phosphorus pentachloride is molecular in the gas phase and in non-polar solvents — and in the solid it is not a molecule at all. It becomes $[\mathrm{PCl_4}]^+[\mathrm{PCl_6}]^-$: a four-coordinate cation and a six-coordinate anion, both of which the molecule prefers to being five-coordinate and neutral.

Phosphorus pentabromide goes further. In the solid it is $[\mathrm{PBr_4}]^+\mathrm{Br}^-$ — the hypervalent partner has been abandoned entirely, and the fifth bromine leaves as a free ion rather than accept its share.

Phosphorus pentaiodide does not exist, and neither does the pentahydride.

That is a graded failure rather than a threshold, and every step of it is the model’s requirement being tested against a ligand less willing to meet it than the last.

The chloride’s behaviour is the most informative row, because the ionic form looks at first like an escape from hypervalency and is not. The $[\mathrm{PCl_6}]^-$ anion is more hypervalent than the neutral molecule — six ligands rather than five, two orphan combinations rather than one — and the arrangement is affordable there for exactly the reason the model gives: the ion carries a whole extra electron, so the charge the ligands are required to take is supplied from outside rather than extracted from the phosphorus.

So the disproportionation is not the molecule avoiding the arrangement. It is the molecule paying for it, by splitting into a piece that has no hypervalency and a piece that has more of it and an extra electron to fund it.

Which is a sharper statement of the argument than the list of stable fluorides gives. Hypervalency is not permitted or forbidden by the ligand’s electronegativity; it is priced by it, and where the price is marginal the compound does not vanish — it finds a different way to pay.

What reading the node as a constraint adds #

The account without d orbitals removes the old explanation, computes the three-centre four-electron system that replaces it, and reduces the six ligand functions of an octahedral molecule to show that only four of them can bond.

Here the same construction is read as a constraint. The non-bonding level’s node on the central atom is not an incidental feature — it is where the charge comes from, it puts at least half an electron on each ligand before any electronegativity enters, and it grows from there. That makes hypervalency a statement about what the ligands will accept, which is why the list of hypervalent compounds is a list of fluorides and oxides and why the hydrides are absent from it. Along the way the computed bond order comes out at $1/\sqrt2$ rather than the half the electron count suggests, so the six lines in a structural formula for SF₆ overstate its bonding by about a third.

The open question is the compounds at the boundary — the ones with ligands that are marginal, where the model says the arrangement is barely affordable and the chemistry says the compounds are barely stable.