Six bonds and four orbitals

Sulfur hexafluoride has six bonds to one sulfur, which is twelve electrons around an atom whose valence shell holds eight. The standard resolutions are to say that the octet rule is a guideline, or that sulfur uses its 3d orbitals, or that the bonds are not ordinary bonds.

The third is right, as hypervalency without d orbitals established, and the reduction formula says so in about six lines of arithmetic — with no energy, no orbital exponent and no fitted parameter anywhere in it.

The count #

Take one σ function on each fluorine — a hybrid pointing at the sulfur, or just the fluorine 2p along the bond; the reduction only counts which are left in place, so the answer is the same either way.

Applying each of Oh’s forty-eight operations and counting unmoved fluorines gives a character, and the reduction formula — character tables and reduction builds it from the group’s own operations — turns those characters into

$$\Gamma_\sigma = a_{1g} \oplus e_g \oplus t_{1u}$$

which is six functions, as it must be: $1 + 2 + 3$.

Now ask what the sulfur has to offer. Its 3s spans a₁g. Its 3p spans t₁u. Those are computed the same way — the characters of an s and a p shell under the group’s operations — and together they supply $1 + 3 = 4$ functions.

Four functions to match six. The a₁g combination pairs with the 3s, the t₁u set pairs with the 3p, and the eg pair has no partner at all: there is nothing in an s-and-p valence shell that transforms as eg in Oh.

What happens to the unmatched pair #

A ligand combination with no central-atom partner does not vanish and does not prevent the molecule existing. It becomes a non-bonding orbital: a combination of fluorine functions that is neither raised nor lowered by the sulfur, because the overlap between them is zero by symmetry.

That zero is the exact kind, not the small kind. The overlap integral between an eg ligand combination and any s or p function on the sulfur vanishes because the integrand is odd under an operation of the group, and the contributions cancel in pairs — the same mechanism exactly zero computes at $10^{-17}$ for a symmetry-forbidden overlap of two atomic functions.

So the twelve electrons are accounted for like this:

• a₁g bonding — 2 electrons, spread over all six bonds
• t₁u bonding — 6 electrons, spread over all six bonds
• eg non-bonding — 4 electrons, on the fluorines only

Eight electrons doing bonding work across six bonds gives a σ bond order of $8/12 = 2/3$ per bond. The remaining four sit on the ligands, so the molecule is held together by fewer bonding electrons than a Lewis structure draws, and the fluorines carry substantial negative charge as a result.

All six carry the same charge, and that is forced rather than assumed: the six fluorines form a single orbit under the group, so no operation distinguishes any of them and every property that can be assigned to an individual atom must be assigned equally to all six. Site symmetry, and what it constrains is the general form of that argument, and it is what makes the per-bond figure above meaningful — a bond order of two thirds spread unevenly would be a different molecule.

The same arithmetic on two more molecules #

The count is not special to an octahedron, and running it on the other two textbook hypervalent species gives a series.

Phosphorus pentafluoride, D3h. Five σ functions span $2a_1' \oplus e' \oplus a_2''$. Phosphorus 3s spans $a_1'$ and 3p spans $e' \oplus a_2''$, so the central atom supplies $a_1'$, $e'$ and $a_2''$ — four functions. One of the two $a_1'$ ligand combinations has no partner. Four bonding orbitals for five bonds: a σ bond order of four fifths.

Xenon tetrafluoride, D4h. Four σ functions span $a_{1g} \oplus b_{1g} \oplus e_u$. Xenon 5s spans $a_{1g}$ and 5p spans $a_{2u} \oplus e_u$, so the matches are $a_{1g}$ and $e_u$ — three functions — and $b_{1g}$ is unmatched. Also unmatched is the xenon’s own $a_{2u}$ p orbital, which points perpendicular to the plane and holds a lone pair. Three bonding orbitals for four bonds: three quarters.

In every case the ligand set spans exactly one species more than an s-and-p shell can supply, and the leftover species is where the electron-rich character lives. That is the symmetry content of the three-centre four-electron picture, arrived at without drawing a single three-centre bond.

The same count for a molecule that is not hypervalent #

The arithmetic is worth running on an ordinary molecule, because a counting argument that only ever produces the answer it was invented for is not an argument.

Methane’s four hydrogen σ functions span $a_1 \oplus t_2$ in Td, which is $1 + 3 = 4$ functions. Carbon’s 2s spans $a_1$ and its 2p spans $t_2$, which is also four. Every ligand combination finds a partner, nothing is left over, and there are four bonding orbitals for four bonds — a σ bond order of one, which is what an ordinary molecule should give.

Water’s two hydrogen functions span $a_1 \oplus b_2$ in C2v; oxygen’s s and p span $2a_1 \oplus b_1 \oplus b_2$. Both ligand combinations are matched, two bonding orbitals for two bonds, and the leftovers are on the central atom rather than on the ligands — an $a_1$ and a $b_1$, which are the two orbitals a Lewis structure calls lone pairs and which water’s lone pairs are not a pair shows the spectrum putting in different species.

So the same reduction distinguishes the two situations cleanly. An ordinary molecule leaves its spare functions on the central atom; an electron-rich one leaves them on the ligands. That is a sharper statement than “the octet is exceeded”, and it is a symmetry statement rather than a count of electrons.

Where the d orbitals would go, exactly #

The reduction says precisely what a d shell would contribute, and it is worth computing rather than dismissing.

Sulfur’s 3d functions span $e_g \oplus t_{2g}$ in Oh. The eg half is exactly the species the ligand σ set has spare. So symmetry permits a d orbital to bond with the leftover combination, and the classical $d^2sp^3$ hybridisation scheme is not a symmetry error.

Xenon’s 5d spans $a_{1g} \oplus b_{1g} \oplus b_{2g} \oplus e_g$ in D4h, and the b₁g half again matches the unmatched ligand combination exactly. Phosphorus’s 3d spans $a_1' \oplus e' \oplus e''$ in D3h, and the spare $a_1'$ is again matched.

Three molecules, three unmatched combinations, and in all three the d shell has the right symmetry. That is not a coincidence: a d shell spans five functions and an s–p shell four, so between them they cover far more than the ligand set can produce, and the leftover is bound to be found somewhere.

What decides the question is therefore not symmetry but size and energy, and this is where the site’s computation stops and the honest statement begins.

The size argument, and its limits #

A hydrogenic 3d orbital’s mean radius is $\langle r\rangle = 10.5/Z_{\text{eff}}$ bohr. Sulfur has no 3d electrons in its ground configuration, so a screening model gives no effective charge for one directly; an electron added to a sulfur 3d would be almost completely screened by everything inside it, which puts $Z_{\text{eff}}$ near one and $\langle r\rangle$ near $10.5$ bohr, or $5.6$ Å.

The S–F distance is $1.560$ Å. Sulfur’s 3p, at its Slater effective charge of $5.45$, has a mean radius of $1.21$ Å.

So a free-atom 3d orbital is more than three times the length of the bond it is supposed to be forming, while the 3p is about the right size. An orbital that diffuse overlaps very little with anything at bonding distance, and the interaction it can supply is correspondingly small.

Symmetry cannot say how much a molecular field contracts it, and that is the honest boundary. A 3d orbital in a molecule with six electronegative ligands is not a free-atom 3d; it contracts, and the modern calculations that settle the question find it contracting enough to matter as a polarisation function — improving the description of the bonding already present — and not enough to make it a bonding partner. Reaching that conclusion requires a variational calculation with two-electron integrals, which is well beyond a counting argument.

What can be said from here is narrower and still useful: the molecule does not need the d orbitals to exist, because four bonding orbitals plus two non-bonding ones account for every electron, and the resulting bond order is what the measured chemistry of these compounds suggests.

What the electron-rich picture gets right #

The three-centre four-electron description of these molecules — a linear F–S–F unit sharing one p orbital on the sulfur between two fluorines, with four electrons in three orbitals — arrives at the same bond order from the other direction.

Three-centre bonding, computed works that case explicitly and finds the non-bonding orbital’s coefficient on the central atom to be exactly zero, by symmetry. Three such units at right angles use three of the sulfur’s p orbitals and produce six bonds, with the sulfur’s s doing separate work — which is the same accounting as the reduction above, in a basis that makes the geometry obvious rather than the symmetry.

Three predictions follow and all three are borne out.

The bonds should be weak individually, since each carries less than a full pair — and the total binding should nevertheless be large, since there are six of them. That is the accounting what one pair can hold together sets out for a ring, arriving here with a central atom in the middle: a conserved total divided among more links.

The ligands should be electronegative. A picture with substantial electron density parked on non-bonding ligand combinations only works when the ligands can hold it, which is why hypervalent compounds are overwhelmingly fluorides, chlorides and oxides, and why SH₆ does not exist.

The central atom need not be from the third row. The usual explanation for the absence of hypervalent second-row compounds is that carbon and nitrogen have no d orbitals available. The electron-rich account gives a different and better reason: a second-row atom is too small to hold six ligands, and the ligand-ligand repulsion rather than any orbital availability is what forbids it.

The table the reduction sums over is generated by closing SF₆’s found operations under multiplication and sorting them into ten conjugacy classes: forty-eight operations, ten species, and the orthogonality relations checked before use. Every multiplicity above is a sum over that table divided by forty-eight, and a fractional answer would mean the table or the character was wrong.

What a descent in symmetry does to the count #

The count depends on the group, so distorting the molecule changes it, and the direction is instructive.

The correlation from Oh to D₄ₕ splits the eg pair into a₁g ⊕ b₁g and t₁u into a₂u ⊕ eu, so a tetragonal distortion turns two of the species above into four. That matters here only as a reminder that the count of bonding combinations is a property of the group rather than of the molecule, and changes when the group does.

Under a tetragonal distortion, the eg pair splits into $a_{1g} \oplus b_{1g}$. The $a_{1g}$ half now has the same symmetry as the central atom’s s orbital and can mix with it, so one of the two non-bonding electron pairs acquires some bonding character while the other does not.

That is the symmetry statement behind axial and equatorial bonds differing in an electron-rich molecule, and it is the same reduction that copper is never quite octahedral uses on a d shell. A distortion is a change of group, and a change of group is a change in which things may mix.

The six fluorines are a single orbit — every operation permutes them among themselves and none distinguishes any of them — which is why the σ reduction has exactly three species in it and not six. Where a molecule has two orbits, the reduction has to be done twice, and the two answers do not merge.

A bond order below one, tested inside a single molecule #

A bond order of two thirds is a structural claim, and the cleanest test of it does not compare two molecules — it compares two kinds of bond inside one, which removes every difference except the one being tested.

Sulfur tetrafluoride is the molecule. Its four bonds are not equivalent: two sit in a plane with the lone pair and two point along an axis through it, and the three-centre account assigns them different orders. The equatorial pair are ordinary two-centre bonds using sulfur’s s and p in the usual way. The axial pair are a single three-centre four-electron system sharing one bonding pair between them, so each carries an order of a half.

The measured lengths are 1.545 ångström equatorial and 1.646 axial. A tenth of an ångström, in one molecule, between two sets of bonds joining the same two elements — and the longer pair are exactly the two the counting argument prices at half.

Nothing about that comparison depends on a scale factor, a reference molecule or a fitted relation. It is an ordering, it is internal, and it comes out the way the arithmetic says.

Sulfur hexafluoride is the harder case and it is worth being honest about it. Its six bonds are 1.564 ångström — between the two values in the tetrafluoride, which is where an order of two thirds should sit, and closer to the equatorial single bond than an interpolation would put it. Taking the tetrafluoride’s two lengths as calibration for orders of one and a half, an order of two thirds predicts something near 1.60, and the molecule comes in four hundredths shorter.

The direction of that discrepancy is informative rather than awkward. The counting argument prices the covalent sharing and contains no electrostatics, and a sulfur in the hexafluoride carries a considerably larger positive charge than one in the tetrafluoride — six electronegative ligands rather than four, and no lone pair to hold density back. A more positive centre has more contracted orbitals and attracts its anionic ligands more strongly, and both effects shorten the bond.

So the two comparisons say different things and both are useful. Within one molecule, where the charge on the central atom is fixed, the bond order predicts the ordering and the size of the difference. Across molecules, where the charge changes, it does not, and the residue is the term a σ-only symmetry count was never going to contain.

That is the ordinary boundary of an argument made from characters. A reduction says which combinations can interact and how many bonding orbitals result; it has no energies in it, no charges and no lengths. What it produced here is a prediction about a ratio of lengths within a molecule, which is the strongest form of statement it is entitled to make — and the molecule obliged.

One further reading of the tetrafluoride’s two lengths is worth recording, because it disposes of a rival account without any calculation. A repulsion argument also predicts that the axial bonds are the longer ones, on the grounds that an axial position has three neighbours at ninety degrees where an equatorial one has two. But that argument predicts a difference in every five-coordinate molecule, including those with no hypervalency in them, and it predicts nothing about the order of the bonds. The three-centre account predicts the same ordering and adds two things the repulsion account cannot: that the axial pair share one bonding orbital, and that they should therefore tolerate being made unequal — which is what the softness of the axial coordinate in these molecules actually shows.

Who found it, and when #

The three-centre four-electron description is George Pimentel’s and Robert Rundle’s, from 1951, and it was available a decade before the $d^2sp^3$ account became standard in teaching. Calculations through the 1980s — notably by Reed and Weinhold, and by Magnusson — established that d functions in these molecules behave as polarisation functions and that removing them entirely changes the computed bonding very little.

The reduction itself is older than either. Applying the character-counting method to a set of ligand functions was routine by the mid-1930s, and the result for an octahedron appears in the earliest treatments of coordination compounds.

What is striking is how long it took for the arithmetic to displace the story. The count above requires no computer, no parameters and about ten minutes with a character table; it says the ligand set spans one species more than s and p can supply; and it has been available and correct since before the compounds’ structures were determined.

Still open: what a d shell contributes #

The argument about hypervalency began by asking whether hypervalent molecules need d orbitals and answering with a calculation of the bonding; the three-centre case was then computed in detail. This essay generalises both to a counting argument that runs on any geometry.

What is left is the part symmetry cannot reach. Whether a particular d shell contributes is a question about radial extent and energy, and answering it needs a variational calculation well beyond this one. The counting argument’s value is that it makes the deferral safe: the molecule is explained without the d orbitals, so their contribution is a correction to a working account rather than the thing holding it up.