Hybridisation does not explain

Methane is tetrahedral. Its four C–H bonds are identical: same length, same strength, related by the symmetry operations of the point group Td. The standard account says carbon forms four equivalent sp³ hybrids, each overlapping with a hydrogen 1s.

Take an ultraviolet photoelectron spectrum of methane and it shows two ionisation energies, around 12.7 and 23 electron volts, in an intensity ratio of about 3 to 1.

Why two bands is a problem #

A photoelectron spectrum measures the energy needed to remove an electron. To a good approximation — Koopmans’ theorem — each band corresponds to removal from one occupied orbital, and orbitals of the same energy give one band.

Four equivalent sp³ bonding orbitals are, by construction, degenerate. Four electron pairs in four identical orbitals should give one band.

The spectrum gives two, in a 3:1 ratio. Something in the account is wrong.

The prediction, made before the spectrum is looked at #

The resolution is that the orbitals a spectrum probes are not the hybrid ones. That sentence is a claim, and it is much stronger as a derivation, because the answer is not fitted to the spectrum — it follows from where the four hydrogens are, and from nothing else.

The argument runs in three steps and none of them mentions an energy.

Generate the group. Methane’s symmetry operations are found by searching the structure for axes and planes that permute the atoms among themselves, and then multiplying what is found together until the set stops growing. It closes at twenty-four operations, which fall into five conjugacy classes when each is conjugated by every other. Only at that point is a character table opened, and it is checked against the classes just counted rather than trusted.

Count what stayed put. A basis function contributes to the character of an operation only if the operation sends it back to itself, which for an s function sitting on an atom means the atom did not move. So the character of the four hydrogen 1s orbitals is a tally of unmoved hydrogens: four under the identity, one under each threefold rotation, none under a twofold, none under an improper fourfold, and two under each dihedral mirror.

Divide. The reduction formula turns that tally into multiplicities, and the multiplicities must come out as whole numbers — which is the check, because a fraction has no meaning and an error anywhere upstream produces one.

The answer is a₁ ⊕ t₂. So the four bonding orbitals must fall into two symmetry species, with degeneracies one and three, and a single band is symmetry-forbidden.

This changes the standing of the argument. The spectrum is no longer an awkward fact that hybridisation has to be excused for; it is the confirmation of a result derived from four coordinates, and the localised picture is refuted by symmetry before any measurement is taken.

Where the threefold degeneracy comes from #

The hydrogen tally says the four bonding orbitals split one-and-three. It does not yet say why the group of three is a group of three, and the same instrument answers that when it is pointed at the other half of the problem.

Run the reduction again, on carbon’s own valence functions rather than on the hydrogens. The three 2p orbitals sit on the atom at the centre, so no operation moves the atom and every character is a trace of what the operation does to a vector.

Carbon’s 2p set spans t₂ exactly, and its 2s, being spherically symmetric about a point every operation fixes, spans a₁. Put the two halves together and the match is complete: a₁ from carbon meets a₁ from the hydrogens, t₂ meets t₂, and every one of the eight valence functions finds a partner of its own species. Nothing is left over and nothing is unpaired.

That is the whole of why methane’s spectrum has the shape it has. Not that carbon “makes four hybrids”, but that a₁ can only interact with a₁ and t₂ only with t₂, and the two species have different energies because they are made of different atomic orbitals — the deeper band from carbon 2s, the shallower threefold one from carbon 2p. The 3:1 ratio and the nine-electronvolt separation are the same fact seen from two sides.

The hybrid construction reproduces the geometry and the bond equivalence correctly and says nothing at all about this. It cannot: four equivalent objects have one energy by construction, and the splitting is precisely the quantity a set of four equivalent objects has been built not to have.

Why both descriptions are correct #

Here is the part that resolves the paradox rather than choosing a side.

The four localised bond orbitals and the four canonical orbitals (a₁ plus t₂) are related by a unitary transformation. They span the same space, they give the same total electron density, and they give the same total energy. A basis is not a thing, and neither set is more real.

That claim is not merely stated here. It is carried out, in coefficients and densities, on this same molecule: four bond orbitals are built, orthogonalised, multiplied by the tetrahedral matrix, and the density that comes out is compared point by point with the density that went in. The two agree to the last bit or two of a double, while the orbitals themselves differ by nearly a factor of four in how many hydrogens each one occupies. The transformation is a rotation, and a rotation changes no total quantity.

What differs is what each description is good for. The localised set matches chemical intuition and makes structure and reactivity easy to think about. The canonical set consists of eigenfunctions of the one-electron Hamiltonian, which is exactly what a photoelectron experiment couples to. A rotation preserves every total and discards every per-orbital quantity, and an ionisation energy is a per-orbital quantity.

So the spectrum does not refute hybridisation. It refutes the claim that hybrid orbitals are what a spectrometer sees, which is a claim hybridisation never made and which teaching routinely implies.

What was actually being claimed #

Worth being fair to Pauling, because the original idea was answering a real question well.

The question in 1931 was: why is carbon tetravalent and tetrahedral, when its ground-state configuration is $2s^2 2p^2$ with only two unpaired electrons? Hybridisation answers it — the four bonds are equivalent because the four orbitals used are equivalent combinations of s and p.

That answer is correct about the geometry and about the equivalence of the bonds, both of which are observable and both of which are true. What it is not is a statement about the energies of the states of the ion, which is what a photoelectron spectrum measures.

Where the teaching goes wrong #

Three specific claims that are made and should not be.

“The electrons are in sp³ orbitals.” They are in a four-dimensional space that can be described by hybrids or by canonical orbitals or by infinitely many other bases. Saying they are in one of them is like saying a vector is really its x-component.

“Hybridisation explains the tetrahedral shape.” It describes it. The angle comes out of the coefficients once four equivalent combinations are chosen, and choosing them was motivated by the observed geometry. The shape is fixed by the total energy, and the hybrid description is fitted to it.

“The atom promotes an electron and then hybridises.” Nothing happens in sequence. There is no energy cost to a change of basis and no process to describe.

The general shape of the error #

This is worth extracting, because the same mistake occurs several times in this subject and each time it looks different.

A description is chosen because it makes something easy. It works. It is taught. Over time the description is promoted into a mechanism — a claim about what nature does — and then it makes predictions it was never entitled to make.

The d-orbital account of hypervalency is the same error: a description that reproduced the geometry, promoted into a claim about which orbitals are occupied, and wrong about the claim.

The lone-pair repulsion account of bond angles is a milder version: a rule that predicts a direction, presented as a mechanism it cannot support.

The remedy is the same in each case, and it is not to abandon the description. It is to be clear about what kind of statement it is.

The same instrument, four more molecules #

Methane is the clearest example and it is not the only one. The reduction above is a general instrument — it takes a set of atoms, a basis on them, and returns the species those functions span — so the honest way to show that the argument is not a trick played on one molecule is to run it again and read what comes out.

Water. Two O–H bonds, equivalent by symmetry, related by the twofold rotation.

Water’s photoelectron spectrum has the bonding pair at about 14.8 and 18.6 electronvolts, nearly four apart, for two bonds that are identical by every structural measure. The same argument one shell up is the familiar one about the lone pairs: the localised picture gives two equivalent rabbit ears in sp³ hybrids, and the canonical orbitals are a pure p perpendicular to the molecular plane at 12.6 electronvolts and an in-plane combination mixed into the a₁ manifold. Two inequivalent lone pairs, not two equivalent ones, and the symmetry says so before any spectrum is taken.

Ammonia. Three bonds, threefold axis, and a result that is neither one band nor three.

The pattern is now visible and it is worth stating before the last two cases rather than after. The number of bands is not the number of bonds, and it is not one. It is the number of symmetry species the bonds’ functions span, which is fixed by the point group and by where the atoms sit, and which nothing about bond equivalence constrains. Four equivalent bonds gave two; two gave two; three gave two.

Ethene. Four equivalent C–H bonds, and the count changes.

Methane and ethene both have four C–H bonds, all equivalent within each molecule, and their hydrogen manifolds could not be more differently arranged: one band of three plus a singleton in the first, four singletons in the second. Nothing about the bonds distinguishes the cases. Everything about the group does — Td has a three-dimensional representation and D₂ₕ has none, so methane is permitted a threefold degeneracy and ethene is forbidden one.

Benzene. Six equivalent C–H bonds, and the count changes again.

Six equivalent bonds, four bands. The series 4→2, 2→2, 3→2, 4→4, 6→4 has no relation to the bond count at all, which is the sharpest form of the essay’s point. A description built on the equivalence of the bonds predicts one band in every one of these molecules and is wrong in every one of them, not because equivalence is a false claim — it is exactly true — but because equivalence is not the quantity a spectrum reports.

And ethene supplies the other half of the general lesson. The localised picture there gives a sigma bond and a pi bond; an equally valid transformation gives two equivalent bent bonds curving above and below the internuclear axis, which Pauling preferred for years. Both describe the same wavefunction and no measurement distinguishes them, because the two sets are related by exactly the kind of rotation that leaves every total alone. It comes up again with resonance structures and with three-centre bonding, and recognising the pattern is more useful than remembering the cases.

What the spectrum does establish #

Positively rather than negatively, since it is easy to lose that.

Methane’s spectrum establishes that the molecule has one totally symmetric valence orbital and a threefold degenerate set, which is a real fact about the electronic structure and follows from the Td symmetry. It also establishes that the canonical orbitals, not the localised ones, are what an ionisation experiment probes.

Both are useful. Neither says anything about whether the bonds are equivalent — they are, by symmetry, and the spectrum is consistent with that.

What it costs #

The symmetry argument costs almost nothing, and it is worth pricing because the alternative — simply stating the result — costs nothing at all and is what most textbook accounts do.

Generating the group is the expensive step, and it is expensive only in the sense that it takes milliseconds. Candidate axes are drawn from the structure, each is tried as a rotation, a mirror and an improper rotation, and what survives is multiplied out to closure. Sulfur hexafluoride is the largest case here and closes at forty-eight operations.

One repair was needed along the way and it is worth recording, because the failure was silent. An operation found by searching is only as accurate as the search, and the search accepts an axis that maps the molecule onto itself to within six hundredths of an ångström — the right tolerance for deciding whether an operation exists and the wrong one for multiplying two together. Ammonia’s threefold axis came out tilted by about $5\times10^{-5}$ radians, the products drifted, no two of them matched, and the group grew without ever closing.

The fix is to stop treating the approximate matrix as the answer and treat it as a question. It still identifies unambiguously which atom goes where, and that permutation is exact — a fact about labels rather than about numbers. The operation is then recomputed as the orthogonal matrix that best realises the permutation, which is accurate to machine precision, and products of such matrices stay accurate to machine precision.

There was a second and better-hidden version of the same problem. A planar molecule’s own plane is a mirror that moves no atom at all, so its permutation is the identity’s — and keying operations on the permutation alone collapsed benzene’s twenty-four operations to twelve, silently, with every one of the twelve correct. That the benzene figure above reads a₁g ⊕ e₂g ⊕ b₁u ⊕ e₁u rather than some six-function span over half a group is the repair, still holding.

Where the model stops #

Three limits, and the second is the one that decides how much the argument is entitled to claim.

Koopmans’ theorem is an approximation. It equates an ionisation energy with an orbital energy and ignores the relaxation of the remaining electrons. It is good enough for the argument here and it is not exact.

The split is computed; the energies are not. Symmetry gives the number of bands and their intensity ratio, from the coordinates alone. It gives no energies whatever, and the 12.7 and 23 electronvolts on the figure above are experimental and marked as such, as are the water and ammonia values quoted in the text. That division is not a shortcoming of this treatment — it is what a symmetry argument is, and the fact that it forbids a single band without knowing a single energy is the reason it is worth making.

A species count is not an ordering. The reduction says how many bands and with what degeneracies; it does not say which comes first. That methane’s a₁ lies below its t₂, rather than above, is a statement about carbon 2s lying below carbon 2p, and it comes from the atom rather than from the group.

Two bands is the weak form of the evidence #

Two bands rather than one refutes the naive reading, and the spectrum says more than that. It gives the two bands in a ratio, and the ratio is an integer the group predicts.

The canonical description puts six of methane’s eight valence electrons in the threefold-degenerate $t_2$ set and two in the totally symmetric $a_1$. So the deeper band should hold a third as many electrons as the shallower one, and the two bands’ intensities should stand in the ratio 3 : 1 — not approximately, since the degeneracy is exact and the electron count is an integer.

They do, and the two ionisation energies are far apart: near 13 to 14 electronvolts for the $t_2$ band and near 23 for the $a_1$, which is a separation of nine electronvolts in a molecule whose four bonds are chemically identical by every other measure.

That turns the argument from a refutation into a positive result. A single band of eight electrons is what four equivalent orbitals would give; two bands in a 3 : 1 ratio, nine electronvolts apart, is what a symmetry classification of the same eight electrons gives, with no free parameters and nothing fitted.

And it sharpens what the localised description is not being blamed for. The localised orbitals reproduce the total — eight electrons, one density, the same total energy — and they do not reproduce the split, because a split is a per-orbital quantity and the localisation transformation is exactly the operation that discards per-orbital quantities. The spectrum is asking a question the description was never able to answer, and answering it in integers.

Where to read on #

The construction being discussed is hybrids are a basis, where the transformation this essay leans on is performed rather than described.

The wider framework is molecular orbital and valence bond theory.

And the standing caution about what an orbital is at all is orbitals are not where the electron is.

What the pictures here cannot show. A photoelectron spectrum is the evidence in this essay and only one figure displays one, because the calculations here give geometry and one-electron functions rather than many-electron ionisation energies. The six reductions show what symmetry requires of such a spectrum before it is measured, which is the part that can be derived.