Molecular Geometry & Bonding
what the shape decides
What is taught wrongly

Hybridisation does not explain

Methane's photoelectron spectrum has two bands, not one. Four equivalent sp³ bonding orbitals cannot produce that, and the resolution is that hybridisation was never a claim about what a measurement would find.
9 min read 6 figures Taught confidently, and wrongA basis is not a thing

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.

methane — TdThe molecule with the point group found from its coordinates rather than looked up: every candidate operation was applied and kept when it permuted the atoms among themselves.HHCHHTdprincipal axis C36 mirror planesno inversion centrecannot be polarcannot be chiralgroup recovered from the coordinates5 atoms
Fig. 1 Methane, point group Td, recovered from its coordinates. Four equivalent bonds by symmetry — and a spectrum with two bands, which four equivalent orbitals cannot produce.

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 electrons pairs in four identical orbitals should give one band.

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

What the canonical orbitals are

The resolution is that the orbitals a spectrum probes are not the hybrid ones.

A molecular orbital calculation on methane gives four occupied valence orbitals, and they are not four equivalent bonds. They are:

  • one orbital of a₁ symmetry, totally symmetric, built from carbon 2s and the in-phase combination of all four hydrogens; and
  • three degenerate orbitals of t₂ symmetry, built from carbon 2p and the three remaining hydrogen combinations.

One and three, in a 3:1 ratio, at different energies. That is the spectrum.

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.

What differs is what each 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.

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.

The sp3 transformationEach row is one hybrid, written in the basis of the atomic orbitals it is made from. The rows are orthonormal, so the matrix is a rotation — and a rotation of a basis changes no observable quantity whatever.spₓp_yp_zhybrid 10.50000.50000.50000.5000hybrid 20.50000.5000-0.5000-0.5000hybrid 30.5000-0.50000.5000-0.5000hybrid 40.5000-0.5000-0.50000.5000every row normalised and every pair orthogonal to 0e+0a change of basis, and nothing moresp3
Fig. 2 The transformation, written out. Four rows, orthonormal, forming a rotation of the atomic basis. The same kind of matrix relates localised bonds to canonical orbitals, and a rotation changes no total quantity.

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 2s22p22s^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.

Two more cases of the same thing

Methane is the clearest example and it is not the only one, and seeing the pattern repeat makes it easier to recognise.

Water’s lone pairs. The localised picture gives two equivalent lone pairs in sp³ hybrids, pointing away from the hydrogens like rabbit ears. Water’s photoelectron spectrum shows the two highest occupied orbitals at clearly different energies — one is a pure p orbital perpendicular to the molecular plane and the other is an sp-hybridised orbital in it. Two inequivalent lone pairs, not two equivalent ones.

water — C2vThe molecule with the point group found from its coordinates rather than looked up: every candidate operation was applied and kept when it permuted the atoms among themselves.HHOC2vprincipal axis C22 mirror planesno inversion centremay be polarcannot be chiralgroup recovered from the coordinates3 atoms
Fig. 3 Water, C₂ᵥ. The symmetry alone says the two highest occupied orbitals must belong to different symmetry species — one symmetric under the twofold rotation and one antisymmetric — so two equivalent lone pairs are not what the canonical orbitals give.

Ethene’s double bond. The localised picture gives a sigma bond and a pi bond. An equally valid transformation gives two equivalent bent bonds, curving above and below the internuclear axis. Both describe the same wavefunction, and Pauling preferred the bent-bond version for years.

ethene — D2hThe molecule with the point group found from its coordinates rather than looked up: every candidate operation was applied and kept when it permuted the atoms among themselves.HHCCHHD2hprincipal axis C23 mirror planeshas an inversion centrecannot be polarcannot be chiralgroup recovered from the coordinates6 atoms
Fig. 4 Ethene, D₂ₕ. Sigma-plus-pi and two bent bonds are the same wavefunction in two bases, and no measurement distinguishes them — which is the general situation this essay is about, met in the case a first course meets earliest.

In every case the pattern is the same: a localised description chosen for its convenience, a canonical description that a spectrum couples to, and a transformation between them that changes nothing measurable. Recognising the pattern is more useful than remembering the individual cases, because it comes up again with resonance structures and with three-centre bonding.

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.

Where the model stops

Two limits.

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.

Nothing on this site computes the spectrum. The ionisation energies quoted come from the experimental literature. This site’s machinery is one-electron and hydrogenic, and the figures show geometry and the transformation matrix, which is what its own computations reach.

The two descriptions, side by side

The transformation at the heart of this essay is worth seeing in both directions.

sp3 hybridsThe directions the hybrids point, with the angle between them computed from the coefficients rather than quoted. The set is an orthogonal transformation of the atomic orbitals, so it describes the same space in different coordinates.sp3109.471°between every pair4 hybridsworst off-diagonal 0e+0an orthogonal transformation of the atomic orbitalsone electron
Fig. 5 The localised picture: four equivalent hybrids at the tetrahedral angle, each overlapping with one hydrogen. Chemically intuitive, matches the structural formula, and not what a photoelectron spectrum couples to.
The sp3 transformationEach row is one hybrid, written in the basis of the atomic orbitals it is made from. The rows are orthonormal, so the matrix is a rotation — and a rotation of a basis changes no observable quantity whatever.spₓp_yp_zhybrid 10.50000.50000.50000.5000hybrid 20.50000.5000-0.5000-0.5000hybrid 30.5000-0.50000.5000-0.5000hybrid 40.5000-0.5000-0.50000.5000every row normalised and every pair orthogonal to 0e+0a change of basis, and nothing moresp3
Fig. 6 The matrix that produces them. An orthogonal transformation of the atomic basis — and the canonical orbitals are reached from the localised ones by another matrix of the same kind.

Neither set is prior. The atomic orbitals give the hybrids by one rotation; the hybrids give the atomic orbitals by its transpose; and the canonical molecular orbitals are related to the localised bonds the same way.

What the spectrum shows is which of the descriptions the measurement singles out, and the answer is the canonical one because ionisation removes an electron from an eigenstate of the one-electron Hamiltonian. That is a fact about the experiment rather than about the molecule.

Where the ladder goes next

The construction being discussed is hybrids are a basis.

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 no figure here displays one, because the site computes geometry and one-electron functions rather than many-electron ionisation energies. The figures show the molecule the argument is about and the transformation that resolves it.

Where this goes next

sp3109.471°between every pair4 hybridsworst off-diagonal 0e+0an orthogonal transformation of the atomic orbitalsone electron Hybrids are a basis An sp³ hybrid set is an orthogonal matrix applied to the atomic orbitals. Rotating a basis changes no observable, so asking whether the electrons are really in hybrids is asking which coordinate system nature prefers. 1s-1sσ overlapS = 0.39002pz-2pzσ overlapS = 0.45852px-2pxπ overlapS = 0.75291s-2pxsymmetry forbids itS = 0 exactlyno splittingoverlaps computed at 2.8 bohr, levels in proportionone electron Molecular orbital and valence bond Two frameworks, taught as rivals, describing the same molecules. One starts from delocalised orbitals and localises; the other starts from localised bonds and delocalises. Pushed far enough they meet. one scale across the plate1s0 radial · 0 angular2s1 radial · 0 angular2pz0 radial · 1 angulareach level solved for separately90% · one electron Orbitals are not where the electron is A many-electron atom has no exact orbitals at all. The orbital picture is a basis for an approximation — an extremely good one — and treating it as a description of reality is the source of most of the confusion in this subject.