Hybrids are a basis

An orbital is a one-electron wavefunction, and combinations of them are still wavefunctions. Four sp³ hybrid orbitals point at the corners of a tetrahedron and subtend 109.471 degrees. That number is not put in anywhere. It comes out of a matrix of ones and halves, and the matrix is the whole of what a hybrid set is.

The construction #

Take one s orbital and three p orbitals on the same atom. Form four combinations:

$$h_1 = \tfrac12(s + p_x + p_y + p_z), \quad h_2 = \tfrac12(s + p_x - p_y - p_z),$$

$$h_3 = \tfrac12(s - p_x + p_y - p_z), \quad h_4 = \tfrac12(s - p_x - p_y + p_z).$$

Each has the same weight of s and the same total weight of p. The signs pick out four directions, and those directions are the alternate corners of a cube — which is to say a tetrahedron.

Why the angle is what it is #

The direction of a hybrid is the direction of its p component. For $h_1$ that is $(1,1,1)$ and for $h_2$ it is $(1,-1,-1)$.

The cosine of the angle between them is their dot product over the product of their lengths: $(1 - 1 - 1)/3 = -1/3$. So the angle is $\arccos(-1/3)$, which is 109.4712 degrees.

Nothing was chosen to make that happen. It follows from having four equally weighted combinations of one s and three p orbitals, and it is the same number that minimising repulsion on a sphere produces by a completely different route — which is a coincidence worth noticing and not a deep connection.

The point: it is a rotation #

Here is the claim this essay exists to make.

The four hybrids span exactly the same space as the four atomic orbitals they were built from. The transformation matrix is orthogonal: its rows are orthonormal, so $C^\mathsf{T}C = I$, and it has determinant $\pm1$. It is a rotation of the basis, and nothing else.

Rotating a basis does not change the object described. The total electron density is identical, the total energy is identical, and every measurable quantity is identical. A calculation done in the hybrid basis and the same calculation done in the atomic basis give the same answers to every question that can be asked of the system.

So “is carbon really sp³ hybridised in methane” is not a question about methane. It is a question about which coordinate system to write the answer in, and nature does not have an opinion about coordinate systems.

What the other sets are #

The same construction with different amounts of p gives the other familiar sets, and the angles follow the same way.

More s character means a larger angle, which is the observation Bent’s rule generalises: sp is 180, sp² is 120, sp³ is 109.5, and pure p would be 90.

That series is genuinely useful, and it is useful as a correlation between composition and direction rather than as a causal claim. An atom does not decide to hybridise and thereby acquire an angle; the angle is what it is, and the hybrid description with the matching composition is the convenient one.

What hybridisation is good for #

Having insisted that it is only a basis, it is worth saying why anybody bothers — because the answer is a good one.

A basis can be well or badly suited to a problem, and the hybrid basis is extremely well suited to describing localised bonds. In the atomic basis, a carbon’s four bonds are each a mixture of contributions from s, $p_x$, $p_y$ and $p_z$, and no single orbital corresponds to any single bond. In the hybrid basis, each bond involves one hybrid, and the description matches the way chemists think and draw.

That is not nothing. A basis in which the answer is simple is worth having, and the entire vocabulary of organic chemistry — sigma frameworks, pi systems, lone pairs in particular orbitals — is the hybrid basis being useful.

The mistake is only to promote the convenience into a physical claim.

The check the figures make #

Every hybrid picture here rests on a transformation that has been checked to be orthogonal.

Each row is checked to be normalised, and every pair of rows checked to be orthogonal, both to better than $10^{-12}$. The inter-hybrid angles are then computed from the coefficients and compared with the value the caption claims, to a hundredth of a degree.

A hybrid set with the right angles and non-orthonormal rows would not be a rotation, and a figure claiming it was would be claiming something false about the mathematics. That has not happened, and the check is what makes the absence of it a fact rather than an assumption.

The transformation runs both ways #

One consequence that makes the “which is real” question look silly.

If hybrids are an orthogonal transformation of the atomic orbitals, then the atomic orbitals are an orthogonal transformation of the hybrids — the inverse of an orthogonal matrix is its transpose. Neither set is prior.

The same applies further along. The canonical molecular orbitals of a molecule — the ones that come out of a calculation, delocalised over the whole framework — are related to a set of localised bond orbitals by another such transformation. Both describe the same wavefunction, and choosing between them is choosing what to look at rather than what is there.

That is the sharpest version of the point, and the photoelectron spectrum of methane is where it becomes experimentally visible.

The composition is continuous #

One more property of the construction that the standard three cases hide.

sp, sp² and sp³ are the equally-weighted cases, but nothing requires the weights to be equal. A general hybrid is $\alpha s + \beta p$ with $\alpha^2 + \beta^2 = 1$, and $\alpha$ can take any value. The label “sp²·³” is perfectly meaningful within a scheme.

That matters because real molecules are almost never at the idealised compositions. Water’s bond angle is 104.5 rather than 109.5, and the hybrid description that matches it has a little more p character in the bonds and a little more s in the lone pairs — which is exactly what Bent’s rule predicts and is a continuous adjustment rather than a switch between named cases.

The continuity is also why “what is the hybridisation of this atom” is not quite a well-posed question. It has an answer within a partitioning scheme, the answer is a real number rather than a category, and different schemes give different numbers.

The rotation, actually performed #

Everything above argues that a change of basis leaves the object alone. That is a theorem and it is one line, and a theorem is not a demonstration. The demonstration was added later, because this site’s whole discipline is that a claim gets a test it could fail.

Methane is the case to do it on. Four bond orbitals are built — an sp³ hybrid pointing at one hydrogen, plus that hydrogen’s 1s — and made orthonormal by Löwdin’s transformation, which is the one orthogonalisation that treats four indistinguishable objects alike rather than privileging whichever is handled first. The tetrahedral matrix is applied. What comes out is looked at rather than arranged.

The bonding density along a carbon–hydrogen line, computed from each of the two descriptions, is one curve rather than two: the largest disagreement across every point sampled on and off the bond axis is around 10⁻¹⁶, which is the arithmetic rather than the physics.

And what does differ is the description rather than the object. The participation ratio counts how many hydrogens each orbital effectively occupies: about 1.07 for a bond orbital and exactly 4 for a canonical one. A factor of nearly four in how the electrons are described, and nothing at all in what they produce.

The last of those three is the essay in one picture. A reader who wanted evidence that “the electrons are in sp³ orbitals” is not a claim about nature has it here: two descriptions that disagree by a factor of four about where the orbitals are, agreeing to fifteen decimal places about where the electrons are.

What it costs, and the check that it can fail #

The demonstration costs three overlap integrals and a four-by-four matrix multiplication, which is to say nothing. What it costs in honesty is worth more attention.

An invariance test that always passes proves nothing, and this one would always pass if the arithmetic were merely reporting that $CC^{\mathsf T}$ equals itself. So the same calculation is handed a mixing that is not orthogonal — one bond orbital with a third of another added to it — and required to report a changed density. It does, by about a fifth. If it did not, the invariance above would be a property of the arithmetic rather than of the transformation, and the figures would be worthless.

Two further things are checked rather than assumed. The four raw bond orbitals must all overlap each other equally, because if they did not they would not be four equivalent bonds and the construction would be describing something else. And the tetrahedral matrix’s orthogonality is checked row by row rather than taken from the algebra, which is the same discipline the figures above apply to the sp³ coefficients themselves.

The honest cost is in what the demonstration is not. There is no Hamiltonian anywhere in it. The basis functions are hydrogen-like with an effective nuclear charge for carbon taken from Slater’s rules, and how much hydrogen enters each bond orbital is a free parameter that nothing here determines — it cannot be determined without a real calculation. Every result is invariant to that parameter, and the invariance is asserted at four separate values rather than hoped for, but the consequence is that none of this is a calculation of methane. It is a statement about bases, which is exactly as strong as it sounds and no stronger.

One detail did surface that a purely algebraic treatment would have missed. A hydrogenic 2s has a radial node, and with carbon’s effective charge that node sits at about 0.6 bohr while the hydrogen is at 2.05 — so across the whole bonding region carbon’s 2s is negative, and a hybrid written with the conventional positive s coefficient overlaps the hydrogen with the wrong sign. The two contributions subtract, and what was meant to be a bond orbital is built from an antibonding interaction. The repair is to take whichever sign makes them add, decided by looking at the two integrals rather than by adopting a convention. Nothing in the density argument depends on it. Every picture does.

The step in the usual account that is not a rotation #

A rotation changes no observable is exact, and it applies to a rotation among orbitals that are equally occupied. The construction as it is usually drawn contains a step that is not one, and that step is the only part of the hybrid picture with an energy attached.

A neutral carbon atom is $2s^2 2p^2$. Its 2s orbital holds two electrons and its three 2p orbitals hold two between them. Those are not equally occupied, so mixing them is not a rotation within an occupied set: it is a rearrangement of which orbitals the electrons are in. Four singly-occupied sp³ hybrids require $2s^1 2p^3$, which is a different configuration of the atom.

That step costs energy and the cost is measured. The state of carbon with one 2s electron and three 2p electrons, all spins parallel, lies 4.18 electronvolts — about 400 kilojoules a mole — above the ground state, and the separation is read off the atomic spectrum rather than estimated.

So the textbook sequence has two operations in it doing quite different work.

Promotion, which is a change of configuration, is real, is not free, and is not a change of basis at all.

Hybridisation, which mixes the promoted atom’s four singly-occupied orbitals into four equivalent ones, is a rotation among equally occupied orbitals and is free — the observable content is identical before and after.

Only the second is what this essay is about, and conflating the two is what makes the whole construction look as though it must be describing something physical: a step with a measured energy has been placed immediately before a step with none, and the energy belongs to the first.

The accounting that follows is the classic argument and it is worth doing, because it is genuinely quantitative and it is the one place the picture earns its keep. Spending 400 kilojoules a mole buys two extra bonds. A carbon–hydrogen bond is worth about 415, so the return is roughly 830 against an outlay of 400 — a clear profit, and the reason carbon is tetravalent rather than divalent.

None of which rescues the reification. The profit is a statement about configurations and bond energies; the four equivalent hybrids that come out of the rotation are one choice of axes among infinitely many, and every one of them describes the same promoted atom equally well.

The useful separation to carry away is therefore a sharper version of the essay’s own. Ask whether a step changes which orbitals are occupied. If it does, it has an energy and can be argued about. If it only changes which combinations the occupied orbitals are written in, it is a rotation, and the question of whether it is real has no content.

Where the model stops #

Three limits, and the second is often forgotten.

Hybridisation is not a physical process. Nothing happens to an atom “before it bonds”. The language of promotion and hybridisation as steps is a pedagogical narrative, not a sequence of events, and there is no energy cost to a change of basis.

The composition is not observable. “Carbon is sp²·³ hybridised here” is a statement about a fitted description. It can be defined precisely within a scheme and it is scheme-dependent, and different partitioning methods give different numbers for the same molecule.

A rotation of the occupied orbitals is not the only freedom there is. The demonstration above uses an orthogonal transformation because that is what relates two orthonormal descriptions. The density is in fact invariant under any non-singular mixing of the occupied orbitals, orthogonal or not, once the metric is handled properly — which is a stronger statement and a less visual one. The orthogonal case is what the figures show because it is the case that keeps both descriptions orthonormal and therefore keeps both interpretable.

Where it came from #

Pauling introduced hybridisation in 1931, and it was an immediate and enormous success, because it answered a question that had no other answer: why is carbon tetravalent and tetrahedral when its ground-state configuration has only two unpaired p electrons?

The hybrid picture gave chemists a way to draw and reason about directional bonds, and The Nature of the Chemical Bond made it universal. It remains one of the most productive ideas in the subject.

The reification came later and from teaching rather than from Pauling. A description repeated often enough acquires the status of a thing, particularly when it is introduced early and never revisited — and the correction is not that hybridisation is wrong but that it is a description, which is a subtler and less satisfying thing to be told.

There is a detail of the chronology that makes the point sharper. Pauling’s construction predates by a decade the experiments that could have tested any claim about orbital energies, and by two decades the photoelectron spectra that eventually did. So for its first twenty years the hybrid picture had no rival account of what a measurement would find, because there were no measurements of that kind to have an account of. The claim it is now criticised for making is one nobody at the time had any occasion to make.

Where to read on #

The experimental case that separates description from claim is hybridisation does not explain, where methane’s photoelectron spectrum shows two ionisation energies rather than one.

The framework this sits inside is molecular orbital and valence bond theory, which are two more bases for the same thing.

And the rule that makes hybrid composition predictive is in why water is bent.

The single sentence version, for a reader who wants one: hybridisation is a rotation, rotations change no observable, and every question of the form “is this atom really sp³” is a question about coordinates rather than about the atom.

What the pictures here cannot show. A hybrid orbital drawn as a direction is a drastic simplification — the actual function has a large lobe, a small opposite lobe, and a shape that depends on the composition. These figures draw where each points, which is what the argument is about, and not what each looks like.