Hybrids that were never orthogonal

A hybrid orbital is a fixed combination of an s function and a p function on one atom, pointing in a chosen direction. Two of them, built the same way and pointing in different directions, overlap by

$$\langle h_1 | h_2\rangle = a + (1-a)\cos\theta$$

where $a$ is the s fraction each of them holds and $\theta$ is the angle between the directions. The s parts overlap completely, because they are the same function; the p parts overlap by the cosine between them, because that is what two p orbitals on one centre do.

Setting that to zero gives one angle for each s fraction. A quarter gives $109.47°$, a third gives $120°$, a half gives $180°$. Those are the three familiar geometries, and the point of writing it this way is that they are the only angles at which the corresponding hybrids are orthogonal.

Most molecules are not at those angles.

What the number means, and why it is not a rounding error #

A basis is a set of functions that describes everything once. Two functions that overlap describe part of the same thing twice, and every computation that treats them as independent is then double-counting by an amount proportional to the overlap.

Water is the ordinary case. Its bonds are at $104.5°$, and two quarter-s hybrids drawn at that angle overlap by 0.062. Six per cent is small enough that a qualitative argument built on it will not usually go wrong, and it is not zero. Ammonia’s is 0.021 at $107.8°$, smaller still.

Hydrogen sulfide is where the ordinary case stops. Its bonds are at $92.1°$, and two quarter-s hybrids there overlap by 0.222. That is a fifth, and the correct response is the one the angle does not fix the hybridisation already reached from the other direction: hydrogen sulfide’s bonds are not built from quarter-s hybrids. Inverting the relation at $92.1°$ gives an s fraction of 0.036, which is sp²⁷ — nearly pure p — and hybrids built at that fraction are orthogonal at exactly the measured angle.

So there are two consistent descriptions and one inconsistent one. Either the label is chosen and the angle follows from it, or the angle is measured and the label follows from it. What cannot be done is to take the angle from experiment and the label from a rule of thumb and expect the result to be a basis.

Cyclopropane, where the label fails outright #

Three carbons at the corners of an equilateral triangle have C–C–C angles of exactly $60°$, and the angle a ring cannot have shows that no geometry available to a three-membered ring changes that by a degree.

Two quarter-s hybrids at $60°$ overlap by 0.625.

Five eighths is not a correction. Two such orbitals share most of their content; a set of four on one carbon would have a metric matrix so far from the identity that no quantity computed from it as though it were a basis would mean anything. And the s fraction that would be orthogonal at $60°$ does not exist — the relation $\cos\theta = -a/(1-a)$ needs $a$ negative for $\theta$ below $90°$, and the function refuses rather than returning one.

There is no pair of equivalent s–p hybrids meeting at less than a right angle. That is a fact about two normalised vectors in a four-dimensional space and it is why the ordinary hybridisation vocabulary simply runs out at cyclopropane.

The repair is old and is a good one. Bent bonds: let the orbitals point away from the internuclear lines, outside the triangle, so that the angle between the orbitals is near $104°$ even though the angle between the bonds is $60°$. At $104°$ two quarter-s hybrids overlap by 0.061 and the set is nearly a basis again. What has been given up is the identification of a bond direction with an orbital direction, which is the assumption that made the label seem to fix the geometry in the first place.

The other end: hybrids that overlap negatively #

The curve crosses zero and keeps going, and the far side is worth a paragraph because it is usually not drawn.

Cyclohexane’s C–C–C angle is $111.4°$, slightly wider than tetrahedral. Two quarter-s hybrids there overlap by −0.024. A negative overlap is as much a failure of orthogonality as a positive one and it has the opposite sign, which matters wherever the overlap enters an expression linearly — a bond order, a population, a cross term in an energy.

The sign has a reading. Two hybrids squeezed below their orthogonality angle have their p lobes pointing partly the same way and their positive overlap comes from the p parts adding to the s parts. Opened past it, the p lobes point apart far enough for their negative overlap to exceed the s parts’ positive contribution.

At $180°$ two quarter-s hybrids overlap by $0.25 - 0.75 = -0.5$, which is why an sp³ description of a linear molecule is not merely unusual but arithmetically impossible: the two orbitals along the axis would be half-parallel.

Where the relation comes from, in four lines #

The relation is Coulson’s and it is worth deriving here, because it is short and because seeing it makes clear that nothing in it is a model.

A hybrid pointing along a unit vector $\hat{u}$ with s fraction $a$ is

$$h = \sqrt{a}\,s + \sqrt{1-a}\,(\hat{u} \cdot \mathbf{p})$$

where $\mathbf{p}$ is the vector of the three p functions. The s function is normalised and orthogonal to every p function; the p functions are orthonormal among themselves. So the overlap of two such hybrids is

$$\langle h_1|h_2\rangle = a\,\langle s|s\rangle + (1-a)\,(\hat{u}_1\cdot\hat{u}_2) = a + (1-a)\cos\theta.$$

There is no approximation anywhere in that. It is the inner product of two vectors in the four-dimensional space spanned by one s and three p functions, and it would hold for any four orthonormal functions with the same angular labels — the same relation governs a set built from Slater functions, from Gaussians, or from numerical atomic orbitals, because none of the radial parts appear.

That is what makes the failure a hard one rather than a modelling choice. A more careful radial function does not move any of the numbers above by anything, because the numbers do not contain a radial function.

The whole overlap matrix, not one pair #

Two hybrids is the simplest case and a molecule has four. The set of four is what has to be a basis, and its metric is a four-by-four matrix with ones on the diagonal and the pairwise overlaps off it.

For water, described as four equivalent quarter-s hybrids with two of them pointing along the bonds at $104.5°$, only the bond-bond entry can be computed from the measurement — the two lone-pair directions are not measured, and different accounts place them differently. Take the usual arrangement, with the four directions tetrahedral except for the two bonds pulled together to the measured angle, and the matrix has one entry at 0.062, four at about $-0.02$, and one at about $-0.04$.

Its eigenvalues then run from about 0.95 to about 1.05. That is the practical meaning of the failure: the set spans a space in which one direction is described with five per cent more weight than another, and every population computed in it is off by that much before any physics enters.

For cyclopropane the same construction gives a matrix with three entries at 0.625, and its smallest eigenvalue is close to zero — meaning the four orbitals very nearly fail to span four dimensions at all. A set whose metric is nearly singular is a set in which almost any computed quantity is unstable, and this is the arithmetic behind the observation that population analyses on strained rings are unreliable.

What “equivalent” is quietly doing #

Every number above assumes the two hybrids being compared hold the same s fraction, and that assumption is doing more work than it looks.

It is exactly right for methane, where symmetry requires the four orbitals to be related by the operations of the group and therefore identical in content. It is exactly right for boron trifluoride and for carbon dioxide, for the same reason.

It is wrong for water, ammonia, and every other molecule with a lone pair, and it is easy to compute how wrong: ammonia’s three bond hybrids hold 0.236 each and its lone pair holds 0.293. Two orbitals with different s fractions $a_1$ and $a_2$ overlap by

$$\sqrt{a_1 a_2} + \sqrt{(1-a_1)(1-a_2)}\cos\theta$$

which reduces to the earlier expression when they are equal, and has a different zero when they are not.

Putting ammonia’s own numbers into it produces the best result in this essay, and it was not what the essay set out to show. The budget gives the three bond hybrids $0.2341$ each and the lone pair $0.2976$. Substituting those, the angle at which a bond hybrid and the lone-pair hybrid are orthogonal comes out at

$$111.094776°.$$

Independently, the geometry fixes the same angle without any orbitals in it: three bonds at $107.8°$ to each other make an angle $\beta$ with the threefold axis where $\cos\theta = (3\cos^2\beta - 1)/2$, and the lone pair lies on the axis, so the bond-to-lone-pair angle is $180° - \beta$. That gives

$$111.094776°.$$

The two agree to every digit, at every bond angle tried from $92°$ to $115°$. One route is a statement that the s fractions of a hybrid set sum to one; the other is spherical trigonometry. So the budget description of a pyramidal molecule is not approximately orthogonal at the measured geometry — it is exactly orthogonal, in all six pairs.

That is the resolution of the whole essay. The equivalent-hybrid description, with four quarter-s orbitals, gives a bond-to-lone-pair orthogonality angle of $109.47°$ against a geometric $111.09°$, and is wrong by more than a degree and a half. The budget description is exact. The criticism here is aimed at the first and does not touch the second, and the two usually travel under one word — which is why “sp³” can be a precise statement in one sentence and a broken one in the next.

What Löwdin orthogonalisation costs #

There is a standard repair for a non-orthogonal set, and it is a familiar one: the localisation transformation builds methane’s four bond orbitals by Löwdin orthogonalisation, multiplying the set by the inverse square root of its overlap matrix.

It works, and it is worth being explicit about what it does to the picture. The transformation produces an orthogonal set that is as close as possible to the original in a least-squares sense — that is its defining property — and “as close as possible” is not “the same”. The orthogonalised hybrids point in different directions from the ones that went in, by an amount that grows with the overlap.

For water’s set, overlapping by 0.062, the directions move by a fraction of a degree and nothing in a picture changes. For a set overlapping by 0.625 the directions move a great deal, and the orthogonal set that comes out is not recognisably the set that went in.

So the repair is available and it does not rescue the label. What it rescues is the calculation: a set of orthogonal orbitals spanning the same space, in which populations and bond orders mean what they usually mean. The directions, which is the part a reader draws, are what was traded away.

Why the label survives anyway #

Given all of the above it is fair to ask why hybridisation is still taught and still used, and the answer is that it is doing a job the criticism does not touch.

The s-character budget is exact. There is one s orbital on an atom, so the s fractions of a complete orthonormal set of hybrids sum to exactly one. Four bonds at the tetrahedral angle each take a quarter and the budget closes; three at $120°$ take a third each and it closes; two at $180°$ take a half each. That relation is arithmetic, has nothing fitted in it, and is the source of Bent’s rule — see Bent’s rule, computed, where the budget is what forces s character into the bonds to electropositive substituents.

The count is right even when the angles are not. Four orbitals, whatever their s content, is four orbitals, and the number of hybrids is the number of directions a valence has to point in. Every argument that turns on the count rather than on the mixing survives the criticism above intact.

What does not survive is the third use, which is the one this essay is about: reading a label as a statement about the geometry, so that “sp³” means “109.47°” and a molecule at another angle is an sp³ molecule with a deviation. There is no such thing as a deviation from an orthogonality condition. Either the set is orthogonal or it is a set that describes some things twice.

Which of the three the reader is being handed #

A practical test, since all three uses appear in the same sentence constantly.

If a text says a carbon is sp³ and infers four bonds, it is using the count and is safe. If it infers a quarter s character in each bond, it is using the budget and is safe as long as the four are equivalent. If it infers 109.47°, it is using the orthogonality relation backwards, and the inference is only as good as the assumption that the four hybrids really are equivalent and really are orthogonal — which is exactly what a measured angle different from 109.47° denies.

Ammonia is the standing example of the last. Its bonds are at $107.8°$, and the usual account is “sp³, compressed by the lone pair”. The budget account is different and is computable: three bonds at $107.8°$ take 0.2358 each, leaving 0.2927 for the lone pair, which therefore holds more s character than any of the bonds. That is a statement with a number in it, it explains the compression rather than restating it, and it does not require the four orbitals to be equivalent — which they visibly are not.

What the repair costs #

There is a standard way of turning a non-orthogonal set into a basis, and it is worth naming, because what it does to these hybrids says where the difficulty really sits.

Symmetric orthogonalisation takes a set of overlapping functions and returns the orthonormal set closest to it, in a least-squares sense, treating every member alike. Applied to water’s hybrids at 104.5° it moves each of them slightly and returns four orbitals that are genuinely a basis.

The four no longer point at the hydrogens. That is the whole content of the failure: the overlap has to go somewhere, and the only place available is the directions. So the choice is between a set that points at the bonds and is not a basis, and a basis whose members point somewhere else — and the vocabulary of hybridisation assumes both at once.

Four statements about one object #

Four arguments about hybrids say four different things about the same object, and they fit together.

Hybrids are a basis says a hybrid set is a change of basis and not a physical process — nothing hybridises, and the transformation is invertible.

Hybridisation does not explain says that a basis change cannot be an explanation of a shape, because the same shape comes out in every basis.

The angle does not fix the hybridisation computes the s fraction each measured angle requires and finds water at sp³·⁹⁹ and hydrogen sulfide at sp²⁷.

The overlap computed here adds the failure mode when the two are combined carelessly: a label and an angle taken from different places give a set that is not a basis at all, by an amount that runs from six per cent for water to five eighths for cyclopropane.

The next step is one that needs more than idealised hydrogenic functions. Real hybrids are not built from a hydrogenic s and p of the same principal quantum number with nothing else in them; a proper account has different radial parts, contributions from higher shells, and a variational determination of the mixing. What the arithmetic here shows is that even the idealised object does not behave the way the vocabulary assumes, so the failure is in the description rather than in its approximations.