The angle does not fix the hybridisation

A hybrid orbital is a linear combination of one s function and some p functions on the same atom. Two such hybrids on one atom must be orthogonal, because the set they belong to is a change of basis and a change of basis is a rotation.

That single requirement, written out, fixes a relation between how much s character each hybrid carries and the angle between them:

$$\cos\theta = -\frac{a}{1-a}$$

with $a$ the fraction of the hybrid that is s. Nothing about bonding is in it. It is the orthogonality of two unit vectors in a four-dimensional space, with the p parts of the two hybrids pointing at an angle $\theta$.

The relation is usually read forwards: assume $a = 1/4$, get $109.47°$. Read backwards it is more useful and much less comfortable, because a measured bond angle then names the s character rather than being explained by an assumed one.

Three exact points #

The relation has three places where it produces the familiar answers, and they are checks rather than results.

At $109.4712°$ — arccos(−1/3), which is what a tetrahedron of directions subtends — the s fraction comes out at exactly $0.2500$. Four such hybrids use $4 \times 0.25 = 1.0000$ of the s orbital: the whole of it, exactly.

At $120°$ the fraction is exactly $1/3$, and three hybrids use $3 \times 1/3 = 1$. The whole of it again.

At $180°$ the fraction is exactly $1/2$, and two hybrids use $1$. Three geometries, three different counts of bonds, and one identity closing in each.

That the three close exactly is not a coincidence and it is not a fit. It is the statement that a complete orthonormal set built from one s and three p functions has s fractions summing to one, because the sum is the squared length of the s function’s expansion in the hybrid basis and a unitary transformation preserves lengths.

There is one s orbital, and the s characters of a full hybrid set add to exactly one. That is the budget this essay is about.

What the measured angles say #

Running the relation backwards on four hydrides with measured angles gives the following.

Water is the mildest case and it is already not sp³. Its bond hybrids carry a fifth of an s function apiece rather than a quarter, and the difference is not a rounding: $0.2002$ against $0.2500$ is a fifth less s character in the bonds, and the twenty per cent that goes missing has to be somewhere.

It is in the lone pairs, which take $0.5995$ between them. Each is sp²·³⁴ — closer to a trigonal hybrid than to a tetrahedral one, and considerably richer in s than the bonds are.

Ammonia behaves the same way with a different count: three bonds at $0.2341$ each spend $0.7024$, and the single lone pair takes the remaining $0.2976$.

The heavy hydrides are not distorted tetrahedra #

The second row is where the usual account stops being a mild misdescription and becomes a wrong one.

Hydrogen sulfide’s angle is $92.1°$, less than two degrees from a right angle. The relation gives an s fraction of $0.0353$: the S–H bonds are ninety-six-and-a-half per cent p. Written as spⁿ that is sp²⁷, and the notation stops being informative long before that — the honest description is that the bonds use the p orbitals and essentially nothing else.

Phosphine at $93.3°$ gives $0.0544$ per bond. Three bonds spend $0.163$ and the lone pair keeps $0.837$: eighty-four per cent s character in one orbital.

That is not a rearrangement of an sp³ centre. It is a different arrangement entirely, and it explains two things that the distorted-tetrahedron story leaves as facts to be memorised.

Phosphine is a far weaker base than ammonia. A lone pair that is eighty-four per cent s is held close to the nucleus, where the s function penetrates — the effect measured in what the screening model cannot see as a fifty-six-fold concentration of s density near the nucleus. A pair held there is less available to anything else.

The angles in the second row and below are all near ninety degrees. H₂S at $92.1$, PH₃ at $93.3$, and the pattern continues down each group. If the bonds are made of p orbitals, ninety degrees is what the p orbitals offer, and the small excess over ninety is the interaction that VSEPR describes.

What this is not #

Three qualifications, because the arithmetic is exact and the interpretation is not.

The relation assumes the two hybrids are equivalent. Two bonds to the same kind of atom are, by symmetry; two bonds to different atoms are not, and then the relation reads with a geometric mean of the two s fractions in place of one. Every molecule in the table above has equivalent bonds, which is why they were chosen.

A hybrid set is a basis and not a set of objects. Hybrids are a basis makes the case in full and the localisation transformation demonstrates it numerically: the canonical orbitals and the localised ones give the identical total density, to $10^{-16}$. So the s fraction computed here is a property of one description among many, and the fact that it comes out non-integral is a fact about the description.

None of this predicts the angle. The angle is the measured input. What the relation supplies is what the angle costs, and the cost is paid out of a budget with one unit in it.

That last point is the one that makes the exercise worth doing. A description that can be given any angle whatever, and that responds by quoting a hybridisation to match, is not making a prediction — and this is exactly the complaint hybridisation does not explain makes. What the budget adds is that the description is not free either: the s character taken by the bonds is unavailable to the lone pairs, and that shows up in properties that can be measured independently.

Bent’s rule is the budget with a preference attached #

The budget says s character is conserved. It does not say where it goes. Bent’s rule does, and it is a ranking rather than a magnitude: s character concentrates in orbitals directed toward the less electronegative substituent.

Bent’s rule, against the substituent series tests that ranking against measured angles across a substituent series. What the present arithmetic adds is the other side of the same ledger — a lone pair is the limiting case of a substituent with no electronegativity at all, so it should take the most s character of anything on the atom, and in every row of the table above it does.

Water: $0.2998$ in each lone pair against $0.2002$ in each bond. Ammonia: $0.2976$ against $0.2341$. Hydrogen sulfide: $0.4647$ against $0.0353$. Phosphine: $0.8367$ against $0.0544$. The inequality holds in all four and it is checked in the direction that could fail, so a sign error in the relation would be caught rather than pass.

A prediction the budget makes on its own #

The budget so far has consumed a measurement and returned a description. It also makes one statement that nothing measured went into, and it is worth pulling out because it is falsifiable.

The two lone pairs of a bent AH₂ molecule are equivalent to each other by the same symmetry that makes the two bonds equivalent, so the identical orthogonality relation applies to them — with their own s fraction, which the budget has already fixed. Water’s lone pairs carry $0.2998$ each, and $\cos\theta = -0.2998/0.7002$ gives

$$\theta_{\text{lone}} = 115.35°$$

against $104.5°$ between the bonds. Hydrogen sulfide’s lone pairs, at $0.4647$ each, come out at $150.22°$ against a bond angle of $92.1$.

Neither number was measured and neither was put in. They follow from one measured angle and the requirement that a set of four orbitals on one atom be orthonormal.

The ordering is the interesting part. In both molecules the lone pairs subtend a wider angle than the bonds, and the excess grows as the bonds become more nearly pure p — eleven degrees for water, fifty-eight for hydrogen sulfide. That is the same conclusion VSEPR reaches by giving a lone pair a larger repulsive weight, arrived at with no repulsion anywhere in the argument, and it is why what a lone pair is worth can fit a weight to each molecule and find the weights refusing to transfer between them: the quantity that transfers is the s budget, and the weight is a proxy for it.

What the prediction is not is a claim that water has two distinguishable lone pairs sitting at $115°$. Water’s lone pairs are not a pair shows the photoelectron spectrum putting the two highest occupied orbitals in different symmetry species with different energies, so the equivalent-pair description is one basis among several and the spectrum is measuring a different one. The angle above is a property of that basis, computed correctly, and stating which basis it belongs to is the whole of the care required.

The arrangement the budget refuses #

An identity that always balances proves nothing, so the budget is asked for something it must reject.

Four bonds at $120°$ would require an s fraction of $1/3$ in each hybrid, and four of those is $4/3$. There is one s orbital. The arrangement is not expensive or unlikely; it does not exist, and the calculation refuses it rather than reporting a hybridisation of sp². The refusal is itself tested.

The same arithmetic bounds the whole problem. Since $a \le 1/2$ for two equivalent hybrids, the relation is defined only from $90°$ to $180°$ — no two orthogonal s–p hybrids can subtend less than a right angle, which is why every strained ring in chemistry with an angle below ninety is described with bent bonds whose maxima do not lie along the internuclear lines.

The angles repulsion produces for two, three and four points on a sphere are 180°, 120° and 109.47°, and they come out of a minimisation with no hybrid in it. Those are the three geometries at which the s budget closes exactly, which is a coincidence worth noticing and not a mechanism.

Two calculations meeting #

The three exact angles are worth one more look, because they arrive twice, from arguments with nothing in common.

VSEPR, computed minimises the mutual repulsion of points on a sphere and finds $180°$, $120°$ and $109.47°$ for two, three and four sites. That calculation contains no orbital, no s function and no orthogonality condition; it is a classical minimisation.

The budget above finds the same three angles as the ones at which a complete hybrid set is used up exactly. That calculation contains no repulsion and no geometry beyond the definition of an angle.

They agree because both are consequences of the same symmetry: an arrangement of $n$ equivalent directions in which every pair subtends the same angle exists only for $n = 2, 3, 4$ in three dimensions, and both calculations are asking about equivalent directions. That is a satisfying agreement and it is also the boundary of it — for five sites there is no such arrangement at all, as five sites are not alike computes, and the hybrid budget cannot be run on a trigonal bipyramid without splitting it into an axial set and an equatorial one.

The tetrahedral case is where the two calculations touch: one distinct angle, four equivalent directions, and a composition of exactly a quarter s in each. Away from it the two accounts have nothing to say to each other, because one is about equivalent objects and the other is not.

Water’s two O–H bonds are equivalent by symmetry, as its recovered group says, and that equivalence is the condition Coulson’s relation was derived under. It is why water can be treated at all — and why the lone pairs, which are not equivalent to the bonds, are outside the relation.

The s character has a second measurement, and it is not an angle #

Everything above infers an s character from a bond angle, through an orthogonality relation. There is an entirely independent route to the same quantity, it is a routine measurement, and having two makes the inference checkable rather than merely self-consistent.

The route is nuclear magnetic resonance. The coupling between a carbon nucleus and a hydrogen bonded directly to it is carried almost entirely by the Fermi contact mechanism, which requires electron density at the nucleus — and only an s orbital has any. A p orbital has a node there and contributes nothing. So the size of the one-bond coupling is a report on how much s character the carbon puts into that bond, and the relation is close to proportional:

$$\%\,s \approx 0.2 \times {}^1\!J(\mathrm{C\!-\!H}) \ \text{in hertz}.$$

The numbers are unusually clean for an empirical relation:

Methane and ethyne land on their ideal values to a tenth of a per cent, which for a one-parameter empirical relation is better than it has any right to be.

Cyclopropane is the row worth reading against this essay’s own arithmetic. Its C–C–C angles are 60°, far below the range in which two equivalent hybrids can be orthogonal at all, so the ring bonds must be built almost entirely from p — and the s character they do not take has to go somewhere. The budget says it goes into the C–H bonds, which should therefore be much richer in s than an ordinary alkane’s.

They are. Cyclopropane’s coupling of 161 hertz puts its C–H bonds at 32 per cent s, against ethane’s 25 — nearly the value of an sp² carbon, in a molecule with no double bond in it. The s-character budget predicted a redistribution and a measurement made on a different physical mechanism reports the same redistribution, with the right sign and roughly the right size.

Two cautions belong with the relation, and both are the ordinary ones.

It is fitted. The coefficient of 0.2 was obtained by assuming the ideal values for a few reference compounds, so the reference rows above are not independent evidence — they are the calibration. What is evidence is everything else the relation then predicts.

And it is specific to C–H. The Fermi contact term dominates for a light nucleus with a compact s orbital; for heavier atoms other mechanisms contribute and the proportionality fails.

What survives both is the important part for this essay. The s character inferred from an angle is not a quantity defined only within the hybrid language, checkable against nothing. It has an independent measurement, the two agree where they can both be evaluated, and cyclopropane — the molecule where the angle argument is under most strain — is the one where they agree most interestingly.

It also supplies the one thing the orthogonality relation cannot: a reading for the bonds it says nothing about. The relation constrains a pair of equivalent hybrids through the angle between them, so it has an answer for the ring bonds of cyclopropane and none for its C–H bonds, which are not equivalent to them and whose angle to the ring is not the quantity the relation contains. The coupling constant reports on exactly those bonds, directly, and it is the half of the budget the arithmetic here can only infer by subtraction.

Who found it, and when #

The relation is Charles Coulson’s, and it appears in Valence (1952) alongside the hybridisation formalism he did as much as anyone to develop. It is sometimes called the directionality relation and sometimes Coulson’s theorem, and both names oversell it slightly: the content is orthogonality, and Coulson’s contribution was to see that it could be inverted and used as a measurement.

The heavy-hydride reading is Henry Bent’s, in the 1961 review that the rule is named for. His argument, stated in terms of the s-character budget rather than of repulsion, was that the near-ninety-degree angles of the heavier hydrides are not distortions at all but the natural consequence of bonds made from p orbitals, and that the s character had gone into the lone pair. That is the reading the arithmetic above supports, and it was a minority position for some time against the distorted-tetrahedron account.

The sp² set is three hybrids at 120° with an s coefficient of 1/√3 each and one p function left entirely out. The excluded p carries the π system in every planar molecule, and its exclusion is a choice of basis rather than a property of the atom.

Still open: predicting an angle #

This essay takes hybridisation as far as arithmetic will take it: given an angle, what a description costs. What it does not do — what no hybridisation argument does — is predict an angle from the electronic structure, because that requires the total energy and therefore the two-electron integrals.

The earlier arguments make the case that the description is a change of basis; this one makes the case that a change of basis still has bookkeeping in it, and that the bookkeeping is checkable against measurements the description was not fitted to. Both are the same warning from opposite sides: the language of hybridisation is precise about relations and silent about causes, and most of the trouble it causes comes from reading a relation as a cause.