Molecular Geometry & Bonding
what the shape decides
Bonding models

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.
11 min read 6 figures A basis is not a thingCounted, not quoted

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.

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. 1 The four sp³ directions, with the angle between them computed from the coefficients rather than quoted. The set is orthonormal to better than one part in 10^15, which is what makes it a rotation rather than an arbitrary construction.

The construction

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

h1=12(s+px+py+pz),h2=12(s+px−py−pz),h_1 = \tfrac12(s + p_x + p_y + p_z), \quad h_2 = \tfrac12(s + p_x - p_y - p_z),

h3=12(s−px+py−pz),h4=12(s−px−py+pz).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.

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 same thing written as a matrix, one row per hybrid. Every row is normalised and every pair of rows is orthogonal, so the matrix is orthogonal — and an orthogonal matrix is a rotation.

Why the angle is what it is

The direction of a hybrid is the direction of its p component. For h1h_1 that is (1,1,1)(1,1,1) and for h2h_2 it is (1,−1,−1)(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(1 - 1 - 1)/3 = -1/3. So the angle is arccos⁡(−1/3)\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 CTC=IC^\mathsf{T}C = I, and it has determinant ±1\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.

sp2 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.sp2120.000°between every pair3 hybridsworst off-diagonal 3e-16an orthogonal transformation of the atomic orbitalsone electron
Fig. 3 Three sp² hybrids: one third s and two thirds p in each, at 120 degrees in a plane. The remaining p orbital is untouched and perpendicular to that plane, which is where a pi bond goes.
sp 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.sp180.000°between every pair2 hybridsworst off-diagonal 0e+0an orthogonal transformation of the atomic orbitalsone electron
Fig. 4 Two sp hybrids: half s and half p, at 180 degrees. Two p orbitals are left over, perpendicular to the axis and to each other.

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, pxp_x, pyp_y and pzp_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 figure on this site verifies the transformation before drawing it.

Each row is checked to be normalised, and every pair of rows checked to be orthogonal, both to better than 10−1210^{-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 αs+βp\alpha s + \beta p with α2+β2=1\alpha^2 + \beta^2 = 1, and α\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 sp2 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 10.57740.816500hybrid 20.5774-0.40820.70710hybrid 30.5774-0.4082-0.70710every row normalised and every pair orthogonal to 3e-16a change of basis, and nothing moresp2
Fig. 5 The sp² matrix: one third s and two thirds p in each row, giving 120 degrees. Change those weights continuously and the angle moves continuously, which is why the three named cases are landmarks rather than a set of options.
sp 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.sp180.000°between every pair2 hybridsworst off-diagonal 0e+0an orthogonal transformation of the atomic orbitalsone electron
Fig. 6 The other extreme, sp: half s and half p, at 180 degrees. Between this and pure p at 90 degrees lies every angle a two-coordinate centre can adopt, and the composition is a description of where it landed.

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.

Where the model stops

Two 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.

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.

Where the ladder goes next

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.

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.

Where this goes next

HHCHHTdprincipal axis C36 mirror planesno inversion centrecannot be polarcannot be chiralgroup recovered from the coordinates5 atoms 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. 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. HHOC2vprincipal axis C22 mirror planesno inversion centremay be polarcannot be chiralgroup recovered from the coordinates3 atoms Why water is bent The standard answer is lone pair repulsion, it predicts the right direction, and it cannot predict the magnitude. A better rule can, and the heavier hydrides show where both accounts run out.