Two bent bonds, or a σ and a π

A carbon–carbon double bond is drawn in two ways that look nothing alike. In one, a σ orbital lies along the axis between the two carbons with a π orbital above and below it, in a different symmetry species, at a different energy. In the other, two identical bonds curve out from each carbon and meet in the middle above and below the plane — the bent or banana bonds, the picture Pauling gave and a great many organic chemists still keep.

The literature treats these as competing claims about what a double bond is. They are not claims at all in the sense that would let one be right.

What they are is two bases for one two-dimensional space, and everything below is the demonstration and the arithmetic of the second basis, which is usually left as a sketch.

The rotation, and the whole of the argument #

Take the two occupied orbitals of the double bond, σ and π. Mix them:

$$\tau_\pm = \frac{\sigma \pm \pi}{\sqrt2}$$

That is a rotation by forty-five degrees in the two-dimensional space they span. It is orthogonal, so the density built from the two mixed orbitals is the density built from the two originals — exactly, not approximately, and for the same reason the localisation transformation leaves methane’s density unchanged.

The invariance is a statement rather than an arithmetic identity, and the way to tell is to break it. Mix the two orbitals with a matrix that is not orthogonal — rows (1, 0.4) and (0.4, 1) — and the density changes by seventeen per cent immediately. The theorem is about orthogonality, and the check is written so that a non-orthogonal mixing fails it.

What the bent pair actually are #

Each bent bond is built on a hybrid on each carbon, and the hybrid follows from the mixing without any further choice. The σ hybrid is an sp² one — one part s to two parts p — and mixing it in equal measure with a pure p halves its s character. So the bent-bond hybrid is one part s to five parts p: sp⁵, exactly, not as an approximation.

Its direction follows too. Its p component is the σ hybrid’s p pointing along the axis plus an equal amount of the perpendicular p, so it points at arctan(1/√(2/3)) = 50.768° off the axis, and the two bent bonds on one carbon are 101.537° apart — wider than tetrahedral, which is the arithmetic behind the usual remark that bent bonds are strained.

The two hybrid sets underneath all of this are the ordinary ones and hybrids are a basis draws them: sp², one part s to two parts p with three of them in a plane, is what the σ half of a double bond is built on, and sp, one part s to one part p with two of them at 180°, is what a triple bond’s σ component uses. The same construction applied to the second gives three bent bonds rather than two, and — the point three bent bonds and the same hybrid is about — the hybrid it gives them is the same sp⁵.

The equivalence of the two bent bonds is enforced rather than chosen. The molecule has a mirror plane through the two carbons and the four hydrogens, and reflection in it takes τ₊ to τ₋: the two members are related by an operation of the group, so any property one has, the other has. That is what makes forty-five degrees the distinguished angle and every other mixing merely a basis.

Where the charge sits, and why a program picks this basis #

The one number in that table that needed an integral rather than algebra is the last: where a bent bond’s centre of charge is.

For either canonical orbital it is on the axis. The σ is symmetric about the plane of the molecule and so is the π, so the expectation of the perpendicular coordinate is zero for both. For the bent pair it is not: τ₊ has its charge 0.235 Å above the plane and τ₋ the same distance below.

That is why an automatic localisation procedure chooses the bent pair. The standard criteria all reward orbitals that are far apart or compact — maximise the separation between the centres of charge, or maximise each orbital’s self-repulsion — and the σ–π pair have coincident centroids while the bent pair are half an ångström apart. Given the freedom to rotate, a localisation converges on the bent bonds every time.

So the bent-bond description is not a chemist’s shorthand for the real thing. It is what the calculation produces when it is asked for the most localised description, and the σ–π pair is what it produces when it is asked for orbitals of definite symmetry. Both are outputs and the question they answer is different.

“Every time” is a claim about a criterion rather than about a molecule, and it has a boundary that can be found.

So “a localisation converges on the bent bonds every time” is true of a homonuclear double bond and is not a theorem. A carbonyl is the case where it stops being true, and the two conventions there are the only two stationary points a Boys functional has — computed in three bent bonds and the same hybrid, where the functional turns out to be a quadratic with no linear term.

The same freedom on methane is where this collection first drew the transformation: the localisation transformation puts the canonical orbitals on one side and the localised ones on the other with the density identical between them. The double bond here is the two-dimensional version, with one mixing angle instead of a four-by-four matrix.

Which one a measurement sees #

Here the symmetry between the two descriptions ends, and the point has been made before in a different molecule.

The canonical orbitals are eigenfunctions of the effective one-electron Hamiltonian and belong to definite symmetry species of the molecule’s group; the bent pair are neither. So the canonical pair have well-defined energies and the bent pair do not — each bent bond is a mixture of two energies.

A photoelectron spectrum measures energies. Ethene’s shows two bands in the region in question, at 10.51 and about 12.85 electronvolts, and their separation is what the σ–π difference is. Two equivalent bent bonds would give one.

What a canonical set is, drawn, is a column of levels: orbitals with energies, ordered, each belonging to a symmetry species — what a photoelectron spectrum measures builds one for this separation and these two overlaps. A localised set has neither of the last two properties, which is exactly why a spectrum can be predicted from one of the two bases and not from the other.

This is the same argument hybridisation does not explain makes for methane — four equivalent sp³ bonds cannot produce a spectrum with two bands, and methane’s has two — and the resolution is the same. A description in equivalent localised orbitals is a legitimate basis and is not a claim about what an ionisation experiment will find.

The general shape of that argument is drawn for water in what a photoelectron spectrum measures: measured bands at their measured energies, beside the symmetry species that account for them. A count of bands is a count of symmetry species and never a count of bonds, and water’s four valence bands against two O–H bonds is the version of the point that is hardest to argue with.

The count of orbitals is the same, and that is not trivial #

One check on the whole construction is that the two descriptions have the same number of orbitals holding the same number of electrons, and it is worth pausing on because the pictures make it look otherwise.

The σ–π picture is usually drawn as one bond plus one bond, and the bent picture as two bonds. Both hold four electrons in two orbitals. What differs is not how much bonding there is but how it is divided up, and the temptation to say a double bond is “a σ bond plus a π bond, so about one and a half bonds’ worth” comes from reading the pictures as inventories.

A double bond is not two single bonds is the essay about that particular error from the energy side — carbon’s single bond is 348 kilojoules a mole and its double is 614, which is not twice anything. This essay is the same point from the orbital side: there is no way to slice the occupied space that makes the two halves independent, because the slicing is a choice of basis and the space is what it is.

The asymmetry is not that one description is more correct. It is that energy is a property of the Hamiltonian’s eigenvectors, and only one of the two bases is made of those. Ask a question about energies and the canonical basis answers it; ask a question about where the charge is and the localised basis answers that one; ask which is real and the question has no content.

The two overlaps, and a limitation worth reporting #

The σ and π parts of a double bond are not equivalent, and the reason they are not is an overlap. The computed values at the measured 1.334 Å carbon–carbon distance are 0.273 for the π pair and 0.324 for the two p functions along the axis — the σ component being the larger, which is why the two are not degenerate and why the bent pair are not eigenfunctions.

The overlap of the two full σ hybrids comes out at 0.013 in this model, which is far too small, and the reason is worth stating rather than hiding: a hydrogenic 2s has a radial node inside the bond, and its contributions to the hybrid–hybrid overlap very nearly cancel against the s–p cross terms. That is a limitation of hydrogenic functions as valence orbitals — the same limitation a Gaussian is the wrong shape is about from the other side — and none of this essay’s argument rests on it. The structural results above come from the mixing, which involves no radial function at all.

The angle nobody chose, and the one everybody did #

There is a nice inversion in the numbers worth drawing out, because it turns a piece of hand-waving into arithmetic.

The bent-bond picture is usually introduced with a remark that the bonds are “strained”, and the evidence offered is that they are curved. That is a statement about a drawing. The quantitative version is the interorbital angle: 101.54°, against the 109.47° two sp³ hybrids want and the 120° an sp² carbon’s other bonds have.

But nothing was chosen to make it 101.54°. The angle is what falls out of requiring the two members to be equivalent, and the requirement is a symmetry of the molecule. So the strain that the picture gestures at is not an input to it — it is a consequence of the molecule having a mirror plane and of the σ hybrid being sp².

That is the same relation the angle does not fix the hybridisation works out in the other direction, where a measured angle is turned into an s character rather than the reverse. Here the s character is fixed by the mixing and the angle follows.

What the choice of basis is actually for #

The freedom being exercised here is not a nuisance to be resolved. It is the reason a chemist and a spectroscopist can both be right about a molecule while describing it differently, and each basis is chosen for what it makes obvious.

The σ–π basis makes the symmetry obvious. The π orbital is antisymmetric about the plane of the molecule and the σ is symmetric, so their behaviour under every operation of the group is immediate, and every selection rule follows without further work. It also makes the energies obvious, because they are the eigenvalues.

The bent basis makes the locality obvious. Each orbital is a bond between two atoms and nothing else, which is what makes bond energies additive to the extent they are, and what makes it possible to talk about a functional group at all — the same locality bond order from the eigenvectors reads off a delocalised calculation after the fact.

Neither makes the other’s virtue available. That is a general property of choosing coordinates, and hybrids are a basis states it: rotating a basis changes no observable, so asking which is real is asking which coordinate system nature prefers.

And the angle is not a fact about double bonds either. It is what the arithmetic gives for a bond whose σ component happens to be sp², and putting the three carbon–carbon bonds side by side shows where it comes from.

That coincidence is worth its own argument, and the reason it is worth noticing here is that it separates the two things a hybrid label carries. One sixth s character is a statement about the component; two components against three is a statement about the arrangement; and the strain a chemist reads off “sp⁵” belongs to the second.

What the bent picture predicts that the σ–π one hides #

There is one place where the bent description earns its keep as more than a preference, and it is worth having because it is a prediction rather than a restatement.

A bent bond’s charge sits off the axis, and that is what makes a double bond’s electron density lie above and below the plane rather than along the line between the nuclei — which is what an electron density map of ethene shows, and what makes the double bond the site of attack for anything electron-poor. The σ–π description gives the same density, of course, but it takes an addition of two orbitals to see it, where the bent description has it in each member separately.

The interorbital angle of 101.5° is the other prediction, and it is the quantitative version of the remark that a double bond is strained. Two hybrids at that angle on a carbon whose other bonds want 120° between them is a geometrical statement that can be compared with the observed bending of the C–H bonds away from the double bond, which is real and is a couple of degrees.

The molecule the arithmetic is about is planar, with its carbons 1.334 Å apart and an H–C–H angle of 117.4° rather than 120°. Those two and a half degrees are the observable the interorbital angle can be set against: two hybrids 101.5° apart, on a carbon whose other two bonds would like 120° between them, leaves the C–H pair squeezed — and squeezed is what they measurably are.

What this says about arguing over pictures #

The pattern recurs and it is worth naming, since three instances of it concern hybrid orbitals alone.

An argument about whether electrons are “really in” hybrids, or “really in” bent bonds, or “really in” delocalised orbitals, is an argument about a basis. The occupied space is an invariant; a basis for it is a choice; and every observable is a property of the space. So the argument has no possible resolution and is not about anything a measurement could settle.

What is settleable is a different question that gets confused with it: which basis has members that are eigenfunctions of the effective Hamiltonian, because those have energies and the others do not. The answer is always the canonical one, and it is the answer to a question about energies rather than about reality.

Water’s lone pairs are not a pair is the third instance in this collection: two equivalent lone pairs is a basis, water’s spectrum shows the two bands the canonical basis predicts, and neither fact refutes the other.

One consequence of that is worth stating for anyone reading a calculation. A published set of localised orbitals and a published set of canonical ones, for the same molecule at the same level of theory, are not two calculations to be compared: they are one calculation printed twice. Any quantity that differs between them is a property of the printing.

One of the two labels needs a symmetry the other does not #

The two descriptions are equivalent for ethene, and the equivalence rests on something ethene has. It is worth saying what, because a great many double bonds do not have it, and there the two descriptions stop being interchangeable — not in their content, but in whether one of them can be written down at all.

The labels σ and π are symmetry species. They mean symmetric and antisymmetric with respect to the molecular plane, and they exist as labels only because ethene is planar and that plane is a symmetry of the molecule. Take the plane away and there is nothing for an orbital to be symmetric or antisymmetric about: the two orbitals mix, neither is σ, neither is π, and the classification has no members.

Twisting a double bond does exactly that. A strained alkene in a small ring, an alkene at a bridgehead, or any molecule in which the two ends of the double bond are rotated relative to one another has no plane containing both carbons and all four substituents — so σ and π are not available labels for it, and a description built from them has to be replaced rather than corrected.

The bent description has no such requirement. It is two equivalent hybrids pointing off the axis, and equivalence between them is enforced by a twofold rotation rather than by a mirror — a symmetry a twisted alkene retains when the plane has gone.

The limiting case makes the point without any subtlety. Twist an ethene through ninety degrees and the two p orbitals become perpendicular: their overlap is zero, the bond that the π label described has ceased to exist, and the molecule is a diradical. That geometry is the transition state for cis–trans isomerisation, and the barrier is measured at about 270 kilojoules a mole — which is, to the accuracy such a comparison allows, what a π bond is worth.

So the essay’s conclusion holds and acquires a boundary. For a planar double bond the two pictures are one occupied space in two bases. For a twisted one only the bent basis exists, and the σ/π vocabulary is not a rival description but an inapplicable one.

What is left #

The mixing angle here is exactly forty-five degrees because the two members are required to be equivalent to each other, and equivalence is what the mirror plane of the molecule enforces. In a molecule without that symmetry — a carbonyl, say, where the two ends of the double bond are different atoms — the localised pair are still available and are no longer equivalent, and the angle that maximises localisation is not forty-five degrees and has to be searched for.

Where two-centre bonding stops is the essay about the cases where no localised description exists at all, and it is worth reading against this one: a double bond has a localised description and a boron cluster does not, and the difference is a property of the occupied space rather than of anyone’s preference.

The triple bond is the case that would test the argument hardest, because there the occupied space is three-dimensional and the localised description is three bent bonds at 120° round the axis rather than two. That mixing is a rotation in three dimensions rather than two, the equivalence is enforced by a threefold axis rather than a mirror, and the s character of the resulting hybrids is a third of the σ hybrid’s rather than a half. None of it is done here.