Three bent bonds, and the same hybrid

A carbon–carbon double bond has two familiar pictures — a σ and a π, or two bent bonds — that are one occupied space in two bases, related by a rotation of exactly forty-five degrees. The density is identical to the last bit a double holds, the bent pair are sp⁵ hybrids at 50.77° to the axis, and their charge sits 0.235 ångström off the plane of the molecule where neither canonical orbital’s does.

The case that tests the argument hardest is the triple bond. A triple bond’s occupied space is three-dimensional rather than two, the localised description is three bent bonds at 120° about the axis rather than two, and the mixing is a rotation in three dimensions rather than in a plane. This essay does it, and the arithmetic produces something nobody put in.

Three dimensions instead of two #

Ethyne’s carbons are sp hybridised: two σ bonds each, one to hydrogen and one to the other carbon, so the s orbital is shared between two hybrids and each carries half of it. The remaining two p orbitals on each carbon are perpendicular to the axis and to each other, and they make two π bonds.

So the occupied space between the carbons is spanned by σ, π$_x$ and π$_y$. Any orthogonal mixing of the three leaves the density alone; the one that makes the three members equivalent to one another is

$$\tau_k = \tfrac{1}{\sqrt{3}}\,\sigma + \sqrt{\tfrac{2}{3}}\left(\cos\varphi_k \,\pi_x + \sin\varphi_k\, \pi_y\right), \qquad \varphi_k = \tfrac{2\pi k}{3},$$

which is the three-dimensional analogue of the forty-five degrees of the double bond. It is a change of basis and nothing more: the projector is unique and the basis is not. Its first row is constant, which is what treats the three alike, and the rest is a rotation.

That the three come out equivalent is not a check, because it was imposed. What is a check is that the mixing is orthogonal, and that is verified on the functions themselves: every mutual overlap between the three τ comes back at zero to about a part in ten thousand, which is the quadrature’s precision rather than the algebra’s, and each is normalised to the same accuracy.

The number that was not put in #

The s character of one bent component follows from the mixing: it is $\tfrac{1}{3}$ of the σ hybrid’s, and the σ hybrid is sp, so it is $\tfrac{1}{3} \times \tfrac{1}{2} = \tfrac{1}{6}$.

The double bond’s bent components have $\tfrac{1}{2}$ of an sp² hybrid’s, which is $\tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}$.

Both are exactly one sixth. Both are what the literature calls sp⁵.

It is arithmetic and not a fact about carbon: a third of a half is a half of a third. But the arithmetic is not empty, because the two factors mean different things. The first is how much s the carbon has left over for this bond, which is set by how many σ bonds the atom is making. The second is how many pieces that share is divided into, which is set by the bond order. The two happen to be complementary — an atom making fewer σ bonds keeps more s for each, and a higher bond order divides it more ways — and they are complementary exactly.

The single bond breaks the pattern, which is what makes the coincidence a coincidence rather than a definition. An sp³ carbon shares a quarter of an s orbital with its neighbour and divides it into one piece, so the answer is a quarter, and 0.25 is not 0.1667.

The angle follows from orthogonality alone #

The interorbital angle needs no calculation either, and its derivation is two lines. Two equivalent hybrids

$$h_{1,2} = \sqrt{s}\,|s\rangle + \sqrt{1-s}\,|p_{\mathbf{d}_{1,2}}\rangle$$

have an overlap of $s + (1-s)\cos\theta$, since the s parts overlap fully and the p parts overlap as the cosine of the angle between their directions. Setting that to zero,

$$\cos\theta = -\frac{s}{1-s}.$$

That is Coulson’s relation, and it has no molecule in it.

The computed angles reproduce it: 109.47122 for sp³, 120 exactly for sp², 101.53696 for both sp⁵ cases. The relation is the reason the coincidence in the previous section propagates from one quantity to the other — once the s characters are equal the angles must be, and no further arithmetic is involved.

What does distinguish them #

Two things do, and it is worth being clear that the equality above is an equality of one property rather than of the bonds.

The total s character in the region. A double bond’s carbon puts a third of its s orbital into the C–C region; a triple bond’s puts a half. That is a real difference and it is the one that shows in the properties the region has — the shorter bond, the higher force constant, the greater acidity of an attached hydrogen. It is also the quantity an angle does not fix, which is why reading a hybridisation off a structure is the wrong direction.

How far off the axis the charge sits. The double bond’s bent components put their charge 0.235 ångström off the molecular plane; the triple bond’s put theirs 0.315 ångström off the axis, a third further. Two identical hybrids, differently arranged, put their charge in different places — which is the sense in which a triple bond is “fatter” than a double one and is a statement about the arrangement rather than about the hybrids.

The σ framework underneath a triple bond is the ordinary sp set — two hybrids along the axis, each half s, with two p orbitals left perpendicular to it — and hybrids are a basis draws all four. It is the perpendicular pair, mixed with the σ bond, that become the three bent components.

What a picture of it can and cannot show #

There is a drawing convention here that deserves a paragraph, because the two descriptions are usually drawn side by side and the drawings are not comparable.

The σ–π picture of a triple bond draws three objects with three different shapes: a cylinder along the axis and two mutually perpendicular pairs of lobes. The bent-bond picture draws three identical objects arranged with threefold symmetry. Neither is more real than the other, and the visual difference between them — one heterogeneous set and one homogeneous set — has no counterpart in anything measurable, because the density is the same.

What is visible in a measurement, and is genuinely different between the two, is nothing at all: every observable is a property of the occupied space. So a reader shown both pictures and asked which is right has been asked a question with no content, and the honest answer is the one that holds for the double bond — the argument has no possible resolution and is not about anything a measurement could settle.

The plane two descriptions of one bond sit in is the two-dimensional version of that; everything here is a point on the analogous object one dimension up, with a three-dimensional occupied space, an orthogonal group acting on it, and two conventional choices of basis both inside the orbit.

The overlaps, and one that is nearly zero for a reason #

The two π overlaps come out equal to nine decimal places, which the axis requires and which is a check on the construction rather than a result. The σ hybrid overlap does not behave the way a first guess expects: it comes out at 0.0226, nearly zero, on a bond of 1.203 ångström.

That is not a statement about triple bonds. It is a statement about hydrogenic 2s functions, which have a radial node inside the bonding region, so the s–s and s–p contributions to a σ hybrid overlap very nearly cancel. The same effect appears in ethene and it is worth repeating here rather than left to be rediscovered — the same node that makes an overlap non-monotonic in the separation: a σ overlap computed from these functions is not a quantity to be compared with a π one.

The rule that follows is worth stating as a rule, because it is easy to violate by accident. A comparison between hybrid overlaps means something when it is between hybrids of the same kind at different separations, and means very little when it is between a σ hybrid built from a nodal 2s and a π function built from a nodeless 2p — the cancellation across the node is a property of the radial function rather than of the bond, so the two numbers are not on one scale.

The quadrature had to be sharpened, and why #

One practical thing had to change, and it belongs in the essay because it looks like a bond-length effect and is not.

The double-bond construction uses this site’s ordinary sixty-point product rule and its normalisations hold to five parts in ten thousand. The triple-bond construction failed the same tolerance: the σ hybrid came back normalised to 0.99938.

The bond is shorter, so the obvious explanation is that the two centres are closer and the integrand sharper. That is the wrong explanation. What changed is the s character: an sp hybrid is half 2s where an sp² hybrid is a third, and it is the 2s radial node that the quadrature resolves least well. The cure is the finer rule used to verify overlaps — eighty points over a wider map — and the norms come back to a part in a thousand.

The distinction is worth having because the two explanations predict different things. A separation effect would get worse for a shorter bond of any kind; a nodal-function effect gets worse whenever more 2s is put in, at any separation. The two can be told apart because both variables are available separately.

The mixing above was chosen by symmetry rather than by a localisation criterion, and it is fair to ask whether a criterion would have chosen it. For the threefold case it would; for a bond with two unlike ends the question has a sharper answer than expected.

So the two conventions are not two points on a continuum with an optimum somewhere between them. They are the only two stationary points there are, which is a stronger statement than the double bond alone supports and is the reason the argument has no resolution rather than an unresolved one.

The overlaps as the bond is stretched #

One thing the model does say quantitatively, and it is worth extracting because it separates the two components of the bond in the way the usual account claims to.

At ethyne’s bond length the π overlap between two carbon 2p functions is 0.3361. At ethene’s longer bond it is smaller, and at ethane’s smaller again — the π overlap falls monotonically with separation over this range, which is the ordinary behaviour of two functions with no radial node between them.

So the ordering “π is weaker than σ” is not what these numbers show, and it is not what the numbers could show, for the reason in the previous section: the σ number is contaminated by a cancellation across the 2s node. What they do show is the π overlap alone, falling with distance, and a triple bond having two of them where a double bond has one.

Closer is not more overlap plots both curves against separation at carbon’s screened charge, and the shapes are the argument: the π curve falls monotonically and the σ curve changes sign inside the range. A comparison between the two at a single separation therefore says less than it appears to, and the number it would produce depends on which side of the sign change the separation falls.

Where the model stops #

Hydrogenic functions at a screened charge. Every integral here is over 2s and 2p functions at carbon’s Slater effective charge of 3.25. The mixing angles and s characters are exact within the model and do not depend on the radial functions at all; the overlaps and the centroid do.

The framework is assumed. That ethyne’s carbon is sp and ethene’s sp² is put in rather than derived — it follows from the number of σ bonds, which follows from the structure, which is an input. Hybrids are a basis rather than a mechanism, and this essay uses the basis without re-deriving why it is the useful one.

Nothing is optimised. The bent components here are the equivalent ones, constructed. A localisation criterion applied to the same occupied space would find its own maximum, and for a system with a threefold axis the two coincide — but that is an argument a localisation transformation could make, and it is not made here.

And the equality is of one number. Two hybrids with the same s character are the same hybrid only in the sense that every property depending on s alone is the same. Their radial parts here are identical because both are built from the same carbon functions; in a real molecule they would not be.

Every mixing in this essay is one matrix of one kind — orthogonal, so the density is untouched — and hybrids are a basis writes the sp case out entry by entry. The one that makes three bent bonds is the three-dimensional member of the same family, and its orthogonality is what the rotation figure above measured.

The coincidence is a quadratic, and it has one more root #

A third of a half is a half of a third explains the equality and leaves it looking like an accident of two small numbers. It is not: the same construction run at any bond order gives a closed form, and the equality is that form’s symmetry.

Fix a carbon whose bond of order $n$ to one neighbour is being localised. The atom makes $5 - n$ σ bonds in all — four for a single bond, three for a double, two for a triple — and its σ hybrids share the one s orbital equally, so each carries

$$s_\sigma = \frac{1}{5-n}.$$

The multiple bond then uses one of those hybrids together with $n-1$ pure p orbitals, and localising that set into $n$ equivalent bent components divides the s character equally among them:

$$s_{\text{bent}} = \frac{s_\sigma}{n} = \frac{1}{n(5-n)}.$$

The denominator is a quadratic in the bond order, and $n(5-n)$ is symmetric about $n = 2.5$. So the double bond and the triple bond, sitting one on each side of that centre, are required to give the same answer — and the answer they give is $1/6$.

The equality is therefore not a numerical coincidence between two molecules; it is the statement that a parabola takes each value twice, and the two bonds happen to be the pair that straddles its vertex.

The table has a fourth row and it is worth reading, with the caveat attached. A quadruple bond, on this arithmetic, gives bent components of exactly a quarter s character — the same as a single bond’s, which is the same coincidence repeated at the other pair of symmetric points. Carbon does not make quadruple bonds and the row is a formal extension rather than a chemical prediction; real quadruple bonds are made between transition metals, where d orbitals join the σ and π sets and the counting above does not apply.

What the row does establish is that the equality found here is not the only one available. Any two bond orders summing to five give bent hybrids of identical s character, and there are exactly two such pairs among the integers — which is why one equality was found and looked singular.

That is the useful form of the result. The bent components of a multiple bond have an s character fixed by the bond order alone, through a formula with no molecule in it, and the two cases this essay computes are the two cases the formula makes equal.

It also says where the s character is largest, which is not where a reader would guess. The denominator $n(5-n)$ is greatest at the middle of the range, so the bent components carry least s character for the double and triple bonds and most for the single one — the opposite of the trend in the σ hybrids, where s character rises with bond order from a quarter to a half. Two quantities that both describe the same atom’s use of one s orbital move in opposite directions as the bond order rises, because one of them is being divided among more components each time.

Which is worth carrying whenever a hybrid label is quoted. sp² and sp⁵ both describe carbons in ethene, in the same molecule, at the same instant, and they are not two opinions: the first names the σ hybrids and the second names the components of the double bond, and the second is a quantity the first has to be divided to obtain.

What the construction requires #

One sixth, exactly, for the triple bond’s bent component, and Coulson’s angle from it.

The equality with the double bond’s, to $10^{-12}$ in the s character and $10^{-9}$ in the angle.

And the refusal, which is the single bond: a quarter and 109.47122 degrees, so the equality is not an artefact of how the quantity was defined. If the construction had been arranged to give one sixth for anything, ethane would have given it too.

The threefold axis, measured rather than imposed: the three bent components put their charge equally far off the axis, to a millionth of an ångström, and that distance is computed from matrix elements of the position operator over the mixed functions rather than from the mixing coefficients.

And the orthogonality, which is what makes the whole thing a change of basis: every pair of bent components overlaps to zero within the quadrature’s precision, and the two exactly-zero overlaps — between σ and each π, which vanish by the mirror symmetry rather than numerically — are held to $10^{-12}$ instead.

Still open: an unequal double bond, and a ring #

The obvious open question is the bond with no symmetry to enforce equivalence. A carbonyl’s double bond joins two different atoms, so the two bent components are still available and are no longer equivalent; the mixing angle that maximises their localisation is not forty-five degrees and has to be searched for, and the two s characters are then different numbers. Whether they still average to a sixth is a question with an answer, and measuring how far a bond’s charge sits off an axis would answer it in the same calculation.

The nearer question is what happens to the coincidence in a ring. Cyclopropane’s carbons make three σ bonds each and its C–C–C angles are sixty degrees, which Coulson’s relation says no set of equivalent hybrids can produce: at any s character between zero and a half the angle is between ninety and a hundred and eighty. So the ring’s bonds are bent in a second and different sense — the hybrids point outside the internuclear lines — and the same relation that fixed the angle here says by how much. The angle a ring cannot have meets that limit from the other side.