What a missing orbital costs

The count of central orbitals a set of ligands can reach needs no character table. In a σ-only model each ligand couples to the centre through a row $(1, x_i, y_i, z_i)$, and the number of orbitals reached is the rank of those rows: four, unless the ligand directions lie in one plane. The Gram matrix of the rows carries a continuous version of the same fact — its smallest eigenvalue goes to zero exactly when an orbital becomes unreachable — and along the path that flattens a tetrahedron into a square it falls to zero where the repulsion energy has its maximum.

That coincidence invites a reading: a small eigenvalue is a nearly-missing orbital, a nearly-missing orbital is a weak bond, and the arrangements chemistry avoids are the ones with small eigenvalues. It is a claim about energy, and the model the count came from can compute energy.

It does not bear the reading out, and the reason is exact.

A bond energy for the model the count belongs to #

The model is the count’s own. There are n ligand σ functions at a common energy, taken as zero. The centre has one s orbital a height $\Delta_s$ above them and three p orbitals at $\Delta_p$, and ligand $i$ couples to them through $\beta S_s$ and $\beta S_p\,\mathbf u_i$, where $\mathbf u_i$ is its direction. Every ligand-based level is doubly occupied — the electron-rich case that the three-centre account of hypervalency is about, in which there are more filled ligand combinations than central orbitals to receive them.

The bond energy is twice the sum of the n lowest eigenvalues of that matrix, and it can be computed exactly for any arrangement. Here the s level sits one unit above the ligands and the p levels one and a half, and the two overlaps are equal; those choices are stated because the result depends on one of them, in a way the essay will name.

It also has a perturbation series, and the series says what the geometry is allowed to enter through before anything is diagonalised.

Second order has no geometry in it #

At second order each ligand level is lowered by the sum of its squared couplings divided by the gaps. Summed over all n ligands, the p part of that is $S_p^2\sum_i |\mathbf u_i|^2/\Delta_p$, and a direction is a unit vector, so the sum is n.

So at second order every arrangement of n ligands has the same bond energy — the tetrahedron, the square, the seesaw, a flat ring. The quantity that enters is the trace of the Gram matrix, and the trace is the sum of the eigenvalues. A zero eigenvalue is paid for by the other three being larger. Nothing is lost; it is moved.

The figure is the proof of it in numbers. At weak coupling every arrangement’s excess over the tetrahedron rises as the fourth power of the coupling: nothing at second order, the first difference at fourth. The dotted line is the fourth-order prediction for the square, and the exact curve lies on it until the coupling reaches about a third.

That removes the naive version of the reading outright. An orbital that the ligands cannot reach is not bonding energy withheld at the order where bonding energy lives. Whatever a missing orbital costs is a higher-order effect.

What fourth order sees #

At fourth order the energy picks up the square of the central Gram matrix, and when the terms are collected there are exactly three: one that depends only on n, and two that depend on the geometry.

$$E^{(4)} = 2\beta^4\left[\frac{n^2 S_s^4}{\Delta_s^3} + c_1\Big|\sum_i \mathbf u_i\Big|^2 + c_2\sum_{ij}(\mathbf u_i\cdot\mathbf u_j)^2\right]$$

with the two geometric coefficients

$$c_1 = S_s^2S_p^2\left(\frac{1}{\Delta_s^2\Delta_p} + \frac{1}{\Delta_p^2\Delta_s}\right),\qquad c_2 = \frac{S_p^4}{\Delta_p^3}.$$

The first geometric term is the squared net direction of the ligands: zero for any arrangement that points nowhere on average, large for one that leans. The second is a sum of squared cosines between every pair, and it is bounded below: $\sum_{ij}(\mathbf u_i\cdot\mathbf u_j)^2 \ge n^2/3$, with equality exactly when $\sum_i \mathbf u_i\mathbf u_i^\mathsf T$ is a multiple of the identity. The excess over $n^2/3$ is the ligands’ departure from a tight frame — from treating the three directions of space equally.

Both coefficients are positive, so both terms cost energy. The arrangement the bond energy prefers at leading order is the one with no net direction and no frame excess: a centred tight frame, which is what geometers call a spherical 2-design — a set of points whose averages of every polynomial up to degree two match the sphere’s.

The check that this is the energy’s series and not a fit is that what it leaves behind is sixth order. On every census arrangement the residual falls by a factor between fifty and seventy when the coupling is halved — sixty-four is the sixth power of two — and nothing in the series was adjusted to the exact numbers.

Two moments, at every order #

The series says geometry enters through two quantities at fourth order. A stronger statement is true, and it needs no series.

The n ligand functions couple to four central orbitals and to nothing else, so the coupling is an n-by-4 matrix V. Every ligand combination orthogonal to V’s columns — there are $n - \operatorname{rank} V$ of them — does not couple to the centre at all and stays exactly at the ligand energy. Those are the orphans the count has been counting, and here they are exact: every census arrangement has precisely $n - \operatorname{rank}$ levels at zero to the ninth decimal. Everything else about the spectrum is fixed by the 4-by-4 matrix $V^\mathsf T V$, whose entries are $n$, the ligands’ summed direction $\sum_i\mathbf u_i$ and their second-moment tensor $\sum_i\mathbf u_i\mathbf u_i^\mathsf T$.

So the σ-only energy depends on the ligand directions through their sum and their second moment, and through nothing else, at every order in the coupling and for any number of electrons. Two arrangements that share those two moments are, to this model, the same molecule.

The octahedron has a twin. A trigonal prism whose six ligands sit at the tetrahedral magic angle from its axis, 54.74°, has zero summed direction and a second moment of exactly two in every direction, as the octahedron does — and its σ energy equals the octahedron’s at every coupling tried, from a tenth to three, to the last digit a double carries. So does every level of the model, filled or empty. The census’s trigonal prism, with its ligands at a different angle, has a different second moment and a higher energy.

Nothing about the two shapes looks alike. The octahedron has three orthogonal axes; the prism has one threefold axis and two eclipsed triangles. What the σ model sees is the average, and the averages are the same.

With the s and p levels at one height the statement becomes a closed form: each reached combination is then a two-level problem with the Gram matrix’s eigenvalue $\lambda_k$ as its squared coupling, and the energy is $\sum_k [\Delta - \sqrt{\Delta^2 + 4\beta^2\lambda_k}]$. That formula does contain the smallest eigenvalue — as one of four terms, each a concave function of its eigenvalue, whose sum is fixed. At fixed trace a concave sum is largest when its arguments are equal and smallest when they are spread, which is the fourth-order result again, now at every order: what costs energy is an uneven spectrum, and a zero is one way among many to have one.

The frame, not the orbital #

The census arrangements, sorted by their exact bond energies at weak coupling, come out in the order of the fourth-order cost in every group, with no exception. The tetrahedron and the octahedron are 2-designs and cost nothing; they are the most stable arrangement of their ligand count at every coupling computed. The trigonal bipyramid is not a 2-design — its axial pair and equatorial triple do not quite average like a sphere — but it comes closest of the five-ligand arrangements, with a cost of 0.099 against the square pyramid’s 0.144.

The smallest eigenvalue does not order them. The pentagonal pyramid has an eigenvalue of 0.19 and the trigonal prism 0.25, so by the reading the pyramid is closer to missing an orbital and should bond more weakly. It bonds more strongly, at every coupling computed, because its frame excess is 1.08 against the prism’s 1.50. The eigenvalue measures the weakest direction; the energy, at the order where geometry first enters, measures the spread of all of them.

A square with an orbital missing beats a seesaw #

The sharpest case is two arrangements of four ligands. The square is flat, reaches three central orbitals and leaves the $p_z$ untouched: its smallest eigenvalue is zero. The seesaw reaches all four, with an eigenvalue of 0.065.

At weak coupling the square is the more stable. Its frame excess is large, 2.67, because it has no extent along one axis at all — but its net direction is zero. The seesaw’s frame excess is smaller, 1.05, and its net direction is 1.12: its ligands lean. With the s level lower than the p levels, the net-direction term carries the larger coefficient, and the seesaw pays more for leaning than the square pays for being flat.

As the coupling grows the ordering reverses, at β = 0.589. By then the reached orbitals are saturated — each is mixed so strongly with the ligand combination it receives that pushing more coupling into it returns less — and an arrangement that spreads its coupling over four orbitals gains on one that concentrates it into three. That is the regime in which the reading would have been right. It is not the regime the count was about, and it is not where a σ-only model’s second-order intuitions come from.

Where the answer depends on the model #

One number in the model decides which arrangement wins at weak coupling, and it is how far the p levels sit above the ligands compared with the s level, $r = \Delta_p/\Delta_s$. With equal overlaps the ratio of the two geometric coefficients is $r^2 + r$, and the square beats the seesaw at weak coupling exactly when that ratio exceeds the ratio of their moments, 1.438.

So the square wins at weak coupling whenever the p levels sit above 0.80 of the s level’s height — which, for a main-group centre whose p orbitals lie above its s, is every case. The coupling at which it gives way grows with $r$: 0.27 when the two sit level, 0.59 at one and a half, 1.41 at three. That is the dependence the model’s constants were stated for. The ordering of the 2-designs does not depend on them at all, since a 2-design has both geometric terms at their minimum whatever the coefficients are.

The flattening path could not have told them apart #

The coincidence that prompted the question survives, and it is worth seeing why it misled. Along the path that flattens a tetrahedron, the bond energy lost, the frame excess and the smallest eigenvalue all turn at the square, and the energy lies almost exactly on the frame excess. The eigenvalue goes the other way, falling to zero.

On a path with one parameter and a symmetry at its midpoint, any two quantities that turn at the midpoint are functions of each other, and nothing measured along the path can say which one the energy is following. The eigenvalue looked like an explanation because it was the quantity the count had just introduced. The census, which visits arrangements that no single path connects, is what separates them — and it separates them cleanly: the cost orders every group, the eigenvalue fails in two of the three groups with four ligands or more.

Where the 2-designs came from before #

The spherical 2-design is not new to this argument. The icosahedron’s indifference to how its half-filled shell was oriented was traced to its vertices forming a spherical 5-design, so that a degree-four localisation functional could not tell one orientation from another. Here the same object appears one degree lower and for a different reason: a σ bond energy’s fourth-order term is a degree-two polynomial in each pair of directions, and a 2-design is the set that averages every such polynomial exactly as the sphere does.

That is the surprising connection in this essay, and it is the kind this argument keeps finding. A property usually quoted as a fact about a shape — the octahedron is the best arrangement of six — turns out to be a fact about which averages the shape gets right, and the degree of the average is set by the order of the theory.

What this model leaves out #

There are no d orbitals and no π. The central atom has s and p only, and the ligands couple through σ alone. That is the model the orphan count lives in, and it is the one the three-centre account of hypervalency uses; it is not a model of any real molecule’s shape.

There is no repulsion between ligands. The bond energy here is the whole energy, so it cannot say what a real molecule adopts. It says what the bonding alone prefers, which is the question the eigenvalue reading asked.

Every ligand level is filled. With fewer electrons the lowest levels would be central, and the geometry would enter at second order through which levels are occupied — the familiar Walsh-diagram situation, and a different question.

The overlaps are equal. With $S_p \ne S_s$ the coefficient ratio becomes $(S_s/S_p)^2(r^2 + r)$, and the boundary moves with it; the 2-design result does not.

How it was computed #

Each census arrangement’s bond energy is twice the sum of the lowest n eigenvalues of the $(n+4)$-dimensional matrix, found by Jacobi diagonalisation, at couplings from 0.03 to 3. The moments are computed from the unit ligand vectors directly, and the fourth-order series from them with no fitting. The checks, made wherever these figures are drawn: the residual after fourth order falls as the sixth power on every arrangement; second order is identical within every ligand count; the tetrahedron and octahedron are 2-designs and the most stable of their counts at every coupling; the square beats the seesaw below a coupling between one half and seven tenths; the pentagonal pyramid beats the trigonal prism at every coupling; and the fourth-order cost orders every group at weak coupling. The magic-angle prism shares every level of the model with the octahedron; every arrangement has exactly $n - \operatorname{rank}$ levels left at zero; and with s and p level the exact energy equals the closed form in the Gram eigenvalues. The refusal is a single ligand, which has no geometry and must give the closed-form two-level energy.

Who counted what #

Perturbative arguments for molecular shape go back to Walsh’s diagrams of the 1950s and their second-order Jahn–Teller refinements, which are about which levels are occupied. The electron-rich three-centre model is Rundle’s and Pimentel’s. Spherical designs were defined by Delsarte, Goethals and Seidel in 1977, and the frame inequality behind the $n^2/3$ bound is older.

What is computed here is the σ-only bond energy of every census arrangement, its series to fourth order in closed form with the two geometric moments identified, the reversal between the square and the seesaw and where it moves with the model, and the comparison of the fourth-order cost with the Gram eigenvalue as predictors of the energy.

Still open: what π adds, and the shapes the model cannot tell apart #

The obvious open question is π. A ligand π function adds two rows per ligand, perpendicular to the bond, and a flat ring of π donors gives the unreached $p_z$ orbital back: for a ring in the xy plane every out-of-plane π function points along z. Whether π also changes which arrangement the energy prefers is a series of the same shape with more terms, and the π rows bring their own net direction and frame excess — which could favour a flat arrangement outright.

The nearer question is which arrangements are twins. Two moments are nine numbers and a direction set is 2n, so for every ligand count past four there is a continuum of shapes the σ model cannot tell apart, of which the octahedron and the tilted prism are one pair. Mapping that continuum for five ligands — where the trigonal bipyramid is not a 2-design and has no reason to be the only shape with its moments — would say how much of what a σ-only model predicts about shape is a prediction about averages, and therefore how much has to come from something the model leaves out.