A filled shell has no shape

A great many orbitals are drawn in these essays, each at a contour enclosing a stated fraction of its density, and every one of them has lobes. It is worth asking what happens to the lobes when the shell they belong to is full.

The answer is that they cancel exactly.

Sum the squared angular parts over a complete shell of angular momentum $l$ and the result is

$$\sum_{m} |Y_{lm}(\theta,\phi)|^2 = \frac{2l+1}{4\pi}$$

which is a constant. Not approximately a constant, not a constant to within the accuracy of some approximation — a constant, independent of direction, exactly. That is Unsöld’s theorem, and computing it over a grid of 768 directions with real spherical harmonics gives a worst departure of $1.1\times10^{-16}$ for the p shell and $4.4\times10^{-16}$ for the d shell.

A filled p shell is spherical. A filled d shell is spherical. A filled f shell is spherical. Whatever the pictures look like, the density of a closed shell has no shape at all.

Why it has to be true #

The theorem sounds like arithmetic luck and is a consequence of something more basic.

A complete shell is a complete set of functions for one value of $l$, and the shell as a whole has to be invariant under rotation — because the set of $2l+1$ functions transforms among itself under any rotation, and a rotation is unitary, so the sum of the squared moduli is preserved.

Turn a p shell by any angle and the three functions become three different combinations of themselves. The sum of their densities is unchanged, by unitarity. And a function that is unchanged by every rotation is a constant.

So Unsöld’s theorem is the statement that a complete shell has no preferred direction, which was obvious before any function was written down. What is not obvious, and is worth the computation, is that the cancellation is exact rather than approximate — the three lobed densities in the figure above add to a constant at every point, with the maxima of one filling the nodal planes of the others precisely.

The consequence for what an orbital picture is #

This site is built on the claim that an orbital picture is worth drawing carefully, and it is worth being exact about what such a picture is a picture of.

It is a picture of a function. The function is one member of a set that spans a shell, and which member it is depends on the basis chosen for the shell.

It is not a picture of a measurable density, when the shell is full. The density of a neon atom is spherical; the density of an argon atom is spherical; every noble gas is spherical, and no experiment on one can detect a lobe, because there is nothing there to detect.

Real and complex harmonics makes the same point from the choice of basis: the complex set $m = -1, 0, +1$ has densities that are rings and a dumbbell, and the real set has three dumbbells, and the two are different bases for one shell. Unsöld says that both sum to the same constant, which is the strongest possible statement that the choice is a choice.

That is not an argument against drawing orbitals. It is an argument for saying what they are, which is the habit of say what it encloses continued in a different currency: a contour of a one-electron function, drawn at a stated fraction of that function’s own density, and not a picture of where the electrons in a closed shell are.

The refusal: a shell one short is not spherical #

A check that only ever sees the complete shell would pass on anything, so the check also removes one function and requires the constancy to break.

It breaks decisively. Take the five d functions and leave out $\mathrm{d}_{z^2}$: the angular density then runs from 0 along the z axis to 0.398 in the xy plane — a spread as large as the whole constant, and a density with a hole through it. Leave out $\mathrm{d}_{xz}$ instead and it runs from 0.265 to 0.398, a spread of a third.

Which function is removed changes the shape and the size of the departure, which is itself a statement: the deficiency has a direction, and the direction is the one belonging to the missing function.

That is where all of coordination chemistry lives. The splitting is a symmetry statement begins from a d shell that is not full, whose density therefore has a shape, and whose shape is what a set of ligands pushes on. A filled d shell would feel no ligand field at all in the sense that matters — its density is spherical, so no arrangement of external charges can lower its energy relative to another arrangement.

That is exactly why $\mathrm{d}^{10}$ ions have no ligand-field stabilisation energy, why zinc(II) shows no preference between geometries where nickel(II) shows a strong one, and why the ligand-field stabilisation curve across a transition series returns to zero at both ends.

The half-filled shell, which is spherical again #

There is a second case where the angular density comes out constant and it is not a filled shell, which is worth separating because it is the one behind a familiar rule.

A half-filled shell with all its electrons of the same spin has one electron in each of the $2l+1$ functions, so the density of that spin is the same sum as before and is again the constant $(2l+1)/4\pi$. A high-spin $\mathrm{d}^5$ ion has a spherical d-electron density; so does a $\mathrm{p}^3$ atom such as nitrogen.

That is the arithmetic behind the special stability usually attributed to half-filled shells, and it is worth being careful about how much it explains. It explains why such an ion has no ligand-field stabilisation energy — a spherical density cannot be lowered by rearranging external charges, so high-spin $\mathrm{d}^5$ sits at zero on the stabilisation curve exactly as $\mathrm{d}^0$ and $\mathrm{d}^{10}$ do. That curve is computed here and the three zeroes are on it.

It does not explain the exchange stabilisation, which is a many-electron quantity about parallel spins and is the larger part of the usual argument. Sphericity is a statement about the density; exchange is a statement about the wavefunction, and the hole that is not repulsion is where this site treats the difference.

So half-filled shells are spherical for the same reason full ones are, and are stable for an additional reason they do not share.

Sphericity is angular and not radial #

One qualification, and it is the reason atoms have any structure at all.

Unsöld’s theorem is about the angular part. The radial part is untouched, and a filled shell’s density still varies enormously with distance from the nucleus — it has a maximum, it decays, and where there are radial nodes it has minima.

So a neon atom is spherical and layered. Its total density, summed over the 1s, 2s and 2p shells, is a function of $r$ alone with structure in it: a sharp peak from the 1s near the nucleus and a broader one from the second shell further out. The radial distribution across the table computes exactly that structure, and it survives the theorem completely.

The distinction is worth keeping because the two are often collapsed into one word. The shells are radial; the lobes are angular. Filling a shell destroys the lobes and leaves the shells, and every chemical property that depends on how far out an atom’s density reaches is a radial property and is unaffected by anything in this essay.

The molecular version, where the moments vanish #

An atom with only closed shells has a spherical density, and a spherical density has no multipole moments of any rank — no dipole, no quadrupole, nothing.

The molecular analogue is a molecule whose group forbids the low moments, and that case is computed here: methane’s Td group forbids both the dipole and the quadrupole, so its first non-vanishing moment is the octupole. The two are the same kind of statement — a symmetry high enough to kill a whole set of moments — arrived at from a group in one case and from a completed sum in the other.

The comparison also shows the limit of the atomic result. A closed-shell atom has every moment zero; a closed-shell molecule has only the moments its group forbids. Sphericity is a much stronger condition than any finite group can impose, which is why the atomic case is exact for all ranks and the molecular case has to be checked rank by rank.

The molecular version of the same idea is a table already computed: methane’s group kills both its low moments and the row is zeroes, while three other molecules keep one. An atom with closed shells only would have a row of zeroes for the same reason and by a much shorter argument — it needs no group at all, only the sum over a shell.

Why the cancellation is exact rather than close #

It is worth looking at how the three p densities manage to add to a constant, because the mechanism is prettier than the statement.

Write the three real functions in terms of direction cosines: $\mathrm{p}_x$, $\mathrm{p}_y$ and $\mathrm{p}_z$ have angular parts proportional to $x/r$, $y/r$ and $z/r$. Their squared sum is proportional to

$$\frac{x^2 + y^2 + z^2}{r^2} = 1$$

which is the whole proof. The cancellation is the Pythagorean identity, and it happens at every point rather than on average.

The d shell is the same trick one rank up. The five real functions’ angular parts are quadratic forms in the direction cosines, and the sum of their squares is proportional to a symmetric combination that reduces to $(x^2+y^2+z^2)^2/r^4$, which is again one.

That is why the departure computed over a grid is $10^{-16}$ rather than $10^{-6}$. There is no integral being evaluated, no series being truncated and no grid resolution entering: the identity holds algebraically at each direction, and the number the computation returns is the accumulated rounding of a few multiplications.

Getting an exact zero out of a numerical calculation is unusual enough to be worth flagging when it happens. It happened once before, for exactly zero — an overlap that symmetry forbids, coming out at $10^{-17}$ — and the reason is the same in both cases: the cancellation is in the integrand or the identity rather than in the arithmetic.

What a contour of a total density would look like #

Since these essays are about contours, the obvious question is what a contour of the total density of a closed-shell atom looks like, and the answer is short: a sphere, of a radius set by the fraction chosen.

That is worth saying because it makes clear what the orbitals index is a list of. Every contour on it is a contour of a one-electron function at a stated enclosed fraction, and none of them is a contour of anything a measurement returns for a filled shell. The index is a catalogue of the one-electron basis, with the level each stated fraction requires.

The one place where a total-density contour has a shape is where the shell is not full — an atom in the middle of a transition series, a molecule with a partly filled valence shell, an excited state. Every one of those is a case where the site’s pictures and a measured density would agree about there being a direction, and would still disagree about how many lobes there are, because the decomposition into orbitals is not something a density carries.

What an experiment on a filled shell actually shows #

If the density of a closed-shell atom is a sphere, it is fair to ask what the experiments that seem to show orbital shapes are showing.

X-ray diffraction on a crystal measures the electron density, and difference maps on molecular crystals do show non-spherical features around atoms. Every one of those atoms is bonded, which means its valence shell is not full: the features are the deformation of the density from a spherical reference, and the reference is spherical for exactly the reason above.

Scanning probe images of molecules on surfaces are maps of a tunnelling current, which depends on the density of states at the surface at a chosen energy, not on the total density. What they resolve is one orbital or a narrow range of them, which is a slice through the decomposition rather than the density itself.

Photoelectron angular distributions measure the angular dependence of ejected electrons, which does carry information about the orbital the electron came from — because the measurement selects one ionisation energy and therefore one orbital, breaking the sum.

The pattern in all three is the same. An experiment sees a shape when it selects part of the shell. Total density, whole shell: a sphere. Any selection — by energy, by bonding, by which electron was removed — and the shape is back.

That is a satisfying resolution rather than a debunking. The pictures here are pictures of the objects those experiments select, drawn at a contour that says what fraction of the selected object they enclose.

The theorem as a list of exact zeros #

The section above answers the objection — what the experiments that seem to show lobes are showing — and leaves the theorem’s own predictions unstated. They are worth listing, because a theorem of this kind predicts zeros rather than values, and a zero is the strongest thing a measurement can be asked to check.

A closed-shell atom has no electric quadrupole moment. A quadrupole moment is an integral of the density against a second-order angular function, and integrating a constant against one of those gives nothing. The prediction is exactly zero rather than small.

It produces no electric field gradient at its own nucleus. So a nucleus that has a quadrupole moment of its own — and most do — shows no quadrupole coupling in a closed-shell atom, and the resonance that would be split is a single line. Put the same nucleus in a bonded environment, where the valence shell is no longer full, and the splitting appears; its size is a measure of how far from full the shell is.

And its polarisability is isotropic. A field applied in any direction produces the same response, so a noble gas scatters light without depolarising it. An open-shell atom or a molecule does depolarise, and the depolarisation ratio measures the anisotropy directly.

All three are consequences of one constant, and all three are checked constantly without anybody calling them checks of Unsöld’s theorem — they are the reason a noble gas is the boring case in every one of those measurements.

Which is the useful way to hold the result. The lobes are not hidden in a closed shell in the way a small effect is hidden below a detection limit. They are absent, by an identity, and three independent kinds of instrument return zero because there is nothing there. The pictures on this page are of a decomposition, and the decomposition is a device; what an instrument can find is exactly what survives the sum, and for a full shell that is a number and a radius and no direction at all.

What a full shell adds to the contour argument #

The essays on contours establish what a contour is, how big an orbital is by four different measures, and what a single level means when several orbitals are compared at it. This asks what survives when the shell is complete, and the answer is nothing angular at all.

Two things follow and they pull in opposite directions.

The pictures are less physical than they look. No closed-shell system has lobes, and a picture of one is a picture of a basis function.

The pictures are more useful than that makes them sound. Chemistry happens in incomplete shells, where the density does have a shape, and the shape is the one the pictures show. A d shell missing one electron is not spherical by a margin of a third; missing three, by a margin as large as the constant itself.

The next step would be the total density of a real atom drawn as a contour — spherical for the noble gases and not for anything else — which needs a many-electron density not computed here. What can be done is the theorem, exactly, and record that the departure from it is $10^{-16}$ where the shell is full and $0.4$ where it is one function short.