Field

Orbitals

A one-electron wavefunction, drawn as a contour surface at a level somebody chose. Nodes, signs, and what the picture is a picture of.

The 2pz orbital. The 2pz orbital at the contour enclosing 90 per cent of its density — a level solved for by integration rather than chosen. The two colours are the two signs of the wavefunction, which is what distinguishes a bonding interaction from an antibonding one. Contours drawn: 2pz at 90% of its density, |ψ| = 9.48e-3.

What an orbital is

Not a region the electron occupies, not a path it follows, and for any atom but hydrogen not an exact anything. An orbital is a one-electron wavefunction, and almost every difficulty in this subject comes from forgetting that.

$Choosing a contour for 3s. The fraction of the density enclosed by a contour, against the contour's level. Picking a level is picking a point on this curve, and the usual practice of picking one that looks right is picking a point without knowing which. Contours drawn: 3s at 50% of its density, |ψ| = 6.24e-3; 3s at 90% of its density, |ψ| = 2.64e-3; 3s at 99% of its density, |ψ| = 7.15e-4.$

Say what it encloses

An orbital picture is a contour at a level somebody chose, and almost no source says which. Two textbooks can draw the same orbital at visibly different sizes with the same caption, and both be printed in good faith.

The radial function of 3s. The radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.

Nodes

An orbital with quantum numbers n and l has exactly n−l−1 radial nodes and l angular ones. That is a count, it is exact, and it is the fastest way to catch a drawing that is wrong.

The radial function of 1s. The radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.

Where the electron is

The wavefunction is largest at the nucleus, the electron is most likely to be found a bohr out, and the ninety-per-cent contour is at 2.66. Three numbers, all correct, all answering different questions.

The radial distribution across the periodic table

A 4s orbital is bigger than a 3d by every measure of size except the one that decides which fills first. Penetration is a feature of a small inner peak, and the periodic table's shape depends on it.

$Orbitals at the 90 per cent contour. Several orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density. Contours drawn: 2pz at 90% of its density, |ψ| = 9.48e-3; 2px at 90% of its density, |ψ| = 9.49e-3; 2py at 90% of its density, |ψ| = 9.49e-3.$

Complex harmonics against real ones

The p orbitals every chemist draws are not eigenfunctions of anything. They are real combinations of the complex solutions, chosen because they point along axes — and the choice is invisible until a magnetic field makes it matter.

4s and 3d from Sc to Zn. The mean radius of the 4s and 3d orbitals across the elements Sc to Zn, at the nuclear charge each shell actually feels. The screening is Slater's, which is fitted; the radius that follows from it is the closed form for a hydrogen-like orbital, which is not.

What an electron actually feels

A 3d orbital is less than half the size of the 4s beside it and fills second anyway. The charge an electron feels is not the nuclear charge, the correction is a fit rather than a derivation, and the two facts together explain the shape of the periodic table.

Four measures of size for 6 orbitals. The most probable radius, the mean radius, the root-mean-square radius and the radius of the sphere holding ninety per cent of the density, for 1s, 2s, 2pz, 3s, 3dz2, 4s. All four are computed from the same radial function, all four are correct, and they are not the same number.

How big is an orbital

Four measures of size, all computed from the same radial function, all correct, and spanning a factor of two and a half for a 1s orbital. The one that governs chemistry is a fifth, and it is not a measure of size at all.

$1s, 2s, 2pz, 3s, 3dz2, 4s at one level and at one fraction. The orbitals 1s, 2s, 2pz, 3s, 3dz2, 4s, each with the contour level that encloses 90 per cent of its own density, and beside it what each one encloses when they are instead all drawn at 1s's level. The first column is a set of pictures that can be compared; the second is what a plate drawn at a single contour value actually shows. Contours drawn: 1s at 90% of its density, |ψ| = 3.94e-2; 2s at 90% of its density, |ψ| = 7.40e-3; 2pz at 90% of its density, |ψ| = 9.48e-3; 3s at 90% of its density, |ψ| = 2.64e-3; 3dz2 at 90% of its density, |ψ| = 3.60e-3; 4s at 90% of its density, |ψ| = 1.24e-3.$

One level is not one comparison

A plate of orbitals drawn at a single contour value looks like a comparison and is not one. At the level that encloses ninety per cent of a 1s, a 2s encloses four per cent, a 3s under one, and a 3d has no surface at all — its wavefunction never reaches that value anywhere in space.

2s and 2p from B to Ne. The mean radius of the 2s and 2p orbitals across the elements B to Ne, at the nuclear charge each shell actually feels. The screening is Slater's, which is fitted; the radius that follows from it is the closed form for a hydrogen-like orbital, which is not.

What the screening model cannot see

Slater's rules put the 2s and the 2p in one group, so they give both orbitals exactly the same effective nuclear charge at every element from boron to neon. The two are separated by several electronvolts in all six, and the model has no term that could produce it.

$Orbitals at the 90 per cent contour. Several orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density. Contours drawn: 2px at 90% of its density, |ψ| = 9.49e-3; 2py at 90% of its density, |ψ| = 9.49e-3; 2pz at 90% of its density, |ψ| = 9.48e-3.$

A filled shell has no shape

Sum the angular densities of a complete p shell and the answer is 3/4π in every direction, to sixteen decimal places. A filled d shell gives 5/4π. The lobes are in the decomposition and not in the density, and nothing that measures a closed-shell atom can see them.

The exponent the molecule chooses, and what it buys. The 1s exponent that minimises the energy of a one-electron diatomic, against the separation of the nuclei, with the binding curves at that exponent and at the free atom's. Held at ζ = 1 the bond comes out at 2.49 bohr and binds 0.0648 hartree; with the exponent free it comes out at 2.00 bohr at ζ = 1.238 and binds 0.0865. The exact answer for this molecule is 2.00 bohr and 0.1026.

The atom does not bring its own orbital

Build a one-electron diatomic from two hydrogen 1s functions and it comes out 25 per cent too long and 37 per cent too weakly bound. Let the molecule choose how large those functions are and the bond length is right to three figures, at an exponent of 1.238 — the orbital contracts by a quarter when the bond forms.

$One isovalue, many fractions — and one fraction, many isovalues. What each of 4 conventional isovalues encloses, for 7 orbitals of hydrogen, and in the last column the level each one needs to enclose 90 per cent. At 0.02 atomic units the fractions run from 0.4 to 96.2 per cent, and the levels in the last column differ by a factor of 32. A plate of orbitals drawn at one value is not a comparison of sizes. Contours drawn: 1s at 90% of its density, |ψ| = 3.94e-2; 2s at 90% of its density, |ψ| = 7.40e-3; 2pz at 90% of its density, |ψ| = 9.48e-3; 3s at 90% of its density, |ψ| = 2.64e-3; 3pz at 90% of its density, |ψ| = 3.03e-3; 3dz2 at 90% of its density, |ψ| = 3.60e-3; 4s at 90% of its density, |ψ| = 1.24e-3.$

The isovalue nobody chose

Every program that draws an orbital asks for a number, and almost every user accepts the default. At 0.02 atomic units that default encloses 96 per cent of hydrogen's 1s, 52 per cent of its 2s and four tenths of one per cent of its 4s — three pictures drawn to one rule, meaning three completely different things.

The wrong shape, fitted as well as it can be. The exact hydrogen 1s orbital and the best sums of one, two, three and six Gaussians, each with its exponents optimised for the energy. Three of them already reproduce the exact function to 99.94 per cent by overlap, which is why the method works at all — and the two places it goes wrong, at the nucleus and far out, are exactly where the other faces of this figure look.

A Gaussian is the wrong shape

Sixty years of molecular calculation are built on functions that get the two ends of an orbital wrong. A Gaussian has no cusp at the nucleus and dies too fast far away, and no number of them fixes either — while three of them already reproduce hydrogen's 1s to better than 99.9 per cent by overlap, and that is why the method works.

The cost of a ratio decided somewhere else. Four bases containing exactly the same six primitives, differing only in how many of the linear coefficients the calculation may choose. The horizontal axis is the effective nuclear charge, which is this one-electron problem's only knob for a different environment; the contraction was fitted at one. At 1.238 — the exponent H₂⁺ chooses when a bond forms — the fully contracted basis is 0.0283 hartree above what the same six functions could give, and one freed coefficient removes most of it.

A contraction is a decision made once

Every published basis set freezes its primitive functions into fixed combinations, on an isolated atom, before any molecule is in sight. Freeing one coefficient recovers three quarters of what that costs — and freeing it at the other end of the basis recovers one per cent.

The surface-honest curve and the page-honest curve. The 2pz orbital in its xz plane, drawn twice. The outer curve is the contour whose SURFACE encloses 90.0 per cent of the density in space — the claim every orbital figure here makes — and in this plane it encloses 96.7 per cent. The inner curve is the one that encloses 90 per cent in the plane, which is what a reader looking at a flat picture would take the caption to mean. The two levels differ by a factor of 1.87 and the curves differ visibly; the caption does not. Contours drawn: 2pz at 90% of its density, |ψ| = 9.48e-3.

A slice is not the surface

The page is flat, so every printed orbital is a slice through a contour rather than the contour itself — and a curve that leaves a tenth of the density outside it in space leaves about a thirtieth outside it on the page. Both claims are true of the same picture and only one of them is ever stated.

Where the electron is, and how fast it is going. The radial distribution in position on the left and in momentum on the right, for the same orbitals. The two run opposite ways: the 1s is the most compact in space and the widest in momentum, and every excited orbital that spreads out in one narrows in the other. Both are normalised, both are the same function, and neither is more fundamental than the other — the transform loses nothing and adds nothing.

The orbital in momentum space

Every orbital has a second picture as complete as the first and almost never drawn. Nothing is added by taking it — it is the same function in the other variable — but the uncertainty product falls out of it, and the functions quantum chemistry is built from turn out to be the only ones that attain the bound.

The cliff, and the slope leading to it. The smallest eigenvalue of the overlap matrix, and what an extra function is worth, as that function is brought towards one already in the basis. Both fall together, and the energy stops improving long before the matrix stops being invertible.

A function that is already there

Adding a function to a basis set can only lower the energy, so a bigger basis is a better one. What that leaves out is that the same act makes the functions less independent: an optimised basis's smallest overlap eigenvalue halves with every function added, and a function placed on top of one already present buys less than a millionth of what a well-placed one buys while driving that eigenvalue to 4×10⁻¹⁰ — past which the calculation is refused outright.

The measurement a basis was not fitted to

Six Gaussians reproduce hydrogen's energy to eleven parts in a hundred thousand and its Compton profile to three parts in a thousand — twenty-five times worse, on a quantity an X-ray scattering experiment measures directly. The gap between the two errors widens as the basis is improved, because the energy is the one property a variational fit is best at.

$A σ contour at 90 per cent, and the two atomic ones. The section through both nuclei of the surface enclosing 90 per cent of the bonding orbital's density at 2 bohr, with circles marking where two atomic contours of the same stated fraction would be. The two pictures are different shapes and enclose different amounts, and the atomic pair encloses 91.70 per cent of the molecular orbital's density. Contours drawn: 1s at 90% of its density, |ψ| = 3.94e-2.$

A bond is not two atoms overlapping

The surface enclosing ninety per cent of a σ orbital's density is one closed surface with both nuclei inside it, at a level of 0.0359. The two atomic surfaces usually drawn instead sit at 0.0394, are a different shape, and enclose 91.70 per cent of the same orbital — and whether the molecular one is one object or two is decided by the fraction the caption claims, anywhere between 2.1 and 5.2 ångström.

One of these is guaranteed to improve, and it is not the one anybody measures. The relative error in the energy and in four properties of the fitted function, against the number of Gaussians. The energy falls at every step, because that is what the variational principle promises. The mean radius is exact for the single-function basis and 332 thousand times worse for the two-function one, and the density at the nucleus is still 6.2 per cent wrong where the energy is wrong by 0.011 per cent.

The property that gets worse

One Gaussian fitted to hydrogen gets the mean radius exactly right — 1.500000, against an exact 1.5. Two Gaussians get it wrong by 1.4 per cent, which is three hundred and thirty-two thousand times further out, while the energy improves fivefold. The variational principle bounds one number and says nothing whatever about any other, and the sequence of errors in everything else need not even be monotone.

The tenth that is not drawn

A ninety per cent contour of a hydrogen 1s orbital is a sphere of radius 2.661 bohr, and two of them stop touching at 5.322 bohr — where the overlap between the two orbitals is still 0.0768 and rising in importance. At a three-ångström contact, 36.9 per cent of the overlap integral lies outside both drawn surfaces, and holding nine tenths of it inside the picture would take a contour enclosing 97.28 per cent.

Where two He atoms stop, with no contact distance put in. The repulsion between two He atoms, computed from the overlap of their filled valence orbitals — four electrons in a bonding and an antibonding pair, of which the antibonding one rises further — against London's dispersion attraction from the measured polarisability and ionisation energy. The minimum is at 3.14 ångström where the tabulated van der Waals contact is 2.8, and nothing anywhere in the calculation is a length.

The radius that was tabulated

A van der Waals radius is a fitted number that every structural argument in chemistry uses. Computing it instead — a repulsion taken from computed overlap integrals, an attraction taken from two measured scalars, and no length anywhere — puts helium's contact at 3.140 ångström against a tabulated 2.80, neon's at 3.226 against 3.08 and argon's at 3.996 against 3.76. And the surface two atoms actually stop at encloses 99.4 per cent of the density, not ninety.

$The fraction a table encloses is not one number. For each ion, the fraction of its own electron density that lies inside its tabulated radius. Along each isoelectronic series the answer runs over eight percentage points, where three neutral atoms at their contact distances spanned three tenths of one. And it peaks at the neutral rather than falling through it, because a van der Waals radius is fitted to the distance between two atoms that are not bonded and an ionic radius is one term of a sum fitted to the distance between two that are.$

The surface a table draws

Three noble gases stop at a surface enclosing between 99.38 and 99.75 per cent of their density — a near-constant, and an argument that a contour is a real boundary. Charge the atoms and it collapses. Across ten electrons the tabulated radius encloses anything from 94.4 to 99.98 per cent, it peaks at the neutral rather than trending through it, and radii built at a fixed enclosure do not add up to a single measured separation.

Two errors, opposite signs, four orders of magnitude apart. A finite basis makes H₂⁺'s binding too large by letting each atom borrow the other's functions, and too small by describing the molecule incompletely. Both are computed here against the exact binding of 0.102634 hartree. The second is thousands of times the first at every basis size, and it is the first that counterpoise removes — so the corrected number is further from the true one than the uncorrected at every row of this table.

The basis the other atom lent

Two atoms in a molecule are described in each other's functions and the separated atoms are not, so the molecule is treated better than the pieces and the binding comes out too large. That is the basis set superposition error, it is removed by a standard correction, and for H₂⁺ in four Gaussians a centre it is six tenths of a microhartree against an incompleteness error of twelve millihartree — a factor of eighteen thousand the other way.

A correction computed at 3 bohr and used everywhere. Three binding curves for H₂⁺ in 2 Gaussians a centre: uncorrected, properly counterpoise corrected at every separation, and corrected once at 3 bohr with that value subtracted throughout. The frozen curve is the uncorrected one shifted down by a constant, so its minimum sits at 2.2270 bohr — exactly where the uncorrected minimum is, and 4.2 millibohr from where the full correction puts it. The depth moves and the structure does not.

A correction computed at one length

The counterpoise correction is expensive, so it is evaluated once at a reference geometry and subtracted across a whole potential surface. A constant does not move a minimum — so a frozen correction returns the uncorrected bond length exactly, at every reference geometry and in every basis, and everything the correction does to a structure is the part that has just been thrown away.

A ninety per cent surface with a neighbour beside it. The contour enclosing 90 per cent of a one-electron ion's density at an effective charge of 1.6, drawn with no field as a circle and in a field of 0.05 atomic units as the closed curve. The surface moves out by 47.3 millibohr on the side the field pulls the density towards and in by the same amount on the far side — 2.84 per cent of its own radius. What the sphere encloses does not change to first order; only where the surface is does.

The surface a neighbour moves

An ion in a crystal sits in the field of the ion next to it, and that field moves its contour. The displacement has a closed form, it is checked against the polarisability it implies, and it turns out to be almost perfectly anti-correlated with the discrepancy it was proposed to explain — the pairs that need the most correction get the least.

A control that ranked better than the mechanism. Rank correlations against the additivity shortfall, over eight ion pairs. The overlap of the two closed shells ranks at 0.8571 — but the cation's formal charge, which cannot be a mechanism, ranks at 0.9524, so the set is confounded: its eight pairs split four and four by charge and everything else rises with it. Held fixed within a charge group the overlap still ranks at 0.80 — and so does the softness, at -1.00. Four pairs cannot separate two candidates.

A control that outranked the mechanism

Ruling polarisation out left one candidate, and the closed-shell overlap ranks at 0.857 against the additivity shortfall — which looked like the answer until the control was read. The cation's formal charge, which cannot be a mechanism, ranks at 0.9524. Eight pairs split four and four by charge cannot separate anything, and within a charge group two candidates both rank perfectly.

$The two halves change places. What fraction of the counterpoise correction belongs to the lighter of two unlike atoms, against their separation. At a bonding distance it is 1.26 per cent — essentially the whole correction is the heavier atom's — and by 9.0 bohr it is 67. The two change places at 6.29 bohr. A symmetric pair's share is exactly a half everywhere, which is what makes one frozen number a complete description there and nowhere else.$

A correction that is two functions

A symmetric pair's counterpoise correction splits exactly in half, at every separation, to the last digit — which is why one frozen number describes it. Give the two atoms different charges and the split runs from 0.03 per cent to 67, changing places at 6.29 bohr: the correction a single number was standing in for is two functions of different shapes.

Eight wells, and where each one puts its pair. The total energy of each pair against separation, with the measured distance marked on every curve. The wells are deep and their minima are in the right region — tenths of an ångström from the measurements — which is what makes the comparison worth making. What they are not is closer to the measurements than the sum of two tabulated radii, and that is the result.

A size a confound cannot supply

A rank correlation of 0.857 was beaten by a control that cannot be a mechanism, so a size is the next thing to ask for: does a closed-shell repulsion of the computed magnitude displace two ions by the tenths of an ångström the additive radii are wrong by. It does not. The balance of a Madelung attraction against six computed repulsions predicts six separations to 0.242 ångström where adding two tabulated radii predicts them to 0.183, and the displacement it produces ranks at 0.14 against the shortfall it was proposed to explain.

One half is a curve and the other is not. The two halves of a counterpoise correction for an unequal pair, against separation, on a logarithmic axis. The heavier centre's falls smoothly over two decades; the lighter centre's scatters over more than one decade between neighbouring points. It is not a rough function — it is a difference of two energies of order a hartree whose difference is a millionth, and the solver does not have seven figures to spare.

The half that cannot be computed

How many points does each half of a counterpoise correction need to interpolate? The natural expectation is two different numbers. The answer is that the question is not yet askable: the lighter centre's half is a difference of two energies agreeing to six figures, its second differences are seven per cent of its own value, and no interpolation of it means anything. The third thing worth checking — the symmetric-pair check — works perfectly.

The trimer's correction, and the sum of its pairs. The counterpoise correction of a three-fragment system computed directly — each fragment's energy alone less its energy in the whole trimer's basis — against the sum of the three pairwise corrections, on a logarithmic axis. The sum is the larger everywhere the difference is above the solver's noise: 30.4 per cent at 1.6 bohr and nothing by six.

The assembly that counts one share twice

A counterpoise correction is divided unequally between its two centres, and the first place that matters is a three-fragment system, where the pairwise corrections are added up and the assembly must double-count one share and undercount another. It does: the heavy centre is over-corrected by seventeen per cent and the light ones under-corrected by two and a half, and the two do not cancel.

Four ways of predicting the same six separations. How far each model's predicted separation is from the measured one, pair by pair. The best additive model misses by 0.0098 ångström on average, the tabulated radii by 0.0237, radii read off the ions' own densities by 0.1828, and the balance of a Madelung attraction against a computed repulsion by 0.2421.

The residue that is two numbers

Is a shortfall between a sum of radii and a measured separation a property of the pairs, or of the compromise a universal table makes? The eight separations are two complete two-by-two blocks, so what no assignment of radii can reproduce is not a residual at all — it is an alternating sum, computed by subtraction, and it comes to three hundredths of an ångström and six.

$The correction collapses and the error made assembling it grows. At a separation of 2 bohr, two quantities against the number of Gaussians a centre. The trimer's own counterpoise correction falls by a factor of 2717 from one function to six — a bigger basis has less to borrow. The fraction by which summing the pairwise corrections overshoots it rises from 5.3 per cent to 38.0. Improving the calculation makes the assembly proportionally worse.$

The correction that gets harder to assemble

Summed across a trimer, pairwise counterpoise corrections come to fifteen per cent more than the trimer's own. Three Gaussians a centre is a small basis, so the natural question was whether the fraction shrinks with a better one or stays put. It does neither. The correction falls by three orders of magnitude and the fraction grows fivefold.

$Both quantities, and the line they have to be read against. At the closest separation, the counterpoise correction itself and the error a pairwise assembly of it makes, against the number of Gaussians a centre. Both fall — the correction by a factor of 3631, the error by 793 — and the fraction rises by exactly the ratio of those two. The dashed line is a kilocalorie a mole. The only basis where the correction is above it and the error below it is two.$

One basis size where it is worth doing

The pairwise assembly's error grows with the basis — 5.25 per cent at one Gaussian a centre, 37.98 at six — while the correction itself falls by a factor of three thousand. Which of the two should a practitioner care about? Drawing a line at a kilocalorie a mole answers it: there is exactly one basis size at which the correction is worth computing and its pairwise assembly is accurate enough to use.

The residue is four times below the model's own error. The quantity whose sign is wanted, beside the accuracy of the numbers it is a difference of. The measured residues are 0.031 and 0.056 ångström; the model gets a single separation right to 0.242 on average. A fourfold alternating difference of quantities known that badly cannot resolve something that small, and that arithmetic was available before any of this was computed.

The residue is below its own noise

Two interaction terms, both negative, leave the sign a coin toss — and a larger block would settle it. There is a larger block: a model that needs no measured separations supplies twenty-one. It cannot settle anything, because a fourfold alternating difference of distances known to a quarter of an ångström cannot resolve three hundredths of one.

Which basis size is usable, against where the line is drawn. The basis size usable at every separation — big enough that the three-body correction matters, small enough that the pairwise assembly reproduces it — against the accuracy line, over two decades. It is not one answer. Four different sizes are the answer over this range, and for much of it there is no answer at all. The standard kilocalorie a mole gives basis 2, and it sits 0.7 per cent above the edge where that answer stops.

The line was holding the answer up

There is exactly one basis size where the three-body correction is worth computing and the pairwise assembly reproduces it, and the accuracy line is the choice that whole picture is most sensitive to. Sweeping the line over two decades gives four different answers, long stretches with no answer at all, and a published window whose lower edge sits seven tenths of a per cent below the standard kilocalorie a mole.

The error runs with the row of the periodic table. The model's relative error on each measured separation, against how many of the two ions have a third-row outermost shell. The three groups do not overlap and they run in order: two second-row ions and the model is about fourteen per cent short, one of each and it is within seven per cent long, two third-row ions and it is eighteen per cent long. The narrowest gap between groups is 11.2 percentage points.

The error was the row, not the charge

An ionic model checked against six measured separations has a wildly uneven error — under one per cent on two pairs, thirteen to eighteen on three others — and the pattern is not obviously size or charge. It is the row of the periodic table. Counting how many of a pair's two ions have a third-row valence shell separates the errors completely, with an eleven-point gap; counting the charge separates nothing.

The overshoot was one arrangement

A fourth fragment leaves four three-body terms out of a pairwise counterpoise assembly as well as the four-body one, so the window in which the assembly is usable was expected to narrow. Asked of three arrangements that each gain one centre, it widens twice and narrows once, the uniform chain loses its answer at a kilocalorie a mole altogether, and the overshoot every earlier calculation reported turns out to belong to the arrangement with the heavy centre inside.

A repair that costs more than the whole

Four fragments are the first system in which a counterpoise correction can be assembled from something between pairs and the whole. Adding the three-body increments leaves what is still missing below a kilocalorie a mole on every cell, widens the usable range for every arrangement and turns a single usable basis size into two — and under a cubic model of cost it is dearer than the full calculation it stands in for until the cluster has eleven fragments. Keeping only the consecutive triples, which pays from five, works for two arrangements and does worse than pairs for the third.

The error falls along the sum of the two exponents. The ionic model's relative error on each of the six checkable separations, against the sum of the two ions' Slater exponents, with the least-squares line through all six. The rank correlation is −0.986 and only 4 of the 720 possible orderings of six points do as well, where the count of third-row ions it replaces is matched by 24. The three pairs with one third-row ion, which the count could not tell apart, fall in the order the line runs.

The sum of the exponents, not the softer ion

The row of the periodic table sorted an ionic model's errors into three groups, and a count that takes three values can say nothing inside a group. Made continuous, the variable the proposed mechanism names — how diffuse the softer ion is — carries no information: an oxide's 2p and a chloride's 3p have the same exponent to a hundredth. The sum of the two exponents carries nearly all of it, orders the middle group, and, asked about that group without having seen it, predicts its spread at twice the size.

Two conditions, and one factor that nearly meets both. The mean error of each group of pairs as the third-row p exponents alone are contracted, with the second-row pairs untouched by construction. One factor has to bring both other groups onto them. The pairs with one third-row ion arrive at 1.376 and potassium chloride at 1.346, 2.2 per cent apart — a test that could have produced two factors nowhere near each other, and did not.

One contraction for two conditions

The ionic model's errors run with the row of the periodic table, and the test proposed for that — stiffen the repulsion and watch for second-row pairs moving out and third-row pairs moving in — produces exactly that pattern from a repulsion that knows nothing about shells. The test that can fail contracts the third-row shells alone and asks one factor to bring two different groups of pairs onto the second-row ones. The two factors needed are 1.35 and 1.38.

Three ions with the same shell, and three different contractions. Each pair with one third-row ion, its error plotted against a contraction of that ion's p exponents alone. Sodium chloride's error falls to the second-row pairs' mean when chloride is contracted by 1.302, potassium fluoride's when potassium is contracted by 1.387, and calcium oxide's when calcium is contracted by 1.441. The single factor that contracts every third-row shell at once, 1.376, is drawn faint: it sits between the three, and the three span 10.7 per cent.

Three contractions for one shell

One contraction of the third-row p shells removed the row pattern from an ionic model's errors, with two conditions met by factors 2.2 per cent apart, and left the order of three pairs untouched. Taken ion by ion, chloride needs 1.302, potassium 1.387 and calcium 1.441 — the pairs' own order — and potassium chloride, fitted on nothing, lands among the second-row pairs. The single factor's two conditions agreed because both were averages of these three.

The same count in both pictures, and no rule between them. Every hydrogenic orbital with a radial node, drawn twice: its position nodes on the left axis and its momentum nodes on the right, at the same nuclear charge. The counts are identical and exact — n − l − 1 in each — because the two radial functions are polynomials of the same degree. The positions are unrelated: a node three quarters of the way out in one picture is not three quarters of the way out, or anywhere in particular, in the other.

The nodes in the other variable

An orbital has n − l − 1 radial nodes, and it has exactly that many in momentum too — the two radial functions are polynomials of the same degree. Nothing pairs one node with another: they are zeros of two different classical families. What is exact is the product over all of them, which is a ratio of factorials and does not depend on the nuclear charge at all.

A bond is a cosine in momentum space. The two factors a two-centre bonding orbital's momentum density is made of. One is the atomic momentum density, unchanged by the bond and falling as the eighth power of the momentum. The other is cos²(q·R/2), the interference between the two centres, whose period is fixed by the bond length and by nothing else. Everything that distinguishes a bond from two atoms in this picture is that cosine — and it has zeros where the atomic factor has none.

Oblate in the picture nobody draws

Every drawing of a σ bond shows a density stretched along the bond, and the position-space calculation agrees: the second moment along the axis is twice the one across it. In momentum the same orbital is flattened in the same direction, because the interference between the two centres cuts the distribution off at π over the bond length — and that cut-off is a zero a measurement could find.

The zero belongs to one determinant

A bonding orbital's momentum profile along the bond is exactly zero at π/R, and that zero reads a bond length with nothing fitted. It is a property of putting every electron into that one orbital. Any antibonding occupation fills it in linearly and drags the minimum outward, a tenth of an electron erases it, and the valence-bond wavefunction built from the same two functions never has one at any separation.

The zero at π/R follows parity, not bonding. The bonding and antibonding combinations of three atomic functions on nitrogen, each along the bond and each normalised to its own largest value, against momentum in units of π/R. For 2s the bonding combination is zero at π/R and the antibonding one is not. For 2p along the bond it is the other way round: the σ bond carries a sine, is zero at the origin and near its largest at π/R, and the antibonding combination carries the cosine. For 2p across the bond the π bond carries the cosine again. The factor is a cosine exactly when the orbital's inversion parity matches the atomic function's.

The zero is a parity, not a bond

A hydrogen-like σ bond has a momentum profile along its axis that vanishes at π/R, and it is natural to read that zero as a bond's signature. Built from 2p functions pointing along the axis, the σ bond carries a sine instead and sits at 83 per cent of its peak there. Which factor an orbital carries is decided by whether its inversion parity matches its atom's, and bonding has nothing to do with it.

Length did not rescue the consecutive triples

On four fragments, keeping only the consecutive triples of a counterpoise assembly worked for two arrangements and did worse than pairs for the uniform chain, and a short chain was the obvious excuse. Carried to eight fragments the excuse fails: the uniform chain's consecutive assembly settles at 48 per cent of the line against 68 for pairs. And the heavy-outside chain turns the lesson over — from five fragments every triple together covers less than the consecutive ones alone.

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