Theme

The thread: Say what it encloses

An orbital picture is a contour at a level somebody chose, and almost no source says which. Every one here is drawn at a level found by integrating the density, and states the fraction it encloses.

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.

Orbitals

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.

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

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.

Orbitals

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.

Orbitals

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.

Orbitals

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.

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

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.

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

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.

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

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 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.

Orbitals

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.

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

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.

Orbitals

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.

Orbitals

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.

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

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.

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.

Orbitals

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.

Orbitals

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.

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.

Orbitals

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.

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.

Orbitals

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 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.

Orbitals

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.

Orbitals

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.

All themes