Choosing a contour for 1s

One of the figures on radial functions: The one-dimensional half of a wavefunction: where the density is, what screening does to it, and how far out an orbital reaches.

14 essays draw this figure, each at the values its own argument needs rather than at the setting shown above. What each one uses it to show is below, in the words of its own caption.

Say what it encloses

The hardest case in the collection, and the one that shows why a level cannot be chosen by eye. A 3s has two radial nodes, so its cumulative integral has two flat stretches, and the half, ninety and ninety-nine per cent levels are separated by nearly a factor of four in radius. Any of the three would be a defensible picture of “the 3s orbital”, and they are three different pictures.

The enclosed fraction against the contour level. Choosing a level is choosing a point on this curve, and the curve is steep enough over the useful range that the choice matters a great deal.

The same construction for an orbital with a radial node. The cumulative integral rises, flattens through the node where there is no density to add, and rises again — so the level that encloses a stated fraction has to be solved for rather than guessed, and a contour drawn at a round number of the wavefunction would fall in the flat region and enclose almost anything.

Where the electron is

The cumulative version: how much of the density is inside a contour at each level. The peak of the distribution and the ninety-per-cent radius are different points on this curve, and quoting either as “the size of the atom” is a choice rather than a fact.

How big is an orbital

The fraction of the density enclosed against the contour level, for a 1s orbital, with three choices marked. The curve has no feature anywhere on it — no knee, no plateau, nothing that would single out one level as the natural one. Picking ninety per cent is picking a point on a smooth curve, and the usual practice of picking a level that looks right is picking a point on it without knowing which.

What the screening model cannot see

The 2s orbital’s enclosed fraction against radius, marked at five, fifty and ninety per cent. The curve leaves the origin with a visible rise before flattening across the node and climbing again: the first few per cent are collected close in.

The same curve for the 2p. It leaves the origin flat, because there is nothing of this orbital near the nucleus at all, and does its whole climb in one stretch.

The isovalue nobody chose

The 4s orbital’s radial density and its cumulative integral. Four shells, three radial nodes, and most of the probability out past ten bohr where the wavefunction is a few thousandths — which is why a contour at two hundredths catches almost none of it.

A Gaussian is the wrong shape

Where the density actually is. The cumulative integral reaches half by 1.4 bohr and ninety per cent by 2.7, so the region a Gaussian gets wrong at the nucleus holds almost nothing and the region it gets wrong far out holds a few per cent. That is the whole reason a wrong shape can give a right energy.

A slice is not the surface

The cumulative probability of a 1s against radius, with the half, ninety and ninety-nine per cent radii marked. The last of these is at 4.2 bohr, more than half again the ninety per cent radius, and the region between them contributes almost nothing to any picture — which is the whole reason the conventional contour sits where it does.

A bond is not two atoms overlapping

The one-centre case for comparison. There the enclosed fraction is a one-dimensional integral of the radial density and the contour is a sphere; here neither is true, and the only thing carried over is the requirement that the figure states what it encloses and that the statement is true.

The tenth that is not drawn

The enclosed probability of a 1s orbital against radius, with three fractions marked. The curve is nearly flat to the right of the first mark, which is exactly the problem: a large change in radius is a small change in enclosed probability, so the region that is left out is enormous and holds very little.

The radius that was tabulated

Where the fractions actually are, for a diffuse orbital. Between the ninety and the ninety-nine per cent surface there is a great deal of space, and it is the space in which two closed shells meet. Everything said about the overlap being drawn outside the picture is the same statement made about the same region.

The surface a table draws

For each species, the fraction of its own electron density that lies inside its tabulated radius. Along each isoelectronic series the answer runs over more than five percentage points — where three neutral atoms at their own contact distances spanned three tenths of one — and both series peak at the neutral rather than trending through it.

Eight rock-salt separations against two ways of building a radius. Shannon’s reproduce them within a per cent and a half, which is a check on the arithmetic rather than a result, because they were fitted to these very numbers. Radii at a fixed ninety-nine per cent enclosure miss them by between six per cent short and thirty-seven per cent long, and the sign of the miss is not random.

The enclosed fraction at which two ions’ own surfaces would just touch at the measured separation, for each of the eight pairs. There is no single value: it falls by eight percentage points across the set, and it falls with the charge on the pair and with the size of the anion. So a crystal does not space its ions by any surface of constant enclosure.

The surface a neighbour moves

Eight rock-salt pairs. The upper bar is the measured shortfall — how far the two computed ninety per cent radii fall short of the measured separation — and the lower bar is what the field supplies. They are almost perfectly opposed.

The other answer to the same problem: the enclosed fraction that would make each pair’s radii add up. It runs over a wide range and has no constant in it, which is the finding polarisation was proposed to explain and does not.

The additivity test itself: a sum of two radii against a measured separation, for two different conventions. The correction this essay computes would move the marks by a few hundredths where they are wrong by tenths, and by tenths where they are nearly right.

A control that outranked the mechanism

Every candidate ranked against the shortfall, over eight pairs. The top bar is a control that cannot be a mechanism, and it is the longest.

The same eight pairs drawn: the shortfall against the overlap squared, on a logarithmic axis. The 1+ pairs are at one end and the 2+ pairs at the other, which is the difficulty.

Where the shortfall itself comes from: radii set so that every ion encloses the same fraction of its own density, added, against the measured separations. The errors run to more than a tenth in both directions.

A size a confound cannot supply

The shortfall and the displacement, pair by pair. The two bars point the same way five times out of eight.

The displacement the repulsion produces as a fraction of the shortfall. One is what an explanation would give; these run from −1.7 to +5.7.