The collection

Every essay — page 41

Page 41 of 41, continuing through the fields in the same order.

Orbitals Where the atoms go Bonding models What symmetry decides Beyond the octet What a spectrum settles When the molecule does not stop What the shape is for What is taught wrongly Series Named objects Orbitals Refutations Search

What is taught wrongly

The explanations that are confident, memorable and false — stated fairly and then tested against a calculation rather than an opinion.

The conventional start misses the lowest mean field on one band, and nowhere else. A four-site ring at every combination of repulsion and staggered site energy on a 12 by 12 grid. A small dot is a system where the conventional broken-symmetry start reaches the lowest mean-field solution; a circle is one where it lands above it, with area proportional to how far. There are 9, every one with a site energy one or two below the repulsion, and the largest miss is 1.124 at a repulsion of six and a site energy of four.

The worst system in the square was the solver's

Every score in the transferability square rests on an unrestricted mean field, and every one was computed from a single conventional start. On a band of nine systems, each with a site energy one or two below the repulsion, that start lands above the lowest solution the method has — by 1.12 at a repulsion of six and a site energy of four, which was the largest error in the whole square. Recomputed on the lowest field, the factor of forty-six that first showed the polarisation to be no rule falls to 2.61 and passes; the grid still fails every candidate, by a factor of fifteen rather than forty-eight; and the typical comparison, which no worst case could show, is far better than chance for every candidate but one.

8 figures $The repulsion each bond needs, and the one Born and Mayer's law supplies. For each bond, the fraction of the ion pair's attraction a repulsion must reach before moving a whole electron stops paying (open circle), and the fraction a Born–Mayer repulsion of range 0.345 Å supplies at the measured length if the ion pair is to sit at its own minimum there, ρ/R (filled). Three bonds need none because they never transferred. Hydrogen fluoride needs 0.351 and is supplied 0.376; hydrogen chloride needs 0.116 and is supplied 0.271; the alkali halides need three quarters and are supplied a fifth or less. Every verdict comes out right — hydrogen fluoride's by a margin of 0.025.$

A repulsion the proton does not have

Asked whether moving one whole electron pays, eight bonds with measured dipoles came out six right and two wrong: hydrogen fluoride and hydrogen chloride, both predicted ionic. The proposed fix was the repulsion between the two ions' closed shells. Add it the way an ion pair demands of itself, and the score becomes eight of eight — but the repulsion that does it is largest for the shortest bonds, and the shortest bonds are the ones whose cation is a bare proton with no shell at all. The term a proton really has, penetration into the anion's cloud, undoes hydrogen chloride and not hydrogen fluoride. And the measured bond energies demand a repulsion only for the three alkali halides.

6 figures

The covalent arrangement the inversion finds is Pauling's mean — except for hydrogen fluoride. For eight bonds, the covalent arrangement's binding found by solving the two-state problem exactly against the measured dipole and bond energy, against Pauling's covalent mean — the geometric mean of the two homonuclear bond energies — which the inversion never sees. The line is equality. HCl 0.97, HBr 0.98, HI 1.03, ClF 1.09 of the mean; HF 1.68. Open circles are the alkali halides, where the ionic weight is near 0.8 and the result is mostly the assumed repulsion.

Two measurements leave nothing to fit

A bond pictured as a mixture of its ionic and covalent arrangements has two unknowns — the coupling between them and the covalent arrangement's own energy — and a dipole fixes only one. Add the bond energy and the two-state problem inverts exactly, with no parameter left. The covalent arrangement it returns can then be held against something it never saw: Pauling's mean of the two homonuclear bond energies. For hydrogen chloride, bromide and iodide it lands on that mean to within three and a half per cent, and chlorine monofluoride within nine. Hydrogen fluoride lands 68 per cent above it. And the couplings rise and fall with length rather than falling — hydrogen fluoride's, on the shortest bond, is the smallest.

7 figures

The chain of four's strongest line is a hole's standing wave read against the singlet's ends. For the half-filled chain of four at infinite repulsion, each removal line's weight is a sum over the four sites of two factors. The first is the hole's standing wave squared on each site: the inner levels at ±0.62 put more of the hole on the end sites, the outer levels at ±1.62 more on the middle ones. The second belongs to the spin chain alone: removing an up electron at an end site leaves the stronger spin doublet with weight (2 + √3)/8 = 0.4665, at a middle site 1/4. Their sums are 0.4067 and 0.3098, which are 1/4 + √3/16 ± √15/80, and the limit of the contrast is their ratio, 1.312498.

The limit was two chains at once

At infinite repulsion every removal line of a half-filled chain is a hole's standing wave read against the spin chain the hole moves through, and every hole level carries exactly half an electron. On the chain of four that makes the contrast's limit (20 + 5√3 + √15)/(20 + 5√3 − √15) = 1.3124978 — not 21/16, which it misses in the sixth figure. On chains of eight and twelve, which no exact spectrum had reached, the limit is 1.0458 and 1.0195, and an energy that vanishes in the limit decides it.

6 figures