Two measurements leave nothing to fit

Four quantities go and one question stays reduced electronegativity to a question with an integer answer: does moving one whole electron from one atom to the other lower the energy? Asked of eight bonds with measured dipoles — a difference in electronegativity does not by itself say how much charge moves — the answer was right for six and wrong for hydrogen fluoride and hydrogen chloride, both predicted ionic. A repulsion the proton does not have found the term that fixes hydrogen chloride — a bare proton inside chloride’s cloud sees less than a whole negative charge — and left hydrogen fluoride on the ionic side of its threshold.

Its lead was the obvious one. Hydrogen fluoride’s ground state is neither H⁺F⁻ nor the neutral pair; it is a mixture, and its dipole of 0.415 of an electron times its length is a statement about the mixing. A two-state model — the ionic and covalent arrangements coupled by one matrix element — has its mixing set by the coupling. Fitted to each hydrogen halide’s dipole it would give a coupling per bond, and whether those couplings are one smooth function of length or four unrelated numbers would say whether the picture is a model or a fit.

The dipole cannot do that on its own, and the reason turns the lead into a sharper test than it asked for.

One measurement, two unknowns #

The two-state model has three energies and one number that mixes them. The ionic arrangement’s energy at the bond length, $E_i$, is what the earlier essays computed: the ionisation energy of the donor less the electron affinity of the acceptor, less the ion pair’s attraction at the measured length, plus the correction that belongs to the pair — the proton’s penetration for a hydrogen halide, a closed-shell repulsion for an alkali halide. The covalent arrangement’s energy, $E_c$, is the neutral atoms bonded without any charge moved, and nothing computed here says what it is. The coupling V mixes them. The ground state is the lower root of a 2×2 matrix, and its weight on the ionic arrangement is the ionic weight c.

A dipole gives c, through the reduction used throughout these essays — the one that cannot tell a bond’s charge from its atoms’ own polarisation: the dipole divided by the bond length and the electron’s charge, the charge a point on each nucleus would need to reproduce it. That is one equation for two unknowns, $E_c$ and V. Any covalent energy can be paired with a coupling that produces the measured weight, so fitting the dipole alone produces a coupling for every assumed $E_c$ — a family, not a number, and the question “is the coupling a function of length” has no answer until $E_c$ is fixed.

The bond energy fixes it. The ground state’s energy relative to the neutral atoms is minus the dissociation energy $D_e$ — the tabulated $D_0$ plus the zero-point energy, half the harmonic wavenumber. With $E_g$, $E_i$ and c known, the 2×2 problem inverts exactly:

$$E_c = E_g + (E_i - E_g)\,\frac{c}{1-c}, \qquad V = -(E_i - E_g)\sqrt{\frac{c}{1-c}}$$

Nothing is left to fit. Every quantity on the right is a measurement or a computation from earlier essays, and the two on the left come out as predictions — which means they can be held against something.

The arrangements, drawn #

The inversion’s output is easiest to read as three levels per bond.

Hydrogen fluoride is the odd one before any comparison is made. Its ionic arrangement, with the proton’s penetration of 1.40 eV counted, lies 4.11 eV below the neutral atoms, and the covalent arrangement the inversion finds lies at −4.71 — only 0.6 eV lower. The two arrangements are nearly degenerate, which is why the mixing is strong and the ionic weight 0.415. In hydrogen chloride the ionic arrangement sits 1.31 eV above the neutral atoms, 4.65 above the covalent one at −3.34, and the ground state is mostly covalent with an ionic weight of 0.181. Hydrogen iodide’s ionic arrangement is higher still, at +4.07, and its ground state is 94 per cent covalent.

That matches what every chemistry course says about the hydrogen halides, and it is worth noticing that it was not put in. The covalent energies were not assumed; they are what the dipoles and bond energies require.

The heavier anions, and a word that was too strong #

The inversion needs the proton’s penetration for all four hydrogen halides, and the essay before computed it only for fluoride and chloride, in closed form for Slater densities with integer principal numbers. Bromide and iodide have outer shells whose Slater principal numbers are 3.7 and 4.0, so the potential of each shell is computed here by radial quadrature — checked against the closed form on fluoride’s and chloride’s integer shells, where it agrees to a part in a billion.

The essay before wrote that hydrogen bromide’s penetration “is certainly larger than chloride’s”. It is — 2.71 eV against 2.62 — but by three per cent, not certainly by much, and hydrogen iodide’s at 2.48 is smaller than both, because iodide’s cloud is more diffuse but its bond is 0.19 Å longer and the proton sits further out in it. What the earlier essay needed from the number holds: hydrogen bromide’s ionic arrangement, which sat 0.05 eV on the covalent side of the threshold, is pushed 2.77 eV above the neutral atoms, far from any verdict the threshold could reverse.

What the covalent arrangement can be held against #

The inversion returns a covalent arrangement for each bond, and there is a number it can be compared with that it knows nothing about. Pauling built his electronegativity scale on the observation that a bond between unlike atoms is stronger than the mean of the two like bonds, and he took the excess — the ionic resonance energy — to measure the difference in electronegativity. The mean he settled on was the geometric one — a mean that errs in a known direction when it is used on electronegativities themselves, and here used on energies: the square root of the product of the two homonuclear bond energies. For hydrogen chloride that is $\sqrt{4.748 \times 2.514}$ = 3.45 eV.

The inversion’s $E_c$ is not built from any homonuclear molecule. It comes from the hydrogen halide’s own dipole, bond energy, and ionic arrangement. If the two-state picture and Pauling’s picture describe the same thing, the covalent arrangement’s binding should be his mean.

For hydrogen chloride it is 3.34 eV against 3.45: 0.966 of the mean. For hydrogen bromide, 3.03 against 3.07: 0.985. For hydrogen iodide, 2.79 against 2.72: 1.025. For chlorine monofluoride, with no correction to its ionic arrangement at all, 2.22 against 2.04: 1.088. Four bonds, three different halogens, two with a proton and one without, and the covalent arrangement the inversion needs is Pauling’s mean to within nine per cent every time and within three and a half for the three hydrogen halides.

Pauling’s ionic resonance energy follows from the same numbers. Where $E_c$ is his mean, the inversion’s ionic stabilisation $E_c - E_g$ is his resonance energy, and for the three heavier hydrogen halides the two agree to 0.12 eV: 1.32 against 1.20 for chloride, 0.93 against 0.89 for bromide, 0.45 against 0.52 for iodide. The square roots his scale takes are 1.15 against 1.10, 0.97 against 0.94, and 0.67 against 0.72 — electronegativity differences from a two-state model with no free parameter, landing on his.

Hydrogen fluoride’s covalent arrangement is 4.71 eV against a mean of 2.81: 1.68 times it. Where Pauling assigns 3.36 eV of hydrogen fluoride’s bond to ionic resonance, the inversion finds 1.46, and the electronegativity difference its square root would give is 1.21 rather than Pauling’s 1.83.

Where the inversion can be trusted #

An exact inverse is only as good as its inputs, and how far errors in them propagate differs by an order of magnitude between these bonds.

The lever on the ionic energy is the odds of the ionic weight, c/(1 − c). For hydrogen iodide it is 0.06: an error of an electronvolt in the penetration moves the covalent arrangement by six hundredths. For hydrogen chloride and bromide it is 0.22 and 0.14, for chlorine monofluoride 0.13 and for hydrogen fluoride 0.71. For the three alkali halides it is 3.6 to 5.3, because their ionic weights are near 0.8, and their ionic arrangements rest on a Born–Mayer repulsion whose range is conventional rather than computed. A tenth of an electronvolt in that repulsion moves lithium fluoride’s covalent arrangement by half an electronvolt. The alkali halides’ points are therefore drawn open and not scored: lithium fluoride lands at 1.12 of the mean, sodium chloride at 1.41 and potassium bromide at 2.30, and those numbers are mostly the repulsion’s.

The other input with a lever is the ionic weight itself. The point-charge reduction ignores the atoms’ own polarisation and the overlap of the two arrangements, and it is not the only reduction available — ionic character depends on how it is weighed. So each well-posed bond is also recomputed with its ionic weight ten per cent smaller. That moves the covalent arrangement by 0.23 eV for hydrogen fluoride, 0.16 for chloride, 0.10 for bromide and 0.05 for iodide. The agreement for chloride, bromide and iodide survives that; so does hydrogen fluoride’s miss, which is 1.9 eV.

The couplings are not a function of length #

The lead’s question can now be answered, because the couplings are fixed.

A coupling between two arrangements of the same electrons is built, in any valence-bond account, from overlaps and exchange integrals, and those fall as the atoms separate. The hydrogen halides’ do not: 1.73 eV for fluoride at 0.917 Å, 2.81 for chloride at 1.275, 2.50 for bromide at 1.415, 1.81 for iodide at 1.609. They rise from the shortest bond to the second and fall after. Leave hydrogen fluoride out and the other three do fall with length, smoothly enough to be one function — but the point that breaks the trend is the one the whole line of essays started from.

So the answer to “model or fit” is split. For three hydrogen halides and chlorine monofluoride the two-state picture is a model: two measurements per bond determine it completely, its covalent arrangement is Pauling’s mean without being told to be, and its couplings fall with length. For hydrogen fluoride it is not the same model.

What hydrogen fluoride would have to be #

A miss of 1.9 eV is too large for the ten-per-cent sensitivity to explain, so the question becomes what single change would close it.

Holding the ionic arrangement, hydrogen fluoride would need an ionic weight of 0.62 instead of 0.415 — a point-charge dipole of 2.73 D against the measured 1.83. Holding the ionic weight, the proton would need to lose 4.08 eV to penetration instead of 1.40. Neither is plausible as a correction to its input: the dipole is measured to three figures, and the penetration is a closed-form integral of a density that already reproduces chloride and bromide.

The textbook explanation of fluorine’s anomalies is its weak homonuclear bond. The fluorine molecule’s dissociation energy, 1.66 eV, is much smaller than chlorine’s although fluorine is the smaller atom, and the usual account blames repulsion between the two atoms’ lone pairs at so short a distance. That depresses Pauling’s mean for every fluoride, and so could make the inversion’s covalent arrangement look too strongly bound by comparison. But the inversion says how much: for Pauling’s mean to equal 4.71 eV, the fluorine molecule would need a dissociation energy of 4.67 eV — nearly three times the measured value, and within a tenth of an electronvolt of the hydrogen molecule’s own 4.75. A weak F–F bond is part of the story and cannot be all of it.

What remains is the reduction of the dipole. Hydrogen fluoride is the bond where the point-charge reading of a dipole is most likely to fail, because it is the shortest, the most strongly mixed, and the one where fluorine’s lone pairs sit closest to the bond axis and contribute their own atomic dipole. Whether a better reduction would move its ionic weight all the way from 0.415 to 0.62 is not computed here, and it is the next thing to compute.

How the claims can fail #

Each statement is checked where the figures are drawn. The non-integer quadrature must reproduce the closed-form potential of fluoride’s and chloride’s shells to a millionth. Every bond’s inversion, solved forwards from the $E_c$ and V it found, must return its measured bond energy and its dipole’s ionic weight to 10⁻⁹, and its ground state must lie below both arrangements — which, without the alkali halides’ repulsion, it would not. Exactly the four hydrogen halides and chlorine monofluoride must be well posed, and the alkali halides’ lever must exceed three. Every well-posed bond but hydrogen fluoride must land within ten per cent of Pauling’s mean, and hydrogen fluoride more than half as high again; for the three heavier hydrogen halides the inversion’s ionic stabilisation must match Pauling’s resonance energy to 0.15 eV, and differ by more than 1.5 for hydrogen fluoride. No ten-per-cent error in any hydrogen halide’s ionic weight may move its covalent arrangement by a quarter of an electronvolt. Bromide’s penetration must exceed chloride’s by under a tenth, and iodide’s fall below both. The shortest hydrogen halide must have the smallest coupling.

The refusals are the forward solve and the closed-form check: if either failed, the numbers above would be the inversion’s arithmetic or the quadrature’s rather than the bonds’.

Where the model stops #

Two arrangements, not three. A hydrogen halide has a second ionic arrangement, H⁻X⁺, and for hydrogen iodide it is the cheaper of the two by the isolated-atom arithmetic, since iodine’s ionisation energy less hydrogen’s affinity is below hydrogen’s ionisation energy less iodine’s affinity. It is left out because the dipole’s sign says which ionic arrangement the ground state leans towards, and a three-state inversion would need a third measurement to determine.

The covalent arrangement is one number. It is taken at the bond length and has no internal structure; a valence-bond covalent function has its own dependence on overlap and exchange, and the agreement with Pauling’s mean is agreement of energies, not a demonstration that the covalent function is the one his picture imagines.

Tabulated inputs. The bond energies, harmonic wavenumbers, homonuclear bond energies, ionisation energies and affinities are standard tabulated values; lithium fluoride’s dissociation energy carries the largest uncertainty, about 0.2 eV, and it is one of the three bonds not scored.

And the point-charge reduction, which is where hydrogen fluoride’s miss most probably lives. The inversion reads the ionic weight off the dipole as if the charge sat on the nuclei, and what a dipole cannot tell apart includes exactly the atomic dipoles that reduction ignores.

A picture with no parameters is a test #

The two-state picture of a polar bond is old enough to be background — molecular orbital and valence-bond descriptions both contain it, and it is usually drawn as a qualitative story with the coupling adjusted to taste. The point of solving it exactly is that the story then has no slack. With two measurements per bond it predicts one number it was not given, and the number is Pauling’s mean — a quantity from a different argument, built from different molecules.

That makes Pauling’s additivity look less like an empirical rule and more like a consequence: where the two-state picture holds, the covalent arrangement is the mean of the like bonds, and the ionic resonance energy his scale is built from is the mixing energy of the two arrangements. And it makes his fluorine look like the case the picture does not hold for — not because the scale is wrong about fluorine’s electronegativity, which every other scale agrees is the largest, but because the number the scale was calibrated on includes something the two-state picture does not.

Still open: the dipole’s reduction, and three arrangements #

The obvious open question is hydrogen fluoride’s ionic weight. The point-charge reduction is the weakest input, and a better one — the ionic weight read from a two-state wavefunction’s own dipole, with the atoms’ lone-pair dipoles computed rather than ignored — would say whether it moves 0.415 towards the 0.62 that would put hydrogen fluoride on Pauling’s mean. There are already three conventions for ionic character that disagree by a factor of six on one wavefunction; the one that makes the inversion consistent across all four hydrogen halides would be a way of choosing among them.

The nearer question is the second ionic arrangement. Hydrogen iodide’s H⁻I⁺ arrangement is the cheaper transfer between free atoms, and a three-state inversion would need one more measurement per bond. The bond’s polarisability along the axis is one candidate: it is a second-order property of the same 2×2 or 3×3 matrix, measured for all four hydrogen halides, and it would test whether hydrogen iodide’s near-perfect agreement survives being asked a question its two-state picture can fail.