A repulsion the proton does not have

Four quantities go and one question stays rewrote electronegativity on the energy the exact theory gives an atom — straight segments between integer electron counts — and found that what survives is a question with an integer answer: does moving one whole electron from one atom to the other lower the energy? On isolated atoms, never; the cheapest transfer the periodic table allows costs 0.729 electronvolts. With the Coulomb attraction of the resulting ion pair at the measured bond length, the answer for eight bonds with measured dipoles is yes for five and no for three. Rounded to the nearest integer, the charges their dipoles imply agree with six of the eight.

The two it got wrong are hydrogen fluoride and hydrogen chloride, the bonds whose partial charges every equalisation scheme has argued over. Both are predicted to move a whole electron; their dipoles imply 0.415 and 0.181 of one. Both sit near the threshold — the attraction beats the transfer cost by 5.51 and 1.31 eV — and the omission the essay named was the obvious one. Two ions at a bond length do not only attract; their filled shells repel, and a repulsion raises the price of the ionic arrangement. It should push both hydrogen halides towards no transfer, and the question was whether it pushes them far enough without pushing an alkali halide across as well.

What an ion pair demands of its own repulsion #

A repulsion has to be given a size, and the point-charge model has no parameter for one. But the ionic hypothesis supplies its own condition. If a bond really were an ion pair sitting at the measured length, it would sit there because that is where the ion pair’s energy is lowest: the repulsion’s slope would exactly cancel the attraction’s. That condition fixes the repulsion at the bond length once its form is chosen.

For the Born–Mayer form, a repulsion falling off exponentially with a range ρ, the condition makes the repulsion at the length exactly ρ/R times the attraction there. For the Born–Landé form, falling as an inverse power n, it makes it exactly 1/n times the attraction. Either way, each bond’s question becomes one inequality: is that fraction larger than the margin divided by the attraction? The margin is what a repulsion has to eat before the transfer stops paying.

Three bonds need nothing, because the attraction never beat the cost: hydrogen bromide, hydrogen iodide and chlorine fluoride. The three alkali halides need three quarters of their attraction or more — 0.784 for lithium fluoride, 0.750 for sodium chloride, 0.809 for potassium bromide — because their transfer costs are small and their margins large. Hydrogen chloride needs 0.116. Hydrogen fluoride needs 0.351.

Born and Mayer’s range for the alkali halides, 0.345 Å, supplies 0.271 to hydrogen chloride, which undoes it, and 0.376 to hydrogen fluoride, which undoes it by 0.025. It supplies a fifth or less to each alkali halide, which leaves them ionic. Eight of eight. The lead’s prediction was right in every number: the repulsion pushes the two hydrogen halides across and no alkali halide.

The window, and the law that matters more than the number #

Eight of eight is not a coincidence of one range. Converting each threshold into the range that reaches it gives a window: every Born–Mayer range from 0.322 Å, hydrogen fluoride’s, to 1.226, lithium fluoride’s, scores all eight. That window is nearly an ångström wide. Its bottom edge sits just under the conventional value, which is the same 0.345 Å that decides whether iodine heptafluoride’s ring puckers — a number that keeps landing on thresholds in problems that have nothing to do with the crystals it was fitted to.

The form, though, is not a detail. A Born–Landé repulsion held to the same condition supplies the same fraction, 1/n, to every bond, whatever its length. Hydrogen fluoride needs 0.351, so it would need an exponent below 2.85 — softer than any closed-shell repulsion anybody has used, and barely steeper than the Coulomb attraction itself. Every Born–Landé exponent from 3 to 8 scores seven of eight, undoing hydrogen chloride and not hydrogen fluoride; every exponent from 9 up scores six, which is the score with no repulsion at all. The exponents conventionally assigned to ions with the electron configurations of neon and argon, seven and nine, straddle the edge between those two.

So the eight of eight belongs to the exponential form specifically, and the exponential form gets it by one property: ρ/R grows as the bond shortens. It hands the largest repulsion to the shortest bonds, and the shortest bonds in the set are exactly the two that needed fixing.

Why the exponential is the right form, where it applies #

The two forms are not equally defensible, and the one that scores eight is the better one. A closed-shell repulsion comes from the exclusion principle: two filled shells that overlap must put their overlapping electrons into combinations of which the antibonding one rises by more than the bonding one falls, and the net cost goes as the square of the overlap. An overlap between two atomic functions decays exponentially with their separation, so its square does too. The contact calculation for the noble gases built its repulsion exactly that way and needed no contact parameter. Born and Mayer’s exponential is the same statement with the range fitted rather than computed; it is why their form displaced Born and Landé’s inverse power, which was chosen for convenience in the lattice sums and fitted to compressibilities.

So the form that scores eight of eight is the form a closed shell actually has. That makes the result more tempting, not less. A reader who knows that the exponential is the physically grounded law, and sees it fix the two failures with the conventional range and break nothing, has every reason to take the fix as a mechanism. What the argument above adds is that the mechanism needs two closed shells, one on each ion, and the two bonds it fixes have one.

The derivation also says what the range means, and it makes the mismatch sharper. The range is the length over which the overlap of the two shells falls by a factor of e — a property of the pair of shells, set by how diffuse each is. For $\mathrm{H^+}$ and $\mathrm{F^-}$ one of the pair does not exist. There is no overlap to decay, so there is no range to assign, and the 0.345 Å that undoes hydrogen fluoride is a range for a pair of shells that is not there.

The shortest bonds are the ones with a proton #

That property deserves a second look, because of which bonds are shortest.

Hydrogen fluoride at 0.917 Å, hydrogen chloride at 1.275 and hydrogen bromide at 1.415 are the three shortest bonds in the set, and their ionic arrangement puts the electron on the halogen and leaves a bare proton as the cation. A proton has no electrons. It has no closed shell, nothing for the anion’s filled shell to overlap with, and nothing for the exclusion principle to act on. A closed-shell repulsion between $\mathrm{H^+}$ and $\mathrm{F^-}$ is not a small term or an uncertain one. It is a term with nothing to be about.

Hydrogen iodide is the exception within the family, and an instructive one. Its cheaper whole-electron transfer runs the other way, from iodine to hydrogen, since iodine’s ionisation energy of 10.45 eV less hydrogen’s affinity of 0.75 is cheaper than hydrogen’s 13.60 less iodine’s 3.06; the ionic arrangement the model considers for it is $\mathrm{I^+H^-}$. It needed no repulsion anyway.

So the fix that scores eight of eight works by giving its largest correction to the bonds that cannot have that correction. The numbers are right and the physics is borrowed. The repulsion that pushes hydrogen fluoride across is the repulsion of a Born–Mayer ion pair, and hydrogen fluoride is not one: a difference does not make a transfer, and an $\mathrm{H^+}$ does not make a closed shell.

What a proton does have #

A proton next to an anion is not a point charge next to a point charge, and the difference is computable. Outside the anion’s electron cloud a proton sees a net charge of −1; inside it, it sees the nucleus less only the electrons closer in than itself. The part of the cloud beyond the proton pulls it outward as much as inward, and counts for nothing. So at short range the attraction is weaker than $e^2/R$. That is penetration, and it is a real term a bare proton has.

For a Slater density the potential of each shell has a closed form, and $\mathrm{F^-}$ and $\mathrm{Cl^-}$ are built from Slater’s rules with the extra electron counted: $\mathrm{F^-}$'s outer shell at an exponent of 2.425, $\mathrm{Cl^-}$'s at 1.917.

At hydrogen fluoride’s length the proton loses 1.40 eV of its 15.71 eV attraction, all of it to the outer shell — the inner shells are far too compact to be reached. The margin is 5.51. Penetration takes off a quarter of what hydrogen fluoride needs. At hydrogen chloride’s longer length the proton sits further into a more diffuse cloud and loses 2.62 eV against a margin of 1.31, twice what is needed. With the term a proton really has, hydrogen chloride comes out right and hydrogen fluoride does not: seven of eight, by a different route from the Born–Landé seven.

And the term the model still leaves out runs the wrong way for hydrogen fluoride. A bare proton polarises the anion it sits beside, drawing the anion’s cloud towards itself, and polarisation always lowers the ion pair’s energy — it makes the transfer more favourable. At 0.917 Å, deep inside a fluoride ion, a point-polarisable estimate of that term is not trustworthy, but its sign is: every correction a hydrogen halide’s ion pair really has, apart from penetration, makes hydrogen fluoride more ionic, not less.

What the measured bond energies demand #

There is a way to ask the measurements directly whether a repulsion is needed, and it gives the sharpest version of the answer.

The ion pair with no repulsion is bound, relative to the separated neutral atoms, by its margin: 7.22 eV for lithium fluoride, 5.51 for hydrogen fluoride. The molecule is bound by its dissociation energy, which is measured. A molecule cannot be less bound than one of its own states, so if the repulsion-free ion pair is bound more strongly than the molecule is, the terms it leaves out must make up at least the difference. If it is bound less strongly, the measurement asks for nothing.

Lithium fluoride is bound by 5.98 eV and its repulsion-free ion pair would be bound by 7.22, so it needs at least 1.24 eV of repulsion. Sodium chloride needs at least 0.30 and potassium bromide 0.20. Born–Mayer at 0.345 Å supplies 2.03, 0.89 and 0.62 — more than each demands, as it should, since the bound is only a floor. Every hydrogen halide and chlorine fluoride asks for nothing: hydrogen fluoride’s ion pair would be bound by 5.51 eV and the molecule is bound by 5.91, so the measurement leaves room to spare, and the others by more.

So the measurements demand a closed-shell repulsion in exactly the three bonds whose cation has a closed shell — $\mathrm{Li^+}$ with helium’s, $\mathrm{Na^+}$ with neon’s, $\mathrm{K^+}$ with argon’s — and in none of the bonds where the cation is a proton. That is the physics the eight of eight had borrowed, stated by the measurements rather than by a model: a repulsion is real where there is a shell to supply it, and nowhere else.

How it was computed #

Every input is quoted: the ionisation energies and electron affinities of the seventeen atoms, the eight bond lengths and dipole moments, and eight dissociation energies from the standard tables — 570.3, 431.4, 366.2 and 298.4 kJ/mol for the hydrogen halides, 577, 412.1 and 379.1 for the alkali halides and 251.5 for chlorine fluoride. The zero-point energies that separate those from the depths of the potential curves are under 0.06 eV for the three alkali halides, which are the only bonds the bound is not vacuous for. The transfer cost for each bond is the cheaper of its two directions; the attraction is 14.40 eV·Å over the length.

The repulsion at each length follows from requiring the ion pair’s energy to be stationary there — for Born–Mayer the repulsion is ρ/R of the attraction, for Born–Landé 1/n — and each bond’s verdict is whether cost minus attraction plus repulsion is negative. The score is the number of verdicts equal to the measured charge rounded to the nearest integer. Penetration is the attraction lost to each shell of the anion, N(1/R − V®) with V the potential of a normalised Slater shell. Its closed form is checked against a direct radial quadrature to 10⁻⁸ in every shell of both anions.

What must hold, and is checked: with no repulsion the score is six of eight and wrong on exactly hydrogen fluoride and hydrogen chloride, which is the refusal — the scoring must reproduce the unrepulsed verdicts before a repulsion is credited with changing them; Born–Mayer at 0.345 Å scores eight; the window of ranges that scores eight runs from hydrogen fluoride’s threshold, between 0.3 and 0.345, to above an ångström; no Born–Landé exponent from 3 up gets hydrogen fluoride right; penetration lives in the outer shell and undoes hydrogen chloride’s transfer and not hydrogen fluoride’s; and the dissociation energies demand a repulsion for the three alkali halides and nothing else, with Born–Mayer supplying more than each demands.

What the point charges still cannot say #

The stationarity condition is the ionic hypothesis taken at its word. It says what repulsion an ion pair would need to sit at the measured length. For the alkali halides that is a fair description of the molecule. For the hydrogen halides it is a description of a state the molecule is mostly not in, and the repulsion it produces is a property of the hypothesis, not of the bond.

Penetration is computed for Slater densities of free anions. A fluoride ion with a proton 0.92 Å from its nucleus is not a free fluoride ion; its cloud is pulled towards the proton, which is the polarisation left out above. The 1.40 eV is the penetration of an undistorted cloud and is a lower bound on how different a proton inside an anion is from a point charge outside one, not a measurement of it.

Bromide and iodide have no penetration here. Hydrogen bromide and hydrogen iodide needed no correction, so their anions were not built; Slater’s rules for their d shells are the least reliable part of the rules, and the argument did not need them.

And the answer is still an integer. Every correction here moves a threshold. None of them produces the 0.415 of an electron that hydrogen fluoride’s dipole implies, because the model’s ground state is one arrangement or the other and never a mixture. A fraction needs the two arrangements coupled, which no amount of repulsion supplies — the same reason an integer is what an oxidation state has and a measurement never does.

A right score from a term the bond does not have #

The score went from six of eight to eight of eight, and the question is what it went there on. It went on a repulsion whose size the ionic hypothesis sets for itself and whose form puts the most of it on the shortest bonds. The two bonds that needed it are the shortest because their cation is a proton, and a proton has none of it. The term the proton does have undoes one of the two. The measured binding energies ask for a closed-shell repulsion in the alkali halides and in no hydrogen halide, which is where the shells are.

A correction that fixes a model’s score has to be a term the failing cases actually have. Otherwise the fix has found a parameter that separates the right answers from the wrong ones, not a mechanism; and a parameter that happens to scale as 1/R will separate short bonds from long ones in any set where the failures are short. Here the failures were short for a reason that is itself the answer. Hydrogen fluoride and hydrogen chloride are not near-ionic bonds with a missing repulsion. They are bonds whose ionic arrangement has no closed shell on one side, and that is why the whole-electron question, asked of them, has no good answer at any threshold.

Still open: the arrangement that is neither #

The obvious open question is the coupling. Hydrogen fluoride’s ground state is not $\mathrm{H^+F^-}$ and not the neutral pair; it is a mixture of the two, and its dipole of 0.415 of an electron is a statement about the mixing — though how much of that fraction sits on each atom depends on how the density is divided. A two-site model that mixes the two arrangements already exists here. A two-state model, the ionic and covalent arrangements coupled by one matrix element, has its crossing where this essay’s margin goes to zero and its mixing set by the coupling. Fitted to the dipole, it would give a coupling for each hydrogen halide, and whether those couplings are one smooth function of the length or four unrelated numbers would say whether the two-state picture is a model or a fit.

The nearer question is the penetration of the heavier anions. Bromide and iodide were not needed here, but hydrogen bromide sits 0.05 eV on the covalent side of its threshold. Its penetration at 1.415 Å is certainly larger than chloride’s, which would push it further the right way, and with an honest d shell it would say whether the whole-electron question’s three correct hydrogen-halide verdicts are robust or lucky.