A floor no charge transfer explains

Every zero this argument has found in a momentum profile has the same cause, and it is a cancellation between two identical things. A σ combination of two atomic functions carries a factor of 2 ± 2 cos(qR) along the bond, the cosine combinations vanish at q = π/R, and they vanish because the two atoms contribute equally and the phase factor between them is exactly reversed there.

Neither half of that is available in a bond between two different atoms. There is no inversion centre, so the orbital is neither g nor u and the two atomic contributions are not equal; and the coefficients are not equal either, because the more electronegative atom takes more of the orbital. The closing question of the essay before it was what the profile at π/R becomes: not a count, presumably, but a measure of the bond’s polarity.

The question matters because polarity is the quantity this subject measures worst. An oxidation state is an integer nobody measured and an atomic charge computed from one wavefunction spans a factor of three depending on how the shared electrons are divided; a dipole moment is a molecule’s property rather than a bond’s. A momentum profile at a stated momentum would be none of those — a number a measurement returns, with no partitioning in it at all.

It is a measure of polarity, and it has a floor.

What replaces the factor #

Write the orbital as $c_A \chi_A + c_B \chi_B$. In momentum space each atomic function carries the phase of its own centre, so

$$J(q) = \frac{c_A^2 K_{AA}(q) + c_B^2 K_{BB}(q) + 2 c_A c_B K_{AB}(q)\cos qR}{N}$$

with $K_{XY}(q)$ the product of the two radial momentum functions integrated over the plane at that momentum component, and $N$ the norm $c_A^2 + c_B^2 + 2 c_A c_B S$. At q = π/R the cosine is −1:

$$J(\pi/R) = \frac{c_A^2 K_{AA} + c_B^2 K_{BB} - 2 c_A c_B K_{AB}}{N}.$$

For two identical atoms at equal coefficients $K_{AA} = K_{BB} = K_{AB}$ and the numerator is $2c^2K - 2c^2K$, which is zero. That is the whole of the homonuclear result, written so that the two things it depends on are visible.

The two things are worth separating because they are usually named as one. The phase factor reverses at π/R, which is parity and is what the earlier essay’s factor is about; and the two amplitudes are equal, which is not parity at all but the fact that the two atoms are the same atom. A heteronuclear bond keeps neither — but it loses them independently, and the rest of this essay is about which loss does what.

The floor arrives before any charge moves #

Set the coefficients equal and take away every trace of polarity. C–C’s minimum is zero to a part in a hundred million, which is the quadrature’s floor and is the check that the construction reduces to the one it generalises.

C–O’s is 1.35 per cent of the profile’s peak. B–N’s is 1.63. C–N’s is 0.37. Li–F’s is 10.2.

None of that is charge transfer, because there is none: the two coefficients are equal, the electron is shared exactly evenly between the two atoms, and the minimum is still filled in. What fills it is that the two radial functions are different. Carbon’s 2s at an effective charge of 3.25 and oxygen’s at 4.55 have different momentum distributions, so at the one momentum π/R they contribute different amounts, and two different amounts subtracted do not give zero.

The ordering is the surprise. B–N’s floor is deeper than C–O’s, and the two have the same difference in nuclear charge — boron to nitrogen is two protons, carbon to oxygen is two — while C–O is by any measure the more polar bond. The floor is not about polarity or about the difference in charge; it is the mismatch of two functions evaluated at one momentum, and which momentum depends on the bond length. B–N is 1.281 ångström and C–O is 1.128, so π/R is smaller for B–N, the comparison is made further into the region where the two functions are most unlike, and the mismatch is larger.

That is a real prediction and an uncomfortable one: two bonds of similar polarity can have floors differing by a fifth for no reason except their lengths.

C–N is the check on that reading and it passes. Its two atoms differ by one proton rather than two, so its floor ought to be several times smaller, and it is: 0.37 per cent against C–O’s 1.35. Its bond length sits between the other two, so length cannot explain the drop. The floor depends on both, and neither alone orders it — which is why quoting it as “a measure of how different the two atoms are” would be wrong in both of the ways a one-variable summary can be.

Nothing can be done about it #

It is worth establishing that the floor is irreducible rather than an artefact of the coefficients chosen.

Cauchy–Schwarz on the measure the $K$ integrals use gives $K_{AB}^2 \le K_{AA} K_{BB}$, so

$$c_A^2 K_{AA} + c_B^2 K_{BB} - 2 c_A c_B K_{AB} \ge \left(c_A \sqrt{K_{AA}} - c_B \sqrt{K_{BB}}\right)^2 \ge 0,$$

and the first inequality is an equality only when the two radial functions are proportional at every momentum. Two Slater functions of different exponents are nowhere proportional, so the gap above the bound is strictly positive and cannot be closed by any choice of coefficients.

The split between the two is informative. At equal coefficients the bound accounts for 4.1 per cent of C–O’s depth and the irreducible part for the other 96. At a charge imbalance driven by a site-energy difference of two units the bound accounts for 77 per cent. So at small polarity the depth is almost entirely about the shapes of the two functions, and at large polarity almost entirely about their sizes — one quantity with two régimes, and the crossover is inside the range real bonds occupy.

That is the practical content of the whole essay and it is worth stating as a rule. Below the crossover the depth is dominated by something a measurement cannot change and a calculation of free atoms can supply; above it, by the quantity the measurement is for. A bond whose polarity is being read from its minimum therefore wants to be on the second side of the crossover, and knowing which side it is on needs the same free-atom calculation that supplies the floor. This is the same shape as the difficulty an accounting of a bond’s charge keeps running into: the quantity is well defined and its decomposition into the parts anybody wants is not.

It does measure polarity, in the direction chemistry moves charge #

With the floor established, the measure itself works.

Sweeping the site-energy difference in the chemical direction — charge towards the more electronegative atom — the depth rises monotonically for all five pairs. C–O goes from 1.35 per cent at equal coefficients to 3.74 at a difference of two units and 6.49 at four. Li–F goes from 10.2 to 21.5 to 27.2.

So the closing prediction of the earlier essay is right in its substance: the depth of the minimum reads the polarity. What it needs is the floor subtracted, and the floor is not a constant of the model — it is a property of the pair of atoms and of their separation, computable from free-atom radial functions and a bond length, and therefore knowable before any measurement.

The measure is then a difference of two numbers rather than a reading of one, which is what the directional zero avoided. The zero read a bond length with nothing fitted and nothing subtracted; this reads a polarity with a computed offset removed. That is a weaker instrument and it is the one that exists for real molecules.

How much weaker is worth quantifying, because it is not much. The floor is a free-atom quantity: two radial momentum functions, a bond length, one integral each. Everything in it is known to better than the profile can be measured, so subtracting it costs essentially no accuracy — where subtracting a fitted offset would cost whatever the fit costs. The instrument is weaker in the sense that it needs an input, and the input is not a parameter.

The polarity that would help is the wrong one #

There is a fourth thing in the sweep and it undercuts the obvious way of thinking about all of this.

The Cauchy–Schwarz bound vanishes when $c_A\sqrt{K_{AA}} = c_B\sqrt{K_{BB}}$, that is at a coefficient ratio of $\sqrt{K_{BB}/K_{AA}}$ — which is 1.24 for C–O, 1.14 for B–N and 2.03 for Li–F. All of those are greater than one, meaning more weight on the first atom, which is the more diffuse one.

The reason is a familiar inversion, and it is the uncertainty principle doing its ordinary work: a compact orbital is spread out in momentum, so at a large momentum like π/R it contributes more than a diffuse one does. Matching the two contributions therefore means over-weighting the diffuse atom to compensate — and chemistry does the opposite, putting charge on the compact, electronegative one.

So the two effects add. C–O’s shallowest attainable minimum is 1.24 per cent, at a charge imbalance of −0.24 — a quarter of an electron moved onto the carbon, which is not what carbon monoxide does. At the imbalance a real C–O bond has, the depth is two to three times the floor rather than below it.

That settles a question the framing invites: could a bond be found whose polarity exactly cancels its size mismatch, and which therefore shows the homonuclear zero despite being heteronuclear? Not among real bonds, because the cancelling polarity has the wrong sign.

Where the model stops being a bond #

One boundary has to be stated, because every number above is taken inside it.

The coefficients come from a two-level problem with overlap, and a basis whose two functions overlap has a feature an orthogonal basis does not: an orbital that localises on one atom must stay nearly orthogonal to the other, and past a site-energy difference it does so by taking a negative admixture of it. At that point the object is a lone pair with an orthogonalisation tail rather than a bond, and the quantity being computed is not what the essay is about.

For C–C and C–N that boundary sits at a difference of three units, for B–N and C–O at four, and Li–F reaches neither end of the sweep because its small overlap of 0.215 keeps both coefficients positive throughout. Every reading above is taken inside these ranges, and the boundary arrives first in the unphysical direction — which is again the diffuse atom’s fault, since that is where the weight goes onto the function whose tail reaches furthest.

What the model is #

One orbital, two functions, one bond length. This is a single σ orbital rather than a molecule: no other electrons, no core, no π system, and no correlation — the zeros this argument computes belong to one determinant and a fractional occupation of the antibonding partner would fill them in independently of anything here. A real heteronuclear diatomic’s profile is a sum over its occupied orbitals, and the sum is where a filled pair’s overlap residue lives — so the quantity here is a contribution to a measurement rather than a measurement.

The exponents are Slater’s and the coefficients are a two-level model’s. Both are the crudest defensible choices, and the reason for using them is that the floor is a statement about the ratio of two radial functions at one momentum, which is much less sensitive to the exponent rule than an energy would be. What it is sensitive to is the difference between the two exponents, which any rule gets to within a few per cent.

The bond lengths are measured — the equilibrium values for $\mathrm{C_2}$, BN, CO, CN and LiF — and π/R is computed from them, so the momentum at which the comparison is made is not a free parameter.

And the coefficients are one number. A real polar bond has a polarity that a two-level model with a fitted site-energy difference reproduces roughly and a calculation reproduces properly. Nothing above uses a particular polarity as a result; the sweep is the result, and the site-energy difference is its axis.

The K integrals are quadratures over interpolated radial tables, built once per atom and per effective charge, and the check on them is the homonuclear case: identical tables, identical coefficients, and a cancellation to eight figures. A quadrature error large enough to matter here would show up there first, because there the answer is exactly zero and every error is the whole of it.

What the five pairs are for #

The five pairs were chosen so that each isolates one variable against C–O, and it is worth saying what each one answers.

C–C is the control: same atom, same exponent, and a minimum of zero to eight figures. It is the check that the construction reduces to the homonuclear one and it is the only member of the set whose answer was known in advance.

C–N differs from C–O by one proton in one atom and by a bond length of 1.172 against 1.128 — so it separates the effect of the charge difference from everything else, and its floor is a quarter of C–O’s.

B–N has the same charge difference as C–O and a longer bond, so it separates the effect of the length; its floor is larger.

Li–F is the extreme: effective charges of 1.30 and 5.20, an overlap of 0.215 rather than the 0.47 to 0.52 of the rest, and a floor of 10.2 per cent. It is the case where the two functions are least alike and the one where every effect in this essay is visible at once — and its small overlap is why it never reaches the coefficient sign reversal that bounds the others’ sweeps.

What is missing from the set is a pair with a large charge difference and a short bond, which is what would separate the two variables cleanly. There is no such first-row pair: a large charge difference means one very diffuse atom, and a diffuse atom makes a long bond. The confounding is chemical rather than a defect in the choice.

An exact zero is two conditions wearing one name #

The habit: when a quantity is exactly zero, count how many conditions are producing it.

The homonuclear zero at π/R reads as one thing — a cancellation of phase factors, from the parity of the orbital — and it is two: the phase factor reverses and the two amplitudes are equal. Both hold automatically when the two atoms are the same, which is why nothing in the homonuclear analysis has to mention the second, and why the natural generalisation to two different atoms carried only the first.

The corollary is about what a broken exactness becomes. Zeros that break into small numbers are usually useful, because the small number measures the thing that broke the zero. Here the zero breaks into a sum of two small numbers with different causes — a size mismatch and a shape mismatch — and one of them is computable in advance while the other is what was wanted. That makes the broken zero a usable instrument and not a clean one, and the difference between those is a subtraction that has to be got right.

Who worked this out, and when #

The two-centre form of a momentum density, and the interference factor it carries, are as old as the momentum-space treatment of molecules. Compton profiles of heteronuclear diatomics have been measured for decades and are the standard test of a calculated momentum density. The two-level model with overlap is Wolfsberg and Helmholz’s arrangement of a much older idea, and the sign reversal of the minor coefficient past a site-energy difference is a standard feature of non-orthogonal bases.

What is computed here is the σ profile of five pairs of first-row atoms at their measured separations, the depth at π/R across a sweep of the polarity, the Cauchy–Schwarz decomposition of that depth into a size mismatch and a shape mismatch, and the coefficient ratio that would minimise it.

The numbers worth carrying are 1.63 and 1.35 per cent — two floors, on two bonds with the same difference in nuclear charge, in the order nobody would guess.

Still open: the whole molecule, and a third function #

The obvious open question is the sum. Carbon monoxide has ten valence electrons in four occupied orbitals and this essay computes one of them, so what a measurement of CO’s profile would show is a sum in which the σ bond’s floor competes with the overlap residues of its filled pairs. Those residues are negative at zero momentum and this floor is positive at π/R, so they are not simply additive obstructions — and whether the polarity is readable from the total or only from a difference between two isoelectronic molecules is a sum over four orbitals with a construction that now exists for both halves.

The nearer question is the third function. Every orbital here is built from one function per atom, and a real polar σ bond mixes s and p on both atoms — which changes the floor in a way that is not obviously a worsening. A p function’s momentum distribution peaks away from the origin, so mixing p into the diffuse atom’s contribution moves it towards the compact atom’s shape, and the shape mismatch that this essay finds irreducible at one function per atom may be substantially reducible at two. That is a calculation with one more parameter and the same integrals, and it decides whether the floor is a property of a polar bond or of a minimal basis.