The zero is a parity, not a bond

A σ bond made of two 1s functions has a momentum profile along its axis that is exactly zero at π/R, and its antibonding partner has its zeros somewhere else — at the origin and at 2π/R. From those two facts it is a short step to a rule: bonding electrons vanish at π/R, antibonding electrons do not, and the profile there counts the antibonding ones. The previous calculation found exactly that for hydrogen, where it turned the depth of the profile at π/R into a linear read of an antibonding occupation.

The step is wrong, and 2p functions are enough to show it.

What a sum and a difference transform to #

Put an atomic function at plus half the bond length and a copy of it, pointing the same way, at minus half. The transform of a translated function is the transform times a phase, so their sum transforms to $2\cos(p_z R/2)$ times the atomic transform and their difference to $-2i\sin(p_z R/2)$ times it. That is true of every atomic function whatever its shape. The only question is which of the two is the bonding combination.

For an s function the sum is bonding: two positive functions add in the region between the nuclei. For a p function across the bond — a $p_x$ on each atom, both pointing the same way — the sum is bonding too, for the same reason: the lobes on each side of the axis match. For a p function along the bond the difference is bonding. Both $p_z$ functions point in the $+z$ direction, so the one on the upper atom presents its negative lobe to the region between the nuclei and the one on the lower atom presents its positive lobe; adding them puts opposite signs in the middle, and it is subtracting them that builds up density there.

So the 2p σ bond carries a sine. In the figure above it is zero at the origin — every $p_z$ profile along $z$ is, because the angular factor vanishes there — and it is near its largest at π/R, at 83 per cent of its own peak. The 2p σ antibonding combination carries the cosine and is exactly zero at π/R. The s pair on the left and the π pair on the right behave as hydrogen’s did; the middle pair is reversed.

The rule parity writes down #

The column that predicts the factor is not the bonding column. It is a comparison of two parities. A homonuclear diatomic has a centre of inversion, so every one of its orbitals is either even under inversion (g) or odd (u); every atomic function has a parity too, $(-1)^l$, even for s and odd for p. The factor is a cosine exactly when the two parities are equal, on all six rows: the 2s σg is g built from g, the 2p σu* is u built from u, the 2p πu is u built from u, and those three are the cosines. The other three mix parities and carry sines.

The reason is short. A momentum function has the same inversion parity as the position function it came from, and a cosine is even in $p_z$ while a sine is odd. The atomic transform contributes its own parity, $(-1)^l$, so the product is even when the trigonometric factor matches the atom and odd when it does not — and the orbital’s parity fixes which of those it has to be. Nothing in that argument knows which combination lowers the energy.

This is the same parity that decides which electronic transitions are allowed, and the same bookkeeping that makes a d–d band weak: a statement about g and u is a statement about inversion and nothing else. Here it decides where a momentum distribution vanishes.

The numbers alongside are the normalisations and the values. The 2s overlap at nitrogen’s bond length is 0.496; the 2p σ overlap, with both functions pointing the same way, is −0.321, and the sign is the whole story of why the difference is bonding; the π overlap is 0.280. The sine combinations sit at 0.218, 0.834 and 0.507 of their peaks at π/R, and the three cosines are zero there to the last bit. The identity that tied hydrogen’s pair together — the two combinations weighted by their normalisations adding to four atoms — holds for all three atomic functions, checked at three momenta each to twelve digits.

What the profile at π/R counts #

If the cosines are exactly zero at π/R and the sines are not, a molecule’s whole profile there is a sum over its electrons in sine combinations and nothing else. That is not the set of antibonding electrons, and the difference is not small.

The six molecules are $\mathrm{Li_2}$, $B_2$, $\mathrm{C_2}$, $\mathrm{N_2}$, $\mathrm{O_2}$ and $\mathrm{F_2}$ at their measured bond lengths, with minimal-basis occupations and hydrogenic functions at Slater’s effective charges. In nitrogen the largest single contribution at π/R is the 2p σ bond, 0.052 per electron of the molecule; the 2s antibonding pair adds 0.015 and the core adds 0.031. Oxygen and fluorine add their π* electrons, which are sines as well, and by fluorine the π* pair is the largest part.

And the core is in every bar. Two 1s functions on atoms two bohr apart barely overlap, but a filled 1s pair is a bonding and an antibonding combination both occupied, and the antibonding one is a sine. Its contribution at π/R is close to two atoms’ worth of 1s profile at that momentum, and it is there in every molecule — which means no first-row molecule’s whole momentum profile along its bond is zero at π/R. Carbon comes closest, at 0.113 of its value at zero momentum; fluorine is furthest, at 0.655.

That settles the question left open by the hydrogen calculation about a second pair sitting under the first. The worry was that a second pair would put a background under the σ pair’s depth. It does, but the more important finding is what the background is made of: the depth at π/R counts sine electrons, and in hydrogen the sine electrons happened to be the antibonding ones only because every function in the basis was an s.

Only one valence shell keeps the zero #

Remove the core and ask the same question of the valence electrons alone, with each molecule’s momentum measured in units of π over its own bond length so that every fringe sits in the same place.

Lithium’s valence profile is zero at π/R and no other molecule’s is. Lithium’s valence shell is a single s bonding pair, which is exactly hydrogen’s situation at a larger scale: one cosine, nothing else occupied. Boron adds the 2s antibonding pair, and its share at π/R becomes 0.059; carbon’s is 0.040; nitrogen’s, with the 2p σ bond occupied, jumps to 0.182; oxygen’s is 0.362 and fluorine’s 0.631.

The order is worth reading against the numbers chemistry usually attaches to these molecules. Their bond orders are one, one, two, three, two and one. Their antibonding valence counts are zero, two, two, two, four and six. The share at π/R orders them lithium, carbon, boron, nitrogen, oxygen, fluorine, which is neither. Nitrogen, with the strongest bond of the six, has more at π/R than carbon or boron, because its extra bond is a p σ bond and a p σ bond is a sine.

So a momentum profile along the bond does not measure bonding at π/R, and it does not measure antibonding either. It measures parity mismatch, electron by electron, weighted by each atomic function’s own profile at that momentum. Where every function is an s, parity mismatch and antibonding coincide, and hydrogen’s result was a true statement about a coincidence.

The overlap sets how loud each sine is #

A count of sine electrons would be a clean thing to read if every sine electron contributed the same amount, and they do not. Each contributes its atomic function’s profile at π/R, times two, over $2 - 2S$ — and both factors change across the series.

The atomic factor is the larger effect. π/R is a momentum set by the bond length, and the atomic profiles are set by the charge, so the question at each molecule is where π/R falls on a profile whose width is about the effective charge over the principal quantum number. Lithium’s bond is long, 5.05 bohr, and its valence charge is small, 1.30: π/R is 0.62 atomic units, well out on a 2s profile that is only about 0.65 wide. Fluorine’s bond is 2.67 bohr and its charge 5.20, so π/R is 1.18 on a profile more than twice as wide, and any sine combination there contributes close to its full atomic value. That is most of why the valence share rises steadily along the second half of the row even though fluorine’s extra electrons are in orbitals that, in energy, look much like oxygen’s.

The overlap is the smaller effect and the more interesting one, because it enters the two combinations oppositely. A sine combination’s weight is $1/(2-2S)$ and a cosine’s is $1/(2+2S)$, so a large positive overlap makes the sine combination louder and a negative one — the 2p σ pair’s, −0.31 across boron, carbon and nitrogen — makes it quieter. Nitrogen’s 2p σ bond is the one sine combination with a negative overlap, and it is still the largest contribution in the molecule, because at 2.07 bohr and charge 3.9 its π/R falls near the peak of a $p_z$ profile.

By fluorine both overlaps have fallen below two tenths, 0.173 and −0.176, and every combination’s normalisation is within a fifth of an atom’s. The interference is still there; it is quieter, and the profile is close to what two separate atoms would give. That is the momentum-space face of fluorine’s weak bond, and it is the one row of the table where the parity bookkeeping matters least.

The last column adds the core. The whole share is above the valence share on every row, and by far the most for lithium, boron and carbon — where the core supplies all, two thirds and two thirds of what sits at π/R, against a tenth for fluorine — because a 1s profile is broad and nearly flat across the momenta in question: on a molecule whose valence profile has fallen far by π/R, a flat core contribution is most of what is left. No row reaches zero in the last column, which is the one statement on the page that does not depend on the basis at all.

Mixing fills whatever is left #

The six molecules above are drawn with no mixing between 2s and 2p, which is the minimal-basis picture every diagram of them starts from and not what their orbitals are. Hybridisation is a change of basis, and in the first-row diatomics the two σg orbitals built from 2s and from 2p mix substantially, which is why nitrogen’s highest occupied σ orbital lies where it does.

A σg orbital with both parts in it is a cosine combination plus a sine combination, and at π/R the cosine is zero, so its value there is the p share times the 2p σ bond’s own value, divided by the mixture’s norm. The cross term between the two parts vanishes at π/R identically — it carries a cosine times a sine — which is why the dots and the line agree. At a p share of three tenths the orbital sits at 0.086 per electron at π/R.

Every σ orbital in a real first-row molecule is mixed in this sense, so even lithium’s zero would not survive a calculation that allowed its 2s to borrow 2p character. The one clean case left standing is the one the whole argument started from: a basis of s functions alone, which is hydrogen.

How the profiles were built #

Each atomic function is hydrogenic, with the charge Slater’s rules give: 3.9 for nitrogen’s 2s and 2p, 6.7 for its 1s. Its momentum function is the Hankel transform of the radial function, computed by quadrature and tabulated, and checked against the closed form for 1s and against the kinetic energy for every orbital. The directional profile along the bond integrates the squared momentum function over the plane perpendicular to it, with the angular factor written out: $1/4\pi$ for s, $3q^2/4\pi p^2$ for a p along the bond, and $3\rho^2/8\pi p^2$ for a p across it after averaging over the azimuth.

The overlaps are three-dimensional quadratures over all of space on a mapped Gauss rule, the same one that draws every overlap figure. A combination’s profile is then $(2 \pm 2\cos qR)$ times the atomic profile over $2 \pm 2S$, with the sign chosen by whether it is a sum or a difference and $S$ the overlap of the two functions pointing the same way.

The molecules’ bond lengths are the spectroscopic equilibrium values, from 2.673 ångström for $\mathrm{Li_2}$ to 1.098 for $\mathrm{N_2}$. The occupations are the minimal-basis ones with the π orbitals below the 2p σ for boron, carbon and nitrogen and above it for oxygen and fluorine, as their photoelectron spectra order them; since both are counted as they are, the order changes nothing at π/R.

The checks, run wherever these figures are drawn: every combination carries one electron along the bond; each cosine combination is zero at π/R and each sine combination is not; the factor is a cosine exactly when the orbital’s parity matches its atomic function’s, on all six; the s bond and the p σ bond carry opposite factors; each atomic function’s two combinations add to four atoms at three momenta; lithium’s valence profile is zero at π/R and the other five are not; every molecule’s whole profile is nonzero there. The refusal is the mixed σ orbital: at zero p share it must keep the s bond’s zero, and at a tenth it must not.

What these functions cannot settle #

Hydrogenic functions at Slater’s charges are a model of an atom, not an atom. A hydrogenic 2s has a node that a real 2s has in a different place, and the 1s and 2s here are not orthogonal to each other because their charges differ. Neither affects the parity rule, which is about phases and holds for any radial function at all. Both affect the numbers: the shares at π/R are this basis’s, and a better basis moves each of them.

The occupations are minimal-basis occupations. A correlated wavefunction puts some of every pair into combinations of the other parity, and the previous calculation showed what that does to a zero. Here there is no zero to fill except lithium’s valence one.

And momentum profiles are measured on whole molecules, usually in the gas phase and so averaged over orientation. A profile along the bond needs oriented molecules, which exist for surfaces and some crystals and not for a gas. The spherically averaged profile of a diatomic still carries the interference, as a weight on $\sin(qR)/qR$ rather than on $\cos qR$, and whether the parity rule survives that average in a recognisable form is a separate calculation.

Bonding was never the variable #

The general point is about what a clean result licenses. Hydrogen’s momentum profile gave a zero, the zero sat where bonding electrons were and antibonding electrons were not, and the obvious generalisation named the correlation it saw. The real variable was a parity that happens to agree with bonding for every s function and disagrees with it for p functions along a bond — and nothing about the hydrogen calculation could have revealed that, because it had no p functions in it.

A regularity found in a basis with one kind of function should be stated in terms the basis cannot vary, and bonding was not one of them. Parity is. The same habit is what separates a symmetry argument from a correlation with a symmetry label attached, and it is cheapest to apply at the moment a result looks general.

Still open: the orientational average, and a heteronuclear bond #

The obvious open question is what survives orientational averaging. A gas-phase Compton profile averages the directional profile over every orientation of the bond, which turns the cosine factor $\cos qR$ into $\sin(qR)/qR$ and the sine factor’s interference term into its negative. Both have zeros, at different places and of much smaller amplitude, so a parity-mismatch count might still be readable from the averaged profile of a first-row molecule — as a sign of the oscillation’s first swing rather than as a value at one momentum. Whether the core pair’s nearly flat contribution swamps it is a quadrature away.

The nearer question is a heteronuclear bond, where there is no inversion centre and no parity to match. Carbon monoxide’s orbitals are neither g nor u, the two atomic transforms have different amplitudes, and the cosine and sine no longer separate — every orbital carries both, weighted by how unequally its two atoms contribute. The prediction from the rule above is that the profile at π/R stops being a clean count of anything and becomes a measure of the orbital’s polarity; checking it is the same construction with two charges instead of one.