Molecular Geometry & Bonding
what the shape decides
Series

Orbital — the series

19 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The 2pz orbital. The 2pz orbital at the contour enclosing 90 per cent of its density — a level solved for by integration rather than chosen. The two colours are the two signs of the wavefunction, which is what distinguishes a bonding interaction from an antibonding one. Contours drawn: 2pz at 90% of its density, |ψ| = 9.48e-3.

    What an orbital is

    Not a region the electron occupies, not a path it follows, and for any atom but hydrogen not an exact anything. An orbital is a one-electron wavefunction, and almost every difficulty in this subject comes from forgetting that.

    part 1 · orbitals
  2. The radial function of 3s. The radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.

    Nodes

    An orbital with quantum numbers n and l has exactly n−l−1 radial nodes and l angular ones. That is a count, it is exact, and it is the fastest way to catch a drawing that is wrong.

    part 2 · orbitals
  3. The radial function of 1s. The radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.

    Where the electron is

    The wavefunction is largest at the nucleus, the electron is most likely to be found a bohr out, and the ninety-per-cent contour is at 2.66. Three numbers, all correct, all answering different questions.

    part 3 · orbitals
  4. The radial function of 3s. The radial part of the wavefunction, which changes sign at each node, and the radial distribution, which is the probability of finding the electron in a shell at that radius. The second vanishes at the nucleus and the first does not.

    The radial distribution across the periodic table

    A 4s orbital is bigger than a 3d by every measure of size except the one that decides which fills first. Penetration is a feature of a small inner peak, and the periodic table's shape depends on it.

    part 4 · orbitals
  5. Orbitals at the 90 per cent contour. Several orbitals drawn at the same enclosed fraction and, unless stated otherwise, at the same scale — so the sizes on the page are the sizes. Each contour was solved for separately by integrating that orbital's own density. Contours drawn: 2pz at 90% of its density, |ψ| = 9.48e-3; 2px at 90% of its density, |ψ| = 9.49e-3; 2py at 90% of its density, |ψ| = 9.49e-3.

    Complex harmonics against real ones

    The p orbitals every chemist draws are not eigenfunctions of anything. They are real combinations of the complex solutions, chosen because they point along axes — and the choice is invisible until a magnetic field makes it matter.

    part 5 · orbitals
  6. 4s and 3d from Sc to Zn. The mean radius of the 4s and 3d orbitals across the elements Sc to Zn, at the nuclear charge each shell actually feels. The screening is Slater's, which is fitted; the radius that follows from it is the closed form for a hydrogen-like orbital, which is not.

    What an electron actually feels

    A 3d orbital is less than half the size of the 4s beside it and fills second anyway. The charge an electron feels is not the nuclear charge, the correction is a fit rather than a derivation, and the two facts together explain the shape of the periodic table.

    part 6 · orbitals
  7. 2s and 2p from B to Ne. The mean radius of the 2s and 2p orbitals across the elements B to Ne, at the nuclear charge each shell actually feels. The screening is Slater's, which is fitted; the radius that follows from it is the closed form for a hydrogen-like orbital, which is not.

    What the screening model cannot see

    Slater's rules put the 2s and the 2p in one group, so they give both orbitals exactly the same effective nuclear charge at every element from boron to neon. The two are separated by several electronvolts in all six, and the model has no term that could produce it.

    part 7 · orbitals
  8. The exponent the molecule chooses, and what it buys. The 1s exponent that minimises the energy of a one-electron diatomic, against the separation of the nuclei, with the binding curves at that exponent and at the free atom's. Held at ζ = 1 the bond comes out at 2.49 bohr and binds 0.0648 hartree; with the exponent free it comes out at 2.00 bohr at ζ = 1.238 and binds 0.0865. The exact answer for this molecule is 2.00 bohr and 0.1026.

    The atom does not bring its own orbital

    Build a one-electron diatomic from two hydrogen 1s functions and it comes out 25 per cent too long and 37 per cent too weakly bound. Let the molecule choose how large those functions are and the bond length is right to three figures, at an exponent of 1.238 — the orbital contracts by a quarter when the bond forms.

    part 8 · orbitals
  9. Where the electron is, and how fast it is going. The radial distribution in position on the left and in momentum on the right, for the same orbitals. The two run opposite ways: the 1s is the most compact in space and the widest in momentum, and every excited orbital that spreads out in one narrows in the other. Both are normalised, both are the same function, and neither is more fundamental than the other — the transform loses nothing and adds nothing.

    The orbital in momentum space

    Every orbital has a second picture as complete as the first and almost never drawn. Nothing is added by taking it — it is the same function in the other variable — but the uncertainty product falls out of it, and the functions quantum chemistry is built from turn out to be the only ones that attain the bound.

    part 9 · orbitals
  10. The same count in both pictures, and no rule between them. Every hydrogenic orbital with a radial node, drawn twice: its position nodes on the left axis and its momentum nodes on the right, at the same nuclear charge. The counts are identical and exact — n − l − 1 in each — because the two radial functions are polynomials of the same degree. The positions are unrelated: a node three quarters of the way out in one picture is not three quarters of the way out, or anywhere in particular, in the other.

    The nodes in the other variable

    An orbital has n − l − 1 radial nodes, and it has exactly that many in momentum too — the two radial functions are polynomials of the same degree. Nothing pairs one node with another: they are zeros of two different classical families. What is exact is the product over all of them, which is a ratio of factorials and does not depend on the nuclear charge at all.

    part 10 · orbitals
  11. A bond is a cosine in momentum space. The two factors a two-centre bonding orbital's momentum density is made of. One is the atomic momentum density, unchanged by the bond and falling as the eighth power of the momentum. The other is cos²(q·R/2), the interference between the two centres, whose period is fixed by the bond length and by nothing else. Everything that distinguishes a bond from two atoms in this picture is that cosine — and it has zeros where the atomic factor has none.

    Oblate in the picture nobody draws

    Every drawing of a σ bond shows a density stretched along the bond, and the position-space calculation agrees: the second moment along the axis is twice the one across it. In momentum the same orbital is flattened in the same direction, because the interference between the two centres cuts the distribution off at π over the bond length — and that cut-off is a zero a measurement could find.

    part 11 · orbitals
  12. An antibonding occupation fills the zero and moves the minimum. The profile along the bond near π/R, per electron, on a logarithmic scale, for five antibonding occupations of the same two orbitals. With nothing in the antibonding orbital the profile is exactly zero at π/R = 1.573. Two hundredths of an electron leave a minimum at 1.608, which reads the separation as 1.954 bohr instead of 1.997. At 0.104 the minimum becomes a flat shoulder, and the Heitler–London bond, at 0.236, has none.

    The zero belongs to one determinant

    A bonding orbital's momentum profile along the bond is exactly zero at π/R, and that zero reads a bond length with nothing fitted. It is a property of putting every electron into that one orbital. Any antibonding occupation fills it in linearly and drags the minimum outward, a tenth of an electron erases it, and the valence-bond wavefunction built from the same two functions never has one at any separation.

    part 12 · orbitals
  13. The zero at π/R follows parity, not bonding. The bonding and antibonding combinations of three atomic functions on nitrogen, each along the bond and each normalised to its own largest value, against momentum in units of π/R. For 2s the bonding combination is zero at π/R and the antibonding one is not. For 2p along the bond it is the other way round: the σ bond carries a sine, is zero at the origin and near its largest at π/R, and the antibonding combination carries the cosine. For 2p across the bond the π bond carries the cosine again. The factor is a cosine exactly when the orbital's inversion parity matches the atomic function's.

    The zero is a parity, not a bond

    A hydrogen-like σ bond has a momentum profile along its axis that vanishes at π/R, and it is natural to read that zero as a bond's signature. Built from 2p functions pointing along the axis, the σ bond carries a sine instead and sits at 83 per cent of its peak there. Which factor an orbital carries is decided by whether its inversion parity matches its atom's, and bonding has nothing to do with it.

    part 13 · orbitals
  14. One zero becomes three, because the average couples them. What the interference factor becomes when the molecule is averaged over every orientation, one curve per angular kind. A directional profile along the bond carries cos(qR) for every orbital, so every cosine combination is exactly zero at q = π/R. The average replaces it by j₀(pR) for an s combination, j₀ − 2j₂ for a pσ one and j₀ + j₂ for a pπ one — because the cross term multiplies the interference by the orbital's own angular density and the average of the product is not the product of the averages. Their first zeros are at 0.66π, π and 1.43π.

    The pair that cancels only at zero overlap

    A directional Compton profile along a bond is exactly zero at π/R for every cosine combination, which makes the depth there a count of parity mismatches. A gas measurement averages over orientations, and the question was whether the count survives. It does not, for a reason nothing in the directional picture shows: the bonding and antibonding combinations of one atomic function are normalised by 2 + 2S and 2 − 2S, so a filled pair leaves a residue proportional to its overlap — and in nitrogen that residue is the largest single term, from a pair whose imbalance is zero.

    part 14 · orbitals
  15. The minimum fills in, and it fills in before any charge moves. The σ profile along the bond for three pairs of atoms, each with equal coefficients on the two atoms — no charge transfer at all. The homonuclear pair has an exact zero at π/R because its two atomic contributions are identical and cancel. The other two do not, because their radial functions differ: the cancellation at π/R needs cₐ gₐ = cᵇ gᵇ everywhere, and two Slater functions of different exponents are nowhere proportional.

    A floor no charge transfer explains

    A σ bond between two identical atoms has a momentum profile that is exactly zero at π/R, because the two atomic contributions are identical and cancel. Two different atoms cannot cancel there at any coefficients, so the depth of the minimum becomes a measure of polarity — and it is one, monotone in the charge imbalance in the direction chemistry moves charge. What it is not is a measure with a zero: it has a floor set by how far the two radial functions are from proportional, and B–N's floor is deeper than C–O's although C–O is the more polar bond.

    part 15 · orbitals
  16. A p function fills the polar bond's minimum, and a pair of them fills it nine times over. Carbon monoxide's σ profile along the bond at equal coefficients, for three bases: one s function on each atom, a quarter p mixed into carbon's, and a quarter p on both atoms in the sense that maximises their overlap. The minimum at π/R is 1.35% of the peak with s alone, 3.63% with p on carbon and 12.53% with p on both. The p function proposed to reduce the floor raises it, and the pair raises it most.

    The third function fills the floor

    A polar σ bond built from one s function per atom has a momentum profile whose minimum at π/R never reaches zero — for carbon monoxide, 1.35 per cent of the peak. The natural hope was that a p function mixed into the diffuse atom would reshape it towards the compact one and lower that floor. It cannot. At π/R an s function on one atom and a p function on the other are exactly out of phase, so a p function on one atom cancels nothing and only adds itself. With p on both atoms in the sense that makes the bond, the pair adds fifteen times as strongly as the s pair cancels: a quarter p takes the minimum to 12.5 per cent. That is eight times the whole range through which polarity moves the s-only bond.

    part 16 · orbitals
  17. Carbon monoxide's three σ orbitals, and what each does at π/R. The momentum profile along the bond of each occupied σ orbital of carbon monoxide, with weights from an extended Hückel calculation over carbon and oxygen 2s and 2pσ, each scaled to its own peak, against momentum in units of π/R. At π/R the lowest, 3σ, reads 2.52% of its peak; the middle, 4σ, 8.40%; and the highest, 5σ, 85.3% — its profile peaks at 0.82 π/R and has no minimum near π/R at all.

    The lone pair sits where the floor was

    A polar σ bond's momentum profile has a minimum at π/R whose floor was proposed as a measure of polarity, and the essay that swept the p weight could not say whether carbon monoxide's own σ bond sat near the s-only floor of 1.35 per cent or the quarter-p value of 12.5. With the weights computed — carbon and oxygen 2s and 2pσ, extended Hückel — the answer is that carbon monoxide has no such bond. Its lowest σ orbital reads 2.5 per cent, its middle one 8.4 per cent of its own peak, and its highest, the orbital with the lone pair's sense, has no minimum at π/R at all: it peaks there. The six σ electrons together have no minimum anywhere, and the polar molecule reads lower at π/R than nitrogen does.

    part 17 · orbitals
  18. Lithium hydride's one σ bond, and the minimum it would need to be covalent to have. The momentum profile along the bond of lithium hydride's occupied σ orbital, from an extended Hückel calculation over lithium 2s and 2pσ and hydrogen 1s, each curve scaled to its own peak, against momentum in units of π/R. As computed the orbital is 91% hydrogen and its profile falls monotonically, reading 10% of its peak at π/R. The same lithium 2s and hydrogen 1s with a quarter of the orbital on hydrogen do have a minimum — at 0.525 π/R, not at π/R, just below lithium's 2s node in momentum at 0.624 π/R.

    The minimum belongs to the heavy atom's node

    Carbon monoxide had no single σ bond for the momentum-space polarity reading to read, so the next place to look was a hydride, where there is one. Lithium hydride's computed σ orbital is ninety-one per cent hydrogen and its profile falls straight through π/R without a minimum. Across all seven first-row hydrides one occupied orbital of thirteen has a minimum at all, and where a lithium-to-boron hydride could have one, it would not be the bond's fringe: it would be the heavy atom's 2s node in momentum, a flat point that a little hydrogen tips over.

    part 18 · orbitals
  19. Lithium's 2s three ways, in momentum: three nodes, one of them past the fringe. The momentum radial functions of the same three lithium 2s orbitals, weighted by p² as a radial density is, each scaled by its largest magnitude and signed to start negative, against momentum in units of lithium hydride's π/R. Every one has exactly one node — the hydrogenic at 0.624 π/R, the Hartree–Fock at 0.889 and the nodeless Slater function, which has no node in position, at 1.080, just past the fringe.

    The node a nodeless orbital has

    Lithium hydride's would-be minimum in momentum was traced to the 2s function's node at 0.624 π/R, with a warning that the position belonged to the hydrogenic model and that a nodeless Slater function would have no node at all. Solving the lithium atom properly moves the node to 0.889 π/R; the nodeless function has one too, at 1.080, because a node in momentum is not a node in position but a disagreement between a function's sum and its slope at the nucleus. Lithium hydride's own bond has no minimum with any of the three. What the function decides is whether a minimum can be born at the node at all.

    part 19 · orbitals

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