Hybridisation — where it appears
Named by 15 essays across 4 fields — each of them below, with the objects they name alongside it.
Hybridisation does not explain
Methane's photoelectron spectrum has two bands, not one. Four equivalent sp³ bonding orbitals cannot produce that, and the resolution is that hybridisation was never a claim about what a measurement would find.
Hybrids are a basis
An sp³ hybrid set is an orthogonal matrix applied to the atomic orbitals. Rotating a basis changes no observable, so asking whether the electrons are really in hybrids is asking which coordinate system nature prefers.
Why water is bent
The standard answer is lone pair repulsion, it predicts the right direction, and it cannot predict the magnitude. A better rule can, and the heavier hydrides show where both accounts run out.
Electronegativity is not one quantity
Four scales, four definitions, four sets of units, and a defence — "they correlate well" — that answers a question nobody asked. Two scales can correlate at 0.99 and still put hydrogen on the wrong side of carbon.
Bent's rule, against the substituent series
More electronegative substituents get more p character, so the angle between them shrinks. The rule predicts a trend, the trend is observed, and the quantity it depends on turns out not to be one quantity.
The angle does not fix the hybridisation
Two equivalent hybrids are orthogonal only at one relation between their s character and the angle between them. Run it backwards on measured bond angles and water's bonds come out sp³·⁹⁹, hydrogen sulfide's sp²⁷, and phosphine's lone pair takes eighty-four per cent of the one s orbital.
Hybrids that were never orthogonal
Two sp³ hybrids are orthogonal at 109.47° and nowhere else. Water's are drawn at 104.5° and overlap by 0.062; cyclopropane's are drawn at 60° and overlap by 0.625, which is not a small correction to a basis but a description that has stopped being one.
Water's lone pairs are not a pair
Every course draws two equivalent lone pairs on water, pointing away from the hydrogens like a pair of ears. Its photoelectron spectrum shows the two bands they would produce at 12.6 and 14.7 electronvolts, two point one apart, in different symmetry species.
The angle a ring cannot have
A closed ring of equal bonds has to turn through a full circle, so its bond angles cannot average more than 180°(n−2)/n. Three, four and five atoms are below the tetrahedral angle at every geometry whatever; six is above it, and reaches it only by leaving the plane.
Three bent bonds, and the same hybrid
A triple bond localises into three bent bonds at 101.537 degrees to one another, each an sp⁵ hybrid with exactly one sixth s character. A double bond's two bent components are sp⁵ hybrids at 101.537 degrees. The two are the same hybrid, built out of different frameworks and in different numbers, and the reason is that a third of a half is a half of a third.
The angle that does not have to be searched for
A carbonyl's two bent components have no symmetry making them equivalent, so the mixing that best localises them looks like something to search for and their s characters look like two different numbers. Neither happens. The localisation functional is a quadratic with no linear term, so its maximum is at forty-five degrees or at the ends — bent bonds or canonical ones, decided by one inequality, with nothing in between.
The criterion that has never said no
The two-orbital mixing criterion is a tautology on a carbonyl's lone pairs, and it is tempting to contrast that with the double bond, where the criterion seems to discriminate. Both halves are wrong. The double bond's margin is wider, ethene's is infinite because symmetry puts both centroids at the same point, and the criterion has never returned a negative on any case it is usually put to — though the same three orbitals supply two pairs it refuses outright.
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.
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.
A verdict that flips and changes nothing
Run the two-orbital mixing criterion along H₂C=CH⁻, H₂C=NH, H₂C=O and H₂C=F⁺ and one pair finally crosses its boundary — the π and the lone pair, a third of the way from oxygen to fluorine. It is the first real test the criterion has had, and the localised orbitals on the far side of it are the ones on the near side: two bent bonds and a lone pair that stays itself.
Named alongside it
The objects these essays reach for when they reach for this one.
Lone pairBasisBond angleLocalisationModel limitElectronegativitys charactersp³ hybridsMolecular orbitalOrthogonalityOverlap integralTetrahedral angle