sp3 hybrids

One of the figures on bonding models: Valence bond, molecular orbital and hybrids, drawn as descriptions of one thing rather than as competing pictures.

Five essays draw this figure, each at the values its own argument needs rather than at the setting shown above. What each one uses it to show is below, in the words of its own caption.

Hybrids are a basis

The four sp³ directions, with the angle between them computed from the coefficients rather than quoted. The set is orthonormal to better than one part in $10^{15}$, which is what makes it a rotation rather than an arbitrary construction.

Three sp² hybrids: one third s and two thirds p in each, at 120 degrees in a plane. The remaining p orbital is untouched and perpendicular to that plane, which is where a pi bond goes.

Two sp hybrids: half s and half p, at 180 degrees. Two p orbitals are left over, perpendicular to the axis and to each other.

Bent's rule, against the substituent series

The four sp³ directions at 109.4712 degrees, measured off the coefficients rather than quoted. Bent’s rule is about what happens when the four hybrids stop being equivalent, and the composition of each is what changes.

The equally-weighted three-fold set, for comparison with the unequal ones above. Every angle here is 120 degrees and every composition is one third s, computed from the coefficients rather than quoted — which is the case Bent’s rule reduces to when the four substituents are alike, and the case from which every departure in this essay is measured.

And two sp hybrids, half s each, at 180. Between pure p at 90 and sp at 180 lies every angle a two-coordinate centre can adopt, so the named cases are landmarks on a continuum rather than a set of options.

Hybrids that were never orthogonal

The overlap between two equivalent hybrids of a stated label, against the angle they are drawn at. Each curve crosses zero once. Molecules sit on their own label’s curve at their measured angle, and the height above the axis is how far their hybrids are from being a basis.

The same three curves with water marked instead of cyclopropane. Water’s angle of 104.5 degrees is below the sp³ crossing rather than far below it, so its bond hybrids are non-orthogonal by a small amount — and the sign of that overlap is the same sign cyclopropane has, which says the two molecules are on the same side of the same relation rather than in different situations.

Two of the curves alone, where the crossings can be read off. The sp³ curve crosses zero at exactly 109.47 degrees and the sp² curve at exactly 120, and both numbers come out of the coefficients rather than being marked on. A molecule at any other angle has hybrids that are not orthogonal, and the amount is the height of the curve at its angle.

The hybrids that point outside the bonds

Why the misalignment costs something: the overlap of two hybrids as they are turned away from each other. Twenty-two degrees is where cyclopropane sits on this curve, and it has lost eight per cent.

The angle that does not have to be searched for

The overlap between two equivalent hybrids as a function of the angle between them, which is the geometrical half of the integral the argument needs. The angular dependence is right; the radial functions are what put the σ overlap two orders of magnitude too low.