Molecular Geometry & Bonding
what the shape decides
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The thread: Counted, not quoted

Node counts, bond angles, overlap integrals and point groups are all computed while the figure is drawn. None of them is a number recalled from a table.
tetrahedral109.47° × 6repulsion minimised, angles measured off the result4 sites Where the atoms go

VSEPR, computed

The tetrahedral angle is not 109.5 degrees because a textbook says so. It is arccos(−1/3), and it falls out of minimising the repulsion of four points on a sphere without ever being written down.

 1s · 1sS = 0.38997sigma interactionseparation 2.8 bohrcontours at 50% of each densitythe signed product integrated over all spaceone electron Bonding models

Overlap decides

Two orbitals interact in proportion to how much they overlap, and the sign of the overlap decides which of the two combinations is the lower in energy. It is one integral, and almost everything about bonding follows from it.

 HHHNC3vprincipal axis C33 mirror planesno inversion centremay be polarcannot be chiralgroup recovered from the coordinates4 atoms What symmetry decides

Point groups from coordinates

A molecule's symmetry is not a label to be looked up. It is decidable from the atom positions by searching for the operations that permute them, and the search either finds an operation or it does not.

 50% of the density|ψ| = 1.48e-190% of the density|ψ| = 3.94e-299% of the density|ψ| = 8.44e-31slevels solved for by integration, drawn at one scale Orbitals

Say what it encloses

An orbital picture is a contour at a level somebody chose, and almost no source says which. Two textbooks can draw the same orbital at visibly different sizes with the same caption, and both be printed in good faith.

 axial — 2 sitesneighbours at 90°, 90°, 90°, 180°equatorial — 3 sitesneighbours at 90°, 90°, 120°, 120°the two are not equivalentneighbour angles measured, not assumed5 sites Where the atoms go

Five sites are not alike

Every other common arrangement has one or two distinct angles. Five has three, because two of its positions are on an axis and three are round an equator — and a molecule built that way does something about it.

 1s · 2pxS = 0 exactlycomputed -9.6e-17 — arithmetic noiseseparation 2.8 bohrcontours at 50% of each densitythe contributions cancel in pairs, so the integral vanishesone electron Bonding models

Exactly zero

Where symmetry forbids an interaction the overlap is not small. It is zero — and computing it and finding arithmetic noise is a different kind of statement from computing it and finding a small number.

 node 1.90node 7.10most probable radius 13.10 a₀R(r)and 4πr²R²r / bohrmeasured off the computed function2 radial nodes · one electron Orbitals

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.

 HHOC2vprincipal axis C22 mirror planesno inversion centremay be polarcannot be chiralgroup recovered from the coordinates3 atoms Where the atoms go

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.

 sp3109.471°between every pair4 hybridsworst off-diagonal 0e+0an orthogonal transformation of the atomic orbitalsone electron Bonding models

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.

 moleculegroupmay be polarmay be chiralwaterC2vyesno2 σcarbon dioxideD∞hnonohas iammoniaC3vyesno3 σmethaneTdnono6 σboron trifluorideD3hnono4 σhydrogen peroxideC2yesyes—bromochlorofluoromethaneC1yesyes—both columns derived from the symbol, not from the bonds What is taught wrongly

The dipole is not a sum of bonds

Adding bond dipoles as vectors gets the easy cases right and rests on a quantity with several incompatible definitions. The symmetry argument is exact, needs no electronegativities, and says when the answer must be zero.

 most probable radius 1.00 a₀R(r)and 4πr²R²r / bohrmeasured off the computed function0 radial nodes · one electron Orbitals

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.

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