Molecular Geometry & Bonding
what the shape decides
Series

Cohesion — the series

11 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. What holds matter together, per pair. Four kinds of interaction between two units of matter, each computed from the model named beside it, on a logarithmic energy scale. The range from top to bottom is a factor of several hundred, which is the number behind why a molecular solid melts hundreds of degrees below a covalent one.

    What holds a solid together

    Four kinds of interaction between two units of matter, each computed from a stated model and put on one logarithmic scale. The ordering is not the one usually taught — an ion pair at contact beats a shared pair, which is a fact about the comparison rather than about the numbers.

    part 1 · solids
  2. What holds matter together, per pair. Four kinds of interaction between two units of matter, each computed from the model named beside it, on a logarithmic energy scale. The range from top to bottom is a factor of several hundred, which is the number behind why a molecular solid melts hundreds of degrees below a covalent one.

    The lattice sum that depends on the order of adding

    An ionic solid's binding is the sum of every pair of charges in it, and the series does not converge absolutely — rearranged, it gives a different answer. That is a genuine mathematical difficulty rather than a technicality, and it is the clearest example of something a real-space, neighbour-by-neighbour method cannot compute at all.

    part 2 · solids
  3. ⟨x²⟩ is the average number of neighbours. For each structure, the mean coordination counted off the edge list beside the mean of x² measured off the eigenvalues. They are equal by an identity about graphs, not by a limit — and the binding each bond supplies, in the last column, obeys no such rule and falls as neighbours are added.

    The bond that weakens as neighbours multiply

    Wrapped structures with two, four and six neighbours per site give a mean of x² of exactly 2, 4 and 6. Their binding per site goes 1.272, 1.611, 1.979 — and their binding per bond falls from 0.636 to 0.330, which is why a metal atom with twelve neighbours has weak bonds and a great many of them.

    part 3 · solids
  4. Five sixths of the bonds, nine tenths of the binding. What a site in the outer layer of an open block keeps, measured three ways: the fraction of its bonds, the fraction the second-moment rule predicts of its binding, and the fraction the calculation gives. The last two agree and the first does not.

    A surface is not a count of broken bonds

    Cut a crystal and every atom in the new face has lost one of its six neighbours. The standard estimate follows immediately: a surface costs one sixth of the cohesive energy per atom exposed. Computed, it costs a little over half that — the atom keeps 91.2 per cent of its binding while keeping only 83.3 per cent of its bonds, because the bonds that survive get stronger when their competitors are removed.

    part 4 · wrong
  5. The same neighbours, and a fifth of the binding between them. five structures in which every site has 4 neighbours. Their second moments are identical — 4 for every one, which is the coordination and is what the band width is read from. Their bindings per site are not: they run from 1.28 to 1.64, and the least bound is the one with the most four-step walks.

    Two structures with the same neighbours

    Five structures in which every atom has exactly four neighbours. Their second moments are 4.000 to nine decimal places, because that identity is the coordination and nothing else. Their bindings per site run from 1.2756 to 1.6363 — a spread of twenty-two per cent — and two of them that agree on the second, third and fourth moments together still differ in the third decimal place.

    part 5 · solids
  6. A mixture is not the average of its ends. The binding per site of a square net whose sites are of two kinds, against how many of each. The straight line is arithmetic rather than a fit: a structure of one kind only has every level shifted by ±δ, so the two ends and the line between them are known before anything is diagonalised. Every mixture lies above it — more bound — by as much as 0.43 per site at the middle, and that departure is the whole of what makes an ordered compound worth forming.

    A mixture is not the average of its ends

    Half of a structure's sites raised and half lowered, and the line between the two pure ends is arithmetic — no diagonalisation needed. Every mixture lies below it, by 0.42987 per site for the ordered arrangement and 0.23690 for the segregated one at the same composition, so composition fixes neither the binding nor the gap. And the departure is not quadratic in the contrast: it grows as its 1.79 power in a chain and its 1.28 power on a square net.

    part 6 · solids
  7. Every arrangement, and the winner is not the one with the most unlike bonds. All 1820 ways of raising 4 of 16 sites on a wrapped square net, at a contrast of 4, each placed by its count of unlike bonds against the binding it gives. The best arrangement has 12 unlike bonds where 16 is available, and it binds at 1.103953 against 1.080031 for the best of those that do have the most. The count and the spectrum are two different orderings.

    The arrangement a count cannot pick

    Three arrangements of one composition came out ordered by their count of unlike bonds, which looked like a rule. Enumerating every arrangement instead of three shows it is not one: it holds at every composition on a square net at a small contrast, fails at four of them at a large contrast, and fails on a triangular net at any contrast at all.

    part 7 · solids
  8. How many places a local search can stop. The number of distinct arrangements a steepest-ascent search settles at, for each net and contrast, with the fraction of starts reaching the best of them written beside it. 16 sites, contrast 1: 2 from 200 starts, best reached 67 per cent of the time; 16 sites, contrast 4: 2 from 200 starts, best reached 59 per cent of the time; 36 sites, contrast 1: 4 from 20 starts, best reached 80 per cent of the time; 36 sites, contrast 4: 12 from 20 starts, best reached 10 per cent of the time. The larger net at the larger contrast is a different kind of landscape.

    Twelve basins where there were two

    Sixteen sites can be searched exhaustively and thirty-six cannot, so the only instrument available past about twenty sites is a local search — and how much weaker it is has never been measured against the answer on a net where both can be run. On sixteen it is barely weaker at all. On thirty-six it fails nine starts in ten, and the arrangement it is beaten by has fewer unlike bonds than the one it starts from.

    part 8 · solids
  9. How often a random start reaches the best of them. The share of random starts that reach the best arrangement found, against the fraction of sites raised, for two nets at two contrasts. Every curve dips in the middle of its left half and recovers: the hard compositions are between a quarter and a third, and the half-filled one — the rightmost point of each curve — is among the easiest. The hardest points are 5 of 16, 6 of 16, 9 of 36, 12 of 36.

    The composition that is hard is not the full one

    A landscape of arrangements measured at one composition on each of two nets raises the question of where the hardest one sits — and the natural guess is the half-filled one, where there is most to arrange. It is the easiest. On thirty-six sites at a contrast of one, half filling has one local optimum and nine billion arrangements, and a quarter filling has six optima and a hundredth as many.

    part 9 · solids
  10. One net has a two-colouring and the other cannot. Sixteen sites wrapped into a square net and into a triangular one, with the wrapping bonds left undrawn. The square net is bipartite: its sites split into two classes with every bond running between them. The triangular net is not, and the obstruction is a triangle — three sites in a cycle of odd length cannot be two-coloured. It has thirty-two of them, two per site, and forty-eight bonds against the square net's thirty-two.

    A net with no two-colouring

    Half filling is the easiest composition to search and the reason given was the two-colouring: on a bipartite net it is the unique arrangement with every bond unlike, so the optimum has nothing competing with it. A triangular net has no two-colouring, its half-filled composition is reached by every one of two hundred starts, and its best arrangement is two bonds short of what counting allows.

    part 10 · solids
  11. On the frustrated net the widest gap wins everywhere but just past a change of winner. Every basin found at half filling on the triangular net, at eleven contrasts, placed by the gap it opens at the Fermi level; the winning basin is filled and the rest open, sized by how many of two hundred starts reach them. Up to a contrast of 2.5 there is one basin. From 3 a second appears, and the winner is the one with the wider gap — except at 4, where the winner has a gap of 4.947 and a runner-up has 5.088. Squares mark basins with thirty-two unlike bonds and circles thirty.

    The gap follows the winner late

    On a triangular net at half filling, the arrangement that binds best was expected to be the one that opens the widest gap at the Fermi level. Across twenty cases it is, eighteen times. The two exceptions are not noise: the frustrated net's winner changes at a contrast of 3.790, from an arrangement with thirty unlike bonds to one with thirty-two, and the gaps of the two do not cross until 4.849. For a whole unit of contrast the better binder has the narrower gap.

    part 11 · solids

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