A cage needs one pair more than it has corners

A closo borane is an ion of formula $\mathrm{B}_n\mathrm{H}_n^{2-}$ whose boron atoms sit at the corners of a deltahedron — a polyhedron with triangular faces — each with one hydrogen pointing outwards. They exist for $n$ from five to twelve, they are remarkably stable, and their electron counts obey one rule with no exceptions.

Do the arithmetic. Each boron brings three valence electrons and each hydrogen one, and the charge adds two, so the ion has $4n + 2$ valence electrons. Each terminal B–H bond takes a pair, using $2n$ of them, and what is left for holding the cage together is $2n + 2$ electrons, which is

$$n + 1 \text{ skeletal pairs.}$$

Twelve vertices, thirteen pairs. Six vertices, seven pairs. That is Wade’s rule and the arithmetic above is the whole of its derivation from the formula — which leaves the interesting question untouched: why should a cage want exactly one more pair than it has corners?

Half of that question has an exact answer and half does not, and separating them is what this essay is for.

Splitting the count into two sets #

Each boron in a cage has, after its terminal hydrogen has taken one, three orbitals available for skeletal bonding. It is convenient to divide them by direction rather than by label: one radial orbital pointing at the centre of the cage, and two tangential orbitals lying in the surface.

That gives $n$ radial orbitals and $2n$ tangential ones, $3n$ in all. The standard account says the radial set supplies one bonding combination and the tangential set supplies n, adding to the $n+1$ that the arithmetic demanded.

The radial half is computable here and is a theorem. The tangential half is not, and saying which is which is the honest form of the result.

The radial half is Perron’s theorem #

The radial orbitals all point at one another through the centre, and their interactions follow the cage’s own connectivity: two of them interact when their vertices are joined. So the matrix governing the radial set is the adjacency matrix of the polyhedron, which is the object Hückel theory diagonalises for every conjugated molecule.

Now Perron and Frobenius. A connected graph’s adjacency matrix has non-negative entries, and for such a matrix the largest eigenvalue is simple — no degeneracy — and its eigenvector has entries all of one sign. So there is exactly one combination of radial orbitals with no node anywhere in it, and every other radial combination changes sign somewhere.

That is the $+1$, and notice what it does not depend on. Not the shape of the cage. Not its symmetry. Not the number of vertices, nor whether the vertices are equivalent, nor whether the faces are triangles. Any connected cage whatever has exactly one nodeless radial combination.

It can be checked directly: for each cage the top eigenvector is required to have one sign throughout, and the gap between the largest eigenvalue and the second is required to be non-zero, since a degenerate top would break the argument. Both hold at every size.

There is a bonus for the cages whose vertices all have the same number of neighbours. The largest eigenvalue of a regular graph is its degree exactly, so the octahedron’s top level is at $+4$ and the icosahedron’s at $+5$, and both come out of the diagonalisation at those values to nine decimal places.

The tangential half, which is not computed here #

The $n$ bonding combinations from the tangential set are the other half of Wade’s rule and they are not computed here.

The obstacle is not difficulty in principle. The tangential orbitals are p functions lying in a curved surface, so the interaction between two of them depends on how the surface twists between their vertices as well as on whether the vertices are joined — it is not the adjacency matrix, and there is no small graph-theoretic object that plays the same role. Writing the calculation would mean building a proper two-orbital-per-site tight-binding model on a sphere, with the local frames transported correctly, and that is a calculation of a different kind from anything else here.

What can be checked is the group-theoretic consequence, in the one case the available character tables reach. For the octahedron the radial set spans $\mathrm{A}_{1g} + \mathrm{E}_g + \mathrm{T}_{1u}$ — one, two and three — and the tangential set spans $\mathrm{T}_{1g} + \mathrm{T}_{2g} + \mathrm{T}_{1u} + \mathrm{T}_{2u}$, of which $\mathrm{T}_{2g}$ and $\mathrm{T}_{1u}$ are bonding, giving six. One plus six is seven, and seven is $n + 1$ for $n = 6$.

That is a count from a character table rather than from a diagonalisation, so it says which combinations exist and not where they sit in energy. Which of them are bonding is the part that is quoted.

The cages come out of a repulsion minimisation #

None of the polyhedra above was typed in. They come from repel, the minimisation that has produced this site’s VSEPR arrangements since its first essays, and the fact that the two coincide is worth pausing on.

Putting $n$ points on a sphere so as to minimise the repulsion between them is the Thomson problem, and for six points the answer is an octahedron and for twelve it is an icosahedron. Putting $n$ boron atoms in a cage produces the same shapes. The two problems have nothing to do with one another — one is a classical electrostatics minimisation, the other is what a set of covalently bonded atoms does — and they give the same answers for seven of the eight closo sizes.

The edges are then found by a rule with a check attached. Euler’s formula for a polyhedron with all-triangular faces gives $e = 3v - 6$ exactly, so the edge count is known before the search: take the $3v - 6$ shortest pairs and require the next-shortest to be meaningfully longer. The gap ratio runs from $1.16$ at five vertices to $1.62$ at twelve, comfortably clear.

Eight is the exception and it is refused. Eight points minimising repulsion give a square antiprism, whose two square faces mean it is not a deltahedron, and the gap ratio comes out at exactly $1.0000$ — the shortest non-edge is the same length as the longest edge. The real $\mathrm{B}_8\mathrm{H}_8^{2-}$ is a dodecahedron rather than an antiprism, so the two arrangements genuinely part company there. The builder refuses rather than accepting a polyhedron it cannot recognise.

What the radial spectra look like across the sizes #

The radial spectra are cheap to compute and the pattern across the eight cage sizes says something the single case does not.

The top level is the vertex degree wherever the cage is regular — four for the octahedron, five for the icosahedron — and where it is not, it sits between the smallest and largest degrees.

That figure is the theorem doing the work it was invoked for. The octahedron and the icosahedron are regular graphs, so their top eigenvalues are their degrees exactly and could have been read off without diagonalising anything; a reader entitled to be suspicious might ask whether the “+1” is a fact about regularity rather than about connectedness. The seven-vertex cage answers it. Its degrees are five and four, its top level is neither, and there is still exactly one of it.

Below the top level the spectra differ in a way that maps onto the symmetry. The octahedron’s three zero levels are its $\mathrm{T}_{1u}$ set, degenerate because the group demands it; the icosahedron’s three at $\sqrt5 = 2.236$ and five at $-1$ are its $\mathrm{T}_{1u}$ and $\mathrm{H}_g$ sets, and the degeneracy of five is the largest this site meets anywhere — larger than any of its thirteen character tables permits, since the biggest dimension in those is three.

That last point is worth a sentence on its own. Why a character table stops where it stops shows that the permitted degeneracies of a group are the dimensions in its table, and that the sum of their squares is the group order. Icosahedral symmetry has order sixty in its rotational form and one hundred and twenty with the improper operations, and its table has a five-dimensional row. A fivefold degeneracy is not available in any of the point groups tabulated here, and the icosahedral cage produces one anyway — from a diagonalisation that knows nothing about groups.

So the spectrum is evidence about a symmetry those tables cannot otherwise reach. It gives the degeneracies without giving their labels, which is the honest half of what a numerical calculation supplies when the group theory is missing.

What the counting rule replaces #

The reason Wade’s rule matters is that the alternative bookkeeping does not work at all.

$\mathrm{B}_6\mathrm{H}_6^{2-}$ has twelve edges and fourteen skeletal electrons — seven pairs for twelve connections. A picture with a two-electron bond along each edge would need twenty-four electrons, so the ion is short by ten. That is what “electron deficient” means, and it is a true statement that predicts nothing.

The pattern appears at a smaller size too. Three-centre bonding, computed shows one pair holding three atoms together, with the non-bonding orbital’s central coefficient coming out at exactly zero by symmetry. What one pair can hold together shows one pair in a ring of any size supplying a total bond order of exactly two, and traces that to a property of the matrix rather than of the graph.

A cage is the same idea at the next dimension up. The pairs are not assigned to edges at all; they occupy combinations that span the whole cage, and the count of them is a count of bonding combinations rather than of connections.

Where the rule generalises, and where it stops #

Wade’s rule as stated here is for the closed cages, and the reason it is famous is that it extends.

Remove one vertex from the deltahedron and keep the same number of skeletal pairs and the result is a nido structure, still with $n + 1$ pairs but now for $n$ atoms in an open basket. Remove two and it is arachno, with $n + 2$ pairs. The rule’s real content is that the pair count is a property of the parent deltahedron rather than of the actual shape, so a fragment carries the count of the cage it was cut from.

That extension is entirely a matter of bookkeeping and this site is not the place to work it. What the essay can say about it is where the exact half survives: the Perron argument holds for the nido and arachno skeletons as well, because they are still connected graphs, so their radial sets still have exactly one nodeless combination. The $+1$ travels with the fragment for the same reason it belonged to the cage.

The rule does stop somewhere. Very large clusters, metal clusters with more than about twelve vertices, and clusters where the vertices contribute different numbers of orbitals all need extensions or corrections, and the literature on them is a subject in its own right. The line to draw is the same one drawn everywhere here: what is computed is a spectrum of a matrix, and what is quoted is which levels are bonding.

The other thing a spectrum shows #

There is a detail in the octahedral spectrum worth reading, because it is the difference between a cage and a ring.

The six radial levels of the octahedron come out at $4, 0, 0, 0, -2, -2$. The gap between the top level and the next is four units, which is comparable with the whole width of the rest of the spectrum. That isolation is what makes the nodeless combination a genuinely separate orbital rather than the top of a band, and it is why filling it with one pair is a meaningful thing to do.

The obvious guess is that the isolation improves as the cage grows, since the top level rises with the degree. It does not, and the numbers say so plainly: the second level rises faster. The gap runs $4.000$ at six vertices, $3.699$ at seven, $3.287$ at nine, $2.828$ at ten, $2.716$ at eleven and $2.764$ at twelve — narrowing by a third across the closo series while the top level rises by only a quarter.

The eleven is the interesting entry in that list and it is worth not smoothing over. The gap does not narrow monotonically: it reaches its smallest value one size before the end of the series and then widens again at twelve. The reason is the icosahedron’s symmetry rather than its size. Eleven vertices cannot be arranged so that every one has the same number of neighbours, so its second level is pushed up by an irregular structure, while the icosahedron is regular of degree five and its second level is held at exactly $\sqrt5$ by the group. A trend read off the two ends of the series would have missed that, and the honest statement is that the gap narrows overall and is not a monotone function of size.

So a cage becomes more band-like as it grows, exactly as the band limit shows a ring doing, and only more slowly. The nodeless combination is decisively separate at six vertices and merely well separated at twelve, and somewhere past the closo sizes it would stop being separate at all.

That is the right way round for the chemistry. The closo series stops at twelve, and one of the reasons usually given is that a larger cage’s skeletal bonding becomes diffuse. The spectra above are a version of that statement with numbers in it, arrived at from a graph rather than from a survey of what has been made.

A note on what a “vertex orbital” is doing #

The division into radial and tangential orbitals is a choice of basis, and several essays here are careful about exactly that.

Nothing about a boron atom in a cage says which of its three skeletal orbitals is radial. The atom has an s function and three p functions; the terminal hydrogen takes one combination of them, and the remaining three span a three-dimensional space with no preferred decomposition. Choosing one radial and two tangential is a choice made because the resulting matrices separate, in the same way and for the same reason that choosing four tetrahedral hybrids on a methane carbon is a choice made because the resulting orbitals are localised.

Hybrids are a basis makes the general point and it applies here without change: the total density, the total energy and every observable are the same whichever decomposition is used, and only the bookkeeping differs. What the radial-tangential split buys is that one of the two blocks becomes the adjacency matrix, which is an object with a theorem attached.

The theorem is then about the block rather than about the molecule, and it is worth being careful that the conclusion survives the choice. It does, and for a reason that is easy to state: the count of bonding combinations in the whole $3n$-dimensional space does not depend on how the space is divided up. Splitting it into $n$ and $2n$ and finding one and $n$ bonding combinations respectively is one route to the total. A different split would give different intermediate numbers and the same total, because the total is a property of the full matrix.

The theorem knows nothing about boron #

The argument is about connected graphs and one nodeless combination, and nothing in it mentions which element sits at a vertex. That is testable, because the same cages can be built out of other things.

They can, and they obey the same count. Replace one or two vertices of a borane cage with carbon and the cluster is a carborane, with a different charge and the same $n+1$ pairs for $n$ vertices. Replace one with a transition-metal fragment and the count still holds, with the metal contributing whatever its own orbitals supply. Take away the boron entirely and build a cage from bare heavy main-group atoms — the anionic clusters of tin, lead and antimony that crystallise from liquid ammonia — and they obey it too.

Four families of compound, three of them containing no boron, one containing no main-group element at the vertex at all, and one count. That is what a theorem about connected graphs predicts and what an observation about boron chemistry could not.

What the count establishes #

The essays on multicentre bonding make one argument: that a bond between two atoms is a special case rather than the base case, and that the general object is a combination over however many centres the structure has.

The cage supplies the counting version of it. A cage’s bonding is not a shortage of two-centre bonds; it is $n+1$ combinations over $n$ centres, one of which exists for a reason that has nothing to do with chemistry at all.

The half that remains uncomputed is named rather than glossed, and it is the same half every time: the energies. The count says how many combinations there are and which are nodeless; it cannot say which of the tangential ones lie below the reference and which above without a model of the surface. That is a bounded gap, it would take a genuinely different calculation to close, and until it is closed the $n$ in $n+1$ is quoted while the $+1$ is proved.