Folding the ring does not give the orbital back

The orphan count prices hypervalency. A main-group centre has four valence orbitals, one s and three p, and a molecule with n σ-bonded ligands and L lone pairs leaves n + L − 4 ligand combinations with no central orbital to bond to — each one a three-centre four-electron system rather than a two-centre bond, with no d orbital needed to explain it. Reduced in each molecule’s own point group, the formula is exactly right for ten molecules as they are.

It is not right for every arrangement of them. Across the census of alternative shapes, three arrangements break it: methane as a square, phosphorus pentafluoride as a flat pentagon and sulfur hexafluoride as a flat hexagon, each orphaning one more combination than the formula says. The explanation offered was airtight as far as it went. A flat ring of ligands has a mirror in its own plane; every ligand σ function is symmetric under that mirror; the p orbital perpendicular to the ring is antisymmetric under it; so the perpendicular p has no partner of its species, and the centre offers three usable orbitals rather than four.

That explanation was then generalised in the obvious direction. Any arrangement whose ligands span fewer than three independent directions has the same problem, it was said, and the flat ones are the only such case. The corollary is that ligands spanning all three directions give the centre all four orbitals. The corollary is the testable part, and it has a direct test: fold the flat ring.

The folded ring is short by the same one #

Take a ring of n ligands round a centre and tilt every ligand by the same polar angle away from the ring’s plane, so the ligands sit on a cone. This is an umbrella: at 90° it is the flat ring, at 100° or 110° it is a shallow pyramid with nothing at the top, and the ligands span all three directions at every angle except 90° exactly. The mirror in the ring’s plane is gone. The group of the folded ring is $C_{nv}$ — an n-fold axis and n vertical mirrors — for every angle off 90°.

The count does not change. A bare ring of four ligands, folded to any angle from 60° to 120°, matches three of the centre’s orbitals and orphans one combination, where the formula for four ligands and no lone pair says none. A ring of five orphans two against one; six, three against two; seven, four against three; eight, five against four. Across five ring sizes and nine polar angles, every folded ring is one over the formula, and at 90°, where the ring is flat and the group is $D_{4h}$, $D_{5h}$ or $D_{6h}$, the flat ring gives the same count it always did.

So folding the ring restores the three dimensions and does not restore the orbital. The flat-ring explanation was correct about the flat ring and wrong as a rule. Spanning three directions is not what a centre needs.

Put a single ligand on the axis over the folded ring and the count returns to the formula at once. An apex over a ring of four is five ligands and orphans one, which is five minus four; over a ring of eight it is nine ligands and orphans five. That holds at every angle, and it holds at 90°, where the ring under the apex is exactly flat: a flat ring with one ligand above it is not short of anything. A single lone pair on the bare folded ring does the same thing. Whatever the folded ring is missing, one ligand or one lone pair on its open side supplies it.

Two ways to waste the same orbital #

The species show what happened. Six ligands flat, in $D_{6h}$, give a σ set of a₁g ⊕ e₂g ⊕ b₁u ⊕ e₁u against the centre’s a₁g for s and a₂u ⊕ e₁u for p. Nothing in the σ set is a₂u, so the perpendicular p is left over.

Fold the six to 110° and the group falls to $C_{6v}$, where the perpendicular p is no longer antisymmetric under anything. It transforms as $A_1$, the totally symmetric species, which is also the species of s. The six ligand σ functions now span $A_1 + B + E_1 + E_2$, where B is one of the two alternating species $B_1$ and $B_2$ — which one depends only on which class of mirrors is named $\sigma_v$. The centre offers two $A_1$ orbitals, s and pz, and the ligands offer one $A_1$ combination — the in-phase sum of all six. One combination can bond with one orbital. The $E_1$ pair matches px and py, and the second $A_1$ orbital goes unused exactly as the perpendicular p did in the flat ring, with B and the $E_2$ pair orphaned.

The unused orbital is a different object in the two cases, and the difference is visible in its shape. In the flat ring it is the pure pz dumbbell, pointing equally above and below the plane. In the folded ring it is the combination of s and pz orthogonal to the one that bonds — a single lobe on the axis, pointing out of the open side of the umbrella, away from the ligands. That is the shape of a lone-pair orbital, and it is exactly where a lone pair goes in a molecule that has one: ammonia is a folded ring of three with its lone pair in that lobe. It is also exactly where an apex ligand goes, which is why capping the umbrella fixes the count.

Flatness is one way to waste an orbital and a single ring on an axis is another. The first wastes an antisymmetric p because no ligand is off the mirror; the second wastes a symmetric hybrid because the ligands form one set. Both leave three orbitals used, and they agree about the number while disagreeing about the species — which is why a census that only looked at numbers could not tell them apart, and why its only examples of the loss were the flat ones.

A table written from the rotation angle #

Nothing above could be computed with the character tables the census used. They reach $C_{4v}$ and stop. The folded rings of five and six, where two of the three flat failures live, need $C_{5v}$ and $C_{6v}$, and seven and eight need $C_{7v}$ and $C_{8v}$. The tabulated groups had already run out once on a twisting path, and a survey of every path found that the pentagonal pyramid had been dropped from the census for the same reason: its group is $C_{5v}$, of order ten, and no table here had order ten.

Every $C_{nv}$ has the same table, written in n, so the tables were generated rather than typed. On a rotation by θ, each two-dimensional representation $E_k$ has character 2cos(kθ); on every mirror, zero. $A_1$ is one on everything and $A_2$ changes sign on the mirrors. For even n there are two more one-dimensional representations whose characters alternate in sign from one n-th of a turn to the next and differ only on the two mirror classes.

The two alternating representations are reduced together. They differ only on the two classes of mirror, and which class is named $\sigma_v$ and which $\sigma_d$ is a convention; no orbital on a main-group centre transforms as either, so whichever of the two a ring of ligands contains is orphaned, and telling them apart would add a name without adding a count.

Two things make the generated tables trustworthy before they are used on anything new. The operations are not assumed: the symmetry finder that located every group in the census is run on each folded ring, and each must return exactly 2n operations, every proper one leaving the axis fixed and every improper one a mirror. And the generated tables are checked against the tabulated ones where both exist. A folded ring of three and of four, each at 70° and at 100°, is found to be $C_{3v}$ and $C_{4v}$ by the finder, and the generated reduction matches the same number of central orbitals as the tabulated one in all four cases. Only then are they trusted for five to eight.

The pentagonal pyramid, counted #

The census’s one uncounted arrangement is sulfur hexafluoride as a pentagonal pyramid: one fluorine on the axis and five in a ring at 100°. It had been set aside as an arrangement next to the flat ones, in the region where the formula was known to fail, which made it the arrangement most likely to add a fourth failure.

It adds none. The six fluorine σ functions span $2A_1 + E_1 + E_2$ in $C_{5v}$; sulfur’s s and p span $2A_1 + E_1$. Both $A_1$ combinations — the apex on its own and the in-phase sum of the ring — find partners in s and pz, the $E_1$ pair finds px and py, and the $E_2$ pair has nothing. Four matched, two orphaned, and n + L − 4 for six ligands and no lone pair is two.

The census now counts twenty-seven arrangements, and the formula is right for twenty-four. The three it misses are still the three flat ones. The pyramid is the capped umbrella of five at one particular angle, so its agreement is not a separate fact: every capped ring agrees at every angle, and this is one of them.

One set of ligands, two symmetric orbitals #

The folded ring turns a vague condition — the ligands must reach the orbital — into one that can be checked on every arrangement at once. It concerns only the totally symmetric species, and it has two halves.

The ligands supply one totally symmetric combination per symmetry-distinct set. The ligands of an arrangement fall into sets that the group’s operations permute among themselves: the six of an octahedron are one set, the apex and the ring of a pyramid are two, the axial and equatorial fluorines of a trigonal bipyramid are two. Summing each set in phase gives one totally symmetric combination, and there are no others.

The centre supplies one totally symmetric orbital for s, plus one for each direction the group leaves unmoved. In a group with no fixed direction, such as an octahedron, a tetrahedron or a flat ring, that is one. In a group with an axis every operation keeps, such as any $C_{nv}$, the p along the axis is totally symmetric too, and it is two.

Run on all twenty-one shapes — the fourteen of the census that have tabulated groups, the pentagonal pyramid, and the folded and capped rings of four, five and six — the first half holds in every row: the number of totally symmetric combinations is the number of ligand sets. A shortfall appears in exactly five rows, and in every one a single ligand set meets a centre with two symmetric orbitals. Three are the bare folded rings. The other two are the bent triatomic and the pyramid of three, which are the shapes of water and ammonia, and the orbital each would waste is the one their lone pairs occupy. In the census they are never short of anything, because the molecules that adopt them have lone pairs to put there.

That is why the census could not see the folded-ring failure. Every polar arrangement with a single ligand set among its shapes was a shape that real molecules adopt only with a lone pair on the open side. The failure needs four or more ligands, a single set, an axis and no lone pair together, and no molecule in the census is built that way.

What was computed, and how #

Every folded ring is a set of unit vectors from the centre, n of them at one polar angle and equally spaced in azimuth, with an optional one on the axis. Bond lengths are all one, because a σ-only symmetry count cannot see a bond length. The symmetry finder that located every group in the census is run on each, and each operation is classified by its determinant and trace.

The ligand σ character on each operation is the number of ligands it leaves in place, and the centre’s p character is the trace of its matrix, with s adding one. Each is reduced by the generated table, every multiplicity is required to be an integer to nine decimal places, and the reductions must add up to n ligand functions and four central orbitals. Matching is by species: each species matches as many copies as both sides have, and the orphan count is n minus the smaller of the matched count and 4 − L, exactly as the census counts.

The checks, run wherever these figures are drawn. The generated and tabulated reductions agree on $C_{3v}$ and $C_{4v}$. The pentagonal pyramid matches four and orphans two. Every bare folded ring of four to eight, at every angle, matches three, orphans one more than the formula, leaves an $A_1$ orbital unused, and agrees with the formula once it has one lone pair. Every capped ring, at every angle including 90°, matches four and agrees. The totally symmetric combinations equal the ligand sets on every arrangement, and every shortfall is one set against two symmetric orbitals. The refusal is the folded ring of three: it has fewer ligands than the centre has orbitals, so it must waste the same $A_1$ orbital as a spare while orphaning nothing and agreeing with the formula’s zero — or the count would be detecting a fold rather than a shortage, and it is not.

Where the count stops #

The count is σ only. A ligand with a π function along the axis could in principle interact with the unused hybrid, and a π-donor apex is chemically different from a σ-only one. The count here is the σ bookkeeping that the three-centre account of hypervalency runs on, and nothing more.

Only the totally symmetric species is analysed in general. The ledger explains every $A_1$ shortfall on these arrangements; it says nothing general about a centre whose p orbitals fall in a species the ligands miss for some other reason. On the arrangements here the flat ones are the only such case, and whether another exists among shapes not built is not settled by these twenty-one.

A capped ring near 109.5° over three ligands is a tetrahedron. The folded ring of three is therefore not capped, since at that angle its group is larger than $C_{3v}$ and the generated table would be the wrong one.

And the folded rings are not molecules anyone has made. Four or more σ ligands on a cone with nothing on the open side and no lone pair is not a shape main-group chemistry adopts, and the count gives a reason it should not — an orbital that could hold a bond or a pair holds neither. Whether that is why it is not adopted is an energy question, and the census’s own energy comparison is a warning against assuming the count explains a preference.

A shape is short of an orbital for a species, not for a dimension #

The flat-ring rule was stated in the language of directions: ligands in fewer than three dimensions leave a direction empty. That is true for the flat ring and it is the wrong generalisation, because a centre does not bond by direction. It bonds by species, and the species a direction’s orbital belongs to depends on what the rest of the group does. In a flat ring the perpendicular p is antisymmetric and has no partner; tilt the ligands and it becomes symmetric and has to share the one partner s already has.

The rule that survives is about supply and demand in a single species. A centre needs as many totally symmetric ligand combinations as it has totally symmetric orbitals, and ligands supply one per set. A flat ring fails for a different species; a folded ring fails for this one; an apex, a second ring or a lone pair fixes the second failure and a lone pair fixes the first. Xenon tetrafluoride is square and not short, and ammonia is folded and not short, for the same reason.

Still open: the tables for every flat ring #

The grid has two empty cells. The flat rings of seven and eight have groups $D_{7h}$ and $D_{8h}$, and neither table exists here, so the count at exactly 90° for those two is not computed. It is not in doubt — the perpendicular p is antisymmetric under the ring’s mirror in any $D_{nh}$, which is the flat-ring argument — but it is argued rather than counted. $D_{nh}$ is $C_{nv}$ with a horizontal mirror, so the generated tables extend to it by one sign, and doing so would complete the grid and put every flat ring through the same arithmetic as the folded ones.

The nearer question is the other species. Every shortfall in the ledger is totally symmetric or is the flat ring’s antisymmetric p. A centre whose px and py are left short while its pz is matched would need ligand sets that miss the $E_1$ species, and whether any three-dimensional arrangement of a single set can do that — perhaps a crown of ligands tilted alternately above and below the plane, whose groups run from D₂d upwards — is the next question the tables can answer now that they can be generated.