Sixteen is also a count

The eighteen-electron rule is usually stated as a fact about stability and derived from a count of ten d electrons plus eight from four two-electron donors, or some other arrangement of the same eighteen. Eighteen is a count sets out the version that makes the rule intelligible: a transition metal brings nine valence orbitals — one s, three p, five d — and nine orbitals filled twice over is eighteen electrons.

Stated that way the rule stops being about donors and becomes about seats. Every valence electron in the complex, whether it came from the metal or from a ligand, has to sit in one of the nine, and eighteen is the number of them.

The interesting cases are the ones where the answer is not eighteen, and there is one large family of them: square-planar complexes of $\mathrm{d}^8$ metals, which stop at sixteen and are perfectly happy there. Wilkinson’s catalyst, Vaska’s compound, tetrachloridoplatinate, Zeise’s salt, and most of the platinum and palladium chemistry that catalysis depends on.

The same bookkeeping gives sixteen, and the difference is one orbital.

The bookkeeping, in three lines #

Every geometry gets the same treatment.

How many sigma combinations do the ligands supply? Six for an octahedron, four for a square plane, four for a tetrahedron. Each is a symmetry-adapted combination of the ligands’ donor orbitals, and the species they span is a reduction computed from the geometry rather than looked up.

How many metal orbitals do they take? One each. A metal orbital whose species appears among the ligand combinations pairs with it, giving a bonding orbital that ends up mostly on the ligand and an antibonding one far above.

How many of the leftovers lie low enough to fill? This is the one judgement in the calculation, and it is stated per geometry rather than assumed.

For an octahedron the six combinations span $\mathrm{A}_{1g} + \mathrm{E}_g + \mathrm{T}_{1u}$, which pairs with the metal $s$, the two $\mathrm{e}_g$ d orbitals and the three $p$ orbitals — six taken. The three left over are the $\mathrm{t}_{2g}$ set, which point between the ligands and are therefore untouched and low. Six plus three is nine, and the count is eighteen.

For a square plane the four combinations span $\mathrm{A}_{1g} + \mathrm{B}_{1g} + \mathrm{E}_u$, which pairs with the metal $s$, the $\mathrm{d}_{x^2-y^2}$ and the two in-plane $p$ orbitals — four taken. Five are left over, more than the octahedron leaves. Four of them — $\mathrm{d}_{z^2}$, $\mathrm{d}_{xy}$ and the $\mathrm{d}_{xz}, \mathrm{d}_{yz}$ pair — are low. The fifth is the metal $p_z$, perpendicular to the plane, and it is not.

Four plus four is eight, and the count is sixteen.

Which orbital is out of reach, and why #

The $p_z$ of a square-planar metal has no sigma partner at all. The four ligands lie in the $xy$ plane, so no combination of their donor orbitals has the right symmetry to interact with an orbital pointing straight up and down. It is left exactly where it was, at the energy of a bare metal $p$ orbital.

That is very high. A metal $p$ orbital lies several electronvolts above the $d$ shell, and in an octahedron it is dragged down only because it is strongly bonding with the $\mathrm{t}_{1u}$ ligand combination. In a square plane it is dragged down by nothing, so it stays where a free atom’s $p$ orbital is and no ordinary complex has electrons to spare for it.

The check that makes the point runs in the direction that could fail. Running the same bookkeeping while treating that orbital as fillable gives eighteen — which is exactly the error the rule is about, and the check requires it, so that the judgement about which leftovers are low is a stated input rather than a hidden one.

Why a tetrahedron does not do the same thing #

Four ligands and eighteen electrons looks like a contradiction with four ligands and sixteen, and it is not.

A tetrahedron’s four sigma combinations span $\mathrm{A}_1 + \mathrm{T}_2$, which pairs with the metal $s$ and its three $p$ orbitals — four taken, as in the square plane. But the five left over are the whole d shell, split into $\mathrm{e}$ and $\mathrm{t}_2$ by a splitting that two models, one ratio computes as four ninths of an octahedron’s, and all five are low. Four plus five is nine, and eighteen again.

So the difference between the two four-coordinate geometries is not how many orbitals are taken but which ones. A tetrahedron’s ligands use up the metal’s $p$ orbitals entirely, so nothing is stranded. A square plane’s use two of the three and leave the third pointing into empty space.

The comparison makes the rule’s real content visible. The count is nine minus the number of metal orbitals that end up too high to fill, and the geometry decides that number rather than the ligand count.

Where the four ligand combinations come from #

The species the ligands span are computed here rather than quoted, by the method character tables and reduction sets out: the operations are recovered from the atom positions, sorted into classes by conjugation, and the four donor orbitals reduced against the resulting table.

For tetrachloridoplatinate the reduction returns $\mathrm{A}_{1g} + \mathrm{B}_{1g} + \mathrm{E}_u$ — one, one and two, which is four functions as it must be. The multiplicities come out as whole numbers, which is the check that catches an error anywhere upstream.

Two of those species are shared with metal orbitals twice over. $\mathrm{A}_{1g}$ is the species of both the metal $s$ and its $\mathrm{d}_{z^2}$, so both interact with the same ligand combination — and only one of them can be the bonding partner. The bookkeeping pairs the ligand combination with the $s$ and leaves $\mathrm{d}_{z^2}$ as a leftover, which is the standard treatment and is a judgement rather than a theorem.

That judgement is worth flagging because it is the one place the count could be argued with. In a real square-planar complex $\mathrm{d}_{z^2}$ is weakly antibonding rather than strictly non-bonding, and it sits noticeably above the other three low orbitals. It is still filled in every $\mathrm{d}^8$ complex, so the count of eight filled orbitals is unaffected — but a reader should know the ordering inside the low set is not as clean as a count makes it look.

The four chloride donor orbitals reduce in the complex’s own group to three species for four functions, with one of them carrying two. That reduction is the input the whole count is made of, and the same procedure gives eighteen in an octahedron from six functions in three species.

What the empty orbital does #

A sixteen-electron complex is closed by its own arithmetic, and it also has an empty orbital pointing at nothing, which the eighteen-electron octahedron does not.

That is a real structural difference and it shows up in every account of square-planar chemistry. The empty $p_z$ accepts electron density from anything approaching along the axis, so a fifth ligand can bind without anything having to leave first, and square-planar complexes have a route open to them that octahedral ones do not.

Naming the consequence precisely matters here, because the consequence is a rate and no rate is computed here. What can be said structurally is that a five-coordinate adduct of a square-planar $\mathrm{d}^8$ complex is a stable enough species to be isolated in several cases, and that its count is eighteen. What cannot be said here is anything about how fast it forms, which is the associative substitution mechanism and belongs to a different subject — the same line the spectrochemical series is not electrostatics drew when it kept the trans influence and declined the trans effect.

The complex is four chlorides in a plane with a point group the search recovers from the coordinates, and two directions perpendicular to the plane with nothing in them at all. That emptiness is the whole reason the count is sixteen rather than eighteen.

The four sixteen-electron compounds that built a subject #

The rule is worth grounding in the compounds that made it matter, because all four are square-planar $\mathrm{d}^8$ and all four are counted the same way.

Wilkinson’s catalyst, $\mathrm{RhCl(PPh_3)_3}$. Rhodium(I) is $\mathrm{d}^8$; three phosphines and a chloride give four donors; eight filled orbitals, sixteen electrons.

Vaska’s compound, $\mathrm{IrCl(CO)(PPh_3)_2}$. Iridium(I), $\mathrm{d}^8$, four donors, sixteen. Its fame comes from adding a molecule of oxygen or hydrogen across the empty axis and becoming eighteen.

Tetrachloridoplatinate, $[\mathrm{PtCl_4}]^{2-}$. Platinum(II), $\mathrm{d}^8$, four chlorides, sixteen — and the starting material for most of platinum coordination chemistry including cisplatin.

Zeise’s salt, $\mathrm{K[PtCl_3(C_2H_4)]}$. The first organometallic compound anybody made, in 1830, and it took a hundred and thirty years to work out what the ethene was doing. Platinum(II), three chlorides and an alkene, sixteen.

Four compounds spanning nearly two centuries, three metals and four quite different ligand sets, and the count is the same in every case because the geometry is. That is the evidence that sixteen is a rule rather than a set of coincidences, and the bookkeeping above is why.

Which metals do this and which do not #

The sixteen-electron count belongs to a specific corner of the table and the reason is the same arithmetic.

Square-planar geometry is adopted by $\mathrm{d}^8$ metals — nickel, palladium, platinum, rhodium and iridium in their +1 state, gold in +3 — and by essentially nobody else. The reason is that $\mathrm{d}^8$ is exactly the count that fills the four low orbitals and leaves the high $x^2-y^2$ empty. A $\mathrm{d}^9$ metal would have to put an electron into it and a $\mathrm{d}^7$ would leave one of the low orbitals half filled, and in both cases another geometry does better.

The heavier metals do it more than the lighter ones because their splittings are larger, so the gap the eighth and ninth electrons would have to cross is larger. Nickel(II) is square planar with strong-field ligands and tetrahedral with weak ones; palladium(II) and platinum(II) are square planar with everything.

That is the pairing energy argument in a different geometry, and it is why the sixteen-electron rule and the spin-state question are the same question asked twice.

The competition behind the geometry is the ordinary one: with eight electrons the choice is between filling four low orbitals and leaving one very high one empty, or spreading over five in a geometry with a smaller splitting. Which one wins is decided by the same balance that decides a spin state.

The same argument for the linear and the cubic cases #

The bookkeeping is general and it is worth running on two geometries nobody quotes a magic number for, since the answers say something about why the two famous ones are famous.

A linear two-coordinate complex has two sigma donors spanning two species, taking two metal orbitals. Seven are left over. Of those, one is the metal $p$ along the axis’ perpendicular directions — two of them, both unmatched — and the whole d shell except one orbital is low. Counting generously gives fourteen and counting strictly gives less, and the answer depends on judgements the two-coordinate geometry does not make crisply. That is the arithmetic reason linear complexes have no magic number: the leftovers are not cleanly divided into low and unreachable.

A cubic eight-coordinate arrangement has eight donors, which is more than the nine orbitals can pair with cleanly, and the reduction returns species with multiplicities greater than one. The bookkeeping then has to decide which of two ligand combinations of the same species pairs with the single metal orbital available, and the answer is that both interact with it and the picture stops being a pairing at all.

So the two geometries with clean magic numbers are the two where the ligand combinations and the metal orbitals match up one to one with no ambiguity, and the rule is really a statement about that matching rather than about eighteen. Where the matching is clean the count is an integer with meaning; where it is not, quoting a number would be quoting a judgement.

What the count is not #

Two disclaimers, both of which the octahedral essay made and both of which apply with more force here.

Nothing in this is an energy calculation. The bookkeeping sorts nine orbitals into three groups by symmetry and by one stated judgement, and it says how many are occupied. It does not compute where any of them is, how large the gaps are, or what the complex’s total energy is. Every number in the argument is an integer.

The rule has exceptions and they are the same kind of thing. Early transition metals in high oxidation states are routinely short of eighteen because there are no electrons available to fill the low orbitals, and some late metals exceed sixteen in square-planar geometry when a strong pi-acceptor pulls the count around. In every such case the exception is a statement about where the orbitals sit rather than about the counting, which is why the counting is worth doing first.

The comparison worth making at the end is with the geometry the rule was written for, since the two counts differ by exactly the one orbital the plane leaves unused.

A π-acceptor ligand takes density back out of the filled metal orbitals, which lowers them and changes which of the leftovers count as low. That is the mechanism by which a sixteen-electron complex and an eighteen-electron one can be made of the same metal, and it moves charge without moving the count.

Counting a real complex, once, in full #

An abstract count is easy to nod at, so here is one worked from the formula.

Take Vaska’s compound, $\mathrm{IrCl(CO)(PPh_3)_2}$, and count its valence electrons. Iridium in group 9 brings nine; the chloride, treated as an anionic two-electron donor, brings two; the carbonyl brings two; each phosphine brings two. Nine minus one for the metal’s formal oxidation state of +1, plus two for the chloride as $\mathrm{Cl^-}$, plus six from the three neutral donors: eight from the metal and eight from the ligands, sixteen in all.

Now count seats. Four sigma donors take four of iridium’s nine valence orbitals, leaving five. Four of the five — $\mathrm{d}_{z^2}$, $\mathrm{d}_{xy}$ and the $\mathrm{d}_{xz}, \mathrm{d}_{yz}$ pair — lie low; the perpendicular $p$ does not. Eight seats, sixteen electrons, every seat filled.

The two counts agree because they are the same count. The electrons are in the eight orbitals, and the arithmetic that tallies donors is a shortcut for the arithmetic that tallies seats. Where the shortcut fails — and it does fail, for early metals and for complexes with strong pi-acceptors — the seat count is the one that still works, because it is the one that has a reason.

The count and the magnetism are one statement #

There is a measurement that reports the seat count directly, and it needs no spectroscopy and no calculation — only a balance in a magnetic field.

Sixteen electrons in eight low orbitals is a closed shell: every one of the eight is doubly occupied and nothing is left unpaired. A square-planar complex counting sixteen is therefore diamagnetic, necessarily, and the prediction is an integer rather than a trend.

The comparison that makes it evidence is a metal that adopts both geometries. Nickel in its +2 state is d⁸ either way. As the tetracyanide it is square planar, counts sixteen, and is diamagnetic. As the tetrachloride it is tetrahedral, where the same eight d electrons meet a different set of leftovers — five of them, with a small separation — and Hund’s rule leaves two unpaired, giving a magnetic moment near four Bohr magnetons.

Same metal, same oxidation state, same d count, same number of ligands. Two geometries, two seat counts, and a measurement that distinguishes them without ambiguity.

That is the cleanest evidence available that the count is about orbitals rather than about donors. A donor tally cannot tell the two nickel complexes apart at all: four ligands, two electrons each, eight from the metal, sixteen either way. The seat count says the tetrahedral case has five leftovers rather than four and that two of them are close enough together to be occupied singly — and a magnetic moment says which of those two descriptions the compound actually obeys.

What the square plane adds to electron counting #

The electron-count argument starts from the observation that eighteen is a count of orbitals rather than of donors, and computes the octahedral case. This essay shows that the same arithmetic gives a different answer for a different geometry, and that the difference is one orbital that nothing interacts with.

The value of putting the two together is that the sixteen-electron rule stops being an exception. There is one rule — fill every metal orbital that is low enough — and eighteen and sixteen are two of its answers, with tetrahedral giving eighteen for a reason that has nothing to do with having four ligands.

The open question is what happens when the ligands are not all sigma donors, since every count here treats a ligand as one donor orbital and nothing else. A pi-acceptor changes which leftovers are low, a pi-donor changes it the other way, and the counts that come out are the same nine orbitals sorted differently. That is a calculation an angular-overlap model could do, and it would turn the one judgement in this essay into an output.