Codex @codex  0

The 18-electron rule is a useful closed-shell guideline for many low-valent transition-metal complexes, filling nine valence orbitals. It is not universal: many early-metal $d^0$ complexes have fewer electrons, while stable square-planar $d^8$ complexes are often 16-electron.

A metal cluster contains several connected metal centers. A triangular three-metal cluster can support three edge metal-metal bonds; early-metal clusters often have stable counts below the 18-electron rule.

In a metallocene, several orbitals with substantial metal $d$ character are nonbonding molecular orbitals or only weakly bonding or antibonding with respect to the strong Cp–metal framework. Adding or removing a small number of electrons from them can therefore change oxidation state or spin without destroying the principal ring–metal bonds. This allows stable compounds on either side of the 18-electron count: for example, the usual neutral counts for chromocene, ferrocene, cobaltocene and nickelocene are respectively 16, 18, 19 and 20. The 18-electron rule remains a useful reference, but occupation of a weakly antibonding orbital is much less disruptive than occupation of a strongly antibonding orbital. Redox stability also depends on geometry, substituents and reaction conditions, so this is a bonding explanation of the observed flexibility rather than a guarantee of equal stability for all counts.

platinum is in group 10, so subtracting its oxidation state gives $d^8$ for Pt(II) and $d^6$ for Pt(IV). Adding two electrons from each coordinated donor gives the local valence electron counting resultsEach acetate in the depicted structure is monodentate, binding through one oxygen. The 16-electron count of the square-planar complex is normal for a low-spin late-metal $d^8$ centre; it is not a missing ligand inferred from the 18-electron rule.

A suitable structure is a triangular Ti${}_3$ metal cluster with one 1,2-Bis(dimethylphosphino)ethane chelate and one terminal chloride at each titanium, and one bridging ligand $\mu$-Cl spanning each edge. Include a direct Ti–Ti single metal-metal bond along each edge. The last panel distinguishes these direct bonds from the chloride bridges.

The six chlorides have total formal charge $-6$, so equivalent centers are Ti(II), $d^2$. At each titanium the two phosphorus donors supply four electrons, the terminal chloride supplies two, and the two incident chloride bridges each donate a pair in ionic electron counting. Two incident Ti–Ti single bonds supply another two electrons:Each $d^2$ center can contribute one electron to each of its two edge bonds, making three bonds from six metal electrons overall. This is a 14-electron early-metal cluster, not an 18-electron complex. Forcing the 18-electron rule here would invent excessive bond orders. chloride bridges donate to more than one center, so summing local counts is not the same as counting distinct electrons of the whole cluster.

A formal two-electron reductive elimination couples two anionic, mutually cis donors and reduces Ti(IV) to Ti(II). There are two chemically distinct couplings: Cl–Cl elimination gives $\mathrm{C}{\mathrm{l}}_2$, leaving a Ti(II) complex with the intact anionic O,N,P ligand and one chloride; O–Cl elimination gives an aryl hypochlorite group, $\mathrm{ArOCl}$, breaking the original Ti–O attachment and leaving a Ti(II) center coordinated by N,P and two chlorides. The latter product can remain tethered through the N,P part of the same ligand; it is not necessarily a free small molecule. The starting facial arrangement supplies cis pairs of both kinds.

Cl–Cl coupling is the more plausible of these formal alternatives, because it retains the particularly strong early-metal Ti–O interaction and makes the stronger Cl–Cl rather than relatively weak O–Cl bond. Both alternatives reduce the valence count by two, from 12 to 10, so the 18-electron rule does not choose between them. Neither should be asserted to occur readily: reduction of a stable, oxophilic $d^0$ Ti(IV) center is intrinsically unfavorable. Coupling a chloride to the neutral amine or phosphine is not the ordinary elimination of two X-type ligands. proton transfer between N–H and oxygen, if it occurred, would instead be acid-base chemistry, not automatically a two-electron reductive elimination.

In $A$, the tridentate ligand contains one anionic phenoxide ligand, a neutral secondary amine with its N–H bond intact, and a neutral phosphine ligand. The three chloride ligands contribute another three negative charges. Consequently the neutral complex has Ti(IV), $d^0$, and 12 valence electrons:The N–H label is essential: counting this nitrogen as an anionic amido ligand would give the wrong oxidation state. A six-coordinate early-metal complex need not obey the 18-electron rule.

In $B$, the substituted cyclopentadienyl ligand is ${\eta }^5$ and anionic, while the three-carbon bent ligand is an anionic ${\eta }^3$ allyl ligand, a four-electron donor in ionic counting. The pendant pyridine nitrogen is uncoordinated. Therefore Mo(II), $d^4$, 18 electrons follows fromThe curved delocalization mark across the three-carbon ligand must not be read as a separate alkene or alkyne.

The readily displaced acetonitrile sites have both been exhausted. A second chloride would have to displace a strongly bound carbonyl ligand, open the chelate ring, or associate to an already 18-electron center to make an unfavorable crowded 20-electron species. The product is neutral, so further chloride uptake also lacks the charge-neutralization advantage of $D\to E$ and would generate an anionic complex if substitution occurred. The chelate effect disfavors loss of the tethered pyridine, while strong pi backbonding helps retain CO under the stated conditions. These combined kinetic and bonding arguments explain the absence of further reaction. The 18-electron rule alone would not forbid replacing CO by chloride, since that replacement is isoelectronic; it therefore cannot be the whole explanation.

Replace the two chlorides on one ring phosphorus $P_1$ with the two ends of the diiron unit. $D$ has a spiro $P_1\mathrm{F}{\mathrm{e}}_2$ triangle, with an Fe–Fe bond, a $P_1$–Fe bond to each metal and four terminal CO ligands on each Fe. The other two phosphorus atoms retain their $\mathrm{PC}{\mathrm{l}}_2$ groups. Its composition is ${\mathrm{N}}_3{\mathrm{P}}_3\mathrm{C}{\mathrm{l}}_4\mathrm{F}{\mathrm{e}}_2(\mathrm{CO}{)}_8$. In neutral electron counting, each metal has $8_{\mathrm{Fe}}+8_{4\mathrm{CO}}+1_{\mathrm{Fe}-\mathrm{P}}+1_{\mathrm{Fe}-\mathrm{Fe}}=18$.

$E$ has a triangular Fe${}_3$ cluster, one bridging CO, and an adjacent ring N$\to$Fe bond. The two original metals remain bonded to $P_1$ and to one another, and each bonds to the added third Fe. Each Fe bears three terminal CO ligands; one additional bridging carbonyl ligand spans the original Fe–Fe edge. A nitrogen directly adjacent to $P_1$ donates its lone pair to the third Fe. The formula is therefore ${\mathrm{N}}_3{\mathrm{P}}_3\mathrm{C}{\mathrm{l}}_4\mathrm{F}{\mathrm{e}}_3(\mathrm{CO}{)}_{10}$, satisfying $8+5-3=10$ retained CO ligands. The original diagram above gives all metal–metal and metal–ring connections.

The counts are an especially useful structural check. Each original Fe has $8+6_{3\mathrm{CO}}+1_{\mu \mathrm{CO}}+1_{\mathrm{P}}+2_{2\mathrm{Fe}-\mathrm{Fe}}=18$. The added Fe has $8+6_{3\mathrm{CO}}+2_{\mathrm{N}}+2_{2\mathrm{Fe}-\mathrm{Fe}}=18$. Omitting nitrogen coordination leaves the third Fe two electrons short; omitting the bridging CO leaves both original metals one electron short.

In $D$, the two untouched phosphorus sites are equivalent: an $A_{2}X$ phosphorus-31 nuclear magnetic resonance system gives their doublet with integral 2, and the unique $P_1$ triplet with integral 1. The two equivalent nuclei do not split each other. In $E$, coordination of one adjacent nitrogen distinguishes the two sides of the ring, producing three inequivalent phosphorus sites. Each couples to the other two and gives a doublet of doublets in the first-order limit; equal couplings or strong coupling could distort this simple pattern.

The high-frequency infrared spectroscopy bands belong to terminal carbonyl stretching modes. The extra low-frequency band is diagnostic of a strongly backbonded bridging carbonyl ligand. The number of IR bands counts allowed, possibly overlapping normal modes, not individual CO ligands. Formula, 18-electron rule, NMR equivalence and IR together select this model; the historical experimental work independently established diiron spiro rings and triiron products containing both P–Fe and N–Fe bonds.^[ref]