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Let $z$ be normal to the Cp ring through its centre. The five carbon $p_z$ orbitals form symmetry-adapted linear combinations with angular patterns $m=0$, $|m|=1$, and $|m|=2$. The all-in-phase $a_1$ combination has sigma symmetry about this axis and can interact with $d_{z^2}$. The $e_1$ pair matches $(d_{xz},d_{yz})$ by pi symmetry, while the $e_2$ pair matches $(d_{xy},d_{x^2-y^2})$ by delta symmetry. Rotating within either degenerate pair changes the drawing but not the physical interaction.

For the usual eta-five geometry, the metal $d$ interactions are ranked qualitatively pi > delta > sigma. The $d_{xz},d_{yz}$ lobes overlap effectively with the occupied ring combinations. The delta interaction is weaker and involves the higher, initially unoccupied Cp combinations; the $d_{z^2}$ interaction with the ring is particularly weak because its axial and equatorial parts contribute little net overlap in this geometry. Thus the statement that a sigma bond is always stronger than a pi bond cannot be transferred indiscriminately to this delocalized ring–metal overlap problem. metal $s$ and $p_z$ orbitals also have the symmetry of the $a_1$ coordination ligand combination and contribute to overall bonding, so weak $d_{z^2}$ overlap does not mean the entire sigma-symmetry channel is absent.

In ionic electron counting, Cp is ${\mathrm{C}}_5{\mathrm{H}}_5^{-}$ with six pi electrons in its three occupied ring combinations, hence a six-electron donor. The same coordination ligand is assigned five electrons in neutral counting.

Use neutral electron counting consistently: each terminal Cp supplies five electrons, while the eight pi electrons of neutral pentalene must be allocated across the shared fused framework rather than counted twice as two independent Cp coordination ligands. One set of 18-electron-consistent proposals is:

The original schematic shows the donor regions and the required iron–iron bond; dashed metal-to-region lines denote delocalized coordination rather than localized carbon–metal single bonds.

These proposals explain the requested electron counts and distance ordering. electron counting alone does not establish the unique ground-state structure: orbital calculations on related pentalene complexes also identify stable formally electron-deficient arrangements. In particular, the formal requirement of a metal–metal bond in the proposed 18-electron iron model should not be converted into a claim that every known iron pentalene complex adopts that model.^[ref]

Four chloride ligands give Ti(IV), $d^0$. Ionic electron counting givestetrahedral molecular geometry separates the four donors without placing them at $90^{\circ }$. A $d^0$ centre has no occupied d orbitals providing a square-planar stabilization; the 18-electron rule is not a requirement for this early-metal compound.

Choose $z$ along M–N–N. The occupied dinitrogen $3{\sigma }_g$ orbital, with amplitude on the terminal N atoms, overlaps an empty metal orbital of sigma symmetry to give N${}_2$-to-metal sigma donation. Filled metal $d_{xz}$ and $d_{yz}$ orbitals separately overlap the two empty N${}_2$ ${\pi }_g^{\ast }$ orbitals, giving metal-to-N${}_2$ pi backbonding. The lower-energy combination of each interacting pair is bonding and the higher is antibonding; each donation channel starts with two electrons in the occupied fragment orbital. The lower panel of the original orbital diagram shows these three channels.

As for a terminal carbonyl ligand, the donor and acceptor interactions reinforce one another: sigma donation helps make electron density available at the metal, while pi backbonding strengthens M–N and weakens N–N. For conventional electron counting an intact end-on neutral N${}_2$ ligand is an L-type two-electron donor; counting the backbonded electrons a second time as extra donated ligand electrons would double count them.

Use ionic electron counting, counting each ordinary donor atom once and keeping ligand charge distinct from denticity. A is 16-electron, B is 14-electron and C is 16-electron, so none is an 18-electron complex under this convention.

A contains a modified, monoanionic PNP pincer: the exocyclic double bond and altered pyridine-ring bond pattern in the drawing matter. Its two phosphines and its central N donor provide six electrons altogether, while neutral N${}_2$ provides two. Neutrality then gives Co(I), $d^8$, and $8+6+2=16$. Treating the modified pincer as a neutral intact pyridine ligand would incorrectly produce Co(0) and a 17-electron count. A four-coordinate approximately square planar molecular geometry in a $d^8$ complex can leave the strongly sigma-antibonding orbital empty and be stable at 16 electrons. The underlying modified-pincer chemistry has been examined using structure, spectroscopy and electronic-structure calculations.^[ref]

B has a neutral tertiary amine donor at the centre of its tripod and three anionic amido arms. Their combined charge is $-3$, so Re is Re(III), $d^4$. The four N donors give eight electrons and N${}_2$ gives two: $4+8+2=14$. Such an electron count is not inherently unexpected in an amido-supported complex: strong amido pi donation and the imposed five-coordinate geometry alter the orbital energies, and the empty orbitals are not nine equivalent slots that must all be filled. Steric protection and chelation also help stabilize this coordinatively restricted structure. The 14-electron figure uses the usual two-electron ionic count for each amido donor; pi bonding should not be appended to it inconsistently.

C has an anionic acyl carbon donor and a neutral phosphine in the chelate, together with two neutral PMe${}_3$ donors. Thus Co(I), $d^8$, receives $2+2+4=8$ ligand electrons, giving $8+8=16$. This is again a reasonable four-coordinate $d^8$ structure. The acyl C=O is part of a carbon-bound acyl ligand, not an additional free CO donor.

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.

B is oxidative addition; D is alkyne insertion into Ni–aryl followed by carbonyl-oxygen coordination; E is carbonyl insertion into Ni–alkenyl. The distinction is visible from the bonds made and from consistent electron counting.

For B the depicted Ni(0) starting complex has two phosphine and two MeCN donors, giving $10+4+4=18$ electrons. Addition across the aryl C–Br bond produces two X-type ligands and Ni(II). The drawn product has lost the two solvent donors, giving $8+4+2+2=16$ electrons. Intrinsic oxidative addition to the unsolvated 14-electron $\mathrm{N}{\mathrm{i}}^0(\mathrm{dppe})$ fragment raises its count to 16; the overall drawn $18\to 16$ change also includes loss of four solvent-donor electrons. A falling displayed count does not turn this oxidative addition into a reductive elimination.

Before D, the aryl–alkyne intermediate is Ni(I), $d^9$, with $9+4_{\mathrm{P},\mathrm{P}}+2_{\mathrm{aryl}}+2_{\mathrm{alkyne}}=17$ electrons. Syn migration couples aryl carbon to one alkyne carbon and leaves a Ni–alkenyl bond. Insertion alone removes the separate two-electron alkyne donor and would give a 15-electron alkenyl fragment. The ester carbonyl oxygen then binds, as drawn, restoring the count to $9+4+2_{\mathrm{alkenyl}}+2_{\mathrm{O}}=17$. Ni remains $+1$.

In E the Ni-bound alkenyl carbon attacks the coordinated ester carbonyl carbon, forming the new ring-closing C–C bond. The C=O pi electrons move to oxygen, and an Ni–O alkoxide bond replaces Ni–C. This is carbonyl migratory insertion, equivalently intramolecular addition to a coordinated carbonyl. A neutral O donor plus an alkenyl X ligand become one alkoxide X ligand, so the drawn count falls $17\to 15$, while Ni remains $+1$. A solvating molecule could fill the resulting vacancy, but it is not shown and should not be silently added to the stated count.

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.

Use an ${\eta }^5$ cyclopentadienyl ligand and four terminal carbonyl ligands at each titanium, with a Ti–Ti single metal-metal bond. The middle panel shows this connectivity. Neutral electron counting assigns five electrons to each Cp and givesEquivalently, the two Cp anions require an average oxidation state Ti(I), hence $d^3$; ionic counting gives $3+6+8+1=18$. The formal Ti–Ti bond order is one. The large coordination envelope can make this bond weak and long; reaching 18 electrons is bookkeeping, rather than proof of a short, strong bond. There is no need to invent extra CO bridges or a higher Ti–Ti bond order.

Place four terminal trifluorophosphine donors on each cobalt and join the cobalt centers by a single metal-metal bond, as in the first panel. The P atoms bind to cobalt; the fluorines remain attached to phosphorus. Each neutral donor supplies two electrons. Neutral electron counting givesThe ionic description is Co(0), $d^9$, with eight electrons from the four donors and one from the Co–Co bond. A Co–Co single bond supplies the missing electron at each center. No bridging PF${}_3$ is required by this count.

For the doubly bridged tetracarbonyl, each iron has one terminal carbonyl ligand, two bridging carbonyl ligands and one cyclopentadienyl ligand. Neutral electron counting gives $8+5+2+2=17$ before direct bonding, so a single Fe–Fe bond gives 18 at each iron. For the triply bridged tricarbonyl, each iron instead has $8+5+3=16$ before direct bonding; formal completion to 18 requires a double Fe–Fe bond. Equivalently, the total fragment counts before metal–metal bonding are 34 and 32 respectively.The increased direct bond order explains the substantial contraction. The pentamethylcyclopentadienyl ligands also alter steric and electronic effects, and bridge geometry changes, so this is not a controlled measurement of bond order alone. Nevertheless, the markedly shorter distance in the less carbonyl-saturated dimer is consistent with stronger direct Fe–Fe bonding. The isolated triply bridged compound was characterized structurally and magnetically in the primary work.^[ref]

The trityl cation acts as an electrophile and removes one metal-bound methyl by methyl abstraction. The balanced conversion isThus $C=[\mathrm{C}{\mathrm{p}}_2\mathrm{ZrMe}{]}^{+}[\mathrm{B}({\mathrm{C}}_6{\mathrm{F}}_5{)}_4{]}^{-}$ and $D=\mathrm{P}{\mathrm{h}}_3\mathrm{CMe}$, 1,1,1-triphenylethane. Transfer is formally of $\mathrm{M}{\mathrm{e}}^{-}$ to carbon, not transfer of a hydride or a methyl radical. zirconium remains Zr(IV), $d^0$. Ionic electron counting changes from $12+4=16$ in the dimethyl to $12+2=14$ at the cation, creating an accessible coordination site. The borate is a weakly coordinating anion, although ion pairing can still occur in solution.

oxidative addition inserts Pt into one B–Br bond. All three B–Br positions of the starting ring are equivalent, so there is no distinct choice among them before the first substitution. The product is the square-planar Pt(II) boryl bromide $\mathrm{trans}-[\mathrm{Pt}(\mathrm{PC}{\mathrm{y}}_3{)}_2(\mathrm{Br})({\mathrm{B}}_3\mathrm{B}{\mathrm{r}}_2{\mathrm{N}}_3{\mathrm{H}}_3)]$, with the two phosphines trans, the Pt–B bond trans to Pt–Br, and the six-membered ring intact. The third panel shows this connectivity. This is a metal–boron sigma bond, not attachment of the whole ring as an ${\eta }^6$ pi ligand. Formal electron counting gives Pt(II), $d^8$, with $8+4_{\mathrm{P}}+2_{\mathrm{Br}}+2_{\mathrm{boryl}}=16$ electrons, appropriate for square-planar coordination. The primary synthesis establishes this trans-bromo(borazinyl) structure.^[ref]

The electron-rich Pt(0) reacts at the polarized B–Br bond, whose electrophilic boron and departing bromide provide a favorable bond-activation pathway. This accounts for B functionalization without needing cleavage of a ring B–N bond or selection of a nitrogen-bound product. It does not imply that arbitrary N–H activation is impossible under every condition.

Replacing an electronegative bromine substituent by a metal fragment changes the boron's electron demand. Relative to the B–Br sites, the metal-bound boron is less effective at accepting N-to-B pi donation, so its adjacent B–N bonds are expected to have less multiple-bond character and be longer; donation toward the remaining B–Br sites produces compensating unequal ring bonds. The original threefold equivalence is lost. The detailed lengths depend on the full electronic structure; a localized drawing with three rigid alternating double bonds would conceal the redistribution that the unequal lengths reveal.

For electron counting, use the ionic method: subtract the metal oxidation state from its group number, then add two electrons for each neutral phosphine ligand or carbonyl ligand and two for each chloride ligand.

$\mathrm{Ni}(\mathrm{PP}{\mathrm{h}}_3)(\mathrm{CO}{)}_3$ is Ni(0), $d^{10}$, 18-electron and tetrahedral. The count is $10+2+3(2)=18$. With only one distinct ligand in an $\mathrm{M}{\mathrm{L}}_3{\mathrm{L}}^{\prime }$ tetrahedral molecular geometry, there is only one geometrical arrangement.

$\mathrm{Rh}(\mathrm{PP}{\mathrm{h}}_3{)}_2(\mathrm{CO})\mathrm{Cl}$ is Rh(I), $d^8$, 16-electron and square planar, with the two phosphines trans. Its count is $8+2(2)+2+2=16$. A square planar molecular geometry is characteristic of this low-spin $4d^8$ configuration. Opposite triphenylphosphine groups avoid their largest mutual steric hindrance; CO and chloride occupy the other trans pair.

$\mathrm{Pd}[\mathrm{P}{(}^{\mathrm{t}}\mathrm{Bu}{)}_2\mathrm{Ph}{]}_2\mathrm{C}{\mathrm{l}}_2$ is Pd(II), $d^8$, 16-electron and square planar, with the bulky phosphines trans. Its count is likewise $8+4+4=16$. The two di-tert-butylphenylphosphine groups are particularly crowded if cis. These predictions concern the favored intact four-coordinate species; ligand dissociation can occur under other conditions.

$E=\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2\mathrm{Cl}$, with chloride replacing the remaining acetonitrile. The pendant pyridine remains attached. The substitution removes the positive electric charge on $D$ but does not change the Mo(II) oxidation state. In ionic electron counting, both neutral acetonitrile and anionic chloride ligand donate two electrons, giving $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{Cl}}=18$ in $E$.

acetonitrile is the remaining weak, exchangeable monodentate donor. Its displacement forms a favorable Mo–Cl bond at a cationic center, retains the intramolecular Mo–N bond, and restores an 18-electron product after substitution. The following formulas specify the coordination changes directly:Tetramethylammonium is a spectator counterion; it is not a carbon donor to Mo.

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]

For a neutral borane ${\mathrm{B}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}$ with one terminal B–H bond assigned to each vertex, the skeletal electron number is $3n+m-2n=n+m$; an overall charge $q-$ adds $q$ electrons. Suitable twelve-electron correspondences areFor C, four BH contributions plus four additional hydrogens give $8+4=12$ electrons. For D, three BH contributions plus five additional hydrogens and the negative charge give $6+5+1=12$. They map to the tetrahedral and triangular low-vertex skeletons respectively. The neutral formal alternative for D is ${\mathrm{B}}_3{\mathrm{H}}_9$, also with twelve skeletal electrons.

The electron counting correspondence alone must not be read as evidence that the neutral boranes are isolable stable bulk materials. Low-vertex hydrogen-rich boranes can be transient; the triangular octahydrotriborate anion is the established practical three-boron example, whereas neutral ${\mathrm{B}}_3{\mathrm{H}}_9$ is a formal analogy rather than an equally established preparative product. The requested skeletal correspondence should be distinguished from a claim about comparable stability or isolation conditions.^[ref]

The wording places both dicarbollide cages on the same iron, leaving the second iron outside the cages. The corresponding proposal is $[L_2\mathrm{Fe}\to \mathrm{Fe}(\mathrm{CO}{)}_4{]}^{2-}$, with the first iron sandwiched between the two open ${\eta }^5$ faces and the second carrying four carbonyls. It is not the symmetric arrangement with one cage on each iron, which would contradict the specified single cage-interacting metal site.

The donor fragment $[\mathrm{F}{\mathrm{e}}^{\mathrm{II}}{L}_2{]}^{2-}$ has $d^6+2(6)=18$ electrons; the external neutral $\mathrm{Fe}(\mathrm{CO}{)}_4$ fragment has $8+4(2)=16$. Metal-to-metal donation therefore provides a natural 18-electron acceptor adduct. In covalent-single-bond oxidation-state bookkeeping, the analogous assignments Fe(III)/Fe(−I) give $d^5+12+1=18$ and $d^9+8+1=18$. In either account the whole species has two negative charges. The outside iron can have an approximately trigonal-bipyramidal five-coordinate arrangement when the Fe–Fe direction and its four CO ligands are counted.

The important structural distinction is a bis(dicarbollide) iron donor joined to an exo tetracarbonyl iron acceptor. Formula and electron counting alone do not uniquely fix all carbonyl angles or prove a particular crystalline conformation.

The pentamethylcyclopentadienyl ligand is the anion ${\mathrm{C}}_5\mathrm{M}{\mathrm{e}}_5^{-}$, normally a six-electron ${\eta }^5$ donor in ionic electron counting. Its methyl substituents change steric hindrance and electron donation relative to an unsubstituted cyclopentadienyl ligand.

The anion formed by N–H deprotonation of pyrrole. It can bind through nitrogen as a two-electron ${\eta }^1$ donor, or through its aromatic ring as a six-electron ${\eta }^5$ donor. Specifying hapticity is essential for electron counting.