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In borazine, each B and N is approximately trigonal planar. The alternating ring contains a sigma framework with perpendicular p orbitals; the nitrogen lone pairs can donate into the boron acceptor orbitals. Six pi electrons can therefore be distributed around the ring, giving partial B–N multiple-bond character rather than six ordinary isolated single bonds. Equal ring distances are compatible with this symmetric delocalization.

The electronegativity difference makes each B–N bond strongly polar, ${\mathrm{B}}^{\delta +}-{\mathrm{N}}^{\delta -}$. The occupied pi orbitals are weighted toward N, and the empty acceptors toward B. In benzene, equivalent carbon centers give much more even pi distribution and strong resistance to addition that destroys the aromatic system. borazine resembles benzene geometrically but has substantially more polarized, more reactive bonding. It presents electrophilic boron sites to nucleophiles and nitrogen donors to Lewis acids. Polar reagents can add across B–N, and water can cleave the ring; benzene ordinarily favors substitution under suitable conditions rather than facile polar addition or hydrolysis. The label “inorganic benzene” should therefore not be interpreted as identical aromatic stabilization or reactivity. nitrogen-bound Lewis-acid adducts and hydrogen-halide additions demonstrate this contrast experimentally.^[ref]

The methyl nucleophile attacks an electrophilic ring boron. That boron becomes tetrahedral, bearing its original methyl, the incoming methyl, and its two ring nitrogen neighbors. This interrupts its p-orbital participation, so the metal attaches to the remaining five-atom conjugated segment. $A$ is the anionic ${\eta }^5$ heptamethylated-ring Cr(CO)${}_3$ complex, paired formally with $[\mathrm{MgBr}{]}^{+}$; solvent ligation and ion pairing at magnesium are not specified by the spectrum. The original connectivity, with the attacked boron labeled $B_1$, is shown in the first panel below.

There are seven methyl groups. The two attached to tetrahedral $B_1$ are inequivalent because the chromium-bound face distinguishes their environments: one signal of relative integral one for each. The other two B–Me groups are equivalent to one another, giving a signal of integral two. The two N–Me groups adjacent to $B_1$ form another equivalent pair, while the N–Me opposite $B_1$ is unique. The predicted methyl-group multiplicities are therefore $1,1,2,2,1$, the same unordered set as the observed ratio. Multiplying by three gives proton integrals totaling 21. This assignment assumes no fast process interchanging the two $B_1$ methyl environments.

The six-electron neutral ${\eta }^6$ ring donor has become an anionic six-electron ${\eta }^5$ donor, so Cr(0) retains an 18-electron count: $6_d+6_{\mathrm{ring}}+3(2{)}_{\mathrm{CO}}=18$. This borazine dearomatization explains both the changed hapticity and the extra methyl without requiring attack at chromium or loss of a carbonyl.

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.

In the chromium complex, Cr(0) is $d^6$ and receives substantial donation from four carbonyl ligands. Its occupied orbitals can give pi backbonding into the boron-rich acceptor orbitals of the B${}_2$N${}_2$ ring, while the nitrogen-rich occupied ring orbitals donate to the metal. This favors binding across the ring face, approximately ${\eta }^4$, with all four ring atoms participating. A nearly planar ring maintains the p-orbital alignment needed for that interaction and accounts for comparable Cr–B and Cr–N distances. A four-electron ring donor also gives the useful count $6_d+8_{\mathrm{CO}}+4_{\mathrm{ring}}=18$.

Ti(IV) in TiCl${}_4$ is $d^0$ and strongly electron deficient. It is a good acceptor for the two nitrogen lone pairs, but has no filled $d$ set for comparable backbonding into the boron acceptors. chromium therefore favors a face-bound pi complex, while titanium favors N,N sigma coordination. Donation makes the nitrogen environments more nearly tetrahedral and permits the 2,4-Diazadiboretidine ring to pucker. The borons are not directly bound in the same way, explaining $d(\mathrm{Ti}\cdots \mathrm{B})>d(\mathrm{Ti}-\mathrm{N})$. The distinction follows from electron availability, rather than simply calling every short metal–ring contact a separate ordinary sigma bond.

Hydride is delivered to steroid C5 and a proton to C4; the C3 ketone is retained. Use the conjugated fragment $\mathrm{O}=C3-C4=C5$. Carbonyl hydrogen bonds from Tyr91–OH and protonated Glu57 lower the barrier to conjugate reduction. Draw a curved arrow from the nicotinamide C4–H bond of NADPH to steroid C5, shift the C4=C5 pi pair toward C3–C4, and shift C3=O pi electrons onto oxygen. The nicotinamide ring rearomatizes to NADP+; the steroid becomes a bound enolate, or its protonated enol equivalent. electron shifts across the nicotinamide ring must accompany C–H bond cleavage rather than leaving an unexplained neutral carbon radical.

Next, reform C3=O and protonate steroid C4 from an appropriately positioned acid, directly or through an enol/proton-relay route. This regenerates the catalytic acid after solvent reprotonation. The enzyme fixes the hydride-accepting face so that the new C5–H has the 5α configuration, giving dihydrotestosterone; delivery to the opposite face would give a different ring-junction configuration. This is not simple 1,2 reduction of the C3 carbonyl to an alcohol. The net local bookkeeping is one hydride plus one proton across C4=C5, with two electrons supplied by NADPH. The original reactive-fragment diagram shows the enolate-forming arrows and the separate protonation step.

The nitrite ion is an ambidentate ligand: its Lewis structures leave donor lone pairs on both nitrogen and oxygen. In the N-bound mode, $\mathrm{M}-\mathrm{N}(\mathrm{O}{)}_2$, the nitrogen lone pair makes a sigma bond to the metal, while a suitable unoccupied nitrite molecular orbital can accept metal $d$ electron density by pi backbonding. Thus N-bound nitro is a sigma donor and pi acceptor. The nitrite group remains bent at nitrogen; N coordination does not make it a linear nitrosyl.

In the O-bound mode, $\mathrm{M}-\mathrm{O}-\mathrm{N}=\mathrm{O}$, an oxygen lone pair makes the metal–oxygen sigma bond. A second occupied oxygen orbital can donate sideways into an available metal orbital: O-bound nitrito is a sigma and pi donor, rather than the strong pi acceptor of the N-bound description. oxygen's greater electronegativity lowers its donor orbital energies, but the occupied oxygen orbitals still explain its pi-donor classification. Actual mixing depends on metal oxidation state and geometry.

A further possibility is bidentate coordination, $\mathrm{M}({\kappa }^2\text{-}O,O\text{-}\mathrm{N}{\mathrm{O}}_2)$, with both oxygen atoms attached to one metal. Each attachment has oxygen-to-metal sigma donation; additional pi donation depends on orbital alignment. nitrite can also be a bridging ligand between metals. Neither alternative should be confused with linkage isomerism, which specifically compares attachment through different atoms at the same binding site.

Both complexes contain cobalt(III), hence a $d^6$ metal ion. The approximately octahedral molecular geometry, relatively high metal oxidation state and five ammonia donors make the octahedral splitting large enough that pairing is favorable. Their low-spin configuration isAll six metal $d$ electrons are paired, and the ligands in coordination chemistry have closed shells, so neither complex has a permanent spin magnetic moment: both exhibit diamagnetism. The difference between N and O attachment is insufficient to change this low-spin assignment.

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.

For the same pair of atoms, the harmonic approximation gives $\tilde{\nu }=(2\pi c{)}^{-1}\sqrt{k/\mu }$, where $k$ is the bond force constant and $\mu$ the reduced mass. The large shifts mainly reflect changes in $k$ from CO pi backbonding; the complexes are not distinguished simply by their total number of valence electrons.

$\mathrm{W}(\mathrm{CO}{)}_6$ has stronger net backbonding than $\mathrm{Cr}(\mathrm{CO}{)}_6$. Both are group-6 M(0), $d^6$, 18-electron complexes. The more spatially extended tungsten $5d$ orbitals provide effective overlap with CO ${\pi }^{\ast }$, consistent with the lower stretching frequency. As a simple same-mode force-constant comparison, $(1977/2000{)}^2=0.977$, about a 2.3% reduction.

$\mathrm{Ni}(\mathrm{CO}{)}_3(\mathrm{PM}{\mathrm{e}}_3)$ has appreciable but less C–O weakening than those hexacarbonyls. Ni(0) is $d^{10}$ and the complex is also 18-electron. The more contracted, lower-energy late-3d occupied orbitals give less effective donation into CO ${\pi }^{\ast }$ than the group-6 carbonyls in this comparison. Trimethylphosphine donates electron density and increases backbonding relative to an otherwise comparable less-donating nickel complex; that effect does not force its frequency below chromium's. Comparing different metals requires both orbital energies and overlap.

$[\mathrm{Cu}(\mathrm{CO}{)}_4{]}^{+}$ has very weak backbonding and a frequency above free CO. Cu(I) is $d^{10}$ and 18-electron here, but positive charge lowers its occupied orbital energies and suppresses backdonation. Sigma donation and the cation's electrostatic polarization can strengthen the internal C–O bond. Consequently the usual free-CO reference, about $2143\mathrm{c}{\mathrm{m}}^{-1}$, lies below this cation's band. A blue shift is compatible with carbon coordination; it does not imply absence of a metal–CO bond. Coupled normal modes and measurement conditions limit literal extraction of a local bond force constant from one band.

A carbonyl ligand binds through carbon, whose occupied $5\sigma$ molecular orbital has a substantial outward-facing carbon lobe. Overlap with an empty metal orbital along the M–C axis gives ligand-to-metal sigma donation. Conversely, an occupied metal $d_{\pi }$ orbital overlaps sideways with an empty CO $2{\pi }^{\ast }$ antibonding molecular orbital, giving metal-to-ligand pi backbonding. There are two mutually perpendicular pi channels. The original orbital schematic below shows one of them; $+$ and $-$ denote orbital phase, not electric charge.

The interactions are synergic: donation makes electron density available at the metal, and backbonding helps accommodate it while strengthening the metal–carbon interaction. Occupying CO ${\pi }^{\ast }$ reduces the C–O bond order, lengthens that bond and lowers its bond force constant. Therefore stronger pi backbonding generally lowers the carbonyl stretching spectroscopic wavenumber even while strengthening the M–C attachment.

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.

$E=\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2\mathrm{Cl}$, 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.

Write $L^{-}$ for the cyclopentadienyl ligand bearing its $\mathrm{C}{\mathrm{H}}_2$–2-pyridyl arm; ${\eta }^5:\kappa N$ means that its ring and pyridyl nitrogen both bind the same metal. $D=[\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2(\mathrm{NCMe}){]}^{+}$. The base removes the pyridinium N–H proton in $C$, restoring a pyridine lone pair. Intramolecular nitrogen coordination then replaces one acetonitrile ligand and closes a chelate ring. Both CO ligands are retained; the molybdenum remains Mo(II), $d^4$.

This is an proton transfer followed by chelation, with metal electron count $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{NCMe}}=18$. The chelate effect, favorable formation of the Mo–N bond, and release of a solvent molecule explain the driving force. Merely invoking a 16-to-18-electron change would be misleading because $C$ is already 18-electron. In the preceding acid treatment, protonation of the ${\eta }^3$ allyl ligand allows release of propene; the other proton resides on the pendant pyridine, accounting for $C$'s overall $2+$ charge. One proton is removed here; the cyclopentadienyl tether is not deprotonated or detached.

The six P–N bonds are not well described as three permanently localized single bonds alternating with three ordinary double bonds. The strongly polar P–N framework has contributions from ionic bonding and nitrogen lone pair donation into neighboring phosphorus-centered acceptor combinations, including P–Cl ${\sigma }^{\ast }$ orbitals. This negative hyperconjugation supplies additional P–N bonding and equivalent environments around the ring. Approximate planarity is compatible with these interactions, but equal lengths and planarity alone do not establish benzene-like aromaticity. Large phosphorus $3d$ participation is not required.^[ref]

Gallium binds to a ring nitrogen, and the two P–N bonds flanking that nitrogen lengthen. Formation of the N$\to$GaCl${}_3$ coordinate covalent bond changes the available nitrogen electron density and local geometry, weakening its contribution to the adjacent P–N interactions. The other four P–N bonds retain greater multiple-bond character and are correspondingly shorter. Loss of the threefold equivalence and altered local geometry permit the ring to pucker. This is a Lewis acid-base adduct, not replacement of a P–Cl bond by GaCl${}_3$. Single-crystal structures and variable-temperature multinuclear nuclear magnetic resonance establish this class of nitrogen-bound group-13 adducts.^[ref]

Secondary amines normally favor non-geminal disubstitution, putting the second amino substituent on a different phosphorus. The first installed $\mathrm{N}{\mathrm{R}}_2$ group donates electron density to the P–N framework and deactivates its own phosphorus toward further attack. It is also bulkier than an initially installed primary-amino group. Crucially, $\mathrm{P}-\mathrm{N}{\mathrm{R}}_2$ has no N–H proton, so the proton-abstraction/chloride-elimination pathway that can direct a primary amine geminally is unavailable. Direct attack at a less-deactivated $\mathrm{PC}{\mathrm{l}}_2$ site is favored. The resulting non-geminal products can have cis or trans relative orientations around the ring; non-geminal substitution alone does not select one of these stereoisomers. These are useful trends, with known substituent-dependent exceptions in experimentally characterized mixed derivatives.^[ref]

Fluoride favors geminal disubstitution: after the first substitution, the second F replaces the other chloride at that same phosphorus, giving a $\mathrm{P}{\mathrm{F}}_2$ unit and two unchanged $\mathrm{PC}{\mathrm{l}}_2$ units. The first highly electronegative F withdraws electron density inductively, maintaining or increasing the electrophilicity of its phosphorus and activating the neighboring P–Cl bond toward another nucleophile. The small fluoride ion also has little steric hindrance to attack there. In this context geminal substitution means both incoming substituents on one phosphorus; it does not describe the relative positions of different ring nitrogen atoms. Electronic activation at the substituted phosphorus competes with effects elsewhere in the ring, so the argument predicts the preferred pathway rather than an unconditional exclusive product.

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]

N${}_2$: Homo ${\sigma }_{2p_z}$; LUMO the degenerate ${\pi }_{2p}^{\ast }$ pair.

O${}_2$: the two ${\pi }_{2p}^{\ast }$ orbitals are the highest occupied, singly occupied spatial orbitals. The next completely empty spatial orbital is ${\sigma }_{2p_z}^{\ast }$. An added electron can instead occupy the opposite-spin vacancy in a singly occupied pi-star orbital; this distinction matters if “LUMO” is being used for an electron-addition spin orbital.

Li${}_2$: Homo ${\sigma }_{2s}$; next empty sigma orbital conventionally denoted ${\sigma }_{2s}^{\ast }$ in this elementary diagram. Be${}_2$: Homo ${\sigma }_{2s}^{\ast }$; the low empty 2p manifold supplies the LUMO candidates, drawn as ${\pi }_{2p}$ in the stated light-diatomic ordering. Mixing makes these labels descriptions of dominant atomic-orbital character, not strictly unmixed atomic functions.

This distinction is the open-shell frontier orbital convention.

Write both electrode reactions first as reductions:The conventional cell reaction runs with oxidation on the left and reduction on the right. Reverse the first equation and multiply the second by four:Four electrons pass through the external circuit per reaction as written. This direction consumes the oxide; deposition requires the reverse electrolysis direction.

Passivation by the oxide film is a likely cause. As poorly conducting CeO${}_2$ covers palladium and thickens, electron transport through the deposit becomes difficult and the interfacial resistance rises. Transport of ions/defects through the film can also become limiting.

Thus abundant cerium in bulk solution does not guarantee continued interfacial deposition. The film can block the electronic or ionic path required to grow more oxide.

This is an oxide-film transport limitation during cerium dioxide electrodeposition.