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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.

Number the fused heterocycle as a 2,3-dihydrobenzofuran: oxygen is position 1 and the acetamido substituent occupies aromatic position 6. Position 5 is the preferred site of electrophilic substitution. It is para to the ring oxygen and ortho to the acetamido group, so both groups stabilize its Wheland intermediate by resonance donation. Position 7 also receives donation, but lies between the two substituents and is more crowded. Position 4 lacks the same combination of directing effects.

The aryl oxygen donates a lone pair into the aromatic system. The acetamido nitrogen can also donate into the ring, although its lone pair is shared with its own carbonyl group, so it is less activating than a free amino group. Comparing resonance-stabilized intermediates and steric accessibility gives the position-5 prediction; simply counting heteroatoms does not.

Pyrrolidine and 2-methylcyclohexanone first form an enamine, through carbonyl addition, carbinolamine formation, dehydration to an iminium ion, and alpha deprotonation. The less hindered enamine carbon is on the side opposite the methyl-bearing alpha carbon. It attacks the terminal beta carbon of methyl vinyl ketone, while the conjugated $\pi$ system shifts to oxygen. Protonation gives an alkylated iminium species; aqueous acid then hydrolyses it back to the ketone.

The result is a Michael addition product: the previously unsubstituted alpha carbon bears $-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COC}{\mathrm{H}}_3$. In drawing the arrows, start the new C–C bond at the enamine carbon, not at nitrogen. The iminium bond is restored during addition and removed by hydrolysis. The sequence is an enamine-mediated conjugate addition, rather than direct addition to the methyl vinyl ketone carbonyl.

The qualitative nucleophilicity order is $p$-OMe $>$ $p$-Me $>$ H $>$ $p$-Cl $\gg$ $p$-NO${}_2$, matching the product-yield order at the stated stopping time. Electron donation raises the availability of the aniline lone pair for attack. A para methoxy group donates by resonance, while methyl is a weaker donor. Chlorine withdraws inductively and gives an aniline less nucleophilic than the unsubstituted compound. A para nitro group strongly withdraws by resonance: electron flow from nitrogen through the ring into the nitro group stabilizes a charge-separated form and greatly reduces nucleophilicity.

The key arrows for the nitroaniline comparison move the nitrogen lone pair into its own aromatic ring, shift the ring $\pi$ bonds toward the para nitro group, and shift a nitro N=O bond toward oxygen. The methoxy donor produces the opposite electronic bias. These are finite-time yields, so they support a kinetic explanation qualitatively; they are not by themselves measured rate constants or equilibrium constants.

The aniline nitrogen attacks the ipso carbon bearing fluorine. The aromatic $\pi$ electrons shift to give a Meisenheimer complex, with negative charge delocalized to the positions ortho and para to the attacked carbon. At either nitro substituent, that charge can be delocalized further onto nitro oxygen. Subsequent proton transfer and fluoride elimination restore aromaticity and give the diarylamine.

Both the ortho and para nitro groups stabilize the addition intermediate in 1-fluoro-2,4-dinitrobenzene. The para-mononitro compound lacks the additional ortho stabilization, so addition is less favourable and generally slower under the stated conditions. This comparison concerns stabilization of the addition pathway; the presence of fluorine does not make the reaction an ordinary aliphatic SN2 reaction.

Base deprotonates the sulfonamide. Its nitrogen attacks one epoxide carbon from the back, breaking the C–O bond at that carbon and producing an adjacent alkoxide. The nitrogen and oxygen attachments on the cyclopentane are therefore trans. The alkoxide then attacks the aromatic carbon bearing the fluorine ortho to the sulfonyl group. Intramolecular nucleophilic aromatic substitution expels fluoride and closes a seven-membered heterocycle with the path $\mathrm{Ar}-\mathrm{S}-\mathrm{N}-{\mathrm{C}}_1-{\mathrm{C}}_2-\mathrm{O}-\mathrm{Ar}$. The para fluorine remains.

This tandem epoxide-opening/ring-closing pathway is required by the formula of F: simple epoxide addition would retain both fluorines, whereas F contains only one. In the second stage, pyrrolidine replaces the remaining para fluorine by another nucleophilic aromatic substitution, producing G. The mass balance is F plus pyrrolidine minus HF, as required by the given formula of G.

The meso epoxide has enantiotopic ends. With achiral reagents, opening at either end gives an enantiomeric product, so F is a racemic pair with the same trans relative configuration; G retains that configuration.

For hydrogen-bond acceptance, the three strongest groups are the terminal carboxylate, the Cys–CO–N(Pro) tertiary-amide carbonyl, and the Pro–CO–NH(Tyr) secondary-amide carbonyl, in that qualitative order. Negative charge makes carboxylate a particularly strong acceptor. Both amide oxygens accept strongly; greater nitrogen alkyl donation generally makes a tertiary amide oxygen more basic than the comparable secondary-amide oxygen. This last ordering is a functional-group estimate, not a precise calculation of a folded tripeptide's site parameters.

For donation, the order is terminal ammonium $>$ tyrosine phenolic OH $>$ the peptide NH between Pro and Tyr. The thiol is a weaker donor. Protonated ammonium has no available lone pair to accept; the proline amide nitrogen has no N–H bond to donate, and amide nitrogen lone pairs are delocalized into carbonyl groups. The carboxylate has no O–H donor in the drawn protonation state.

At pH 7, a glutamate residue is normally carboxylate. Its most natural role is general-base activation of the attacking water. Draw the carboxylate oxygen lone pair toward a water H, the water O–H pair toward its oxygen, and that oxygen's lone pair toward C3 while the conjugated pi system moves toward the carbonyl oxygen:This can be concerted, avoiding a freely diffusing hydroxide intermediate. The protonated glutamate subsequently returns its proton to the enolate system, directly or through solvent, regenerating carboxylate. A favorable arrangement is needed; ordinary carboxylate does not quantitatively deprotonate bulk water in a prior equilibrium.

The side-chain amide of glutamine is neutral. Its N–H can donate a hydrogen bond to the carbonyl oxygen and stabilize developing enolate charge, while its amide oxygen can accept a hydrogen bond to orient attacking water. Draw $\mathrm{Gln}-\mathrm{CON}{\mathrm{H}}_2\cdots \mathrm{O}=C1-C2=C3$, with the dotted interaction from amide N–H to substrate oxygen. This assists polarization and transition state stabilization; the amide nitrogen is not an ordinary amine base because its lone pair is delocalized into its own carbonyl.

The guanidinium group of arginine is positively charged. It can stabilize developing negative charge by hydrogen bonding and electrostatic attraction, drawn $\mathrm{Arg}-\mathrm{guanidiniu}{\mathrm{m}}^{+}\cdots {\mathrm{O}}^{\delta -}-C1$. Its delocalized cation is not the preferred neutral general base at pH 7. Productive catalysis requires stronger stabilization of the reacting arrangement than of the unreactive ground-state complex.

Phenylalanine and leucine have hydrocarbon side chains and provide no conventional acid/base or hydrogen-bond donor/acceptor group here. They can position the enone, exclude unproductive solvent and select the attacked face through hydrophobic effect and steric hindrance. Those binding effects can assist catalysis but should not be drawn as an imaginary side-chain proton relay. The original schematic depicts the Glu water relay and the Gln/Arg oxyanion-stabilizing contacts.

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.

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.

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]

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]

Draw a curly arrow from the nitrogen lone pair to methyl carbon, and a second from the C-I bond to I:This is concerted SN2 quaternization. The product is choline iodide; nitrogen has four carbon substituents and positive charge, while the alcohol remains OH.

The reacting Homo is the tertiary-amine nitrogen lone pair. The relevant LUMO is the C-I antibonding orbital, ${\sigma }_{\mathrm{C}-\mathrm{I}}^{\ast }$, of MeI. Backside lone-pair donation into that orbital forms N-C while weakening C-I. The neutral OH group is not first converted to an alkoxide in the stated reaction.

The expected order is B-F > C-F > N-F.

B-F has a large favorable polar/ionic contribution because electron-deficient boron and fluorine have a large electronegativity difference. Fluorine lone-pair donation into boron's empty 2p orbital can additionally strengthen bonding in planar BF${}_3$. C-F is a strong, polar second-row sigma bond, but tetrahedral CF${}_4$ has no equivalent empty carbon 2p acceptor for that pi donation.

N-F has less favorable bond polarity and substantial repulsion between nitrogen and fluorine lone pairs, making it weaker. Invoking occupied boron d orbitals is unnecessary and inappropriate for this second-row comparison.

C-C is strongest; N-N is much weaker despite nitrogen's smaller size. Carbon's effective nuclear attraction and good overlap of compact second-row valence orbitals favor a strong C-C sigma bond. The larger, more electron-deficient boron centres give less effective B-B bonding in the stated molecule, despite its relatively high-s-character trigonal bonding.

In hydrazine, the two neighboring nitrogen lone pairs strongly repel one another, destabilizing the N-N bond. Bond strength therefore cannot be predicted from atomic size alone; lone-pair repulsion explains the especially low N-N value.

This illustrates bonding trends in second-period single bonds.

The configurations are $\mathrm{C}:1s^{2}2s^{2}2p^2$ and $\mathrm{N}:1s^{2}2s^{2}2p^3$. A simple complete-core-screening estimate subtracts the two $1s$ electrons, giving $Z_{\mathrm{eff}}\approx 4$ for carbon and $5$ for nitrogen. This neglects screening by the other valence electrons, and is an explicitly rough estimate.

A more systematic estimate using Slater's rules includes $0.85$ for each $1s$ electron and $0.35$ for each other electron in the $2s,2p$ group:Both models predict the important trend: nitrogen's valence shell experiences a greater effective nuclear charge. There is no unique numerical effective nuclear charge without specifying an orbital and a screening prescription; the one-electron formula in (a) does not by itself define neutral-atom orbital energies.^[ref]

The $2s$ atomic orbital is spherically symmetric and has one radial node. A real $2p$ atomic orbital has two lobes of opposite quantum phase, one atomic-orbital nodal plane through the nucleus and no radial node. The three spatial orientations $2p_x,2p_y,2p_z$ remain degenerate in a spherically symmetric atom when external fields and fine structure are neglected.

In a many-electron atom, $2s$ penetrates the screened core more effectively than $2p$, experiences a stronger attraction and is lower in energy: ${\epsilon }_{2s}<{\epsilon }_{2p}$. The hydrogen-like ion degeneracy therefore does not survive screening. Both valence orbital types are generally more tightly bound and spatially contracted in nitrogen than in carbon, because of the larger effective nuclear charge. These are central-field orbital statements; the detailed atomic multiplet structure also involves electron repulsion and Hund's rules.