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One consistent localized orbital hybridization description uses the Lewis structuresFor methyldiazonium, take C as $s{p}^3$ and both N atoms as $sp$. For diazomethane, the double-bond resonance contributor suggests C as $s{p}^2$, central N as $sp$, and terminal N as $s{p}^2$. The central N uses two opposite $sp$ hybrids for the linear C–N–N sigma bond framework. In the localized double-bond picture, terminal N uses one $s{p}^2$ hybrid for its sigma bond, two for lone pairs, and its remaining p atomic orbital for a pi bond.

These labels specify an approximate bonding construction, not uniquely measurable atomic properties. An alternative representation takes terminal N as $sp$, with one $sp$ lone pair and one p lone pair; the latter can participate in the delocalized C–N–N $\pi$ system. It describes the same resonance more explicitly. Neither construction requires an unpaired electron or a bent central N.

The products are hexanenitrile, $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}\equiv \mathrm{N}$, and ethyl triphenylmethyl ether, $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The nitrile carbon comes from cyanide, so replacing chloride in the five-carbon chain gives a six-carbon product.

For the primary chloride, the carbon-end lone pair of $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted nucleophilic substitution step forms the C–C bond and releases $\mathrm{C}{\mathrm{l}}^{-}$. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$. An ethanol oxygen lone pair then attacks the carbocation, giving $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-{\mathrm{O}}^{+}(\mathrm{H})\mathrm{Et}$. Finally pyridine accepts that O–H proton, with the O–H bond pair returning to oxygen, giving the neutral ether and pyridinium chloride.

Sodium borohydride is chemoselective for the ketone; lithium aluminium hydride reduces both the ketone and the lactam. In an amide, donation of the N lone pair into the carbonyl $\pi$ system reduces the carbonyl carbon's electrophilicity. It is therefore substantially less susceptible to the milder hydride donor.

Lithium aluminium hydride supplies more strongly reducing hydride equivalents, and the strong affinity of aluminium for oxygen assists removal of the amide oxygen through the iminium pathway. This difference explains both the unchanged lactam carbonyl in B and the absence of that oxygen in C. Both reactions are interpreted after the usual protonating workup.

B is 4-hydroxy-1-methylpyrrolidin-2-one; C is 1-methylpyrrolidin-3-ol. Their ring connectivities are shown below. Each ring remains five membered; the attached N-methyl group is retained.

For B, a B–H bond of sodium borohydride supplies a hydride equivalent to the ketone carbonyl carbon, and the carbonyl $\pi$ pair moves to oxygen. This gives an alkoxide, whose protonation during the reaction or workup gives the secondary alcohol. The lactam carbonyl remains. Converting one $\mathrm{C}=\mathrm{O}$ to $\mathrm{CH}(\mathrm{OH})$ adds two H atoms and gives ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{N}{\mathrm{O}}_2$.

For C, lithium aluminium hydride reduces that ketone in the same net way and also reduces the lactam. Coordination of the amide oxygen to aluminium assists hydride attack at its carbonyl carbon, giving a tetrahedral intermediate. The N lone pair then forms a C=N bond while the C–O bond breaks to an aluminium-bound oxygen group. A resulting iminium ion accepts a second hydride at carbon, yielding the cyclic tertiary amine. Aqueous workup protonates the ketone-derived alkoxide.

The lactam conversion replaces $\mathrm{N}-\mathrm{C}(=\mathrm{O})$ by $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2$; its oxygen is removed rather than retained as another alcohol. Together with ketone reduction, the net change is $+4\mathrm{H}-\mathrm{O}$, giving ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{NO}$. No stereochemistry is specified: the new alcohol stereocentre is formed as a racemic mixture in an achiral environment.

D is $(\mathrm{MeO}{)}_2\mathrm{CHC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Me}$, and E is $\mathrm{OHCC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$. D results from transesterification; E results from acid-catalysed acetal hydrolysis.

For D, methoxide attacks the ester carbonyl carbon and moves the carbonyl $\pi$ pair to oxygen. Collapse of the tetrahedral reaction intermediate reforms C=O and expels ethoxide. Proton transfer with the methanol solvent regenerates methoxide. Excess methanol favours replacement of the ethyl ester by a methyl ester; the dimethyl acetal survives these basic conditions.

For E, protonate one acetal oxygen, then break its C–O bond with the pair going to oxygen, releasing methanol and forming an oxocarbenium ion. Water attacks that carbon. Deprotonation gives a hemiacetal; protonation of its remaining methoxy oxygen permits loss of a second methanol. The hydroxyl oxygen's lone pair forms C=O during this loss, and deprotonation gives the aldehyde, regenerating the acid catalyst.

The spectroscopy is consistent with these products. D has one ester carbonyl near $1745\mathrm{c}{\mathrm{m}}^{-1}$ and six carbon-13 NMR environments: acetal CH, two distinct chain $\mathrm{C}{\mathrm{H}}_2$ sites, carbonyl, ester $\mathrm{OC}{\mathrm{H}}_3$, and the two equivalent acetal $\mathrm{OC}{\mathrm{H}}_3$ groups. Thus seven carbon atoms give six resonances. E has separate ester and aldehyde carbonyl bands, approximately $1745$ and $1730\mathrm{c}{\mathrm{m}}^{-1}$, and six distinct carbon atoms: two carbonyls, two chain methylenes and the two ethyl carbons.

For E, each BH supplies two skeletal electrons and each phosphorus vertex supplies three after retaining a lone pair. Hence $10(2)+2(3)=26$ electrons, or thirteen pairs, matching a twelve-vertex closo cluster. Place the two adjacent P vertices on a regular icosahedron. The remaining ten boron vertices fall into four symmetry classes: two sites adjacent to both P vertices, four adjacent to exactly one, two adjacent to neither but opposite the common-neighbour pair, and two opposite the P pair. The populations are $2:4:2:2$, so four boron resonances are expected. Interchanging the identical phosphorus vertices is allowed when determining chemical equivalence.

For F, a ten-vertex bicapped square antiprism offers a consistent closo model. There are eight BH vertices, one B vertex bearing the $\mathrm{NE}{\mathrm{t}}_3$ donor instead of H, and one P vertex. Their skeletal contributions are respectively $16$, $3$ and $3$, giving 22 electrons, eleven pairs. The donor-bearing B contribution follows from B plus the donated lone pair minus the exoskeletal B–N bond, $3+2-2=3$.

One explicit arrangement consistent with six boron signals puts P at one axial cap and the $\mathrm{NE}{\mathrm{t}}_3$ substituent on one B vertex of the nearer square. The remaining mirror plane through that B, the opposite vertex of its square, and the cap axis leaves three B sites individually fixed: the donor-bearing B, its opposite square vertex and the other axial cap. The other six B sites form three exchanged pairs. Thus the nine borons give three singleton environments and three two-site environments, six signals in all. The original compound has a crystallographically characterized positional assignment; the spectrum alone asks for a compatible structure and does not establish a unique positional isomer. A single-crystal X-ray diffraction determination or assigned heteronuclear correlations would distinguish alternatives.^[ref]

${\mathrm{S}}_4^{2+}$ is approximately square planar; $\mathrm{S}{\mathrm{n}}_5^{2-}$ is a closo trigonal bipyramid. Both have 22 valence electrons, $4(6)-2=22$ and $5(4)+2=22$, but their different local orbital requirements prevent inferring identical shapes from this total alone.

For the sulfur ring, four S–S sigma bonds use eight electrons and four in-plane lone pairs use another eight, leaving six electrons in the perpendicular cyclic pi system. The resulting filled bonding and degenerate nonbonding pi levels give a six-pi-electron aromatic description of a four-membered ring, consistent with equivalent S–S links in the ideal square. The out-of-plane p orbitals are retained; this is not a tetrahedral four-vertex closo cage.

For tin, allocate one mainly s-rich lone pair to each atom. The remaining $22-5(2)=12$ electrons form six skeletal pairs, exactly $n+1$ for $n=5$. Wade's rules therefore predict a trigonal bipyramid, with three equatorial vertices and two axial vertices. The axial pair is not joined by a polyhedral edge; its bonding is part of the delocalized three-dimensional framework. The contrast is a planar aromatic sigma/pi ring versus a three-dimensional multicentre-bonded Zintl cluster, rather than cationic charge alone imposing planarity or anionic charge alone imposing a polyhedron.

A phosphine ligand such as $\mathrm{P}{\mathrm{R}}_3$ donates a phosphorus lone pair. It can also accept pi backbonding into orbitals with substantial P–R antibonding character; phosphorus $3d$ orbitals need not be essential acceptors. Alkyl substitution generally increases sigma donation relative to electron-withdrawing substituents.

A phosphite ligand such as $\mathrm{P}(\mathrm{OR}{)}_3$ binds through a phosphorus lone pair and can accept pi backbonding. Its donation and acceptance differ from those of both phosphine ligands and carbonyl ligands.

Overlap between adjacent p-type atomic orbitals permits delocalized electronic states across more than one chemical bond. A cyclic path with suitable occupancy can support aromaticity; an external lone pair can act as a donor without contributing an extra independent pair to every ring count.

An electron-counting approach relating skeletal electron pairs to a cluster vertex count $n$. A closo deltahedron has $n+1$ pairs; nido and arachno frameworks have $n+2$ and $n+3$. The count excludes electrons in externally directed bonds or lone pairs. It is a structural guide with a defined domain, not a rule that every molecule with the same total valence count has the same shape.

Interconversion of the two spatial orientations of a three-coordinate pyramidal centre through a flatter geometry. Quaternary ammonium nitrogen has no lone pair and cannot follow the ordinary three-coordinate amine pathway.

${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}$, an aromatic nitrogen heterocycle whose nitrogen lone pair can accept a proton.

A five-membered aromatic heterocycle containing NH and four CH groups. The nitrogen lone pair participates in its six-electron aromatic system.

A six-membered aromatic cation containing oxygen. One oxygen lone pair participates in the pi system while the other remains outside it.

A single nitrate Lewis structure singles out one N=O connection, yet the three N–O chemical bonds of the isolated symmetric ion are equivalent. Chemical resonance reconciles that symmetry with the integer electron pair allocations of a Lewis structure. The equal chemical bond lengths in nitrate are discussed in The Open University, Discovering chemistry: resonance structures.

Chemical resonance uses several Lewis structures with the same chemical atom connections and total electric charge to describe electronic structure that one allocation cannot fully express. The chemical atoms stay in place; the positions of bond pairs, lone pairs, and formal charges differ between drawings. A two-headed arrow $\leftrightarrow$ connects the drawings and is distinct from a chemical equilibrium arrow $\rightleftharpoons$. See the IUPAC Gold Book, resonance form.

In each nitrate resonance structure, use the existing formal charge convention $\mathrm{FC}=v-n-b/2$:Here $v$ counts neutral-chemical atom valence electrons, $n$ counts nonbonding electrons, and $b$ counts incident bonding electrons. The sum is $+1-1-1+0=-1$, agreeing with the nitrate ion's net electric charge. The formal charge rules and their distinction from actual atomic electric charge are given in Flowers et al., Chemistry 2e, 7.4: formal charges and resonance.

Keep one labelled oxygen fixed while reading the three nitrate resonance structures. Its connection is double in one drawing and single in two. Equal weighting within this three-contributor Lewis structure model gives the average bond orderThe same calculation applies to all three N–O connections. Averaging the drawn formal charges similarly gives $-2/3$ per oxygen, while nitrogen retains $+1$; $+1+3(-2/3)=-1$. These fractions summarize the chosen contributors. They are not a measurement of partial atomic charge, a literal fraction of an electron sitting on an oxygen, or time spent in one drawing.

The resonance hybrid has one spatially continuous electron density extending over the ion. None of its N–O chemical bonds is permanently selected as the double connection of one contributor. The dashed marks above are symbols; electrons do not travel along them or make the marks float between chemical atoms. For isolated symmetric nitrate, symmetry makes the three N–O chemical bonds equivalent; an environment that distinguishes the oxygen sites can remove that equivalence. The fraction $4/3$ alone predicts neither a chemical bond length nor a dissociation energy. The general relation between multiple chemical bonds, lengths, and strengths is discussed in 7.5 Strengths of Ionic and Covalent Bonds; its tabulated averages are not a nitrate-specific calibration of $4/3$.

Ozone provides a useful two-contributor comparison. Its $3\times 6=18$ valence electrons give three bond pairs and six lone pairs. One resonance structure has a double O–O connection on the left and a single one on the right; the other reverses these allocations while retaining the same chemical atom positions and connections. The central oxygen has one lone pair and formal charge $+1$; the singly connected terminal oxygen has three lone pairs and formal charge $-1$; the doubly connected terminal oxygen has two lone pairs and formal charge zero. Thus $6+12=18$ electrons and zero net electric charge are conserved. The equal-contributor bond order count is $(2+1)/2=3/2$ for either connection. See MIT 5.111, lecture 10, pp. 4–5: ozone resonance structures.

For carbon dioxide, $\mathrm{O}=\mathrm{C}=\mathrm{O}$ and ${\mathrm{O}}^{-}-\mathrm{C}\equiv {\mathrm{O}}^{+}$ share sixteen valence electrons and the same O–C–O connections. The mirror drawing ${\mathrm{O}}^{+}\equiv \mathrm{C}-{\mathrm{O}}^{-}$ is also possible. The two charge-separated drawings are equivalent to one another for an isolated symmetric carbon dioxide molecule, but neither is equally dominant with $\mathrm{O}=\mathrm{C}=\mathrm{O}$ in the elementary formal charge comparison.

Chemical resonance does not mean that a molecule rapidly flips between drawings, nor that separate molecules at chemical equilibrium carry the individual drawings. The actual electron density belongs to one combined electronic description. In a quantum mechanics treatment, contributors enter a linear combination of many-electron wavefunctions; a molecular orbital treatment can describe the extended electronic arrangement through molecular orbitals. The drawings alone provide neither a measured density nor numerical weights. Their usefulness is to expose the limitations of one localized allocation without pretending that every algebraically allowed Lewis structure contributes equally.

A resonance structure is one Lewis structure contributing to a chemical resonance description. The contributors retain the same atomic nuclei in the same positions, the same connections, and the same total valence electron count and electric charge; allocations of bond pairs, lone pairs, and formal charges differ. They are alternative representations of one electronic structure, not distinct substances with different atomic arrangements. See Chemistry 2e, 7.4: resonance forms.

The number of electrons allocated to the connected cluster framework after separating external bonds and appropriate lone pairs. For ${\mathrm{B}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}^{\mathrm{q}-}$ with one terminal B–H bond per vertex it is $n+m+q$.

In tetrahedral molecular geometry, four neighbouring chemical atoms surround a central chemical atom in the directions of the vertices of a tetrahedron. In the regular idealization, the central atomic nucleus lies at the centre of a regular tetrahedron, and every pair of chemical bonds makes the same bond angle. Methane, ${\mathrm{CH}}_4$, provides the standard example: its carbon has four bonding electron domains and no lone pair. See Chemistry 2e, §7.6: tetrahedral methane.

The exact ideal bond angle follows from four directions proportional to $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. Each has squared Euclidean norm $3$, and each distinct pair has dot product $-1$. HenceThis is a statement about a regular tetrahedron, rather than a universal exact value for every molecule labelled tetrahedral. It also explains why a cross-shaped flat drawing with $90^{\circ }$ gaps is misleading: four equal directions can spread out more evenly in three dimensions.

The same vectors sum to zero. Consequently, four equal bond-dipole contributions along those directions cancel. The symmetric methane molecule is nonpolar, but a tetrahedral shape alone does not imply a nonpolar molecule: replacing one neighbour can make the electronic distribution and bond contributions unequal. Chemical polarity depends on both molecular geometry and electron density. See McMurry, Organic Chemistry, §2.2: molecular dipoles and methane.

A thioether linkage $\mathrm{R}-\mathrm{S}-R\prime$. sulfur lone pair participation can generate cyclic sulfonium intermediates during substitution.