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A bond pair is an electron pair allocated to a connection between two chemical atoms in a Lewis structure. A single line stands for one bond pair, or two electrons. A double connection contains two bond pairs, or four electrons; a triple connection contains three bond pairs, or six electrons. A bond pair is shared bookkeeping for the combined arrangement, rather than two distinguishable electrons each permanently owned by one endpoint.

In water, two bond pairs use $2\times 2=4$ of the eight valence electrons. In Ammonia, three bond pairs use $3\times 2=6$ of the eight valence electrons. In $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each C=O connection contains two bond pairs, so there are four bond pairs in the entire carbon dioxide molecule. Counting connections and counting bond pairs therefore give different answers for multiple connections.

The same bond pair is treated differently for different questions: both endpoints count its two electrons for the octet rule, whereas each endpoint receives just one electron for formal charge. Neither convention asserts equal physical electron density at the endpoints. For electronic shapes, occupancy, and energy, see molecular orbital theory rather than treating a bond pair line as an entire wavefunction.

A covalent bond joins chemical atoms through shared electrons. In the familiar two-centre description, one bond pair contributes to a connection between two atomic nuclei. Sharing means that the electrons belong to the combined arrangement and interact with both atomic nuclei; it does not mean that identifiable dots shuttle along drawn paths. The ^[ref] emphasizes the concentration of electrons between atomic nuclei and the resulting attraction and characteristic separation. The pair picture is especially useful for simple molecules, but is not a universal description of every covalent bond.

A Lewis structure turns this picture into practical bookkeeping. Count the valence electrons, indicate which chemical atoms are connected, and allocate two electrons to each line. A bond pair is shared; a lone pair is assigned to one chemical atom without connecting it to another. A double line represents two bond pairs. Dots and lines record allocations, rather than positions, trajectories, or a measured three-dimensional arrangement. ^[ref]

For molecular hydrogen, ${\mathrm{H}}_2$, each neutral hydrogen chemical atom supplies one valence electron. Thus $1+1=2$ valence electrons make one bond pair, represented by $\mathrm{H}-\mathrm{H}$, with no lone pairs. Each hydrogen counts the same shared pair when checking its two-electron arrangement. The molecule still contains two electrons, not four: counting a pair around both endpoints does not create extra electrons.

For water, ${\mathrm{H}}_2\mathrm{O}$, neutral oxygen supplies six valence electrons and the two hydrogen chemical atoms supply one each: $6+1+1=8$. Two O–H connections consume four electrons as two bond pairs. The remaining $8-4=4$ electrons form two lone pairs on oxygen. Around oxygen, the two bond pairs and two lone pairs count as eight electrons; each hydrogen counts two. The whole water molecule has ten electrons, because two additional oxygen electrons are omitted from this valence electron drawing.

For Ammonia, ${\mathrm{NH}}_3$, the total is $5+3\times 1=8$ valence electrons: five from nitrogen and three from hydrogen. Three N–H connections use six electrons as three bond pairs, leaving two electrons as one lone pair on nitrogen. Thus water and Ammonia have the same total of eight valence electrons but different allocations: two plus two pairs versus three plus one. This small calculation already distinguishes their connectivity without a detailed quantum mechanics calculation.

For molecular oxygen, ${\mathrm{O}}_2$, $6+6=12$ valence electrons give the conventional $\mathrm{O}=\mathrm{O}$ Lewis structure. Its two bond pairs use four electrons, leaving eight as four lone pairs, two on each oxygen. Each oxygen consequently counts eight electrons locally. However, actual molecular oxygen has two unpaired electrons and exhibits paramagnetism: it is attracted into a stronger magnetic field. The all-paired Lewis structure cannot capture this observation. Molecular orbital theory describes the unpaired electrons; the successful counting diagram alone is insufficient. ^[ref]

Why does sharing produce a stable separation? For molecular hydrogen, the electronic arrangement responds as the atomic nuclei approach. Their interactions can lower the energy, but pushing them too close raises it. A curve of potential energy against separation therefore has a local minimum, rather than falling forever. Attraction and repulsion balance there. This is a statement about the combined system, not a claim that stationary electrons sit between otherwise motionless atomic nuclei. ^[ref]

The following Morse potential makes those features calculable. Set $x=r/r_e$, where $r$ is separation and $r_e$ its equilibrium value, and let $D$ be the well depth. With the self-chosen parameter $a=2$, define the dimensionless quantitiesHere $F=-dV/dr$ is the net separating force, obtained from the derivative of the potential energy. Positive $f$ pushes the chemical atoms apart; negative $f$ draws them together. These are illustrative curves with $r_e=D=1$ in chosen units, not measured molecular hydrogen data or fitted parameters.

At $x=1$, $u=-1$ and $f=0$. The second derivative is $u^{\prime \prime }(1)=8>0$, confirming a local minimum. At $x=0.5$, $u=(1-e{)}^2-1\approx 1.9525$; at $x=1.5$, $u=(1-e^{-1}{)}^2-1\approx -0.6004$. As $x$ increases without bound, $u$ approaches zero from below. Thus moving from the well bottom to separated chemical atoms requires an increase of $D$ in this model's potential energy. An actual quantum mechanics dissociation threshold also depends on the initial motion of the atomic nuclei, so the well depth alone is not every measured dissociation energy.

The force panel separates $f$ into a positive term $4e^{-4(x-1)}$ and a negative term $-4e^{-2(x-1)}$. At $x=1$ they are $+4$ and $-4$, cancelling exactly; nearer separations favour the positive term. These are algebraic components of the chosen Morse potential, not separately measured microscopic forces. Their usefulness is to show how a balance arises from the slope of one energy curve.

The beauty of the pair picture is its economy: a few marks summarize connectivity and constrain what a proposed molecule can contain. Its historical anchor is Gilbert N. Lewis's 1916 paper ^[ref], also available as an ^[ref]. Lewis structures remain useful because they answer counting questions quickly; quantum mechanics supplies the richer account when energy, shape, or unpaired electrons matter.

The electron density describes the expected number of electrons per unit volume at position $\mathbf{r}$. For a system with a fixed number $N$ of electrons, $\int n(\mathbf{r})d^3\mathbf{r}=N$. A Lewis structure allocates electron counts without specifying this spatial density.

Electronegativity describes a chemical atom's tendency to attract electron density towards itself in a chemical bond. It helps answer a useful question: when two different chemical atoms share electrons, which end of the chemical bond tends to be richer in electron density? For hydrogen chloride, the answer is chlorine: the covalent bond is a polar covalent bond, conventionally written ${\mathrm{H}}^{\delta +}-{\mathrm{Cl}}^{\delta -}$. The symbols indicate partial atomic charges, not isolated ions carrying whole elementary electric charges.^[ref]

Electronegativity is a comparison scheme, not the number of electrons on a chemical atom. A partial atomic charge depends on how the electron density is partitioned; formal charge instead follows an integer allocation in a Lewis structure. Neither is numerically equal to electronegativity. Nor is electronegativity identical to electron affinity, which concerns adding an electron to an isolated species. Different electronegativity scales use different measured inputs and conventions; compare numbers only after specifying the scale.^[ref]

For a concrete comparison, the Royal Society of Chemistry lists Pauling electronegativities ${\chi }_{\mathrm{P}}(\mathrm{H})=2.20$ and ${\chi }_{\mathrm{P}}(\mathrm{Cl})=3.16$. Their difference is $0.96$, so the electronegativity rule predicts bond polarization towards chlorine in hydrogen chloride. An isolated molecular hydrogen molecule has identical ends and zero electronegativity difference. These examples explain a direction of unequal sharing, not a quantitative partial atomic charge.^[ref][ref]

A larger electronegativity difference often suggests stronger bond polarization, but there is no universal numerical boundary where a covalent bond suddenly becomes ionic bonding. Tables of cutoffs are classification aids with exceptions. Hydrogen fluoride, for example, is described by a polar covalent bond even when its electronegativity difference exceeds a textbook's nominal ionic threshold. The actual electron density, physical setting, and chemical substance's structure matter.^[ref]

The distinction from ionization energy is equally useful: an ionization energy measures the cost of removing an electron from a specified isolated species, whereas electronegativity compares attraction for electron density in a bonded environment. Neither the removal cost alone nor the attachment release alone supplies a unique bonded-atom electronegativity. A scale combines selected measured inputs with a model and a convention.^[ref]

For an original relative comparison, stipulate fictitious bonded sites $A,B,C$ on one chosen dimensionless electronegativity scale:The electronegativity heuristic predicts bond polarization towards $B$ in $A-B$, towards $C$ in $B-C$, and towards $C$ in $A-C$. Shifting ${\chi }_0$ leaves every comparison unchanged. Although the $A-C$ difference is twice the $A-B$ difference, this does not make its partial atomic charge or bond dipole twice as large: those require additional information about electron density and bond length. These symbols describe a stipulated ordering, not a table of measured chemical elements.

Keep the chemical context attached to that ordering. A chemical atom in a molecule has neighbours and an electron configuration; an isolated-atom energy scale and a bonded-atom comparison are different constructions. The chemical environment dependence of electronegativity explains why a table is a useful approximation rather than an immutable property of every site. Furthermore, an applied electric field can induce bond polarization even between identical chemical atoms, through electric polarizability; equal tabulated electronegativity does not forbid that response.

An electron pair consists of two electrons with opposite spins sharing the same spatial part of their quantum state. The Lewis structure distinction between a bond pair and a lone pair records whether that pair is assigned to a connection between chemical atoms or to one chemical atom.

The formal charge of a chemical atom in a Lewis structure assigns it all of its nonbonding electrons and half of the electrons in its connections:Here $v$ is the neutral chemical atom's valence electron count, $n$ is the number of nonbonding electrons assigned to it, and $b$ is the number of bonding electrons incident on it. Count electrons, not electron pairs: two lone pairs mean $n=4$, and two double connections mean $b=8$. The number $\mathrm{FC}$ gives electric charge in units of the positive elementary charge. Equal splitting is a bookkeeping convention, regardless of the actual electron density. See the IUPAC Gold Book, formal charge.

For $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each oxygen has $v=6$, $n=4$, $b=4$, giving formal charge $6-4-4/2=0$. Carbon has $v=4$, $n=0$, $b=8$, giving formal charge $4-0-8/2=0$. For ${\mathrm{O}}^{-}-\mathrm{C}\equiv {\mathrm{O}}^{+}$, the left oxygen has formal charge $6-6-2/2=-1$, carbon still has formal charge zero, and the right oxygen has formal charge $6-2-6/2=+1$. The net electric charge is zero in either drawing.

The Lewis structure with no separated formal charges is the dominant contributor in this elementary comparison; the charge-separated drawing is a less important contributor, rather than an equally dominant alternative. This is a qualitative selection rule, not a calculation of energy or numerical chemical resonance weights. See Flowers et al., Chemistry 2e, 7.4: comparing carbon dioxide structures. A zero formal charge does not imply zero partial atomic charge: the latter depends on partitioning the actual electron density, which a Lewis structure does not supply.

As a short consistency derivation, summing $n+b/2$ over all chemical atoms counts every drawn electron once: a nonbonding electron is assigned once, and a bonding electron contributes half at each of its two endpoints. Thuswhere $N_{\mathrm{valence}}$ is the total drawn valence electron count. The right side is the net electric charge in units of elementary charge. This catches a wrong budget even when individual octet rule checks look plausible.

A hydrogen bond acceptor supplies the electron-rich atom or group $Y$ interacting with the attached hydrogen of a hydrogen bond donor in $X-\mathrm{H}\cdots Y$. An oxygen lone pair in water is the familiar example. Acceptance describes the association, not necessarily acquisition of a proton. A single molecule, such as water or Ammonia, can offer both hydrogen bond donor and hydrogen bond acceptor sites. See Glasfeld: water and nitrogen-containing accepting sites.

A lone pair drawn in a Lewis structure is a useful clue, not a universal test. At an ordinary neutral amide nitrogen, the nominal lone pair participates in chemical resonance with the carbonyl group, so this nitrogen is ordinarily excluded from lists of hydrogen bond acceptors; the carbonyl group's oxygen remains an accepting site. Protonation changes the test again: ${\mathrm{NH}}_4^{+}$ has four N–H covalent bonds and no nitrogen lone pair, whereas Ammonia has three and one. These distinctions follow the actual chemical form, and accepting ability also depends on the partner and whether it can approach the site. See McMurry: amide resonance and protonation and Glasfeld: limitations of nitrogen accepting sites.

Conversely, an accepting group need not supply an atom-centred lone pair. The following phenol–benzene experiment probes an O–H donor interacting with the ring's distributed $\pi$ electron density.

Saggu, Levinson and Boxer (2011) used infrared spectroscopy to compare $200\mathrm{mmol}{\mathrm{L}}^{-1}$ phenol in different solvents. The O–H stretching peak was at $3611\mathrm{c}{\mathrm{m}}^{-1}$ in ${\mathrm{CCl}}_4$ and $3555.9\mathrm{c}{\mathrm{m}}^{-1}$ in benzene. To distinguish association from a general solvent effect, they diluted the ring-containing chemical solutions with ${\mathrm{CCl}}_4$: two peaks appeared, with the free peak nearly fixed while the associated peak depended on the partner. Replacing the O–H hydrogen with deuterium reproduced the ordering of shifts for different partners. Thus the perturbed O–H group identifies the hydrogen bond donor; the ring supplies the hydrogen bond acceptor. A peak shift alone does not specify a unique molecular geometry. ^[ref]

The roles make recognition between molecules intelligible: exposed hydrogen bond donors need compatible hydrogen bond acceptors in reachable positions. A count of sites is a starting description, not a prediction of binding strength. In water, forming a contact between two partners can also replace their existing hydrogen bonds to water; the Gibbs free energy of association concerns the whole change. See Glasfeld, pp. 15–16: competition with solvent interactions.

For the routine examples here, assign all bonding electrons between different chemical elements to the more electronegative partner; divide bonding electrons between identical chemical elements equally. Retain each lone pair on its original chemical atom. Subtract the assigned valence electron count $N_i^{\mathrm{assigned}}$ from the neutral chemical atom's valence electron count $v_i$. The IUPAC convention uses Allen electronegativity, based on free-chemical atom valence electron energies. Close electronegativity comparisons and some reversibly bound chemical compounds require care beyond this simple rule.^[ref]

The examples explain exceptions to memorized rules:

These drawings record allocations, not molecular geometry or a physical process of breaking the chemical bonds. In particular, the O–O line in the molecular oxygen panel only marks the connection; it does not assert a complete Lewis structure or spin state.

Before calculating electronic energy, ask a simpler question: where can the available valence electrons go? A Lewis structure makes that budget visible, and nitrate shows why a correct budget can still require several resonance structures.

A Lewis structure records which chemical atoms are connected in a molecule or ion and how its valence electrons are allocated. One line represents one bond pair, two lines two bond pairs, and three lines three bond pairs; two dots beside a chemical atom represent one lone pair. The marks are a counting language, rather than measured electron positions. See the IUPAC Gold Book, Lewis formula.

For water, ${\mathrm{H}}_2\mathrm{O}$, the budget is $6+2\times 1=8$ valence electrons. Two O–H lines use four electrons; the other four form two lone pairs on oxygen. For Ammonia, ${\mathrm{NH}}_3$, $5+3\times 1=8$ valence electrons give three N–H lines and one lone pair on nitrogen. Both molecules contain four electron pairs, but their allocations differ. Each shared bond pair is counted once in the total budget, even though both endpoints count it when checking the octet rule.

A practical construction method is:

For carbon dioxide, $4+2\times 6=16$ valence electrons. Starting with two single C–O connections leaves twelve electrons, enough for three lone pairs on each oxygen, but only four electrons around carbon. Converting one lone pair from each oxygen into a second bond pair gives $\mathrm{O}=\mathrm{C}=\mathrm{O}$. Eight electrons are in the four bond pairs, and eight are in four lone pairs, two on each oxygen. Carbon has no lone pair; all three chemical atoms satisfy the octet rule, and all three formal charges vanish.

For the nitrate ion, ${\mathrm{NO}}_3^{-}$, the valence electron budget isNitrogen contributes five, the three oxygen chemical atoms contribute eighteen, and the negative electric charge adds one electron. Start with nitrogen connected to each oxygen by one bond pair: those three lines use six electrons. Put the remaining eighteen electrons into three lone pairs on each oxygen. Each oxygen now satisfies the octet rule, but nitrogen counts only six electrons. Reassign one oxygen lone pair as a second bond pair on its N–O connection. This changes the allocation without adding an electron; it applies the construction procedure in Flowers et al., Chemistry 2e, 7.3: drawing Lewis structures.

The resulting nitrate Lewis structure has one N=O connection and two N–O connections. Its four bond pairs use eight electrons, and its eight lone pairs use sixteen: $8+16=24$. The doubly connected oxygen has two lone pairs; each singly connected oxygen has three; nitrogen has none. All four chemical atoms satisfy the octet rule. Any of the three oxygen chemical atoms can supply the additional bond pair, giving the three equivalent resonance structures below.

Why not draw three N=O connections to make the ion look symmetric? With two lone pairs on each oxygen, that drawing still spends $12+12=24$ electrons and has the right total formal charge, but places twelve electrons around nitrogen. It fails the octet rule for this second-period chemical element. Three equivalent resonance structures express the symmetry without that invalid local count.

The economy is beautiful: a handful of dots and lines rules out many impossible counts and identifies lone pairs that can participate in acid-base chemistry. A Lewis structure helps choose rational candidate connections before a detailed quantum chemistry calculation. It does not specify a complete many-electron wavefunction, actual electron density, or molecular geometry. Adding the separate VSEPR model, described in Chemistry 2e, 7.6: molecular structure and polarity, predicts bent water and pyramidal Ammonia; the layouts above alone make no such prediction. Molecular orbital theory supplies a richer electronic description, including the unpaired electrons of molecular oxygen that an all-paired Lewis structure misses.

In 1916 Gilbert N. Lewis proposed two-dot notation for the shared electrons of a chemical bond and fuller dot formulae for the outer electrons, explicitly relating a shared pair to the existing single-line notation. See Lewis, The Atom and the Molecule, JACS 38, 762–785 (1916), pp. 777–778, available as an ^[ref]; the relevant passage is also readable in a ^[ref]. The modern quantum mechanics interpretation should be distinguished from his historical model of fixed outer electrons.

A lone pair is an electron pair assigned to one chemical atom without a connecting line to another chemical atom in a Lewis structure. It is usually shown as two dots. Water has two lone pairs on oxygen, accounting for four valence electrons; Ammonia has one lone pair on nitrogen, accounting for two. In $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each oxygen has two lone pairs, and carbon has none.

A lone pair can be reactive. In protonation of Ammonia, the nitrogen lone pair becomes a new bond pair to a proton:The proton supplies no electron, so the eight valence electrons are retained: three bond pairs plus one lone pair become four bond pairs. The resulting nitrogen has formal charge $5-0-8/2=+1$; each hydrogen has formal charge zero. This is why identifying an available lone pair helps explain the chemical base behaviour of Ammonia, rather than merely decorating its formula.

Lewis already used the Ammonia to ${\mathrm{NH}}_4^{+}$ example in The Atom and the Molecule (1916), pp. 777–778. In modern terms the dots identify an allocation of electrons, not two stationary particles or an isolated patch unaffected by the rest of the molecule. A molecular orbital description can spread electronic amplitude over several chemical atoms while a localized Lewis structure remains useful for acid-base chemistry.

Why do methane, Ammonia, and water have different shapes when each central chemical atom is surrounded by eight valence electrons? The answer separates the positions of atomic nuclei from the electron domains used in a VSEPR model.

Molecular geometry describes the three-dimensional arrangement of the atomic nuclei in a molecule, including bond lengths and bond angles. For a larger molecule, a shape name often describes just the neighbours of one central chemical atom. A Lewis structure records connections and electron bookkeeping; its flat layout does not fix molecular geometry.

In the VSEPR model, electron pair geometry includes both bonding electron domains and lone pairs, whereas molecular geometry counts only the chemical atoms. In methane, four bond pairs and no lone pairs give tetrahedral molecular geometry. In Ammonia, three bond pairs and one lone pair give trigonal pyramidal molecular geometry. In water, two bond pairs and two lone pairs give bent molecular geometry. All three use the same ideal tetrahedral electron pair geometry; this does not make their actual bond angles equal. See Flowers, Theopold, Langley and Robinson, Chemistry 2e, §7.6: electron-pair and molecular geometries.

A static drawing also needs a physical interpretation. The atomic nuclei undergo oscillations; an arrangement at a minimum of potential energy and an experimentally averaged arrangement need not have identical bond lengths or bond angles. This distinction matters when comparing measurements with quantum mechanics calculations; see the IUPAC Gold Book: experimental bond-length conventions. Gas electron diffraction gives a concrete route from a laboratory signal to molecular geometry rather than inferring shape from a drawing alone.

Molecular orbital theory describes the electronic arrangement of a molecule using molecular orbitals: one-electron spatial wavefunctions extending over the combined atomic nuclei. Filling bonding molecular orbitals can stabilize a chemical bond; filling antibonding molecular orbitals opposes that stabilization. A molecular orbital is a building block of a many-electron description, rather than the entire many-electron wavefunction. See Robert W. Field, MIT 5.61, lecture 24, pp. 1–4: molecular orbitals and the LCAO approximation.

First make the interference visible with an exactly normalized one-dimensional toy. Replace the three-dimensional $1s$ atomic orbitals of the hydrogen atom by two Gaussian functions centred at $x=\pm a$, with dimensionless $x$ and $a>0$:The normalized sum and difference are ${\psi }_{\pm }=({\varphi }_A\pm {\varphi }_B)/\sqrt{2(1\pm S)}$, following the same linear combination of atomic orbitals algebra. At $a=1$, the orbital overlap is $S=e^{-1}\approx 0.367879$. At the midpoint, ${\varphi }_A(0)={\varphi }_B(0)\approx 0.455581$, givingBoth probability densities have total integral one: interference redistributes probability, rather than creating an extra electron. These are toy wavefunctions, not actual hydrogen atom orbitals or predictions of molecular hydrogen energies.

Now fill the two $1s$-derived molecular orbitals in the minimal model. The lower bonding molecular orbital is labelled ${\sigma }_g$; the upper antibonding molecular orbital is ${\sigma }_u^{\ast }$. The Pauli exclusion principle permits at most two electrons in each spatial molecular orbital, with opposite spin projections along a chosen axis. For the lowest electron configuration in this model, fill ${\sigma }_g$ first. The subscripts $g$ and $u$ denote even and odd behaviour under inversion through the midpoint; the star marks an antibonding molecular orbital. See Flowers et al., Chemistry 2e, 8.4: occupancy and bond order and Field, MIT 5.61, lecture 25, pp. 1–3: bonding, antibonding, and symmetry labels.

With $N_b$ electrons in bonding molecular orbitals and $N_a$ in antibonding molecular orbitals, the model's bond order is $b=(N_b-N_a)/2$. The three elementary counts are:

The zero bond order says this minimal model gives no net covalent bond for ${\mathrm{He}}_2$. It does not rule out every bound pair of helium chemical atoms: a very weak Van der Waals force can bind a neutral helium dimer, which has been observed experimentally. See Zeller et al., Imaging the He2 quantum halo state using a free electron laser, PNAS 113, 14651–14655 (2016), p. 14651: van der Waals binding and Luo et al., The weakest bond: Experimental observation of helium dimer, Journal of Chemical Physics 98, 3564–3567 (1993). The bond order count does not include that weak binding mechanism.

The useful surprise is that sharing alone does not determine a chemical bond: the relative quantum phase and the occupancy matter together. The same two shapes explain why adding an electron to a bonding molecular orbital increases the model's bond order, while adding one to an antibonding molecular orbital decreases it. Molecular orbital theory also accounts for the unpaired electrons and paramagnetism of molecular oxygen, which its all-paired Lewis structure misses. See Chemistry 2e, 8.4: the molecular orbital description of oxygen.

The nitrate ion, ${\mathrm{NO}}_3^{-}$, is an anion with one nitrogen connected to three oxygen chemical atoms and net electric charge $-1$ in units of elementary charge. Its 24 valence electrons admit three equivalent octet resonance structures; see Lewis structure for their construction and chemical resonance for their interpretation. The N–O connections are covalent bonds within the ion. See 7.2 Covalent Bonding.

The octet rule is a useful empirical guide: many Lewis structures of chemical compounds containing carbon, nitrogen, and oxygen give each of those chemical atoms eight electrons when nonbonding electrons and all incident bonding electrons are counted. Hydrogen instead normally counts two electrons. These are local counts, not independent budgets: a shared bond pair counts at both endpoints but only once in the molecule's valence electron total.

In water, oxygen counts $4+4=8$ electrons; in Ammonia, nitrogen counts $2+6=8$; in $\mathrm{O}=\mathrm{C}=\mathrm{O}$, carbon counts $0+8=8$ and each oxygen counts $4+4=8$. The charge-separated carbon dioxide drawing also satisfies the octet rule, so the octet rule alone cannot select the dominant contributor. Formal charge provides another check.

The octet rule is a heuristic rather than a theorem about every chemical bond. Odd valence electron totals cannot be partitioned entirely into electron pairs, and some valid Lewis structures have fewer or more than eight electrons around a chemical atom. See Chemistry 2e, 7.3: exceptions to the octet rule. The capacity of the $s$ and $p$ electron subshells, $2+6=8$, helps motivate the guide for these chemical elements, but does not prove universal compliance. Lewis discussed a stable outer group of eight, together with a two-electron arrangement for hydrogen, in The Atom and the Molecule (1916), pp. 768 and 774.

An oxidation state is an electron-bookkeeping assignment to a chemical atom: its electric charge number after the ionic approximation of bonds. It helps identify oxidation and reduction and determine stoichiometry, even for covalent bonds without separate charged ions. In water, each hydrogen has oxidation state $+1$ and oxygen has $-2$; this does not assert that a water molecule contains two free protons and a free oxide ion.^[ref]

Three conventions answer different questions. Formal charge divides each bonding electron pair equally in a chosen Lewis structure. Oxidation state uses the ionic approximation of bonds. Partial atomic charge instead partitions the actual electron density and depends on that partition. In the ordinary water Lewis structure, every formal charge is zero although the oxidation states are not. Neither bookkeeping convention supplies numerical partial atomic charges.

A partial atomic charge assigns an electric charge to a chemical atom by partitioning the actual electron density. The result depends on the partitioning convention and need not be an integer. It differs from formal charge, which divides bonding electrons equally in a specified Lewis structure.

Write $t=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3{)}_3$. D is the sodium enolate $\mathrm{t}-\mathrm{C}({\mathrm{O}}^{-})=\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{a}}^{+}$. Its two resonances in the alkene-proton region belong to the two hydrogens of the terminal methylene group. One is cis to oxygen and the other trans to oxygen, so they are not related by a symmetry operation even though both belong to the same carbon. Replacing one or the other gives different alkene geometries, so they are diastereotopic. Their weak or unresolved geminal coupling is consistent with the reported singlet descriptions. Without an additional spatial assignment, it is not justified to decide uniquely which numerical shift belongs to which of those two orientations. The nine-proton singlet is the tert-butyl group. The formula ${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{ONa}$ follows from removing one alpha hydrogen from pinacolone.

E is the enol-dominated 1,3-diketone $\mathrm{tCO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COt}$, observed as $\mathrm{tC}(\mathrm{OH})=\mathrm{CH}-\mathrm{COt}$. The very downfield broad signal belongs to the enol OH in a six-membered intramolecular O–H···O chelate. The one-proton signal in the vinyl region is its central enol CH; a predominantly diketo structure would instead require a two-proton central CH2 signal and no enol OH. The eighteen-proton singlet accounts for both tert-butyl groups. Rapid exchange between the two degenerate enol orientations can average their different local environments; accidental unresolved equality cannot be excluded by the supplied spectrum alone. Thus one should not infer a permanently symmetric single enol Lewis structure from that singlet. The total integration is $1+1+18=20$, agreeing with ${\mathrm{C}}_{11}{\mathrm{H}}_{20}{\mathrm{O}}_2$.

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.

The ions are acetylide ${\mathrm{C}}_2^{2-}$ and linear tricarbide ${\mathrm{C}}_3^{4-}$. Charge balance gives $\mathrm{M}{\mathrm{g}}^{2+}[{\mathrm{C}}_2{]}^{2-}$ and $(\mathrm{M}{\mathrm{g}}^{2+}{)}_2[{\mathrm{C}}_3{]}^{4-}$.

The acetylide has ten valence electrons, two more than neutral ${\mathrm{C}}_2$. They fill the p-derived sigma bonding orbital, giving the formal configuration ${\pi }_u^4{\sigma }_g(2p{)}^2$ above the cancelling $2s$ pair. Its bond order is three, with the Lewis structure ${}^{-}\mathrm{C}\equiv {\mathrm{C}}^{-}$ and a lone pair at each end; each carbon is approximately $sp$ hybridized.

The tricarbide has $3(4)+4=16$ valence electrons and is isoelectronic with carbon dioxide. The cumulene contributor ${}^{2-}\mathrm{C}=\mathrm{C}={\mathrm{C}}^{2-}$ has a central $sp$ carbon, two sigma bonds and two perpendicular pi bonds. The end carbons each have two lone pairs and can be described with an $s{p}^2$ localized basis. Its two C–C bonds have formal order two in this model; the central C–C–C angle is $180^{\circ }$. Other resonance contributors help describe its protonation chemistry.

There are $\boxed{4+2(1)+2(5)=16}$ valence electrons, all paired, so diazomethane exhibits diamagnetism. A localized orbital occupancy consistent with the double-bond construction in 2(c)(i) isHere $c,t$ mean central and terminal. Each listed occupied spatial orbital contains two electrons of opposite spin. The central N's two p atomic orbitals are perpendicular, so they can support the two perpendicular pi bonds. No central-N lone pair is left over: its positive formal charge accompanies four bonding pairs in this Lewis structure.

Allowing resonance mixes a terminal-N lone pair into the C–N–N $\pi$ orbitals. In the alternative terminal-$sp$ basis, the same electron count is four occupied $\sigma$ orbitals ($8$ electrons), one terminal-$sp$ lone pair ($2$), one perpendicular N–N $\pi$ orbital ($2$), and two occupied orbitals of a three-centre C–N–N $\pi$ system ($4$). The $16$-electron count and the paired ground state survive this change of description; the localized picture does not assert two perfectly isolated double bonds.