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A scale proportional to the average spectroscopic one-electron binding energy of a free ground-state atom’s valence electrons. For an s,p atom, $\chi =C(n_s{\epsilon }_s+n_p{\epsilon }_p)/(n_s+n_p)$, with positive binding energies and a scale constant $C$. It distinguishes s and p binding rather than assigning both one Slater charge.^[ref]

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.

$\mathrm{S}{\mathrm{n}}_5^{2-}$, a five-vertex trigonal-bipyramidal Zintl cluster. Removing five mainly s-rich lone pairs from its 22 valence electrons leaves 12 skeletal electrons.

An electron in an inner electron shell, contrasted with a valence electron.

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 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 periodic group is a vertical column of the periodic table. For example, fluorine and chlorine are in periodic group 17, with seven valence electrons in each neutral ground state chemical atom. A shared outer electron configuration supports similarities in chemical reactions, rather than equality of every property.

In a one-carrier isotropic Drude model, $R_H=1/(nq)$ and an inferred carrier density is $n_H=1/(e|R_H|)$. Dividing by the atomic number density gives an effective Hall count per atom. A noninteger result need not mean a literal fractional count of valence electrons: multiple energy bands, anisotropic Fermi surfaces and different charge carrier mobilities can change the Hall coefficient. In a two-carrier isotropic approximation,Its sign identifies the dominant Hall response rather than necessarily the most numerous charge carrier.

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.

The neutral molecule Li-two. In the elementary valence molecular orbital description, its two valence electrons occupy the bonding sigma-two-s orbital, giving bond order one and a closed-shell ground state.

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.

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.

Sum oxidation states over all chemical atoms of the specified molecule, ion or formula unit, including multiplicities:Here $q$ is the net electric charge and $e$ the positive elementary charge. For a neutral entity the sum is zero; for an ion it equals its signed electric charge number $z$. The reason is simple: the allocation assigns every valence electron exactly once, so the sum is the neutral electron budget minus the actual electron count.

For water, $2(+1)+(-2)=0$; for hydrogen peroxide, $2(+1)+2(-1)=0$. In permanganate, ${\mathrm{MnO}}_4^{-}$, assigning each oxygen $-2$ gives $x+4(-2)=-1$, so manganese has oxidation state $+7$. This is not a free ${\mathrm{Mn}}^{7+}$ ion: the entire permanganate anion has net electric charge $-e$. The sum rule supplies a constraint; without additional assignments it cannot determine every individual oxidation state.

In ionic counting, two ${\eta }^5$ cyclopentadienyl anions donate $12$ electrons. Therefore ferrocene, cobaltocene and nickelocene contain 18, 19 and 20 valence electrons, respectively: Fe(II) is $d^6$, Co(II) $d^7$ and Ni(II) $d^8$. ferrocene fills the bonding and approximately nonbonding levels while leaving the next metal–ring antibonding pair empty. cobaltocene and nickelocene add one and two electrons to that antibonding set.

Increasing antibonding occupation lowers the net metal–ring bond order. It therefore explains the simultaneous increase in M–C distance and decrease in dissociation enthalpy from Fe to Co to Ni. The trend does not follow the shrinking size of the isolated metal ions; the orbital occupation is decisive here. The specified enthalpy separates the entire sandwich into a gaseous metal dication and two gaseous ring anions. It measures that complete ionic dissociation process, not ten independent localized M–C bond energies, so dividing it by ten is not a justified bond-energy determination.

BeLi has three valence electrons: two occupy a bonding sigma orbital and the third its higher partner. The elementary bond order is therefore $1/2$, with one unpaired electron: expect a relatively weak, paramagnetic, doublet diatomic molecule. Because Be attracts valence charge more strongly than Li, the bond is polar toward Be. This is a qualitative prediction, not a numerical dissociation energy.

This is the elementary picture of the lithium beryllium diatomic molecule.

The four sigma bonds are the two C–H connections, C–N and N–N, containing eight electrons. A terminal-nitrogen axial lone pair contains two more. Of the remaining six electrons, two occupy an N–N pi bond lying in the molecular plane, while four occupy a perpendicular three-centre four-electron pi bond extending over C–N–N. There are therefore 16 valence electrons in total.

For orientation, a three-site model with equal on-site energy $\alpha$ and adjacent interaction $\beta <0$ gives perpendicular pi levels at $\alpha +\sqrt{2}\beta$, $\alpha$, and $\alpha -\sqrt{2}\beta$. The lower two are doubly occupied and the upper is empty. The middle level has a terminal-end combination with little central-atom amplitude; unequal C and N orbital energies change its coefficients and make it only approximately nonbonding.

The highest occupied molecular orbital is the upper occupied perpendicular pi orbital, with substantial carbon amplitude, broadly above the strongly bonding sigma framework and the lower occupied pi orbitals. It is not the terminal axial nitrogen lone pair. Real diazomethane has C–N bonding and N–N antibonding character in this frontier orbital, rather than being an exactly nonbonding equal-site model. Photoelectron and Rydberg-state experiments support this assignment.^[ref]

Li${}_3$ can retain a bound triangular framework, but an equilateral triangle with three valence electrons is Jahn-Teller unstable. The six core electrons are effectively spectators. In the simplest triangular valence model, two electrons occupy the fully bonding level and one occupies the degenerate upper pair. Making one edge different splits that pair; the single electron occupies the lower component, giving an electronic stabilization that can outweigh the cost of a small distortion. An isosceles triangle is sufficient to lift the degeneracy, so three different edge lengths are not required.

The two 2.8-Å contacts are close to, but slightly longer than, the 2.7-Å dimer bond, while the 3.5-Å edge is much weaker. Delocalization and partial antibonding occupation explain why these cannot be treated as three independent full-strength Li–Li single bonds. lithium's nearby valence $2p$ functions provide additional polarization and mixing channels absent from a hydrogen $1s$-only description, allowing a more flexible multicentre electronic structure. The existence of a bound minimum, unlike ground-state H${}_3$, is an energetic fact rather than a consequence of three-electron counting alone. Calculated Li${}_3$ surfaces show bound states, a conical intersection and Jahn-Teller coupling.^[ref]