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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.

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.

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 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.