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Acid-base chemistry studies how chemical acids and chemical bases exchange protons or electron pairs. Here the focus is transfer of protons in water: a chemical acid donates a proton, and a chemical base accepts it. The resulting chemical equilibrium determines pH, while a buffer solution can resist changes in pH. This connects microscopic chemical reactions with a compact, measurable description of a chemical solution.

The important distinction is between how much chemical acid is present and how readily it transfers a proton. A weak acid and a strong acid at the same total concentration need not have the same pH. Conversely, two buffer solutions can have the same pH while containing very different total quantities of their chemical acids and chemical bases.

A chemical acid can donate a proton, or accept an electron pair to form a covalent bond. The pH and buffer solution examples below use the former meaning. See IUPAC Gold Book, acid.

A chemical base can accept a proton, using an available electron pair to form a covalent bond; more generally it can donate an electron pair to form a covalent bond with another species. See IUPAC Gold Book, base.

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.

An electron domain is a region of electron density counted around a central chemical atom in the VSEPR model. One lone pair counts as one domain; a bond to one neighbour counts as one domain even when its bond order is two or three. Domain count is therefore different from simply counting electron pairs.

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.

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.

Light initiates methane chlorination by photodissociation of $\mathrm{C}{\mathrm{l}}_2$ through homolysis. The elementary bookkeeping isHere $h\nu$ denotes the photon energy. The first propagation step replaces a chlorine radical with a methyl radical; the second regenerates a chlorine radical. Adding the two propagation equations cancels both free radicals and gives $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2⟶\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$: chloromethane and hydrogen chloride are made while the chain carrier is regenerated.^[ref]

For example, recombination $2{\mathrm{C}{\mathrm{H}}_3}^{\bullet }⟶{\mathrm{C}}_2{\mathrm{H}}_6$ of methyl radicals makes ethane and a covalent bond from two unpaired Electrons, removing two carriers. Recombination of two ${\mathrm{Cl}}^{\bullet }$ also terminates carriers; in the gas phase, stabilization of the product requires disposal of released energy, for example to a third collision partner. The figure omits that partner. Termination competes with propagation, and further chlorination can consume chloromethane, so this cycle alone does not predict a pure product or a measured reaction rate.^[ref]

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 radical has one or more unpaired Electrons. In a simple atomic orbital or molecular orbital occupancy picture, an unpaired Electron occupies a spatial orbital without an opposite-spin partner in that orbital. A dot such as ${\mathrm{H}}^{\bullet }$ marks this electronic property; it does not mean a missing Electron or positive electric charge. Free radicals can be neutral or charged. Terminology also has a boundary: metal ions exhibiting paramagnetism are not normally called free radicals just because they have unpaired Electrons.^[ref]

Compare three species containing the same single proton. The hydrogen atom ${\mathrm{H}}^{\bullet }$ has one Electron in its $1s$ atomic orbital, and is a neutral radical. The hydrogen cation ${\mathrm{H}}^{+}$ has no Electrons, so no unpaired Electron. The hydrogen anion ${\mathrm{H}}^{-}$ has two Electrons paired in its $1s$ atomic orbital. It is charged but is not a radical in its ground state. With $N_p$ protons, $N_e$ Electrons, and positive elementary charge $e$, the independent electric charge count isThe electron pair and electric charge questions require different information. An odd $N_e$ ensures that not all Electrons can be paired; an even $N_e$ alone does not ensure pairing, as the molecular orbital description of molecular oxygen illustrates.

The distinction matters whenever a chemical reaction mechanism uses dots and charge signs together. Homolysis gives one bonding Electron to each fragment, while heterolysis assigns both to one fragment. A radical chain reaction can pass an unpaired Electron from one species to another without generating net electric charge. Detection then exploits the unpaired Electron's spin and magnetic moment, rather than merely measuring whether the sample carries an electric charge.

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.