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

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

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

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

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

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

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

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

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

In trigonal pyramidal molecular geometry, three neighbouring chemical atoms form a triangular base and the central chemical atom lies outside its plane. The pyramid has four atomic nuclei; its central chemical atom is the apex, not the centre of the base. Ammonia, ${\mathrm{NH}}_3$, has this arrangement. The ^[ref] lists an H–N–H bond angle of $106.67^{\circ }$, commonly rounded to $106.7^{\circ }$ or about $107^{\circ }$.

The Lewis structure of ammonia allocates eight valence electrons as three bond pairs and one lone pair. Four electron domains therefore suggest tetrahedral electron pair geometry; removing the lone pair marker leaves three hydrogen chemical atoms below nitrogen. The VSEPR model treats the lone pair as more space-demanding and predicts compression relative to $109.4712^{\circ }$, but it does not calculate the precise $106.67^{\circ }$ value. See Chemistry 2e, §7.6: lone-pair effects.

Three equal bond-dipole contributions in this pyramid cancel sideways, but retain a component along its axis. By contrast, three equal contributions lying in one plane at $120^{\circ }$ would cancel completely. This shows how a small change in molecular geometry can permit a permanent electric dipole moment. Ammonia is a polar molecule; NIST lists a measured magnitude of about $1.476$ debye in the same ^[ref]. The vector argument predicts allowed direction and cancellation, not the full measured magnitude.

A valence electron belongs to the electronic arrangement used to describe a chemical atom's chemical bonds and ordinary chemical reactions. For the neutral hydrogen, carbon, nitrogen, and oxygen chemical atoms used here, the valence electrons occupy the outermost occupied electron shell. Their respective electron configurations are $1s^1$, $1s^{2}2s^{2}2p^2$, $1s^{2}2s^{2}2p^3$, and $1s^{2}2s^{2}2p^4$, giving one, four, five, and six valence electrons. See Chemistry 2e, 6.4: electron configurations and valence electrons.

Lewis structures show these valence electrons while omitting the inner Electrons. Consequently, eight drawn valence electrons in water do not mean eight total Electrons: neutral water has ten, with two inner oxygen Electrons omitted. Neutral ammonia likewise has ten total Electrons but eight valence electrons; neutral carbon dioxide has twenty-two total Electrons but sixteen valence electrons.

This distinction makes the small counts useful: connections and possible lone pairs can be organized without representing every Electron or solving the entire many-Electron wavefunction. The outermost-electron shell shortcut applies to the examples above, rather than serving as a universal definition for all chemical elements; the electronic arrangement relevant to chemical bonds can also involve other electron subshells.

The VSEPR model predicts approximate molecular geometry by arranging the electron domains around a central chemical atom to reduce their mutual repulsion. Start with a Lewis structure, count the domains, and then distinguish the resulting electron pair geometry from the positions of the chemical atoms themselves. A lone pair counts as a domain even though it is not another chemical atom.^[ref]

Three worked counts have the same total of eight valence electrons. In methane, carbon's four plus the contributions of four hydrogen chemical atoms give $4+4=8$: four bond pairs, no lone pairs, and tetrahedral molecular geometry. In ammonia, nitrogen's five plus three from hydrogen give $5+3=8$: three bond pairs and one lone pair, producing trigonal pyramidal molecular geometry. In water, oxygen's six plus two from hydrogen give $6+2=8$: two bond pairs and two lone pairs, producing bent molecular geometry. All three have four electron domains; their molecular geometries differ.

The regular tetrahedral directions can be represented by the unit vectorsEach pair has dot product $-1/3$, so the ideal bond angle is $\theta =\arccos (-1/3)\approx 109.47^{\circ }$. This is a geometric calculation, rather than a prediction that every four-domain molecule has that exact bond angle. Lone pairs and different kinds of chemical bonds distort the ideal arrangement; water's H–O–H angle is about $104.5^{\circ }$.^[ref]

The model is attractive because small-number bookkeeping produces a spatial prediction without first solving for every molecular orbital. Shape then helps explain chemical polarity and which hydrogen bond donor or hydrogen bond acceptor sites are accessible. It is a qualitative model, not a full calculation of the electronic energy or a universal solution for every chemical compound. Gillespie and Nyholm's 1957 review ^[ref] is a historical anchor for its development.