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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.

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.

The periodic table arranges chemical elements in increasing atomic number, across successive periodic table periods. Its columns are periodic groups, numbered 1–18. The layout places recurring outer electron configurations above one another, making similarities in chemical bonds and chemical reactions easier to recognize. Ordering is by proton count, not by mass; the mass number of a particular isotope does not determine its box. See the official IUPAC periodic table and numbering conventions.

For the first eighteen neutral chemical atoms in their ground states, a compact description counts Electrons in successive electron shells. Thus carbon has $(2,4)$, meaning two Electrons in the first electron shell and four in the second; the sum $2+4=6$ agrees with its atomic number. Sodium has $(2,8,1)$, whose sum is $11$. These are occupancy counts, not circular trajectories of Electrons.

Reading each row from left to right gives:

The counting outline is short: the occupied electron subshells in this range are $1s$, $2s$, $2p$, $3s$, and $3p$, filled in that order for these neutral ground states. Each $s$ electron subshell holds up to two Electrons and each $p$ electron subshell up to six, giving $2$, $2+6=8$, and $2+6=8$ along these three rows. This $2,8,8$ summary applies only through argon; it is not a general rule that each electron shell fills completely before the next begins, nor a claim that the third electron shell can hold only eight Electrons. See Flowers et al., Chemistry 2e, electronic structure and electron configurations.

The recurring outer counts make the arrangement useful. Lithium and sodium each have one valence electron, whereas fluorine and chlorine each have seven valence electrons. Such similarities help anticipate chemical reactions, without making all properties identical throughout a periodic group. Helium is an instructive exception to an eight-Electron shortcut: its two Electrons complete the first electron shell, so it belongs above neon and argon in periodic group 18. Hydrogen shares column 1 with lithium and sodium but has its own distinct chemistry. See Flowers et al., Chemistry 2e, periodic-table organization.

Consider the reacting zinc and copper species in the zinc-copper displacement experiment. Split the net chemical equation into two half-reactions:

The zinc oxidation state rises from $0$ to $+2$; the copper oxidation state falls from $+2$ to $0$. Add the half-reactions: two released Electrons match two accepted Electrons. Cancelling $2e^{-}$ from both sides gives^[ref]Thus the closed net chemical reaction has zero net production of free Electrons. Cancellation does not destroy Electrons: it removes a matched internal exchange from the written chemical equation. The half-reactions alone are incomplete descriptions of the whole reacting system. In a galvanic cell, the exchange can pass through a wire; in the zinc-copper displacement experiment, the reacting materials meet at the same metal surface. The equations do not require a reservoir of free Electrons in the bulk chemical solution.

Check both kinds of balance. The oxidation line has net electric charge $0$ on the left and $(+2-2)e=0$ on the right; the reduction line has $(+2-2)e=0$ on the left and $0$ on the right. The net chemical equation contains one zinc and one copper chemical atom on each side and has net electric charge $+2e$ on each side, where $e$ is the positive elementary charge. The written reacting species need not be electrically neutral as a group: the bulk chemical solution also contains compensating anions.

There is an even more literal check. Zinc has atomic number $30$ and copper $29$, so the initial pair contains $30+(29-2)=57$ Electrons. The final pair contains $(30-2)+29=57$ Electrons. Only their distribution has changed. Conserved chemical element counts give the usual chemical mass conservation inventory; changing the counts of neutral and charged species does not create or remove their atomic nuclei.

Matching Electrons is necessary but does not replace balancing every chemical element. The acidic permanganate-iron balance illustrates unequal one- and five-Electron exchanges, with water and protons also needed. Conversely, a balanced chemical equation alone establishes neither a reaction rate nor a detailed mechanism. Its usefulness is that it fixes the stoichiometry even when those additional questions remain open.

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.