Covalent bond

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.

A bond with formal bond order three. A localized orbital description usually contains one sigma and two pi components.

A bond with formal bond order one, conventionally represented by one line. Delocalization can make the actual bond order differ from the integer in one Lewis structure.

A formal bond order of five. The common d-metal model has one sigma, two pi and two delta bonding orbitals occupied, ${\sigma }^2{\pi }^4{\delta }^4$, with the corresponding antibonding orbitals empty.

A formal fourfold bond. In the familiar d-metal example, one sigma bond, two pi bonds and one delta bond are occupied, giving the configuration ${\sigma }^2{\pi }^4{\delta }^2$ and formal bond order four.

A delta bond involves sideways orbital overlap with two nodal planes containing the internuclear axis. In suitable metal-metal bonds, d-orbital overlap can supply this component in addition to sigma and pi bonding.

A coordinate covalent bond is described as forming by donation of a lone pair from one partner into an acceptor orbital of the other. This formation description does not imply that the final electrons remain distinguishable by origin.

An electron-pair interaction represented as donation from a metal-base fragment into an orbital of a metal-acid fragment. electron counting must avoid adding a new pair twice to an already occupied donor orbital.

Donation from an occupied ligand orbital into a metal orbital of pi symmetry relative to the metal–ligand axis. It differs from metal-to-ligand pi backbonding.

Pi backbonding is donation from occupied metal orbitals into ligand acceptor orbitals with pi symmetry about the metal–ligand axis. In a carbonyl ligand, occupation of CO ${\pi }^{\ast }$ weakens C–O while strengthening metal–carbon bonding. Lower CO stretching spectroscopic wavenumber is therefore a useful comparative probe, with electrostatic and mode-coupling qualifications.

Sigma donation transfers electron density from an occupied ligand orbital into a metal acceptor orbital with the appropriate symmetry along the bond axis. CO uses an occupied carbon-centered $5\sigma$ orbital. Backdonation can occur simultaneously.

A covalent bond formed through side-on overlap of compatible atomic orbitals, with a nodal plane containing the internuclear axis.

A delocalized pi system made from three compatible p orbitals and four electrons. In the simple orbital construction the bonding and nonbonding combinations are occupied and the antibonding combination is empty.

A covalent bond with bonding electron density of cylindrical symmetry around the internuclear axis; end-on orbital overlap is a common localized construction.

A glycosidic bond links a sugar to another group through its anomeric centre. In a nucleotide, rotation about the base–sugar linkage contributes to anti and syn conformations; it is distinct from the phosphate links joining successive nucleotides.

A double bond is a covalent bond of bond order two. Restricted rotation and substituent geometry at a carbon–carbon double bond can affect acyl-chain packing in a lipid bilayer.

A phosphodiester bond connects two groups through a phosphate diester. In a nucleic acid backbone, the bridging phosphate is connected through oxygen to the $3^{\prime }$ carbon position of one sugar and the $5^{\prime }$ carbon position of another. These covalent bonds connect successive nucleotides along a strand; they are distinct from the hydrogen bonds between paired nucleobases.

A disulfide bond is the single covalent bond between the two adjacent sulfur chemical atoms in a disulfide, conventionally written $\mathrm{R}-\mathrm{S}-\mathrm{S}-{\mathrm{R}}^{\prime }$. Here $R$ and $R^{\prime }$ stand for the attached organic groups; they may belong to the same molecule or to initially separate molecules. A disulfide bond is a chemical bridge, rather than a hydrogen bond or a mere contact between nearby groups. Two thiols, $\mathrm{R}-\mathrm{SH}$ and ${\mathrm{R}}^{\prime }-\mathrm{SH}$, can supply its two sulfur endpoints. ^[ref]

The stoichiometry of forming a disulfide bond is especially revealing. Written as an oxidation half-reaction, it isThe reverse reduction half-reaction consumes two electrons and two hydrogen ions to restore two thiols. Both sides contain two sulfur chemical atoms; the two hydrogen chemical atoms have moved into the two hydrogen ions, and the net electric charge remains zero because $+2-2=0$. Thus one disulfide bond corresponds to two thiol groups and two transferred electrons. These half-reactions describe redox bookkeeping, not a complete standalone chemical reaction: an accompanying reactant must accept or supply the electrons, and pH affects the actual protonation of the groups. Mercaptoethanol can promote reduction through exchange of disulfide bonds, but adequate excess and appropriate conditions are needed; adding it does not by itself prove complete reduction. ^[ref]

In a protein disulfide bond, the endpoints are normally the amino acid side chains of two cysteine amino acid residues. An intrachain protein disulfide bond closes an intramolecular bridge within one polypeptide chain; an interchain protein disulfide bond connects different polypeptide chains, and can arise by intermolecular crosslinking of initially separate molecules. Forming or reducing these bridges changes covalent bond connectivity and can change protein structure, while preserving the order of amino acid residues and the peptide bonds in the protein backbone. A complete chemical account of protein primary structure includes crosslinks as well as sequence, so preservation of sequence does not imply preservation of every covalent bond. Mature insulin has two interchain protein disulfide bonds and one intrachain protein disulfide bond. ^[ref]

A bridge is useful because it constrains which parts of a polypeptide chain can move apart without breaking a covalent bond. Its placement therefore matters as well as its presence. The ribonuclease A refolding experiment provides the existing experimental example: breaking and reforming four protein disulfide bonds can accompany loss and recovery of function. Counting four bridges alone cannot tell whether their endpoints have the native pairing. This is an attractive connection between simple chemical counting and the richer problem of protein folding.

How can an experiment reveal an initially hidden disulfide bond? Ellman's reagent, also called DTNB, measures free thiols by producing a yellow absorbing species. Its reactive disulfide bond undergoes exchange with the conjugate base $\mathrm{R}-{\mathrm{S}}^{-}$ of a thiol, producing a mixed disulfide attached to $R$ and one released 2-nitro-5-thiobenzoate species, abbreviated TNB. At the assay pH, the strongly colored form is the dianion $\mathrm{TN}{\mathrm{B}}^{2-}$: one negative charge is on sulfur and the other is on the carboxylic acid-derived $\mathrm{CO}{\mathrm{O}}^{-}$ group. With excess DTNB and complete reaction, one released TNB reports one reacted thiol. George L. Ellman's 1959 paper introduced the tissue assay; it was subsequently reexamined quantitatively. ^{[ref] [ref]}

The modern apparatus can be simple: tubes, pipettes, a buffer solution, cysteine standards, DTNB, and an instrument that measures absorbance through a cuvette at wavelength $412\mathrm{nm}$. The Pierce protocol uses $50\mu \mathrm{L}$ of a $4\mathrm{mg}\mathrm{m}{\mathrm{L}}^{-1}$ DTNB stock, $2.50\mathrm{mL}$ of buffer solution, and $250\mu \mathrm{L}$ of sample. Its buffer solution is $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ sodium phosphate at pH 8.0 with a metal-binding additive. After 15 minutes at room temperature, measure absorbance against a buffer solution plus DTNB blank; run known cysteine concentrations in parallel. The observable is increased yellow color and increased blank-corrected absorbance as more accessible thiols react, rather than direct imaging of disulfide bonds. These are published protocol conditions, not measurements performed for this article. ^[ref]

For a before/after test, split a purified protein sample into matched aliquots. Measure accessible free thiols in one aliquot; subject the other to complete reduction of its disulfide bonds, remove the reducing reagent, then measure the resulting thiols with fresh DTNB. Process the unreduced aliquot through a matching cleanup. Normalize to recovered protein and known dilution, since cleanup can change both. To estimate all reducible protein disulfide bonds, expose buried groups using matching protein denaturation conditions in both aliquots and calibrate in that same chemical solution; a native-accessibility measurement answers a different question. A larger corrected reading after reduction supports conversion of previously unreactive groups into detectable thiols. Attribution to disulfide bonds requires excluding other reducible sulfur modifications. This is a proposed comparison based on the detection principles, not a reconstruction of Ellman's original tissue measurements. ^[ref]

Under the Beer-Lambert law, the blank-corrected absorbance is $A=\epsilon lc$, where $l$ is the cuvette path length, $c$ is the released TNB concentration, and $\epsilon$ is its molar absorption coefficient. Riddles, Blakeley and Zerner measured ${\epsilon }_{412}=14150\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ in dilute aqueous salt chemical solutions, confirming the value independently using cysteine. It must not be applied indiscriminately to other solvents or protein denaturation mixtures. ^[ref] For aliquots compared on the same final volume and protein concentration basis, the ideal estimate isThe factor of two follows from the reduction stoichiometry; ${\bar{n}}_{\mathrm{S}-\mathrm{S}}$ is the number of disulfide bonds per protein molecule, or an average for a mixed population. If the aliquots have different dilutions or recoveries, convert each reading to a common sample basis before subtracting.

As a hypothetical calculation, choose $c_{\mathrm{protein}}=10\mu \mathrm{mol}{\mathrm{L}}^{-1}$ in each final cuvette, two free thiols and two disulfide bonds per protein molecule, and $l=1\mathrm{cm}$. Before reduction, $c_{\mathrm{SH}}=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$ gives $A=0.283$; complete reduction adds four thiols per protein molecule, so $c_{\mathrm{SH}}=60\mu \mathrm{mol}{\mathrm{L}}^{-1}$ gives $A=0.849$. The difference $0.566$ gives $c_{\mathrm{S}-\mathrm{S}}=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$ and ${\bar{n}}_{\mathrm{S}-\mathrm{S}}=2$. The published tube volumes imply a dilution factor $2.80/0.250=11.2$; reversing that dilution multiplies both thiol and protein concentrations by 11.2 and leaves the bridge count per protein molecule unchanged.

Controls determine whether that arithmetic has chemical meaning. Use fresh cysteine standards and a processed blank without protein to detect color generated by residual reducing reagent. Even a reducing reagent containing no thiols can interfere: Han and Han showed that tris(2-carboxyethyl)phosphine, abbreviated TCEP, reduces DTNB to two TNB equivalents per TCEP. ^[ref] In the hypothetical model, $5\mu \mathrm{mol}{\mathrm{L}}^{-1}$ residual TCEP in only the reduced cuvette would add $A=0.1415$ and falsely add $0.5$ bridge per protein molecule. A stable endpoint and agreement after a further reduction treatment help check completion. A sample without DTNB checks its own color; cloudy samples can also attenuate light without producing TNB. Match pH, temperature, accessibility and calibration chemical solution, and minimize delay that permits reoxidation. If two already-free but buried thiols become accessible only in the reduced aliquot, they mimic one new disulfide bond per protein molecule. Finally, bulk absorbance supplies a count under these assumptions: it does not locate the cysteine endpoints, distinguish intrachain from interchain bridges, or establish the native pairing of a protein disulfide bond.

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 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.

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.

Electron pair geometry describes the arrangement of all electron domains in the VSEPR model, including lone pairs. Molecular geometry instead describes the arrangement of chemical atoms. Four domains have a tetrahedral ideal arrangement, but keeping only the bonded neighbours gives the different molecular geometries of methane, Ammonia and water.

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.

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.

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 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.

Resonance donation is electron-pair delocalization from a substituent into an adjacent conjugated system. Drawing electron-pair arrows into the system gives contributing charge-separated structures; their stabilization can change reaction rates and preferred sites.

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.

A resonance hybrid is the combined electronic description expressed using several resonance structures. The word does not describe a mixture of separate molecules or ions: a nitrate ion has one electron density, rather than three species in chemical equilibrium. Equal weighting is appropriate for the equivalent nitrate contributors in the elementary model; the unequal carbon dioxide contributors illustrate why it is not a general rule. See Chemistry 2e, 7.4: resonance hybrids do not fluctuate between forms.

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.