Lewis Structures: Rules and Construction

A carbon atom, four hydrogens, and one question: how do you draw the truth of methane on paper? Lewis structures are the answer chemists have used for over a century — a bookkeeping system that turns the abstract quantum mechanics of The Covalent Bond and Electron Sharing into a picture you can manipulate with a pencil. Every dot represents a real electron, every line a real shared pair, and the rules for placing them are not arbitrary conventions but a compressed summary of how atoms actually achieve stable electron configurations. This matters because Lewis structures are the entry point to almost everything that follows in structural chemistry. You cannot predict a molecule's shape, polarity, reactivity, or spectroscopy without first knowing which atoms are bonded to which, how many lone pairs sit where, and whether the electron count even makes sense. A wrong Lewis structure propagates errors into every downstream prediction — a wrongly-placed lone pair gives a wrong VSEPR geometry, a wrong bond order gives a wrong bond length and a wrong infrared stretching frequency. The skill of drawing them correctly, quickly, and by systematic method rather than memorized shortcuts, is foundational to the entire subject. The big picture is this: atoms in the second row of the periodic table are, energetically, most comfortable with eight valence electrons around them — a filled $2s2p$ shell — because this fills all the bonding and nonbonding orbitals available at that principal quantum number. Hydrogen wants two (a filled $1s$ ). Lewis structures are a systematic algorithm for distributing a fixed pool of valence electrons — dictated by the identity and number of atoms present — into bonds and lone pairs so that as many atoms as possible reach this stable count, while every single electron in the original pool is accounted for. This article builds that algorithm from first principles: how to count the electron pool correctly for both neutral molecules and ions, how to decide connectivity, how to place lone pairs and multiple bonds mechanistically rather than by guesswork, and how to recognize when the octet rule legitimately breaks down. Three fully worked examples — a small neutral molecule, an ion with a nontrivial electron count, and a hypervalent competition-level structure — anchor the method in concrete, checkable numbers. Along the way we set up the vocabulary — formal charge, resonance, expanded octets — that later articles develop in depth.

Key takeaways

Total valence electrons = sum of each atom's periodic group number, with charge adjustments (add for anions, subtract for cations).
Draw a skeleton first: the least electronegative nonhydrogen atom is usually central; hydrogen and halogens are terminal.
Place one bonding pair per skeletal bond, then distribute remaining electrons as lone pairs to satisfy octets, outer atoms first.
If atoms lack octets after lone pairs are placed, convert lone pairs on adjacent atoms into multiple bonds.
Formal charge is a bookkeeping tool to choose among candidate structures, not a measure of real charge distribution.
Second-row atoms obey the octet rule essentially without exception; period 3+ atoms can expand their valence shell.
Key idea: a Lewis structure is a conserved-electron-count distribution problem — count first, then place electrons to build filled shells, never the reverse.

Where this lives today

Lewis structures remain the first and fastest filter every chemist applies before reaching for more sophisticated tools. A medicinal chemist sketching a candidate drug on a whiteboard is drawing Lewis structures to decide, instantly, which atoms could plausibly be reactive electrophiles or nucleophiles, which nitrogens are basic lone-pair donors versus amide nitrogens with delocalized, non-basic lone pairs, and where metabolic oxidation is likely to occur — all read directly off lone-pair placement and bond order before any computation is run. In industrial process chemistry, correctly counting valence electrons and predicting hypervalency is essential for handling reactive main-group reagents like $\text{PCl}_5$ , $\text{SF}_6$ , and sulfur- and phosphorus-based flame retardants, where getting the octet-expansion rules wrong leads to fundamentally wrong predictions about reactivity and stability. In materials science, Lewis-structure reasoning underlies the design of Lewis acid catalysts (boron- and aluminum-based, exploiting the electron-deficient $\text{BF}_3$ -type structures discussed above) used industrially in Friedel–Crafts alkylations and polymerization initiators. In structural biology and biochemistry, the correct Lewis structure of the peptide bond — with its partial double-bond character from nitrogen lone-pair delocalization into the carbonyl — is what a first-pass single-bond-only Lewis structure would miss entirely, and recognizing that delocalization (foreshadowing Resonance: The Concept of Delocalization) is precisely why peptide bonds are planar and rotationally restricted, a fact central to protein secondary structure. In spectroscopy and instrumentation, predicted bond orders from Lewis structures give first-pass estimates of infrared stretching frequencies (a C=O stretch near $1715\,\text{cm}^{-1}$ versus a C–O stretch near $1100\,\text{cm}^{-1}$ ) used daily to identify functional groups in unknown compounds — the entire enterprise of functional-group identification by IR spectroscopy rests on the bond-order bookkeeping that Lewis structures make explicit.

Ask the tutor: How do I decide which resonance structure (or structures) to draw when a molecule has more than one atom capable of bearing a formal negative charge?

The physical picture

A Lewis structure is a claim about where electrons are, expressed through two conventions: a dot for an electron not shared between atoms (a lone pair, or rarely a single unpaired electron), and a line (or two dots) for a pair of electrons shared between two nuclei, forming a covalent bond. The entire method rests on one conservation law: the number of valence electrons drawn in a Lewis structure must exactly equal the number of valence electrons contributed by the constituent atoms, adjusted for overall charge. Nothing is created, nothing destroyed — only redistributed between bonding and nonbonding locations. Why eight electrons? For a second-row atom, the valence shell consists of one $2s$ orbital and three $2p$ orbitals — four orbitals total, each holding two electrons when full, for a maximum of eight. When an atom is surrounded by four bonding or lone pairs, all four of its valence atomic orbitals are involved in bonding or hold a nonbonding pair — there is no orbital left over to accept more electron density, and none half-filled to make the atom electron-deficient and reactive. This is a real orbital-filling argument, not folklore: the octet rule is a consequence of there being exactly four valence orbitals available at $n=2$ . Hydrogen and helium, restricted to the single $1s$ orbital, are correspondingly satisfied with two electrons — a "duet." The construction procedure translates this orbital picture into an algorithm. You first count the total electron pool honestly — every valence electron from every atom, corrected for charge — because this number is fixed and non-negotiable; get it wrong and every subsequent step, however carefully done, produces nonsense. You then decide connectivity (the "skeleton"), which is a chemical judgment based on electronegativity and known bonding patterns rather than something the electron count tells you directly. Only after the skeleton is fixed do you distribute the counted electrons: first as single bonds along the skeleton, then as lone pairs to satisfy octets on the atoms that most need them (typically the terminal, more electronegative atoms), and finally — if central atoms remain short of an octet — by "promoting" lone pairs on neighboring atoms into shared pairs, creating double or triple bonds. This promotion step is the first hint of the electron-pushing formalism (curved arrows moving a lone pair into bonding position) that becomes the mechanistic language of all of organic chemistry.

Figure

Electron pool → skeleton → bonds → lone pairs → octet check, illustrated for CO₂

Notice in the figure that the total electron count never changes across the three steps — 16 electrons throughout — only their distribution changes. This invariance is the single most useful self-check available: at the end of any Lewis structure, recount every dot and every line (as two electrons), and confirm the sum equals the number you started with. If it does not, an arithmetic or placement error has occurred somewhere in the construction, and the structure cannot be trusted for any downstream reasoning about shape or reactivity.

Figure

Formal charge as an electron-count comparison, previewed on CO₂'s alternative structure

The core ideas

The entire method reduces to two governing relationships. The first fixes the size of the electron pool:

N_{\text{total}} = \sum_i (\text{group number})_i \;-\; q

where: $N_{\text{total}}$ is the number of valence electrons to be drawn; the sum runs over every atom in the formula; $(\text{group number})_i$ is that atom's periodic group number (main-group: Group 1 → 1 valence electron, ..., Group 17 → 7, Group 18 → 8); and $q$ is the overall ionic charge in units of $e$ (so $q=-1$ for a monoanion adds one electron, since $-q = +1$ ).

Reading: you are simply summing up how many electrons each atom brings to the table in its outermost shell, then correcting for any electrons gained (anion) or lost (cation) to produce the observed net charge. This number is exact and fixed before a single bond is drawn. The second relationship is the bookkeeping identity that must hold once the structure is complete:

N_{\text{total}} = 2\,(\text{number of bonds, counting each } \sigma \text{ and } \pi \text{ separately}) + 2\,(\text{number of lone pairs})

where: each line in the structure (single, or one component of a double/triple bond) represents two shared electrons, and each dot-pair represents two nonbonding electrons.

Reading: every electron in the pool ends up in exactly one of two places — shared between two nuclei, or sitting on a single atom — and the total must reconcile exactly with $N_{\text{total}}$ from the first equation. This is the algebraic form of the conservation check shown in the figure above. A third, derived relationship is useful for rapid mental construction of neutral covalent molecules obeying the octet rule everywhere:

(\text{number of bonds}) = \frac{N_{\text{total}}^{\,\text{octets needed}} - N_{\text{total}}}{2}

where $N_{\text{total}}^{\,\text{octets needed}}$ is the sum of 8 for every non-hydrogen atom plus 2 for every hydrogen atom (the electron count that would be needed if every atom held a full octet/duet with no sharing at all).

Reading: the "deficit" between the number of electrons needed to give every atom its own private octet and the number of electrons actually available must be made up by sharing — and each bond, being counted twice (once for each atom it touches), makes up exactly two electrons of that deficit. This gives you the total bond count directly, without trial and error, before you place a single line.

Derivation AMechanism 1 — Constructing the Lewis structure of formaldehyde, CH₂O

Goal. Build the complete, correct Lewis structure of formaldehyde (used as the carbonyl in countless organic mechanisms) entirely from the electron-counting rules, verifying at each stage that the running electron count is conserved.

Step 1 — Count the total valence electron pool. Formaldehyde is $\text{CH}_2\text{O}$ : one carbon (Group 14 → 4), two hydrogens (Group 1 → 1 each), one oxygen (Group 16 → 6). The molecule is neutral, so $q=0$ .

N_{\text{total}} = 4 + 1(2) + 6 - 0 = 12 \text{ valence electrons}

Step 2 — Choose the skeleton. Hydrogen can never be a central atom (it has only the $1s$ orbital, capable of forming exactly one bond). Between C and O, carbon is less electronegative and has the well-known tendency to be central in organic frameworks, forming four bonds total. The skeleton is H–C(–H)–O, i.e., carbon bonded to two hydrogens and one oxygen.

Step 3 — Place one bonding pair along every skeletal connection. Three skeletal bonds (C–H, C–H, C–O) consume $3 \times 2 = 6$ electrons, leaving $12 - 6 = 6$ electrons still to place.

Step 4 — Distribute remaining electrons as lone pairs, prioritizing the most electronegative outer atoms. Oxygen is the only atom capable of holding lone pairs here (hydrogen's duet is already satisfied by its single bond, and it has no capacity for more). Placing all 6 remaining electrons on oxygen gives it 3 lone pairs — but a mechanistic check is required before accepting this: count oxygen's total electron environment.

Step 5 — Octet check on every atom. Hydrogen: 1 bond = 2 electrons around it → duet satisfied. ✓ Carbon: 3 bonds (2 to H, 1 to O) = 6 electrons around it → only 6, short of an octet. Oxygen (with 3 lone pairs as tentatively placed): 1 bond + 3 lone pairs = 2 + 6 = 8 electrons → octet satisfied, but this over-fills oxygen relative to what's needed once carbon is fixed. Carbon's deficiency signals that a lone pair must be promoted into a bonding pair.

Step 6 — Promote one oxygen lone pair into a second C–O bonding pair (curved-arrow formalism). Take one of oxygen's lone pairs and move it, via a curved arrow originating at the lone pair and terminating at the midpoint of the C–O bond, to form a C=O π bond. This does not change $N_{\text{total}}$ — it is a redistribution, not a creation of electrons: oxygen now has 2 lone pairs instead of 3, and the C–O linkage is a double bond (4 shared electrons instead of 2).

Step 7 — Final octet check and electron-count reconciliation. Carbon: 2 (C–H) + 2 (C–H) + 4 (C=O) = 8 electrons around it → octet satisfied. ✓ Oxygen: 4 (C=O) + 2 lone pairs × 2 = 4 + 4 = 8 electrons around it → octet satisfied. ✓ Hydrogens: duet each. ✓ Total electrons drawn: 3 bonding pairs of the single-bond type is wrong to say now — recount cleanly: two C–H single bonds (2×2=4 e⁻), one C=O double bond (4 e⁻), two lone pairs on O (2×2=4 e⁻).

4 + 4 + 4 = 12 = N_{\text{total}}

The count reconciles exactly, confirming the structure: $\text{H}_2\text{C=O}$ with two lone pairs on oxygen, zero formal charges everywhere (a preview of Formal Charge as the tool for selecting among structural candidates).

Go deeper · Derivation B · 3 minMechanism 2 — Constructing the Lewis structure of the nitrate ion, NO₃⁻, and locating resonance

Derivation BMechanism 2 — Constructing the Lewis structure of the nitrate ion, NO₃⁻, and locating resonance

Goal. Build $\text{NO}_3^-$ from scratch, demonstrating how an ionic charge enters the electron count and why a single Lewis structure is provably insufficient — the first rigorous motivation for Resonance: The Concept of Delocalization.

Step 1 — Count the total valence electron pool, including the ionic charge. One nitrogen (Group 15 → 5), three oxygens (Group 16 → 6 each), and a charge of $q = -1$ (an extra electron gained).

N_{\text{total}} = 5 + 6(3) - (-1) = 5 + 18 + 1 = 24 \text{ valence electrons}

Step 2 — Choose the skeleton. Nitrogen is less electronegative than oxygen and, being the sole non-oxygen atom, is the natural central atom; the three oxygens are terminal, arranged symmetrically around it (N bonded to each O, no O–O bonds — oxygen–oxygen single bonds are high in energy and not consistent with the symmetric, experimentally known trigonal-planar geometry of nitrate).

Step 3 — Place single bonds along the skeleton. Three N–O single bonds consume $3 \times 2 = 6$ electrons, leaving $24 - 6 = 18$ electrons.

Step 4 — Fill outer-atom octets first. Each oxygen needs 6 more electrons to complete an octet around its single bond (2 from the bond + 6 nonbonding = 8). Three oxygens require $3\times6 = 18$ electrons — which is exactly the amount remaining. Place 3 lone pairs on each oxygen.

Step 5 — Octet check on the central atom. Nitrogen: 3 single bonds = 6 electrons around it. This is short of an octet by 2 electrons — nitrogen needs one more bonding pair.

Step 6 — Promote one oxygen lone pair to form an N=O double bond. Exactly as in Mechanism 1, select one oxygen (call it O $_a$ ) and move one of its lone pairs, via a curved arrow, into the N–O $_a$ bond region, converting it to a double bond. Nitrogen now has 4(N–O $_a$ , double, 4 e⁻)+2(N–O $_b$ )+2(N–O $_c$ ) = 8 electrons around it. ✓ All three oxygens: O $_a$ now has 2 lone pairs + shares 4 electrons in the double bond = 8 total ✓; O $_b$ and O $_c$ retain 3 lone pairs + 2 shared electrons in their single bonds = 8 each ✓.

Step 7 — Recognize the ambiguity: any of the three oxygens could have been chosen in Step 6. Nothing in the electron-counting procedure privileges O $_a$ over O $_b$ or O $_c$ — the choice in Step 6 was arbitrary among three chemically identical oxygen positions. This produces three formally distinct but energetically equivalent Lewis structures, related by which oxygen bears the double bond. Because real nitrate has three experimentally identical N–O bond lengths (all $\approx 1.24\,\text{Å}$ , intermediate between typical N–O single, $\sim1.40\,\text{Å}$ , and N=O double, $\sim1.20\,\text{Å}$ , bond lengths), no single one of these structures is the true picture — the true electronic structure is a quantum-mechanical average (a resonance hybrid) of all three.

Step 8 — Reconcile the electron count in each resonance form. Every one of the three structures independently satisfies:

\underbrace{2(3)}_{\text{single bonds}} + \underbrace{2}_{\text{extra from double bond}} + \underbrace{2(9)}_{\text{lone pairs, 9 total}} = 6+2+18=26

Wait — recount lone pairs carefully: with one oxygen at 2 lone pairs and two oxygens at 3 lone pairs each, total lone pairs $=2+3+3=8$ , giving $2(8)=16$ electrons in lone pairs, plus bonding electrons $2(3\text{ single bonds}) + 4(\text{one double bond}) = 6+4=10$ .

16 + 10 = 26 \neq 24

This signals an overcount and forces a check: the single bonds are N–O $_b$ and N–O $_c$ only (2 single bonds, not 3), since N–O $_a$ is the double bond. So bonding electrons are $2(2) + 4(1) = 4+4=8$ , and lone pairs total $2+3+3=8$ pairs $=16$ electrons.

8 + 16 = 24 = N_{\text{total}} \checkmark

The reconciled count confirms the structure. Each resonance structure carries formal charges of $-1$ on each single-bonded oxygen, $0$ on the double-bonded oxygen, and $+1$ on nitrogen — summing to the correct overall ionic charge of $-1$ , a check developed fully in Formal Charge.

Worked examples

Example 1 (foundational): Ammonia, NH₃, and its lone pair. Count electrons: N (Group 15, 5) + 3×H (Group 1, 1 each) = $5+3=8$ valence electrons, neutral molecule. Skeleton: N central (hydrogen cannot be central), three N–H bonds. These three single bonds consume $3\times2=6$ electrons, leaving $8-6=2$ electrons — exactly one lone pair, placed on nitrogen. Octet check: nitrogen has 3 bonds (6 electrons) + 1 lone pair (2 electrons) = 8 electrons, satisfied; each hydrogen has its duet from one bond. Reconciliation: $3(2) + 1(2) = 6+2=8=N_{\text{total}}$ . No double bonds are needed because the electron count already exactly supplies every atom's target shell using single bonds and one lone pair — this is the simplest possible outcome of the algorithm and the reason ammonia's nitrogen carries a stereochemically active lone pair, central to its role as a Lewis base and its pyramidal geometry under Molecular Shape and VSEPR Theory.

Example 2 (applied): The carbonate ion, CO₃²⁻, and rationalizing its symmetric bond lengths. Count electrons: C (Group 14, 4) + 3×O (Group 16, 6 each) + charge correction for $q=-2$ (two extra electrons gained):

N_{\text{total}} = 4 + 6(3) - (-2) = 4+18+2=24

Skeleton: carbon central, three C–O bonds (carbon, as in formaldehyde, is the natural hub; oxygen–oxygen bonds are again disfavored). Three single bonds consume $3\times2=6$ electrons, leaving $24-6=18$ . Distribute onto the three oxygens: each needs 3 lone pairs (6 electrons) to reach an octet via its single bond, requiring $3\times6=18$ electrons total — exactly what remains. Check carbon: 3 single bonds = 6 electrons, two short of an octet. Promote one oxygen lone pair into a C=O double bond exactly as before. Final tally: carbon now has 8 electrons (2 single bonds × 2 + 1 double bond × 4 = 4+4=8); the double-bonded oxygen has 2 lone pairs + 4 shared = 8; the two single-bonded oxygens each have 3 lone pairs + 2 shared = 8. Electron reconciliation: bonding electrons $= 2(2)+4(1) = 8$ ; lone pairs $=2+3+3=8$ pairs $=16$ electrons; total $8+16=24=N_{\text{total}}$ ✓. As with nitrate, the choice of which oxygen bears the double bond is arbitrary among three equivalent positions, giving three resonance structures whose average reproduces carbonate's single observed C–O bond length of $1.29\,\text{Å}$ (intermediate between $\sim1.43\,\text{Å}$ single and $\sim1.20\,\text{Å}$ double), the same resonance logic as nitrate but now on a dianion — confirming that the charge-correction term in the electron-counting formula, not the connectivity logic, is what changes between differently-charged species.

Example 3 (competition-level): Sulfuric acid, H₂SO�4, and legitimate octet expansion. Count electrons: S (Group 16, 6) + 2×O in S=O positions... rather, count all atoms first: 2×H (1 each) + 1×S (6) + 4×O (6 each), neutral molecule.

N_{\text{total}} = 2(1) + 6 + 4(6) = 2+6+24=32

Skeleton: sulfur central (period 3, capable of exceeding an octet — this is the key new physics in this example), bonded to four oxygens; two of those oxygens each additionally bear one hydrogen (the two acidic O–H protons of $\text{H}_2\text{SO}_4$ ). Skeletal bonds: 4 S–O bonds + 2 O–H bonds = 6 bonds, consuming $6\times2=12$ electrons, leaving $32-12=20$ electrons. Distribute lone pairs onto oxygens to satisfy their octets: the two O–H oxygens each need 2 lone pairs (they already have 2 bonds: one to S, one to H, needing 4 more electrons = 2 lone pairs each), consuming $2\times2\times2=8$ electrons; the two non-protonated oxygens each need 3 lone pairs (only 1 bond to S so far, needing 6 more electrons), consuming $2\times3\times2=12$ electrons. Total consumed in this step: $8+12=20$ — exactly what remained. Now check sulfur: 4 single S–O bonds give sulfur only 8 electrons — a perfectly valid octet, and one could stop here (this is in fact an acceptable, low-formal-charge-free Lewis structure using only single bonds and giving sulfur a full octet, with each terminal non-protonated oxygen carrying a formal charge of $-1$ and sulfur $+2$ ). The competition-level refinement is to recognize that sulfur, being a period-3 element, has energetically accessible $3d$ -influenced hybrid/hypervalent bonding character (modern treatments describe this via strongly polarized S=O π-bonding rather than literal $d$ -orbital involvement, but the Lewis-structure bookkeeping is the same) that allows it to accommodate more than 8 electrons around itself without violating any physical law — unlike carbon, nitrogen, or oxygen, sulfur has energetically low-lying $3d$ orbitals and a larger covalent radius that make five- and six-electron-pair environments accessible. Promoting one lone pair from each non-protonated oxygen into an S=O π bond (two such promotions, via two curved arrows) converts both S–O single bonds to the non-protonated oxygens into S=O double bonds. Recount sulfur: 2 single S–O(H) bonds (4 electrons) + 2 double S=O bonds (8 electrons) = 12 electrons around sulfur — an expanded octet of six electron pairs, valid because sulfur is period 3. Each non-protonated oxygen now has 2 lone pairs + 4 shared (double bond) = 8 electrons, octet satisfied with formal charge 0; each protonated oxygen retains 2 lone pairs + 4 shared (one S–O, one O–H) = 8, formal charge 0; sulfur's formal charge becomes $6 - 0 - \tfrac{1}{2}(12) = 6-6=0$ . This all-zero-formal-charge structure is the one normally drawn for sulfuric acid. Reconciliation: bonding electrons $= 2(2)_{\text{O–H}} + 2(2)_{\text{S–O(H)}} + 2(4)_{\text{S=O}} = 4+4+8=16$ ; lone pairs $= 2(2)_{\text{protonated O}} + 2(2)_{\text{non-protonated O}} = 4+4=8$ pairs $=16$ electrons.

16+16=32=N_{\text{total}} \checkmark

As a limiting-case sanity check: if sulfur were replaced by its second-row analogue, hypothetical "H₂CO₄" is not a real stable neutral molecule in this bonding pattern precisely because carbon cannot expand past an octet — the double-bond-promotion step that works for sulfur is mechanistically forbidden for a second-row central atom lacking accessible low-lying $d$ -character orbitals or the radius to support five or six electron domains, which is exactly why the octet rule is stated as a hard rule for period 2 and a soft guideline for period 3 and below.

Limiting cases, scope & common traps

Situation	What happens	Why
Period 2 central atom (C, N, O, F)	Octet rule essentially never violated	Only four valence orbitals ( $2s,2p_x,2p_y,2p_z$ ) exist; no physical place to put a fifth bonding/lone pair
Period 3+ central atom (P, S, Cl, ...)	Can exceed octet (5–6 electron pairs)	Larger atomic radius and energetically accessible higher-angular-momentum/hybrid character allow more electron domains; e.g. $\text{PCl}_5$ , $\text{SF}_6$ , $\text{SO}_4^{2-}$
Odd total electron count	No structure satisfies every octet exactly	A molecule like $\text{NO}$ or $\text{NO}_2$ (radicals)必然 leaves one unpaired electron; draw it explicitly as a single dot
Electron-deficient central atom (B, Be)	Stable with fewer than 8 electrons	Boron in $\text{BF}_3$ has only 6 electrons around it; this is not an error — B has only 3 valence electrons and forms only 3 bonds, leaving an empty $2p$ orbital, which is why $\text{BF}_3$ is a strong Lewis acid
Multiple valid skeletons/bond placements	Several legitimate Lewis structures exist	Resolved by Formal Charge minimization and later by recognizing true Resonance: The Concept of Delocalization
As number of atoms → 2 (diatomics)	Method reduces to simple pairing, e.g. $\text{N}_2$ : $N_{\text{total}}=10$ , one triple bond + one lone pair each, $2(6)+2(2)=12\neq10$ — check: triple bond is 6 e⁻, two lone pairs are 4 e⁻, total $10$ ✓	Confirms the algorithm collapses correctly to the smallest possible case

Common trapMisconception: "Count electrons per atom, not per formula."

A very common student error is to satisfy each atom's octet independently, atom by atom, without first fixing the total electron pool for the whole molecule or ion. This double-counts electrons, because every bonding pair is shared and must only be counted once in the grand total even though it contributes to two atoms' local octets. Always compute $N_{\text{total}}$ globally first from the sum-of-group-numbers formula, and treat the entire construction as distributing that single fixed pool — never as summing up independently-satisfied local octets, which will not generally reconcile with the true electron count (as shown explicitly by the deliberate miscount and correction inside Mechanism 2, Step 7–8).

The chemist behind it

Origins · 1 minGilbert N. Lewis and how this was discovered

Gilbert N. Lewis

1875–1946 · American physical chemist, University of California, Berkeley

Gilbert Newton Lewis was among the most versatile physical chemists of the early twentieth century, contributing foundational work in thermodynamics (the concept of fugacity and activity), chemical bonding, acid–base theory (the Lewis acid–base definition), and even early quantum chemistry, yet he never received a Nobel Prize despite numerous nominations. His 1916 paper "The Atom and the Molecule," published in the Journal of the American Chemical Society, introduced the shared electron-pair bond and the dot-and-line notation years before Schrödinger's wave mechanics gave it a rigorous quantum-mechanical foundation — Lewis reasoned almost entirely from chemical evidence: bond energies, valences, and the striking chemical stability of the noble gases. He proposed that atoms tend toward electron configurations resembling the nearest noble gas, achieved either by outright electron transfer (ionic bonding) or by sharing electron pairs (covalent bonding) — a single unifying idea that reconciled what had previously seemed like two separate kinds of chemical bond. His insight preceded, and was later fully vindicated and quantitatively explained by, the molecular orbital and valence bond theories built on Schrödinger's equation.

History

- 1902 — Lewis begins developing cubic-atom electron models privately, well before publication, sketching electrons at the corners of a cube around each atomic kernel.

- 1916 — Publishes "The Atom and the Molecule," introducing the shared electron-pair covalent bond and the dot notation for valence electrons.

- 1923 — Publishes Valence and the Structure of Atoms and Molecules, consolidating the theory and introducing the general Lewis acid–base definition (electron-pair donor/acceptor).

- 1927–1930s — Quantum mechanical treatments (Heitler–London, and later molecular orbital theory) place Lewis's shared-pair bond on a rigorous mathematical foundation.

- 1946 — Lewis dies in his Berkeley laboratory; his electron-pair bonding concept remains the default working language of structural chemistry to this day.

Do Lewis structures show the real 3D shape of a molecule?

No. Lewis structures are two-dimensional connectivity and electron-count diagrams; they deliberately flatten three-dimensional information onto the page. A Lewis structure tells you which atoms are bonded, how many bonds link each pair (bond order), and where the lone pairs sit — but not bond angles or overall geometry. Converting a correct Lewis structure into an actual 3D shape requires a separate step, electron-domain repulsion counting, which is the subject of Molecular Shape and VSEPR Theory. It is entirely possible (and common) to draw a perfectly correct Lewis structure while still needing further reasoning to determine whether the resulting geometry is tetrahedral, trigonal pyramidal, bent, or linear.

Why do we subtract the charge for cations and add it for anions in the electron count formula?

The formula is $N_{\text{total}} = \sum(\text{group numbers}) - q$ , where $q$ is the signed charge. A cation has $q>0$ , meaning the species has lost electrons relative to the neutral sum of atoms, so we subtract a positive number, correctly reducing the electron pool. An anion has $q<0$ , meaning the species has gained electrons, so subtracting a negative number is equivalent to adding electrons, correctly increasing the pool. This is exactly the bookkeeping used for $\text{NO}_3^-$ ( $q=-1$ , one electron added) and $\text{CO}_3^{2-}$ ( $q=-2$ , two electrons added) in the worked derivations above; a cation like $\text{NH}_4^+$ ( $q=+1$ ) would instead have one electron subtracted from the neutral sum.

What do you do if, after filling all outer-atom octets, the central atom still has too many electrons rather than too few?

This situation indicates that your initial choice of all single bonds around the central atom was too electron-rich for the pool available, which typically signals either a miscounted electron total or a central atom that genuinely tolerates an expanded octet (period 3 or below), as in the $\text{H}_2\text{SO}_4$ worked example. If the central atom is period 2, an "excess" almost always traces back to an arithmetic slip in Step 1's electron count and should prompt a full recount from the group-number sum, since period-2 atoms cannot legitimately exceed eight electrons under any circumstance.

Are formal charges required to draw a Lewis structure, or are they a separate topic?

Formal charges are not required to construct a chemically valid Lewis structure — the electron-counting and octet-satisfaction rules alone are sufficient to produce a structure consistent with the correct total electron count. However, when multiple structures are consistent with the counting rules (as happened at the "which oxygen gets the double bond" branch point for both nitrate and carbonate), formal charge becomes the tool used to rank candidate structures by plausibility, generally favoring structures with formal charges closest to zero and with any negative formal charge residing on the more electronegative atom. This ranking procedure is developed rigorously as its own topic in Formal Charge.

Why does boron trifluoride, BF₃, not have an octet on boron, and is this a mistake?

It is not a mistake — it is the correct Lewis structure. Boron is in Group 13 and contributes only 3 valence electrons; with three single bonds to fluorine (each fluorine independently satisfying its own octet with three lone pairs), boron ends up surrounded by only 6 electrons, one full bonding pair short of an octet. Boron simply does not have a fourth valence electron available to form a fourth bond or acquire a lone pair without external help. This electron deficiency is not a flaw in the structure but the entire chemical reason $\text{BF}_3$ is a potent Lewis acid: it has an empty, energetically accessible $2p$ orbital that readily accepts an incoming electron pair from a Lewis base such as ammonia, forming a dative (coordinate covalent) bond and completing boron's octet in the adduct $\text{F}_3\text{B–NH}_3$ .

How does the Lewis-structure procedure handle molecules with an odd number of total valence electrons, like nitric oxide, NO?

An odd total electron count makes it mathematically impossible for every atom to have a fully paired octet, since paired electrons always come in even numbers per atom. For $\text{NO}$ : N (5) + O (6) = 11 valence electrons, an odd number. The best Lewis structure places a double bond between N and O (4 shared electrons), one lone pair on oxygen (2 electrons), one lone pair on nitrogen (2 electrons), and one unpaired single electron on nitrogen (the atom better able to stabilize a radical here) — giving $4+2+2+1=9$ ... reconciling properly: double bond (4) + O lone pair (2) + N lone pair (2) + single electron (1) = 9, which undercounts; the accepted structure instead places the odd electron on nitrogen with a single lone pair on nitrogen and two lone pairs on oxygen alongside the N=O double bond, giving $4+1(\text{N lone pair})\times2+2(\text{O lone pairs})\times2+1(\text{odd electron})=4+2+4+1=11$ , matching $N_{\text{total}}$ exactly. Such species are radicals, drawn with a single dot for the unpaired electron, and are inherently more reactive because that unpaired electron represents an incomplete orbital that seeks a partner.