Chemical Bonding and Molecular Structure · TOPIC LESSON
Lewis Structures and Octet Rule
Account for every valence electron before interpreting bonds, charge or polarity.
JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections
BY THE END, YOU CAN
Construct valid Lewis structures for neutral molecules and polyatomic ions.
Calculate formal charges and separate them from oxidation states and partial charges.
Recognize octet exceptions without assigning impossible electron counts to second-period atoms.
Use resonance contributors to explain equivalent bonds and delocalized charge.
Compare ionic and covalent character with polarization and energetic qualifications.
Distinguish bond polarity from the vector sum that determines molecular polarity.
BEFORE YOU START
Valence electrons and periodic-table groups.
Ionic charges and electron configurations.
Basic geometry and vector cancellation.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. What a Lewis drawing can and cannot show
A Lewis structure is an electron-accounting model. A line between two atoms represents a shared pair, and a lone pair represents two valence electrons assigned locally in the drawing. The model is powerful because it tracks connectivity, electron counts and formal charges with little machinery. It is not a photograph of electrons, a complete orbital wavefunction or an automatic three-dimensional shape prediction.
Main-group valence counts usually follow the group: carbon contributes four, nitrogen five, oxygen six and a halogen seven. Hydrogen contributes one and normally completes a two-electron shell. Add one electron for each negative unit of overall charge and subtract one for each positive unit. The resulting total is a strict budget. Moving an existing lone pair into a bond does not create electrons.
The octet guideline describes a common stable valence arrangement for many main-group atoms, particularly carbon, nitrogen, oxygen and fluorine. It is a pattern with exceptions, not a claim that every stable atom in every compound has eight nearby electrons. A valid drawing must conserve the total electron budget and respect the capabilities of the atoms. An aesthetically symmetrical diagram that contains the wrong number of electrons is invalid.
Bond order in a localized Lewis contributor counts lines: single, double or triple. A multiple bond includes more electron density between the same two nuclei, but is one connection between them. This distinction matters later in geometry: a double bond is one electron-domain direction, not two separate bonds pointing toward two different atoms.
02 / UNDERSTAND & APPLY
2. A reliable construction sequence
Begin with a plausible atom skeleton. Hydrogen is terminal, and fluorine is terminal in ordinary molecular Lewis structures. The less electronegative atom often occupies a central position, but molecular formula alone does not always identify connectivity. HCN is H–C–N, while molecules with several carbon atoms can have multiple constitutional isomers. Use a stated name or structural information rather than inventing uniqueness.
Connect the skeleton with single bonds, subtracting two electrons per bond from the budget. Complete the octets of terminal atoms other than hydrogen, then place any remaining electrons on the central atom. If a second-period central atom still lacks an octet, consider converting a neighboring lone pair into a multiple bond. Each conversion shifts an existing pair, leaving the total unchanged.
Finish with three audits: count every electron, inspect allowed local shells, and calculate formal charges. A polyatomic ion should be enclosed conceptually in brackets with the overall charge identified. The individual formal charges must add to that overall charge. If they do not, the arithmetic or the drawing is wrong even if the bond pattern looks familiar.
For CO₂, sixteen valence electrons are available. Two initial C–O single bonds plus terminal oxygen octets use all sixteen but leave carbon with four electrons. Turning one lone pair from each oxygen into a bonding pair gives O=C=O: carbon and both oxygens have octets and all formal charges vanish. This construction explains why the double bonds are introduced; they are not inserted merely because the formula was memorized.
ORIGINAL WORKED EXAMPLE
Build a charged structure
Construct a Lewis structure for NH₄⁺ and identify the nitrogen lone pairs and formal charge.
Total valence electrons = 5 + 4(1) − 1 = 8.
Place N centrally and form four N–H single bonds, using all eight electrons.
Each H has a duet and N has eight shared electrons around it. There are no nitrogen lone pairs.
Nitrogen is assigned half of eight bonding electrons, or four. Its formal charge is 5 − 4 = +1; each H has formal charge zero.
Answer Four N–H single bonds, no lone pair on N, total ion charge +1.
03 / UNDERSTAND & APPLY
3. Three different meanings of charge
Formal charge assigns each atom all its nonbonding electrons plus half its bonding electrons. It therefore treats every bond as equally shared for bookkeeping. For an atom with V valence electrons, L nonbonding electrons and B bonding electrons, FC = V − L − B/2. A line contributes one assigned electron to each of its two atoms; a lone pair contributes two to its atom.
Oxidation state uses a different convention: assign a heteronuclear bond’s electrons to the more electronegative atom, with specified rules for equal atoms and unusual cases. Partial charge describes a real or model-derived uneven electron distribution and usually is not an integer. Formal charge, oxidation state and partial charge answer different questions. A formally neutral carbon in CO₂ has oxidation state +4 and carries electron deficiency relative to oxygen.
Among plausible Lewis contributors, minimizing unnecessary charge separation often helps identify important ones. For comparable octet-satisfying structures, a negative formal charge is usually better accommodated by a more electronegative atom. These are ranking guides, not permission to violate an octet on carbon or nitrogen. Carbon monoxide illustrates the limitation: its important octet structure has a triple bond with C⁻ and O⁺ formal charges even though oxygen is more electronegative.
Never infer molecular ionic charge by counting only one charged atom in a contributor. Nitrate has a positive formal charge on nitrogen and negative formal charges on two oxygens in a standard contributor, summing to −1. The charge labels within a drawing must be summed before comparing with the bracketed ion charge.
FC = V − L − B/2
V: neutral-atom valence count; L: nonbonding electrons; B: all bonding electrons around that atom.
ORIGINAL WORKED EXAMPLE
Formal versus oxidation charge
For one octet-satisfying NO₃⁻ contributor with one N=O and two N–O bonds, calculate all formal charges and nitrogen’s oxidation state.
N has no lone pairs and eight bonding electrons: FC(N) = 5 − 0 − 4 = +1.
The double-bonded O has four nonbonding and four bonding electrons: 6 − 4 − 2 = 0.
Each single-bonded O has six nonbonding and two bonding electrons: 6 − 6 − 1 = −1.
Total formal charge = +1 + 0 − 1 − 1 = −1. For oxidation state, let N be x: x + 3(−2) = −1, so x = +5.
Answer Formal N charge +1; nitrogen oxidation state +5; total nitrate charge −1.
04 / UNDERSTAND & APPLY
4. Resonance: several drawings, one electronic structure
Sometimes no single localized Lewis drawing represents the observed equivalent bonds adequately. Resonance contributors use the same atomic connectivity and positions but differ in electron placement. They are alternative bookkeeping pictures contributing to one delocalized electronic structure. The molecule does not alternate between them as if it were performing a chemical reaction.
In nitrate, the N=O position can be drawn toward any of three equivalent oxygen atoms. Each contributor satisfies the same electron budget and overall charge. The actual free ion has equivalent N–O bonds. The elementary resonance-average bond order is (2 + 1 + 1)/3 = 4/3 for each bond. This is a useful qualitative representation of intermediate bonding, not a claim that electrons divide into literal one-third bond fragments.
Equivalent contributors have equal roles by symmetry. Inequivalent contributors generally do not have equal weights, so simply averaging all drawn bond orders can be unjustified. A carbonyl group can be represented by a neutral C=O contributor and a charge-separated C⁺–O⁻ contributor. The latter highlights bond polarity, but it does not make the bond a 50:50 mixture of a full single and full double bond.
Resonance changes electron placement, not atom connectivity, total electron number or total charge. Moving a hydrogen to another atom creates a different species or tautomer, not a resonance contributor. Curved-arrow reasoning must begin at an electron pair or bond and end where that electron pair is placed. It must not give second-period atoms more than an octet.
Delocalization helps explain intermediate bond lengths and the stabilization of ions such as carboxylates. The relevant comparison is with a hypothetical localized structure of the same connectivity; resonance stabilization is not an extra bond added to the electron count. In an unsymmetrical environment, equivalent-looking bonds can become measurably different, so symmetry assumptions should be stated.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Lewis resonance contributors; dots are lone-pair electrons. The text separators identify resonance and emphasize that atom positions stay fixed. The drawings represent one delocalized ion, not separate species in chemical equilibrium.
ORIGINAL WORKED EXAMPLE
Compare carbonate bond descriptions
A carbonate ion is described by three equivalent octet contributors, each with one C=O and two C–O⁻ bonds. What are carbon’s formal charge and the elementary average C–O bond order?
Carbon has four bond lines in each contributor and no lone pairs. FC(C) = 4 − 4 = 0.
Each oxygen position is double-bonded once and single-bonded twice across the equivalent drawings.
Its average is (2 + 1 + 1)/3 = 4/3.
The actual ion is one resonance hybrid; no oxygen is permanently designated the double-bonded oxygen in the free symmetric ion.
Answer Carbon formal charge 0 in those contributors; average C–O bond order 4/3.
05 / UNDERSTAND & APPLY
5. Electron-deficient, odd-electron and hypervalent cases
Electron-deficient compounds can have a central atom with fewer than eight valence-shell electrons in a useful Lewis model. BF₃ has three B–F bonds and six electrons around boron in its neutral simple structure. Its electron deficiency helps explain acceptance of a lone pair from a donor such as NH₃. BeCl₂ provides another common elementary electron-deficient example, although actual aggregation and phase must be considered when discussing real structures.
Odd-electron species have an odd total valence electron count, so pairing every electron into an ordinary closed-shell Lewis picture is impossible. NO has eleven valence electrons and NO₂ has seventeen. At least one unpaired electron remains in a simple description. These radicals show why octet counting alone cannot fully describe magnetism and electronic structure; molecular-orbital ideas are needed for a more complete account.
Period-three and heavier central atoms can be represented by expanded-valence Lewis structures, as in PCl₅ or SF₆. Treat these as useful bookkeeping representations. A modern bonding explanation does not require the simplistic claim that low-lying d orbitals must form five or six equivalent hybrids. Multicenter bonding, ligand contributions and molecular orbitals provide a more appropriate framework for hypervalent molecules.
Second-period atoms such as C, N, O and F do not expand their valence octets in ordinary Lewis structures. Drawing five bond lines around nitrogen merely to eliminate formal charge is not a valid repair. The octet restriction is often a stronger constraint than the preference for zero formal charge. Also distinguish a polyatomic ion’s local bonding from the ionic attraction between that ion and its counterion.
ORIGINAL WORKED EXAMPLE
Reject an attractive but impossible drawing
A proposed nitrate drawing has nitrogen double-bonded to two oxygens and single-bonded to a third, giving five bond lines around N. Can it be accepted because its formal charges look smaller?
Five shared pairs place ten bonding electrons around nitrogen.
Nitrogen is a second-period atom and does not support an expanded octet in an ordinary Lewis description.
The valid octet contributors use one double bond and two single bonds, with N⁺ and two O⁻ formal charges.
Formal-charge minimization cannot overrule this shell constraint.
Answer Reject the ten-electron nitrogen drawing; retain octet-satisfying resonance contributors.
06 / UNDERSTAND & APPLY
6. Donation of a pair does not create a new bond category forever
A coordinate covalent bond is described as forming when one partner supplies both electrons of a shared pair. In NH₃ + H⁺ → NH₄⁺, nitrogen donates its lone pair to the proton. This formation story is chemically useful for identifying a Lewis base as an electron-pair donor and a Lewis acid as an electron-pair acceptor.
Once an isolated ammonium ion has formed, its four N–H bonds are equivalent in its tetrahedral environment. One is not permanently tagged as a weaker or differently counted bond because its electron pair was donated in the construction. Lewis structures count the completed bond like any other two-electron covalent bond, and formal-charge calculations divide it equally.
The same distinction clarifies compounds containing both ionic and covalent interactions. Ammonium chloride contains covalent N–H bonds within NH₄⁺ and electrostatic lattice interactions between NH₄⁺ and Cl⁻. Saying “NH₄Cl is ionic” refers to its constituent ions and bulk lattice; it does not mean every atom pair inside it is connected by a purely ionic interaction.
Electron-pair donation can also connect neutral partners, as in an NH₃→BF₃ adduct. Electron counting then gives boron an octet; the donor and acceptor acquire formal charges in the simple bond representation. These formal charges need not match the sign or magnitude of actual partial charges on every atom. Always distinguish the formation arrow, the final connectivity and the chosen charge convention.
ORIGINAL WORKED EXAMPLE
Follow a donated pair
Compare nitrogen’s lone-pair count in NH₃ and NH₄⁺. Does the fourth N–H bond in NH₄⁺ use an extra pair created from nowhere?
NH₃ has 5 + 3 = 8 valence electrons: six in three bonds and two in one nitrogen lone pair.
H⁺ contributes a nucleus but no electron to the formation step.
The existing lone pair becomes the fourth N–H bonding pair. NH₄⁺ still has eight total valence electrons.
Nitrogen has no remaining lone pair in NH₄⁺.
Answer One lone pair becomes a bond; total valence electrons remain eight.
07 / UNDERSTAND & APPLY
7. Ionic and covalent character form a continuum
An ideal ionic model describes attraction between oppositely charged ions; an ideal nonpolar covalent model describes equal electron sharing. Real bonds often fall between these limits. Electronegativity differences can guide a comparison of polar character, but there is no universal numerical cutoff that transforms every bond from covalent to ionic. Bonding also depends on environment, structure and the electronic response of the partners.
Forming separated gaseous ions is not the whole energetic story of an ionic solid. Ionization costs energy, electron attachment can release or require energy, and lattice formation stabilizes the assembled ions. A salt’s favorable overall formation cannot be justified merely by declaring that atoms “want octets.” Likewise, not every ionic solid dissolves readily in water: dissolution reflects the balance of lattice disruption, solvation and entropy.
Polarization describes distortion of an anion’s electron cloud by a cation. In the qualitative Fajans picture, a smaller, more highly charged cation has greater polarizing power, while a larger, more diffuse anion is more polarizable. Greater distortion commonly means more covalent character. Thus comparing LiCl with NaCl highlights cation size, and comparing LiI with LiF highlights anion size.
These comparisons work best within reasonably comparable series. Cation electronic configuration, coordination, crystal structure and solvent interactions can complicate simple rankings. Do not turn a qualitative polarity trend into an exact prediction of melting point or solubility without considering those other factors. Aluminum chloride’s strong covalent character and molecular aggregation also show why “metal plus nonmetal” is only a first classification guide.
Charge neutrality fixes ionic formula ratios: Mg²⁺ and Cl⁻ give MgCl₂, while Al³⁺ and O²⁻ give Al₂O₃. It does not determine a unique isolated molecular geometry for the solid. In a lattice, each ion interacts with multiple neighbors rather than belonging to one isolated molecule represented by the empirical formula.
ORIGINAL WORKED EXAMPLE
Compare polarization carefully
In the usual qualitative polarization model, which has greater covalent character: LiF or LiI? State the controlled comparison and its limitation.
The cation is Li⁺ in both, so its charge and approximate size are held comparable.
I⁻ is larger and has a more diffuse electron cloud than F⁻, making it more polarizable.
Its electron distribution is distorted more strongly by the small Li⁺ cation.
The trend predicts greater covalent character for LiI; it is not a complete numerical calculation of lattice energy or solubility.
Answer LiI has greater covalent character in this comparison.
08 / UNDERSTAND & APPLY
8. Polar bonds can cancel in a molecule
Unequal sharing creates a bond dipole, represented in introductory chemistry as pointing toward the more electronegative end. The molecular dipole is a vector result: direction matters as much as bond polarity. For a simple pair of equal and opposite bond dipoles, the sum vanishes. Therefore a molecule can contain polar bonds and still have zero permanent molecular dipole.
CO₂ is linear, so its two C–O bond dipoles cancel. H₂O is bent, so its O–H contributions do not cancel. In tetrahedral CCl₄, four equal C–Cl contributions cancel by symmetry. Replacing one chlorine by hydrogen breaks that equivalence and removes the symmetry-based cancellation guarantee. Do not decide polarity from formula alone without considering geometry.
A simple two-charge dipole has magnitude μ = qr. If μ is expressed in C m and the charge separation r in metres, q follows in coulombs. Dividing an observed diatomic dipole by the hypothetical full-charge value er gives a rough percent-ionic-character measure for that simple model. It is not an exact universal partition of a chemical bond into ionic and covalent percentages, especially for polyatomic systems.
Formal charge does not substitute for this vector analysis. A neutral Lewis structure can describe a polar molecule; a polyatomic ion’s net charge does not make its internal bond-dipole cancellation a simple neutral-molecule problem. For the common examination-style neutral examples, identify geometry, compare equivalent bond vectors and only then determine whether symmetry forces zero dipole.
μ = qr
Two-charge model; use SI units consistently.
μ_molecule is a vector sum, not the sum of bond-dipole magnitudes
Geometry and equivalent substituents determine cancellation.
ORIGINAL WORKED EXAMPLE
A model estimate of ionic character
A hypothetical diatomic AB has bond length 1.50 × 10⁻¹⁰ m and dipole moment 8.00 × 10⁻³⁰ C m. Estimate ionic character relative to full charge separation using e = 1.60 × 10⁻¹⁹ C.
The full-charge model gives μ_ionic = er = (1.60 × 10⁻¹⁹)(1.50 × 10⁻¹⁰) = 2.40 × 10⁻²⁹ C m.
The ratio is 8.00 × 10⁻³⁰ / 2.40 × 10⁻²⁹ = 1/3.
Multiply by 100 for the conventional estimate.
The result describes this simple dipole model, not a uniquely measured electron-transfer fraction.
Answer Approximately 33.3% ionic character in the stated model.
09 / UNDERSTAND & APPLY
9. Original structural reasoning practice
These problems are original. A complete answer should conserve electrons and total charge before invoking a trend. Describe connectivity explicitly when a drawing is not available: specify single or multiple bonds, lone pairs, and any formal charges. Resonance arrows connect alternative contributors; reaction arrows connect different chemical species.
If a calculation produces a plausible number but an impossible local shell, revise the drawing rather than trust the number. If a polarity argument ignores geometry, it is incomplete. Use each model for the question it can answer, and identify where a more detailed orbital or energetic description would be required.
ORIGINAL PRACTICE · 1
How many total valence electrons does SO₄²⁻ have?
Reveal answer & reasoning
32.
Sulfur contributes 6 and four oxygens contribute 24; the 2− charge adds two more. This total must be preserved in any chosen valid contributor.
ORIGINAL PRACTICE · 2
In H₃O⁺ with three O–H single bonds and one lone pair on O, find FC(O).
Reveal answer & reasoning
+1.
Oxygen has V=6, L=2 and B=6, so FC=6−2−3=+1. Each hydrogen is formally neutral.
ORIGINAL PRACTICE · 3
A student draws NH₃ with three bonds but no lone pair. What is missing?
Reveal answer & reasoning
One lone pair on nitrogen.
NH₃ has eight valence electrons. Three bonds account for six, leaving two. Including that pair gives nitrogen its octet and formal charge zero.
ORIGINAL PRACTICE · 4
Are CH₃–CH₂–OH and CH₃–O–CH₃ resonance contributors?
Reveal answer & reasoning
No; they are constitutional isomers.
Their atom connectivity differs. Resonance retains the atomic skeleton and changes only electron placement; these two structures place oxygen in different connections.
ORIGINAL PRACTICE · 5
Why is a single fixed N=O bond inadequate for the free nitrate ion?
Reveal answer & reasoning
It fails to represent the three equivalent N–O bonds.
Three equivalent octet contributors place the double bond at different oxygens. The hybrid is delocalized; the ion does not switch among separate localized structures.
ORIGINAL PRACTICE · 6
Which is more strongly polarizing in a comparable ionic environment, Na⁺ or Mg²⁺?
Reveal answer & reasoning
Mg²⁺.
It has greater positive charge and a smaller ionic radius in comparable coordination. Both favor stronger distortion of an anion. This is a qualitative trend rather than a complete bonding calculation.
ORIGINAL PRACTICE · 7
Do polar C–O bonds make linear CO₂ polar?
Reveal answer & reasoning
No; its permanent molecular dipole is zero by symmetry.
The two equal bond dipoles point oppositely along the molecular axis and cancel. A scalar sum of their magnitudes would give the wrong conclusion.
ORIGINAL PRACTICE · 8
Why can NO not be represented with every valence electron paired in an ordinary closed-shell Lewis structure?
Reveal answer & reasoning
It has eleven valence electrons, an odd number.
Nitrogen contributes five and oxygen six. Pair-only allocation consumes electrons in twos and cannot exhaust an odd total; an unpaired electron is required.
KEEP THE BIG IDEAS
Quick revision
Count total valence electrons first and conserve that budget in every drawing.
Formal charge is an equal-sharing convention, distinct from oxidation state and partial charge.
Resonance preserves atom connectivity and charge; equivalent contributors explain equivalent bonds.
Octet exceptions exist, but second-period atoms do not acquire expanded octets to improve a drawing.
Coordinate bonding describes pair donation during formation, not permanently distinguishable equivalent bonds.
Ionic character depends on polarization and environment as well as electronegativity.
Determine molecular polarity from geometry and vector cancellation.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.