Lewis Structures, Formal Charge, and Resonance: Step-by-Step Worked Examples
A worked-example walkthrough of drawing Lewis structures, assigning formal charges, and recognizing resonance for the molecules students get wrong most often on exams — CO2, NO3−, SO4²−, N3−, ozone, and CO.
Learning Objectives
- ✓Draw Lewis structures using the systematic five-step approach.
- ✓Compute formal charges to choose between competing valid structures.
- ✓Recognize resonance and use it to predict observed bond lengths and reactivity.
1. Direct Answer: The Five-Step Lewis Structure Approach
Step 1: Count total valence electrons (sum over all atoms; add for anionic charge, subtract for cationic charge). Step 2: Place the least electronegative atom as the central atom (H and F never central; carbon usually central). Step 3: Connect peripheral atoms to the central atom with single bonds. Step 4: Distribute remaining electrons as lone pairs to satisfy octets on peripheral atoms first, then the central atom. Step 5: If the central atom lacks an octet, convert peripheral lone pairs into double or triple bonds until the central atom has an octet. After this, compute formal charges to verify the structure makes chemical sense.
Key Points
- •Step 1: total valence electrons (add for anion, subtract for cation).
- •Step 2: least electronegative atom central (H and F never).
- •Step 3-4: bond peripherals then distribute lone pairs.
- •Step 5: multiple bonds if central atom lacks octet.
2. Formal Charge: The Decision Rule
Formal charge = valence electrons of free atom - (lone pair electrons + bonding electrons / 2). When two valid Lewis structures exist, the preferred structure has (1) formal charges closest to zero on all atoms, (2) negative formal charges on the more electronegative atoms, (3) the smallest absolute sum of formal charges. Formal charge does not represent actual charge distribution — it is a bookkeeping tool to select the most stable Lewis structure.
Key Points
- •FC = valence - (lone pair e⁻ + bonding e⁻ / 2).
- •Prefer structure with FCs closest to zero.
- •Negative FC should reside on more electronegative atom.
3. Worked Example: CO2
Total valence electrons = 4 + 2(6) = 16. Carbon central with O on each side. After single bonds and lone pairs on oxygens: C has only 4 electrons. Convert two lone pairs (one per O) into double bonds: O=C=O with two lone pairs on each O. Now C has 8 electrons. Formal charges: C = 4 - (0 + 8/2) = 0; each O = 6 - (4 + 4/2) = 0. All zero — best structure. CO2 is linear with two equivalent C=O double bonds of 1.16 Å (bond order 2). No resonance needed because the symmetric structure is best.
Key Points
- •CO2: 16 valence electrons, two C=O double bonds.
- •All formal charges zero — single best Lewis structure.
- •Linear geometry with bond order 2 throughout.
4. Worked Example: NO3⁻ (Resonance)
Total valence electrons = 5 + 3(6) + 1 = 24 (added 1 for negative charge). Nitrogen central with 3 oxygens. After single bonds and lone pairs, N has only 6 electrons. Convert one O lone pair to a double bond: but which O? All three are equivalent. Three valid Lewis structures exist with the double bond on each O in turn — these are resonance structures. The actual molecule is the average: all three N-O bonds equivalent at 1.26 Å (intermediate between single 1.36 Å and double 1.22 Å). Formal charges in any one resonance structure: N = 5 - (0 + 8/2) = +1; double-bonded O = 6 - (4 + 4/2) = 0; single-bonded O = 6 - (6 + 2/2) = -1 (each). Average over resonance: each O carries -2/3 formal charge.
Key Points
- •NO3⁻: 24 valence electrons, three equivalent resonance structures.
- •Observed bond length intermediate between single and double.
- •Negative charge delocalized across all three oxygens.
5. Worked Example: SO4²⁻ and Expanded Octets
Total valence electrons = 6 + 4(6) + 2 = 32. Sulfur central, 4 oxygens. After single bonds and lone pairs on each O, S has 8 electrons — octet satisfied. Formal charges: S = 6 - (0 + 8/2) = +2; each O = 6 - (6 + 2/2) = -1. Sum of |FCs| = 2 + 4 = 6. To reduce formal charges, convert two single S-O bonds to double bonds (expanding sulfur’s octet to 12 — allowed for period 3+ elements with empty d-orbitals). New formal charges: S = 6 - (0 + 12/2) = 0; double-bonded O = 6 - (4 + 4/2) = 0; single-bonded O = -1 each. Sum of |FCs| = 2. This structure is preferred because formal charges are minimized. Four resonance structures exist with the double bonds on different O pairs. Observed bond length 1.49 Å (intermediate between single and double S-O).
Key Points
- •Period 3+ elements can exceed the octet (use empty d-orbitals).
- •Choose structure that minimizes formal charges.
- •SO4²⁻ has 4 equivalent resonance structures with 2 S=O double bonds each.
6. Worked Example: Ozone (O3)
Total valence electrons = 3(6) = 18. Center O bonded to two terminal O. After single bonds and lone pairs, center O has only 6 electrons. Convert one terminal O lone pair to double bond: now have O=O-O. But the other arrangement O-O=O is equally valid. These are the two resonance structures. Actual molecule: bent geometry (117° angle, not 120° due to lone pair on central O), both O-O bonds equivalent at 1.28 Å (intermediate between single 1.48 and double 1.21). Formal charges in one resonance structure: central O = 6 - (2 + 6/2) = +1; double-bonded terminal O = 6 - (4 + 4/2) = 0; single-bonded terminal O = 6 - (6 + 2/2) = -1. Average over resonance: terminal Os each -1/2, central +1.
Key Points
- •Ozone has two resonance structures.
- •Both O-O bonds observed equivalent at 1.28 Å.
- •Bent geometry due to lone pair on central O.
7. Worked Example: CO (Carbon Monoxide)
Total valence electrons = 4 + 6 = 10. C and O bonded. Try single bond: C has only 4 electrons (4 bonding + 0 lone pair), O has 8 electrons. Try double bond C=O: C has 6 electrons, O has 8. Try triple bond C≡O: C has 8 electrons, O has 8. Triple bond satisfies octet rule. Formal charges: C = 4 - (2 + 6/2) = -1; O = 6 - (2 + 6/2) = +1. This is unusual — the negative formal charge is on the LESS electronegative atom (C). But the structure is correct because the triple bond is the only way to give both atoms a complete octet with 10 electrons. The observed dipole moment is small (0.122 D, almost negligible) and points C toward O — consistent with the negative formal charge on C cancelling much of the expected O-electronegativity polarization. CO is a key counterexample to the "negative FC on more electronegative atom" rule.
Key Points
- •CO has a triple bond (C≡O).
- •Formal charges: C = -1, O = +1 (negative FC on less electronegative atom).
- •Counterexample to the standard formal charge preference rule.
8. Using ChemistryIQ for Lewis Structure Practice
Snap a photo of any Lewis structure problem and ChemistryIQ identifies the central atom, counts valence electrons, constructs the structure step by step, computes formal charges on every atom, and shows all valid resonance structures with bond-order calculations. The app handles octet exceptions (expanded octets for period 3+, electron-deficient compounds like BF3) and provides VSEPR geometry predictions from the resulting structure.
Key Points
- •Automated Lewis structure construction with formal charge analysis.
- •Resonance structure enumeration.
- •Octet exception handling (expanded octets, electron-deficient).
High-Yield Facts
- ★Five-step Lewis structure approach: count e⁻, place central atom, bond, distribute, multiple bonds.
- ★Formal charge = valence - (lone pair e⁻ + bonding e⁻ / 2).
- ★Prefer structure with formal charges closest to zero.
- ★Period 3+ atoms can exceed octet (expanded octets).
- ★Resonance structures: equivalent valid Lewis structures with different electron arrangements.
- ★Observed bond lengths in resonance hybrids are intermediate between single and double.
Practice Questions
1. Draw the Lewis structure for NO2⁻ and explain its resonance.
2. For NH4+, total valence electrons and formal charges?
3. Why does SF6 exist if S has only 6 valence electrons?
FAQs
Common questions about this topic
No. Formal charge assumes all bonds are perfectly covalent (electrons shared equally). Oxidation state assigns bonding electrons to the more electronegative atom (ionic limit). They give different numbers for the same atom in the same molecule. Formal charge guides Lewis structure selection; oxidation state predicts redox behavior.
When you can draw multiple equally valid Lewis structures by moving only electrons (no atoms move) and where the alternative position of a double bond is between equivalent atoms, you have resonance. Common signals: multiple oxygens or other equivalent terminal atoms on a central atom, conjugated pi systems (alternating single/double bonds), aromatic rings. The molecule's actual electronic structure is the average of all valid contributors.
Yes — they are called free radicals. Examples: NO (11 electrons), NO2 (17 electrons), ClO2 (19 electrons). They cannot satisfy the octet rule on every atom because the electron count is odd. Draw the best structure that maximizes octets, leaving one atom with 7 electrons (the radical center). These molecules are typically reactive.
Three categories: (1) Incomplete octets — H (2 electrons), Be (4 electrons), B and Al (6 electrons each in BF3, AlCl3). (2) Expanded octets — period 3+ atoms can exceed 8 (PCl5 = 10, SF6 = 12, IF7 = 14). (3) Odd-electron radicals — odd total electron count makes octet impossible on every atom.
Snap a photo of any Lewis structure problem and ChemistryIQ constructs the structure step by step, computes formal charges, identifies resonance, and handles octet exceptions. The app also generates VSEPR geometry from the resulting structure. This content is for educational purposes only and does not constitute chemistry advice.