4.2 Lewis Structures, Formal Charges & Resonance

Key Takeaways

  • Drawing valid Lewis structures requires tabulating total valence electrons, selecting the least electronegative atom as central, completing terminal octets, and shifting lone pairs to satisfy electron deficits.
  • Formal charge (FC = V - N - 0.5B) evaluates electron distributions; dominant Lewis structures minimize formal charge magnitudes and place negative charges on the most electronegative elements.
  • Resonance structures depict electron delocalization across equivalent or non-equivalent frameworks; the true molecule is a stable resonance hybrid with intermediate fractional bond orders.
  • Molecules with an odd count of valence electrons form free radicals containing an unpaired electron, imparting paramagnetism and high chemical reactivity.
  • Incomplete octets occur in electron-deficient Group 2 and 13 species, whereas expanded octets occur strictly in Period 3+ elements possessing accessible d orbitals and larger atomic radii.
Last updated: September 2026

Lewis Structures, Formal Charges & Resonance

Quick Summary: Lewis dot structures represent valence electrons organized into shared covalent bonds and nonbonding lone pairs. Formal charge calculations identify dominant electronic configurations by assessing deviations from neutral atomic valences. Delocalized π\pi electron systems are described through resonance hybrids exhibiting fractional bond orders. Key exceptions to the octet rule include odd-electron free radicals, electron-deficient Group 2 and 13 molecules, and hypervalent Period 3+ elements with expanded valence shells.


1. Protocol for Constructing Lewis Structures

Constructing a valid Lewis formula follows a systematic six-step procedure:

  1. Tabulate Total Valence Electrons (NvalN_{\text{val}}): Sum the valence electrons from all neutral atoms. Add one electron for each negative ionic charge; subtract one electron for each positive ionic charge.
  2. Identify the Central Atom: Place the least electronegative element at the center (excluding hydrogen, which is always terminal). Carbon is virtually always central; halogens are typically terminal.
  3. Form Single (σ\sigma) Bonds: Connect each terminal atom to the central atom using a single covalent bond, subtracting 22 electrons per bond from NvalN_{\text{val}}.
  4. Satisfy Terminal Octets: Distribute remaining electrons as lone pairs to give terminal atoms an octet (duet for hydrogen).
  5. Assign Remaining Electrons to the Central Atom: Place any surplus electrons on the central atom as nonbonding pairs.
  6. Form Multiple Bonds to Complete Octets: If the central atom lacks an octet, convert nonbonding lone pairs from terminal atoms into double or triple bonds.

2. Formal Charge Calculation & Structure Evaluation

Formal charge (FC) evaluates electron allocation assuming equal sharing across all covalent bonds:

FC=V−N−12B=V−lone pair electrons−number of bonds\text{FC} = V - N - \frac{1}{2}B = V - \text{lone pair electrons} - \text{number of bonds}

where VV is valence electrons in the isolated atom, NN is nonbonding electrons, and BB is shared bonding electrons. The sum of all formal charges equals the overall species charge (∑FCi=Qnet\sum \text{FC}_i = Q_{\text{net}}).

Formal Charge Criteria for Dominant Structures

  1. Structures minimizing formal charge magnitudes (00 preferred over ±1\pm 1, ±2\pm 2 disfavored) dominate.
  2. Negative formal charges must reside on the most electronegative atoms.
  3. Like formal charges on adjacent bonded atoms indicate electrostatic destabilization.

Diagnostic Comparison: The Cyanate Ion (OCN−\text{OCN}^-, 16 Valence Electrons)

StructureFC(O)\text{FC}(\text{O})FC(C)\text{FC}(\text{C})FC(N)\text{FC}(\text{N})Evaluation
I: [:O¨=C=N¨:]−[:\ddot{\text{O}}=\text{C}=\ddot{\text{N}}:]^-6−4−2=06 - 4 - 2 = 04−0−4=04 - 0 - 4 = 05−4−2=−15 - 4 - 2 = -1Minor contributor (charges 0 and -1)
II: [:O≡C−N¨:]−[:\text{O}\equiv\text{C}-\ddot{\text{N}}:]^-6−2−3=+16 - 2 - 3 = +14−0−4=04 - 0 - 4 = 05−6−1=−25 - 6 - 1 = -2Heavily disfavored (+1 on O, -2 on N)
III: [:O¨−C≡N:]−[:\ddot{\text{O}}-\text{C}\equiv\text{N}:]^-6−6−1=−16 - 6 - 1 = -14−0−4=04 - 0 - 4 = 05−2−3=05 - 2 - 3 = 0Dominant (places -1 on more EN oxygen)

Because oxygen (EN 3.44) is more electronegative than nitrogen (EN 3.04), Structure III is the dominant contributor.


3. Resonance Structures & the Resonance Hybrid

When a species can be represented by multiple valid Lewis structures differing solely in electron positions, these representations are resonance structures.

  • The Resonance Hybrid: The molecule does not oscillate among forms; it exists as a single, static resonance hybrid with delocalized π\pi electrons, lowering potential energy by the resonance stabilization energy.
  • Carbonate Ion (CO32−\text{CO}_3^{2-}, 24 e−e^-): Three equivalent resonance structures distribute one double bond and two single bonds across three oxygen atoms. All three C−O\text{C}-\text{O} bonds are identical in length (128 pm128\text{ pm}, between single 143 pm143\text{ pm} and double 122 pm122\text{ pm}). The calculated bond order is: Bond Order=Total Bonding Pairs in LinkageResonance Positions=1+1+23=43≈1.33\text{Bond Order} = \frac{\text{Total Bonding Pairs in Linkage}}{\text{Resonance Positions}} = \frac{1 + 1 + 2}{3} = \frac{4}{3} \approx 1.33 Each oxygen bears an effective formal charge of −23-\frac{2}{3}.
  • Nitrate (NO3−\text{NO}_3^-) and Ozone (O3\text{O}_3): Exhibit fractional bond orders of 1.331.33 and 1.51.5, respectively.

4. Exceptions to the Octet Rule

Three systematic classes of stable molecules depart from the octet rule:

1. Odd-Electron Molecules (Free Radicals)

Species with an odd total number of valence electrons possess an unpaired electron, imparting paramagnetism and high reactivity:

  • Nitric Oxide (NO\text{NO}, 11 e−e^-): :N˙=O¨::\dot{\text{N}}=\ddot{\text{O}}:, with the unpaired electron localized on nitrogen (less electronegative than oxygen).
  • Nitrogen Dioxide (NO2\text{NO}_2, 17 e−e^-): Paramagnetic brown gas that readily dimerizes to diamagnetic N2O4\text{N}_2\text{O}_4.

2. Incomplete Octets (Electron-Deficient Compounds)

Compounds of Group 2 (Be\text{Be}) and Group 13 (B,Al\text{B}, \text{Al}) are stable with fewer than eight valence electrons:

  • BeCl2\text{BeCl}_2 (vapor): Beryllium forms two single bonds (44 valence electrons).
  • BF3\text{BF}_3: Boron forms three single bonds (66 valence electrons). Forming a B=F\text{B}=\text{F} double bond would place an unfavorable +1+1 formal charge on fluorine.
  • Lewis Acid Behavior: Electron-deficient centers act as strong Lewis acids, reacting with Lewis bases like NH3\text{NH}_3 to form adducts via coordinate covalent bonds (F3B←NH3\text{F}_3\text{B}\leftarrow\text{NH}_3).

3. Expanded Octets (Hypervalent Molecules)

Central atoms from Period 3 and below (P,S,Cl,Br,I,Xe\text{P}, \text{S}, \text{Cl}, \text{Br}, \text{I}, \text{Xe}) accommodate 1010, 1212, or 1414 valence electrons:

  • PCl5\text{PCl}_5: 5 single bonds (1010 valence electrons).
  • SF6\text{SF}_6: 6 single bonds (1212 valence electrons).
  • XeF4\text{XeF}_4: 4 single bonds + 2 lone pairs (1212 valence electrons).
  • I3−\text{I}_3^-: 2 single bonds + 3 lone pairs (1010 valence electrons).

Why Period 2 Elements NEVER Expand Their Octets

Second-period elements (C,N,O,F\text{C}, \text{N}, \text{O}, \text{F}) cannot expand their octets because their valence shell (n=2n=2) contains only 2s2s and 2p2p subshells (maximum 8 electrons). The 3d3d subshell is separated by an enormous quantum energy gap, and their small atomic radii make coordinating five or six ligands sterically impossible.

Test Your Knowledge

What is the formal charge on the sulfur atom in a Lewis structure of the sulfite ion, SO₃²⁻, in which sulfur forms three single bonds to oxygen atoms and retains one nonbonding lone pair?

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Test Your Knowledge

Which of the following resonance contributors represents the dominant (most stable) Lewis structure for the cyanate ion, OCN⁻, and why?

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Test Your Knowledge

Experimental measurements show that all three carbon-oxygen bonds in the carbonate ion, CO₃²⁻, are identical in length (128 pm) and strength. What explains this observation?

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Test Your Knowledge

Why does sulfur readily form sulfur hexafluoride, SF₆, whereas oxygen cannot form oxygen hexafluoride, OF₆?

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