Chemical Bonding I: Lewis Structures & Molecular Geometry

Lewis Symbols, the Octet Rule, and Ionic Bonding

A Lewis symbol shows an element's chemical symbol surrounded by dots representing valence electrons only, which lets us track bonding without worrying about core electrons. Most main-group atoms are most stable with eight valence electrons (an octet), matching the electron configuration of the nearest noble gas; hydrogen and helium instead aim for a duet. Ionic bonding arises from the complete transfer of one or more electrons from a low-ionization-energy metal to a high-electron-affinity nonmetal, generating oppositely charged ions that attract each other electrostatically in a crystal lattice. The strength of that attraction is quantified by lattice energy, the energy released when gaseous ions come together to form one mole of solid ionic compound.

• •Lattice energy increases (more exothermic/more negative) as ionic charges increase — Coulomb's law depends on q₁q₂.
• •Lattice energy increases as ionic radii decrease, since electrostatic attraction falls off with distance (1/r).
• •MgO (2+/2−, small ions) has a much larger lattice energy than NaCl (1+/1−, larger ions).
• •Born–Haber cycles use Hess's law to relate lattice energy to sublimation, ionization, dissociation, electron affinity, and formation enthalpies.

Covalent Bonding, Bond Length, and Bond Energy

A covalent bond forms when two atoms share one or more pairs of electrons, lowering the system's potential energy relative to separated atoms. As more electron pairs are shared between the same two atoms, the bond gets shorter and stronger: single < double < triple in both bond order and bond energy, but longer in bond length in the reverse order. Bond energy is the energy required to break one mole of a particular bond in the gas phase, and it can be used in Hess's-law-style estimates of reaction enthalpy. Electronegativity, the tendency of an atom to attract shared electrons toward itself, determines how symmetrically the electron pair is shared.

• •Bond order 1 (single) → longest, weakest; bond order 3 (triple) → shortest, strongest for a given pair of atoms.
• •Nonpolar covalent: ΔEN ≈ 0 (e.g., Cl–Cl); polar covalent: 0 < ΔEN < ~2.0 (e.g., H–Cl); ionic: ΔEN ≳ 2.0 (e.g., Na–Cl), though the cutoffs are approximate guidelines, not hard limits.
• •Dipole moment (μ) measures the size of the bond/molecular dipole; it depends on both charge separation and distance.
• •Electronegativity increases up and to the right on the periodic table (excluding noble gases); F is the most electronegative element.

The Lewis Structure Algorithm

Drawing a correct Lewis structure is a reliable, stepwise process rather than guesswork, and mastering the steps prevents most common errors on exams. Begin by summing all valence electrons (add electrons for a negative ion charge, subtract for a positive charge). Next, arrange atoms into a skeleton structure, placing the least electronegative atom (other than hydrogen, which is always terminal) in the center, connected to the other atoms by single bonds. Then distribute remaining electrons to give outer (terminal) atoms complete octets first, place any leftover electrons as lone pairs on the central atom, and finally, if the central atom still lacks an octet, convert lone pairs on terminal atoms into additional shared pairs (double or triple bonds) until every atom (except recognized exceptions) has eight electrons.

• •Step 1: Count total valence electrons, adjusting for ionic charge.
• •Step 2: Build the skeleton — least electronegative atom central, H and F always terminal.
• •Step 3: Give terminal atoms octets using lone pairs.
• •Step 4: Place leftover electrons on the central atom.
• •Step 5: If the central atom is short of an octet, form multiple bonds by converting terminal lone pairs into bonding pairs.

Formal Charge and Choosing the Best Structure

When more than one valid Lewis structure can be drawn for the same skeleton, formal charge helps identify which structure best represents the real distribution of electron density. Formal charge assigns electrons as if bonds were shared perfectly equally, then compares that count to the free-atom valence electron count. The best structure minimizes formal charges overall, places any negative formal charge on the more electronegative atom, and avoids like-charge adjacent atoms and unnecessary charge separation. Formal charge is a bookkeeping tool, not a measure of real charge, but it is an excellent, exam-tested way to rank competing Lewis structures.

• •Preferred structures have formal charges as close to zero as possible.
• •If nonzero formal charges are unavoidable, negative formal charge should sit on the more electronegative atom.
• •Example: for CO₂, O=C=O (all formal charges zero) beats structures with a C≡O and a charged terminal O.
• •Formal charges must sum to the overall ionic charge of the species.

Resonance and Delocalization

Some molecules and ions cannot be described accurately by a single Lewis structure because equivalent alternative placements of multiple bonds and lone pairs exist. These resonance structures differ only in electron placement, not atom position, and the true structure is a resonance hybrid — a blend that gives bond lengths and strengths intermediate between the individual contributing structures. Experimentally, this shows up as all C–O bonds in the carbonate ion being identical and intermediate in length between a C–O single and C=O double bond, rather than two long and one short. Resonance structures are connected in Lewis diagrams by double-headed arrows (⇌), signaling they are not separate species in equilibrium but different depictions of one real structure.

• •Classic resonance examples: O₃, NO₃⁻, CO₃²⁻, SO₂, benzene.
• •Equivalent resonance structures (same energy) contribute equally to the hybrid.
• •Delocalization spreads electron density over multiple atoms, generally increasing stability.
• •Resonance structures are not rapidly interconverting isomers — they are alternate depictions of the same real molecule.

Exceptions to the Octet Rule

While the octet rule is an excellent guideline, several categories of molecules systematically violate it and must be recognized rather than forced into an octet. Electron-deficient (incomplete octet) molecules occur mainly with boron and beryllium, whose central atoms are stable with only six or four valence electrons, respectively, because they lack enough electrons to reach an octet even after bonding. Odd-electron (free-radical) species like NO and NO₂ have a total valence-electron count that is odd, making a full octet on every atom impossible. Expanded-octet species occur only for central atoms in period 3 or below, which have accessible empty d-orbitals (or sufficient orbital space) allowing them to accommodate more than eight electrons.

• •Incomplete octet examples: BF₃ (6 e⁻ on B), BeCl₂ (4 e⁻ on Be) — both are strong Lewis acids because of this deficiency.
• •Odd-electron example: NO has 11 valence electrons, so one atom must have an unpaired electron.
• •Expanded octet examples: PCl₅ (10 e⁻ on P), SF₆ (12 e⁻ on S), ClF₃, XeF₄ — never occurs for period-2 elements (C, N, O, F).
• •Expanded octets are only possible when the central atom is from period 3 or higher.

VSEPR Theory: Electron-Domain and Molecular Geometry

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts 3-D molecular shape by assuming that electron domains (bonding pairs, whether single/double/triple counts as one domain, and lone pairs) around a central atom arrange themselves to minimize mutual repulsion. The number of electron domains determines the parent electron-domain geometry, while the arrangement of atoms only (ignoring lone pairs) gives the molecular geometry, which is what we report as the shape. Lone pairs occupy more space than bonding pairs because they are held closer to just one nucleus, so they compress adjacent bond angles below the ideal value and are placed in equatorial positions when possible (trigonal bipyramidal) to minimize 90° repulsions.

• •2 domains → linear (180°); 3 domains → trigonal planar (120°); 4 domains → tetrahedral (109.5°); 5 domains → trigonal bipyramidal (90°/120°/180°); 6 domains → octahedral (90°/180°).
• •Repulsion strength order: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair.
• •In trigonal bipyramidal geometry, lone pairs always occupy equatorial positions first (less 90° crowding).
• •Double and triple bonds count as a single electron domain for VSEPR purposes, though they do occupy slightly more space than single bonds.

Full VSEPR Reference Table (Notation AXₙEₘ)

Chemists use the shorthand AXₙEₘ, where A is the central atom, X is a bonded atom, and E is a lone pair, to catalog every possible geometry systematically. Memorizing this table by domain count is one of the highest-yield study tasks for this unit because exam questions routinely give a molecular formula and ask for the electron-domain geometry, molecular geometry, and bond angle together.

• •AX₂ (2 domains, 0 LP): linear, 180° — e.g., CO₂, BeCl₂.
• •AX₃ (3 domains, 0 LP): trigonal planar, 120° — e.g., BF₃, SO₃.
• •AX₂E (3 domains, 1 LP): bent, ~118° — e.g., SO₂, O₃.
• •AX₄ (4 domains, 0 LP): tetrahedral, 109.5° — e.g., CH₄.
• •AX₃E (4 domains, 1 LP): trigonal pyramidal, ~107° — e.g., NH₃.
• •AX₂E₂ (4 domains, 2 LP): bent, ~104.5° — e.g., H₂O.
• •AX₅ (5 domains, 0 LP): trigonal bipyramidal, 90°/120°/180° — e.g., PCl₅.
• •AX₄E (5 domains, 1 LP): seesaw, ~90°/~117°/~173° — e.g., SF₄.
• •AX₃E₂ (5 domains, 2 LP): T-shaped, ~90°/~175° — e.g., ClF₃.
• •AX₂E₃ (5 domains, 3 LP): linear, 180° — e.g., XeF₂, I₃⁻.
• •AX₆ (6 domains, 0 LP): octahedral, 90° — e.g., SF₆.
• •AX₅E (6 domains, 1 LP): square pyramidal, ~90° — e.g., BrF₅.
• •AX₄E₂ (6 domains, 2 LP): square planar, 90° — e.g., XeF₄.

Predicting Molecular Polarity

A molecule is polar overall only if it has polar bonds AND those individual bond dipoles do not cancel by symmetry; a molecule can contain polar bonds yet be nonpolar overall if its geometry is symmetric enough. To decide, draw the correct Lewis structure and geometry, sketch each bond dipole vector pointing toward the more electronegative atom, and determine whether the vector sum is zero (nonpolar) or nonzero (polar). Lone pairs on the central atom also contribute to the net dipole and typically prevent cancellation, which is why bent and pyramidal molecules are almost always polar even with identical terminal atoms.

• •CO₂ (linear, symmetric): dipoles cancel → nonpolar overall despite polar C=O bonds.
• •H₂O (bent) and NH₃ (trigonal pyramidal): dipoles do not cancel → polar.
• •CCl₄ (tetrahedral, symmetric): dipoles cancel → nonpolar; CHCl₃ (asymmetric): dipoles do not cancel → polar.
• •Symmetric AXₙ molecules with identical terminal atoms and no lone pairs on A are generally nonpolar.

Estimating Reaction Enthalpy from Bond Energies

Because breaking bonds requires energy input and forming bonds releases energy, the overall enthalpy change of a gas-phase reaction can be estimated as the energy needed to break all bonds in the reactants minus the energy released forming all bonds in the products. This bond-energy method uses tabulated average bond energies (which vary slightly by molecular environment, so results are approximations, not exact thermodynamic values) and is especially useful for reactions where standard enthalpies of formation are not readily available.

• •ΔH_rxn ≈ Σ(bond energies broken, reactants) − Σ(bond energies formed, products).
• •A negative ΔH_rxn indicates the products have stronger/more bonds overall — an exothermic reaction.
• •This method gives only an estimate because tabulated bond energies are averages across many compounds.
• •Higher bond order corresponds to higher bond energy, so breaking a triple bond (e.g., N≡N) costs much more energy than breaking a single bond.

Draw the Lewis structure for the carbonate ion, CO₃²⁻, including formal charges, and note its resonance.

1. 1Count valence electrons: C(4) + 3×O(6) + 2 (for the 2− charge) = 4 + 18 + 2 = 24 electrons.
2. 2Build the skeleton: C is less electronegative and becomes the central atom, bonded by single bonds to 3 O atoms (6 electrons used, 18 remain).
3. 3Give each terminal O three lone pairs to complete its octet: 3 O × 6 e⁻ = 18 electrons used — all 24 electrons are now placed, but C only has 6 electrons (no octet).
4. 4Convert one O lone pair into a second bond to C, forming one C=O double bond; this satisfies C's octet using only atoms already present.
5. 5Calculate formal charges: C = 4 − 0 − ½(8) = 0; double-bonded O = 6 − 4 − ½(4) = 0; single-bonded O's = 6 − 6 − ½(2) = −1 each.
6. 6Formal charges sum to 0 + 0 + (−1) + (−1) = −2, matching the ion's charge — structure is valid.
7. 7Because any of the three O atoms could bear the double bond equally, draw three resonance structures connected by ⇌; the real structure is a hybrid with all three C–O bonds equal and intermediate in length.

CO₃²⁻: central C double-bonded to one O and single-bonded to two O⁻ atoms, three equivalent resonance structures, all formal charges 0 for C, 0 for =O, and −1 for each −O.

Determine the electron-domain geometry, molecular geometry, and bond angle for SF₄.

1. 1Count valence electrons: S(6) + 4×F(7) = 6 + 28 = 34 electrons.
2. 2Build skeleton: S central, single-bonded to 4 F atoms (8 electrons used, 26 remain).
3. 3Give each F three lone pairs to complete octets: 4 × 6 = 24 electrons used (30 total so far), leaving 2 electrons.
4. 4Place the last lone pair on S, since S is period 3 and can hold an expanded octet.
5. 5Count domains on S: 4 bonding domains + 1 lone pair = 5 total domains → electron-domain geometry is trigonal bipyramidal.
6. 6Place the lone pair in an equatorial position (minimizes 90° lone-pair/bonding-pair repulsions) → molecular geometry AX₄E is 'seesaw'.
7. 7Bond angles are compressed from ideal: ~173° (axial–axial), ~117° (equatorial–equatorial), ~90° (axial–equatorial).

SF₄: electron-domain geometry trigonal bipyramidal; molecular geometry seesaw (AX₄E); angles ≈ 90°, ≈117°, ≈173° due to equatorial lone-pair compression.

Is SO₂ polar or nonpolar? Justify using Lewis structure and geometry.

1. 1Valence electrons: S(6) + 2×O(6) = 18. Central S bonded to two O atoms.
2. 2Best Lewis structure (minimizing formal charge): one S=O double bond, one S–O single bond, and a lone pair on S, with resonance between the two possible placements.
3. 3Count domains on S: 2 bonding domains + 1 lone pair = 3 total domains → electron-domain geometry trigonal planar.
4. 4Molecular geometry (AX₂E, ignoring the lone pair for shape) is bent, with a bond angle near 118°.
5. 5Each S–O bond is polar because O is more electronegative than S, and the bent shape (plus the lone pair) prevents the two bond dipoles from canceling.
6. 6Because the bond dipoles reinforce rather than cancel, SO₂ has a net nonzero dipole moment.

SO₂ is polar: bent (AX₂E) geometry means the two polar S–O bond dipoles do not cancel, giving a net molecular dipole.

Use average bond energies to estimate ΔH for H₂(g) + Cl₂(g) → 2 HCl(g), given BE(H–H)=436 kJ/mol, BE(Cl–Cl)=243 kJ/mol, BE(H–Cl)=431 kJ/mol.

1. 1Identify bonds broken (reactants): 1 mol H–H and 1 mol Cl–Cl.
2. 2Energy to break bonds = 436 + 243 = 679 kJ (energy absorbed, positive).
3. 3Identify bonds formed (products): 2 mol H–Cl.
4. 4Energy released forming bonds = 2 × 431 = 862 kJ.
5. 5Apply ΔH ≈ ΣBE(broken) − ΣBE(formed) = 679 − 862 = −183 kJ.
6. 6Negative sign indicates the reaction is exothermic, consistent with forming a stronger set of bonds overall.

ΔH ≈ −183 kJ (exothermic).

Explain, using formal charge, why :N≡C–O: (formal charges N=−1, C=0, O=+1... check) is disfavored versus the standard cyanate structure for OCN⁻, and identify the best Lewis structure.

1. 1Total valence electrons for OCN⁻: O(6)+C(4)+N(5)+1(charge) = 16 electrons.
2. 2Possible skeletons keep C central (least electronegative, excluding H): O=C=N⁻ type structures with varying multiple bonds.
3. 3Structure 1: O=C=N⁻ (double bond each side): FC(O)=6−4−2=0, FC(C)=4−0−4=0, FC(N)=5−4−2=−1. Sum=−1 ✓.
4. 4Structure 2: ⁻O–C≡N (single bond to O, triple to N): FC(O)=6−6−1=−1, FC(C)=4−0−4=0, FC(N)=5−2−3=0. Sum=−1 ✓.
5. 5Structure 3: O≡C–N²⁻ type placing triple bond on O and single on N gives large formal charges (e.g., N = −2), which is disfavored.
6. 6Compare structures 1 and 2: both have all-small formal charges: structure 2 places the single negative charge on O (more electronegative than N), which is generally preferred.
7. 7Best structure is ⁻O–C≡N (cyanate ion), matching known experimental bond lengths; resonance with O=C=N⁻ contributes some character too.

The best Lewis structure for OCN⁻ is ⁻O–C≡N, since it places the formal negative charge on the more electronegative oxygen atom while keeping all formal charges minimal.

Q1. Which pair of ions would be predicted to have the largest lattice energy?

Q2. How many total valence electrons should be used to draw the Lewis structure of SO₄²⁻?

Q3. What is the formal charge on nitrogen in NH₄⁺, where N forms four single bonds to H and has no lone pairs?

Q4. Which species is a well-known exception to the octet rule due to having an odd number of valence electrons?

Q5. Which of the following can have an expanded octet on its central atom?

Q6. A central atom with 4 bonding domains and 1 lone pair (AX₄E) has what molecular geometry?

Q7. Which molecule is nonpolar despite containing polar bonds?

Q8. What is the ideal bond angle in a trigonal planar electron-domain geometry?

Q9. Why do lone pairs compress bond angles below the ideal tetrahedral value in H₂O (104.5° vs. 109.5°)?

Q10. In the best Lewis structure for CO₂, what is the formal charge on each oxygen atom?

Q11. Which best describes a resonance hybrid?

Q12. Using bond energies BE(N≡N)=941 kJ/mol, BE(H–H)=436 kJ/mol, BE(N–H)=391 kJ/mol, estimate ΔH for N₂(g) + 3H₂(g) → 2NH₃(g).

Q13. An AX₃E₂ molecule (5 total electron domains, 2 lone pairs) has what molecular geometry?

Q14. Which statement about BF₃ is correct?

Q15. Which of the following molecules is predicted to be polar?