Molecular and Ionic Compound Structure — AP Chemistry Chem Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Lewis structures with formal charge calculations, VSEPR theory for molecular geometry prediction, orbital hybridisation, and molecular polarity plus intermolecular force classification.

You should already know: High-school chemistry, Algebra 2.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Molecular and Ionic Compound Structure?

Molecular and ionic compound structure is the study of how atoms bond and arrange in 2D and 3D space to form stable compounds, and how these arrangements dictate all physical and chemical properties from boiling point to reaction reactivity. This content makes up 7-9% of your total AP Chemistry exam score, per the College Board CED, and is a foundational prerequisite for nearly all later units including thermodynamics, kinetics, and organic chemistry. You will frequently see both multiple-choice and free-response questions asking you to justify property differences using structural concepts covered in this guide.

2. Lewis structures and formal charge

Lewis structures are 2D representations of a compound’s valence electrons, showing bonding pairs (shared between two atoms) and lone pairs (localized on a single atom). The octet rule states that most main-group atoms (period 2 and above, except H) tend to form bonds to reach 8 valence electrons, though there are three key exceptions: hydrogen only needs 2 valence electrons, period 3 and higher elements can have expanded octets (>8 electrons) due to available d-orbitals, and radical species have an odd number of total valence electrons.

Formal charge (FC) is used to identify the most stable Lewis structure for a compound with multiple possible resonance forms. It is calculated using the formula: $$FC=V-\left(LP+\frac{1}{2}BP\right)$$ Where $V$ = number of valence electrons of the neutral free atom, $LP$ = number of lone pair electrons on the atom in the structure, and $BP$ = number of bonding electrons attached to the atom. The most stable resonance structure has formal charges closest to 0, with any negative formal charges located on the most electronegative atoms.

Worked Example

Draw the most stable Lewis structure for the sulfate ion ($S{O}_4^{2-}$):

1. Total valence electrons: $6(S)+4\times 6(O)+2(negativecharge)=32$
2. Test structure 1: S with 4 single bonds to O: FC on S = $6-(0+8/2)=+2$, FC on each O = $6-(6+2/2)=-1$, total charge = $+2+4(-1)=-2$, which matches the ion charge but has a high positive FC on S.
3. Test structure 2: S with 2 double bonds to O and 2 single bonds to O: FC on S = $6-(0+12/2)=0$, FC on double-bonded O = $6-(4+4/2)=0$, FC on single-bonded O = $-1$, total charge = $2(-1)=-2$. This structure has formal charges closest to 0, so it is the most stable. Note that S is in period 3, so its expanded octet (12 electrons) is allowed.

3. VSEPR theory and molecular geometry

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the 3D shape of molecules based on the principle that negatively charged electron domains (bonding pairs, where single/double/triple bonds all count as 1 domain, plus lone pairs) repel each other to maximize the distance between them.

• Electron Domain Geometry (EDG): The shape formed by all electron domains (bonding + lone pairs) around the central atom.
• Molecular Geometry: The shape formed only by the bonding domains (ignoring lone pairs), which describes the actual 3D arrangement of atoms in the molecule.

Lone pair repulsion is stronger than bonding pair repulsion, following the order: $LP-LP>LP-BP>BP-BP$. This causes bond angles to be smaller than the ideal EDG angle when lone pairs are present.

Worked Example

Predict the EDG, molecular geometry, and bond angle for ammonia ($N{H}_3$):

1. Count electron domains on central N: 3 single bonds to H + 1 lone pair = 4 total domains.
2. EDG for 4 domains is tetrahedral, with an ideal bond angle of 109.5°.
3. Molecular geometry for 3 bonding domains and 1 lone pair is trigonal pyramidal. The lone pair repels the bonding pairs more strongly, so the actual H-N-H bond angle is ~107°.

4. Hybridisation

Hybridisation describes the mixing of pure atomic orbitals (s, p, d) to form identical hybrid orbitals that match the electron domain geometry of the central atom, explaining why bonds to identical terminal atoms have equal energy and bond length. Hybridisation correlates directly to the number of electron domains around the central atom:

Hybrid orbitals only form sigma (σ) bonds (single bonds, or the first bond in a double/triple bond). The remaining unhybridized p or d orbitals form pi (π) bonds, which are the second and third bonds in double and triple bonds respectively.

Worked Example

Identify the hybridisation of each C atom in ethene ($C_2{H}_4$), which has a C=C double bond:

1. Each C atom has 3 electron domains: 2 single bonds to H, 1 double bond to C.
2. 3 electron domains correspond to $s{p}^2$ hybridisation.
3. Each C has 1 unhybridized p orbital, which overlaps side-by-side to form the pi bond in the double bond.

5. Polarity and intermolecular forces

Molecular Polarity

A covalent bond is polar if the electronegativity difference between the two atoms is >0.4, creating a dipole moment (partial positive charge on the less electronegative atom, partial negative on the more electronegative atom). A molecule is polar if it has a net dipole moment, which requires:

1. At least one polar covalent bond
2. Asymmetric molecular geometry that prevents individual bond dipoles from canceling out.

Symmetric geometries (linear with two identical terminal atoms, trigonal planar with three identical terminals, tetrahedral with four identical terminals, square planar with four identical terminals) will always cancel dipoles even if individual bonds are polar.

Intermolecular Forces (IMFs)

IMFs are attractive forces between molecules, weaker than intramolecular covalent or ionic bonds, that determine physical properties including boiling point, melting point, solubility, and vapor pressure. They are listed below from strongest to weakest:

1. Ion-dipole: Attraction between a charged ion and a polar molecule, strongest IMF, found in ionic solutions (e.g. $N{a}^{+}$ and water molecules in salt water).
2. Hydrogen bonding: A special type of dipole-dipole attraction that occurs only when a hydrogen atom is covalently bonded directly to N, O, or F (highly electronegative small atoms) and interacts with a lone pair on a separate N/O/F atom. It is responsible for water’s unusually high boiling point.
3. Dipole-dipole: Attraction between the partial positive end of one polar molecule and the partial negative end of another polar molecule.
4. London Dispersion Forces (LDF): Temporary attractive force caused by random fluctuations in electron clouds that create temporary dipoles. LDFs are present in all molecules, and are the only IMF in nonpolar molecules and noble gases. LDF strength increases with molar mass/number of electrons, as larger electron clouds are more easily polarized.

Key Rule

"Like dissolves like": Polar solutes dissolve in polar solvents, and nonpolar solutes dissolve in nonpolar solvents, because solute-solvent IMFs must be strong enough to overcome solute-solute and solvent-solvent IMFs to form a solution.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting period 3 and higher elements can have expanded octets when drawing Lewis structures, leading to unnecessarily high formal charges. Why: Students memorize the octet rule as an absolute. Correct move: Always check the period of the central atom first; period ≥3 elements can hold >8 valence electrons to minimize formal charge.
• Wrong move: Confusing electron domain geometry and molecular geometry on VSEPR questions. Why: Students skip counting lone pairs and only look at bonding atoms to determine shape. Correct move: Always calculate total electron domains first to get EDG, then adjust molecular geometry for the number of lone pairs.
• Wrong move: Assuming all molecules with polar bonds are polar. Why: Students ignore geometric symmetry that cancels dipole moments. Correct move: Draw the 3D VSEPR shape first, then sum dipole vectors to check for a net dipole before labeling a molecule as polar.
• Wrong move: Counting double/triple bonds as multiple electron domains for VSEPR or hybridisation calculations. Why: Students confuse number of bonds with number of domains. Correct move: Each multiple bond counts as exactly 1 electron domain, regardless of bond order.
• Wrong move: Misidentifying hydrogen bonding in molecules where H is bonded to C, not N/O/F. Why: Students see H in a molecule and assume H-bonding exists. Correct move: Only H covalently bonded directly to N, O, or F can act as a hydrogen bond donor, and the acceptor must have a lone pair on N/O/F.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following is the most stable resonance structure for the nitrate ion ($N{O}_3^{-}$)? A) N with 3 single bonds to O, FC on N = +2, each O = -1 B) N with 1 double bond and 2 single bonds to O, FC on N = 0, double-bonded O = 0, single-bonded O = -1 each C) N with 2 double bonds and 1 single bond to O, FC on N = -1, double-bonded O = 0, single-bonded O = -1 D) N with 3 double bonds to O, FC on N = -3, each O = +1

Solution

Correct answer: B. Total valence electrons for $N{O}_3^{-}$ = $5+3\times 6+1=24$. Option B has a total formal charge sum of -1, matching the ion charge, with formal charges closest to 0. N is in period 2 so it cannot have an expanded octet, eliminating options C (9 electrons on N) and D (12 electrons on N). Option A has a high +2 formal charge on N, making it less stable than B.

Question 2 (Short Free Response)

A student claims that $CHC{l}_3$ (trichloromethane) is nonpolar because it has a tetrahedral electron domain geometry, which is symmetric. (a) Identify the error in the student’s reasoning. (b) State the molecular geometry of $CHC{l}_3$ and whether it is polar or nonpolar, justifying your answer.

Solution

(a) The student confuses overall geometric symmetry with symmetric distribution of electron density. Tetrahedral molecules only cancel dipole moments if all four terminal atoms are identical, which is not the case for $CHC{l}_3$. (b) Molecular geometry is tetrahedral (4 bonding domains, 0 lone pairs on central C). It is polar: the C-Cl bonds are significantly more polar than the C-H bond, so individual bond dipoles do not cancel out, resulting in a net molecular dipole.

Question 3 (Long Free Response)

The boiling points of three compounds are given below:

• Methane ($C{H}_4$): -161.5 °C
• Methanol ($C{H}_{3}OH$): 64.7 °C
• Fluoromethane ($C{H}_{3}F$): -78.4 °C (a) Identify the strongest intermolecular force present in each pure compound. (b) Justify the difference in boiling points between methanol and fluoromethane, using your answers from part (a). (c) Explain why methane has the lowest boiling point of the three.

Solution

(a) $C{H}_4$: London Dispersion Forces (nonpolar molecule, no polar bonds). $C{H}_{3}OH$: Hydrogen bonding (H is covalently bonded directly to O). $C{H}_{3}F$: Dipole-dipole forces (polar molecule, but H is bonded to C, not F, so no hydrogen bonding). (b) Hydrogen bonding is significantly stronger than dipole-dipole forces. Stronger intermolecular forces require more thermal energy to overcome, so methanol has a much higher boiling point than fluoromethane. (c) Methane is nonpolar, so it only exhibits weak London Dispersion Forces, which are weaker than both hydrogen bonding and dipole-dipole forces. It also has a very low molar mass (16 g/mol), leading to even weaker LDFs, so it requires the least energy to boil.

8. Quick Reference Cheatsheet

9. What's Next

This unit is the foundational building block for nearly all subsequent AP Chemistry topics. You will use VSEPR and polarity to predict reaction mechanisms in inorganic and organic chemistry units, intermolecular forces to explain thermodynamics of solution formation, phase changes, and colligative properties, and hybridisation to analyze bonding in complex coordination compounds later in the syllabus. 30% of AP Chemistry free-response questions ask you to justify physical property differences, so mastering the content in this guide will directly boost your exam score.

If you struggle with any of the concepts in this guide, from formal charge calculations to intermolecular force identification, you can get personalized, step-by-step help from Ollie, our AI tutor, at any time. You can also find more practice questions and full-length timed mock AP Chemistry exams on the homepage to test your mastery before test day.