Intermolecular forces — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: London dispersion forces, dipole-dipole interactions, hydrogen bonding, ion-dipole forces, Coulomb’s law applications to intermolecular attractions, and relative strength comparisons for predicting bulk material properties.

You should already know: Polar vs nonpolar covalent bond classification. Net molecular dipole moment calculation. Coulomb's law for electrostatic interactions.

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 / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Intermolecular forces?

Intermolecular forces (often abbreviated IMFs) are electrostatic attractive forces that act between separate discrete molecules, atoms (for monatomic species like noble gases), or ions, distinct from intramolecular bonds (covalent, ionic, metallic) that hold atoms together within a single chemical unit. All intermolecular attractions arise from Coulombic interactions between partial or full charges, and are universally weaker than intramolecular chemical bonds: typical IMFs range from 1 to 100 kJ/mol, while covalent bonds exceed 150 kJ/mol. This difference means phase changes (melting, boiling) only overcome IMFs, not break intramolecular bonds. A common synonym is van der Waals forces, though this term usually refers only to IMFs between neutral molecules, excluding ion-dipole forces. According to the AP Chemistry CED, Unit 3 (Intermolecular Forces and Properties) makes up 18-22% of total exam score, with IMFs as the core foundational concept for this unit. IMFs are tested in both MCQ and FRQ sections, most frequently as the reasoning behind predictions of bulk properties like boiling point, solubility, and vapor pressure.

2. London Dispersion Forces

London dispersion forces (LDF) are the weakest class of intermolecular forces, arising from temporary induced dipoles formed by random fluctuations in electron cloud distribution. LDF are present between all atoms, molecules, and ions, regardless of polarity, because every species has an electron cloud that can fluctuate. Strength of LDF depends on polarizability (how easily the electron cloud is distorted), which increases with total number of electrons (and thus molar mass). For molecules of the same molar mass, LDF strength also depends on molecular shape: linear, unbranched molecules have more surface area for intermolecular contact than spherical branched isomers, leading to stronger LDF. The potential energy of LDF attraction follows: $$V\propto -\frac{{\alpha }_1{\alpha }_2}{r^6}$$ where ${\alpha }_1,{\alpha }_2$ are polarizabilities of the two interacting species, and $r$ is the distance between them. The inverse $r^6$ relationship means LDF attraction drops off very quickly as molecules move apart.

Worked Example

Arrange the following species in order of increasing strength of London dispersion forces at the same temperature and pressure: neon (Ne), n-pentane (${\text{C}}_5{\text{H}}_{12}$), neopentane (${\text{C}}_5{\text{H}}_{12}$), and krypton (Kr). Explain your reasoning.

1. First calculate total electron count for each species, the base determinant of polarizability: Ne (10 e⁻), Kr (36 e⁻), both pentane isomers (42 e⁻).
2. Higher total electron count gives higher polarizability, so pentane isomers will have stronger LDF than Kr, which is stronger than Ne.
3. Between the two pentane isomers (same electron count/molar mass), n-pentane is unbranched and linear, giving more surface area for intermolecular contact than spherical, branched neopentane.
4. Final order from weakest to strongest LDF: $\text{Ne}<\text{Kr}<\text{neopentane}<\text{n-pentane}$.

Exam tip: Always check for same-molar-mass isomers first when comparing LDF strength — if molar mass is identical, the difference comes from surface area, not another type of IMF.

3. Dipole-Dipole Interactions and Hydrogen Bonding

Dipole-dipole interactions are attractive forces between the permanent dipoles of polar neutral molecules. Because dipoles are permanent (not temporary induced dipoles), dipole-dipole forces are stronger than LDF for molecules of similar molar mass (similar polarizability). The potential energy of dipole-dipole attraction follows: $$V\propto -\frac{{\mu }_1^2{\mu }_2^2}{r^{6}kT}$$ where $\mu$ is the permanent dipole moment of each molecule, $k$ is Boltzmann’s constant, and $T$ is absolute temperature. Larger dipole moments give stronger attraction, and higher temperature disrupts dipole alignment, weakening net attraction.

Hydrogen bonding is a special, exceptionally strong subclass of dipole-dipole interactions. Hydrogen bonding only occurs when a hydrogen atom is covalently bonded to N, O, or F (highly electronegative, small atomic radius), creating a very large partial positive charge on H. This partially positive H is attracted to a lone pair of electrons on a N, O, or F in a neighboring molecule. The small size of H lets the two charged species get very close, leading to much stronger attraction (20-40 kJ/mol) than regular dipole-dipole (1-5 kJ/mol).

Worked Example

Which of the following molecules can form hydrogen bonds with other identical molecules? For each that can, state the requirement it meets. ${\text{CH}}_3{\text{OCH}}_3$ (dimethyl ether), ${\text{CH}}_3{\text{CH}}_2\text{OH}$ (ethanol), $\text{HI}$, ${\text{NH}}_3$.

1. Hydrogen bonding between identical molecules requires at least one H covalently bonded directly to N, O, or F (the donor) and a N/O/F with a lone pair (the acceptor) on the other molecule.
2. Dimethyl ether has O (acceptor) but no H bonded directly to O (all H are bonded to C), so it cannot form hydrogen bonds between identical molecules.
3. Ethanol has a terminal -OH group, with H bonded directly to O, so it can form hydrogen bonds between molecules: the H of one ethanol is attracted to the O of another.
4. HI has H bonded to I, which is not N/O/F, so it cannot form hydrogen bonds.
5. Ammonia (${\text{NH}}_3$) has all H bonded directly to N, so it can form hydrogen bonds between molecules. Only ethanol and ammonia can form hydrogen bonds between identical molecules.

Exam tip: For hydrogen bonding between two different molecules (e.g. ethanol and water), you only need a donor H (bonded to N/O/F) on one molecule and an acceptor N/O/F on the other — both do not need a donor H.

4. Ion-Dipole Forces

Ion-dipole forces are the strongest class of intermolecular forces, arising from electrostatic attraction between a full charged ion (cation or anion) and a permanent dipole of a polar neutral molecule. Ion-dipole forces range up to ~100 kJ/mol, comparable to weak intramolecular bonds, and are the key force driving the dissolution of ionic compounds in polar solvents like water. The potential energy of ion-dipole attraction follows: $$V\propto -\frac{z\mu }{r^2}$$ where $z$ is the charge of the ion, $\mu$ is the dipole moment of the polar molecule, and $r$ is the distance between the ion and dipole center. Higher ion charge and larger dipole moment give stronger attraction, and because attraction follows an inverse $r^2$ relationship (instead of $r^6$ for neutral IMFs), ion-dipole attraction falls off much more slowly with distance, leading to a stronger net interaction.

Worked Example

Which arrangement gives the strongest ion-dipole attraction between a ${\text{Ca}}^{2+}$ cation and a water molecule? Explain your reasoning: (A) ${\text{Ca}}^{2+}$ near the partially positive H end of ${\text{H}}_2\text{O}$, (B) ${\text{Ca}}^{2+}$ near the partially negative O end of ${\text{H}}_2\text{O}$, (C) ${\text{Ca}}^{2+}$ halfway between the O end and the H end, perpendicular to the H-O-H bond angle.

1. Ion-dipole attraction follows Coulomb’s law: opposite charges attract, like charges repel.
2. Water is a polar molecule with a partial negative charge on the more electronegative O atom, and partial positive charges on the two H atoms.
3. ${\text{Ca}}^{2+}$ is a positively charged cation, so it will experience the strongest attraction to the partially negative end of the water dipole, which is the O atom.
4. Arrangement B gives the strongest ion-dipole attraction.

Exam tip: When asked to explain why an ionic compound dissolves in water, always explicitly name ion-dipole forces between ions and water molecules as the key stabilizing interaction, do not only write "like dissolves like".

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming branched alkanes have stronger London dispersion forces than straight-chain alkanes of the same molecular formula. Why: Students confuse branching with higher polarity, forgetting branching reduces surface area contact between molecules. Correct move: For isomers with identical molar mass, always assign stronger LDF to the less branched (more linear) isomer.
• Wrong move: Classifying hydrogen bonding as an intramolecular covalent bond on the AP exam. Why: Organic chemistry courses discuss intramolecular hydrogen bonding, leading students to confuse it with a full chemical bond. Correct move: Unless the question explicitly specifies intramolecular hydrogen bonding, classify hydrogen bonding as a special type of dipole-dipole intermolecular force.
• Wrong move: Claiming any molecule containing N, O, or F can form hydrogen bonds. Why: Students memorize that N/O/F are required for hydrogen bonding and forget the requirement that H must be covalently bonded directly to N/O/F. Correct move: Always check the bonding of H before confirming hydrogen bonding; if H is bonded to C, no hydrogen bond can form even if N/O/F is present elsewhere.
• Wrong move: Ranking IMF strength by only checking for hydrogen bonding, ignoring that a large nonpolar molecule can have stronger LDF than a small polar molecule with hydrogen bonding. Why: Students memorize the "hydrogen bonding > dipole-dipole > LDF" rule and forget it only applies to similar molar mass. Correct move: Always compare molar mass first when ranking IMF strength; LDF strength increases with molar mass and can exceed weaker IMFs in smaller molecules.
• Wrong move: Claiming CH2F2 is polar and can form hydrogen bonds because it contains F. Why: Students see F and automatically assume hydrogen bonding, ignoring the bonding of H. Correct move: In CH2F2, all H are bonded directly to C, not F, so CH2F2 cannot form hydrogen bonds, only dipole-dipole interactions.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following lists correctly ranks the compounds in order of increasing boiling point, where boiling point increases with increasing intermolecular force strength? A) ${\text{F}}_2<{\text{Cl}}_2<{\text{Br}}_2<{\text{I}}_2$ B) ${\text{I}}_2<{\text{Br}}_2<{\text{Cl}}_2<{\text{F}}_2$ C) $\text{HF}<\text{HCl}<\text{HBr}<\text{HI}$ D) ${\text{CH}}_4<{\text{CH}}_3{\text{CH}}_3<{\text{CH}}_3{\text{CH}}_2\text{OH}<{\text{CH}}_3\text{OH}$

Worked Solution: All halogens are nonpolar diatomic molecules, so only London dispersion forces are present. LDF strength increases with increasing molar mass, so boiling point increases from F2 (lowest molar mass) to I2 (highest molar mass), matching order A. Check other options: Option C is incorrect because HF has hydrogen bonding, giving it a much higher boiling point than the other hydrogen halides, opposite the given order. Option D is incorrect because CH3CH2OH has a higher molar mass than CH3OH (both have hydrogen bonding), so CH3CH2OH has a higher boiling point, making the order wrong. The correct answer is A.

Question 2 (Free Response)

The table below gives data for four organic compounds:

Worked Solution: (a) Compound 1 (n-butanal): has a polar aldehyde group with a permanent dipole, so the strongest IMF is dipole-dipole interactions. Compound 2 (2-methylpropanal): also has a polar aldehyde group, so the strongest IMF is dipole-dipole. Compound 3 (n-butanol): has H bonded directly to O in the -OH group, so the strongest IMF is hydrogen bonding. Compound 4 (diethyl ether): has O but no H bonded directly to O, so the strongest IMF is dipole-dipole. (b) n-butanal (compound 1) has the higher boiling point. Both compounds have the same molar mass and the same strongest IMF (dipole-dipole), so the difference comes from London dispersion forces. n-butanal is straight-chain, while 2-methylpropanal is branched, so n-butanal has more surface area for intermolecular contact, leading to stronger LDF. Stronger IMFs require more energy to overcome, resulting in a higher boiling point. (c) n-butanol has a hydroxyl group with H covalently bonded directly to O, allowing it to form strong hydrogen bonds between molecules. Diethyl ether has O but no H bonded to O, so it cannot form hydrogen bonds, and only has weaker dipole-dipole interactions. Stronger IMFs require more energy to overcome, so n-butanol has a significantly higher boiling point even with the same molar mass.

Question 3 (Application / Real-World Style)

In biochemistry, the 3D folded shape of a protein is stabilized by intermolecular interactions between the protein and surrounding water. Serine is an amino acid with a polar side chain of $-{\text{CH}}_2\text{OH}$, while valine has a nonpolar side chain of $-\text{CH}({\text{CH}}_3{)}_2$. Which amino acid is more likely to be found on the surface of a protein folded in aqueous solution? How many total hydrogen bonds can a single serine side chain form with surrounding water molecules?

Worked Solution: Polar side chains form favorable intermolecular interactions with polar water molecules, while nonpolar side chains do not. Since the protein surface is in direct contact with water, the polar serine is more likely to be found on the surface (nonpolar valine tends to be buried in the protein interior to avoid water). For hydrogen bonding: the serine side chain has one O-H bond. The oxygen atom has two lone pairs, so it can accept two hydrogen bonds from water molecules. The H bonded to O can donate one hydrogen bond to a water molecule. Total hydrogen bonds = 2 + 1 = 3. In context: This ability to form multiple hydrogen bonds with water makes serine hydrophilic (water-loving), which explains its preference for the protein surface.

7. Quick Reference Cheatsheet

8. What's Next

Intermolecular forces are the foundational prerequisite for all remaining topics in Unit 3: Intermolecular Forces and Properties. Next, you will apply IMF strength rules to predict bulk properties like boiling point, vapor pressure, solubility, and viscosity, which make up the majority of Unit 3 exam questions. Without mastering the classification and relative strength of different IMFs, you will not be able to write the justification-based reasoning required for most Unit 3 FRQ questions, which almost always require linking observed bulk properties to underlying IMF strength. This topic also feeds into concepts across the rest of the course: IMFs are critical for understanding colligative properties of solutions, acid-base interactions, organic reactivity, and biological macromolecule structure.

Bulk properties of liquids and solids Solutions and solubility Colligative properties Biological macromolecule intermolecular interactions