Intermolecular Forces

AP Chemistry· AP Chemistry CED — Intermolecular Forces and Properties· 14 min read

1. What Are Intermolecular Forces?★★☆☆☆⏱ 2 min

Intermolecular forces (abbreviated IMFs) are electrostatic attractive forces that act between separate discrete molecules, monatomic atoms (like noble gases), or ions. They are 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.

2. London Dispersion Forces★★☆☆☆⏱ 3 min

📘 Definition

London Dispersion Forces (LDF)

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.

Example:

Present between all molecules, including nonpolar species like $O_{2}$ and noble gas atoms.

The strength of LDF depends on polarizability, or how easily the electron cloud is distorted. Polarizability 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.

V \propto -\frac{\alpha_1 \alpha_2}{r^6}

📐 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 ( $C_{5} H_{12}$ ), neopentane ( $C_{5} H_{12}$ ), 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:
5
$\text{Ne} < \text{Kr} < \text{neopentane} < \text{n-pentane}$

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★★★☆☆⏱ 4 min

📘 Definition

Dipole-Dipole Interactions

Attractive forces between the permanent dipoles of polar neutral molecules. Stronger than LDF for molecules of similar molar mass, because dipoles are permanent rather than temporary induced.

The potential energy of dipole-dipole attraction follows the relationship below, where $μ$ 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.

V \propto -\frac{\mu_1^2 \mu_2^2}{r^6 kT}

📘 Definition

Hydrogen Bonding

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.

Hydrogen bonds have strengths of 20-40 kJ/mol, compared to 1-5 kJ/mol for regular dipole-dipole interactions.

📐 Worked Example

Which of the following molecules can form hydrogen bonds with other identical molecules? For each that can, state the requirement it meets. $CH_{3} OCH_{3}$ (dimethyl ether), $CH_{3} CH_{2} OH$ (ethanol), $HI$ , $NH_{3}$ .

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 ( $NH_{3}$ ) 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★★★☆☆⏱ 3 min

📘 Definition

Ion-Dipole Forces

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. They are the key force driving dissolution of ionic compounds in polar solvents like water.

Ion-dipole forces range up to ~100 kJ/mol, comparable to weak intramolecular bonds. The potential energy of ion-dipole attraction follows the relationship below, where $z$ is the charge of the ion, $μ$ 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.

V \propto -\frac{z \mu}{r^2}

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 $Ca^{2 +}$ cation and a water molecule? Explain your reasoning: (A) $Ca^{2 +}$ near the partially positive H end of $H_{2} O$ , (B) $Ca^{2 +}$ near the partially negative O end of $H_{2} O$ , (C) $Ca^{2 +}$ 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
$Ca^{2 +}$ 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. AP-Style Worked Practice Questions★★★★☆⏱ 4 min

📐 Worked Example

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) $F_{2} < Cl_{2} < Br_{2} < I_{2}$ B) $I_{2} < Br_{2} < Cl_{2} < F_{2}$ C) $HF < HCl < HBr < HI$ D) $CH_{4} < CH_{3} CH_{3} < CH_{3} CH_{2} OH < CH_{3} OH$

1
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 $F_{2}$ (lowest molar mass) to $I_{2}$ (highest molar mass), matching order A.
2
Check other options to confirm: 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 $CH_{3} CH_{2} OH$ has a higher molar mass than $CH_{3} OH$ (both have hydrogen bonding), so $CH_{3} CH_{2} OH$ has a higher boiling point, making the order wrong.
3
The correct answer is A.

📐 Worked Example

The table below gives data for four organic compounds:

Compound	Molar Mass (g/mol)	Structure
1	72	n-butanal, $CH_{3} CH_{2} CH_{2} CHO$
2	72	2-methylpropanal, $(CH_{3})_{2} CHCHO$
3	74	n-butanol, $CH_{3} CH_{2} CH_{2} CH_{2} OH$
4	74	diethyl ether, $CH_{3} CH_{2} OCH_{2} CH_{3}$

(a) Identify the strongest type of intermolecular force present in each pure compound. (b) Predict which compound (1 or 2) has the higher boiling point. Justify your answer. (c) Explain why n-butanol has a significantly higher boiling point than diethyl ether, despite having the same molar mass.

1
(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.
2
(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.
3
(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.

📐 Worked Example

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 $- CH_{2} OH$ , while valine has a nonpolar side chain of $- CH (CH_{3})_{2}$ . 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?

1
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).
2
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.
3
This ability to form multiple hydrogen bonds with water makes serine hydrophilic (water-loving), which explains its preference for the protein surface.

6. Common Pitfalls

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 $CH_{2} F_{2}$ 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 $CH_{2} F_{2}$ , all H are bonded directly to C, not F, so $CH_{2} F_{2}$ cannot form hydrogen bonds, only dipole-dipole interactions.

7. Quick Reference Cheatsheet

IMF Class / Property	Rule / Relationship	Key Notes
London Dispersion Force	Strength $\propto$ molar mass (electron count) $\propto$ surface area contact	Present between all molecules/atoms; stronger for linear vs branched isomers of same molar mass
Dipole-Dipole Interaction	Strength $\propto$ product of permanent dipole moments	Only present between polar molecules; stronger than LDF for similar molar mass
Hydrogen Bonding	Donor: H covalently bonded to N/O/F; Acceptor: N/O/F with lone pair	Special strong subclass of dipole-dipole; only one donor/acceptor needed for mixed systems
Ion-Dipole Force	$V \propto - \frac{z μ}{r ^{2}}$	Strongest IMF class; key to dissolution of ionic compounds in polar solvents
IMF Strength Order (same molar mass)	Ion-dipole > H-bonding > dipole-dipole > LDF	Does not apply to different molar mass; large LDF can exceed small molecule H-bonding
Boiling Point Trend	Boiling point $\propto$ IMF strength	Higher IMF = more energy to separate molecules = higher boiling point
"Like Dissolves Like" Solubility	Polar solutes dissolve in polar solvents; nonpolar in nonpolar	Solubility is favorable when new solute-solvent IMFs match strength of original IMFs
Vapor Pressure Trend	Vapor pressure $\propto 1/$ IMF strength	Stronger IMFs = fewer molecules escape to gas phase = lower vapor pressure

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Rank IMF strength for isomers
2022 · FRQ
Explain boiling point differences
2021 · MCQ
Identify hydrogen bonding capability

What's Next

Intermolecular forces are the foundational prerequisite for all remaining topics in Unit 3: Intermolecular Forces and Properties. Mastery of IMF strength rules and identification is critical for every subsequent topic in this unit, as all bulk properties of materials directly depend on the strength and type of intermolecular attractions between particles. Next, you will apply your understanding of IMFs to predict key bulk properties including boiling point, vapor pressure, solubility, and viscosity, which make up the majority of Unit 3 exam questions. You will also extend IMF concepts to analyze solutions, colligative properties, and gas behavior later in the course.