Intermolecular Forces and Properties — AP Chemistry Chem Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Molecular interpretation of states of matter, ideal and real gas behavior, solutions and solubility rules, UV-Vis and IR spectroscopy basics, and chromatography and distillation separation techniques.

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 Intermolecular Forces and Properties?

Intermolecular Forces and Properties is the study of attractive and repulsive forces between molecules, and how these forces determine the macroscopic physical characteristics of substances including phase changes, gas behavior, solubility, and response to electromagnetic radiation. It makes up 18-22% of your AP Chemistry exam score (Unit 3 of the College Board CED), and ties directly to foundational topics like atomic structure and bonding, as well as advanced topics like reaction kinetics and thermodynamics. Common synonyms for exam use include intermolecular interactions, bulk physical properties, and condensed matter behavior.

2. Solid, liquid, gas — molecular interpretation

All three states of matter are defined by the relative strength of intermolecular forces (IMFs) between particles and the average kinetic energy (KE) of the particles themselves:

• Solids: IMFs are strong enough to lock particles into fixed, ordered lattice positions. Particles only exhibit vibrational motion, so solids have fixed shape and volume, with negligible compressibility.
• Liquids: IMFs are strong enough to keep particles in close contact, but not strong enough to lock them in place. Particles have translational and rotational motion, so liquids have fixed volume but take the shape of their container, with very low compressibility.
• Gases: IMFs are negligible relative to particle kinetic energy. Particles move freely and are spaced ~10x further apart than in condensed phases, so gases have no fixed shape or volume, and are highly compressible.

Worked Example: Compare the molecular behavior of H₂O(s), H₂O(l), and H₂O(g) at 1 atm.

Answer: Ice (solid) water molecules are held in a hexagonal hydrogen-bonded lattice, where average particle KE is lower than the strength of hydrogen bonds. In liquid water, hydrogen bonds are constantly breaking and reforming as molecules move past each other, with average KE roughly equal to IMF strength. In water vapor, hydrogen bonds are essentially non-existent between molecules, average KE is far higher than IMF strength, and particles move independently of each other.

Exam Tip: Examiners require explicit reference to specific IMF types (hydrogen bonding, dipole-dipole, London dispersion) when explaining phase change temperatures for full credit, so avoid generic references to "strong forces".

3. Ideal and real gases

An ideal gas is a hypothetical model that simplifies gas behavior calculations, based on four core assumptions:

1. Gas particles have negligible volume relative to the total volume of the container
2. No IMFs exist between gas particles
3. All collisions between particles and container walls are perfectly elastic (no kinetic energy loss)
4. Average particle KE is directly proportional to absolute temperature (Kelvin)

The ideal gas law describes the relationship between state variables for ideal gases: $$PV=nRT$$ Where $P$ = pressure (atm or kPa), $V$ = volume (L), $n$ = moles of gas, $R$ = gas constant ($0.0821\text{ Lcdotpatm/(molcdotpK)}$ or $8.314\text{ J/(molcdotpK)}$), and $T$ = absolute temperature (Kelvin).

Real gases deviate from ideal behavior at two extreme conditions:

• High pressure: Particles are packed close enough that their own volume is no longer negligible, and IMFs become significant, reducing the pressure exerted on container walls
• Low temperature: Particles move slowly enough that IMFs alter collision trajectories, further reducing measured pressure relative to ideal predictions

The van der Waals equation corrects for these deviations, with constant $a$ accounting for IMF strength (higher $a$ = stronger IMFs) and constant $b$ accounting for particle volume (higher $b$ = larger molecules): $$\left(P+a\frac{n^2}{V^2}\right)(V-nb)=nRT$$

Worked Example: A 1.00 mol sample of CO₂ is held at 300 K in a 0.500 L container. Calculate the ideal pressure, and explain why the real pressure is lower than this value.

Calculation: Rearrange the ideal gas law: $P=\frac{nRT}{V}=\frac{(1.00\text{ mol})(0.0821\text{ Lcdotpatm/(molcdotpK)})(300\text{ K})}{0.500\text{ L}}=49.3\text{ atm}$

Explanation: At this high pressure, CO₂ molecules are close enough that dipole-dipole IMFs attract each other, reducing the force of collisions with the container walls. The real measured pressure is ~45 atm, lower than the ideal value.

Exam Tip: You do not need to memorize the van der Waals equation, but you must know what $a$ and $b$ represent, and the conditions that cause real gas deviations.

4. Solutions and solubility

A solution is a homogeneous mixture of a solute (minor component) dissolved in a solvent (major component). Solubility is the maximum amount of solute that can dissolve in a fixed amount of solvent at a given temperature, governed by the "like dissolves like" rule: polar solutes dissolve in polar solvents, and nonpolar solutes dissolve in nonpolar solvents, determined by matching IMF types between solute and solvent. Dissolution is energetically favorable if the energy released from forming solute-solvent IMFs offsets the energy required to break solute-solute and solvent-solvent IMFs.

Key solubility rules:

• For solid solutes: Solubility generally increases with increasing temperature (most dissolution reactions are endothermic)
• For gaseous solutes: Solubility decreases with increasing temperature, and increases with increasing partial pressure of the gas above the solution, described by Henry's Law: $$C_g=k_H{P}_g$$ Where $C_g$ = dissolved gas concentration, $k_H$ = Henry's constant for the specific gas-solvent pair, and $P_g$ = partial pressure of the gas above the solution.

Worked Example: Explain why ethanol (CH₃CH₂OH) is miscible with water, but hexane (C₆H₁₄) is not.

Answer: Ethanol has a polar hydroxyl (-OH) group that can form hydrogen bonds with water molecules. The energy required to break water-water hydrogen bonds is offset by the energy released from forming water-ethanol hydrogen bonds, making dissolution favorable. Hexane is nonpolar, only has weak London dispersion forces, so it cannot form energetically favorable interactions with polar water molecules, so the two liquids separate into layers.

Exam Tip: Always reference solute-solute, solvent-solvent, and solute-solvent IMFs in solubility explanations, rather than only stating "like dissolves like" to earn full marks.

5. Spectroscopy (UV-Vis, IR) basics

Spectroscopy is the study of how matter interacts with electromagnetic radiation, used to identify molecular structure and measure solute concentration:

• Infrared (IR) spectroscopy: Uses infrared radiation (2.5-25 μm wavelength) to excite molecular vibrations (bond stretching and bending). Each bond type has a characteristic absorption frequency (measured in wavenumbers, cm⁻¹) that you will need to recognize for the exam:
• O-H (hydroxyl) stretches: 3200-3600 cm⁻¹, broad peak (from hydrogen bonding)
• C=O (carbonyl) stretches: ~1700 cm⁻¹, sharp, strong peak
• C-H (alkyl) stretches: 2800-3000 cm⁻¹, sharp peaks
• UV-Vis spectroscopy: Uses ultraviolet and visible light (200-700 nm wavelength) to excite electrons from the ground state to excited state in molecules with conjugated pi systems or transition metal complexes. Absorbance is proportional to solute concentration per the Beer-Lambert Law: $$A=\epsilon bc$$ Where $A$ = absorbance (unitless), $\epsilon$ = molar absorptivity (L/(mol·cm), constant for a given substance at a specific wavelength), $b$ = path length of the sample cuvette (cm), and $c$ = solute concentration (mol/L).

Worked Example: A 0.020 M solution of a blue copper complex has an absorbance of 0.56 at 620 nm in a 1.0 cm cuvette. Calculate the molar absorptivity of the complex.

Calculation: Rearrange the Beer-Lambert Law: $\epsilon =\frac{A}{bc}=\frac{0.56}{(1.0\text{ cm})(0.020\text{ mol/L})}=28\text{ L/(molcdotpcm)}$

6. Chromatography and distillation

These are common separation techniques that rely on differences in IMF strength between mixture components:

• Chromatography: Separates components based on their relative affinity for a stationary phase (solid or liquid supported on a solid) and mobile phase (liquid or gas that moves through the stationary phase). The retention factor ($R_f$) for each component is calculated as: $$R_f=\frac{\text{distance moved by solute}}{\text{distance moved by solvent front}}$$ Polar solutes have lower $R_f$ values when using a polar stationary phase (e.g., silica gel), because they form stronger IMFs with the stationary phase and move slower with the mobile phase.
• Distillation: Separates liquid mixtures based on differences in boiling point (directly related to IMF strength). Simple distillation is used for liquids with boiling point differences >30°C, while fractional distillation uses a fractionating column to repeatedly condense and vaporize the mixture for separation of liquids with smaller boiling point differences. Lower boiling point liquids (with weaker IMFs) distill first.

Worked Example: A mixture of ethanol (boiling point 78°C) and water (boiling point 100°C) is separated using fractional distillation. Which liquid distills first, and why?

Answer: Ethanol distills first. Ethanol only has one hydroxyl group per molecule, while water has two, so water forms stronger hydrogen bonding networks, leading to a higher boiling point. The lower boiling point ethanol vaporizes first, rises up the fractionating column, and is condensed into pure liquid first.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Confusing intermolecular forces (between molecules) with intramolecular covalent bonds (within molecules) when explaining boiling point. Why it happens: Both involve attractive forces, so terms are often mixed up. Correct move: Always specify "intermolecular hydrogen bonds" when discussing phase changes, because phase changes only break IMFs, not covalent bonds.
• Pitfall 2: Using Celsius instead of Kelvin in gas law calculations. Why it happens: Celsius is more commonly used in daily life, so students forget the ideal gas law uses absolute temperature. Correct move: Convert all temperatures to Kelvin by adding 273 before plugging into any gas law formula, and double-check units before solving.
• Pitfall 3: Stating that all gases behave ideally at STP. Why it happens: STP is taught as the standard condition for ideal gas calculations, leading to the misconception that real gases are perfect at STP. Correct move: Note that STP is the condition where deviations are smallest, but ideal behavior is only a hypothetical limit; real gases deviate slightly even at STP, and significantly at high pressure/low temperature.
• Pitfall 4: Only stating "like dissolves like" when explaining solubility without referencing IMF types. Why it happens: The rule is short and easy to remember, but examiners require evidence of understanding the mechanism. Correct move: Always list the IMFs present in the solute, solvent, and solute-solvent pair to justify solubility, even if you mention the like dissolves like rule.
• Pitfall 5: Confusing IR and UV-Vis spectroscopy uses. Why it happens: Both are spectral techniques, so students mix up their applications. Correct move: Remember IR = identify functional groups (bond vibrations), UV-Vis = measure concentration of colored/transition metal/conjugated compounds (electron transitions).

8. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following gases has the largest deviation from ideal gas behavior at 200 atm and 25°C? A) He B) H₂ C) CO₂ D) NH₃

Solution: Correct answer D) NH₃. Deviation from ideal behavior increases with stronger intermolecular forces and larger molecular size. NH₃ has hydrogen bonding, the strongest IMF of the options, so it has the highest van der Waals $a$ constant, leading to the largest deviation from ideal gas behavior. A and B are incorrect because He and H₂ have very weak London dispersion forces and small size, leading to near-ideal behavior. C is incorrect because CO₂ has dipole-dipole forces which are weaker than hydrogen bonding in NH₃.

Question 2 (Free Response)

A student separates a mixture of three amino acids using paper chromatography with a polar stationary phase and a nonpolar mobile phase. The Rf values for the three amino acids are 0.12, 0.68, and 0.85. a) Identify which amino acid is the most polar, justify your answer. b) If the solvent front moved 8.2 cm, calculate the distance moved by the least polar amino acid.

Solution: a) The most polar amino acid is the one with Rf = 0.12. Polar solutes have stronger attractive interactions with the polar stationary phase, so they move slower with the nonpolar mobile phase, resulting in a lower Rf value. b) The least polar amino acid has the highest Rf = 0.85. Rearrange the Rf formula: $$\text{Distance moved by solute}=R_f\times \text{distance moved by solvent front}=0.85\times 8.2\text{ cm}=7.0\text{ cm (2 significant figures)}$$

Question 3 (Free Response)

A 0.15 M solution of a red iron complex has an absorbance of 0.42 at 510 nm in a 1.0 cm cuvette. a) Calculate the molar absorptivity of the complex. b) If a sample of the same complex has an absorbance of 0.28 under the same conditions, what is its concentration?

Solution: a) Use the Beer-Lambert Law: $$\epsilon =\frac{A}{bc}=\frac{0.42}{(1.0\text{ cm})(0.15\text{ mol/L})}=2.8\text{ L/(molcdotpcm)}$$ b) Rearrange the Beer-Lambert Law to solve for concentration: $$c=\frac{A}{\epsilon b}=\frac{0.28}{(2.8\text{ L/(molcdotpcm)})(1.0\text{ cm})}=0.10\text{ mol/L}$$

9. Quick Reference Cheatsheet

10. What's Next

This unit is foundational to almost all remaining content in the AP Chemistry syllabus. The IMF principles you learned here explain reaction rate trends (stronger IMFs between reactants reduce collision frequency), thermodynamics of dissolution and phase changes (enthalpy of solution depends on the relative strength of solute-solute, solvent-solvent, and solute-solvent IMFs), and acid-base behavior (polarity of O-H bonds directly determines acid strength, for example). Understanding gas behavior is also critical for calculating reaction yields for gas-phase reactions in stoichiometry, and solubility rules are required for predicting precipitation reaction products in chemical reactions.

If you’re stuck on any subtopic, from calculating Rf values to interpreting IR spectra, you can ask Ollie, our AI tutor, for personalized explanations and extra practice problems tailored to your weak spots. You can also access more AP Chemistry study guides and full-length practice exams on the homepage to test your mastery of this unit before moving on to Unit 4: Chemical Reactions.