Intermolecular forces (IMFs) are the attractive forces between molecules (or formula units). They are much weaker than intramolecular forces (ionic, covalent, metallic bonds within molecules), but they determine critical physical properties such as boiling point, melting point, viscosity, and surface tension.
Hydrogen bonding between water molecules — the partial positive H of one molecule is attracted to the lone pair on the O of another. (Wikimedia Commons)
Types of IMFs (Weakest to Strongest)
IMF Type
Present In
Strength
Origin
London Dispersion Forces (LDF)
ALL molecules and atoms
Weakest (but increases with molar mass and surface area)
Temporary, instantaneous dipoles caused by random electron motion inducing dipoles in neighbors
Dipole-Dipole
Polar molecules
Moderate
Permanent partial charges (δ+ and δ−) on polar molecules attract each other
Hydrogen Bonding
Molecules with H bonded to N, O, or F
Strongest IMF (special case of dipole-dipole)
Very electronegative N, O, or F creates a large δ+ on H, which is attracted to lone pairs on N, O, or F of another molecule
Ion-Dipole
Ionic compound + polar solvent
Strongest of all (not technically IMF between molecules)
Full charge on ion attracts partial charge on polar molecule
How to identify IMFs present in a substance:
1. Is it ionic? → Ion-dipole when dissolved; strong lattice forces when solid
2. Does the molecule have H bonded to N, O, or F? → Hydrogen bonding + dipole-dipole + LDF
3. Is it a polar molecule? → Dipole-dipole + LDF
4. Is it nonpolar? → LDF only
Remember: ALL molecules have LDF. Polar molecules have both LDF and dipole-dipole. H-bonding molecules have all three.
LDF strength depends on:
• Molar mass: More electrons = more polarizable = stronger LDF. This is why I₂ is a solid but F₂ is a gas at room temperature.
• Shape/surface area: Long, extended molecules have stronger LDF than compact, spherical ones (more points of contact). Example: n-pentane (bp 36°C) vs. neopentane (bp 10°C) — same formula C₅H₁₂, different shapes.
IMFs and Physical Properties
Boiling point: Stronger IMFs = higher boiling point (more energy needed to separate molecules into gas phase).
Surface tension: Stronger IMFs = higher surface tension (molecules at surface pulled inward more strongly).
Viscosity: Stronger IMFs = higher viscosity (greater resistance to flow).
Vapor pressure: Stronger IMFs = LOWER vapor pressure (fewer molecules escape to gas phase).
3.2 — Properties of Solids
The properties of a solid depend on the type of particles present and the forces between them. The AP exam requires you to classify solids and predict their properties.
Solid Type
Particles
Forces
Melting Pt
Conductivity
Hardness
Examples
Ionic
Cations & anions
Electrostatic (lattice energy)
High
No (solid); Yes (liquid/dissolved)
Hard, brittle
NaCl, MgO, CaCl₂
Molecular
Molecules
IMFs (LDF, dipole-dipole, H-bonding)
Low
No
Soft
Ice, sugar, dry ice
Covalent network
Atoms linked by covalent bonds throughout
Covalent bonds in 3D network
Very high
No (except graphite)
Very hard
Diamond, SiO₂, SiC
Metallic
Cations in electron sea
Metallic bonds
Variable (often high)
Yes
Variable
Fe, Cu, Au, W
Covalent network solids explained:
• Diamond: Each C is sp³ hybridized and bonded to 4 other C atoms in a tetrahedral network. Extremely hard (hardest natural substance), very high mp (3550°C), does NOT conduct electricity (all electrons in bonds).
• Graphite: Each C is sp² hybridized in flat hexagonal sheets. Delocalized π electrons between layers allow conductivity. Layers slide over each other easily → soft and slippery (lubricant).
• SiO₂ (quartz): Each Si bonded to 4 O atoms in a 3D network. Very high mp, very hard, insulator.
Amorphous vs. Crystalline Solids
Crystalline solids: Regular, repeating 3D arrangement of particles. Sharp melting point. Examples: NaCl, diamond, ice.
Amorphous solids: Irregular, disordered arrangement. Soften gradually over a range of temperatures. Examples: glass, rubber, plastic.
3.3 — Solids, Liquids, and Gases
Matter exists in three common phases: solid, liquid, and gas. The phase depends on the balance between kinetic energy (which tends to separate particles) and intermolecular forces (which tend to hold them together).
Heating Curves
Heating curve for water showing the five regions: solid, melting, liquid, boiling, and gas. Temperature is constant during phase changes. (Wikimedia Commons)
Key features of a heating curve:
• Sloped regions: Temperature is rising. Only ONE phase is present. Use q = mcΔT.
• Flat regions (plateaus): Temperature is constant. TWO phases coexist. Use q = nΔHfus or q = nΔHvap. All added energy goes to breaking IMFs, not increasing temperature.
• The boiling plateau is longer than the melting plateau because ΔHvap > ΔHfus (more IMFs must be broken to go from liquid to gas than from solid to liquid).
Phase Diagrams
A generic phase diagram. Lines represent phase boundaries; the triple point is where all three phases coexist; the critical point marks the end of the liquid-gas boundary. (Wikimedia Commons)
Phase diagram features:
• Triple point: The unique T and P where solid, liquid, and gas all coexist in equilibrium.
• Critical point: Beyond this T and P, the liquid and gas phases become indistinguishable (supercritical fluid).
• Normal boiling point: Temperature at which vapor pressure = 1 atm (intersection with P = 1 atm line).
• Normal melting point: Temperature at which solid and liquid coexist at 1 atm.
Phase Changes
Phase Change
Direction
Energy
ΔH
Melting (fusion)
Solid → Liquid
Absorbs energy (endothermic)
ΔHfus > 0
Vaporization
Liquid → Gas
Absorbs energy (endothermic)
ΔHvap > 0
Sublimation
Solid → Gas
Absorbs energy (endothermic)
ΔHsub = ΔHfus + ΔHvap
Freezing
Liquid → Solid
Releases energy (exothermic)
−ΔHfus
Condensation
Gas → Liquid
Releases energy (exothermic)
−ΔHvap
Deposition
Gas → Solid
Releases energy (exothermic)
−ΔHsub
3.4 — Ideal Gas Law
The ideal gas law combines several gas laws into one equation that relates pressure, volume, temperature, and amount of gas. It assumes gas particles have no volume and no intermolecular forces.
PV = nRT
P = pressure (atm) | V = volume (L) | n = moles | R = 0.08206 L·atm/(mol·K) | T = temperature in Kelvin
Always convert temperature to Kelvin: K = °C + 273.15
Component Gas Laws
Law
Relationship
Conditions
Formula
Boyle’s
P and V are inversely proportional
Constant n, T
P₁V₁ = P₂V₂
Charles’s
V and T are directly proportional
Constant n, P
V₁/T₁ = V₂/T₂
Gay-Lussac’s
P and T are directly proportional
Constant n, V
P₁/T₁ = P₂/T₂
Avogadro’s
V and n are directly proportional
Constant P, T
V₁/n₁ = V₂/n₂
Worked Example: A 2.50 L container holds 0.100 mol of an ideal gas at 300 K. What is the pressure?
P = nRT / V = (0.100)(0.08206)(300) / 2.50 = 0.985 atm
Dalton’s Law of Partial Pressures
Ptotal = PA + PB + PC + ...
Each gas in a mixture exerts pressure independently. Partial pressure: PA = χA × Ptotal where χA = mole fraction = nA / ntotal
Example: A mixture contains 0.30 mol N₂ and 0.10 mol O₂ at 2.0 atm total pressure.
χN₂ = 0.30 / 0.40 = 0.75 → PN₂ = 0.75 × 2.0 = 1.5 atm
Molar Volume and Gas Density
At STP (0°C, 1 atm), one mole of any ideal gas occupies 22.4 L. The density of a gas can be found from:
d = PM / RT where M = molar mass (g/mol)
Heavier gases are denser at the same T and P.
3.5 — Kinetic Molecular Theory
The Kinetic Molecular Theory (KMT) provides a model to explain the behavior of ideal gases at the molecular level. Its postulates are the foundation for all gas law behavior.
The Five Postulates of KMT
Gas particles are in continuous, random, straight-line motion.
The volume of individual gas particles is negligible compared to the container volume.
Gas particles exert no attractive or repulsive forces on each other.
Collisions between particles (and with container walls) are perfectly elastic — total kinetic energy is conserved.
The average kinetic energy of gas particles is directly proportional to the absolute temperature (in Kelvin).
Maxwell-Boltzmann distribution of molecular speeds at different temperatures. Higher T → broader curve, peak shifts right. (Wikimedia Commons)
Average kinetic energy: KEavg = (3/2) kBT (per molecule) or KEavg = (3/2) RT (per mole)
Root-mean-square speed: urms = √(3RT / M) where M = molar mass in kg/mol
Key insight: At the same temperature, all gases have the same average KE. Lighter gases move faster (higher urms).
Maxwell-Boltzmann Distribution
At higher temperatures: curve broadens, peak shifts to higher speeds, peak height decreases.
At lower temperatures: curve is narrower, peak is taller, speeds are clustered more tightly.
Lighter gases at the same T: curve is broader with peak shifted right (faster speeds).
The area under the curve always equals the total number of molecules (remains constant).
Effusion and Graham’s Law
Graham’s Law: rate₁ / rate₂ = √(M₂ / M₁)
Lighter gases effuse (escape through a tiny hole) and diffuse faster. Example: H₂ (M = 2) effuses √(32/2) = 4 times faster than O₂ (M = 32).
3.6 — Deviation from Ideal Gas Law
Real gases deviate from ideal behavior because molecules do have volume and do exert intermolecular forces. The van der Waals equation corrects for these two flawed assumptions of KMT:
(P + an²/V²)(V − nb) = nRT
• a correction (added to P): Accounts for intermolecular attractions. Real molecules attract each other, reducing the force of collisions with walls. The term an²/V² corrects P upward. Larger a = stronger IMFs.
• b correction (subtracted from V): Accounts for the finite volume of molecules. Real molecules take up space, so the actual free volume is less than V. Larger b = larger molecule size.
When Do Gases Deviate Most?
High pressure: Molecules are forced close together → their volume becomes significant relative to container volume, and IMFs become stronger.
Low temperature: Molecules move slowly → IMFs have more influence (particles don’t have enough KE to overcome attractions).
Ideal behavior is most closely approached at:
• Low pressure (particles far apart → volume negligible, IMFs minimal)
• High temperature (high KE overcomes IMFs)
• Small, nonpolar molecules with weak IMFs (e.g., He, H₂)
AP Exam Tip
When comparing two gases at the same conditions, the gas with stronger IMFs (larger a) will have a lower actual pressure than predicted by the ideal gas law, because intermolecular attractions reduce the force of wall collisions.
3.7 — Solutions and Mixtures
A solution is a homogeneous mixture of a solute dissolved in a solvent. The dissolution process involves breaking solute-solute and solvent-solvent interactions and forming new solute-solvent interactions.
“Like Dissolves Like”
Polar solutes dissolve in polar solvents (e.g., NaCl in water, sugar in water).
Nonpolar solutes dissolve in nonpolar solvents (e.g., oil in hexane, I₂ in CCl₄).
Polar and nonpolar substances generally do NOT mix (e.g., oil and water).
Dissolution of Ionic Compounds
When an ionic compound dissolves in water, the process involves:
Breaking the crystal lattice (requires energy = lattice energy; endothermic)
Hydration of ions (releases energy = hydration energy; exothermic). Water molecules surround each ion — δ− oxygen faces cations, δ+ hydrogen faces anions.
If hydration energy ≥ lattice energy, dissolution is energetically favorable.
Concentration Units:
• Molarity (M) = moles of solute / liters of solution
• Molality (m) = moles of solute / kg of solvent (not on the AP exam but useful to know)
Dilution formula: M₁V₁ = M₂V₂ (moles of solute before = moles after)
Example: How many mL of 12.0 M HCl are needed to prepare 500 mL of 0.500 M HCl?
(12.0)(V₁) = (0.500)(500) → V₁ = 20.8 mL
3.8 — Representations of Solutions
Solutions can be described at three levels of representation, and the AP exam frequently asks you to translate between them:
Macroscopic: What you observe in the lab — a clear, homogeneous liquid (e.g., saltwater looks like water). Cannot distinguish individual particles.
Particulate: A diagram showing individual ions or molecules surrounded by solvent molecules. This is very commonly tested on the AP exam.
Symbolic: Chemical equations with state symbols, e.g., NaCl(s) → Na⁺(aq) + Cl⁻(aq).
Reading particulate diagrams on the AP exam:
• Strong electrolytes (strong acids, strong bases, soluble salts) dissociate completely into ions. Show Na⁺ and Cl⁻ separately, NOT NaCl units.
• Weak electrolytes (weak acids, weak bases) partially dissociate. Show mostly intact molecules with a few ions.
• Nonelectrolytes (molecular compounds like sugar, ethanol) dissolve as intact molecules. Show individual C₁₂H₂₂O₁₁ molecules, NOT broken up into atoms or ions.
• The number of particles in the diagram should reflect the correct stoichiometric ratios (e.g., CaCl₂ produces 1 Ca²⁺ and 2 Cl⁻).
3.9 — Separation of Solutions and Mixtures
Mixtures can be separated using physical methods that exploit differences in the physical properties of the components. The choice of technique depends on the properties of the substances in the mixture.
Technique
Separates Based On
How It Works
Example
Filtration
Particle size
Porous barrier traps solid particles while allowing liquid to pass through
Sand from saltwater
Distillation
Boiling point differences
Heat mixture; component with lower bp evaporates first, then condenses and is collected separately
Ethanol from water; desalination
Chromatography
Differential affinity for mobile vs. stationary phase
Components travel at different rates based on polarity/size; more attracted to stationary phase = travels slower
Separating pigments, amino acids, or drug components
Evaporation
Volatility
Heat to evaporate solvent, leaving dissolved solid behind
Recovering NaCl from saltwater
Chromatography details for AP exam:
• Stationary phase: Does not move (e.g., paper, silica gel, packed column).
• Mobile phase: Moves (e.g., solvent flowing through the stationary phase).
• Rf value (retention factor) = distance traveled by substance / distance traveled by solvent front. Each substance has a characteristic Rf under the same conditions.
• Components with greater affinity for the mobile phase travel farther (higher Rf).
• Components with greater affinity for the stationary phase travel less (lower Rf).
3.10 — Solubility
Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature. It depends on the nature of the solute and solvent, temperature, and (for gases) pressure.
Temperature Effects on Solubility
Most solid solutes: Solubility increases with temperature (dissolving is usually endothermic — adding heat shifts equilibrium toward dissolved state by Le Chatelier’s principle).
Some solids (exceptions): Li₂SO₄ and Ce₂(SO₄)₃ decrease in solubility with temperature.
Gases: Solubility decreases with temperature (dissolving gas is exothermic; higher T drives gas out of solution). This is why warm soda goes flat faster.
Pressure Effects on Solubility (Henry’s Law)
Henry’s Law: C = kH × Pgas
The solubility of a gas is directly proportional to the partial pressure of the gas above the solution. Higher pressure → more gas dissolves. This applies only to gases, not solids or liquids.
Example: Carbonated drinks are bottled under high CO₂ pressure. When opened, pressure drops, CO₂ solubility decreases, and bubbles form.
Types of Solutions by Saturation
Unsaturated: Contains less solute than the maximum; more can dissolve.
Saturated: Contains the maximum amount of solute at that temperature. A dynamic equilibrium exists between dissolving and crystallization.
Supersaturated: Contains MORE solute than should dissolve at that temperature. Unstable — crystallization can be triggered by adding a seed crystal or scratching the container.
3.11 — Spectroscopy and the Electromagnetic Spectrum
Electromagnetic radiation interacts with matter in ways that depend on the energy (frequency) of the radiation. Different regions of the EM spectrum provide different types of information about molecular structure.
Key equations:
• E = hν (energy of a photon; h = 6.626 × 10⁻³⁴ J·s, ν = frequency in Hz)
• c = λν (c = 3.00 × 10⁸ m/s, λ = wavelength, ν = frequency)
• E = hc / λ (combining both)
Higher frequency (ν) = shorter wavelength (λ) = higher energy
Types of Spectroscopy on the AP Exam
Type
EM Region
What It Probes
AP Relevance
UV-Vis
Ultraviolet & visible
Electronic transitions (electrons jumping between energy levels)
Beer-Lambert Law, colored solutions
IR
Infrared
Molecular vibrations (bond stretching and bending)
Identifying functional groups (O—H, C=O, N—H peaks)
Microwave
Microwave
Molecular rotations
Less commonly tested
Why substances appear colored:
A substance absorbs certain wavelengths of visible light. The color you see is the complementary color of the absorbed light. For example, a solution that absorbs red light appears green.
3.12 — Photoelectric Effect
The photoelectric effect occurs when light shining on a metal surface ejects electrons. It provided key evidence for the particle nature of light (photons) and was explained by Einstein.
Key equation: KEelectron = hν − Φ
• hν = energy of the incoming photon
• Φ (work function) = minimum energy needed to eject an electron from the metal surface
• KEelectron = kinetic energy of the ejected electron
Threshold frequency (ν₀): The minimum frequency of light needed to eject an electron. At ν₀, KE = 0, so Φ = hν₀.
Key Observations
Below the threshold frequency, no electrons are ejected regardless of light intensity (brightness). This contradicts the wave model of light.
Above the threshold frequency, increasing intensity ejects more electrons (more photons hit the surface) but does NOT increase their KE.
Increasing frequency (above threshold) increases the KE of ejected electrons.
Electrons are ejected instantly when light of sufficient frequency hits the surface — no time delay.
Example: A metal has a work function of 4.50 × 10⁻¹⁹ J. Light with ν = 1.00 × 10¹⁵ Hz strikes the surface. What is the KE of the ejected electrons?
Ephoton = hν = (6.626 × 10⁻³⁴)(1.00 × 10¹⁵) = 6.626 × 10⁻¹⁹ J
KE = 6.626 × 10⁻¹⁹ − 4.50 × 10⁻¹⁹ = 2.13 × 10⁻¹⁹ J
3.13 — Beer-Lambert Law
The Beer-Lambert Law (also called Beer’s Law) relates the absorbance of light by a solution to the concentration of the absorbing species. It is the basis for using spectrophotometry to determine unknown concentrations.
A = εbc
• A = absorbance (unitless, measured by a spectrophotometer)
• ε = molar absorptivity (L/(mol·cm)) — a constant specific to the substance and wavelength
• b = path length (cm) — the distance light travels through the solution (usually 1 cm in standard cuvettes)
• c = molar concentration (mol/L)
Key point: Absorbance is directly proportional to concentration (at a given wavelength with constant ε and b).
Calibration Curves
A calibration curve (standard curve) is a graph of absorbance vs. concentration using solutions of known concentrations (standards). The resulting line should be linear (following Beer’s Law). You can then use this line to determine the concentration of an unknown solution by measuring its absorbance and finding the corresponding concentration.
How to use a calibration curve:
1. Prepare several standard solutions of known concentration.
2. Measure the absorbance of each at the same wavelength (usually λmax, the wavelength of maximum absorbance).
3. Plot A vs. c — should be linear through the origin.
4. Measure absorbance of unknown solution.
5. Use the line (or equation y = mx) to find the unknown concentration.
Example: A calibration curve gives A = 125c (where c is in mol/L). An unknown solution has A = 0.500. What is its concentration?
c = A / 125 = 0.500 / 125 = 0.00400 mol/L = 4.00 × 10⁻³ M
Absorbance vs. Transmittance
Transmittance (T) is the fraction of light that passes through the solution. A = −log(T). Higher concentration → higher absorbance → lower transmittance (more light absorbed). On the AP exam, you primarily work with absorbance.
📝 Practice Multiple-Choice Questions
Test your knowledge of Unit 3. Click “Show Answer” to reveal the correct choice and explanation.
1. Which intermolecular force is primarily responsible for the high boiling point of water compared to H₂S?
(A) London dispersion forces
(B) Dipole-dipole interactions
(C) Hydrogen bonding
(D) Ion-dipole forces
Answer: (C)
Water (H₂O) can form hydrogen bonds because H is bonded to the highly electronegative O atom. H₂S cannot form H-bonds (S is not electronegative enough). Despite being lighter than H₂S, water has a much higher boiling point (100°C vs. −60°C) due to hydrogen bonding.
2. Which type of solid would you expect to have the highest melting point?
(A) Molecular solid (ice)
(B) Ionic solid (NaCl)
(C) Covalent network solid (diamond)
(D) Metallic solid (sodium)
Answer: (C) Diamond is a covalent network solid where every carbon is bonded to four others by strong covalent bonds throughout the entire structure. Breaking these requires enormous energy. Diamond’s mp is ~3550°C, far higher than NaCl (801°C), Na (98°C), or ice (0°C).
3. On a heating curve, what happens to the temperature during a phase change?
(A) Temperature increases rapidly
(B) Temperature decreases
(C) Temperature remains constant
(D) Temperature oscillates
Answer: (C)
During a phase change, temperature remains constant even though heat is being added. All the energy goes toward overcoming intermolecular forces (breaking IMFs), not increasing kinetic energy. This is why heating curves have flat plateaus at the melting and boiling points.
4. A mixture of gases contains 2.0 mol N₂ and 3.0 mol O₂ at a total pressure of 5.0 atm. What is the partial pressure of O₂?
5. At the same temperature, which gas has the highest root-mean-square speed?
(A) O₂ (M = 32)
(B) N₂ (M = 28)
(C) He (M = 4)
(D) CO₂ (M = 44)
Answer: (C)
urms = √(3RT/M). At the same T, lighter gases move faster. Helium (M = 4 g/mol) is the lightest, so it has the highest urms.
6. Under which conditions does a real gas behave MOST like an ideal gas?
(A) High pressure and high temperature
(B) High pressure and low temperature
(C) Low pressure and low temperature
(D) Low pressure and high temperature
Answer: (D)
At low pressure, molecules are far apart (volume is negligible, IMFs are minimal). At high temperature, molecules have high KE that overcomes any IMF attractions. These are the conditions closest to ideal behavior.
7. When NaCl dissolves in water, which interaction is primarily responsible for stabilizing the ions in solution?
(A) London dispersion forces
(B) Hydrogen bonding
(C) Ion-dipole interactions
(D) Covalent bonding
Answer: (C)
When ionic compounds dissolve, ion-dipole interactions form between the ions (full charges) and the polar water molecules (partial charges). The δ− oxygen of water faces Na⁺; the δ+ hydrogen faces Cl⁻. This hydration energy helps overcome the lattice energy.
8. The solubility of CO₂ gas in water decreases when the temperature is increased. Which of the following best explains this?
(A) CO₂ decomposes at high temperatures
(B) Dissolving CO₂ is exothermic, so increasing T shifts equilibrium toward less dissolved gas
(C) Water molecules move too fast to interact with CO₂
(D) The polarity of water decreases at higher temperatures
Answer: (B)
Dissolving a gas in a liquid is generally exothermic. By Le Chatelier’s principle, increasing temperature shifts the equilibrium toward the endothermic direction (gas coming OUT of solution), decreasing solubility.
9. Light with a frequency below the threshold frequency strikes a metal surface. What happens?
(A) Electrons are ejected with low kinetic energy
(B) Electrons are ejected but only after a delay
(C) No electrons are ejected regardless of light intensity
(D) More electrons are ejected if intensity is increased
Answer: (C)
Below the threshold frequency, individual photons do not carry enough energy to overcome the work function (Φ). No electrons are ejected, regardless of how bright (intense) the light is. This is because energy comes in discrete packets (photons), and each photon must individually have enough energy.
10. A solution of known concentration has an absorbance of 0.800 at a path length of 1.00 cm and a molar absorptivity of 200 L/(mol·cm). What is the concentration?
(A) 4.00 × 10⁻⁴ M
(B) 4.00 × 10⁻³ M
(C) 2.50 × 10⁻³ M
(D) 1.60 × 10² M
Answer: (B)
A = εbc → c = A / (εb) = 0.800 / (200 × 1.00) = 4.00 × 10⁻³ M.
💡 Study Tip for Unit 3
Unit 3 is the largest unit on the AP exam (18–22%). Master identifying IMFs from molecular structure — this skill appears on virtually every AP Chemistry exam. Know the ideal gas law cold and practice Dalton’s Law problems. Understand heating curves inside and out (why plateaus are flat, how to calculate energy for each segment). For spectroscopy, focus on Beer’s Law and calibration curves — they are high-yield FRQ topics. Always connect molecular-level explanations (IMFs, KMT) to macroscopic observations (boiling points, pressure, solubility).
