What is AP Chem unit 3?
Unit 3 asks you to explain properties of matter by reasoning from the particle level upward. Every macroscopic observation, whether a high boiling point, a gas that deviates from ideal behavior, or a colored solution that absorbs light, has a particle-level explanation rooted in forces, motion, or energy.
Unit 3 covers how intermolecular forces determine solid and liquid properties, how the ideal gas law and kinetic molecular theory describe gas behavior, how solutions are prepared and represented, how separation techniques exploit IMF differences, and how spectroscopy and Beer-Lambert law connect light absorption to molecular structure and concentration.
Intermolecular forces drive physical properties
London dispersion, dipole-dipole, hydrogen bonding, and ion-dipole forces determine boiling point, melting point, vapor pressure, and solubility. Stronger forces mean higher boiling points and lower vapor pressures. Polarizability, molecular size, and the presence of N-H, O-H, or F-H bonds determine which forces are present and how strong they are.
Gas laws connect P, V, T, and n
PV = nRT describes ideal gas behavior. Dalton's law adds partial pressures for mixtures. Kinetic molecular theory explains pressure and temperature at the particle level using KE = 1/2 mv² and the Maxwell-Boltzmann distribution. Real gases deviate from ideal behavior at low temperature and high pressure due to intermolecular attractions and finite particle volume.
Solutions and spectroscopy require quantitative reasoning
Molarity M = n/L and dilution M1V1 = M2V2 are the core solution calculations. Solubility follows the like-dissolves-like principle. Chromatography and distillation separate mixtures using IMF differences. Beer-Lambert law A = εbc links absorbance to concentration, and photon energy E = hν links light frequency to electronic, vibrational, or rotational transitions.
Structure determines properties at every scaleWhether you are comparing boiling points of alkanes, explaining why a real gas compresses less than predicted, or reading a calibration curve, the reasoning chain is the same: identify the particles, identify the forces or energy relationships between them, and connect those to the observable property. Unit 3 builds this reasoning pattern across solids, liquids, gases, solutions, and light-matter interactions.
Unit 3 review notes
3.1
Intermolecular Forces
Intermolecular forces (IMFs) are attractions between separate particles. Their strength determines physical properties like boiling point, melting point, and vapor pressure. For any comparison question, identify which forces are present, then rank their strength based on molecular structure.
- London dispersion forces: Present in all molecules; arise from temporary fluctuating dipoles. Strength increases with molecular size, number of electrons, and contact surface area. Branched molecules have lower boiling points than straight-chain isomers because of reduced contact area.
- Dipole-dipole interactions: Occur between polar molecules with permanent dipole moments. Stronger than London dispersion for small polar molecules. Orientation-dependent: positive end of one molecule attracts negative end of another.
- Hydrogen bonding: A strong dipole-dipole interaction requiring a hydrogen atom bonded directly to N, O, or F interacting with a lone pair on another N, O, or F. Explains the unusually high boiling point of water and the properties of alcohols and amines.
- Ion-dipole forces: Attractions between an ion and a polar molecule. The strongest IMF type; critical for explaining why ionic compounds dissolve in water (hydration of ions).
- Polarizability: How easily an electron cloud distorts. Increases with more electrons and larger electron clouds; enhanced by pi bonding. Higher polarizability means stronger London dispersion forces.
Given two molecules, identify all IMFs present in each and predict which has the higher boiling point, explaining your reasoning in terms of force type and strength.
| Force type | Molecules involved | Relative strength | Structural requirement |
|---|
| London dispersion | All molecules | Weakest to strongest (size-dependent) | All molecules; increases with electron count and surface area |
| Dipole-dipole | Polar molecules | Moderate | Permanent dipole moment (asymmetric polar bonds) |
| Hydrogen bonding | Molecules with N-H, O-H, or F-H | Strong | H bonded to N, O, or F; lone pair acceptor nearby |
| Ion-dipole | Ion + polar molecule | Strongest IMF | Ion present; polar solvent molecule |
3.2
Solid Types and the Three Phases
Solids are classified by the particles present and the forces holding them together. Each type has a predictable set of macroscopic properties. The three phases differ in particle arrangement, freedom of motion, and compressibility.
- Ionic solids: Held together by strong electrostatic attractions between oppositely charged ions in a crystal lattice. High melting points, low vapor pressure, brittle (like-charge repulsion when layers shift), conduct electricity only when molten or dissolved.
- Covalent network solids: Atoms connected by covalent bonds throughout the entire structure (e.g., diamond, graphite, SiO2). Extremely high melting points. Diamond is hard and nonconducting; graphite conducts electricity due to delocalized electrons between layers.
- Metallic solids: Metal cations surrounded by a sea of delocalized valence electrons. Conduct electricity and heat, are malleable and ductile. Alloys (substitutional or interstitial) alter properties by changing the lattice.
- Molecular solids: Discrete molecules held together by IMFs (London dispersion, dipole-dipole, or hydrogen bonding). Generally low melting points and higher vapor pressures than ionic or network solids.
- Phase particle models: Solids: particles in fixed positions, vibrational motion only. Liquids: particles in contact but mobile, translational motion. Gases: particles far apart, rapid random translational motion, highly compressible.
Given a substance's melting point, conductivity, and hardness, identify its solid type and explain the particle-level basis for each property.
| Solid type | Particles | Forces | Melting point | Conductivity |
|---|
| Ionic | Cations and anions | Electrostatic (Coulombic) | High | Only when molten or dissolved |
| Covalent network | Atoms | Covalent bonds | Very high | Usually none (diamond); yes (graphite) |
| Metallic | Metal cations + electron sea | Metallic bonds | Variable (generally high) | Yes (solid and liquid) |
| Molecular | Molecules | IMFs (LDF, dipole, H-bond) | Low to moderate | No |
3.4
Ideal Gas Law and Partial Pressures
The ideal gas law PV = nRT relates the four macroscopic properties of a gas sample. For mixtures, Dalton's law of partial pressures applies: each gas contributes independently to the total pressure based on its mole fraction.
- PV = nRT: P is pressure (atm), V is volume (L), n is moles, R is 0.08206 L·atm·mol⁻¹·K⁻¹, T is temperature in Kelvin. Always convert Celsius to Kelvin before calculating.
- Dalton's law of partial pressures: Ptotal = PA + PB + PC + ... Each gas in a mixture exerts pressure independently. Partial pressure PA = XA × Ptotal, where XA is the mole fraction of gas A.
- Mole fraction: XA = moles of A / total moles. Dimensionless; used to find partial pressure from total pressure or to find composition from partial pressures.
- Gas law graphs: P vs. 1/V is linear at constant T and n (Boyle's law). V vs. T is linear at constant P and n (Charles's law). P vs. T is linear at constant V and n (Gay-Lussac's law). V vs. n is linear at constant P and T (Avogadro's law).
A container holds 0.50 mol N2 and 0.25 mol O2 at a total pressure of 3.0 atm. Calculate the partial pressure of each gas and the mole fraction of N2.
3.5
Kinetic Molecular Theory and Real Gas Deviations
Kinetic molecular theory (KMT) explains ideal gas behavior at the particle level. Real gases deviate from ideal predictions when intermolecular attractions or finite particle volumes become significant.
- KMT assumptions: Gas particles are in constant random motion, have negligible volume compared to the container, undergo elastic collisions, and have no intermolecular attractions. Average kinetic energy is proportional to Kelvin temperature.
- KE = 1/2 mv²: Average kinetic energy of a particle depends on mass and velocity. At the same temperature, heavier molecules move more slowly than lighter ones because average KE is the same for all gases at a given T.
- Maxwell-Boltzmann distribution: A graph showing the distribution of particle speeds or kinetic energies at a given temperature. At higher T, the curve shifts right and flattens (broader distribution, higher average speed). Heavier molecules have a curve shifted left compared to lighter molecules at the same T.
- Real gas deviations: At low temperature, intermolecular attractions pull particles together, reducing pressure below ideal predictions. At very high pressure, finite particle volume becomes significant, increasing pressure above ideal predictions.
Sketch two Maxwell-Boltzmann curves: one for N2 at 300 K and one at 600 K. Describe how the peak position and curve shape change and explain why.
| Condition | Dominant effect | Deviation from ideal | Explanation |
|---|
| Low temperature | Intermolecular attractions | Pressure lower than ideal | Particles slow down and attract each other before hitting walls |
| High pressure | Finite particle volume | Volume larger than ideal | Particle volume is no longer negligible relative to container volume |
| High temperature, low pressure | Neither effect significant | Behaves nearly ideally | Particles move fast and are far apart |
3.7
Solutions, Mixtures, and Particulate Representations
A solution is a homogeneous mixture with uniform macroscopic properties throughout. Molarity is the primary concentration unit. Particulate diagrams show both the relative number of solute particles (concentration) and how solute and solvent particles interact.
- Molarity: M = moles of solute / liters of solution. Use total solution volume, not just solvent volume. Units are mol/L or M.
- Dilution equation: M1V1 = M2V2. Moles of solute are conserved when a solution is diluted by adding solvent. Useful for preparing solutions from stock solutions.
- Homogeneous vs. heterogeneous mixtures: Homogeneous mixtures (solutions) have uniform properties throughout. Heterogeneous mixtures have properties that vary by location. Solutions can be solid, liquid, or gas.
- Particulate models of solutions: Diagrams must show correct relative particle counts for concentration comparisons and correct interactions: ion-dipole forces for ionic solutes in water, hydrogen bonding for polar molecular solutes, water molecules oriented with O toward cations and H toward anions.
Draw a particulate diagram showing NaCl dissolved in water. Label the ion-dipole interactions and show the correct orientation of water molecules around Na+ and Cl-.
3.9
Separation Techniques and Solubility
Separation of mixtures and solubility both depend on differences in intermolecular interactions. The like-dissolves-like principle governs which solutes dissolve in which solvents, and the same logic explains how chromatography and distillation separate components.
- Like dissolves like: Polar and ionic solutes dissolve in polar solvents (e.g., NaCl in water) because solute-solvent interactions (ion-dipole or dipole-dipole) are comparable in strength to the forces being broken. Nonpolar solutes dissolve in nonpolar solvents via London dispersion forces.
- Chromatography: Separates components based on differential IMF strength between each component and the stationary phase versus the mobile phase. More polar components interact more strongly with a polar stationary phase and travel more slowly (lower Rf). The chromatogram reveals relative polarities.
- Distillation: Separates liquids based on differences in vapor pressure (and therefore boiling point), which reflect differences in IMF strength. The component with weaker IMFs has higher vapor pressure and distills first.
- Filtration limitation: Filtration cannot separate the components of a true solution because solute particles are too small to be captured by a filter. Chromatography or distillation is required.
Two compounds are separated by paper chromatography. Compound A has a higher Rf than compound B. Which is more polar? Explain using IMF interactions with the stationary phase.
| Technique | Basis of separation | IMF principle |
|---|
| Chromatography | Differential interaction with stationary vs. mobile phase | More polar compound interacts more strongly with polar stationary phase |
| Distillation | Differences in vapor pressure / boiling point | Weaker IMFs give higher vapor pressure; that component distills first |
| Filtration | Particle size | Does not separate true solutions; only removes suspended solids |
3.11
Spectroscopy, Photon Energy, and the Electromagnetic Spectrum
Different regions of the electromagnetic spectrum interact with matter in different ways depending on the energy of the photons. The two key equations are c = λν (connecting wavelength and frequency) and E = hν (connecting frequency and photon energy).
- Spectral region to transition type: Microwave radiation causes transitions in molecular rotational energy levels. Infrared radiation causes transitions in molecular vibrational energy levels. Ultraviolet/visible radiation causes transitions in electronic energy levels.
- c = λν: Speed of light c = 3.00 × 10^8 m/s equals wavelength (λ, in meters) times frequency (ν, in Hz). Shorter wavelength means higher frequency.
- E = hν: Photon energy equals Planck's constant h (6.626 × 10^-34 J·s) times frequency. Higher frequency (shorter wavelength) means higher energy per photon. UV photons carry more energy than IR photons.
- Absorption and emission: When a photon is absorbed, the atom or molecule increases in energy by exactly the photon's energy. When a photon is emitted, the species decreases in energy by the same amount. Energy levels are quantized, so only specific photon energies are absorbed or emitted.
A photon has a wavelength of 500 nm. Calculate its frequency and energy. Identify which type of molecular transition this photon could cause.
| Spectral region | Transition type | Relative photon energy |
|---|
| Microwave | Molecular rotation | Lowest |
| Infrared | Molecular vibration | Moderate |
| Ultraviolet/Visible | Electronic transitions | Highest |
3.13
Beer-Lambert Law
The Beer-Lambert law A = εbc quantitatively relates the absorbance of a solution to the concentration of the absorbing species, the path length of the light through the solution, and the molar absorptivity. It is the basis for using a spectrophotometer to determine unknown concentrations.
- A = εbc: A is absorbance (unitless), ε is molar absorptivity (M⁻¹cm⁻¹, a property of the specific molecule at a specific wavelength), b is path length (cm), c is concentration (M). Absorbance is directly proportional to concentration when b and λ are held constant.
- Calibration curve: A graph of absorbance (y-axis) vs. known concentration (x-axis) for standard solutions. The linear relationship allows you to read off the concentration of an unknown solution from its measured absorbance.
- Wavelength of maximum absorbance: The spectrophotometer is set to λmax, the wavelength at which the species absorbs most strongly. This maximizes sensitivity and the linear range of the Beer-Lambert relationship.
- Molar absorptivity: ε is a constant for a given chemical species at a given wavelength. A higher ε means the species absorbs light more intensely; a small amount of it produces a large absorbance signal.
A solution has an absorbance of 0.45 at λmax. The molar absorptivity is 150 M⁻¹cm⁻¹ and the path length is 1.00 cm. Calculate the concentration of the absorbing species.
Practice AP Chem unit 3 questions
Try stimulus-based AP practice questions and written prompts after you review the notes.
Thin-layer chromatography is performed on a mixture of pentane (C5H12), 1-pentanol (C5H11OH), and pentanal (C4H9CHO) using polar silica gel and nonpolar hexane. The chromatogram is shown.
QuestionWhich spot most likely represents 1-pentanol, and why?
Spot C, because it forms strong hydrogen bonds with the polar silica.
Spot A, because it forms strong hydrogen bonds with the polar silica.
Spot C, because it has the strongest dispersion forces with the solvent.
Spot A, because it has the strongest dispersion forces with the solvent.
QuestionAt 20∘C, the solubility of sodium chloride (NaCl) is 36.0 g per 100.0 g of water. A student adds 50.0 g of NaCl to 200.0 g of water at 20∘C. How much additional NaCl must be added to create a saturated solution?
3. Answer the following questions about carbon disulfide, CS₂, and carbonyl sulfide, COS.
Carbon disulfide, CS₂, and carbonyl sulfide, COS, are both liquids at room temperature. The molar mass of CS₂ is 76.14 g/mol, and the molar mass of COS is 60.07 g/mol. The Lewis electron-dot diagrams for both molecules are shown in Figure 1.
Figure 1. Lewis electron‑dot diagrams for CS₂ and COS (linear molecules)
i. Identify the types of intermolecular forces present in liquid CS₂ and liquid COS.
ii. The boiling point of CS₂ is 319 K, and the boiling point of COS is 223 K. Explain this difference in terms of the intermolecular forces identified in part A(i).
A student places a 15.2 g sample of CS₂(l) in a rigid, evacuated 5.00 L container at 350 K. All of the liquid vaporizes.
i. Calculate the pressure, in atm, of the CS₂ gas in the container. (Assume R = 0.08206 L·atm/(mol·K))
ii. The actual pressure of the CS₂ gas in the container is found to be slightly lower than the value calculated in part B(i). Explain this deviation using principles of non-ideal gas behavior.
The student compares the motion of particles in samples of CS₂(g) and COS(g) at 350 K.
i. Which gas particles, CS₂ or COS, have the higher average kinetic energy? Justify your answer.
ii. Which gas particles, CS₂ or COS, have the higher root-mean-square speed? Justify your answer.
The student prepares a solution by dissolving iodine, I₂(s), in liquid CS₂.
The student uses a spectrophotometer to analyze the solution of I₂ in CS₂. A calibration curve is generated using standard solutions, as shown in Figure 2.
Figure 2. Beer’s Law calibration curve for iodine (I₂) dissolved in CS₂
6. A scientist is investigating the physical properties of dichloromethane, CH2Cl2, and carbon tetrachloride, CCl4. The Lewis electron-dot diagrams for the molecules are shown in Figure 1.
Table 1. Physical properties of dichloromethane and carbon tetrachloride
Substance | Molar Mass (g/mol) | Boiling Point (°C) | Enthalpy of Vaporization (kJ/mol) |
|---|
CH2Cl2 | 84.93 | 39.6 | 28.0 |
CCl4 | 153.82 | 76.7 | 30.0 |
Figure 1. Lewis electron‑dot diagrams for dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4).
5. A physical chemist is investigating the properties of fluoromethane (CH3F) and chloromethane (CH3Cl). The Lewis diagrams for the molecules are shown in Figure 1, and selected physical property data are provided in Table 1.
Table 1. Physical properties of fluoromethane and chloromethane
Compound | Molar Mass (g/mol) | Dipole Moment (D) | Boiling Point (K) |
|---|
CH3F | 34.03 | 1.85 | 195 |
CH3Cl | 50.49 | 1.87 | 249 |
Figure 1. Lewis diagrams for fluoromethane (CH3F) and chloromethane (CH3Cl)