Difficulty: Intermediate | Prerequisites: Polarity, electronegativity, Lewis structures
Intermolecular forces (IMFs) govern physical properties like boiling point, melting point, viscosity, and solubility. The AP exam frequently asks you to identify which types of IMF are present in a substance and then use physical data (boiling points, condensation temperatures) to rank their relative strengths. This is one of the most predictable free-response topics, and the scoring rubric is quite specific about the language it expects.
All molecules have London dispersion forces. Polar molecules also have dipole-dipole forces. Molecules with N-H, O-H, or F-H bonds have hydrogen bonding (the strongest common IMF). You can compare IMF strength between two substances by looking at which one condenses or boils at a higher temperature: stronger IMFs mean higher boiling/condensation points.
Intermolecular forces (IMFs)
Forces of attraction between separate molecules (as opposed to intramolecular bonds within a molecule). Think of them as the glue that holds molecules near each other in a liquid or solid.
London dispersion forces (LDFs)
Weak, temporary attractive forces caused by momentary fluctuations in electron density, creating instantaneous dipoles. Present in all molecules and atoms. In simple terms, electrons slosh around randomly and create brief, weak attractions. Larger molecules with more electrons have stronger LDFs.
Dipole-dipole forces
Attractive forces between the positive end of one polar molecule and the negative end of another. Think of them as magnets lining up, positive to negative. They only exist in polar molecules.
Hydrogen bonding
An especially strong type of dipole-dipole interaction occurring when hydrogen is bonded to nitrogen, oxygen, or fluorine. In simple terms, it is the strongest common IMF (short of ionic interactions) and is responsible for water's unusually high boiling point.
Condensation point
The temperature at which a gas transitions to a liquid. A higher condensation point indicates stronger intermolecular forces, because more energy must be removed before the molecules slow down enough to stick together.
HCl is a polar molecule (electronegativity difference between H and Cl creates a permanent dipole).
It therefore has dipole-dipole forces.
Like all molecules, it also has London dispersion forces.
It does not have hydrogen bonding, because H is bonded to Cl, not to N, O, or F.
Complete answer: dipole-dipole forces and London dispersion forces.
The problem states that at 70°C, CCl₄ condenses (becomes liquid) while HCl remains gaseous.
This means CCl₄ has a higher condensation point than HCl.
A higher condensation point indicates stronger intermolecular forces.
Therefore, the IMFs among CCl₄ molecules are stronger than those among HCl molecules.
CCl₄ is nonpolar (tetrahedral symmetry cancels the bond dipoles), so it has only London dispersion forces.
HCl is polar and has both LDFs and dipole-dipole forces.
Yet CCl₄ still has stronger total IMFs. This is because CCl₄ is a much larger molecule with more electrons (total electrons: 74) compared to HCl (18 electrons).
More electrons means stronger London dispersion forces, and in this case, CCl₄'s very strong LDFs outweigh HCl's dipole-dipole forces combined with its weaker LDFs.
London dispersion forces scale with molecular size and electron count.
For small polar molecules vs large nonpolar molecules, the LDFs of the larger molecule can dominate.
The AP exam tests whether you can reason from physical evidence (boiling point, condensation temperature) to relative IMF strength, not just list the types.
No formulas for this topic. The key reasoning chain is:
Higher condensation/boiling point → stronger IMFs → more energy needed to overcome attractions between molecules
IMF hierarchy (weakest to strongest):
London dispersion < dipole-dipole < hydrogen bonding
But LDFs increase with molecular size, so large nonpolar molecules can have stronger total IMFs than small polar ones.
This is why cooking oil (large nonpolar molecules with strong LDFs) has a much higher boiling point than rubbing alcohol (smaller, polar, with hydrogen bonding). It is also why noble gases with more electrons (xenon vs helium) have higher boiling points, even though neither is polar.
Students often assume that because CCl₄ is nonpolar, its IMFs must be weaker than those of polar HCl. Size matters: CCl₄'s large electron cloud gives it very strong London dispersion forces.
A frequent error is listing "van der Waals forces" as a single type. On the AP exam, you should specify which type: London dispersion, dipole-dipole, or hydrogen bonding. "Van der Waals" is an umbrella term and does not earn full credit on its own.
Some students claim HCl has hydrogen bonding because it contains hydrogen. Hydrogen bonding requires H bonded to N, O, or F specifically.
When comparing IMF strength, students sometimes just list the types without using the physical evidence given. The rubric requires you to cite the condensation/boiling point data as evidence.
⚠️ "Identify all types" means list every IMF present, not just the strongest one. For HCl, you need both dipole-dipole and London dispersion.
⚠️ "Justify in terms of the information above" means cite the specific data from the problem (e.g. "CCl₄ condenses at 70°C while HCl remains gaseous"). Do not just state a rule.
⚠️ Know that hydrogen bonding requires H-N, H-O, or H-F bonds specifically. HCl does not qualify.
True or false: All molecules have London dispersion forces.
Fill in the blank: Hydrogen bonding occurs when H is bonded to ________, ________, or ________.
True or false: A nonpolar molecule always has weaker IMFs than a polar molecule.
Fill in the blank: A substance with a higher boiling point has ________ intermolecular forces.
True or false: HCl exhibits hydrogen bonding.
Answers: 1. True. 2. Nitrogen, oxygen, fluorine. 3. False. 4. Stronger. 5. False.
Q: Identify all types of intermolecular forces present in liquid HCl.
A: Dipole-dipole forces and London dispersion forces. HCl is polar (permanent dipole), and all molecules experience London dispersion forces. It does not have hydrogen bonding because H is bonded to Cl, not to N, O, or F.
Q: CCl₄ condenses at 70°C while HCl remains gaseous at the same temperature. What does this tell you about their relative IMF strengths?
A: The intermolecular forces among CCl₄ molecules are stronger than those among HCl molecules. CCl₄ condenses at a higher temperature, meaning more thermal energy must be removed before its molecules can be held together as a liquid, indicating stronger attractions.
Q: Explain why CCl₄ has stronger IMFs than HCl, even though CCl₄ is nonpolar.
A: CCl₄ is a much larger molecule with 74 electrons, compared to HCl's 18. The London dispersion forces increase with the number of electrons and the size of the electron cloud. CCl₄'s very strong LDFs exceed the combined dipole-dipole and London dispersion forces in the smaller HCl molecule.
Q: Would you expect Br₂ or HF to have a higher boiling point? Explain.
A: HF (bp 19.5°C) has hydrogen bonding, which is strong. However, in practice, the answer depends on which effect dominates. Br₂ (bp 59°C) actually has the higher boiling point because its much larger electron cloud produces very strong London dispersion forces that outweigh HF's hydrogen bonding, illustrating the same principle as the CCl₄ vs HCl comparison.
IMFs connect to phase diagrams and heating curves (the energy input at phase transitions reflects IMF strength). They also relate to solution chemistry: "like dissolves like" is really about compatible IMFs between solute and solvent. Colligative properties (boiling point elevation, freezing point depression) build on the same foundation of molecular interactions.
intermolecular forces, IMF, London dispersion forces, van der Waals, dipole-dipole, hydrogen bonding, boiling point, condensation point, polarity, nonpolar, CCl₄, HCl, electron cloud, molecular size, AP Chemistry free response, phase transitions, physical properties, IMF comparison