Intermolecular Forces Practice #2- Advanced Problems
Intermolecular Forces Practice #2: Advanced Problems
If you're here, you probably already know the basics of intermolecular forces. Hydrogen bonding, dipole-dipole, London dispersion—the usual suspects. This post skips the intro lecture and jumps straight into problems that'll actually test your understanding.
These aren't the "identify the strongest IMF" questions from your first exam. These require you to compare forces across different molecules, predict physical properties, and explain why certain substances behave the way they do.
The Four Forces You Need to Know Cold
Before diving in, make sure you have these down cold. If any of these definitions feel fuzzy, go back and review. There's no hiding weak fundamentals at this level.
- London Dispersion Forces (LDF): Temporary dipoles caused by electron unevenness. Every molecule has them. They're the weakest IMF and get stronger as molecular weight increases.
- Dipole-Dipole: Permanent dipole interactions between polar molecules. Stronger than LDF but still weak compared to hydrogen bonding.
- Hydrogen Bonding: Occurs when H is bonded to N, O, or F and interacts with another N, O, or F. This is a big deal—it's significantly stronger than regular dipole-dipole.
- Ion-Dipole: The force between an ion and a polar molecule. This is what happens when NaCl dissolves in water.
Comparing the Forces: Strength Overview
Use this table to quickly check relative strengths. Memorize this, but also understand why each force has the strength it does.
| Force Type | Relative Strength | Present In |
|---|---|---|
| London Dispersion | Weakest | All molecules (nonpolar and polar) |
| Dipole-Dipole | Weak to Moderate | Polar molecules only |
| Hydrogen Bonding | Strong | Molecules with H-N, H-O, or H-F bonds |
| Ion-Dipole | Strong to Very Strong | Ions + polar molecules (solutions) |
Problem 1: Boiling Point Comparison
Predict the order of boiling points for CH₃OH, CH₃SH, and CH₃OH₂. Explain your reasoning.
First, identify the IMF in each:
- CH₃OH: Can hydrogen bond. The H attached to O interacts with another molecule's O.
- CH₃SH: No hydrogen bonding. S isn't electronegative enough. Only has LDF and weak dipole-dipole.
- CH₃OH₂: This is actually CH₃NH₂ (methylamine). The N allows for hydrogen bonding, but it's weaker than O because N is less electronegative.
The order: CH₃SH < CH₃NH₂ < CH₃OH
CH₃SH boils lowest because it lacks hydrogen bonding entirely. CH₃NH₂ can hydrogen bond but weaker than CH₃OH. CH₃OH has the strongest hydrogen bonding of the three.
Problem 2: Solubility Predictions
Will dimethyl ether (CH₃OCH₃) be more soluble in water or in hexane (C₆H₁₄)?
Dimethyl ether is polar—it has a dipole moment from the C-O-C arrangement. It cannot hydrogen bond with itself because there's no H attached to O.
In water: Water can hydrogen bond with dimethyl ether. This dipole-dipole interaction makes it somewhat soluble. But without H-bonding capability, it's not highly miscible.
In hexane: Both are nonpolar-ish. Hexane only has LDF. Dimethyl ether is polar, so LDF alone won't create strong attractions.
Answer: More soluble in water. The dipole-dipole interactions and ability to form some H-bonds with water outweigh the weak interactions with hexane.
Problem 3: Vapor Pressure Reasoning
Propionaldehyde (C₂H₅CHO) and acetone (CH₃COCH₃) have similar molecular weights. Acetone has a higher vapor pressure. Explain why.
Vapor pressure inversely relates to IMF strength. Weaker forces = easier to escape = higher vapor pressure.
Acetone is a ketone with carbonyl on the terminal carbon. Propionaldehyde is an aldehyde with the carbonyl at the end of a chain. Both are polar and have dipole-dipole interactions.
The difference: Propionaldehyde has slightly stronger dipole-dipole because of how the electrons are distributed. It also has a larger surface area, meaning more LDF opportunities.
Acetone's higher vapor pressure comes from weaker overall intermolecular forces. The carbonyl group's position in acetone creates less efficient packing and slightly weaker attractions compared to propionaldehyde.
Problem 4: Phase Change Prediction
At room temperature, Cl₂ is a gas while Br₂ is a liquid. Explain this using intermolecular forces.
Both are diatomic halogens with only London dispersion forces. No dipole-dipole, no hydrogen bonding.
The difference is molecular weight and electron cloud size. Br₂ (160 amu) is much heavier than Cl₂ (71 amu). Larger electron clouds create more temporary dipoles, meaning stronger LDF.
Stronger LDF = more energy needed to separate molecules = higher boiling point. Room temperature provides enough energy to vaporize Cl₂ but not Br₂.
Problem 5: Identifying Incorrect Statements
Which statement is false? "All molecules containing hydrogen can form hydrogen bonds."
This is false. Hydrogen bonding requires specific conditions:
- H must be attached to N, O, or F (highly electronegative atoms)
- The N, O, or F must have a lone pair to accept the H
Hydrocarbon hydrogens (C-H) cannot hydrogen bond. HCl cannot hydrogen bond. The H must be bonded directly to N, O, or F.
Getting Started: How to Approach IMF Problems
Follow this checklist when you encounter any IMF question:
- Identify all molecules involved. Write out the structures if needed.
- Determine polarity. Is the molecule polar? If yes, dipole-dipole exists.
- Check for N, O, or F. Is H bonded to any of these? If yes, hydrogen bonding is possible.
- Consider molecular weight and size. Larger molecules have stronger LDF.
- Compare apples to apples. When ranking, only change one variable at a time if possible.
Common Mistakes That Cost Points
Confusing molecular polarity with hydrogen bonding. Just because a molecule is polar doesn't mean it hydrogen bonds. Acetone is polar but cannot hydrogen bond with itself.
Ignoring molecular size. LDF scales with surface area and electron count. A large nonpolar molecule can have stronger LDF than a small polar molecule.
Forgetting that all molecules have LDF. Even molecules with strong hydrogen bonding still have LDF. It's always there.
Misidentifying H-bond donors and acceptors. Both the donor (H-N/O/F) and acceptor (lone pair on N/O/F) need to be present for hydrogen bonding to occur between two molecules.
Quick Reference for Exam Day
When comparing two substances:
- Hydrogen bonding > dipole-dipole > LDF (usually)
- Size matters when comparing same-type forces
- Phase at room temperature tells you about IMF strength immediately
- Boiling point ranking follows IMF ranking
If you can explain why each force exists and how it affects physical properties, you're ready. These problems aren't about memorization—they're about understanding the underlying electrostatic attractions that hold the molecular world together.