Chemistry • Year 11 • Module 1 • Lesson 10
Intermolecular Forces and Physical Properties
Apply your understanding of IMF type and strength to real data, ranking tasks, a cause-and-effect chain, and a real-world scenario.
1. Interpret boiling-point data — Group 17 hydrogen halides
The table below gives boiling points and molecular masses for the hydrogen halides. 8 marks
| Molecule | Molecular mass (g mol−1) | Boiling point (°C) | IMFs present |
|---|---|---|---|
| HF | 20 | 19.5 | |
| HCl | 36 | −85 | |
| HBr | 81 | −67 | |
| HI | 128 | −35 |
1.1 Complete the “IMFs present” column for each hydrogen halide. 4 marks (1 per row)
1.2 Describe the general trend in boiling point from HCl to HI and explain it in terms of IMFs. 2 marks
1.3 Identify which molecule is anomalous in the series and explain why it does not follow the general trend. 2 marks
2. Interpret graph — boiling points of alkanes
The graph below shows boiling points for the first six straight-chain alkanes. 7 marks
Figure 2. Boiling points of straight-chain alkanes (C1–C6). Data from NIST WebBook. IMF present: dispersion forces only.
2.1 Describe the trend in boiling point as the number of carbon atoms increases. 2 marks
2.2 Explain the trend at the molecular level, using the terms electron count, dispersion forces, and energy in your response. 3 marks
2.3 Using the graph, estimate the boiling point of heptane (C7H16). State one assumption you are making. 2 marks
3. Compare IMF types across five features
Complete the three-column table below. For each feature, write a concise description for each IMF type. 12 marks (1 per cell; 4 rows × 3 columns)
| Feature | Dispersion forces | Dipole-dipole forces | Hydrogen bonding |
|---|---|---|---|
| Which molecules? | |||
| Relative strength | |||
| Cause / origin | |||
| Named example molecule |
4. Cause-and-effect chain — why ethanol has a higher boiling point than dimethyl ether
Ethanol (C2H5OH, MW 46) and dimethyl ether (CH3OCH3, MW 46) are isomers with identical molecular masses, yet ethanol has a boiling point of 78 °C while dimethyl ether boils at −24 °C. Complete the cause-and-effect chain below. 5 marks
| Cause 1: Ethanol has an O–H bond; dimethyl ether has only C–O and C–H bonds with no H attached to O. | → | Effect 1: |
| Cause 2: Hydrogen bonding is approximately 5–10× stronger than dipole-dipole forces. | → | Effect 2: |
| Cause 3: Both molecules have similar molecular mass (both 46 g mol−1) so their dispersion forces are similar. | → | Effect 3: |
| Overall outcome (so…): | ||
5. Predict and justify — a WA petroleum refinery scenario
The Kwinana refinery in Western Australia processes crude oil by fractional distillation: the crude mixture is heated and different fractions collected at different temperatures. Fractions collected at lower temperatures have shorter carbon chains; fractions collected at higher temperatures have longer chains. 4 marks
5.1 Predict which alkane fraction — short-chain (e.g. propane, C3H8) or long-chain (e.g. octane, C8H18) — would be collected first from the top of the distillation column (lowest temperature zone) and justify your prediction in terms of IMFs. 2 marks
5.2 A student claims that water contamination in the crude oil would collect near the top of the distillation column because “water has a small molecular mass of 18.” Is this correct? Explain why or why not using IMFs. 2 marks
Q1.1 — IMFs present in hydrogen halides
HF: Dispersion forces + hydrogen bonding (F–H···F; F is electronegative enough). HCl: Dispersion forces + dipole-dipole forces (polar H–Cl bond; Cl not electronegative enough for H-bonding). HBr: Dispersion forces + dipole-dipole forces. HI: Dispersion forces + dipole-dipole forces (very weak dipole-dipole; large dispersion forces dominate).
Q1.2 — General trend HCl to HI (2 marks)
Boiling point increases from HCl (−85 °C) to HBr (−67 °C) to HI (−35 °C) [1]. This is because molecular mass and electron count increase down the series → stronger dispersion forces → more energy needed to vaporise → higher BP [1].
Q1.3 — Anomalous molecule (2 marks)
HF is the anomalous molecule [1]. Despite being the lightest hydrogen halide, it has the highest boiling point (19.5 °C) because fluorine’s very high electronegativity enables strong F–H···F hydrogen bonding between molecules. This hydrogen bonding far outweighs the dispersion forces in the larger, heavier HI, explaining the anomalously high BP [1].
Q2.1 — Alkane boiling point trend (2 marks)
Boiling point increases steadily (approximately linearly) as the number of carbon atoms increases from 1 to 6 [1]. Each additional carbon adds to the molecular size, raising the boiling point by approximately 25–35 °C per –CH2– unit [1].
Q2.2 — Molecular-level explanation (3 marks)
Each additional –CH2– unit increases the electron count of the molecule by approximately 8 electrons [1], making the electron cloud larger and more polarisable → stronger temporary dipoles → stronger dispersion forces between adjacent molecules [1]. More energy is required to overcome these stronger dispersion forces during boiling, so the boiling point is higher [1].
Q2.3 — Heptane BP estimate (2 marks)
Estimated BP ≈ 95–100 °C (actual: 98 °C) [1]. Assumption: the increase in boiling point is approximately constant (linear) with each additional CH2 group, so adding one more carbon from hexane (BP 69 °C) should give an increase of approximately 28–32 °C [1].
Q3 — IMF comparison table
Which molecules? Dispersion: ALL molecules (polar and non-polar). Dipole-dipole: polar molecules only. Hydrogen bonding: molecules with N–H, O–H, or F–H bonds.
Relative strength: Dispersion: weakest (but increases with molecular size). Dipole-dipole: moderate (stronger than dispersion in small molecules). H-bonding: strongest (≈5–10× dipole-dipole for small molecules).
Cause / origin: Dispersion: instantaneous temporary dipoles from electron cloud fluctuations. Dipole-dipole: permanent partial charges (δ+/δ−) from electronegativity differences in bonds. H-bonding: highly polar H (bonded to F/O/N) interacting with lone pair on F/O/N of adjacent molecule.
Named example: Dispersion: CH4 / Cl2 / noble gases. Dipole-dipole: HCl / SO2 / CH3Cl. H-bonding: H2O / HF / NH3 / alcohols.
Q4 — Cause-and-effect chain (5 marks)
Effect 1: Ethanol can form hydrogen bonds (O–H···O) between molecules; dimethyl ether cannot (no H attached to O or N/F), so it has only dispersion and dipole-dipole forces [1]. Effect 2: Much more energy is required to separate ethanol molecules (overcome H-bonds) than dimethyl ether molecules (overcome dipole-dipole forces only) [1]. Effect 3: The stronger IMFs in ethanol cannot be attributed to larger dispersion forces (same MW), so the difference is entirely due to the H-bonding in ethanol [1]. Overall outcome: Ethanol has a significantly higher boiling point (78 °C) than dimethyl ether (−24 °C) despite identical molecular masses, demonstrating that IMF type (here, H-bonding vs dipole-dipole) can override molecular mass in determining BP [2 marks for a full, linked overall outcome].
Q5.1 — Kwinana distillation (2 marks)
Propane (C3H8) would be collected first, from the top (lowest temperature zone) [1]. Short-chain alkanes have fewer electrons → weaker dispersion forces → less energy needed to vaporise → lower boiling point. They therefore rise to the top of the column where temperatures are lowest and are collected first. Octane has a much longer chain, stronger dispersion forces, higher BP, and condenses lower in the column where it is hotter [1].
Q5.2 — Water BP in refinery (2 marks)
The student is incorrect [1]. Water (BP 100 °C) has a much higher boiling point than short-chain alkanes like propane (BP −42 °C) because water forms strong hydrogen bonds (O–H···O). Despite its small molecular mass (18 g mol−1), water’s H-bonds require far more energy to break than the dispersion forces in light alkanes. Water would therefore condense lower in the column (higher temperature zone), not near the top [1].