Imagine all substances are held together with Velcro. Some molecules have industrial-strength Velcro (hydrogen bonds), some have standard Velcro (dipole-dipole), and some have the weakest possible fuzzy fabric (dispersion forces only). The strength of the "Velcro" between molecules determines the boiling point, viscosity, surface tension — everything. Learning to identify and rank IMFs is the final key to predicting physical properties.
Core Content
| IMF Type | Present in | Relative strength | Cause | Example molecules |
|---|---|---|---|---|
| Dispersion forces | ALL molecules (polar and non-polar) | Weakest (increases with size) | Instantaneous temporary dipoles from electron cloud fluctuations | CH₄, CO₂, I₂, noble gases, hydrocarbons |
| Dipole-dipole forces | Polar molecules only (non-polar molecules excluded) | Moderate (in addition to dispersion) | Permanent partial charges (δ+/δ−) due to electronegativity differences in bonds | HCl, SO₂, CHCl₃, acetone (CH₃COCH₃) |
| Hydrogen bonding | Molecules with N–H, O–H, or F–H bonds | Strongest IMF (~5–10× dipole-dipole) | Highly polar H···X bond (X = F, O, N); lone pair on adjacent F, O, or N | H₂O, HF, NH₃, alcohols, DNA base pairs |
Even in non-polar molecules, electrons are constantly moving. At any instant, the electron distribution may be asymmetric — creating a temporary (instantaneous) dipole. This temporary dipole induces a dipole in the adjacent molecule. The result is a weak, fleeting attraction.
Two conditions must BOTH be met:
| Substance | IMFs present | BP (°C) | Expected BP without H-bonding | Anomaly |
|---|---|---|---|---|
| H₂O | Dispersion + H-bonding (O–H···O) | 100 | ~−80 (extrapolating H₂S trend) | +180°C anomaly |
| HF | Dispersion + H-bonding (F–H···F) | 19.5 | ~−100 (extrapolating HCl trend) | +120°C anomaly |
| NH₃ | Dispersion + H-bonding (N–H···N) | −33 | ~−100 (extrapolating PH₃ trend) | +67°C anomaly |
| H₂S | Dispersion + dipole-dipole only | −60 | — | No anomaly (no H-bonding) |
Insert diagram showing 5–6 water molecules with H-bonds drawn as dashed lines between O of one molecule and H of adjacent molecule. Label: O–H covalent bond (solid line), hydrogen bond O–H···O (dashed line), δ+ on H, δ− on O. Show that each O can accept 2 H-bonds and each H can donate 1 H-bond, giving a maximum of 4 H-bonds per water molecule.
Worked Examples
| Step 1 — Identify IMFs in each molecule
A — Propane (C₃H₈): No polar bonds, no N/O/F–H → only dispersion forces. MW = 44, 26 electrons. B — Propan-1-ol (C₃H₇OH): Has O–H bond → hydrogen bonding. Also has dispersion forces. MW = 60, 34 electrons. C — Chloromethane (CH₃Cl): C–Cl bond is polar (Cl is electronegative) → dipole-dipole + dispersion. No O/N/F–H bond → no hydrogen bonding. MW = 51, 26 electrons. |
Always check for H-bonding first (N–H, O–H, F–H). Then check for polarity (dipole-dipole). Then assess size for dispersion. This is the three-step framework. |
| Step 2 — Rank IMF strength for each
A: Dispersion only (weakest overall) → lowest BP expected C: Dispersion + dipole-dipole; similar size to propane → moderate BP B: Dispersion + hydrogen bonding; larger molecule too → highest BP expected |
When A and C have similar size/electron counts, the extra dipole-dipole forces in C push it above A. B wins because H-bonding is far stronger than dipole-dipole, and B is also the largest molecule. |
| Step 3 — Confirm with actual BPs
A — Propane: BP −42°C C — Chloromethane: BP −24°C B — Propan-1-ol: BP 97°C |
The actual BPs confirm the prediction. B's hydrogen bonding gives it a BP ~140°C higher than A, despite only being ~16 Da heavier. This illustrates how dramatically hydrogen bonding elevates BP compared to dispersion and dipole-dipole forces. |
| Compound | MW (g mol⁻¹) | BP (°C) |
|---|---|---|
| H₂O | 18 | 100 |
| H₂S | 34 | −60 |
| H₂Se | 81 | −41 |
| H₂Te | 130 | −2 |
| Identify the general trendH₂S → H₂Se → H₂Te: BP increases from −60°C to −41°C to −2°C. This is because MW increases down the group → more electrons → stronger dispersion forces → higher BP. This is the expected trend for molecules with only dispersion and weak dipole-dipole forces. | Extrapolating this trend downward would predict H₂O should have a BP of approximately −80°C — much lower than 100°C. The massive 180°C deviation is the anomaly. |
| Explain the H₂O anomalyH₂O has a dramatically higher BP (100°C) than predicted by the trend (~−80°C extrapolated). This is because oxygen is highly electronegative (χ = 3.5), making O–H bonds strongly polar. H₂O can form strong hydrogen bonds (O–H···O) between molecules. Hydrogen bonding is far stronger than the dispersion forces that govern H₂S, H₂Se, H₂Te — requiring much more energy to overcome. | S, Se, Te are not electronegative enough to create H-bonds. H₂S has only dispersion + weak dipole-dipole forces despite having an H–S bond. The electronegativity difference is key: χ(O) = 3.5, χ(S) = 2.6 — oxygen forms H-bonds; sulfur doesn't. |
Activities
1 Identify all IMFs present in each molecule: (a) CH₄ (methane), (b) HCl, (c) CH₃OH (methanol), (d) NH₃.
2 Rank these molecules in order of increasing boiling point and justify: butane (C₄H₁₀, non-polar, MW 58), dimethyl ether (CH₃OCH₃, polar, MW 46), ethanol (C₂H₅OH, has O–H, MW 46).
3 Two students disagree. Student A says HF has the highest BP of the hydrogen halides (HF, HCl, HBr, HI) because "fluorine forms the strongest bonds." Student B says HI should have the highest BP because "it has the most electrons." Who is correct, and why?
| Alkane | Formula | MW (g mol⁻¹) | BP (°C) |
|---|---|---|---|
| Methane | CH₄ | 16 | −162 |
| Ethane | C₂H₆ | 30 | −89 |
| Propane | C₃H₈ | 44 | −42 |
| Butane | C₄H₁₀ | 58 | −1 |
| Pentane | C₅H₁₂ | 72 | 36 |
| Hexane | C₆H₁₄ | 86 | 69 |
A Describe the trend in boiling points in the alkane series. What type of IMF is responsible for this trend, and why does it increase with each additional –CH₂– group?
B Pentanol (C₅H₁₁OH, MW 88, has O–H bond) has a BP of 138°C. Compare this to pentane (C₅H₁₂, MW 72, BP 36°C). Both have 5 carbon chains. Explain the large difference in BP despite the similar size.
C Using the alkane data, predict the approximate BP of heptane (C₇H₁₆, MW 100). Explain your reasoning and state any assumptions.
Multiple Choice
Click to check. One attempt only.
1. Which of the following molecules has hydrogen bonding as one of its intermolecular forces?
2. The boiling points of the noble gases increase from He (−269°C) to Xe (−108°C). This trend is best explained by:
3. Which correctly ranks these molecules from lowest to highest boiling point: methane (CH₄), ethanol (C₂H₅OH), chloroethane (C₂H₅Cl)?
4. The BP of HF (19°C) is anomalously high compared to HCl (−85°C), even though HCl is a larger, heavier molecule. The best explanation is:
5. A researcher compares two isomers of C₅H₁₂: n-pentane (straight chain, BP 36°C) and neopentane (tetrahedral/branched, BP 9.5°C). Both are non-polar. The difference in BP is best explained by:
Short Answer
6. Explain why the boiling point of water (100°C) is much higher than that of hydrogen sulfide (H₂S, −60°C), even though H₂S is a larger, heavier molecule. In your answer, identify all the IMFs present in each substance and explain which is stronger. 4 MARKS
7. A student is given three substances with BP data: fluoromethane (CH₃F, BP −78°C), methanol (CH₃OH, BP 65°C), and methane (CH₄, BP −162°C). (a) Identify the IMFs present in each substance. (b) Explain the BP trend in terms of IMF strength. 5 MARKS
8. Evaluate the statement: "The boiling point of a substance depends entirely on the type of intermolecular force present — a substance with hydrogen bonding will always have a higher boiling point than one with only dispersion forces." Is this statement correct? Provide evidence and reasoning to support your answer. 4 MARKS
1. (a) CH₄: dispersion forces only (non-polar, no N/O/F–H bonds). (b) HCl: dispersion + dipole-dipole (polar C–Cl bond; no H-bonding as Cl not electronegative enough). (c) CH₃OH: dispersion + hydrogen bonding (O–H bond → O–H···O H-bonds). (d) NH₃: dispersion + hydrogen bonding (N–H bond → N–H···N H-bonds).
2. Ranking (lowest → highest): Butane (C₄H₁₀) < Dimethyl ether (CH₃OCH₃) < Ethanol (C₂H₅OH). Butane: non-polar → dispersion only; MW 58 but only dispersion → BP −1°C. Dimethyl ether: polar (C–O bonds) → dispersion + dipole-dipole; MW 46; BP −24°C. Despite being lighter, ether is higher than methane-sized molecules; but lower than ethanol. Ethanol: has O–H → hydrogen bonding is the dominant IMF; BP 78°C despite being same MW as ether (46) — H-bonding is far stronger than dipole-dipole, explaining the ~100°C difference.
3. Student A is correct — but for the right reason: HF has the highest BP (19°C) among the hydrogen halides because F is the most electronegative atom, enabling strong F–H···F hydrogen bonding between HF molecules. HI has more electrons (larger dispersion forces) and its BP (−35°C) is higher than HBr (−67°C) and HCl (−85°C) but lower than HF. The hydrogen bonding in HF easily outweighs the increased dispersion forces in the larger, heavier HI. Student B's reasoning is partly correct (more electrons → stronger dispersion) but fails to account for the dominant H-bonding effect in HF.
A: BP increases steadily as chain length increases (from −162°C for CH₄ to 69°C for C₆H₁₄). The IMF responsible is dispersion forces — the only type present in non-polar alkanes. Each additional –CH₂– group adds approximately 8 electrons (6 from C + 2 from H bonds), increasing the size and polarisability of the electron cloud → stronger temporary dipoles → stronger dispersion forces → more energy to vaporise → higher BP.
B: Pentanol (BP 138°C) is 102°C higher than pentane (BP 36°C) despite only 16 Da difference in MW. Pentanol has an O–H group → it forms strong hydrogen bonds (O–H···O) between molecules, requiring much more energy to separate them. Pentane is non-polar with only dispersion forces — relatively weak. Hydrogen bonding in pentanol is approximately 5–10 times stronger than the dispersion forces in pentane, producing the ~100°C BP elevation.
C: Predicted BP ~98°C (actual = 98°C). Reasoning: each step in the series adds ~CH₂ (MW +14) and increases BP by approximately 25–35°C. From hexane (C₆H₁₄, BP 69°C), adding one CH₂ unit to give heptane (C₇H₁₆) should increase BP by ~28–30°C → predicted 97–99°C. Assumption: the trend is linear (constant increment per CH₂), which is approximately true for mid-range alkanes but slightly changes for very short or very long chains.
1. C — NH₃ has N–H bonds → hydrogen bonding. HCl: H–Cl, no H-bonding. CH₄: no polar bonds, no H-bonding. CO₂: no N/O/F–H bonds (C=O bonds but no H attached to O).
2. B — Noble gases are monatomic and non-polar → only dispersion forces. Larger atoms have more electrons → stronger dispersion → higher BP.
3. A — CH₄: dispersion only, very small → lowest. C₂H₅Cl: dispersion + dipole-dipole → moderate. C₂H₅OH: dispersion + H-bonding → highest. Actual BPs: CH₄ −161°C, C₂H₅Cl 12°C, C₂H₅OH 78°C.
4. D — HF forms F–H···F hydrogen bonds due to F's very high electronegativity. HCl cannot H-bond (Cl insufficient electronegativity). H-bonding in HF dominates over the greater dispersion forces of larger HCl.
5. C — Same formula, same MW, same non-polar nature → same type of IMFs (dispersion only). The difference must be shape: n-pentane's linear form has more surface contact area → stronger total dispersion. Neopentane's compact sphere has minimal contact area → weaker dispersion. This shape effect on BP is an important concept.
Q6 (4 marks): H₂O has dispersion forces AND hydrogen bonding (O–H···O). Oxygen's high electronegativity (χ = 3.5) makes O–H bonds strongly polar, creating a large δ+ on H — this H forms a hydrogen bond with the lone pair on O of an adjacent water molecule (1 mark). H₂S has only dispersion forces and weak dipole-dipole forces — sulfur (χ = 2.6) is not electronegative enough to enable hydrogen bonding, so no H···S hydrogen bonds form (1 mark). Hydrogen bonds in water are far stronger (5–10×) than the dispersion and dipole-dipole forces in H₂S, requiring much more energy to break on boiling (1 mark). Despite H₂S being larger and heavier (MW 34 vs 18), its weaker IMFs mean it boils at −60°C; water's strong H-bonding elevates its BP to 100°C, a difference of 160°C (1 mark).
Q7 (5 marks): CH₄ IMFs: dispersion forces only (non-polar, no polar bonds, no N/O/F–H) (1 mark). CH₃F IMFs: dispersion + dipole-dipole forces (C–F bond is polar: F is electronegative → permanent dipole; no H bonded to F, so no H-bonding) (1 mark). CH₃OH IMFs: dispersion + hydrogen bonding (O–H bond present; O is highly electronegative → H forms H-bonds with lone pairs on O of adjacent molecules) (1 mark). BP trend: CH₄ lowest (−162°C) — weakest IMFs (dispersion only, small molecule) (1 mark). CH₃F is higher (−78°C) — dipole-dipole forces in addition to dispersion, despite similar size to CH₄; polar C–F bond adds extra attractive force between molecules. CH₃OH highest (65°C) — hydrogen bonding is far stronger than dipole-dipole, creating a much larger energy requirement to vaporise (1 mark).
Q8 (4 marks): The statement is incorrect as an absolute claim (1 mark). While hydrogen bonding is the strongest type of IMF, boiling point depends on the total strength of all IMFs — which is influenced by both the type of IMF AND the molecular size (number of electrons → dispersion force strength) (1 mark). A counterexample: hexadecane (C₁₆H₃₄, non-polar, BP 287°C) has only dispersion forces yet boils far higher than water (BP 100°C, which has hydrogen bonding). The very large dispersion forces from 130+ electrons in hexadecane dominate over the H-bonding in water's 10-electron molecule (1 mark). A more accurate statement would be: "For molecules of similar size, a substance with hydrogen bonding will generally have a higher boiling point than one with only dispersion forces. However, for molecules of very different sizes, the stronger dispersion forces in a much larger molecule can exceed the hydrogen bonding in a smaller molecule" (1 mark).
Tick when you've finished all activities and checked your answers.