Edexcel International A Level Chemistry: Intermolecular forces (YCH11) – Revision Notes

Start with the intermolecular forces study guide if you need the ideas explained from scratch; this page is for quick recall.

These notes cover Topic 7: Intermolecular Forces (outcomes 7.1 to 7.6) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017. The topic is part of Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, an IAS compulsory unit.

When you are ready, try the practice questions. Course hub: Edexcel International A Level Chemistry. Checklist: printable checklist. Diagnostics: free 10-minute diagnostics.

The three forces at a glance (7.1)

Force	Found between	Origin	Gets stronger with
London forces	All molecules	Instantaneous dipole in one molecule induces a dipole in a neighbour	More electrons; larger contact area
Permanent dipole-permanent dipole	Polar molecules	δ+ end of one molecule attracts δ− end of another	Greater overall molecular dipole
Hydrogen bond	H on N, O or F, with a lone pair on N, O or F nearby	δ+ H attracted to a lone pair on a very electronegative atom	Number of hydrogen bonds each molecule can form

Strength order for a single interaction: London < permanent dipole < hydrogen bond, all much weaker than a covalent bond. In larger molecules the total London forces can outweigh permanent dipole effects.

Key definitions to write word for word:

• London force: an attraction between an instantaneous dipole in one molecule and the dipole it induces in a neighbouring molecule.
• Hydrogen bond: an attraction between a δ+ hydrogen atom bonded to N, O or F and a lone pair on an N, O or F atom of another molecule. The X-H···Y arrangement is close to linear (about 180°).

Hydrogen bonding in H₂O, NH₃ and HF (7.2 and 7.4)

Molecule	H on N/O/F	Lone pairs	What limits hydrogen bonding
H₂O	2	2	Nothing; balanced, so four per molecule in ice
NH₃	3	1	Only one lone pair to accept
HF	1	3	Only one H to donate; molecules form zig-zag chains

Boiling order: H₂O > HF > NH₃. Water wins on the number of hydrogen bonds, not on the strength of each one.

Method: does it hydrogen bond with itself?

1. Draw or picture the displayed formula.
2. Find any H bonded directly to N, O or F. If none, the answer is no.
3. Check for a lone pair on N, O or F. If present, the answer is yes.
4. For “with water” questions, a lone pair on O or N is enough, because water supplies the H.

Quick sort: yes for alcohols, carboxylic acids, amines with N-H, NH₃, HF, H₂O₂. No for ethers, aldehydes, ketones, halogenoalkanes, CH₃F and tertiary amines (all their H atoms are on carbon).

Water’s anomalous properties (7.3)

• High boiling and melting temperatures. Group 16 hydrides H₂S to H₂Te follow a London-force trend (more electrons, higher boiling temperature). Water has the fewest electrons yet boils highest, because extra energy is needed to break its hydrogen bonds.
• Ice is less dense than water. Each molecule in ice is hydrogen bonded to four others in a tetrahedral arrangement. The near-linear, relatively long hydrogen bonds hold molecules apart in an open structure with empty space. On melting, some hydrogen bonds break, the structure partly collapses and the molecules move closer.
• Ice at 0 °C: 0.917 g cm⁻³. Liquid at 0 °C: 1.00 g cm⁻³. Volume rises by about 9% on freezing.

Boiling temperature trends (7.5)

Comparison	Example data	Reason
Longer alkane chain	Propane 231 K, pentane 309 K	More electrons and more contact: stronger London forces
Branched isomer	Pentane 309 K, 2,2-dimethylpropane 283 K	Same electrons; compact shape, less contact, weaker London forces
Alcohol vs alkane with equal electrons	Ethanol 352 K, propane 231 K (26 electrons each)	Similar London forces; alcohol also has hydrogen bonds
HCl to HI	HCl 188 K, HBr 206 K, HI 238 K	More electrons, stronger London forces outweigh falling dipole
HF	293 K	Hydrogen bonding

Method: explaining any boiling temperature difference

1. Name the forces present in each substance.
2. Compare electron numbers (for London forces) and say whether they are similar or different.
3. Name the extra or stronger force in the higher-boiling substance.
4. Finish with energy: “more energy is needed to overcome the intermolecular forces”.

Solvents (7.6)

Solute	Good solvent	Why
Many ionic solids (e.g. KCl)	Water	Hydration: δ− O attracted to cations, δ+ H attracted to anions; energy released compensates for breaking the lattice
Short-chain alcohols	Water	Hydrogen bonds form between the O-H group and water
Halogenoalkanes	Non-aqueous (e.g. ethanol, other halogenoalkanes)	Polar, but cannot hydrogen bond with water; breaking water’s hydrogen bonds is not repaid
Non-polar molecules (I₂, alkanes, fats)	Hexane, cyclohexane	Similar London forces in solute and solvent

Rule of thumb: a substance dissolves well in a solvent whose intermolecular forces are similar to its own.

Suggested practicals (reminders)

• Enthalpy change of vaporisation of water: ΔvapH = energy supplied (power × time) ÷ moles evaporated (mass lost ÷ 18.0). Because some heat escapes rather than evaporating water, expect an answer above the true value.
• Temperature change on dissolving: q = m × c × ΔT using the mass of water; ΔH = q ÷ moles of solute, with sign from the direction of temperature change.
• Solubility of simple molecules in different solvents: match the forces.

Must-know distinctions

• Intramolecular vs intermolecular: boiling a simple molecular substance overcomes forces between molecules only.
• Polar molecule vs hydrogen bonding: CH₃Cl is polar but cannot hydrogen bond.
• Strength of one hydrogen bond vs number of hydrogen bonds: explains H₂O > HF.
• Electron number vs contact area: chain length changes both; branching changes only contact area.
• Accepting vs donating: methoxymethane cannot hydrogen bond with itself, but its oxygen lone pairs can accept hydrogen bonds from water, which helps it dissolve.

Quick self-test

1. Which type of intermolecular force acts between all molecules?
2. Which intermolecular force acts in liquid ICl but not in liquid Br₂?
3. State the two conditions needed for hydrogen bonding between molecules of one substance.
4. Can propanone, CH₃COCH₃, form hydrogen bonds between its own molecules?
5. Explain why liquid NH₃ forms fewer hydrogen bonds per molecule than water.
6. Hexane boils at 342 K and 2-methylpentane at 333 K. Give the reason in one sentence.
7. How many electrons does a molecule of ethanol, C₂H₅OH, have?
8. Why does HI boil higher than HCl even though HCl is more polar?
9. Using densities 0.917 g cm⁻³ (ice) and 1.00 g cm⁻³ (water), what fraction of a floating block of ice is below the surface of fresh water?
10. Why does 1-bromobutane not dissolve in water?
11. What attractions form when potassium chloride dissolves in water?
12. Over 300 s, a 400 W heater evaporates 52.0 g of boiling water. Calculate ΔvapH in kJ mol⁻¹.

Answers

1. London forces.
2. Permanent dipole-permanent dipole interactions (London forces act in both).
3. An H atom bonded directly to N, O or F, and a lone pair on an N, O or F atom of another molecule.
4. No. Its O has lone pairs, but every H is bonded to carbon.
5. NH₃ has only one lone pair, so it can accept only one hydrogen bond; water has two H atoms and two lone pairs.
6. Branching reduces surface contact between molecules, so London forces are weaker.
7. 26 (2 × 6 + 6 × 1 + 8).
8. HI has more electrons (54 vs 18), so its London forces are stronger, and this outweighs HCl’s larger dipole.
9. 0.917 / 1.00 = 0.917, so about 92% is below the surface.
10. It cannot form hydrogen bonds with water, so the energy needed to break water’s hydrogen bonds is not released again by new attractions.
11. Ion-dipole attractions (hydration): δ− O of water around K⁺, δ+ H of water around Cl⁻.
12. Energy = 400 × 300 = 120 000 J = 120 kJ; moles = 52.0 / 18.0 = 2.89 mol; ΔvapH = 120 / 2.89 = +41.5 kJ mol⁻¹.

Where marks are usually lost

• Writing “van der Waals forces” when the question asks you to name London forces precisely.
• Describing a London force without the word “induced” or without saying the dipole is instantaneous.
• Saying hydrogen bonding happens “because there is hydrogen in the molecule”.
• Drawing a hydrogen bond to an H atom rather than to a lone pair, or with a bent X-H···Y angle.
• Forgetting to compare electron numbers before arguing about hydrogen bonds in alcohol vs alkane questions.
• Writing that ice has “more hydrogen bonds so it is heavier” instead of describing the open structure.
• Explaining the HCl to HI trend using dipoles alone.
• Saying a halogenoalkane is non-polar to explain its insolubility in water.
• Missing the positive sign on an enthalpy change of vaporisation.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11), Specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 7: Intermolecular Forces, Unit 2.