Edexcel International A Level Chemistry: Intermolecular forces (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers 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 content is in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, an IAS compulsory unit. Questions 9 to 12 are longer problems that combine outcomes.

Related: the study guide or the revision notes. Course hub: Edexcel International A Level Chemistry. Checklist: printable checklist. Diagnostics: free 10-minute diagnostics.

Questions

1. Describe how London forces arise between two molecules of nitrogen, N₂. [3]

2. State whether each substance forms hydrogen bonds between its own molecules. Give a reason each time.

(a) Methylamine, CH₃NH₂ [1] (b) Trimethylamine, (CH₃)₃N [1] (c) Ethanoic acid, CH₃COOH [1] (d) Chloromethane, CH₃Cl [1]

3. This question is about liquid hydrogen fluoride.

(a) Describe, or sketch, a hydrogen bond between two HF molecules. Show partial charges and the lone pair involved. [2] (b) State the F-H···F bond angle and explain why it takes this value. [2]

4. Boiling temperatures: H₂O 373 K, HF 293 K, NH₃ 240 K. Fluorine is the most electronegative of the three elements. Explain why water, not hydrogen fluoride, has the highest boiling temperature. [4]

5. Three isomers of C₅H₁₂ have these boiling temperatures: pentane 309 K, 2-methylbutane 301 K, 2,2-dimethylpropane 283 K.

(a) How many electrons does each isomer molecule contain? [1] (b) Explain the order of boiling temperatures. [3]

6. Butan-1-ol, CH₃CH₂CH₂CH₂OH, boils at 391 K. Pentane boils at 309 K. Explain why butan-1-ol is much less volatile than pentane. [4]

7. Use these boiling temperatures: HCl 188 K, HBr 206 K, HI 238 K.

(a) Explain the trend from HCl to HI. [3] (b) Hydrogen fluoride boils at 293 K. Explain why this is much higher than the value predicted from the HCl to HI trend. [2]

8. A block of ice of volume 2.40 m³ floats in a freshwater lake. Densities: ice 0.917 g cm⁻³, liquid water 1.00 g cm⁻³. The fraction of a floating object below the surface equals its density divided by the density of the liquid.

(a) Calculate the volume of the block above the water surface. [2] (b) Use ideas about hydrogen bonding to account for the block floating. [3]

9. A student shakes three liquids separately with water.

(a) 1-chlorobutane forms a separate layer, even though its molecules are polar. Explain this. [3] (b) Propan-1-ol mixes completely with water. Explain this. [2] (c) Suggest a better solvent than water for iodine, I₂, and explain your choice. [2]

10. Boiling temperatures of the Group 15 hydrides: NH₃ 240 K, PH₃ 185 K, AsH₃ 211 K, SbH₃ 256 K.

(a) Explain the trend from PH₃ to SbH₃. [2] (b) From PH₃ to SbH₃ the boiling temperature rises by about 2 K for each extra electron. Use this to estimate the boiling temperature NH₃ would have without hydrogen bonding. [2] (c) Explain why the actual boiling temperature of NH₃ is higher than your estimate, and why each NH₃ molecule forms fewer hydrogen bonds than each H₂O molecule. [3]

11. A 60.0 W heating coil keeps water boiling in an open vacuum flask that stands on a balance. Over 900 s the balance reading drops by 22.6 g.

(a) Calculate the enthalpy change of vaporisation of water, in kJ mol⁻¹. [4] (b) Use this value for the true enthalpy change of vaporisation: +40.7 kJ mol⁻¹. Suggest why the experimental value is higher. [1] (c) State which forces are overcome when water boils, and which bonds are not broken. [2]

12. A student stirs 7.58 g of potassium nitrate, KNO₃ (Mr = 101.1), into 60.0 g of water in a polystyrene cup. The thermometer reading drops by 10.4 K. Use c = 4.18 J g⁻¹ K⁻¹ and treat the 60.0 g of water as the mass heated.

(a) Calculate the enthalpy change of solution of potassium nitrate. Give the sign. [4] (b) Describe how water molecules interact with the ions as the solid dissolves. [2] (c) Explain why potassium nitrate does not dissolve in hexane. [2]

Answers

1. Electrons move randomly, so at one instant the electron cloud of one N₂ molecule is uneven, creating an instantaneous dipole [1]. This dipole distorts the electron cloud of a neighbouring molecule, creating an induced dipole [1]. The δ+ end of one dipole attracts the δ− end of the other [1]. [3] Examiner insight: Use both words, “instantaneous” and “induced”; “temporary charges” alone is too vague.

2. (a) Yes: H is bonded directly to N, and N has a lone pair [1]. (b) No: N has a lone pair, but every H is bonded to carbon [1]. (c) Yes: the O-H group provides δ+ H, and O atoms have lone pairs [1]. (d) No: polar, but no H bonded to N, O or F [1]. [4] Examiner insight: “No” or “yes” alone is not enough when a reason is asked for; point to the specific bond.

3. (a) F-H···F-H, with δ+ on H and δ− on F in each molecule [1]; the hydrogen bond drawn from the H of one molecule to a lone pair on F of the other [1]. (b) About 180° [1]. The two electron-rich F atoms repel, so they sit as far apart as possible, in line with H [1]. [4] Examiner insight: Draw the hydrogen bond as a dashed or dotted line ending at a lone pair, not at the F nucleus in general.

4. All three form hydrogen bonds [1]. Each H₂O molecule has two δ+ H atoms and two lone pairs, so on average it forms twice as many hydrogen bonds per molecule as HF [1]. HF has three lone pairs but only one H, so it forms fewer hydrogen bonds, even though each one is stronger [1]. More hydrogen bonds per molecule means more energy is needed to separate water molecules [1]. [4] Examiner insight: The key idea is the number of hydrogen bonds; electronegativity alone predicts the wrong order.

5. (a) 42 electrons in each (5 × 6 + 12 × 1) [1]. (b) Only London forces act between alkane molecules [1]. The more branched the molecule, the more compact and spherical it is, so the surface contact between neighbouring molecules is smaller [1]. Weaker London forces need less energy to overcome, so boiling temperature falls: pentane > 2-methylbutane > 2,2-dimethylpropane [1]. [4] Examiner insight: Because the electron count is the same, any answer based on “fewer electrons” or “smaller mass” earns nothing here.

6. Both molecules have 42 electrons [1], so their London forces are similar [1]. Butan-1-ol also has an O-H group, so it forms hydrogen bonds between molecules [1]. Extra energy is needed to overcome these, so butan-1-ol has a higher boiling temperature and lower volatility [1]. [4] Examiner insight: State the equal electron count explicitly; it is what makes the comparison fair.

7. (a) The number of electrons increases (18, 36, 54) [1], so London forces get stronger [1]. The bonds become less polar, so permanent dipole interactions weaken, but the increase in London forces is larger, so boiling temperature rises [1]. (b) HF forms hydrogen bonds (H bonded to F, which has lone pairs) [1]; breaking these takes far more energy than overcoming the London and dipole attractions in HCl [1]. [5] Examiner insight: Mention that dipoles weaken down the group, then say London forces outweigh this.

8. (a) Fraction below = 0.917 / 1.00 = 0.917, so fraction above = 0.083 [1]. Volume above = 2.40 × 0.083 = 0.199 m³ [1]. (b) In ice each molecule forms four hydrogen bonds, arranged tetrahedrally [1]. This holds the molecules in an open structure with empty space between them [1]. On melting, some hydrogen bonds break and the structure partly collapses, so molecules pack closer together in the liquid [1]. [5] Examiner insight: Give the unit, and describe the open arrangement; “more hydrogen bonds” alone explains nothing.

9. (a) 1-chlorobutane cannot form hydrogen bonds with water (no H on N, O or F) [1]. To mix, water molecules would have to break hydrogen bonds with each other [1], and only weaker permanent dipole and London attractions would form with 1-chlorobutane, so too little energy is released and the liquids separate [1]. (b) The O-H group of propan-1-ol forms hydrogen bonds with water molecules [1]; these replace the hydrogen bonds broken in each pure liquid [1]. (c) Hexane (or cyclohexane) [1]: iodine and hexane both have only London forces, so similar attractions form between solute and solvent [1]. [7] Examiner insight: In (a), “polar” is a trap; explain the failure to hydrogen bond, not “non-polar”.

10. (a) Electron number increases from PH₃ to SbH₃ [1], so London forces strengthen and boiling demands more energy [1]. (b) NH₃ has 10 electrons, 8 fewer than PH₃ [1]. Estimate = 185 − 8 × 2 ≈ 169 K (about 170 K) [1]. (c) NH₃ forms hydrogen bonds (H on N, lone pair on N), which need extra energy to overcome, so it boils about 70 K above the estimate [1]. Each NH₃ has three H atoms but only one lone pair [1], so it can accept only one hydrogen bond, while H₂O, with two H atoms and two lone pairs, can form more [1]. [7] Examiner insight: Show the working behind an estimate; a bare number is hard to credit.

11. (a) Energy = 60.0 × 900 = 54 000 J = 54.0 kJ [1]. Moles = 22.6 / 18.0 = 1.256 mol [1]. ΔvapH = 54.0 / 1.256 [1] = +43.0 kJ mol⁻¹ [1]. (b) Some energy is lost to the surroundings and does not evaporate water, so the energy per mole is overestimated [1]. (c) Hydrogen bonds (and London forces) between water molecules are overcome [1]; the covalent O-H bonds within molecules are not broken [1]. [7] Examiner insight: Vaporisation is endothermic, so give the positive sign; convert J to kJ before dividing.

12. (a) Moles KNO₃ = 7.58 / 101.1 = 0.0750 mol [1]. q = 60.0 × 4.18 × 10.4 = 2608 J [1]. ΔH = 2.608 / 0.0750 = 34.8 kJ mol⁻¹ [1]. The temperature fell, so the process is endothermic: ΔH = +34.8 kJ mol⁻¹ [1]. (b) The δ− oxygen atoms of water molecules are attracted to K⁺ ions, and the δ+ hydrogen atoms to NO₃⁻ ions [1]. Forming these ion-dipole attractions (hydration) releases energy that compensates for breaking the lattice [1]. (c) Hexane is non-polar and has only London forces [1], so it cannot hydrate the ions; too little energy would be released to overcome the strong ionic attractions in the lattice [1]. [8] Examiner insight: Use the mass of water, not the mass of solid, in q = mcΔT, and decide the sign from the temperature change.

Where marks are usually lost

• Saying a compound hydrogen bonds because it contains hydrogen, without checking H is on N, O or F.
• Arguing about boiling temperatures without first comparing electron numbers.
• Using “fewer electrons” to explain branching, when isomers have the same number.
• Explaining the HCl to HI trend with dipoles, which weaken down the group.
• Describing ice’s lower density without the tetrahedral, open structure.
• Saying halogenoalkanes are non-polar to explain their poor solubility in water.

Next steps

• Revision notes and study guide.
• Course hub: Edexcel International A Level Chemistry.
• Printable checklist: Edexcel A Level Chemistry checklist.
• Find your gaps with all free 10-minute diagnostics.
• Book a free trial class.

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.