These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. No calculator is needed. Where boiling points are given, they are measured at 1 atm and rounded to the nearest degree. Electrons per atom: H 1, C 6, N 7, O 8.
Question 1 (multiple choice · foundation)
Which substance can form hydrogen bonds between its own molecules?
- (A) Ethanal, CH₃CHO
- (B) Fluoroethane, CH₃CH₂F
- (C) Methoxyethane, CH₃OCH₂CH₃
- (D) Ethane-1,2-diol, HOCH₂CH₂OH
Answer and explanation
Answer: (D). Ethane-1,2-diol has two O–H groups. Each H on O is a donor, and each O has lone pairs to act as an acceptor, so its molecules hydrogen bond with each other.
- (A) has an O with lone pairs, but every H is bonded to C. It can accept a hydrogen bond (from water, say) but cannot donate one to another ethanal molecule.
- (B) contains F, but the F is bonded to C, not to H. No H is on N, O or F.
- (C) is an ether. Like dimethyl ether, all its H atoms are on carbon.
Question 2 (multiple choice · core)
Krypton boils at a higher temperature than argon. Which statement best explains this?
- (A) Krypton atoms have more electrons in a larger cloud, so the cloud is more polarizable and the dispersion forces between atoms are stronger.
- (B) Krypton atoms form weak covalent bonds with each other, but argon atoms do not.
- (C) Krypton atoms have a permanent dipole because of their larger number of protons.
- (D) Krypton atoms are heavier, so they move more slowly and are harder to separate.
Answer and explanation
Answer: (A). Noble gas atoms are nonpolar, so dispersion is the only force between them. A larger, more polarizable electron cloud gives larger temporary dipoles and stronger attractions.
- (B) Noble gases do not form covalent bonds with each other; boiling separates atoms, it does not break bonds.
- (C) A single atom has a symmetrical cloud on average. The number of protons does not create a permanent dipole.
- (D) uses mass as the cause. Mass tracks electron count here, but the attraction comes from the electron cloud, not from slower motion.
Question 3 (multiple choice · core)
For which pair is a dipole–induced dipole force the strongest interaction besides London dispersion forces?
- (A) A hydrogen chloride molecule, HCl, and an argon atom
- (B) A sodium ion, Na⁺, and a water molecule
- (C) A methane molecule, CH₄, and an ethane molecule, C₂H₆
- (D) An ammonia molecule, NH₃, and a water molecule
Answer and explanation
Answer: (A). HCl is polar and argon is nonpolar. HCl’s dipole distorts argon’s electron cloud and the induced dipole is attracted to it.
- (B) An ion and a polar molecule: the main extra force is ion–dipole.
- (C) Both molecules are nonpolar, so there are dispersion forces only.
- (D) Both molecules are polar and both have H on N or O with lone pairs on N or O, so they hydrogen bond with each other.
Question 4 (multiple choice · core)
Hexane, CH₃(CH₂)₄CH₃, and 2,2-dimethylbutane, (CH₃)₃CCH₂CH₃, have the same formula, C₆H₁₄. Hexane boils at 69 °C and 2,2-dimethylbutane at 50 °C. Which statement best explains the difference?
- (A) Hexane’s straight chains can lie alongside each other, giving more contact area and stronger dispersion forces.
- (B) Hexane has more electrons per molecule, so its dispersion forces are stronger.
- (C) Hexane molecules form hydrogen bonds, but 2,2-dimethylbutane molecules do not.
- (D) The branched molecule is polar, and dipole–dipole forces are weaker than dispersion forces.
Answer and explanation
Answer: (A). The two molecules have the same 50 electrons, so polarizability is similar. The compact, branched molecule touches its neighbours at fewer points, so its dispersion forces are weaker.
- (B) Same formula means the same number of electrons (6 × 6 + 14 × 1 = 50).
- (C) Neither molecule has H on N, O or F.
- (D) Both are hydrocarbons and essentially nonpolar. Also, dipole–dipole forces add to dispersion forces; they do not replace them.
Question 5 (explanation · core)
Propylamine, CH₃CH₂CH₂NH₂, and trimethylamine, (CH₃)₃N, are isomers with the formula C₃H₉N. One boils at about 48 °C and the other at about 3 °C.
(a) Identify all the intermolecular forces between molecules of each substance. (b) Match each substance to its boiling point and justify your answer. (c) Describe how two propylamine molecules line up when they hydrogen bond.
Worked solution and suggested Marlbridge rubric
(a) Both molecules are polar (the N atom has a lone pair and polar bonds that do not cancel). Propylamine: dispersion, dipole–dipole and hydrogen bonding, because it has two H atoms bonded to N. Trimethylamine: dispersion and dipole–dipole only. Its N has a lone pair, but all nine H atoms are bonded to C, so it has no donor H.
(b) Propylamine boils at about 48 °C, trimethylamine at about 3 °C. Both have 34 electrons, so their dispersion forces are of similar size. Propylamine has hydrogen bonding in addition, so more energy is needed to separate its molecules.
(c) An H atom on the N of one molecule (δ+) points towards the lone pair on the N of the other molecule (δ−). The N–H···N arrangement is close to a straight line.
| Point | What earns it |
|---|---|
| 1 | Identifies hydrogen bonding in propylamine, linked to H bonded to N |
| 1 | States trimethylamine has no hydrogen bonding because no H is on N (dipole–dipole and dispersion only) |
| 1 | Correct match and uses equal electron count (similar dispersion) so that hydrogen bonding decides |
| 1 | Correct orientation: δ+ H on N of one molecule towards the lone pair on N of the other |
Do not award the third point for “propylamine has stronger forces” without saying why the dispersion forces are comparable.
Question 6 (constructed response · core)
A protein is one long chain. When it folds, three kinds of contact form:
- Contact 1: an –OH group on a side chain lies next to a C=O group in a distant part of the same chain.
- Contact 2: two nonpolar –CH(CH₃)₂ side chains lie next to each other inside the folded protein.
- Contact 3: a positively charged –NH₃⁺ group on the surface is surrounded by water molecules.
(a) Name the main interaction in each contact. (b) For Contact 3, describe how the water molecules are oriented. (c) Explain why the folded shape can be changed by gentle heating even though no covalent bonds in the chain break.
Worked solution and suggested Marlbridge rubric
(a) Contact 1: hydrogen bonding (H on the O of –OH to a lone pair on the O of C=O). Contact 2: London dispersion forces. Contact 3: ion–dipole forces between –NH₃⁺ and water.
(b) The water molecules point their δ− oxygen atoms (with lone pairs) towards the positive –NH₃⁺ group, with their H atoms pointing away.
(c) The shape is held by many noncovalent interactions between different parts of the chain. Each one is much weaker than a covalent bond. A small amount of energy can overcome them and let the chain rearrange, while the covalent bonds along the chain stay intact.
| Point | What earns it |
|---|---|
| 1 | Hydrogen bonding for Contact 1 and dispersion for Contact 2 (both needed) |
| 1 | Ion–dipole for Contact 3 |
| 1 | Oxygen (δ−) of water faces the –NH₃⁺ group |
| 1 | Shape held by noncovalent forces that are much weaker than covalent bonds, so they are overcome first |
Accept “hydrogen bonding” for Contact 3 as well as ion–dipole if the student notes that the N–H atoms of –NH₃⁺ can donate to water’s O; the orientation in (b) must still put water’s O towards the group.
Question 7 (argumentation · stretch)
Data (boiling points at 1 atm): methanol, CH₃OH, 65 °C; water, H₂O, 100 °C; octane, C₈H₁₈, 126 °C.
A student claims: “Hydrogen bonding is the strongest intermolecular force, so any substance whose molecules hydrogen bond will boil at a higher temperature than any substance whose molecules do not.”
(a) Use the data to evaluate the claim. (b) Water and methanol both hydrogen bond, and methanol has more electrons. Suggest why water still boils at a higher temperature.
Model answer and suggested Marlbridge rubric
(a) The claim is not supported. Methanol hydrogen bonds, but octane does not, yet octane boils about 61 °C higher. Octane has 66 electrons against methanol’s 18, and its long chains have a large contact area. Its dispersion forces add up to more than methanol’s combined dispersion, dipole–dipole and hydrogen bonding forces. A single hydrogen bond is stronger than a single dipole–dipole or dispersion interaction, but the total attraction also depends on how much dispersion acts over the whole molecule. The claim only works for molecules of similar size.
(b) A water molecule has two H atoms on O and two lone pairs on O, so each molecule can take part in up to four hydrogen bonds. Methanol has only one H on O, so it forms fewer hydrogen bonds per molecule. The extra hydrogen bonding in water outweighs methanol’s slightly larger dispersion forces (18 electrons against 10).
| Point | What earns it |
|---|---|
| 1 | States the claim is wrong, citing octane (no hydrogen bonding) boiling above methanol (hydrogen bonding) |
| 1 | Explains with octane’s much larger electron count or contact area giving stronger dispersion forces |
| 1 | Water has more hydrogen bond donors (two O–H) than methanol (one O–H) |
| 1 | Links more hydrogen bonds per molecule to stronger total attraction, outweighing the small difference in dispersion forces |
How did you do?
- Q1 or Q5 wrong: re-read the two rules for hydrogen bonding in the study guide.
- Q2, Q4 or Q7(a) wrong: revisit “What makes dispersion forces stronger” and Worked example 2.
- Q3 or Q6 wrong: practise naming forces between two different species (Worked example 3).
Then tick off the topic checklist.