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. Melting and boiling points are at 1 atm. Data marked “invented” are fictional and only for practice.
Question 1 (multiple choice · foundation)
A grey solid (invented data) melts at 2700 °C. It does not conduct electricity as a solid or when molten, does not dissolve in water, and scratches steel. Which kind of solid is it most likely to be?
- (A) Ionic
- (B) Metallic
- (C) Molecular
- (D) Covalent network
Answer and explanation
Answer: (D). A very high melting point means strong bonds must break. No conduction in any state means no mobile ions or electrons. Hardness fits a rigid 3D network of covalent bonds.
- (A) A molten ionic compound conducts, because its ions can move.
- (B) A metal conducts as a solid.
- (C) Molecular solids melt at low temperatures, because only intermolecular forces hold them.
Question 2 (multiple choice · core)
Diamond and graphite are both pure carbon. Which statement best explains why graphite is soft enough to use in pencils, while diamond is extremely hard?
- (A) The covalent bonds in graphite are much weaker than those in diamond.
- (B) Graphite consists of layers held together by weak dispersion forces, so the layers slide; diamond is a 3D network in which every atom is held by covalent bonds at fixed angles.
- (C) Graphite is a molecular solid made of separate C₆ rings, while diamond is a covalent network.
- (D) Graphite contains mobile electrons that act as a lubricant between atoms.
Answer and explanation
Answer: (B). Softness depends on the weakest link. In graphite that is the dispersion force between sheets; in diamond there is no weak link, because covalent bonds run in all three directions.
- (A) The bonds within a graphite layer are strong. The softness comes from the forces between layers.
- (C) Graphite has no separate molecules; each layer is a giant sheet of bonded carbon atoms.
- (D) Mobile electrons explain graphite’s conductivity, not its softness.
Question 3 (multiple choice · core)
Which liquid has the highest vapor pressure at 20 °C?
- (A) Water, H₂O
- (B) Ethanol, CH₃CH₂OH
- (C) Acetone, (CH₃)₂CO
- (D) Propan-1-ol, CH₃CH₂CH₂OH
Answer and explanation
Answer: (C). Acetone is polar but has no H bonded to O, so its molecules cannot hydrogen bond with each other. The other three all have O–H groups and hydrogen bond. The weakest total attractions give the most molecules escaping, so the highest vapor pressure.
- (A) Water has the most hydrogen bonding per molecule and a low vapor pressure.
- (B) and (D) hydrogen bond; propan-1-ol also has more electrons than ethanol, so it has the lower vapor pressure of the two. Choosing (B) because it is “smaller” ignores hydrogen bonding.
Question 4 (multiple choice · core)
Steel is iron containing a small percentage of carbon atoms in the gaps between iron atoms. Compared with pure iron, steel is
- (A) less malleable, and still a good electrical conductor.
- (B) less malleable, and a poor electrical conductor.
- (C) more malleable, and still a good electrical conductor.
- (D) more malleable, and a poor electrical conductor.
Answer and explanation
Answer: (A). The interstitial carbon atoms get in the way when layers of iron cations try to shift, so the lattice is more rigid. The sea of mobile electrons remains, so steel still conducts.
- (B) An alloy keeps its delocalized electrons; it does not become an insulator.
- (C) and (D) Interstitial atoms reduce malleability; they do not increase it.
Question 5 (constructed response · core)
Iodine, I₂, is a grey-black solid that melts at 114 °C. It does not conduct electricity.
A student draws a particulate model of solid iodine as a regular grid of single I atoms, all equally spaced, with every atom joined to its four neighbours by solid lines.
(a) Identify the error in the student’s model and describe a correct particulate model. (b) Use your model to explain why iodine melts at a relatively low temperature. (c) Use your model to explain why iodine does not conduct electricity.
Worked solution and suggested Marlbridge rubric
(a) The student has drawn a covalent network. Solid iodine is a molecular solid. A correct model shows separate I₂ molecules (pairs of I atoms joined by one covalent bond), arranged in a regular pattern, with a key showing weaker intermolecular attractions (dashed lines) between molecules. The gaps between molecules are larger than the I–I bond length.
(b) Melting overcomes only the dispersion forces between I₂ molecules; the I–I covalent bonds do not break. These forces are fairly strong for a molecular substance (106 electrons per molecule, a large polarizable cloud), but far weaker than the bonds in a network or ionic solid.
(c) Each molecule’s valence electrons are held in its covalent bond and lone pairs, and there are no ions, so no charged particles are free to move.
| Point | What earns it |
|---|---|
| 1 | Identifies that iodine is molecular: separate I₂ molecules, not a network |
| 1 | Model distinguishes the I–I covalent bond from weaker forces between molecules |
| 1 | Melting overcomes only intermolecular (dispersion) forces; covalent bonds stay intact |
| 1 | No mobile electrons or ions: electrons localized in bonds and lone pairs |
Question 6 (constructed response · core)
Two invented polymers each consist of long chains.
- Polymer P: chains made only of –CH₂– units. Softens at 120 °C (invented).
- Polymer Q: chains in which every few carbon atoms there is a –C(=O)–NH– group. Softens at 250 °C (invented).
(a) Identify the strongest interaction between neighbouring chains in each polymer. (b) Explain the difference in softening temperatures. (c) Neither polymer conducts electricity. Explain why.
Worked solution and suggested Marlbridge rubric
(a) P: London dispersion forces only (nonpolar hydrocarbon chains). Q: hydrogen bonding, because the N–H hydrogen on one chain is attracted to a lone pair on the C=O oxygen of a neighbouring chain. Q also has dispersion and dipole–dipole forces.
(b) Softening needs neighbouring chains to slide past each other, which means overcoming the forces between them. In Q, many hydrogen bonds between chains add to the dispersion forces, so more energy is needed: a higher softening temperature.
(c) Both are molecular substances. Their electrons are held in covalent bonds and lone pairs within each chain, and there are no ions, so no charged particles can move through the material.
| Point | What earns it |
|---|---|
| 1 | Dispersion only for P; hydrogen bonding (N–H to O=C) for Q |
| 1 | Softening means overcoming forces between chains, not breaking covalent bonds |
| 1 | Q’s hydrogen bonds act in addition to dispersion forces, so the attraction is stronger |
| 1 | No mobile charged particles: electrons localized in bonds and lone pairs, no ions |
Accept “dipole–dipole” for Q in (a) only if hydrogen bonding is also named.
Question 7 (argumentation · stretch)
The two xylenes are isomers with the formula C₈H₁₀ (a benzene ring with two –CH₃ groups).
| Isomer | Position of –CH₃ groups | Melting point | Boiling point |
|---|---|---|---|
| ortho-xylene | next to each other | −25 °C | 144 °C |
| para-xylene | opposite each other | 13 °C | 138 °C |
A student claims: “para-Xylene has the stronger intermolecular forces, because it melts at a temperature 38 °C higher.”
(a) Evaluate the student’s claim. (b) Suggest why para-xylene has the higher melting point.
Model answer and suggested Marlbridge rubric
(a) The claim uses the wrong evidence. Boiling overcomes intermolecular forces completely, so boiling point is the direct measure. ortho-Xylene boils 6 °C higher, so it has slightly stronger intermolecular forces. This fits the structures: both have 58 electrons and similar dispersion forces, but ortho-xylene has a small dipole (its two –CH₃ groups are on the same side), while the symmetrical para-xylene is nonpolar.
(b) Melting only rearranges molecules. para-Xylene is symmetrical, so its molecules pack neatly and regularly into a crystal, making the solid harder to disorder. The less symmetrical ortho-xylene packs less efficiently.
| Point | What earns it |
|---|---|
| 1 | States the claim is not supported because melting point is not a direct measure of force strength |
| 1 | Uses the boiling points to conclude ortho-xylene has (slightly) stronger forces |
| 1 | Structural reason: similar dispersion (same electrons) plus a small dipole in ortho; para is nonpolar |
| 1 | Melting point depends on packing; symmetrical para-xylene packs better in the solid |
How did you do?
- Q1 or Q4 wrong: revisit the summary table and “Metallic solids” in the study guide.
- Q2 or Q5 wrong: compare Figures 1 and 2 and re-read Worked example 2.
- Q3, Q6 or Q7 wrong: re-read “Boiling point and vapor pressure”, “Melting point is more subtle” and Worked example 3.
Then tick off the topic checklist.