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. Data for every question: molar masses C 12.01, Fe 55.85, Ni 58.69, Cu 63.55, Zn 65.38 g mol⁻¹; radii Fe 126, Ni 124, Cu 128, Zn 134 pm (metallic radii) and C 75 pm (covalent radius).
Question 1 (multiple choice · foundation)
Which description best represents the bonding in solid potassium?
- (A) A regular array of K⁺ ions surrounded by valence electrons that are free to move through the whole solid
- (B) A regular array in which K⁺ ions alternate with K⁻ ions
- (C) Potassium atoms joined to their neighbours by shared pairs of electrons in fixed bonds
- (D) Separate neutral potassium atoms held together only by weak attractions between atoms
Answer and explanation
Answer: (A). This is the electron-sea model: positive cores in an array, held together by their attraction to delocalized valence electrons.
- (B) describes an ionic solid. A solid metal contains no anions.
- (C) describes covalent bonding, where electrons are localized between two particular atoms. In a metal the electrons are not fixed in bonds.
- (D) describes a molecular or noble-gas solid. Weak attractions alone could not explain why metals are solids with strong bonding throughout.
Question 2 (multiple choice · core)
A fictional metal Q has atoms of radius 140 pm. Four fictional elements, R, S, T and U, have atomic radii of 138 pm, 60 pm, 145 pm and 130 pm. Which element is most likely to form an interstitial alloy with Q?
- (A) R
- (B) S
- (C) T
- (D) U
Answer and explanation
Answer: (B). An interstitial alloy needs atoms that are much smaller than the host atoms, so they fit into the gaps. S differs from Q by |140 − 60| ÷ 140 × 100 = 57.1 %.
- (A) R differs by only 1.4 %. It is almost the same size, so it would substitute for Q.
- (C) T differs by 3.6 % and is slightly larger than Q. It could not fit into the gaps; it would substitute.
- (D) U differs by 7.1 %. “Smaller” is not enough; it is still comparable in size, so it would substitute rather than sit in the gaps.
Question 3 (multiple choice · core)
A copper wire can be bent and drawn thinner without breaking, but a crystal of sodium chloride shatters when struck. Which statement best explains the behaviour of the copper?
- (A) When layers of copper cores slide past each other, the delocalized electrons still surround the cores, so no strong repulsions develop.
- (B) Metallic bonds are always much weaker than ionic bonds, so the layers separate easily without breaking the metal.
- (C) Copper atoms are not arranged in any regular pattern, so there are no layers to break apart.
- (D) Each valence electron stays attached to its own copper atom, keeping every atom neutral as it moves.
Answer and explanation
Answer: (A). The cores are all positive and all identical. When one layer moves, the cores sit in new positions within the same sea of electrons, and the attraction between cores and sea is kept.
- (B) is not the reason. Many metals have strong metallic bonding and high melting points; malleability comes from the non-directional bonding, not weakness.
- (C) is false. Metal cores are in a regular array; the layers slide rather than break.
- (D) contradicts the model. The valence electrons are delocalized, not attached to particular atoms.
Question 4 (multiple choice · core)
A copper–zinc alloy contains 1 zinc atom for every 3 copper atoms. What is the percent by mass of zinc in the alloy?
- (A) 25.0 %
- (B) 25.5 %
- (C) 33.3 %
- (D) 74.5 %
Answer and explanation
Answer: (B). Take 1 mol Zn and 3 mol Cu. Mass of Zn = 65.38 g. Mass of Cu = 3 × 63.55 = 190.65 g. Percent Zn = 65.38 ÷ (65.38 + 190.65) × 100 = 25.5 %.
- (A) is the atom percent (1 atom in 4). It ignores the slightly larger molar mass of zinc.
- (C) treats the ratio 1 : 3 as one part in three.
- (D) is the percent by mass of copper, not zinc.
Question 5 (calculation · core)
A sample of steel contains 0.45 % carbon by mass; the rest is iron.
(a) Calculate the number of carbon atoms for every 100 iron atoms. (b) A student draws a particulate model of this steel with one carbon atom in the gap beside every second iron atom. Use your answer to (a) to comment on the model.
Worked solution and suggested Marlbridge rubric
(a) Take 100 g: 0.45 g C and 99.55 g Fe. n(C) = 0.45 ÷ 12.01 = 0.03747 mol. n(Fe) = 99.55 ÷ 55.85 = 1.7825 mol. Carbon atoms per 100 iron atoms = 100 × 0.03747 ÷ 1.7825 = 2.10.
(b) The student’s model shows 1 carbon atom per 2 iron atoms, which is 50 per 100 iron atoms. The real steel has only about 2 per 100, so roughly one carbon atom for every 48 iron atoms (1.7825 ÷ 0.03747 = 47.6). The model has far too many carbon atoms. A better model shows many iron atoms with an occasional small carbon atom in a gap. (The idea that carbon sits in the gaps is correct.)
| Point | What earns it |
|---|---|
| 1 | Moles of C and Fe from the mass percentages |
| 1 | 2.1 carbon atoms per 100 iron atoms |
| 1 | States the model has too many C atoms, comparing with the calculated ratio |
Common error: using the mass ratio as an atom ratio gives 0.45 per 100, which forgets that a carbon atom is much lighter than an iron atom.
Question 6 (constructed response · core)
(a) A student draws a particulate model of solid magnesium containing 9 magnesium atoms. Describe what the model should show, stating the number and charge of the cores and the number of delocalized electrons. (b) Explain, using your model, why magnesium conducts electricity. (c) Steel is an interstitial alloy of iron and carbon. Explain why steel is less malleable than pure iron, yet still conducts electricity.
Worked solution and suggested Marlbridge rubric
(a) Each magnesium atom has 2 valence electrons and gives both to the sea. The model shows 9 Mg²⁺ cores of equal size in a regular array, with 18 electrons spread between and around them, not attached to any core. Total charge: 9(+2) + 18(−1) = 0.
(b) The delocalized electrons are free to move through the whole array. When a voltage is applied, they move towards the positive terminal, carrying charge, so the metal conducts.
(c) In steel, small carbon atoms sit in the gaps between iron atoms. They get in the way when layers of iron cores try to slide, making the lattice more rigid, so steel is harder to bend or reshape. The iron cores are still surrounded by a sea of delocalized electrons, so mobile charge carriers are still present and steel conducts.
| Point | What earns it |
|---|---|
| 1 | 9 Mg²⁺ cores and 18 delocalized electrons, in a regular array |
| 1 | Conductivity linked to mobile delocalized electrons |
| 1 | Carbon atoms in the gaps hinder the sliding of layers (more rigid lattice) |
| 1 | Electron sea still present in the alloy, so it conducts |
Do not award the first point for “Mg⁺ cores” or for electrons drawn attached to individual cores.
Question 7 (argumentation · stretch)
A copper–nickel alloy contains 75 % copper and 25 % nickel by mass.
(a) Use the radii given to classify the alloy as interstitial or substitutional. Show your reasoning. (b) Calculate the fraction of atoms that are nickel atoms, and state how many nickel atoms a 20-atom particulate model should contain. (c) A student argues: “Nickel atoms are smaller than copper atoms, so they must sit in the gaps between the copper atoms.” Evaluate this argument.
Model answer and suggested Marlbridge rubric
(a) Percent difference = |128 − 124| ÷ 128 × 100 = 3.1 %. The atoms are of comparable size, so nickel atoms take the places of some copper atoms in the lattice: a substitutional alloy.
(b) Take 100 g: n(Cu) = 75 ÷ 63.55 = 1.1802 mol; n(Ni) = 25 ÷ 58.69 = 0.4260 mol. Fraction of Ni atoms = 0.4260 ÷ (1.1802 + 0.4260) = 0.265. In a 20-atom model: 20 × 0.265 = 5.3, so about 5 nickel atoms and 15 copper atoms, all at normal lattice positions.
(c) The argument is not valid. Being smaller is not enough to make an interstitial alloy; the added atoms must be much smaller than the host atoms to fit into the gaps, as carbon (75 pm) is compared with iron (126 pm). A nickel atom is only 3 % smaller than a copper atom. It is far too large to fit between copper atoms, but it fits well in a copper atom’s place, so it substitutes.
| Point | What earns it |
|---|---|
| 1 | Substitutional, supported by the small size difference (about 3 %) |
| 1 | Fraction of Ni atoms 0.265 (or 26.5 %) with working |
| 1 | About 5 Ni atoms (and 15 Cu) in a 20-atom model |
| 1 | Rejects the argument: interstitial atoms must be much smaller; a nearly equal-sized Ni atom replaces Cu instead |
Accept 0.27 in (b) if the working is shown.
How did you do?
- Q1, Q3 or Q6 wrong: re-read “Drawing the electron-sea model” and “What the model explains” in the study guide.
- Q2 or Q7(a, c) wrong: practise classifying alloys by relative size (Worked example 1).
- Q4, Q5 or Q7(b) wrong: go back through moles, as in Worked examples 2 and 3.
Then tick off the topic checklist.