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, effective ionic radii in pm: Li⁺ 76, Na⁺ 102, K⁺ 138, Mg²⁺ 72, Ca²⁺ 100, F⁻ 133, Cl⁻ 181, Br⁻ 196, I⁻ 220, O²⁻ 140. Take the distance between ion centres as d = r₊ + r₋.
Question 1 (multiple choice · foundation)
Which description gives the best particulate model of solid lithium fluoride, LiF?
- (A) Separate Li–F pairs, each with a small Li⁺ touching a larger F⁻, with gaps between the pairs
- (B) A continuous array in which Li⁺ and F⁻ alternate in every direction, the F⁻ ions are drawn larger, and the numbers of Li⁺ and F⁻ are equal
- (C) A continuous array in which Li⁺ and F⁻ alternate in every direction, the Li⁺ ions are drawn larger, and the numbers of Li⁺ and F⁻ are equal
- (D) Rows that contain only Li⁺ ions alternating with rows that contain only F⁻ ions, with the ions in each row touching
Answer and explanation
Answer: (B). An ionic solid is an extended array; each ion touches only opposite charges; Li⁺ (76 pm) is much smaller than F⁻ (133 pm); and LiF is 1 : 1.
- (A) shows molecules. In the solid each ion is attracted to all its neighbours, not to one partner.
- (C) reverses the sizes. A cation has lost its outer electrons, so Li⁺ is the smaller ion.
- (D) puts like charges in contact along each row. That arrangement would have strong repulsions, which the real array avoids.
Question 2 (multiple choice · core)
In which compound is the attraction between the ions strongest?
- (A) NaCl
- (B) KBr
- (C) CaO
- (D) NaI
Answer and explanation
Answer: (C). By Coulomb’s law the attraction depends on |q₊q₋| / d.
| Compound | Charge product | d (pm) | Relative to NaCl |
|---|---|---|---|
| NaCl | 1 | 283 | 1.00 |
| KBr | 1 | 334 | 0.85 |
| CaO | 4 | 240 | 4.72 |
| NaI | 1 | 322 | 0.88 |
CaO has 2+ and 2− ions, so its product of charges is four times larger than the others, and its ions are also small.
- (A) is the strongest of the 1+/1− compounds, but charge outweighs the small size difference.
- (B) has the largest distance between ion centres, so it has the weakest attraction here.
- (D) has the large I⁻ ion, so its d is long. Choosing it confuses “more electrons” with “stronger attraction”.
Question 3 (multiple choice · core)
The four potassium halides KF, KCl, KBr and KI all contain ions with charges of +1 and −1. Which is predicted to have the highest melting point?
- (A) KF
- (B) KCl
- (C) KBr
- (D) KI
Answer and explanation
Answer: (A). The charges are the same, so the distance between ion centres decides. F⁻ is the smallest halide ion, so KF has the shortest distance (138 + 133 = 271 pm) and the strongest attraction. The distances for KCl, KBr and KI are 319, 334 and 358 pm.
- (B) and (C) have larger anions than KF, so their ions are further apart.
- (D) has the largest anion and the weakest attraction. It would be the lowest, not the highest. Picking it suggests the idea that a heavier compound must melt at a higher temperature, which is not the reasoning used here.
Question 4 (multiple choice · core)
A crystal of calcium fluoride shatters when it is struck sharply with a hammer. Which statement best explains this?
- (A) The blow moves one layer of ions relative to the next, bringing ions of the same charge next to each other, and their repulsion splits the crystal.
- (B) The blow breaks the covalent bonds between calcium and fluorine atoms inside each CaF₂ molecule.
- (C) The blow knocks the delocalized electrons out of the crystal, so the positive ions repel each other.
- (D) The attractions between the ions are weak, so very little energy is needed to separate them.
Answer and explanation
Answer: (A). In the undisturbed array each ion faces opposite charges. A shift of one ion spacing lines up like charges, and repulsion pushes the layers apart. This is brittleness.
- (B) treats CaF₂ as a molecular, covalent substance. It is an ionic array with no molecules.
- (C) describes a metal. An ionic solid has no sea of delocalized electrons.
- (D) contradicts the model: the attractions are strong, which is why ionic solids have high melting points. Brittleness comes from repulsion after a shift, not weak attraction.
Question 5 (constructed response · core)
A student is asked to draw a two-dimensional particulate model of solid magnesium chloride, MgCl₂.
(a) The student’s drawing contains 12 ions in total. How many should be Mg²⁺ ions and how many Cl⁻ ions? Show that your answer gives a neutral drawing. (b) Describe how the sizes of the two kinds of ion should compare in the drawing, and explain why. (c) A second student draws 6 Mg²⁺ and 6 Cl⁻ ions. Explain why this model is not consistent with the formula or with the charges of the ions. (d) State one feature of the arrangement that both students’ drawings must show to be consistent with Coulomb’s law.
Worked solution and suggested Marlbridge rubric
(a) The formula MgCl₂ gives a ratio of 1 Mg²⁺ : 2 Cl⁻. 12 ions ÷ 3 = 4 groups, so 4 Mg²⁺ and 8 Cl⁻. Charge check: 4(+2) + 8(−1) = +8 − 8 = 0.
(b) The Cl⁻ ions should be drawn much larger than the Mg²⁺ ions (181 pm against 72 pm, about 2.5 times the radius). Mg²⁺ has lost its outer shell (two electrons), and it has 12 protons pulling on 10 electrons, so it is small. Cl⁻ has gained an electron into its outer (third) shell, so it is large.
(c) 6 Mg²⁺ and 6 Cl⁻ is a 1 : 1 ratio, not 1 : 2. The total charge would be 6(+2) + 6(−1) = +6, so the drawing is not neutral, but a real ionic compound is.
(d) Any one: each Mg²⁺ is surrounded by Cl⁻ ions; ions of the same charge do not touch; the pattern repeats as an extended array rather than separate units.
| Point | What earns it |
|---|---|
| 1 | 4 Mg²⁺ and 8 Cl⁻ with the charge total shown as zero |
| 1 | Cl⁻ drawn larger than Mg²⁺, with a reason based on lost or gained electrons (or the radii given) |
| 1 | Identifies the wrong ratio and the non-zero total charge (+6) in (c) |
| 1 | A correct arrangement feature linked to attraction of opposite charges or avoidance of like-charge repulsion |
Question 6 (calculation and explanation · core)
Potassium fluoride, KF, and calcium oxide, CaO, are both white ionic solids.
(a) Calculate the distance between ion centres, d, for each compound. (b) Calculate |q₊q₋| / d for each compound, and the ratio of the CaO value to the KF value. (c) K⁺ and Ca²⁺ both have 18 electrons. Explain why Ca²⁺ is the smaller ion. (d) Predict which compound has the higher melting point. Justify your prediction using Coulomb’s law, and say which factor, charge or distance, contributes more.
Worked solution and suggested Marlbridge rubric
(a) d(KF) = 138 + 133 = 271 pm. d(CaO) = 100 + 140 = 240 pm.
(b) KF: |(+1)(−1)| / 271 = 1 ÷ 271 = 0.00369 pm⁻¹. CaO: |(+2)(−2)| / 240 = 4 ÷ 240 = 0.01667 pm⁻¹. Ratio CaO : KF = 0.01667 ÷ 0.00369 = 4.52.
(c) Calcium has 20 protons and potassium has 19. With the same 18 electrons in the same shells, the extra proton in Ca²⁺ pulls the electrons closer to the nucleus, so Ca²⁺ is smaller (100 pm against 138 pm).
(d) CaO has the higher melting point. Its ions carry double the charge, so the product of charges is 4 times larger, and its ions are closer together (240 pm against 271 pm). Both effects make the attraction stronger, so more energy is needed to disrupt the array. Charge contributes more: it gives a factor of 4, while the shorter distance gives only 271 ÷ 240 = 1.13.
| Point | What earns it |
|---|---|
| 1 | Both distances correct |
| 1 | Both values of |q₊q₋| / d and the ratio 4.5 |
| 1 | Ca²⁺ smaller because of more protons with the same number of electrons |
| 1 | CaO predicted higher, linked to stronger Coulombic attraction |
| 1 | Identifies charge as the larger factor, with numbers (4 against about 1.1) |
Accept the ratio from unrounded values (4.52) or rounded values (4.5). A student who compares only the charge products, and says CaO is 4 times stronger, earns the prediction point but not the “which factor” point unless distance is also considered.
Question 7 (argumentation · stretch)
A student tests a sample of potassium bromide, KBr, with a simple conductivity tester. Results: the dry solid does not conduct; the molten salt does conduct; a solution of KBr in water does conduct.
The student claims: “Solid KBr does not conduct because it contains no ions. The ions only form when the solid melts or dissolves.”
Evaluate the claim. Use a particulate model of solid KBr in your answer.
Model answer and suggested Marlbridge rubric
Model answer. The claim is incorrect. Solid KBr already consists of K⁺ and Br⁻ ions arranged in a repeating three-dimensional array, with each K⁺ surrounded by Br⁻ ions and each Br⁻ surrounded by K⁺ ions. The ions are held in fixed positions by strong attractions between opposite charges; they can only vibrate. A current needs charged particles that can move through the sample, so the solid does not conduct. When the solid melts, or dissolves in water, the same ions are no longer locked in the array and can move, so the liquid and the solution conduct. The results show that ions must be mobile to conduct, not that ions are created by melting.
| Point | What earns it |
|---|---|
| 1 | States that the solid already contains K⁺ and Br⁻ ions in an extended array (claim rejected) |
| 1 | Explains that in the solid the ions are fixed in position by strong attractions, so they cannot carry charge |
| 1 | Explains that melting or dissolving lets the existing ions move, which is why both conduct |
Do not award the first point for “the claim is wrong” without describing the ions present in the solid.
How did you do?
- Q1 or Q5 wrong: re-read “Drawing a particulate model” and Worked example 1 in the study guide.
- Q2, Q3 or Q6 wrong: practise ranking with |q₊q₋| / d, charges first (Worked example 2).
- Q4 or Q7 incomplete: connect each property to what the ions are doing: fixed, mobile, or pushed next to like charges.
Then tick off the topic checklist.