Question 1Multiple choice · 1 markA* challenge
Which row describes the shape and bond angle(s) of NH3?
| shape | bond angle | |
|---|---|---|
| A | trigonal planar | 107° |
| B | trigonal pyramidal | 107° |
| C | trigonal planar | 120° |
| D | trigonal bipyramidal | 90° and 120° |
Show answer
Answer: B
10 free AS Level Chemistry Physical Chemistry questions written in the style of Cambridge Chemistry papers: 5 multiple-choice and 5 structured, each with its answer and mark scheme. 4 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Atomic structure, Atoms, molecules & stoichiometry, Chemical bonding, States of matter, Chemical energetics, Electrochemistry, Equilibria, Reaction kinetics. Work through the questions below, then generate a full Physical Chemistry paper with a fresh set every time.
Generate a full Physical Chemistry paper →Which row describes the shape and bond angle(s) of NH3?
| shape | bond angle | |
|---|---|---|
| A | trigonal planar | 107° |
| B | trigonal pyramidal | 107° |
| C | trigonal planar | 120° |
| D | trigonal bipyramidal | 90° and 120° |
Answer: B
A 2.5 g sample of impure magnesium carbonate contains 60% MgCO3 by mass. The impurity does not react. The sample reacts with excess hydrochloric acid.
What volume of carbon dioxide is produced at room conditions?
Answer: B
Which substance has a giant molecular (covalent) lattice?
Answer: B
At equilibrium [H2] = 0.3, [I2] = 0.68 and [HI] = 0.51 mol dm−3. What is Kc?
Answer: B
50 g of solution rises by 3.3 K when 0.099 mol of reactant reacts. What is ΔH in kJ mol−1?
Answer: C
A 2.00 g sample of limestone (impure CaCO3, with unreactive impurities) is added to 50.0 cm3 of 1.00 mol dm−3 HCl, which is in excess. When the reaction has finished, 25.0 cm3 of the remaining solution is titrated with 0.250 mol dm−3 NaOH. The mean titre is 20.08 cm3.
Calculate the amount of HCl present initially, and the amount remaining in 25.0 cm3 of the final solution.
Calculate the percentage by mass of CaCO3 in the limestone.
Explain why this back-titration method is used rather than titrating the limestone directly.
Calculate the volume of CO2 produced at room conditions.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.05 mol; 5.02 × 10−3 mol 1 × 0.050 B1 0.25 × 20.08 ÷ 1000 B1 | 2 |
| (b) | 100% HCl remaining in whole solution (× 2) C1 HCl reacted = initial − remaining C1 ÷ 2 and × Mr (100.1) C1 100 A1 | 4 |
| (c) | limestone is insoluble/reacts slowly so end-point cannot be seen directly; excess acid ensures complete reaction CaCO3 insoluble / slow reaction B1 complete reaction ensured by excess acid; excess measured accurately B1 | 2 |
| (d) | 480 cm3 mol CO2 = mol CaCO3 C1 480 A1 | 2 |
Electron pair repulsion theory predicts the shapes of molecules.
State the shape and bond angle of SF₆, PCl₅ and H₂O.
Explain why the bond angle in H₂O is smaller than in CH₄.
BF₃ reacts with NH₃ to form F₃B–NH₃. Explain how this bond forms.
Explain why HF has a much higher boiling point than HCl.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | octahedral 90°; trigonal bipyramidal 90° and 120°; bent 104.5° SF₆ octahedral, 90° B1 PCl₅ trigonal bipyramidal, 90° and 120° B1 H₂O non-linear, 104–105° B1 | 3 |
| (b) | lone pairs repel more H₂O has two lone pairs M1 lone pairs repel more strongly than bonding pairs, so the H–O–H angle is reduced M1 | 2 |
| (c) | dative bond from N lone pair to empty B orbital the lone pair on N is donated M1 into the empty orbital on B, forming a dative covalent bond M1 | 2 |
| (d) | hydrogen bonding HF forms hydrogen bonds between molecules M1 which are much stronger than the id–id and pd–pd forces in HCl M1 | 2 |
The first ionisation energies of the elements show trends across periods and down groups.
Define the first ionisation energy.
Explain why the first ionisation energy of oxygen is lower than that of nitrogen.
Write the full electronic configuration of Fe³⁺.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | energy to remove one electron from each atom in one mole of gaseous atoms energy required to remove one electron B1 from each atom in one mole of gaseous atoms B1 to form one mole of gaseous 1+ ions B1 | 3 |
| (b) | spin-pair repulsion in 2p oxygen has a pair of electrons in one 2p orbital B1 repulsion between paired electrons makes one easier to remove B1 | 2 |
| (c) | 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ B1 | 1 |
Electrolysis of aqueous copper(II) sulfate uses copper electrodes.
Write the half-equations at the anode and cathode.
A current of 0.50 A is passed for 30 minutes. Calculate the mass of copper deposited. (F = 96 500 C mol−1)
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | Cu → Cu²⁺ + 2e⁻; Cu²⁺ + 2e⁻ → Cu anode: Cu → Cu²⁺ + 2e⁻ B1 cathode: Cu²⁺ + 2e⁻ → Cu B1 | 2 |
| (b) | 0.296 g Q = It = 900 C C1 mol Cu = 900/(2 × 96 500) C1 0.296 g A1 | 3 |
Ammonia, NH₃, and boron trifluoride, BF₃, have different shapes.
Predict the shape and bond angle of each molecule, explaining your answers using VSEPR.
Explain why ammonia is very soluble in water.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | NH₃ trigonal pyramidal 107°; BF₃ trigonal planar 120° NH₃: 3 bond pairs + 1 lone pair B1 trigonal pyramidal, about 107° B1 BF₃: 3 bond pairs, no lone pairs B1 trigonal planar, 120° B1 | 4 |
| (b) | hydrogen bonds with water ammonia forms hydrogen bonds with water molecules B1 lone pair on N and δ+ H B1 | 2 |
In the 9701 syllabus, Physical Chemistry covers Atomic structure, Atoms, molecules & stoichiometry, Chemical bonding, States of matter, Chemical energetics, Electrochemistry, Equilibria, Reaction kinetics.
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in