CAIE A Level 9701

Chemistry 2025 · May/June · Paper 22

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This CAIE A Level Chemistry Paper 22 covers a wide spread of core first-year chemistry ideas, especially bonding and structure, periodicity, atomic structure, organic mechanisms, energetics, and rates. Students can expect a mix of structured short-answer questions, data-based calculations, explanations of observations, and mechanism-drawing tasks, with some questions requiring precise chemical equations, definitions, and interpretation of tables or graphs.…

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Questions & Model Answers 22 Qs

Q1a(i)
Short Answer 1 mark
Diamond and graphite are both crystalline solids made from carbon atoms. Graphite conducts electricity. Diamond does not conduct electricity. Name the type of lattice structure present in the crystalline solids diamond and graphite.
Giant covalent lattice
Both diamond and graphite consist of a giant covalent network of carbon atoms.
Q1a(ii)
Short Answer 1 mark
Explain how graphite conducts electricity.
Graphite has delocalised electrons that are free to move through the structure and carry charge.
The mobile delocalised electrons allow electrical conduction.
Q1a(iii)
Short Answer 1 mark
Explain why diamond does not conduct electricity.
Diamond has no delocalised electrons / all outer electrons are used in covalent bonds, so there are no mobile charge carriers.
Without mobile electrons or ions, diamond cannot conduct electricity.
Q1b(i)
Short Answer 2 marks
Separate samples of phosphorus(V) chloride and silicon(IV) chloride are each added to an excess of cold water. Write an equation for each reaction.
PCl5 + 4H2O -> H3PO4 + 5HCl
SiCl4 + 2H2O -> SiO2 + 4HCl
Q1b(ii)
Short Answer 2 marks
Describe the appearance of phosphorus(V) chloride and silicon(IV) chloride at room temperature.
phosphorus(V) chloride: white solid
silicon(IV) chloride: colourless liquid
Q1b(iii)
Short Answer 2 marks
Compare the appearance of the mixtures produced when each reaction is complete.
PCl5 gives a white mist / white fumes due to HCl gas; SiCl4 gives a white precipitate / white solid (silica) in a colourless solution.
Q1c(i)
Short Answer 1 mark
Some oxides are amphoteric. Describe what is meant by amphoteric.
Able to react as both an acid and a base.
Q1c(ii)
Short Answer 1 mark
Identify the formula of a Period 3 oxide that is amphoteric.
Al2O3
Q1d(i)
Short Answer 3 marks
Complete Table 1.1 by identifying the strongest force of attraction in each oxide that is broken during melting. Use the abbreviations i.d. = instantaneous dipole-induced dipole, p.d. = permanent dipole-permanent dipole, H = hydrogen bond, C = covalent bond, I = ionic bond.
SO2: p.d.
H2O: H
SiO2: C
Na2O: I
MgO: I
Q1d(ii)
Short Answer 2 marks
A student suggests the following hypothesis. The stronger the covalent bond between atoms in non-metal oxides, the higher the melting point. Use Table 1.1 to deduce if this hypothesis is true or false or if there is not enough information to make a conclusion. Explain your answer.
Not enough information to make a conclusion.
Table 1.1 compares different types of bonding/structure, not just covalent bond strength in non-metal oxides. Melting point depends on more than covalent bond strength alone.
Q2a(i)
Short Answer 2 marks
Define relative atomic mass.
The weighted mean mass of an atom of an element compared with 1/12 of the mass of an atom of carbon-12.
Q2a(ii)
Short Answer 2 marks
Table 2.1 shows the abundances of two of the isotopes present in the sample of iron. Use Table 2.1 to calculate the relative isotopic mass of the third isotope of iron in the sample. Show your working.
56
Using the weighted average: 55.8 = (53.9×6.0 + 55.9×91.9 + x×2.1)/100, giving x ≈ 56.
Q2b
Short Answer 1 mark
Deduce the number of pairs of electrons in the 3d sub-shell in an iron(II) ion.
2
Q2c
Short Answer 1 mark
Sketch the shape of the lowest energy orbital in the shell with principal quantum number n = 2.
A spherical 2s orbital.
Q2d
Short Answer 2 marks
Complete Table 2.2 to show information about particles in one ion of 56Fe3+.
electrons = 23
protons = 26
Q2e
Short Answer 2 marks
The atomic radius of iron is 1.26 × 10^-10 m. Suggest the change to the radius, if any, after an iron atom reacts to produce an Fe3+ ion. Explain your answer.
The radius decreases.
Electrons are lost to form Fe3+, reducing electron-electron repulsion and increasing the effective attraction between the nucleus and remaining electrons.
Q3a(i)
Short Answer 1 mark
C3H6 reacts with HBr(g) in an addition reaction. Define addition reaction.
A reaction in which two molecules combine to form one product molecule.
Q3a(ii)
Structured 4 marks
Complete Fig. 3.1 to show the mechanism for the addition reaction between C3H6 and HBr to produce 2-bromopropane. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.
The C=C bond attacks Hδ+ of HBr; the H–Br bond breaks heterolytically to Br−; a secondary carbocation forms; Br− attacks the carbocation to form 2-bromopropane.
Q3a(iii)
Short Answer 2 marks
Explain why the major product of this reaction is 2-bromopropane rather than 1-bromopropane.
The reaction proceeds via the more stable secondary carbocation intermediate, which is formed preferentially, so 2-bromopropane is the major product.
Q3b(i)
Short Answer 1 mark
H2(g) and nickel are added to alkene X. Fig. 3.2 shows how the concentration of X changes with time. Use Fig. 3.2 to identify the limiting reagent in this reaction. Explain your answer.
Alkene X is the limiting reagent, because its concentration falls to zero.
Q3b(ii)
Short Answer 1 mark
Use Fig. 3.2 to describe how the gradient changes as the reaction proceeds. State what this shows about the rate during this reaction.
The gradient becomes less steep with time and eventually becomes zero; the rate decreases and then becomes zero as the reaction finishes.
Q3c
Short Answer 2 marks
Hydrocarbon Y contains two C=C groups. There are no other functional groups present. Y reacts with an excess of H2(g) to produce hexane, C6H14. Use Table 3.1 to calculate the enthalpy change per mole of C6H14 produced in this reaction.
enthalpy change = -240 kJ mol^-1
Breaking 2 C=C and 2 H-H; forming 2 C-C and 4 C-H gives ΔH = (2×610 + 2×436) - (2×350 + 4×410) = -240 kJ mol^-1.
Paper Details
Board CAIE
Level A Level
Subject Chemistry
Code 9701
Year 2025
Session May/June
Paper No. Paper 22
Q&A 22 questions
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