EJC Prelim H2 Chemistry Paper 1 Worked Solution (updated)

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 General Certificate of Education Advanced Level Higher 2 CHEMISTRY Paper 1 Multiple Choice 9729/01 19 September 2025 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, civics group and registration number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this question paper. The use of an approved scientific calculator is expected, where appropriate. This document consists of 13 printed pages and 3 blank pages.

2 © EJC 9729/01/J2PE/25 1 Three particles approach an electric field at the same speed. They are deflected as they pass through the electric field. What could be the identities of particles X, Y and Z? X Y Z A 35 17Cl– 37 17Cl– 23 11Na+ B 6 3Li+ 7 3Li+ 1 1H– C 37 17Cl– 35 17Cl– 23 11Na+ D 7 3Li+ 6 3Li+ 1 1H– Answer: A Since particles X and Y are deflected to the positive terminal, they are negatively charged. Since particle Z is deflected to the negative terminal, it is positively charged. Hence, X and Y must be Cl− while Z must be Na+ (either option A or C is correct). Between particles X and Y, the larger angle of deflection of particle X indicates that its 𝑒 𝑚 ratio is larger than that of Y. Hence, particle X should have a smaller mass number than particle Y since both X and Y have the same charge. Hence, X is Cl – 17 35 while Y is Cl – 17 37 . + – X Y Z

3 © EJC 9729/01/J2PE/25 [Turn over 2 Which of the following about calcium and copper is correct? A The outermost orbital of the atoms of both elements has the same shape. B Atom of both elements have orbitals of only 2 different shapes of various sizes. C Atom of both elements have the same number of electrons in the outermost shell. D Both elements form ions of 2+ charge with the electronic configuration of [Ar]. Answer: A Ca: 1s2 2s2 2p6 3s2 3p6 4s2 Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Ca2+: 1s2 2s2 2p6 3s2 3p6 or [Ar] Cu2+: 1s2 2s2 2p6 3s2 3p6 3d9 or [Ar] 3d9 A ✓: Both atoms have 4s orbital as the outermost filled orbital. Thus, they have the same shape. B ×: Ca has s and p orbitals (2 different shapes), but Cu has s, p and d orbitals (3 different shapes). C ×: Ca has 2 electrons in the outmost shell (4s) while Cu has only 1 electron. D ×: Thus, the ions do not have the same electronic configuration. 3 The following diagrams show the structures of an element, one of its oxides and its halides. What could the element be? A aluminium B carbon C phosphorus D silicon Answer: D Based on the first diagram on the left, the element forms 4 single bonds. Based on the last diagram on the right, the element forms 4 single bonds with the halogen atoms. Hence, it can be concluded that the element is in Group 14.

4 © EJC 9729/01/J2PE/25 4 The boiling point of water (100 ºC) is greater than that of HF (20 ºC). Which statement is a correct explanation of the above? A Each hydrogen bond formed between water molecules is stronger than that formed between HF molecules. B There are more atoms in a water molecule than there are in an HF molecule, resulting in stronger intermolecular forces in water. C There are, on average, more hydrogen bonds between water molecules than there are between HF molecules. D The water molecule has stronger permanent dipole –dipole interactions than the HF molecule. Answer: C A ×: Intermolecular hydrogen bonds for HF is stronger than that of H2O. This is because F is more electronegative than O, hence H–F bond is more polar than H–O bond, resulting in stronger hydrogen bonding. B ×: The electron cloud for both are comparable and hence have similar strength of intermolecular instantaneous dipole–induced dipole. C ✓: Each H₂O can form two hydrogen bonds; HF tends to form with fewer hydrogen bonds per molecule. D ×: HF has the larger dipole, but dipole –dipole differences don’t outweigh water’s extensive hydrogen -bond network.

5 © EJC 9729/01/J2PE/25 [Turn over 5 For a fixed mass of an ideal gas, which of the following graphs does not have the same general shape as the rest? ( = density of the gas; M = molar mass of gas) A p  against T B pV against M T C p against ρT D T p against V Answer: B A: mPV nRT PV RT M mPM RT PM RTV PR TM   =  = =  = = B: 1() TPV nRT PV mR M PV mR M T =  = = C: mPV nRT PV RT M mPM RT PM RTV RPT M   =  = =  = = D: 1 PV nRT T VP nR = = P  0 T grad = R/M M/T PV 0 grad = R/M 0 T P 0 V T/P grad = (nR)−1

6 © EJC 9729/01/J2PE/25 6 Which option correctly describes the species in terms of its behaviour as a Lewis base and as an Arrhenius acid? species Lewis base Arrhenius acid A HCl no yes B AlH3 yes no C NH3 no no D O2− yes yes Answer: A A ✓: HCl is not a Lewis base and is an Arrhenius acid (produces H⁺/H₃O⁺ in water) B ×: AlH₃ is a Lewis acid (electron ‐pair acceptor), not a Lewis base; it’s also not an Arrhenius acid. C ×: NH₃ is a Lewis base and an Arrhenius base (gives OH⁻), not an Arrhenius acid. D ×: O²⁻ is a very strong base (Lewis and Arrhenius base), not an acid. 7 Use of the Data Booklet is relevant to this question. Based on its position in the Periodic Table, which properties will indium, In, be expected to possess? 1 In the vapour state, the chloride dimerises to form In2Cl6. 2 Its oxide dissolves in both acids and alkalis. 3 Its ionic salts are typically coloured. A 1 only B 1 and 2 C 2 and 3 D 1, 2 and 3 Answer: B Option 1 ✓: Group 13 trichlorides (Al, Ga, In) are covalent and dimerise to form a general formula of M₂Cl₆. Option 2 ✓: In2O3 is amphoteric (reacts with both acid and base). Option 3 ×: In is not a transition metal, hence its salts are usually white.

7 © EJC 9729/01/J2PE/25 [Turn over 8 Metal peroxides decompose when heated to form metal oxides and oxygen gas. Which factor contributes to solid BaO₂ being more thermally stable than solid MgO₂? A The hydration enthalpy of Mg²⁺ ion is more exothermic than that of Ba²⁺ ion. B The lattice energy of BaO₂ is more negative than that of MgO₂. C The charge density of Ba²⁺ ion is lower than that of Mg²⁺ ion. D The O–O bond in O 2– 2 is weaker than the O=O bond in O2. Answer: C A ×: Hydration enthalpy is more exothermic for Mg² ⁺ than Ba²⁺ and however it does not account for the solid-state thermal stability of peroxides. B ×: Lattice energies of Group-2 peroxides become less negative down the group (|LE|: MgO₂ > BaO₂). A more negative BaO₂ lattice energy is not true and wouldn’t explain BaO₂’s greater stability. C ✓: Ba2+ has lower charge density → lower polarising power → less weakening of the O–O bond in O 22−, so BaO ₂ is more thermally stable (decomposes at a higher temperature) than MgO₂. D ×: This is a general reason peroxides decompose (forming strong O=O), but it does not explain why BaO₂ is more stable than MgO₂. 9 F2 reacts with BrO – z ions in a 2 : 1 molar ratio to form F− and BrO – 4 ions. What is the value of z? A 1 B 2 C 3 D 5 Answer: B [R]: 2F2 + 4e– → 4F– Reacting ratio between F2 and BrOx− is 2:1.  1 mole of BrOx− loses 4 moles of electron to produce BrO4−. Oxidation state of Br in BrO4− = +7  Oxidation state o