Prelim Paper 1 QP (with suggested solutions) ACJC H2 Chem

1 © ACJC 2025 9729/01/Prelim/2025 [Turn over Anglo-Chinese Junior College JC2 Preliminary Examination Higher 2 CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 9729/01 17 September 2025 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and index number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions in this section. 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 booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 16 printed pages.

2 © ACJC 2025 9729/01/Prelim/2025 [Turn over 1 The most common oxidation state of americium, Am, in aqueous solution is +3. Recently, Cu3+ has been shown to quantitatively oxidise Am3+(aq) in dilute HNO3, while itself is reduced to Cu2+. In an experiment, 20.0 cm 3 of 0.0120 mol dm –3 Am3+(aq) was found to require 24.00 cm3 of 0.0300 mol dm–3 Cu3+ for complete oxidation. What is the formula of the americium-containing species formed? A Am2O22+ B AmO22+ C AmO2+ D AmO+ n(electrons) = n(Cu3+) = 7.20 X 10-4 change in oxidation state of Am = (7.20 X 10-4) / (20.0 X 0.0120 / 1000) = +3 Initial oxidation state of Am = +3 +3 = +6 Correct option is AmO22+ 2 Use of the Data Booklet is relevant to this question. The table shows the fifth, sixth, seventh, eighth, ninth and tenth ionisation energies of an element (Z ≤ 20) in the Periodic Table. 5th 6th 7th 8th 9th 10th ionisation energy / kJ mol–1 7975 9590 11343 14944 16964 48610 What can be inferred about the element from the above data? A It is in the third period of the Periodic Table. B It is in Group 2 of the Periodic Table. C It is likely to form an ionic compound when reacted with oxygen. D Its 6th and 7th electrons are removed from different subshells. C: There is a large electronegativity difference between the Group 1 element and oxygen (Group 16). Hence, the compound formed will be ionic in nature. A: There is insufficient data to conclude that the element lies in the third period. Since Z ≤ 20, this Group 1 element can be in the fourth period. B: The largest increase in successive ionisation energies (IE) occur between 9th and 10th IE. This implies that the 10th electron is removed from an inner shell. Since Z ≤ 20, only s and p subshells exists. Hence, each shell can only accommodate up to a maximum of 8 electrons. This would imply that the 2nd to 9th electron also lies in another inner shell. There will be only one electron in the valence shell. This element therefore belongs to Group 1 (either Na or K). D: As a Group 1 element, the 6th and 7th electrons are removed from the same p-subshell.

3 © ACJC 2025 9729/01/Prelim/2025 [Turn over 3 Particle R has a proton number n and forms a stable monoatomic ion of charge ‒1. Particle S has a proton number of ( n+2) and it forms a stable monoatomic ion which is isoelectronic with the ion of R. Which statement is correct? A Ion of S has a smaller ionic radius than ion of R. B R has a larger atomic radius than S. C Ion of S requires less energy than ion of R when an electron is removed from each particle. D Ion of R releases more energy than ion of S when an electron is added to each particle. nR gains 1 electron to form stable n −R , which has (n+1) electrons  R is in Group 17. 2n+ S forms a stable ion which is isoelectronic with n −R . Hence 2n+ S must have lost 1 electron, to form 2n + + S , which has (n+2)–1 = (n+1) electrons  S is in Group 1 of the next period. A✓ Since n −R and 2n + + S are iso-electronic, 2n + + S with a higher nuclear charge will have a smaller ionic radius as the effective nuclear charge experienced by the valence e–s is higher. B As S is an element in the next period, with one additional filled principal quantum shell, S has a larger atomic radius despite the higher nuclear charge. C 2 22nn e+ − + ++ −→SS will be more endothermic than nn e−−−→RR since the e– is being removed from positively charged 2n + + S . D 2 nn e− − −+→RR will be endothermic due to repulsion of the incoming e–; 22nn e+− ++ +→SS will be exothermic due to attraction of the incoming e–. 4 Which statement about the trend in the property of the halogens down the group is correct? A The electronegativity increases. B The volatility increases. C The enthalpy change of reaction with hydrogen becomes less exothermic. D The reactivity as reducing agents increases. A: Electronegativity decreases down any Group. B: IDID gets stronger down Group 17. Volatility should decrease. C: Halogens react with hydrogen in the gaseous phase to give hydrogen halides. As reactivity/oxidising power of the halogens decreases down the group, the vigour of the reaction also decreases down the group. Hence, the enthalpy change of reaction with hydrogen becomes less exothermic. D: Halogens are usually oxidising agents, not reducing agents.

4 © ACJC 2025 9729/01/Prelim/2025 [Turn over 5 Use of the Data Booklet is relevant to this question. Which sequence is correct in terms of increasing radius? A Rb+ < Sr2+ < As3− < Se2− B Sr2+ < Rb+ < Se2− < As3− C As3− < Se2− < Sr2+ < Rb+ D Se2− < Sr2+ < Rb+ < As3− All ions are isoelectronic, but different number of protons. Sr2+ Rb+ Se2− As3− Proton / Electron Ratio 38 ÷ 36 37 ÷ 36 34 ÷ 36 33 ÷ 36 Attraction Strongest Weakest 6 Two bulbs are connected as shown in the diagram below. The bulbs are connected by a narrow tube of negligible volume. When the tap is opened, the two gases mix. The connected bulbs were then allowed to cool to room temperature. What was the final pressure, in kPa, in the connected bulbs? A 13.9 B 24.3 C 37.5 D 64.7 When the connected bulbs were allowed to cool to room temperature, water vapour condensed to give liquid water, which occupy negligible volume. Hence, we only need to consider the amount of neon gas present in the connected bulbs. PV = nRT 100000 x (3 x 10-3) = n x 8.31 x (180 + 273) → n = 0.07969 mol (amount of neon) When the tap is opened, and the total volume is 8 dm3, PV = nRT P x (8 x 10-3) = 0.07969 x 8.31 x 293 P = 24254 Pa = 24.3 kPa tap water vapour 5 dm3 100 kPa 180 oC neon 3 dm3 100 kPa 180 oC

5 © ACJC 2025 9729/01/Prelim/2025 [Turn over 7 (CH3)2S.BCl3 is a solid that is commonly used in laboratories as a convenient source of BC l3. When heated, it reversibly decomposes to (CH3)2S and BCl3. Which statement is true? A The dative bond is formed using the 2p orbitals of boron and sulfur. B (CH3)2S and BCl3 act as the Lewis acid and Lewis base respectively in the formation of (CH3)2S.BCl3. C The dative bond is from boron to sulfur. D The C-S-C bond angle decreases when the solid decomposes. A The dative bond is formed using the hybrid sp3 orbitals of boron and sulfur. B (CH3)2S and BCl3 act as the Lewis base and Lewis acid respectively in the formation of (CH3)2S.BCl3. D The C-S-C bond angle decreases when the solid decomposes. C The dative bond is from sulfur to boron. 8 The compound Bi2Sr2Ca2Cu3O10 is a superconductor. In this compound, the oxidation number of bismuth is +3, strontium and calcium is +2 and oxygen is −2. What are the possible oxidation numbers of the three copper atoms