2024 Prelims ACJC H2 Chem P2 (Ans)

This paper consists of 22 printed pages and 2 blank pages. Anglo-Chinese Junior College JC2 Preliminary Examination Higher 2 CANDIDATE NAME Answers FORM CLASS TUTORIAL CLASS INDEX NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional materials: Data Booklet 9729/02 21 August 2024 2 hours READ THESE INSTRUCTIONS FIRST Write your form class, index number and name in the spaces provided at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner's Use Question no. Marks 1 / 12 2 / 11 3 / 10 4 / 10 5 / 10 6 / 22 Presentation of answers TOTAL / 75

© ACJC2024 9729/Mid-Year Learning Checkpoint/2024 Section B: Structured Questions (75 marks) Answer all the questions in the spaces provided. 1 Y and Z are two elements found in Period 3. Their fifth to eighth ionisation energies are given in Table 1.1. Table 1.1 successive ionisation energies / kJ mol–1 5th 6th 7th 8th Y 6530 9353 11019 33606 Z 7004 8496 27107 31719 (a) (i) State and explain the group number of Y. [1] Y belongs to Group 17. There is a large jump between the 7th and 8th ionisation energies, so the 8th electron to be removed comes from an inner (principal quantum) shell. [1] Comments • Generally well done. However, many students misspelt ‘principal’ as ‘principle’. Also, the terms ‘principal quantum shell’ and ‘subshell’ should not be used interchangeably. (ii) Complete the electronic configurations of Y+ and Z+. Y+: 1s22s22p63s23p4 Z+: 1s22s22p63s23p3 Comments • Students should write out the electronic configuration of elements Y and Z first before removing an electron. (iii) From your answer in (a)(ii), explain why the second ionisation energy of Y is expected to be lower than Z. [1] The second ionisation energy for element Y involves removing a paired electron from the 3p subshell that has interelectronic repulsion while the second ionisation energy for element Z involves removing an unpaired electron from the 3p subshell. Comments • A significant number of students were able to identify that there was greater interelectronic repulsion in Y but did not mention this was due to the paired electrons. • For the 2 nd IE of Z, only the electrons in the 3p subshell are unpaired, the inner electrons are still paired so it is incorrect to say that there are no paired electrons in Z. (iv) However, it turns out that the second ionisation energies of Y and Z are 2300 kJ mol–1 and 2260 kJ mol–1 respectively. Suggest why Y has a higher second ionisation energy than Z. [1]

© ACJC2024 9729/Mid-Year Learning Checkpoint/2024 The effect of the higher (effective) nuclear charge of Y is more dominant than the interelectronic repulsion between the paired electrons, so the 3p electron to be removed is held more tightly by the nucleus. [1] Comments • Many students struggled to adequately account for the higher second IE of Y. Students should recognise that there are two factors to consider: 1. effective nuclear charge and 2. interelectronic repulsion . The predominant factor is determined based on data given. Some managed to identify the predominant factor of higher effective nuclear charge but did not link it to the valence electron being more tightly held by the nucleus. (b) (i) The atomic radii of some Period 4 elements are shown in Fig. 1.1. Fig. 1.1 With reference to Fig. 1.1, describe and explain the shape of the graph from • V to Zn • Ga to Br [4] From V to Zn, the atomic radii remains relatively invariant . While nuclear charge increases from V to Zn , electrons are added to the inner shell / 3d subshell , increasing shielding effect on valence electrons. Increase in effective nuclear charge is minimal/effective nuclear charge remains relatively constant. From Ga to Br, the atomic radii of the elements decreases across the Period. Across the Period, increasing nuclear charge and approximately constant shielding effect causes effective nuclear charge to increase . The valence electrons experience stronger nuclear attraction and are pulled closer to the nucleus. Hence, atomic radius decreases across the Period. Comments • This question tested two main learning outcomes: 1. Describing and explaining the general trend in atomic radius across the Period (Ga to Br) in terms of shielding effect and nuclear charge

© ACJC2024 9729/Mid-Year Learning Checkpoint/2024 2. Explain why atomic radii of transition elements are relatively invariant • Many students had the general idea of at least one of these learning outcomes, but explanations were lacking. (ii) With reference to the Data Booklet or otherwise, sketch the ionic radii of the Period 4 elements on Fig. 1.1. You may assume that V to Zn form doubly charged cations. [2] • V to Zn line: lower values of decreasing or constant trend • Ga to Br line: anions of As, Se and Br must be higher with decreasing trend & Ga3+ and Ge2+/Ge4+ must be lower with decreasing trend Comments • Poorly done. Some students only sketched the ionic radii from V to Zn. • Students should recognise that cation ic radii are always smaller than their respective atomic radii as there is one shell of electrons less (V 2+ to Ge2+/Ge4+). Sharper students extracted the actual cationic radii from the Data Booklet. • Students should recognise that anionic radii are larger than their respective atomic radii as there is greater interelectronic repulsion when more electrons are added (As3-, Se2-, Br-). (c) Draw and label the d orbitals in cobalt ion. [2]

© ACJC2024 9729/Mid-Year Learning Checkpoint/2024 Comments • Most students were able to get at least 1m. When labelling orbitals, students should follow the convention (e.g. dyz and not dzy). [Total: 12] 2 The hydroboration-oxidation reaction of an alkene bond using borane, BH3, provides a useful method for hydration. Using this reaction, 1 -methylcyclohexene can be converted to trans-2- methylcyclohexanol as shown in Fig. 2.1. Fig. 2.1 (a) (i) Explain the role of borane in Fig. 2.1. [1] BH3 is a lewis acid as it accepts an electron pair from the alkene. OR BH3 is an electrophile as borane is electron-deficient OR alkene is electron-rich. Comments • Some students merely stated the role of borane without any explanation. (ii) Suggest why the trans configuration is favoured in the product formed in Fig. 2.1. [1] Steric hindrance is minimised/ There is less interelectronic repulsion when the bulky methyl and BH3– groups are at opposite sides of the ring. OR Cis configuration is not favoured because the bulky methyl and BH3– groups are on the sam