TMJC 2025 H2 Chem P2 Answer

CANDIDATE NAME CIVICS GROUP H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 18 September 2025 2 hours Candidates answer on the Question Paper. Additional materials: Data Booklet This document consists of 24 printed pages. READ THESE INSTRUCTIONS FIRST Write your name and civics group in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams and graphs. Do not use 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. TAMPINES MERIDIAN JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION For Examiner’s Use Percentage Paper 1 / 30 / 15 Paper 2 1 / 7 2 / 16 3 / 11 4 / 12 5 / 16 6 / 13 Paper 2 Total / 75 / 30 Paper 3 / 80 / 35 Paper 4 / 55 / 20 Grand Total / 100

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry 1(a) The graph in Fig. 1.1 shows the atomic radii of some Period 4 metals. Fig. 1.1 (i) With reference to Fig 1.1, explain briefly why the atomic radius of Ca is smaller than K. [1] (ii) With reference to Fig 1.1, describe and explain why the atomic radii from Cr to Ni are relatively constant. [2] 0 50 100 150 200 250 K Ca Sc Ti V Cr Mn Fe Co Ni atomic radii/ pm Elements From Cr to Ni, • Nuclear charge increases. • Electrons are added to the inner 3d orbitals and provide shielding for the 4s / valence electrons. • Effective nuclear charge varies slightly OR Increase in nuclear charge is offset by the increase in shielding effect. Atomic radius remains relatively invariant/ relatively constant. Ca has greater nuclear charge than K while shielding effect remains relatively constant. Hence, Ca has a stronger electrostatic forces of attraction between nucleus and valance electrons than K, and the valence electrons of Ca are closer to the nucleus.

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry (b) (i) Shibuichi is an alloy of copper and silver and is used in traditional Japanese sword fittings. A particular sample of Shibuichi produced the following peaks in its mass spectrum as shown in Fig 1.2. Fig. 1.2 Calculate the average Ar of copper from these data. [1] (ii) With reference to Fig. 1.2, state the number of protons and neutrons present in the atom with nucleon number 109. Number of protons: ___________ Number of neutrons: ___________ [1] 61.7 26.3 6.2 5.8 0 10 20 30 40 50 60 70 63 65 107 109 % abundance Nucleon number 47 62 Ar = [63 × 61.7]+[65 × 26.3] 88 = 63.6

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry (c) Table 1.2 shows the successive ionisation energies of element B. Table 1.2 1st 2nd 3rd 4th 5th 6th 7th 8th I.E. / kJ mol-1 900 1800 3250 4900 6050 12690 14200 16250 (i) State which group does element B belongs to. [1] (ii) Both element A and element B are consecutive elements in the Periodic Table and that element A is positioned to the left of element B. Explain why the 3rd ionisation energy of element B is lower than element A. [1] [Total: 7] Group 15 The 3rd I.E. of element B is lower as the np electron to be removed from B has higher energy compared to the ns electron to be removed from element A. Less energy is needed to remove 3rd electron of element B.

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry 2 Adipic acid, HOOC(CH 2)4COOH, is a flexible food additive used as a gelling aid, firming and buffering agent and can be found in many foods. It is a dibasic acid that ionises in 2 stages. (a) (i) Write an expression for the first acid dissociation constant of adipic acid, Ka1. [1] (ii) Suggest a reason why the value of Ka2 is lower than that of Ka1. [1] A mixture of adipic acid and its potassium salt can function as a buffer. (iii) With the aid of a chemical equation, briefly explain how this mixture can act as a buffer when a small amount of base is added. The effects of Ka2 can be ignored in any buffer action. [2] More energy is required to remove a H + from a negatively charged anion after the 1st H+ is removed from the acid. When a small amount of OH– is added: OH– + HOOC(CH2)4COOH → HOOC(CH2)4COO – + H2O (or OH– + H2A → HA– + H2O) The added OH– is removed as H2O. [H+] is slightly changed hence pH remains fairly constant. Ka1 = − + 2 2 4 2 2 2 4 2 [HO C(CH ) CO ][H ] [HO C(CH ) CO H] Stage 1 Stage 2

6 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry (iv) Determine the volume of 0.200 mol dm-3 KOH(aq) that needs to be added to 100 cm3 of 0.200 mol dm-3 adipic acid to form a buffer of pH 4.7. [2] 3-hexenedioic acid can be converted to adipic acid and 3-bromohexanedioic acid which serve as precursors or intermediates for organic synthesis. (b) (i) Propose a 2 -step reaction synthesis to convert 3 -hexenedioic acid to D. Show clearly all reagents and conditions and the structures of intermediates for each step. O OH O OH O O OH O 3-hexenedioic acid D [3] Let the volume of KOH be v dm3 and total buffer volume be vt dm3. After neutralisation with KOH, the remaining excess acid will form a buffer with the conjugate base of the salt formed. No of mol of salt = no of mol of KOH added = 0.200 v No of mol of remaining acid = 0.200 (0.100 – v) pH = −lg Ka1 + lg [salt] [acid] 4.7 = 4.43 + lg 0.2𝑣 𝑉𝑡 0.2(0.1−𝑣) 𝑉𝑡 0.27 = lg 0.2𝑣 𝑉𝑡 0.2(0.1−𝑣) 𝑉𝑡 = lg 𝑣 (0.1−𝑣) ➔ 100.27 = 𝑣 (0.1−𝑣) = 1.862 v = 0.1862 – 1.862v ➔ 2.862v = 0.1862 Volume of KOH added, v = 0.0651dm3 O OH O OH O O OH O steam, conc H3PO4, high T and P conc H2SO4 heat O OH O OH OH

7 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry (ii) Another possible product that could be produced from the synthesis method in (i) is shown below. Suggest why D is more likely to be formed than E. O O OH O E [1] (c) A sample of 3-bromohexanedioic acid was able to rotate plane polarised light. After reacting with hot NaOH(aq) followed by acidification , the product obtained was no longer able to rotate plane polarised light. The reaction between 3-bromohexanedioic acid and NaOH(aq) is shown in the equation below. (i) Describe the mechanism for the reaction between 3-bromohexanedioic acid and NaOH(aq). Include all relevant lone pairs, dipoles, curly arrows and charges. [3] Compound B will experience ring strain / angle strain due to creation of a 4 - membered ring and is thus less stable to be formed.

8 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry (ii) State the type of isomerism displayed by the products of the above reaction. [1] (iii) Draw the isomers produced in the above reaction. [2] [Total: 16] enantiomerism

9 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry 3(a) Chlorate(V), ClO3–, reacts with chloride ions according to the equation as shown below. 2ClO3–(aq)