TMJC 2025 H2 Chem P1 Answer

___________________________________________________________________ H2 CHEMISTRY 9729/01 Paper 1 Multiple Choice 25 September 2025 1 hour Additional materials: Multiple Choice Answer Sheet Data Booklet _________________________________________________________________________ READ THESE INSTRUCTIONS FIRST Write in soft pencil. There are thirty questions in 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 use of the Answer Sheet very carefully. You are advised to fill in the Answer Sheet as you go along . No additional time will be given for the transfer of answers once the examination has ended. 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. Use of the Answer Sheet Ensure you have written your name, class, date and subject on the Answer Sheet. Shade the last four digits of your centre/index number on the Multiple Choice Answer Sheet. (e.g. if your centre/index number is 30541234, shade 1234). Use a 2B pencil to shade your answers on the Answer Sheet; erase any mistakes cleanly. Multiple shaded answers to a question will not be accepted. TAMPINES MERIDIAN JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry 1 Use of the Data Booklet is relevant to this question. In an experiment, a sample containing phosphorus and an unknown element A is vapourised, ionised and passed through an electric field as shown below. Given that the extent of deflection for A+ is smaller than P+, which could be the identity of element A? A sulfur B silicon C sodium D nitrogen Answer: A Angle of deflection  charge mass Since the extent of deflection of X+ is smaller, this implies that X+ has a larger mass than P+. 2 Use of the Data Booklet is relevant to this question. Which particle contains the largest number of unpaired electrons? A O B Cl– C K+ D Fe Answer: D 8O: 1s22s22p4 (2 unpaired electrons) 17Cl–: 1s22s22p63s23p6 (0 unpaired electrons) 19K+: 1s22s22p63s23p6 (0 unpaired electrons) 26Fe: 1s22s22p63s23p63d64s2 (4 unpaired electrons) beam of P+ and A+ + – P+ A+

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry [Turn over 3 In a microwave oven, the microwave produced is absorbed by polar molecules. Which molecules would absorb microwave energy? 1 SO3 2 CH2F2 3 CH3CH2OH A 3 only B 1 and 2 only C 2 and 3 only D 1, 2 and 3 Answer: C SO3 ⇒ trigonal planar, non–polar CH3F ⇒ tetrahedral, polar C–F bond CH3CH2OH ⇒ tetrahedral, polar O–H bond 4 The melting point of potassium is lower than the melting point of magnesium. Which statement is most relevant in explaining the difference? A Potassium ion has a smaller radius than magnesium ion. B Potassium ion has a lower charge than magnesium ion. C Potassium atom contains fewer electrons than magnesium atom. D Potassium atom is heavier than magnesium atom. Answer: B Both potassium and magnesium contain giant metallic structures. Strength of metallic bond is affected by: (i) no. of valence electrons delocalised (K < Mg) (ii) charge density (K+ < Mg2+)

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry 5 Which of the following cannot be explained by hydrogen bonding? A The existence of hydrogen–difluoride anion, HF2 – . B The difference in volatility between pentan–1–ol and hexan–1–ol. C The difference in melting point between 2–nitrophenol and 4–nitrophenol. D The relative molecular mass of ethanoic acid in organic solvent is higher than expected. Answer: B Option A: hydrogen bond Option B: Pentan–1–ol is more volatile (or has a lower boiling point) because it has a smaller electron cloud size than hexan –1–ol. Hence there is a smaller extent of distortion of electron cloud, resulting in weaker instantaneous dipole –induced dipole attraction between pentan–1–ol molecules. Option C: Due to the proximity of –NO2 and –OH group in 2–nitrophenol, intramolecular hydrogen bonding will occur. Hence less extensive intermolecular hydrogen bonding will be formed between 2–nitrophenol molecules, resulting in lower melting point compared to 4–nitrophenol. Option D: Ethanoic acid molecules can form a dimer via hydrogen bond in organic solvent. hydrogen bond

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry [Turn over 6 Which diagram correctly describes the behaviour of a fixed mass of an ideal gas? A B C D Answer: C Option C is correct pV = nRT p = nRT ( 1 V) since nRT is constant ⇒ p = k ( 1 V), where k is a constant 0 V pV constant T 0 V p constant T 0 V pV constant T 0 V p constant T 0 V p constant T

6 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry Option A and B is incorrect since nRT is constant ⇒ pV = k, where k is a constant 7 Use of the Data Booklet is relevant to this question. Sodium percarbonate, (Na2CO3)xy(H2O2), is an oxidising agent used in laundry cleaning products. On acidification, 10.0 cm3 of 0.100 mol dm -3 sodium percarbonate releases 48.0 cm 3 of carbon dioxide at room temperature and pressure. An identical sample, on titration with 0.0500 mol dm -3 KMnO4, requires 24.0 cm 3 before the first pink colour appears. 2 moles of KMnO4 reacts with 5 moles of H2O2. What is the ratio of y/x? A 1 3 B 2 3 C 3 2 D 3 Answer: C (Na2CO3)x.y(H2O2)  xNa2CO3  xCO2 Na2CO3 = 10.0 ×0.1001000 = 0.001 CO2 = 48.0 241000  = 0.002  x = 2 2KMnO4  5H2O2 (Na2CO3)x.y(H2O2)  yH2O2 KMnO4 = 24.0 ×0.05001000 = 0.0012 H2O2 = 5×0.00122 = 0.003  y = 3 0 V pV constant T

7 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry [Turn over 8 G, H and J are three elements found in Period 3 of the Periodic Table. Among the elements in Period 3, • the melting point of G is the highest. • the electrical conductivity of H is the highest. • the melting point of the oxides of J is the highest. Which of the following elements is not represented by G, H or J? A Na B Mg C Al D Si Answer: A Among the Period 3 elements: • Silicon has the highest melting point. Hence, G is silicon. • Aluminium has the highest electrical conductivity. Hence, H is aluminium. • Magnesium oxide has the highest melting point. Hence, J is magnesium. 9 Which equation defines standard enthalpy change of formation correctly? A Na(s) + Cl(g) → NaCl(s) B 2H2(g) + O2(g) → 2H2O(g) C Mg2+(g) + O2–(g) → MgO(s) D H2(g) + S(s) + 2O2(g) → H2SO4(l) Answer: D Standard enthalpy change of formation of a substance is the energy change when 1 mole of the substance is formed from its elements under standard conditions of 298K and 1bar.

8 Tampines Meridian Junior College 2025 JC2 Preliminary Exa