2024 H1 Chem JC2 Prelim P1 (Q&A) JPJC

© Jurong Pioneer Junior College [Turn over Name:____________________________________ Class:_____________ JURONG PIONEER JUNIOR COLLEGE JC2 Preliminary Examination 2024 CHEMISTRY 8873/01 Higher 1 2024 Paper 1 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, class and exam index number on the Answer Sheet in the spaces provided unless this has been done for you. There are 30 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 11 printed pages and 1 blank page.

2 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2024 1 In two separate experiments a beam of protons and a beam of electrons, travelling at the same velocity, pass through an electric field as shown. Which statements are correct? 1 The electrons are deflected to a larger extent than the protons. 2 The electron beam is deflected in the opposite direction to the proton beam. 3 The proton beam travels in a straight path towards the negatively charged plate. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 3 only 1 Answer: B (1 and 2 only) ✓1 Electron has the same magnitude of charge but smaller mass than proton. Since  of deflection  charge mass , electron is deflected to larger extent than proton. ✓2 The relative charge of electron is −1 while that of proton is +1. Hence, electron is deflected towards the positive plate while proton is deflected towards the negative plate. 3 Under the influence of uniform electric field, the proton beam travels in a curved path. The particle travels in straight path only when outside of electric field. 2 The 7th, 8th, 9th and 10th ionisation energies of four consecutive elements in the Periodic Table, P, Q, R and S are shown. element 7th ionisation energy / kJ mol-1 8th ionisation energy / kJ mol-1 9th ionisation energy / kJ mol-1 10th ionisation energy / kJ mol-1 P 9941 18580 21610 25180 Q 10530 12140 22480 25860 R 9520 12810 14530 26740 S 10300 11690 15280 17110 Which element has the highest first ionisation energy? A P B Q C R D S

3 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2024 [Turn over 2 Answer: B For each element, locate the sharp increase in IE which indicates the number of valence electron (and hence, the Group number) each element has. Element P is from Group 17 (sharp  from 7th to 8th IE). Element Q is from Group 18 (sharp  from 8th to 9th IE). Element R is from Group 1 of the next Period (sharp  from 9th to 10th IE). Element S is from Group 2 of the next Period (since W, X Y and Z are consecutive elements). Since 1st IE increases across the Period and decreases down the Group, element Q has the highest 1st IE. 3 A gaseous mixture containing 10 cm 3 of C 3H8 and 90 cm 3 of oxygen was sparked and C3H8 undergoes complete reaction. The resultant mixture was passed through excess NaOH(aq). What is the final volume of gas remaining after cooling back to room temperature? A 20 cm3 B 40 cm3 C 60 cm3 D 80 cm3 3 Answer: B 5O2(g) + C3H8(g) 3CO2(g) + H2O(l) Initial vol./ cm3: 90 10 0 - Change: -50 -10 30 - Final vol./ cm3: 40 0 30(absorbed by NaOH) So final volume of gas (O2) left = 40 cm3 4 Use of the Data Booklet is relevant to this question. HFC-134a is a molecule that was developed during the 1980s to replace CFCs that were causing great damage to the ozone layer. The percentage composition by mass of HFC-134a is: C, 23.5%; H, 2.0%; F, 74.5%. Which structural formula could be that of HFC-134a? A CH2FCF3 B CHF2CF3 C CHF3 D CH2F2 4 Answer: A C H F Mole ratio: 23.5/12 =2 2/1 = 2 74.5/19 = 4 Empirical formula: CHF2 Option A has same empirical formula.

4 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2024 5 A 25.0 cm3 sample of 0.20 mol dm−3 TlNO3 required 25.0 cm3 of 0.10 mol dm−3 acidified KMnO4 to oxidise it to Tl3+ in solution. What is the oxidation state of the manganese in the reduced form? A +2 B +3 C +4 D +5 5 Answer: B Tl+ ⟶ Tl3+ + + 2e- Amt of Tl+ = 0.20 x 25.0/1000 = 5.00 x 10-3 mol Amt of electron lost = 5.00 x 10-3 x 2 = 1.00 x 10-2 mol Amt of MnO4- = 0.10 x 25.0/1000 = 2.50 x 10-3 mol Thus, MnO4- : e- = 2.50 x 10-3 : 1.00 x 10-2 = 1 : 4 Thus, the oxidation state of Mn = (+7) + 4(-1) = +3 6 Silicon carbide, SiC, has the same structure as silicon(IV) oxide, SiO2. Hence, SiC is used as a major industrial abrasive, and a refractory material which is resistant to decomposition by heat. Which type of structure explains these properties? A A simple molecular structure with covalent bonds between silicon and carbon atoms. B A layered structure with covalent bonds between silicon and carbon atoms and weak instantaneous dipole-induced dipole attraction between the layers. C A giant covalent structure with strong covalent bonds between silicon and carbon atoms forming a 3-dimensional network. D A giant ionic lattice with strong ionic bonds holding the oppositely charged ions together. 6 Answer: C SiO2 exists as giant molecular structure with atoms arranged in tetrahedral manner. 7 Consider the following four compounds.

5 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2024 [Turn over 1 (CH3)3CH 2 CH3CH2CH2CH3 3 CH3CH2CH2OH 4 CH3CH2Cl What is the order of decreasing boiling point of these compounds? A 1 → 2 → 4 → 3 B 3 → 4 → 1 → 2 C 3 → 4 → 2 → 1 D 4 → 3 → 2 → 1 7 Answer: C 3: Hydrogen bonding between molecules. 4: Pd-pd 2: Straight chain, bigger surface area of contact between molecules, strong id-id 1: branched, smaller surface area, weaker id-id between molecules 8 Which of the following statement explains why the boiling point of water (100 °C) is higher than that of ammonia (–33 °C)? A There are, on average, more hydrogen bonds between water molecules than there are between ammonia molecules. B The Mr of water is greater than that in ammonia, so instantaneous dipole-induced dipole forces of attraction between water molecules is stronger. C Ammonia has intramolecular hydrogen bonds, which water does not have. D The O–H bond in water is stronger than the N–H bond in ammonia. 8 Answer: A compound Boiling Point /C No. of lone pairs No. of delta positive H) No. of Hydrogen Bonds formed per molecule (on average ) Ammonia N H H H .. -33.3 1 3 1 water O H H .. .. 100 2 2 2 (greatest no.)

6 © Jurong Pioneer Junior College 8873/01/J2 PRELIMINARY EXAMINATION /2024 hydrogen fluoride FH .. .. .. 19.5 3 1 1 Why does H2O have a higher boiling point than NH3? =>because H2O on average, forms 2 hydrogen bonds per molecule whereas NH3 forms 1 hydrogen bond per molecule. H2O forms more extensive hydrogen bonds between H2O molecules, which need more energy to overcome 9 Phosphorus reacts with chlorine to form both PCl3 and PCl5. However, n