2024 Prelims TJC H2 Chem P1 (Ans)

1 9729 / TJC PRELIM / 2024 TEMASEK JUNIOR COLLEGE 2024 JC2 PRELIMINARY EXAMINATION Higher 2 CHEMISTRY 9729/01 Paper 1 Multiple Choice 12 September 2024 1 hour Additional Materials: Multiple Choice Answer Sheet (OMS) Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. There are thirty questions In this paper. Answer all questions. For each question there are four possible answers A, B, C, 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. Write your name & Civics Group on the Answer sheet. Shade your index number in the appropriate boxes. 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 14 printed pages and 2 blank pages. Write your name and Civics Group Index number (refer to entry proof)

2 9729 / TJC PRELIM / 2024 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 (OMS). 1 Use of the Data Booklet is relevant to this question. Chlorine radicals is one of the main chemical species responsible for the depletion of ozone from the stratosphere. What does 35Cl contain? protons neutrons electrons A 18 35 17 B 18 35 18 C 17 18 17 D 17 18 18 Answer: C Chlorine radical is a chlorine atom. The proton number = electron number = 17. The neutron number = nucleon number – proton number = 35 – 17 = 18 2 Which compound is composed of cation and anion that has the same number of electrons as each other? 1 K2O2 2 NaN3 3 NH4F 4 K2CO3 A 1 and 3 B 1 and 4 C 2 and 3 D 2 and 4 Answer: A Cation Number of e- Anion Number of e- 1 K+ 19 – 1 = 18 O22- (2 x 8) + 2 = 18 2 Na+ 11 – 1 = 10 N3- 3(7) + 1 = 22 3 NH4+ 7 + 4 – 1 = 10 F- 9 + 1 = 10 4 K+ 19 - 1 = 18 CO32- 6 + 3(8) + 2 = 32

3 9729 / TJC PRELIM / 2024 [Turn over 3 Use of the Data Booklet is relevant to this question. The sixth ionisation energies of four different elements are shown below. The elements are arsenic, selenium, antimony and tellurium (though not necessarily in that order) from Groups 15 and 16. Which of the following shows the sixth ionisation energy of antimony? A 12300 B 10400 C 7880 D 6820 Answer: B Arsenic (As) and Antimony (Sb) are in Group 15 and have 5 valence electrons in their valence shell. The sixth ionisation energies of As and Sb involve the removal of electron from an inner electron shell, hence the values are large → options A and B. Sb is in a Period below As, the electron to be removed from Sb is further from the nucleus and hence less strongly attracted. Hence, the sixth ionisation energy indicated in B is for Sb. 4 Trimethoprim (TMP) is used for the treatment of urinary tract infections. It has the following structure: Which are the correct bond angles w, x, y and z? w x y z A 90 180 120 90 B 90 105 118 107 C 109.5 105 120 107 D 109.5 180 118 120 Answer: C w: 4 bond pairs 0 lone pair (tetrahedral)

4 9729 / TJC PRELIM / 2024 x: 2 bond pairs 2 lone pairs (bent) y: 3 bond pairs 0 lone pair (trigonal planar) z: 3 bond pairs 1 lone pair (trigonal pyramidal 5 Beryllium chloride, BeC l2, reacts with methylamine, CH 3NH2 to form a compound. Which of the statements is incorrect? A The compound is formed from 1 mole of BeCl2 and 2 moles of CH3NH2. B The Be-N bond formed is chemically similar to a covalent bond. C The beryllium atom in beryllium chloride is electron deficient. D The compound is capable of forming only two hydrogen bonds per molecule. Answer: D A & C Only 4 e around Be on BeCl 2, hence electron deficient. It can accommodate another 2 pairs of e from N to achieve octet. B A dative bond is the same as a single covalent bond (strength and length) D The compound formed has no more lone pairs of e on N available for hydrogen bonding. 6 In which of the following pairs of compounds would the first compound have a higher melting point than the second compound? A K2O, Na2O B AlCl3, AlF3 C NH2CH2CO2H, HOCH2CO2H D , Answer: C A: incorrect. 2 2 MO MO LE qq rr +− +−  + . Since K Narr++ , thus Na2O has a higher melting point. B: Incorrect. AlF3 is ionic while A lCl3 is a simple molecular structure  AlF3 higher melting point as large amount of energy is needed to break the strong ionic bond compare ti breaking weak id-id between AlCl3 molecules. C: Correct. NH2CH2CO2H exists as zwitterions 3 2 2H NCH CO+− held together by strong electrostatic forces of attraction whereas HOCH2CO2H molecules are held together by weaker

5 9729 / TJC PRELIM / 2024 [Turn over hydrogen bonding. Hence, NH 2CH2CO2H has a higher melting point. Energy needed to break the ionic bond is large than breaking H-bond, Hence, mpt of NH2CH2CO2H is higher. D: Incorrect. forms more extensive intermolecular hydrogen bonds , which requires more energy to overcome, hence higher melting point. On the other hand, due to intramolecular hydrogen bonding, forms less extensive intermolecular hydrogen bonds , which requires less energy to overcome, hence lower melting point. 7 Use of the Data Booklet is relevant to this question. A solution containing 25 cm3 of 0.1 mol dm-3 VO2+ reacts completely with 0.245 g of zinc. Which of the following can be the vanadium-containing species in the product? A VO3– B VO2+ C V3+ D V2+ Answer: D Number of moles of VO2+ = 325 0.1 2.5 10 mol1000 − =  Number of moles of Zn = 30.245 3.75 10 mol65.4 −= 2VO2+  3Zn  6 e– VO2+  3 e– Oxidation state of V in product = +2, hence V2+ is formed. δ+ δ- hydrogen bond

6 9729 / TJC PRELIM / 2024 8 Which graph is a correct representation of Charles’ Law? A B C D Answer: C Charles Law states that V  T, hence the graph should be a straight line with positive gradient. V will be 0 at 0K, not at 0°C, so the graph of V against T/°C does not pass through the origin. Option A shows the correct relationship between P and 1/V, as indicated by Boyle’s Law. 9 The three minerals below are obtained from mines around the world. Each one behaves as a mixture of two carbonate compounds. They can be used as fire retardants because they decompose in the heat, producing CO2. This gas smothers the fire. Barytocite BaCa(CO3)2 Dolomite CaMg(CO3)2 Huntite Mg3Ca(CO3)4 What is the order of effectiveness as fire retardant, from best to worst? best worst A huntite dolomite barytocite B huntite barytocite dolomite C dolomite huntite barytocite D dolomite barytocite huntite P/ Pa 1 V / m-3 0 0 P/ Pa 1 V / m-3 0 0 V/ m3 T/°C 0 0 V/ m3 T/°C 0 0

7 9729 / TJC PRELIM / 2024 [Turn over Answer: A Each mineral behaves as a mixture of two carbonate compounds. mineral formula behaves as a mixture of barytocite BaCa(CO3)2 1 mol BaCO3 and 1 mol CaCO3 dolomite CaMg(CO3)2 1 mol CaCO3 and 1 mol MgCO3 huntite Mg3Ca(CO3)4 3 mol MgCO3 and 1 mol CaCO3 The effectiveness of each of the three minerals as fire retardant is dependent on its ease of thermal decomposition to produce CO2, which smothers the fire. The ease of thermal decomposition of the minerals is dependent on the charge density and hence the polarising power of the respective Group II metal ions (Ba 2+, Ca2+ and Mg 2+). The order of effectiveness as fire retardant, from best to worst, corresponds to t