2023 TMJC H2 Chem Prelim P1 (Ans)

2023 H2 Chem Prelim P1 (Ans) Answer: C Let % abundance of 11B be m%. 10.75 = (m)(11)+(100-m)(10) 100 m = 75 2 Use of Data Booklet is relevant to this question. On losing an electron, which particle would have the greatest number of unpaired electrons? A C– B Fe3+ C Ti2+ D P Answer: B C– forms C : [He] 2s22p2 (2 unpaired electrons) Fe3+ forms Fe4+ : [Ar] 3d4 (4 unpaired electrons) Ti2+ forms Ti3+ : [Ar] 3d1 (1 unpaired electrons) P forms P+: [Ne] 3s23p2 (2 unpaired electrons) 3 Particle X has a charge of +1 and a proton number, n. Particle Y has a proton number of (n+1) and is isoelectronic with X. Which of the following statements correctly describes X and Y? A X and Y are isotopes. B The atoms of X and Y have the same electronic configurations. C Y has a charge of +1 and same charge density as that of X. D Y has a charge of +2 and smaller ionic radius than X. 1 A sample of boron contains two naturally occurring isotopes, 10B and 11B. This sample has a relative atomic mass value of 10.75. What is the percentage of the isotope 11B in the sample? A 20 B 25 C 75 D 80

2 2023 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry Answer: D X Y Proton number, n n n + 1 No of electrons n – 1 n – 1 Charge +1 +2 Option A is wrong X and Y have different number of protons. Hence they cannot be isotopes. Option B is wrong X and Y have different number of protons. X atom has n electrons while Y has (n + 1) electrons. Hence X and Y atoms have different full electronic configurations. Option C is wrong Y has charge of +2 while it has same number of electrons as that of X. Hence, Y has smaller ionic radius. This leads to higher charge density. Option D is correct Y has charge of +2 while it has same number of electrons as that of X. Hence, Y has smaller ionic radius. 4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory is used to predict the shapes of molecules or polyatomic ions. Which of the following pairs of species are planar? A F2O and PF5 B SO32– and ClF3 C BrF4– and NO2 D ICl5 and CO2 Answer: C F2O 2 bp 1 lp Bent planar PF5 5 bp 0 lp Trigonal bipyramidal Non-planar SO32– 3 bp 1 lp Trigonal pyramidal Non-planar ClF3 3 bp 2 lp T-shaped Planar BrF4– 4 bp 2 lp Square planar Planar NO2 2 bp 1 lone e - Bent Planar ICl5 5 bp 1 lp Square pyramidal Non-planar CO2 2 bp 0 lp linear Planar

3 [Turn over Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry 5 Which of these phenomena cannot be explained by hydrogen bonding? 1 NH4Br has a higher boiling point than HBr. 2 Ice is less dense than water. 3 Dimerisation of NO2 to form N2O4. 4 Ethanoic acid forms dimers which dissolved in organic solvents A 1, 3 and 4 only B 1 and 3 only C 2 and 4 only D 1 only Answer: B Larger amount of energy is needed to overcome the strong ionic bonds in NH4Br than the weaker intermolecular permanent dipole-permanent dipole attraction in HBr. When dissolved in organic solvents such as hexane, ethanoic acid form dimers by forming intermolecular hydrogen bonds between two ethanoic acid monomers. Dimerisation of NO2 radicals to N2O4. Ice has a three-dimensional tetrahedral lattice with an open structure held by intermolecular hydrogen bonds. With the open structure, ice has less mass per unit volume (i.e. lower density) than water.

4 2023 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry 6 A 2.50 dm3 vessel containing ethane gas at a pressure of 1.50 x 105 Pa is connected to a 7.50 dm 3 vessel containing methane gas at a pressure of 2. 50 x 10 5 Pa. The gases are allowed to mix freely. The temperature of the system is then raised from 100 °C to 350 °C. (Note that there is no chemical reaction between both gases) What is the final pressure of the system? A 2.25 x 105 Pa B 3.76 x 105 Pa C 6.68 x 105 Pa D 7.88 x 105 Pa Answer: B Pethane = 1.50 x 105 x 2.50/10 = 3.75 x 104 Pa Pmethane = 2.5 x 105 x 7.50/10 = 1.875 x 105 Pa Ptotal = 3.75 x 104 + 1.875 x 105 = 2.25 x 105 Pa @ 100K 12 12 5 2 pp=TT p2.25 x 10 =(100 + 273) (350 + 273) p2 = 3.76 x 105 Pa 7 Which statements concerning only the elements in the third period, sodium to argon, are correct? 1 The element with the highest pH for its chloride in water is sodium. 2 The element with the highest electrical conductivity is aluminium. 3 The element with the highest melting point for its oxide is silicon 4 The element that has eight atoms in its molecule is sulfur. A 1 and 4 only B 2 and 3 only C 1, 2 and 4 only D 1, 3 and 4 only Answer: C All statements are correct except for option 3. MgO has highest melting point among Period 3 oxides.

5 [Turn over Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry 8 The equations for three reactions are given below: Cl2(g) + H2S(g) → 2HCl(g) + S(s) SO2(g) + 2H2S(g) → 2H2O(l) + 3S(s) Cl2(g) + 2H2O(l) + SO2(g) → 2HCl(aq) + H2SO4(aq) Which is the correct order of strength of the three reacting gases as oxidising agents? Strongest Weakest A Cl2 SO2 H2S B Cl2 H2S SO2 C H2S SO2 Cl2 D SO2 H2S Cl2 Answer: A Cl2(g) + H2S(g) → 2HCl(g) + S(s) SO2(g) + 2H2S(g) → 2H2O(l) + 3S(s) Cl2(g) + 2H2O(l) + SO2(g) → 2HCl(aq) + H2SO4(aq) From equation (1) and (2), chlorine gas can oxidise both H2S and SO2, thus it is the strongest oxidizing agent. From equation (1) and (2), hydrogen sulfide get oxidised by SO2 and HCl, thus it is the weakest oxidising agent. -2 0 -2 0 +4 +6

6 2023 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry 9 The equation for the reaction of nitrogen monoxide with oxygen is shown below. 2NO(g) + O2(g) ⎯⎯ → 2NO2(g) From initial rate experiments, the following rate equation was derived. Rate = k[NO]2[O2] The results of the initial rates experiments are shown. initial [NO] / mol dm−3 initial [O2] / mol dm−3 initial rate of formation of NO2/ mol dm−3 s−1 0.0010 0.002 7.00 x 10−6 0.0020 0.003 a b 0.004 1.26 x 10−4 What are the missing values a and b? a b A 1.9 x 10−5 0.0030 B 1.9 x 10−5 0.0045 C 4.2 x 10−5 0.0030 D 4.2 x 10−5 0.0045 Answer: C Rate = k[NO]2[O2] Comparing 1st and 2nd experiment [NO] doubled and [O2] x1.5 so rate increases 22 = 4 and increases another x1.5. So total rate increases 4 x 1.5 = 6  Initial rate increases by 7 x 10−6 x 6 = 4.2 x 10−5 Comparing 1st and 3rd experiment Initial rate increased by (1.26 x 10−4) / (7 x 10−6) = 18 Since [O2] doubled, means the rate must have increased by 18/2 = 9 times due to the increase in [NO].  [NO] must have increased by 9 = 3 times  Value of [NO] is 0.003

7 [Turn over Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry 10 Caffeine is a drug that increases the activity of the brain and nervous system. Its elimination from the body follows a first-order kinetics with a half-life of 5.0 h. How long will it take for someone who took 50 m g of caffeine, to have 10 m g of caffeine left in his system? A 5.0 h B 11.6 h C 25.0 h D 58.0 h Answer: B For first order kinetics, t ½ = 5.0 h [A]t = [A]o  (½)n 10 = 50 x (½)n  n = 2.32 time = 2.32 x 5 = 11.6 h 11 A student carried out an experiment shown below to determine the enthalpy change of combustion of propan-1-ol (Mr = 60.0) . It was found that the combustion of 1.0 g of propan-1-ol raises the temperature of 200 g of water by 40 °C. Given that the enthalpy change of combustion of propan-1-ol is −2200 kJ mo