2020 YIJC Prelim Exam Paper 1 (Suggested Answers)

2020 JC2 H2 Chemistry 9729 Preliminary Examination 1 H2 CHEMISTRY 9729 Paper 1 Multiple Choice Questions 1 2 3 4 5 6 7 8 9 10 B A D C A C B A B C 11 12 13 14 15 16 17 18 19 20 C D D D B A B A D C 21 22 23 24 25 26 27 28 29 30 B B D D D A A B D C 1 Answer: B At ground state, the electronic configuration of C is 1s2 2s2 2p2.  The occupied orbital of the highest energy is the 2p orbital which is dumb-bell in shape.  The occupied orbital of the lowest energy is the 1s orbital which is spherical in shape.  There are 2 electrons in the p orbitals and 4 electrons in the s orbitals (1s and 2s).  The electrons are present in 3 different energy levels. 2 Answer: A The angle of deflection is proportional to charge/mass of the particle. The angle of deflection of X+ is smaller than that of P +. Since both particles have the same charge, X+ must have a larger mass than P+. 3 Answer: D C2H4: trigonal planar, 120o HCHO: trigonal planar, 120o NH4+: tetrahedral, 109.5o NH3: trigonal pyramidal, 107o 4 Answer: C C2H2: 5  bonds C2H5OH: 8  bonds CH3CHO: 6  bonds CH3CO2H: 7  bonds 5 Answer: A Since the flasks are identical, the volumes of the 3 gases are the same. The masses of the 3 gases are also the same. pV = nRT and n = m/Mr pV = (m/Mr)RT Since V, m and R are constant, Mr  T/P Mr(X) = k(t/p); Mr(Y) = 0.5k(t/p); Mr(Z) = 2k(t/p) 6 Answer: C Amount of CO2 at r.t.p. = 500 ÷ 24000 = 0.020833 mol In 1 molecule of CO2, there are 3 atoms. Amount of atoms = 3 0.020833 = 0.0625 mol Number of atoms = 0.0625 6.02 1023 = 3.76 1022 7 Answer: B H2S(g) + 1 2O2(g)  S(s) + H2O(g) ΔH = [ΔHf(products)] – [ΔHf(reactants)] ΔH = [0 + (243.0)] – [(20.5) + 0] = 222.5 kJ mol1 8 Answer: A ΔGo = ΔHo −TΔSo = (+61) – 298 (−30×10−3) = + 69.9 kJ mol−1  ΔGo is positive.  A positive ΔGo indicates that dissolution is not spontaneous, hence silver chromate is almost insoluble in water.  Since ΔS is negative, at a higher temperature, ΔG tends towards a more positive value. Dissolution becomes less spontaneous at higher temperature.

2020 JC2 H2 Chemistry 9729 Preliminary Examination 2 9 Answer: B  At low [substrate], the graph shows a straight line (similar to y = mx graph) shows that the rate is directly proportional to [substrate]. Hence the order of reaction is first order with respect to [substrate].  At high [substrate], the graph shows a horizontal line (similar to y = k graph) shows that the rate is independent of [substrate]. Hence the order of reaction is zero with respect to [substrate].  The enzyme-substrate is an intermediate of the reaction and hence should not appear in the rate equ ation since rate equation is expressed in terms of the concentrations of the reactants. 10 Answer: C 4X  2X  X  0.5X 3 half-lives = 3 5 min = 15 mins Y  0.5Y 1 half-life = 15 mins 11 Answer: C  When the pressure is increased, the position of the equilibrium shifts to the side with fewer number of moles of gaseous molecules to decrease the additional pressure. Hence the equilibrium shifts to the right and the colour becomes lighter.  When the temperature is increased, the position of the equilibrium will shift towards the endothermic reaction to absorb the additional heat. Hence the equilibrium shifts to the left and the colour becomes darker. 12 Answer: D There is no sharp increase in the pH of a weak acid – weak base titration. Hence there is no suitable indicator for such a titration. 13 Answer: D A mixture of benzoic acid and sodium benzoate forms an acidic buffer solution. pH = pKa + log [salt] [acid] = log(6 105) + log (0.1/0.01) = 5.22 14 Answer: D For precipitation to occur, ionic product = Ksp PbX2(s) ⇌ Pb2+(aq) + 2X(aq)  PbBr2 [Pb2+][1 103]2 = 4.0 105 [Pb2+] = 40.0 mol dm3  PbF2 [Pb2+][1 103]2 = 2.7 108 [Pb2+] = 0.027 mol dm3  PbI2 [Pb2+][1 103]2 = 7.1 109 [Pb2+] = 0.00710 mol dm3 PbSO4(s) ⇌ Pb2+(aq) + SO42(aq)  PbSO4 [Pb2+][1 103] = 1.6 108 [Pb2+] = 0.000016 mol dm3 PbSO4 requires the lowest concentration of Pb2+(aq) to be precipitated. Hence, PbSO 4 will be the one that precipitate out first. 15 Answer: B Going across period 3,  Na, Mg and Al are metals, Si is metalloid, P4, S8 and Cl2 are non-metals.  The maximum oxidation states exhibited in the compounds of the elements corresponds to their group number increasing for +1 in Na to +7 in chlorine.  The melting points of the elements does not decreases consistently across the period. melting point Na Mg Al Si P S Cl

2020 JC2 H2 Chemistry 9729 Preliminary Examination 3 16 Answer: A Rn (Group 18 of Period 6), Fr (Group 1 of Period 7), Ra (Group 2 of Period 7) The shielding effect in Rn is the lowest since it has fewer inner shell electrons than Fr and Ra. Hence it will require the most amount of energy to remove the valence electron. Both Fr and Ra belongs to the same period. The nuclear charge of Fr is lower than that of Ra and they both have approximately the same shielding effect. Hence the effective nuclear charge of Fr is lower than Ra and requires a lower amount of energy to remove the valence electron. 17 Answer: B CCl4 does not dissolve in water because is non- polar and cannot form hydrogen bonds with water molecule. It also does not react with water because the C atom does not have any energetically accessible d-orbitals. 18 Answer: A Going down Group 17, (A) the strength of the instantaneous dipole – induced dipole increases as the electron cloud size gets bigger , hence boiling increases; (B) the bond length increases as the size of the p-orbitals overlapping increases and thus the effectiveness of the orbital overlapping decreases. (C) the electronegativity decreases; (D) the ionic radius increases as the increase in the number of inner electron shell and thus shielding effect outweighs the increase in the nuclear charge; 19 Answer: D  a five-membered ring is a cyclopentane;  a ketone has 2 R groups bonded to the C=O;  a tertiary alcohol has 3 R groups bonded to the carbon with the –OH group. In this case, the type of constitutional isomerism involved in positional and chain isomerism. 20 Answer: C Amount of C = 0.66 ÷ 44.0 = 0.0150 mol Amount of H = 2 (0.36 ÷ 18.0) = 0.0400 mol Mole ratio of C : H = 0.0150 : 0.0400 = 3 : 8 Empirical formula of Y is C3H8 21 Answer: B Alkanes are unreactive as the C-H bonds are not polar and hence they will not be attacked by electrophiles and nucleophiles.

2020 JC2 H2 Chemistry 9729 Preliminary Examination 4 22 Answer: B type of H atom number of atoms expected ratio primary, HP 9 9 1 = 9 secondary, HS 2 2 4 = 8 tertiary, HT 1 1 6 = 6 23 Answer: D (A) is incorrect as pentaerythritol only reacts with HBr(g) or conc. HBr. (B) is incorrect as both the empirical and molecular formulae is C5H12O4. (C) is incorrect as the central C atom, which is the adjacent C atoms to all the C with –OH group does not contain any H for elimination to occur. (D) is correct. Since 1 mole of –OH reacts with sodium to form 0.5 mole of hydrogen gas, the presence of 4 –OH groups in pentaerythritol will produce 2 moles of hydrogen gas. 24 Answer: D 1 CH3)2C=CHCH2CH2CH=CH2 will oxidise to form (CH 3)2C=O and HO2CCH2CH2CO2H and CO2 2 CH3(CH2)4CH=CHCH2OH will oxidise to form CH3(CH2)4CO2H and ( CO2H)2. (CO2H)2 will further oxidise to form CO 2 and H 2O so it forms only one organic compound in the end. 3 CH3(CH2)2CH