RI 2021 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2021 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B D C B C D B D D A C A B B A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A C B D C A A A B B D B C C D Q1(B) 1 Correct. Angle of deflection is proportional to q m . Since m electron < m proton, electrons have a greater angle of deflection and are deflected to a larger extent than protons. 2 Correct. Electrons are attracted to the positive plate while protons are attracted to the negative plate i.e. the electron beam are deflected in the opposite direction to the proton beam. 3 Incorrect. The proton beans will travel in a curve path towards the negative plate. Q2(D) Q in the ionic nitrate, QNO3, exists as Q+. Since Q+ has 80 electrons, Q has 81 electrons and 81 protons. From the Periodic table, Q is the element thallium which belongs to group 13. The nucleon number of Q is 81 + 122 = 203. Q3(C) A Incorrect. 2nd IE of F : F+ → F2+ + e– [He]2s22p4 [He]2s22p3 3rd IE of Al : Al2+→ Al3++ e– [Ne]3s1 [Ne] 2nd IE of F is greater because the electrons are removed from an inner electronic shell. B Incorrect 3rd IE of electron removed from F 2p Ne 2p Na 2p Mg 2p Al 3s C Correct. 4th IE of Na : Na3+ → Na4+ + e– [He]2s22p4 [He]2s22p3 3rd IE of Ne : Ne2+→ Ne3+ + e– [He]2s22p4 [He]2s22p3 Na3+ and Ne 2+ have the same electronic configuration and experience the same shielding effect. 4 th IE of Na is greater due to the greater nuclear charge of Na which causes the 2p electrons of Na3+ to experience a greater attraction to the nucleus, requiring more energy to remove. D Incorrect. Successive IE’s always increase even if the electrons are removed from different shells. This is because the nuclear charge remains the same, but the number of electrons and shielding effect experienced by the remaining electrons decreases. Hence the electrostatic attraction between the nucleus and the outermost electron increases, resulting in an increase in energy required to remove each subsequent electron. Q4(B) Since there is a large jump between the 7 th and 8th IE for element W, the 8 th electron is removed from an inner shell i.e. W has 7 valence electrons and is from group 17. Since W, X, Y and Z are consecutive elements, X is from group 18, while elements Y and Z are from groups 1 and 2 of the next period respectively. X, the group 18 element, has a higher first IE than W, the group 17 element of the same period since IE increases across the period. X also has the higher first IE then Y and Z since Y and Z are from the next period and their valence electrons are further away from the nucleus and experience greater shielding effect.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q5(C) compound C2H6 C2H4 C2H2 carbon- carbon bond C–C C=C C≡C bond energy → Bond energy increases due to increased number of shared electrons and increased attraction between bonding electrons and nuclei. bond length → Bond length decreases. In general, the stronger the bond, the shorter the bond length. Q6(D) molecule molecular shape polarity A BCl3 trigonal planar  polar  non-polar B NCl3 trigonal pyramidal  non-polar  polar C SO2 linear  bent non-polar  polar D CHCl3 tetrahedral  polar  Q7(B) Since all 4 compounds have similar M r, their strengths of id-id interactions are similar. M, P and Q can form stronger intermolecular hydrogen-bonding (due to the presence of – OH groups) compared to the weaker pd- pd interactions in N. Hence, N has a lower bp than M, P and Q. M has 3 –OH groups and an average of 3 hydrogen bonds per molecule, resulting in more extensive hydrogen bonding compared P and Q which have 1 –OH group each, forming an average of 1 hydrogen bond per molecule. Hence, M has a higher BP than P and Q. Due to presence of an addition electron- withdrawing C=O group in Q, the –OH group in Q is more polar, forming stronger intermolecular hydrogen bonds compared to P. Hence Q has a higher BP than P. Q8(D) A Incorrect. This is Avogadro’s Law. B Incorrect. This is an application of Dalton’s Law, not the definition of Dalton’s Law. C Incorrect. The partial pressure of a gas is given by the product of its mole fraction and the total pressure. Even then, this is not Dalton’s Law. D Correct. Q9(D) Experiment 1 – SiO2 solid does not dissolve in water i.e. SiO2 solid remains. Experiment 2 – SiO2 solid does not react with, and hence does not dissolve in HCl(aq) i.e. SiO2 solid remains. Experiment 3 – SiO2 solid does not react with, and hence does not dissolve in NaOH(aq) i.e. SiO2 solid remains. Note: S iO 2 only reacts with concentrated NaOH. Q10(A) No. of molecules = Amount in moles x Avogadro’s constant A Ethyl methanoate, CH3CH2O–CHO (Mr = 74.0) No. of molecules = 2.00 74.0 x 6.02 x 1023 = 1. 63 x 1022 B Br2(l) (Mr = 159.8) No. of molecules = 4.00 159.8 x 6.02 x 1023 = 1.51 x 1022 C No. of molecules = 550 24000 x 6.02 x 1023 = 1 .38 x 1022

This document is copyrighted, please do not reproduce it without permission © Raffles Institution D No. of molecules = 1.55 x 1022 Q11(C) Amount of H2SO4 = 20.0 1000 x 5.00 = 0.100 mol Amount of NaOH = 20.0 1000 x 5.00 = 0.100 mol H2SO4 + 2NaOH → Na2SO4 + 2H2O Since H2SO4 reacts with NaOH in a 1:2 ratio, NaOH is limiting. q = mc∆T = (20.0 + 20.0)(4.18)(50.0 – 25.0) = 4180 J ∆H = − q nNaOH = −4180/0.100 = −41800 J mol–1 = −41.8 kJ mol–1 Q12(A) This is a graph of rate of forward reaction against time and the rate equation for the forward rate is rate = k(pCO2 ). At time t, the pressure i.e. pCO2 is lowered, causing a decrease in the forward rate just after time t as seen in options A and B. The pressure was then allowed to return to atmospheric pressure, so p CO2 increases back to the initial pressure and the forward rate increases back to the original rate as seen in option A. Q13(B) 1 Correct. For the hydrogen and bromine reaction, HBr appears in the denominator of the rate equation. When [HBr] increases, rate decreases i.e. the formation of HBr slows down the rate of reaction. 2 Correct. The rate equation for H 2 and Br 2 involves many species. It is unlikely that so many species will collide and be involved in a single step reaction. Also, the stoichiometry of the reactants reaction is not the same as the orders of reaction in the rate equation. This is not likely to be a single step reaction. Since the rate equation involves 1 mole of H 2 and 1 mole of I2 which is the same as the stoichiometry of the reaction between H2 and I2, it could be a single step reaction. 3 Incorrect. Using the rate equation for reaction 1, when [Br 2] x 2, rate x 2 1.5 = 2.8 i.e. the rate of reaction 1 is not doubled when [Br 2] is doubled. Q14(B) Stereoisomers = cis-trans isomers and enantiomers Max no. of stereoisomers = 2m+n m = no. of double bonds that can undergo cis-trans isomerism n = no. of chiral centres No. of stereoisomers = 21+1 = 4 No. of stereoisomers = 23 = 8 Q15(A) A Correct. In the propagation steps, methylproprane reacts with the X● radical generated in the initiation step. (CH3)3CH + X● → (CH3)3C● + HX -- (a) (CH3)3C● + X2 → (CH3)3CX + X● (CH3)3CX i.e. C 4H9X is generated in one of the propagation steps. B Incorrect. In the