RI 2024 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2024 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 D D C D B B A D C C D A C B A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 D C D D B C D C B D A C B C B Q 1(D) Options A and B are incorrect as X is a cation while both options show anions. No. of e− in Ne atom = 10 protons neutrons electrons 23Na+ 11 12 10 24Mg2+ 12 12 10 T hus, cation X is 24Mg2+ since it has the same number of neutrons as protons. Q2(D) Statement 1 is incorrect as a longer covalent bond generally results in a smaller bond energy. Statement 2 is incorrect as I in H I has a larger nuclear charge than Cl in HC l but HI has a lower bond energy than HCl. Statement 3 is correct as more energy is required to overcome a greater attraction between shared pair of electons and the nuclei of the bonded atoms. Q3(C) Statement 1 is incorrect as H 2S has stronger permanent dipole-permanent dipole forces of attraction than H 2Se and hence a higher boiling point. This is so as the H−S bond is more polar than the H−Se bond due to greater electronegativity difference between H and S than H and Se since S is more electronegative than Se. Statement 2 is correct as H2O has less number of electrons and hence a smaller and less polarizable electron cloud, resulting in weaker instantaneous dipole-induced dipole forces of attraction than H2Se. Statement 3 is correct as the H−O bond is more polar than the H−S bond due to greater electronegativity difference between H and O than H and S since O is more electronegative than S. Q4(D) There are four bond pairs of electrons on N and no lone pair of electrons, hence the bond angle is 109. There are four bond pairs of electrons on C in CH 2 and no lone pair of electrons, hence the bond angle is 109. There are three bond pairs of electrons on N and no lone pair of electrons, hence the bond angle is 120. Q5(B) T otal 10 single unpaired electrons 1 unpaired electron 1 unpaired electron 0 unpaired electron 2 unpaired electrons 3 unpaired electrons 2 unpaired electrons 1 unpaired electron 0 unpaired electron

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q6(B) Option A is incorrect as the highest possible oxidation state of P is +5 and PC l5 contains 85.1% by mass of chlorine. Option B is correct as the highest possible oxidation state of Si is +4 and SiO2 contains 53.2% by mass of oxygen. Option C is incorrect as the highest possible oxidation state of P is +5 and P4O10 contains 56.3% by mass of oxygen. Option D is incorrect as the highest possible oxidation state of S is +6 and SO3 contains 59.9% by mass of oxygen. Q7(A) Ar = 69.3  58 + 26.7  60 + 1.20  61 + 3.80  62 + 1.00  64 69.3 + 26.7 + 1.20 + 3.80 + 1.00 = 58.8 Note: Do not assume that relative abundances are percentages and do not assume that the relative abundances will always add up to 100. They do not add up to 100 in this case and students who made these assumptions would have incorrectly chosen option C. Q8(D) T he least negative value (exothermic) for lattice energy will belong to a compound with smallest cationic and anionic charges as well as the largest cationic and anionic radii. Since K + has a smaller charge than Ca 2+ and peroxide anion, O 2 2- , has a larger anionic radius than oxide anion, O 2−, potassium peroxide (option D) will have the least negative value for lattice energy. Q9(C) Since [SOC l2] changed from 0.016 mol dm −3 to 0.002 mol dm −3 (0.016 ÷ 2 ÷ 2 ÷ 2) after 24 hours, each half-life is 24 ÷ 3 = 8 hours. A total of 4 half- lives are required to decrease [SOC l2] from 0.016 mol dm−3 to 0.001 mol dm −3 (0.016 ÷ 2 ÷ 2 ÷ 2 ÷ 2), thus a total of 32 hours is needed. Q10(C) When Q is in large excess, the reaction exhibits pseudo-order kinetics and given that the graph of [P] against time has a constant half-life, this means that the reaction is pseudo-first-order of reaction with respect to P. When [P] remained constant and the [Q]  2.4, the initial rate of reaction  2.4 2 approximately. Hence, the reaction is 2nd order with respect to Q. rate = k[P]1[Q]1, substituting in values for 1st expt, 4.76  10−4 = k [0.02][0.025]2 k = 38.1 Q11(D) Activation energy remains constant even as temperature changes. As temperature decreases, the maximum of the Boltzmann distribution curve displaces to the left and takes on a smaller value. Thus, the number of particles with kinetic energy greater than or equals to activation energy will be lower. Q12(A) Statement 1 is incorrect as the heterogenous catalyst (Fe) forms weak bonds with the reacting molecules and these bonds are very much weaker than the covalent bonds within the reacting molecules. Statement 2 is incorrect as Fe is acting as a heterogenous catalyst and its mode of action does not change oxidation state unlike the mode of action for homogeneous catalysts

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q13(C) Since the vessel has a fixed volume of 1.00 dm 3, concentration is directly proportional to amount of gas. H2 + CO 2 ⇌ H2O + CO initial / mol dm−3 0.01 0.01 0 0 cha nge / mol dm−3 −x −x +x +x eq uilibrium / mol dm−3 0.01−x 0.01−x x x Kc = x2 (0.01−x)2 = 0.123 x = 0.00026 mol dm−3 [H2(g)]eqm = 0.01−x = 0.01−0.00026 = 0.0074 mol dm−3 Q14(B) Electronegativity increases across the period, thus P has the highest electronegativity than Si than Al than Mg. Atomic radius decreases across the period, thus P has the smallest atomic radius (0.110 nm) than Si (0.117 nm) than Al (0.143 nm) than Mg (0.160 nm). Since the atomic radius of Si is closer to P than Al, option D is incorrect and option B is correct. Q15(A) A reducing agent undergoes oxidation and hence the metal will be the reducing agent in the reaction and since all elements have oxidation state = 0, a greater change in the magnitude of the oxidation number for the reducing agent depends on the magnitude of oxidation state of the cation. Option A is correct as the cation has a large magnitude of charge than the anion. Q16(D) Statement 1 is incorrect as a pH of 13 at the end of experiment would mean [OH −] = 0.1 mol dm −3 which is the same as the initial [OH−] before mixing. This is not possible as some of the OH− would have reacted and there is also mixing with another solution, causing total volume to increase. When calculated, the pH at the end of the experiment is 12.2. Statement 2 is correct as pH = pKa + lg [salt] [acid] an d [salt] [acid] = 1 since half the acid has been transformed to an equal amount of salt. Q17(C) KOH will react with some of the C 2H5COOH to give C2H5COO− resulting in a buffer solution. pH = −lg(1.35  10−5) + lg( 10.0 1000 0.20 25.0 1000 0.15 − 10.0 1000 0.20 ) = 4.93 Q18(D) The reaction of a negatively charged carbon atoms with a carbonyl compound, to give an alcohol with more carbon atoms is nucleophilic addition, similar to the reaction of :CN− with carbonyl compounds. While the C 2H5− anion functions as a nucleophile, option D is the closest as a Lewis base also donates it’s lone pair electrons. A Incorrect. C 2H5− is not electrophile, as it is donating, not accepting an electron pair. B Incorrect. C 2H5− does not generate OH − in this reaction as there is no water. C Incorrect. C2H5− is not a H+ acceptor. D Correct, C2H5− donates its lone pair electrons in the reaction. Q 19(D) Options A and B are incorrect as there is no change in the number of H atoms when it should have increased by two when 2 moles of HBr iadded to ß−selinene. There are 3 possible products that are tertiary bromoalkanes:

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q20(B) As the 2o carbocatio