RI 2022 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2022 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C C D C A B A B B D D C B B C 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C A C B C A B B B D D A A D D Q1(C ) The ionic bonds in NaF requires the most energy to overcome while the instantaneous dipole- induced dipole in CH3CH2CH2CH3 requires the least energy to overcome, thus NaF has the highest boiling point while CH3CH2CH2CH3 has the lowest boiling point. Both CH 3CH2NH2 and CH 3CH2OH can form an average of one intermolecular hydrogen bond per molecule. The more polar O−H bond in CH3CH2OH results in stronger hydrogen bonds between CH3CH2OH molec ules compared to the less polar N−H bond in CH 3CH2NH2, thus CH 3CH2OH has a higher boiling point than CH3CH2NH2. Q2(C ) A Incorrect. Br is less electronegative than Cl. B Incorrect. Having the same outer shell electronic configuration does not explain why the Br−Cl bond is polar. C Correct. Since Br has one additional electronic shell than Cl, its outer shell electrons are more shielded from the nuclear charge and hence the shared pair of electrons in the Br−C l bond is less attracted to Br, making Br less electronegative than C l, resulting in a polar Br−Cl bond. D Incorrect. Since Br and Cl have the same outer shell electronic configuration, the repulsion between electrons in the outer shell will be similar in both atoms. Q3(D ) 1 Incorrect. Ethanoic acid has a higher pKa and hence smaller K a, thus is a weaker acid than thioacetic acid. S o H+ is not more easily removed from ethanoic acid. 2 Incorrect. Ethanoic acid has a higher p Ka and hence smaller Ka than thioacetic acid. 3 Correct. Thioacetic acid has a smaller p Ka and hence higher K a, is a stronger weak acid than ethanoic acid. The weaker S−H bond in thioacetic acid allows for greater extent of ionisation, and at the same concentration, thioacetic acid will form a higher concentration of H +, giving a solution of lower pH than ethanoic acid. Q4(C ) At the same temperature and pressure, since there is same number of moles of O2 and N2O in entonox gas, PO2 = PN2O. Since PT = 3.55 × 107Pa, PN2O = 3.55 × 107 2 = 1.775× 107Pa Assuming N2O behaves as an ideal gas, pV = nRT = M Mr RT Mass of N2O = pVMr RT = 1.775× 107× 5 1000 × ( 14.0 × 2 + 16.0) 8.31 × ( 273+20) = 1604 g ≈ 1.60 kg Q5(A ) 1 Correct. AlCl3 hydrolyses in water to give a solution of pH 3 , which will cause a vigorous effervescence when added to Na2CO3. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) 2 Incorrect. While Mg Cl2 hydrolyses slightly in water to give a weakly acidic solution of pH 6.5, it will not give a vigorous effervescence when added to Na2CO3 due to the lower concentration of H +, resulting in a slower rate of reaction. MgCl2(s) + 6H2O(l) → [MgH2O)6]2+(aq) + 2Cl–(aq) [Mg(H2O)6]2+(aq) ⇌ [Mg(H2O)5(OH)] +(aq) + H+(aq) 3 Incorrect. NaCl does not undergo hydrolysis in water to produce H + and hence would not react with Na2CO3. Q6(B ) Since X is more electronegative than A rsenic, X is P (adjacent and above Arsenic in Group 15) as electronegativity decreases down a group . Since Y is more electronegative than X, Y is S (adjacent and in the same period as P) as electronegativity increases across a period. Thus proton number of Y is 16.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q7(A) In 1 g of solder glass, there is 0.16 g of B2O3 and 0.84 g of PbO. nB in 0.16 g of B2O3= 0.16 (10.8 × 2 + 16.0 × 3) × 2 = 0.0045977 mol nPb in 0.84 g of PbO = 0.84 (207.2 + 16.0) = 0.0037634 mol nPb nB = 0.0037634 0.0045977 = 0.8185 ≈ 0.82 Q8(B) Since more energy is required to remove an electron from an inner electronic shell and J and M have a higher sixth ionisation energy than G and H, J and M must be from Group 15 while G and H must be from the Group 16. Since sixth ionisation energy decreases down the group and H has a higher sixth ionisation energy than J, so H must be above J in the Periodic Table and thus in Period 3. Q9(B) Outershell electronic configuration of I−: 5s25p6 Outershell electronic configuration of Xe: 5s25p6 Outershell electronic configuration of Cs+: 5s25p6 Since I−, Xe and Cs + are isoelectronic, they have the same shielding effect . As n uclear charge increases from I− to Xe to Cs +, more energy is required to remove the valence electron in Cs+ than Xe than I− due to strong er electrostatic attraction between valence electron and nucleus. Hence, ∆H1 > ∆H3 > ∆H2. Q10(D) Since and all the cations have the same charge and all the anions have the same charge, the solid chloride with the smaller cationic radius will have the most exothermic lattice energy as chloride has a smaller anionic radius. From the Data Booklet, the cation radius of Pb 2+ = 0.120 nm while that of Zn2+ = 0.074 nm. Thus, ZnCl2 will have the smallest interionic radii and most exothermic lattice energy. Q11(D) The increase in the number of moles of gaseous particles in the reaction causes ∆S > 0. Since, ∆G = ∆H −T∆S and ∆S > 0, ∆H < 0, ∆G < 0 at all temperatures, thus the reaction is spontaneous at all temperatures. Q12(C) Since lead( IV) oxide remained chemically unchanged and increased the rate of reaction, it is acting as a catalyst. The activation energy of experiment 1 (uncatalysed) will be higher than experiment 2 (catalysed) and the rate constant in experiment 2 will be higher as the catalyst increases the rate constant by lowering the activation energy (k = Ae -Ea RT ). Q13(B) 2SO2(g) + O2(g) ⇌ 2SO3(g) initial amt. / mol 2.00 2.00 0 change / mol −1.80 −0.90 +1.80 eqm. amt. / mol 0.20 1.10 1.80 Converting number of moles to concentration, K c = ( 1.80 0.500) 2 ( 0.20 0.500) 2 × ( 1.10 0.500) ≈ 36.8 Q14(B) pH = pKa + lg ( [HCO3 –] [H2CO3 ]) 7.4 = −lg (2.5 × 10−4) + lg ( [HCO3 –] [H2CO3 ]) 7.4 + lg (2.5 × 10−4) = lg ( [HCO3 –] [H2CO3 ]) 10 7.4 + lg (2.5 × 10-4) = [HCO3 –] [H2CO3 ] [H2CO3 ] [HCO3 – ] = 1.6 × 10−4 Q15(C) AgCl(s) ⇌ Ag+(aq) + Cl−(aq) ------ (1) Solubility of silver chloride will increase when NH3(aq) is added, due to the formation of soluble Ag[(NH3)2]+ complex ion, which will shift the position of equilibrium of (1) to the right. Solubility of silver chloride will decrease when NaCl(aq) is added, due to the presence of the common ion C l− which increases the concentration of Cl−, shifting the position of equilibrium of (1) to the left.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q16(C ) Ag2SO4(s) ⇌ 2Ag+(aq) + SO42−(aq) Let the solubility of Ag2SO4 in water be s mol dm–3. At equilibrium in the saturated solution, [Ag +] = 0.032 mol dm–3 [SO42−] = 0.032/2 = 0.016 mol dm–3 Ksp = [Ag+]2[SO42−] = 0.0322 × 0.016 Ksp = 1.6384 × 10−5 mol3 dm−9 Let the solubility of Ag2SO4 in 0.50 mol dm–3 Na2SO4 solution be y mol dm–3. Na2SO4 (aq) → 2Na+(aq) + SO42−(aq) At equilibrium in the saturated solution, [Ag+] = y mol dm-3 [SO42−] = y 2 + 0.50 mol dm-3 Ksp = [Ag+]2[SO42−] = y2 × ( y 2 + 0.50) = 1.6384 × 10−5 S ince Ag2SO4 is sparingly soluble in water and the presence of SO42− ions from Na2SO4 further suppresses its solubility, y 2 << 0.50. Thus, ( y 2 + 0.50) ≈ 0.50. y2 × (0.50) = 1.6384 × 10−5 y = 5.7 × 10−3 mol dm−3 Hence, solubility of Ag 2SO4 in 0.50 mol dm –3 Na2SO4 = 5.7 × 10−3 mol dm−3. Q17(A ) 1 Incorrect. (CH3)3C Cl (CH3)3C Cl 2 Correct. (CH3)3C Cl (CH3)3C Cl carbocation 3 Correct. (CH3)3C OH 2 (CH3)3C carbocation H2O 4 Incorrect. (CH3)2C Cl O Cl(CH3)2C O Q18(C ) 1 Correct. Enantiomers i