RI 2023 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2023 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B C C D B D A C A A D D C C B 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C C C A C A B A B D B D A B C Q1(B ) species proton no. no. of protons + neutrons no. of e– no. of neutrons Nd 60 145 60 145 – 60 = 85 Nd2+ 60 145 58 85 Pm 61 145 61 145 – 61 = 84 Pm3+ 61 145 58 84 Q2(C) The greater the difference in electronegativity between X/Y and C l/O, the greater the ionic character. compound difference in electronegativity XCl2 3.0 – 1.2 = 1.8 Y2O 3.5 – 0.9 = 2.6 XO 3.5 – 1.2 = 2.3 YCl 3.0 – 0.9 = 2.1 ZO2 3.5 – 1.8 = 1.7 ZCl4 3.0 – 1.8 = 1.2 Q3(C) A While C l is more electronegative than Br, causing CH 3Cl to be more polar and have stronger pd -pd interactions than CH 3Br, the significantly larger electron cloud side of Br causes the id- id interactions and hence the total IMF of CH 3Br to be stronger than that of CH3Cl. B There are no H -bonds between molecules of each compound. C Correct. Explanation in option A. D The relative boiling points is dependent on the IMF between the molecules and not the strength of the covalent bonds within the molecules. Q4(D) Shared electrons are circled. Q5(B) molecule structure polar? 1,1-difluoroethene Yes cis-1,2-difluoroethene Yes trans-1,2-difluoroethene No tetrafluoroethene No

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q6(D) Period 3 element Outermost shell Na Mg Al Si P S Cl Ar Elements with only 1 orbital i n outermost shell which contains just one electron = Na, Al & Cl Elements with only 1 orbital i n outermost shell which contains a pair of electrons = Mg, Al, Si & P Q7(A) 1 Correct. Down group 2, the E values become more negative i.e. the position of equilibrium lies more to the left, indicating that the group 2 elements are more readily oxidised and become stronger reducing agents. 2 Correct. Down group 2, the size of the 2+ cations increase, causing a decreasing in the charge density and polarizing power of the cations. The C –O bond of the carbonates are polarized to a lesser extent, requiring more energy to break, increasing the thermal stability of the metal carbonate. 3 Incorrect. See option 2. Q8(C) Let x be the percentage of 29Si in the sample. (92.23/100)(28) + (x/100)(29) + (0.0777–x)(30) = 28.10 x = 5.54. Q9(A) H2SO4 + 2NaOH → Na2SO4 + 2H2O Amt of H2SO4 = (50.0/1000)(2.00) = 0.100 mol Amt of NaOH = (100/1000)(1.00) = 0.100 mol NaOH is the limiting reagent. ⇒ No. of moles of water formed = 0.100 mol q = (100 + 50)(4.18)(29.0 – 20 .0) = 5643 J ∆𝐻𝐻 = − 5643 0.100 = −56.4 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚−1 Q10(A) CaCl and CaCl2: Since Ca2+ has a higher charge and smaller cationic radius than Ca+, the magnitude of the lattice energy of CaC l2 is greater than that of CaCl. MgCl2 and CaC l2: Since Mg 2+ has a smaller cationic radius than Ca 2+, the magnitude of the lattice energy of MgCl2 is greater than that of CaCl2. Hence, |LE(CaCl)| < |LE(CaCl 2)| < |LE(MgCl2)| Q11(D) At low [substrate], not all of the active sites are occupied. In this case, rate ⍺ [substrate] and t he reaction is first order with respect to the substrate.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution At high [substrate], all the active sites are occupied i.e. the active sites of the enzyme become saturated with substrate. In this case, any increase in [substrate] will not have any effect on the reaction rate. The reaction is zero order with respect to the substrate. Q12(D) A Incorrect. Adding a catalyst increases both the rate of the forward and backward reactant by the same extent, causing no change to the value of Kp. B Incorrect. K p is only affected by changes in temperature. Hence, changing the pressure will not affect the value of K p for all gas phase reactions. C D Correct. Q13(C) Since equilibrium partial pressure of NH 3 = 92 atm and total pressure is 100 atm, the sum of partial pressure of N2 and H2 = 100 – 92 = 8 atm. The mole ratio and hence the partial pressure ratio of N2 : H2 remains as 1 : 3 as the initial ratio and stoichiometric ratio are the same. Hence, equilibrium partial pressure of N 2 = ¼ × 8 = 2 atm while equilibrium partial pressure of H 2 = ¾ × 8 = 6 atm. 𝐾𝐾𝑝𝑝 = 922 (2)(63) = 19.6 Q14(C) Total amt of Br2 = (30.0/1000)(0.500) = 0.0150 mol Amt of Br– = [2.98 / (39.1 + 79.9)] = 0.0250 mol Reduction: Br2 + 2e– → 2Br– (from Data Booklet) Amt of e– gained = amt of Br– = 0.0250 mol Amt of Br2 reduced = 0.0250/2 = 0.0125 mol Let k be the stoichiometric coefficient of e – Oxidation: Br2 → 2BrOx– + ke– (unbalanced) Amt of Br2 oxidised = 0.015 – 0.0125 = 0.00250 mol Amt of BrOx– = amt of Br2 oxidised × 2 = 0.00250 × 2 = 0.00500 mol Since in a redox reaction, the amount of e– gained = amount of e– lost, mole ratio of BrOx– : e– = 0.00500 0.0250 = 2 𝑘𝑘 , k = 10 Since 1 mol of Br2 lost 10 mol of e–, 1 mol of Br atom lost 5 mol of e– and the oxidation state of Br increases from 0 in Br2 to + 5 in BrOx–. Thus, x = 3 as the oxidation state of Br in BrO3– = +5. Q15(B) Mixing CH3COONa and AgNO3 gave a white ppt of CH3COOAg. ⇒ CH3COOAg is insoluble in water (1 is correct) Mixing white ppt of CH 3COOAg and KBr causes CH3COOAg to dissolve to form a cream ppt of AgBr. ⇒ Initially, CH3COOAg ⇌ CH3COO– + Ag+. When Br– was added, the IP of AgBr exceeded its Ksp despite the low [Ag+], implying that the Ksp of AgBr is very low and easily exceeded i.e. AgBr is less soluble than CH 3COOAg. (2 is correct) No further change upon addition of CH3COONa to AgBr ⇒ no reaction took place i.e. 3 is incorrect. Q16(C) hybridisation no. of p orbitals used for hybridisation no. of s orbitals used for hybridisation sp 1 1 sp2 2 1 sp3 3 1 The 3 carbons in propane, CH 3CH2CH3, are sp 3 hybridised. Hence, 3 x 3 = 9 p orbitals were used for hybridization. The 6 carbons in benzene, C 6H6, are sp2 hybridised. Hence, 6 x 2 = 12 p orbitals were used for hybridization. The 2 carbons in ethene, H– C≡C–H, are sp hybridised. Hence 2 x 1 = 2 p orbitals were used for hybridization. In total, 9 + 12 + 2 = 23 p orbitals were used for hybridization.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q17(C) The lack of reactivity of chloroethene to nucleophiles is similar to the case of chlorobenzene where the overlap of the p- orbital on C l with the π electron cloud of C=C or benzene allows the lone pair on C l to be delocalized into the C=C or benzene, giving rise to partial double bond character in the respective C–Cl bonds. Q18(C) Non-cyclic constitutional isomers of C5H10 Q19(A) Since X proceeds to give a 1,2 -disubstituted product, X should contain a 2- directing group i.e. X should be chlorobenzene (the nitro group is 3-directing). Since Y proceeds to give a 1,3 -disubstituted product, Y should contain a 3- directing group. The methyl group that is present is 2,4- directing and can be oxidised to give –COOH which is 3-directing. Hence Y is benzoic acid. Q20(C) A Incorrect. Benzene undergoes electrophilic substitution with the given reagents and conditions. B Incorrect. Hydrogenation of benzene requires high pressure, high temperature and the presence of a catalyst. C Correct. A carbocation acts as an electrophile in Friedel-Craft alkylation. D Incorrect. In benzene, single and double carbon-carbon bonds do not exist . Instead, the delocalisation of the 6 π electrons give rise to bond lengths t