NJC 2023 A Level Chem Solution

H2 Chemistry TYS 2023 Paper 1 1 H2 Chemistry TYS 2023 suggested answers Paper 1 Answer Key 1 B 11 D 21 A 2 C 12 D 22 B 3 C 13 C 23 A 4 D 14 C 24 B 5 B 15 B 25 D 6 D 16 C 26 B 7 A 17 C 27 D 8 C 18 C 28 A 9 A 19 A 29 B 10 A 20 C 30 C 1 Ans: B Nucleon number Proton Neutron Electron Nd2+ 145 60 85 58 Pm3+ 145 61 84 58 2 Ans: C Ionic compound has a large electronegativity difference between the elements while covalent compound has small difference in electronegativity. A X and Cl Electronegativity difference = 1.8 Y and O Electronegativity difference = 2.6 B X and O Electronegativity difference = 2.3 Y and O Electronegativity difference = 2.6 C Y and Cl Electronegativity difference = 2.1 Z and O Electronegativity difference = 1.7 D Z and Cl Electronegativity difference = 1.2 X and Cl Electronegativity difference = 1.8 3 Ans: C Both CH3Cl and CH3Br are covalent compound with simple molecular structure. Both are polar molecules with instantaneous dipole-induced dipole (id-id) and permanent dipole-permanent dipole (pd-pd) between molecules. As CH3Br has a larger electron cloud size, its electron cloud is more easily distorted, and dipoles are more easily induced. The strength of id-id is stronger for CH3Br, leading to its larger boiling point.

H2 Chemistry TYS 2023 Paper 1 2 4 Ans: D NH4+ 4 single bonds, 8 shared e− CO2 2 double bonds, 8 shared e− HCHO 1 double bond + 2 single bonds 8 shared e− 5 Ans: B 1,1-difluroethene cis-1,2-difluroethene trans-1,2-difluroethene tetrafluroethene 6 Ans: D Period 3 elements with only one orbital in the outermost shell which contains just one electron: 1) Na : [Ne] 3s1 2) Al : [Ne] 3s2 3p1 3) Cl : [Ne] 3s2 3p5 (one of the 3p orbital contains only 1 e−) Period 3 elements with only one orbital in the outermost shell which contains a pair of electrons: 1) Mg : [Ne] 3s2 2) Al : [Ne] 3s2 3p1 3) Si : [Ne] 3s2 3p2 (e− occupies the p orbital singly first) 4) P : [Ne] 3s2 3p3 (e− occupies the p orbital singly first)

H2 Chemistry TYS 2023 Paper 1 3 7 Ans: A Statement 1 is correct. Going down Group 2, Eꝋ(M2+/M) becomes more negative. This shows that Ba(s) undergoes oxidation more readily than Mg(s). Hence Ba(s) is a stronger reducing agent than Mg(s). Statement 2 is correct. Down the group, cationic radius increases and charge density of M2+ decreases. This leads to a decrease in polarising power of the cations. The electron cloud of the CO32− anion is being distorted to a smaller extent down the group, decreasing the extent of weakening of C–O bonds within the CO32− anion. More energy is required to break the C–O bonds within the CO32− anion. Therefore, thermal stability of the Group 2 carbonates increases down the group. Statement 3 is incorrect. Down the group, cationic radius increases and charge density of M2+ decreases. 8 Ans: C The least complicated way for this question is to work out the Ar of the Si samples in all 4 options. 28Si 29Si 30Si Ar A 92.23% 2.23% 5.54% 28.13 B 92.23% 3.89% 3.89% 28.12 C 92.23% 5.54% 2.23% 28.10 D 92.23% 7.77% 0% 28.08 9 Ans: A H2SO4 + 2NaOH ⎯→ Na2SO4 + 2H2O Amount of H2SO4 used = 50.0 1000 × 2.00 = 0.1 mol Amount of NaOH used = 100.0 1000 × 1.00 = 0.1 mol NaOH is the limiting reagent, amount of H2O produced = 0.1 mol Hn = − mcT nH2O = − 150 × 4.18 × (29.0 - 20.0) 0.1 = − 56430 J mol−1  −56.4 kJ mol−1

H2 Chemistry TYS 2023 Paper 1 4 10 Ans: A |L.E.| ∝ | q+× q- r++ r- | CaCl contains Ca+ and Cl−, hence the smallest magnitude of L.E Ionic radius of Ca2+ is smaller than Mg2+, hence CaCl2 has a smaller magnitude of L.E. than MgCl2. 11 Ans: D At low [substrate], not all of the active sites are occupied. The rate of reaction increases proportionally with substrate concentration. It is first order with respect to the substrate. At high [substrate], all the active sites of the enzyme are saturated. Any increase in the substrate concentration cannot increase the rate of reaction. The rate of reaction no longer depends on the substrate concentration and is zero-order with respect to the substrate. 12 Ans: D Statements A, B and C are incorrect. Kp value is only affected by change in temperature. Statement D is correct. As temperature increases, by Le Chatelier’s Principle, the equilibrium system would partially remove the excess heat by favouring the reverse endothermic reaction. At the new equilibrium, the ratio of product reactant decreases, hence the value of Kp decreases. 13 Ans: C N2(g) + 3H2(g) 2NH3(g) I /mol x 3x 0 C / mol −y −3y +2y E / mol x−y 3x−3y 2y PN2 + PH2 + PNH3 = 100 atm PN2 + PH2 = 100 − 92 = 8 atm As equilibrium amount of N2 : H2 = 1 : 3, and pV = nRT PN2 : PH2 = 1 : 3 PN2 = 2 atm, PH2 = 6 atm, PNH3 = 92 atm Kp = (PNH3)2 PN2 × (PH2)3 = (92)2 2 × (6)3 = 19.6 atm−2

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H2 Chemistry TYS 2023 Paper 1 6 14 Ans: C Br2 + KOH ⎯→ KBr + KBrOx + H2O (not balanced) Amount of Br2 used = 30.0 1000 × 0.500 = 0.015 mol Amount of KBr formed = 2.98 39.1 + 79.9 = 0.02504 mol Method 1: Total amount of Br in reactant = 0.015 × 2 = 0.030 mol (2 Br in a Br2 molecule) Amount of Br in KBrOx = 0.030 − 0.02504 = 0.00496 mol Mol ratio Br2 : KBr : KBrOx = 3 : 5 : 1 Balancing the equation, 3Br2 + 6KOH ⎯→ 5KBr + KBrOx + 3H2O x = 3 Method 2: [R] Br2 + 2e− ⎯→ 2Br− [O] Br2 ⎯→ 2BrOx− + ne− (not balanced) From the reduction half equation, 0.0125 mol of Br2 gains 0.0250 mol of e− to give 0.0250 mol of Br−. Amount of Br2 that undergoes oxidation = 0.0150 − 0.0125 = 0.00250 mol During oxidation, 0.0025 mol of Br2 lose 0.0250 mol of e− Mol ratio Br2 : e− = 1 : 10 During oxidation, oxidation state of EACH Br increases by 5 units from 0 (in Br2) to +5 (in BrOx−) Hence x = 3. 15 Ans: B Observations from first student: Reaction 1.1 : AgNO3 + CH3COO−Na+ ⎯→ CH3COO−Ag+ (white ppt) + NaNO3 Reaction 1.2 :CH3COO−Ag+ (white ppt) + KBr ⎯→ AgBr (cream ppt) + CH3COO−K+ Observations from first student: Reaction 2 :AgNO3 + KBr ⎯→ AgBr (cream ppt) + KNO3 AgBr + CH3COO−Na+ ⎯→ no reaction Statement 1 is correct. We observed that white ppt of CH3COO−Ag+ is formed in reaction 1.1 Statement 2 is correct. We observed that cream ppt of AgBr is formed in reaction 1.2 Statement 3 is wrong. Br− is not oxidised to Br2 in the above reactions. 16 Ans: C Propane : 3 sp3 hybridized C atoms (9 × 2p orbitals hybridized with 2s) Benzene : 6 sp2 hybridized C atoms (12 × 2p orbitals hybridized with 2s) Ethyne : 2 sp hybridized C atoms (2 × 2p orbitals hybridized with 2s)

H2 Chemistry TYS 2023 Paper 1 7 Total : 23 × 2p orbitals hybridized with 2s

H2 Chemistry TYS 2023 Paper 1 8 17 Ans: C Note: Halogen directly bonded to C=C does not undergo Nucleophilic Substitution due to the same reasons as halogenoarene. 1. The p-orbital of the halogen atom overlaps with the  electron system of C=C. The lone pair electrons on halogen is delocalized into C=C. As a result, there is some double bond character in the C–X bond. The C–X bond is shorter and stronger than that in halogenoalkane and thus is very difficult to break. 2. The C in +C–X bond is less accessible to the nucleophilic reagents that attack halogenoalkanes since the electron rich C=C repels the approaching nucleophile. 18 Ans: C CCH H C H CC H H H H H H H CCH H C H CC H H H H H H H pent-1-ene pent-2-ene CCH H C CH3 C H H H H H 2-methylbut-1-ene 3-methylbut-1-ene CCH H C H C H CH3 H H H CCH H C CH3 C H H H H H 2-methylbut-2-ene 19 Ans: A Refer to the Data Booklet for directing effect of substituents towards subsequent electrophilic sub. −Cl is 2,4-directing −NO2 is 3-directing −CH3 is 2,4-directing −COOH is 3-directing Cl Cl NO2 X CH3 Y COOH COOH H3C

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