NYJC 2025 H2 Chem prelim 9729 Answers (updated for 2026 syallbus)

Nanyang Junior College 2025 J2 H2 Chemistry Prelim Exam Answers 1 Paper 1 Answer Key Paper 2 Answers 1 (a) (i) [1] for all 3 plots and ensure that the plot at proton number 20 is lower than proton number 12. (ii) • The number of protons increase and nuclear charge increase. • Successive electrons are added to the same valence shell, shielding effect remains approximately constant. • Effective nuclear charge increases and electrostatic attraction between the nucleus and valence electron increases. • More energy is required to remove the valence electron (iii) Proton number 15: 1s2 2s2 2p6 3s2 3px1 3py1 3pz1 Proton number 16: 1s2 2s2 2p6 3s2 3px2 3py1 3pz1 [1] for both electronic configurations. Reject students who write 1s2 2s2 2p6 3s2 3px4 for proton number 16. Element with proton number 16, the valence electron to be removed is from the paired electrons in 3px orbital. Hence, it experiences inter-electronic repulsion arising from the two electrons occupying the same 3px orbital and requires less energy to remove. [1] 1 B 11 B 21 B 2 B 12 A 22 D 3 A 13 D 23 C 4 D 14 A 24 C 5 B 15 C 25 D 6 C 16 A 26 C 7 D 17 B 27 C 8 B 18 A 28 D 9 A 19 C 29 D 10 B 20 B 30 A

Nanyang Junior College 2025 J2 H2 Chemistry Prelim Exam Answers 2 (b) (i) [1] Copper has a giant metallic structure with strong metallic bonding (or electrostatic attraction) between copper ion and sea of delocalised electrons. The presence of sea of delocalised electrons acts as mobile charge carriers, hence making copper a good conductor of electricity. [1] (ii) • Both Na2O and MgO have giant ionic structure held together by strong ionic bonding (or electrostatic attraction) between positively charged metal cations and negatively charged oxide anions. • |LE|  |q+q−| r+ + r • Both Na2O and MgO have the same oxide anion, but Mg 2+ is doubly charged while Na+ is singly charged. • The ionic radius of Na+ > Mg2+. The magnitude of LE is larger for MgO than Na2O. • Ionic bonding of MgO is stronger than Na 2O. More energy is required to break the stronger ionic bonds between Mg 2+ and O2– in MgO than ionic bonds between Na + and O2– in Na2O. • Melting point of MgO is higher than Na 2O and hence able to withstand higher temperature in case of fire. 2 (a) (i) The order of reaction with respect to a given reacta nt is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. [1] (ii) When [NaOH] decreases, gradient remains constant, hence rate of reaction is constant. Rate of reaction is independent of [NaOH], hence the reaction is zero order wrt [NaOH]. [1] When [CH3CHBrCH2CH3] = 0.25 mol dm−3 initial rate = 0.0048 / 30 = 1.60 x 10−4 mol dm−3 min−1 When [CH3CHBrCH2CH3] = 0.50 mol dm−3 initial rate = 0.0095 / 30 = 3.16 x 10−4 mol dm−3 min−1 When [CH3CHBrCH2CH3] doubles (0.50/0.25), the gradient doubles, showing that the rate of reaction doubles from 1.60 x 10−4 to 3.16 x 10−4 mol dm−3 min−1. [1] Hence the reaction is first order wrt [CH3CHBrCH2CH3]. (iii) rate = k [CH3CHBrCH2CH3] 1.60 x 10−4 = k (0.25) k = 6.40 x 10−4 min−1 or 1.07 x 10−5 s−1 [1]

Nanyang Junior College 2025 J2 H2 Chemistry Prelim Exam Answers 3 (iv) Br Nucleophilic Substitution (SN1) + − + slow + Br- + OH- OH [1] [1] (v) The OH− nucleophile can attack the trigonal planar carbocation centre from either the top or bottom of the plane with equal probability, producing equal amounts of the pair of enantiomers (a racemic mixture). [1] The product mixture is optically inactive (i.e. does not rotate plane -polarised light) as each enantiomer rotate plane -polarised light in the opposite direction by the same magnitude hence the rotating powers of the enantiomers cancel out. [1] (b) (i) Stereoisomerism is defined as compounds having the same molecular formula but different spatial formulae / different spatial arrangement of atoms. [1] (ii) CH2CH3 CH3 H Br CH2CH3 HBr CH3 mirror plane [1] (iii) [1] (iv) 3 (a) [1]

Nanyang Junior College 2025 J2 H2 Chemistry Prelim Exam Answers 4 (b) Cathode: Co2+(aq) + 2e → Co(s) [1] Anode: H2O(l) → O2(g) + 4H+(aq) + 4e [1] (c) n(Co) coated = (8.90 x 0.2 x 10–1 x 10) / 58.9 = 0.0302 mol [1] n(O2) = (0.0302) / 2 = 0.0151 mol Volume of gas = 0.0151 x 24 = 0.363 dm3 (or 363 cm3) [1] (d) Co2+(aq) + 2e → Co(s) Q = It = nzF 0.5  t = 0.0302  2  96500 t = 11657.2s = 3.24 h [1] unit must be in h (e) Co2+ + 2e ∏ Co −0.28 V Pb2+ + 2e ∏ Pb −0.13 V O2 + 4H+ + 4e ∏ 2H2O +1.23 V The mass of the lead electrode will decrease. Eѳ(Pb2+/Pb) is more negative than Eѳ(O2/H2O), Pb will be preferentially oxidised to form Pb2+ in aqueous solution. [1] Dull gray solid Pb will be deposited at the leaf. Eѳ(Pb2+/Pb) is less negative than Eѳ(Co2+/Co), Pb2+ will be preferentially reduced. [1] 4 (a) (i) Acidity depends on the stability of the conjugate base. In phenoxide, the p -orbital of oxygen atom overlaps with the π electron cloud of the benzene ring. Lone pair of electrons of the oxygen atom delocalised into the benzene ring. The negative charge on the oxygen atom is dispersed into the benzene ring and stabilising the phenoxide. [1] In carboxylate ion, the p -orbital of oxygen atom overlaps with the π electron cloud of the C=O bond. Lone pair of electrons of the oxygen atom delocalised into the C=O bond. The negative charge on the oxygen atom is dispersed over the carbon atom and two e lectronegative oxygen atoms and stabilising the carboxylate ion. The charge dispersion is more effective than that in phenoxide due to the two electronegative oxygen atoms and hence carboxylate ion is more stable than phenoxide ion. [1] Hence, carboxylic acid is a stronger acid than phenol. (ii) COO - OH [1] Reject -O-C4H6-COOH as the COOH should deprotonate first since it has a lower pKa. Reject -O-C4H6-COO- as it differs from 2 -hydroxybenzoic acid by 2 H + hence not a conjugate base of the acid. The conjugate base of 2 -hydroxybenzoic acid can form intramolecular hydrogen bonding that helps to further stabilize the conjugate base. [1]

Nanyang Junior College 2025 J2 H2 Chemistry Prelim Exam Answers 5 (iii) [4-hydroxybenzoic acid] = 2.76 138.0 ÷ 100 1000 = 0.2000 mol dm-3 [1] [H3O+] = √10-4.58 × 0.2000 = 2.293 × 10-3 mol dm-3 pH = -lg (2.293 × 10-3) = 2.639 = 2.64 [1] (iv) n4-hydroxybenzoic acid = 2.76 138 = 0.02000 mol nNaOH = 0.35 40.0 = 8.750 × 10-3 mol [1] Since the neutralisation between 4 -hydroxybenzoic acid and NaOH is 1:1 mole ratio, NaOH is a limiting reagent. n4-hydroxybenzoic acid left = 0.02000 – (8.750 × 10-3) = 0.01125 mol n4-hydroxybenzoate formed = 8.750 × 10-3 mol [1] Since there is a mixture of 4 -hydroxybenzoic acid and 4-hydroxybenzoate, solution Q is an acidic buffer. pH = pKa + lg [4-hydroxybenzoate] [4-hydroxybenzoic acid] = 4.58 + lg 8.750 × 10-3 100 1000 ⁄ 0.01125 100 1000 ⁄ = 4.470 = 4.47 [1] (v) COO - OH COO -- O 2-hydroxybenzoic acid at pH 12 4-hydroxybenzoic acid at pH 12 position on Fig. 4.1: 3 position on Fig. 4.1: 4 [1] for each correct structure and position (vi) Any pH between 4.58 and 9.51. [1] (b) (i) O Cl [1] (ii) Electrophilic Addition slow COOHO O COOHO O Br H Br + − COOHO O + COOHO O + + Br− Br− [1] for step 1 [1] for step 2

Nanyang Junior College 2025 J2 H2 Chemistry Prelim Exam Answers 6 (iii) Formation of T involves a less stable carbocation that is adjacent to an electron withdrawing ester group, which destabilises the carbocation. Hence, a smaller amount of that carbocation intermediate is formed, and T is a minor product. [1] COOHO O + (T forms a less stable carbocation than S) (c) (i) Fe(OH)3 [1] (ii) A complex is a species that contains a central me