2024 Prelims NYJC H2 Chem P1 (Ans)

J2 Prelims 2024 H2 Chemistry Paper 1 Answers and Comments Page 1 of 8 2024 NYJC H2 Chem P1 (Ans) Nanyang JC J2 Preliminary Examinations 2024 H2 Chemistry 9729/01 Paper 1 MCQ Answers and Comments Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 D 6 B 11 B 16 D 21 D 26 C 2 B 7 D 12 D 17 C 22 C 27 D 3 D 8 C 13 C 18 C 23 A 28 C 4 A 9 A 14 D 19 A 24 A 29 A 5 D 10 B 15 A 20 B 25 B 30 B 1 D mod N22P1Q6 ∡ of deflection, θ = k × charge mass (NB: Elements V and W are not their usual symbol in the Periodic Table.) Both Ca and V form cations with +2 charge. Hence, since θ(Ca2+) > θ(V2+),  Ar(Ca) < Ar(V), V must be below Ca in Group 2. Therefore element V is Sr and W is either Rb or Y. From Ca2+, we can determine a value for k. k = 240 (or 240.6 using Ar) Using k, we can calculate the charge mass ratio for W, charge mass (𝐖) = 0.0337 Lastly, compare the charge mass ratio calculated against those of Rb+ and Y3+, charge mass (Rb+) = 0.0117 and charge mass (Y3+) = 0.0337, we can deduce the identity of element W as Yttrium (proton number 39). 2 B mod NY20 Promo P1Q3 Deduce the group number of elements G and H by looking at the largest jump in IEs. G: largest jump at 4th IE, 4th electron is removed from inner shell, 3 valence electrons, G is Group 3. H: largest jump at 7th IE, 7th electron is removed from inner shell, 6 valence electrons, H is Group 16. From the Data Booklet, we can conclude • G is likely to be Al • H is an element below S in Group 16. (1st IE of H < 1st IE of S. Down the group, 1st IE decrease.) H: [noble gas] ns2 npx2 npy1 npz1 Statement 1 is correct. There is only one p orbital with paired electrons in H. G is Period 3, H is at least Period 4. Statement 2 is correct. H has a larger nuclear charge than G. The factors that affect IE are i) nuclear charge (IE increase with increasing nuclear charge) ii) shielding effect (IE decrease with increasing shielding effect) iii) distance from nucleus (IE decrease with increasing distance from nucleus) Hence the only factor that can explain why the 3rd IE of H is larger than 3rd IE of G is the nuclear charge. H is below S in Group 16 as the 1st IE of H (940) is smaller than the 1st IE of S (1000). However, as we do not have the IEs of Group 16 elements below S, we cannot positively identify which of Se, Te or Po is H. Hence statement 3 cannot be deduced from the data. (FYI, H is Se) (For H, we can eliminate Group 6 as 1st IE of Cr is < 940 kJ mol −1. Elements below Cr in Group 6 will also have 1st IE < 940.) 3 D To deduce bond angles fast in MCQs, use the following shortcut. A: 7 B: 7 C: 7 D: 9

J2 Prelims 2024 H2 Chemistry Paper 1 Answers and Comments Page 2 of 8 2024 NYJC H2 Chem P1 (Ans) • The number of bond pairs around the central atom is given by the number of non-central atoms in the formula • To determine the number of lone pairs, give the charge to the central atom. Hence if charge is −, give one electron to central atom. If charge is +, take one electron away from central atom. Next, subtract the valence electrons of the central atom that are used to bond to each non-central O (need 2 electrons each) or X ( need 1 electron each). Lastly pair the remaining electrons around the central atom to obtain the number of lone pairs. A) NO3− BeCl2 BP−LP 3 − 0 2 − 0 bond angle 120° < 180° B) BrO3− PCl4+ BP−LP 3 – 1 4 − 0 bond angle 107° < 109° C) AsF6− XeF4 BP−LP 6 – 0 4 – 2 bond angle 90° = 90° D) SO2 SO42− BP−LP 2 – 1 4 − 0 bond angle 118° > 109° 4 A mod N21P1Q5 Shapes of hybrid orbitals sp sp2 sp3 50% s character. more spherical 75% p character. more elongated shortest bondlength longest bondlength strongest bond weakest bond 5 D In explaining solubility, we compare • Energy required to break bonds between solute particles and bonds between solvent particles • Energy evolved when bonds are formed between solute and solvent particles The strongest bond required to be broken is the hydrogen -bonding between water molecules. The most extensive/dominant bonds formed between water and hexanol molecules (due to the long alkyl chain of hexanol) is the instantaneous dipole – induced dipole interactions between hexanol alkyl chains and water molecules. Therefore, water and hexanol do not mixed because the energy evolved when id -id interactions formed between water and hexanol is not sufficient to compensate for the energy required to break hydrogen bonding in water molecules. 6 B A. Na2O(s) dissolve in water to form NaOH(aq). Since dilute NaOH(aq) is not saturated, Na2O can still dissolve. B. MgO is an ionic oxide that is basic. It will not react with alkaline NaOH(aq). MgO is also insoluble, giving only a weakly alkaline solution of pH 9 with water. C. Al2O3 is an ionic oxide with high covalent character. It is amphoteric and will dissolve in NaOH(aq) to form a soluble complex NaAl(OH)4. D. P4O10 is a covalent oxide that is acidic . It reacts with NaOH(aq) to form a soluble salt Na3PO4(aq). 7 D Down Group 2 , cationic radius increase while cationic charge is constant. Charge density increases, polarising power becomes stronger. CO32− electron cloud is less polarised . C−O bond in CO 32− is weakened to a smaller extent , more energy is required to break the C−O bond. Thermal decomposition temperature increase. 8 C Cl2(aq) + 2I−(aq) → l2(aq) + 2Cl−(aq)

J2 Prelims 2024 H2 Chemistry Paper 1 Answers and Comments Page 3 of 8 2024 NYJC H2 Chem P1 (Ans) Cl2, stronger oxidising agent, will oxidise I− to iodine. l2(aq) ∏ l2(org) (2) brown purple A. False. I2 is purple in non -polar organic hexane solvent. B. False. I2 is a non-polar molecule with very low solubility in water. The equilibrium(2) lies almost completely to the product side. C. Correct. Down Group 17, incoming electron added to valence shell further away from nucleus. Hence attraction of Cl nucleus to incoming electron is stronger. D. False. I− is a stronger reducing agent since its valence electrons are less readily attracted to the nucleus. 9 A nReO4− : ne− : nZn amt(mol) 0.008 0.0319 OA:RA ratio 1 4 Zn : e− 2 1 LCM Zn 1 8 4 Change in OS = Final OS – Initial OS −8 = Final OS − (+7) Final OS = −1 10 C A is wrong. 2 mol of water is formed in the equation. B is correct. Both Na(s) and O2(g) are in their elemental states. C is wrong. ∆Hhyd is for gas ion to aq ion. D is wrong. Equation given is −LE. LE is defined for ionic solid formed from constituent ions. 11 B mod N21P1Q11 ∆S < 0 as t here are less number of gas particles in the product. Less ways of arranging gas particles in system. Entropy of product decrease. Using ∆G = ∆H −T∆S, as T↓, |T ∆S| < | ∆H| such that ∆G becomes more negative. Hence reaction is more spontaneous when temperature decrease. 12 D Use the equation, [Rn]t = [Rn]0 x (½)n [Rn]0 = initial conc, [Rn]t = conc at any point in time, n = number of half-lives. 0.4 = 15 x (½)n n = 5.228 time taken = 5.228 x 3.82 = 19.97 days 13 C 1 is wrong. Use of a catalyst does not affect relative rates of forward and backward reactions. Both rates increase to the same extent. No change in position of equilibrium. 2 is correct. A medium high temperature of 500 °C (774 K) is used. Low temperatures 227 °C (500 K), although favouring the forward exothermic reaction to produce heat, results in a low rate of reaction, hence are not used. 3 is correct. A high pressure of 2.03 x 107 Pa (200 atm) is used to shift the eqm position to the right by reducing total number of gas particles. 14 D Analysis of the graph shows: • VNaOH required to reached second equivalence volume is larg