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