2025 Prelim P1 Teaching Solutions_H2 Chem_HCI

2025 HCI C2 H2 Chemistry Prelim Exam Teaching Solutions/Paper 1 1 HWA CHONG INSTITUTION 2025 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION MARK SCHEME / SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 D D D C A B A A B A 11 12 13 14 15 16 17 18 19 20 B D D C B C B A C A 21 22 23 24 25 26 27 28 29 30 A A B D D C A B C B Comments 1 D The largest difference in consecutive ionisation energies occurs between the 5 th and 6th ionisation energies, implying that the 6th electron is removed from an inner quantum shell and element X has 5 valence electrons. Since element X belongs to Group 15, it is likely to form a chloride of the formula XCl5. 2 D Isotopes have the same number of protons but different number of neutrons. The number of protons can be determined from the number of electrons and charge. Once the number of protons are determined, the number of neutrons can be found from the nucleon number. E.g., in A, H has 10 electrons and a charge of −2, so H has 8 protons. Thus, H has 10 neutrons. G H no. of protons no. of neutrons no. of protons no. of neutrons A 8 10 8 10 B 18 19 16 18 C 16 20 18 20 D 20 22 20 20 In D, G and H have the same number of protons (20) but different number of neutrons (22 and 20). Note that A is incorrect as G and H have the same number of protons and neutrons. 3 D The shape of a species depends on the number of lone pairs and bond pairs of electrons around the central atom. Refer to Topic 2 Chemical Bonding Sec. 6.1. A AlCl3 PCl3 3 b.p. and 0 l.p. – trigonal planar 3 b.p. and 1 l.p. – trigonal pyramidal B BF3 NH3 3 b.p. and 0 l.p. – trigonal planar 3 b.p. and 1 l.p. – trigonal pyramidal C SF4 XeF4 4 b.p. and 1 l.p. – see saw (expanded octet) 4 b.p. and 2 l.p. – square pyramidal (expanded octet) D PH4+ BF4ˉ 4 b.p. and 0 l.p. – tetrahedral 4 b.p. and 0 l.p. – tetrahedral

2025 HCI C2 H2 Chemistry Prelim Exam Teaching Solutions/Paper 1 2 4 C Using nt = ns + nb Since T is constant, PtVt = PsVs + PbVb After connection: Vt = Vs + Vb Pt = (50)(10) + (120)(30) (10+30) = 102.5 kPa = 103 kPa 5 A This question tests the definition of relative masses found in Topic 1 & 4. Option B is wrong as the relative atomic masses must take into account all the isotopes of the element, hence the term ‘average mass’ must be used. Option C is wrong as the reference should be 12C & not 1H. Option D is wrong as the reference should be to 1/12 the mass of one atom of 12C. 6 B Always check if the sum of the abundances of the isotopes add up to 100. In the case, they add up to 101.32. Ar (Ge) = ( 19.01 101.32 ×70) + ( 24.96 101.32 ×72) + ( 7.76 101.32 ×73) + ( 41.0 101.32 ×74) + ( 8.59 101.32 ×76) = 72.8 7 A Use the equation method to obtain the answer: • Flip the first equation to get the MgO on the right side like in the combustion equation. Remember to flip the sign on its H value. • Half the third equation to ensure that th