RVHS_2023_J2_H2_CM_Prelim_P1_-_P3_Suggested_Solutions

RIVER VALLEY HIGH SCHOOL 2023 JC 2 H2 Chemistry 9729 Preliminary Examination (Paper 1 - 3) Suggested Solutions Paper 1 Worked solutions 1 D Statement A: Incorrect. This statement would explain why the 3rd IE of aluminium is higher than that of magnesium. Statement B: Incorrect. Ionisation energy is dependent on nuclear charge and shielding effect, not octet configuration. Statement C: While it is a factual statement, it does not explain the significant difference in 3rd IE between the two species. Statement D: Al2+: 1s2 2s2 2p6 3s1 vs Mg2+: 1s2 2s2 2p6 Electron to be removed from Mg2+ is from n = 2 principal quantum shell and is closer and more strongly attracted to the nucleus due to a significantly smaller shielding effect, while that to be removed from Al2+ is from n = 3 principal quantum shell. 2 C charge mass ratio of proton (1H+) = 1 1 = 1 charge mass ratio of particle to be deflected = 1 × 8.012.0 = 2 3 charge mass ratios of: 12C3 : 3 12 = 1 4 (incorrect) 10B2 : 2 10 = 1 5 (incorrect) 6Li4: 4 6 = 2 3 (correct) 3He : 1 3 (incorrect) 3 D Tritium isotope: 1p, 1e, 2n Helium isotope: 2p, 2e, 1n Option 1 is incorrect. Option 2 is incorrect. Option 3 is incorrect (charged sub-atomic particles refer to protons and electrons). 4 B

A B C D 5 B A is correct. Ice cannot conduct electricity. B is incorrect. Water has a simple covalent structure. When it is in the form of ice, hydrogen bonds are formed between the molecules of water. However, it is not a giant covalent structure. C is correct. Each O atom in water molecule has 4 bond pairs. So bond angle is 109.5° D is correct. Density of ice is lower due to its open structure. 6 B = Let initial volume be 3V2 => final volume is V2 = (Note: T needs to be converted to K) p2 = 34.1 atm 7 C When temperature increases from 25 oC to 62 oC, Kw increases, signifiying that there is more dissociation of water (to form H+ and OH−). Since H2O  H+  OH−, n(H+) = n(OH−)  value of pH = value of pOH At 62 oC, pKw = −lg(1.00  1013) = 13  pH = 6.5 and pOH = 6.5 (i.e., [H+] = [OH−] = 3.162  107 mol dm3) 8 A S < 0 since amount of gaseous molecules decrease from 1.5 to 1. H = 396  (297) =  99 kJ mol1 < 0 9 B Statement A: Incorrect. Energy level of products is higher than that of reactants. Statement B: Correct. Hsoln = LE + Hhyd Both LE and Hhyd are negative values, so LE > 0, Hhyd <0. For Hsoln to be > 0, magnitude of LE has to be > magnitude of Hhyd. Statement C: Reaction is endothermic, heat is taken in. Statement D: Incorrect definition of Hsoln 10 A Statement 1: Correct. H = 12/18  44 = 29.3 kJ mol1 Statement 2: Incorrect. G = 0 during phase change. Statement 3: Correct. G = HTS, S = (29.3)/(373) = 78.6 J K1 11 B CH4 + 2O2  CO2 + 2H2O HCHO + O2  CO2 + H2O COV 2 = 10 cm3 = Vorg total OV 2 left = 18 – 10 = 8 cm3 O