RVHS Prelim_H1_Chemistry_P1_Soln

River Valley High School 8873/01/PRELIMS/25 2025 JC2 Preliminary Examination [Turn over RIVER VALLEY HIGH SCHOOL JC2 H1 CHEMISTRY 8873 PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 1 worked solutions 1 D 6 C 11 D 16 A 21 A 26 C 2 B 7 B 12 C 17 A 22 B 27 B 3 D 8 D 13 A 18 B 23 A 28 B 4 C 9 D 14 B 19 A 24 B 29 B 5 D 10 A 15 C 20 D 25 C 30 C 1 D Calculate the charge/mass ratio to each option and the one with the largest ratio will be deflected by the largest angle. A: q/m = 3/11 = 0.273 B: q/m = 2/9 = 0.222 C: q/m = 0/1 = 0 (i.e., won’t be deflected) D: q/m = 2/4 = 0.5 (Answer) 2 B Identify the elements in each option and choose the element that will have either ns2 or ns2 np5 after removing the first electron. Element Electronic configuration Species after removal of 1st electron Electronic configuration of new species A: Mg 1s22s22p63s2 Mg+ 1s22s22p63s1 B: Al 1s22s22p63s23p1 Al+ 1s22s22p63s2 C: Si 1s22s22p63s23p2 Si+ 1s22s22p63s23p1 D: P 1s22s22p63s23p3 P+ 1s22s22p63s23p2 For Al, it will have a higher 2nd IE than Mg (to its left) and Si (to its right). For Al+ vs Mg+, Al+ has a greater effective nuclear charge and smaller ionic radius, hence the IE to remove the next electron from Al+ is higher. For Al+ vs Si+, the 3p electron in Si+ is higher in energy than the 3s electron in Al+, so the IE to remove the next electron from Si+ is lower, thus the IE to remove the next electron from Al+ is higher.

2 River Valley High School 8873/01/PRELIMS/25 2025 Preliminary Examination 3 D In 1 cm3 of air, there is 8.00 × 10−9 g of ozone. Mr of ozone is 16.0 × 3 = 48.0 Amount of ozone = 8.00 × 10−9 / 48 No. of ozone molecules = (8.00 × 10−9 / 48) × L 4 C CxHy + (x+ 𝑦 4) O2 → xCO2 + 𝑦 2 H2O 20 cm³ 60 cm³ KOH absorbs CO2 ⇒ VCO₂=60 cm³. Thus, x=60/20=3 VCxHy + Vinitial O2 = 20 + 120cm³ = 140 cm³ After combustion, Vexcess O2 + VCO₂ = 140 – 50 = 90 cm³ Vexcess O2 = 90 – 60 = 30 cm³ VO2 consumed during combustion = 120 – 30 = 90 cm³ x+ 𝑦 4 = 90/ 20 = 4.5 y = 6 Answer: C 5 D NH3 + HCl → NH4Cl Dative bond is formed in NH4+ 6 C Number of electrons in HS = 17; number of electrons in F2 = 18 As HS has similar number of electrons in the molecules as F 2, and that HS is polar and F2 is non-polar, the main reason for HS to have a higher boiling point will be due to the presence of an overall polarity of its molecules. Hence, more energy is required to overcome the stronger permanent dipole - permanent dipole interactions (and also instantaneous dipole -induced dipole interactions) between HS molecules than the weaker instantaneous dipole - induced dipole interactions between F2 molecules. 7 B The option that fits the characteristics of a simple molecular solid will be one with relatively low melting point, generally poor solubility in water (as compared to an ionic compound), and highly likely to have good solubili