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 solubility in organic solvents.

3 River Valley High School 8873/01/PRELIMS/25 2025 Preliminary Examination [Turn over 8 D Amount of CH3CH2OH = 0.86 46 = 0.0187 mol Heat released from combustion of ethanol = 0.0187  1367 = 25.56 kJ Heat absorbed by water = 300  4.18  18 = 22570J = 22.57 kJ Process efficiency = 22.57 25.56  100% = 88.0% Answer: D 9 D Answer: D +- +- q × qLattice Energy α r + r Between RbF and Rb 2O, O2− has a bigger charge and hence, Rb 2O will have a more negative lattice energy. Between KF and K2O, O2− has a bigger charge and hence, K 2O will have a more negative lattice energy. Between Rb2O and K2O, K+ has a smaller ionic radius and hence, K 2O will have a more negative lattice energy. 10 A Answer: A (1) Neutralisation is exothermic. (2) Bond breaking requires energy hence it is endothermic. (3) Removal of electron needs energy to overcome attraction forces between electrons and nucleus. (4) Combustion of C is exothermic. 11 D Let half-life of Q be T, and half-life of P be 2T. Let initial number of atoms of P = NP and initial number of atoms of Q = NQ = 4Np. After two half-lives of P, time elapsed = 4T -> this is equivalent to four half-lives of Q For P, number of atoms after 4T = NP / 4 For Q, number of atoms after 4T = NQ / 16 = 4NP / 16 = NP / 4 Hence the ratio of number of P to number of Q after two half-lives of P = 1

4 River Valley High School 8873/01/PRELIMS/25 2025 Preliminary Examination 12 C Total number of molecules, at either T1 or T2, is represented by: P + Q Number of molecules with energy equal to or greater than activation energy at T2: Q + R Therefore the fraction will be represented by . 13 A Statement 1: Correct A catalyst provides an alternative reaction pathway of a lower activation energy, which means altering the reaction mechanism. Statement 2: Correct With a lowered activation energy, the proportion of molecules with energies above the lowered activation energy increases. Statement 3: Incorrect Since temperature is not changed in this scenario, the average kinetic energy of the molecules stays the same. 14 B Statement A: Incorrect. Adding a solid does not affect position of equilibrium. Statement B: Correct Reduce the temperature → shifts to the left Lowering temperature favours the exothermic direction (reverse reaction), so position of equilibrium shifts left. Statement C: Incorrect Adding more product shifts the position of equilibrium left to decrease the concentration of W, not right. Statement D: Incorrect. Increasing volume, decreases the pressure of the system →favours the direction which produces more moles of gas (left side, 2 Y vs. right side, 1 W) to counteract the decrease in pressure. so position of equilibrium shifts left. QR PQ + +

5 River Valley High School 8873/01/PRELIMS/25 2025 Preliminary Examination [Turn over 15 C Statement A: Incorrect. Low pressure (1 atm) reduces SO₃ yield as position of equilibrium shifts left (more gas moles on the left). 1–2 atm is used for practical/economic reasons, not to increase % of SO₃. Statement B: Incorrect. 450 °C increases rate of reaction. As the reaction is exothermic (ΔH < 0), higher temperature actually shifts position of equilibrium to the left to absorb more heat, reducing the equilibrium SO₃ yield Statement D: Incorrect. A catalyst (V ₂O₅) does not change equilibrium composition ; it speeds up both forward and reverse reactions equally. Statement C: Correct. 17 A Option A : Incorrect. The pH of the solutions of chlorides generally decrease across the period due to increasing extent of hydrolysis of the chlorides. Option B : Correct. Option C : Correct. The white fumes is the hydrogen chloride. Option D : Correct. The white ppt is A l(OH)3 which dissolves in excess NaOH to form soluble Al(OH)4‾. 16 A Answer: A Reducing power: As the atomic size increases down the group, the attraction of nucleus for electrons decreases and hence the ease of losing electrons increases. Thus the reducing power increases. Ionic radius: Down the group, each element has 1 more principal quantum shell, and hence bigger ionic radius. Electronegativity: As the atomic size increases down the group, the attraction of nucleus for electrons decreases, and hence the electronegativity decreases.

6 River Valley High School 8873/01/PRELIMS/25 2025 Preliminary Examination 18 B Option A : Boiling and melting point overcome intermolecular forces of attraction.. Option B : HCl → H2 + Cl2 H-Cl bond is broken. HCl has the highest bond energy, so it is most stable Option C : it is a covalent molecule. Not ionic compound. Option D : bond length to relate to bond strength. 19 A Option 1 correct: ionic radius of Na+, Mg2+, Al3+ decrease across the period. Option 2 correct: pH decreases across the period as pH of oxides beco