RVHS 2026 Prelim P1 Ans

2026 RVHS Preliminary Examination H2 Chemistry Paper 1 Worked Solutions 1 N−  1s22s22p4 → lose one e− to become: 1s22s2p3 S  1s22s22p43s23p4 → lose one e− to become: 1s22s23s23p3 Cl+ 1s22s22p43s23p4 → lose one e− to become: 1s22s23s23p3 Answer: D 2 Based on the significant increase in 4 th ionisation energy from the 3 rd one, element X is group 13 (e.g. Al) Such an element will likely form an oxide with the formula X2O3 (e.g. Al2O3) 3 Option A is incorrect (hence this is the answer). While it is true that first ionisation energy of Mg is higher than that of Al, this is due to the fact that the electron to be removed from A l is at a higher energy level, and not because Mg has a higher nuclear charge. Option B, C and D are correct and true, hence they are not the answer. 4 Option A is incorrect. In ice, hydrogen bonding arranges water molecules into an open tetrahedral lattice, spacing them further apart than in liquid water. This gives ice a lower density than water at 0°C. Option B is correct. In CH ₃F, hydrogen atoms are bonded to carbon, which is not electronegative enough for hydrogen bonding to occur. Neither CH ₃F nor HC l can form hydrogen bonds. The higher boiling point of CH ₃F is due to its stronger permanent dipole - dipole interactions between molecules, not hydrogen bonding. Option C is incorrect. In 2 -nitrophenol, the ortho −NO₂ group forms an intramolecular hydrogen bond with the −OH group, preventing intermolecular hydrogen bonding and lowering the boiling point. In 4 -nitrophenol, no such intramolecular hydrogen bond is possible since the para −NO₂ group and −OH group are far apart, so intermolecular hydrogen bonds form freely, raising the boiling point. Option D is incorrect. In non -polar benzene, two ethanoic acid molecules form a dimer via two hydrogen bonds between their −COOH groups. This doubles the effective molecular mass from 60 to 120. 5 Option C is correct. Three possible structures of C 2N exist. In either structure, there are a total of 2 lone pairs of electrons and 1 unpaired electron.

2 River Valley High School 9476/01/Prelims/26 6 Option A is incorrect. At constant V, the gradient of the graph is nR. The gradient for L should be greater since the number of moles of L is greater than the number of moles of M. Option B is incorrect. At constant p, V = T p nR , gradient is positive. Option C is incorrect. At constant T, V = T p nR , so graph should be a straight line passing through the origin. The gradient for L should be greater. Option D is correct. At constant T, pV = constant. The pV value for L is greater as the number of moles of L is greater than the number of moles of M. 7 Option A is incorrect as H2S is a proton donor and hence acting as a Bronsted-Lowry acid. Option B is incorrect as H2S is a proton donor and hence acting as a Bronsted-Lowry acid. Option C is correct as H 2S is a proton acceptor and hence acting as a Bronsted -Lowry base. Option D is incorrect as H2S is a proton donor and hence acting as a Bronsted-Lowry acid. 8 Since this is a strong acid and strong alkali neutralisation, this reaction is exothermic, resulting in temperature rise. Option 1: correct, as a lower ambient temperature will cause greater heat loss during the experiment, which will result in overall lower temperature change/ rise Option 2: correct, as lower concentration of reagents (with all other factors being the same) will result in less heat energy produced (as there are less amounts of reagents reacting). Option 3: incorrect. Assuming that all other experimental factors are the same, higher [NaOH] can only result in: 1) An increase in temperature change if NaOH was the limiting reagent in the original experiment 2) The same temperature change if NaOH was in excess in the original experiment. 9 3.6 g of ice/ water = 0.200 mol of ice/ water For 1 mol of melting ice at 0 °C G = 0 H – TS = 0 (6010) – (273)S = 0 S = 22.0 J mol−1 K−1 For 0.2 mol of melting ice: S = (0.2)(22.0) = 4.4 J mol−1 K−1

3 River Valley High School 9476/01/Prelims/26 [Turn over 10 2Fe(s) + 3O2(g) → 2Fe2O3(s) H < 0, S < 0 Hence, based on G = H – TS, G will be negative only at low enough T such that |TS| < |H| 11 CxHy(g) + (x + y 4 )O2(g) → xCO2(g) + y 2 H2O(l) Volume of CO2 = 65 – 25 = 40 cm3 2 xy CO CH V 40 = = 4V 10 , x = 4 Volume of O2 left = 25 cm3 Volume of O2 reacted = 90 – 25 = 65 cm3 2 xy O reacted CH V 65 = = 6.5V 10 x + y 4 = 6.5, y = 10 Hence the hydrocarbon is C4H10. 12 Mass of pure MCO3 = 1.25 × 96.0% = 1.20 g Amount of CO2 = 288 24000 = 0.0120 mol Amount of MCO3 = amount of CO2 = 0.0120 mol Mr of MCO3 = 1.20 0.0120 = 100.0 g mol−1 Ar(M) + 12.0 + 3(16.0) = 100.0 Ar(M) = 40.0 Hence, M is Ca.

4 River Valley High School 9476/01/Prelims/26 13 Answer: D Option A (incorrect): Catalyst lowers the activation energy for both forward and backward reactions. More particles with sufficient energy for effective collisions. Therefore, both kf and kb increase. Option B (incorrect) and D (correct): Rate constant does not change with concentration. Option C (incorrect): Increase temperature increases both forward and backward reactions due to more particles with sufficient energy for effective collisions. Therefore, both kf and kb increase. 14 Answer: B Experiment 1: initial rate = 40 cm3 min−1 = k[A][B]2 Experiment 2: rate = k  (2  [A])  (½  [B])2 = ½  k[A][B]2 = 20 cm3 min−1 15 2SO2(g) + O2(g)  2SO3(g) Initial partial pressure/ atm x x 0 Change in partial pressure/ atm −2.4 −1.2 +2.4 Eqm partial pressure/ atm x−2.4 x−1.2 (5)(0.48) = 2.4 (x − 2.4) + (x − 1.2) + 2.4 = 5 x = 3.1 Hence, Kp = 3 22 2 SO 2 SO O () ( ) ( ) p pp = 2 2 (2.4) (0.7) (1.9) = 6.19 atm−1

5 River Valley High School 9476/01/Prelims/26 [Turn over 16 Ksp = [Ba2+][SO42−] 1.1  10−10 = [Ba2+](0.5) [Ba2+] required to precipitate BaSO4 = 2.2  10−10 mol dm−3 Ksp = [Ba2+][CO32−] 2.6  10−9 = [Ba2+](1.0) [Ba2+] required to precipitate BaCO3 = 2.6  10−9 mol dm−3 Ksp = [Ba2+][IO3−]2 4.0  10−9 = [Ba2+](1.5)2 [Ba2+] required to precipitate Ba(IO3)2 = 1.8  10−9 mol dm−3 Since the [Ba2+] required to precipitate the barium salts increases from BaSO4 to Ba(IO3)2 to BaCO3, BaSO4 is the first to precipitate followed by Ba(IO3)2 then BaCO3. 17 Option 1 is correct. Since the Ksp of AgCl is larger than AgBr, the ionic product (IP) of AgC l will drop below its Ksp causing it to dissolve. For AgBr, the IP will drop but it is still higher than its Ksp value and hence it does not dissolve. Option 2 is incorrect. [Ag+] decreases as the complex is formed. This will cause the IP for both AgCl and AgBr to decrease. Option 3 is incorrect. When dilute NH 3(aq) is added to separate solutions of each silver halide, the NH3 will react with the Ag+(aq) in both solutions to form the Ag(NH3)2+ complex. 18 Option A is incorrect. pH = pKa + lg = –lg(1.4  10−4) + lg = 3.38 Option B is incorrect. Upon dilution with some water, the pH of the buffer solution should remain the same as the amount of lactic acid and sodium lactate in the buffer solution remains unchanged (although their concentrations decrease). pH = pKa + lg = pKa + lg = pKa + lg Hence, pH of the buffer solution will not decrease upon dilution with water. Option C is incorrect. For a given volume, a more diluted buffer solution will have a lower buffering capacity due to smaller amount of lactic acid and lactate ions to remove any added OH− and H+ ions respectively. Option D is correct. Since the buffer solution has a greater amount of lactic acid (weak acid) than the lactate ions (conjugate base), the buffer solution is more effective in buffering the effect of addition of small amounts of base than acid.

6 River Valley High School 9476/01/Prelims/26 19 Option A is incorrect. Since the pH at equivalence point is above 7