YIJC 2025 Prelim P1 Answers (for exchange) H2 Chem

1 ©YIJC [Turn over YIJC 2025 H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 B D D B B 6 7 8 9 10 B D C B C 11 12 13 14 15 A D B C B 16 17 18 19 20 D C A C C 21 22 23 24 25 D C D C B 26 27 28 29 30 B D C A D 1 Answer: B Negatively charged ions attracted to positive plate, positive ions to negative plate. D must be negatively-charged, while E and F are positively-charged. Angle of deflection is proportional to charge/mass ratio. Since D and E have roughly the same angle of deflection but different polarity, D and E must have roughly the same charge/mass ratio. F has almost double the angle of deflection as E, so the charge/mass ratio of F must be almost double that of E. particles 14N‾ 28Si2+ 14C2+ charge / mass 1/14 2/28 = 1/14 2/14 = 1/7 2 Answer: D The sharp rise between 2 nd and 3rd I.E. shows that there are 2 valence electron s in the outermost shell. Hence G is in Group 2. The outermost electronic configuration is ns2. It cannot be in Period 3 because Mg (Group 2 in Period 3) has only 12 electrons, and so cannot have 13 successive IEs. The next inner quantum shell contains 8 electrons, as seen by the 8 IEs before the next sharp rise. For a d-block element, the next inner quantum shell should contain 3s, 3p and 3d electrons, which would be more than 8. It is not Al as Al has 3 valence electrons.

2 ©YIJC [Turn over 3 Answer: D While alcohols have intermolecular hydrogen bonding between molecules, as the carbon chain length of an alcohol increases, the boiling point increases as the id-id interactions become stronger due to the larger and more polarisable electron cloud. Hydrogen chloride dissociates and releases an H+ that forms dative bond, not hydrogen bond, with water molecules to produce H3O+. CH3CHO has a higher boiling point than CH 3CH2CH3 due to stronger permanent dipole-permanent dipole attraction between CH3CHO molecules, and not intermolecular hydrogen bonding. In ice, each water molecule forms 4 hydrogen bonding with 4 other water molecules. These hydrogen bonds are relatively long , giving rise to an open structure . I n liquid, water molecules aggregate together via hydrogen bonding. Hence, ice has a lower density than water at 0 °C. 4 Answer: B Option 2 is non-polar molecule. AlCl3 is trigonal planar in shape and there is no net dipole moment in the molecule as all the dipole moments cancel out each other. Option 3 is non-polar molecule. CO2 is linear in shape and there is no net dipole moment in the molecule as all the dipole moments cancel out each other. Option 1 and 4 are polar as there is net dipole moment in the molecule. 5 Answer: B Option A is correct as Al is a metal (solid) that conducts electricity. Option B is wrong as Al2O3 is insoluble in water. Option C is correct as Al3+ has high charge density so it can undergo hydrolysis in water to form an acidic solution. Option D is correct as Al2O3 is amphoteric. 6 Answer: B Option 1 is correct: The valence orbital gets larger / more diffuse as we go down Group 17, and the overlap of the orbital with 1s orbital of H gets less effective. Thus, the H–X bond strength decreases. Option 2 is correct: The thermal stability of HX decreases down Group 17 due to the weaker H –X bond, which is easier to break. Option 3 is incorrect: Decomposition of HX involves breaking covalent bond, not intermolecular forces of attraction which is influenced by electron cloud size.

3 ©YIJC [Turn over 7 Answer: D Statement 1: Correct Statement 2: Incorrect. Relative molecular mass is the average mass of one molecule, and not an atom in the molecule. Statement 3: Incorrect. One mole of a compound contains as many units of that compound as there are atoms in 12.00 g of carbon-12. A compound is made up of at least 2 atoms / ions. Hence, there will not be the same number of atoms as there are atoms in 12.00 g of carbon-12. 8 Answer: C At room temperature and pressure, all the water vapour has become liquid water. Thus, the 0.00208 mol of gas refers to N2O(g). Amount of N2O(g) formed = 0.00208 mol Amount of NH4NO3 decomposed = 0.00208 mol Mass of NH4NO3 decomposed = 0.00208 * (14+4+14+3*16) = 0.167 g Percentage decomposed = 0.167 / 0.2 x 100% = 83.3% 9 Answer: B 1.00 g of propan-1-ol = 0.016667 mol Heat absorbed by water, q = mcT = (200)(4.18)(39.5) = 33022 J Since efficiency of heat transferred is 90%, heat released by combustion of propan-1-ol = 33022 x (100/90) = 36691 J Hc = – 36691 / 0.016667 = –2200

4 ©YIJC [Turn over 10 Answer: C The least negative lattice energy (i.e., weakest ionic bond) is for the compound with: • Low ionic charge • Large ionic radii → sodium azide Ions Charges Radii Expected Lattice Energy Mg²⁺ and N3– Higher Small cation but larger anion more negative Mg²⁺ and N3– Higher Small cation and anion most negative Na⁺ and N3– Lower Large cation and larger anion least negative Na⁺ and N3– Lower Large cation but smaller anion less negative 11 Answer: A Statements 1 and 2 are correct. 1 Between 0 K and 195 K, entropy is low as the NH3 molecules are held in their fixed positions in the solid state. Statement 1 is correct. 2 Between 195 K and 240 K, there is an increase in the number of ways to distribute energy among NH3 molecules. Statement 2 is correct. 3 At 240 K, there is an increase in the number of NH3 molecules and hence number of ways to arrange NH3 molecules. Statement 3 is incorrect, as there is no addition of NH3 molecules to the system, although the number of gaseous NH3 has increased. 12 Answer: D rate = k[sucrose][acid] When [acid] is doubled, rate is doubled. However, since acid is a catalyst, [acid] is constant. So, rate = k’[sucrose], where k’ = k[acid]. This is a pseudo first-order reaction. Half-life, t1/2= (ln 2) / k’ = (ln 2) / (k[acid]) When [acid] is doubled, (ln 2) / (k[acid]) is halved since [acid] appears in the denominator.

5 ©YIJC [Turn over 13 Answer: B The distribution curve does not change since temperature is kept constant. However, with a catalyst, EA is lowered. 14 Answer: C equilibrium 1 CO2 + H2O ⇌ HCO3− + H+ K1 equilibrium 2 HCO3− ⇌ CO32− + H+ K2 equilibrium 2 (reversed) CO32− + H+ ⇌ HCO3− 1 / K2 equilibrium 3 (add eqm 1 and eqm 2(reversed)) CO2 + H2O + CO32− ⇌ 2HCO3− K3 = K1 × (1 / K2) 15 Answer: B When temperature increases, the system will want to decrease the temperature by favouring the endothermic reaction to absorb heat. Since backward reaction is endothermic, reaction will shift to the left. This will result in more reactants than products, hence Kp will decrease. 16 Answer: D H2PO4– + H2O ⇌ HPO42– + H3O+ Ka (of H2PO4–) Kw = Ka (acid) x Kb (conjugate base) i.e., 1.00 x 10–14 = Ka (of H2PO4–) x Kb (of HPO42–) 1.00 × 10−14 6.3 × 10−8 = 𝐾𝑤 𝐾𝑎(H2PO4−) = Kb (conjugate base of H2PO4– ) = Kb (HPO42–) number of particles energy EA B A D C EA(cat)

6 ©YIJC [Turn over 17 Answer: C PbCl2(s) ⇌ Pb2+(aq) + 2Cl−(aq) ------ (1) A True. Ksp is only dependent on temperature. B True. As more Cl− is added, the position of equilibrium (1) will shift to the left, causing the solubility of PbCl2 to decrease. This is known as the common ion effect. C False. At M, the concentration of Cl− is not necessarily twice that of Pb2+ as Cl− is also contributed by the addition of KCl. D True. Pb2+(aq) + 4Cl−(aq) ⇌ [PbCl4]2−(aq) When even more Cl− is added, a soluble complex [PbCl4]2− is formed. The formation of this complex decreases the [Pb2+], causing the position of equilibrium (1) to shift to the right. This increases the solubility of PbCl2. 18 Answer: A Option 1 is correct. (Refer to diagram) Option 2 is correct. (Refer to diagram) Option 3 is incorrect as there’s o