2024 H1 Chem Prelim P1 Ans VJC

1 VICTORIA JUNIOR COLLEGE 2024 JC2 H1 CHEMISTRY PRELIMINARY EXAMINATION PAPER 1 ANSWERS 1 A 7 A 13 D 19 A 25 D 2 B 8 C 14 A 20 B 26 C 3 C 9 D 15 C 21 A 27 C 4 B 10 A 16 C 22 B 28 A 5 D 11 C 17 D 23 C 29 B 6 C 12 B 18 D 24 A 30 D 1 A particle electronic configuration A C– 1s22s22p3 B N2– 1s22s22p5 C F+ 1s22s22p4 D Ne 1s22s22p6 Hence, C– has a half-filled set of p orbitals. 2 B Element X has the highest 6 th ionisation energy, it means that the sixth electron removed from X is from its next inner shell. Hence, X has five valence electrons and X is in Group 1 5. As W, X, Y and Z are four consecutive elements in the Periodic Table, W is in group 14. Hence, the formula of the fluoride of W is WF4. 3 C (1 and 3 only) Option 1: Correct A dative bond is formed when the lone pair on O in CH3OCH3 is donated to the vacant orbital of B in BF3. Option 2: Incorrect The bond angle around B changes from 120° in BF3 (3 bond pairs, trigonal planar), to 109.5° in Q (4 bond pairs, tetrahedral). Option 3: Correct In CH3OCH3, there are 2 b ond pairs and 2 lone pairs around O (bent). In Q, there are 3 bond pairs and 1 lone pair around O (trigonal pyramidal). 4 B Option A: Incorrect NH3: Trigonal pyramidal All dipole moments are not cancelled out, so the molecule is polar. CCl4: Tetrahedral All dipole moments are cancelled out, so molecule is non-polar. Option B: Correct CO and HF: linear Since electronegativity difference is greater between H and F than between C and O, so HF is more polar than CO. Option C: Incorrect H2O and H2S: bent. Since O is more electronegative than S, so H2O is more polar than H2S since O-H bond is more polar than S-H bond. Option D: Incorrect CO2: linear; BF3: trigonal planar Dipole moments in these two molecules cancel out and so these molecules are non-polar. 5 D (CH3)2CHOH and CH 3CH2CH2OH form hydrogen bonds between molecules. However, CH3CH2CH2OH, with its straight-chain structure, has greater surface area for molecular interaction, leading to greater extent and stronger instantaneous dipole –induced dipole interactions than (CH3)2CHOH, which is a branched molecule having smaller area for molecular interaction. Hence, more energy is required to overcome the stronger id-id interactions between CH3CH2CH2OH and CH 3CH2CH2OH has a higher boiling point than (CH3)2CHOH. There are only weaker permanent dipole -permanent dipole interactions between CH 3CH2CHO molecules, hence, CH3CH2CHO has the lowest boing point among three molecules. 6 C (2 and 3 only) Bronsted-Lowry base is a proton acceptor. Option 1: Incorrect NH3 donated a lone pair to CH 3Cl to form CH 3NH2. There is no transfer of proton in this reaction. Option 2: Correct HNO3 accepted a proton to form H2NO3+ Option 3: Correct H2O accepted a proton to form H3O+ 7 A Since Kw increases when temperature increases, the dissociation of water is endothermic. Hence, at higher temperature, more H + and OH – are formed in equal amounts. So, [H+] = [OH –] but pH decreases due to higher [H+]. 8 C Concentration of dilute HNO3 = (10.0 × 0.010)  100 = 0.00100 mol dm–3 HNO3, being a strong acid, will fully dissociate in water. Hence, [H+] = [HNO3]. pH = –log[H+] = –log(0.00100) = 3 9 D Option A: Incorrect Solution B is a monoprotic base which its initial pH (before adding acid solution A) is 11.10. pOH = 14 – 11.10 = 2.90 [OH–] = 10–2.90 = 1.26 × 10–3 mol dm–3 < 0.125 mol dm–3 Hence, solution B is a weak base. Option B: Incorrect Since solution B is a weak base, given that pH at equivalence point is 5.22 (less than 7), solution A must be a strong acid.

2 Option C: Incorrect Since monoprotic base (solution B) reacts with monobasic acid (solution A), 1 mol of A will react with 1 mol of B at the equivalence point. Amount of B reacted = (25.0 × 10–3) × 0.125 = 3.13 × 10–3 mol Hence, amount of A reacted at the equivalence point = 3.13 × 10–3 = 12.50 × 10–3 × Concentration of A Concentration of A = 0.250 mol dm–3 Option D: Correct From graph, equivalence point has pH 5.22 which falls within the working range of bromocresol green (pH range 3.8 – 5.4). Hence, bromocresol green is a suitable indicator to detect the endpoint of this titration. 10 A Option A: Incorrect statement pH of blood will remain relatively constant only when a small amount of H + or OH – is added to it. However, when a large amount of H+ is added, there is insufficient HCO3– in the buffer and the pH decreases significantly. Option B: Correct statement A decrease in [H+] of blood will shift equilibrium position of H2CO3(aq) ⇌ H+(aq) + HCO 3–(aq) to the right, resulting in more hydrogencarbonate ions to be formed. Options C and D: Correct statement A mixture of H 2CO3 and HCO 3– is a buffer present in blood to control its pH. 11 C Option A: Incorrect Electronegativity of the Period 3 elements increases across the period. Hence, the difference in the electronegativities of Period 3 elements and oxygen becomes smaller, causing the bonds formed in the oxides to be more covalent. Option B: Incorrect Sulfur can also form more than one oxide, which are SO2 and SO3. Option C: Correct Electronegativity of the Period 3 elements increases across the period. Hence, the difference in the electronegativities of Period 3 elements and chlorine becomes smaller, causing the bonds formed in the chlorides to be more covalent. Option D: Incorrect NaCl dissolves in water to give a neutral aqueous solution which will not turn blue litmus red. 12 B (1 and 2 only) Options 1 and 2: Correct The reducing power of Group 1 elements increases down the group . With increasing number of electronic shells, the valence electrons are further away from nucleus. Hence, t he nuclear attractions between nucleus and valence electrons get weaker down the group, enabling valence electrons to be given out more easily. As such, rubidium reduces water more readily than sodium and caesium loses electrons more easily than potassium. Option 3: Incorrect Group 1 element such as caesium is not an oxidising agent, it is a reducing agent. 13 D Option A: Correct Down Group 17, the halogen molecules have greater number of electrons , which results in stronger instantaneous dipole -induced dipole attractions between molecules. Hence, melting point of Group 17 elements increases down the group. Since iodine is a solid, Singaporium should be a solid at r.t.p. too. Option B: Correct Increase in atomic radius of halogen (X) down the group leads to poorer extent of overlap between the orbitals of halogen atoms in the halogen molecules, this leads to weakening of X –X bond. Hence, the bond energy of Sg–Sg is less endothermic than that of I–I. Option C: Correct Oxidising power of halogens decreases down the group. Hence, Sg2 being a weaker oxidising agent than I2, is unable to oxidise iodide to iodine. Option D: Incorrect Increase in atomic radius of halogen (X) down the group leads to poorer extent of overlap between the orbital of H and the orbital of the halogen, this leads to weakening of H –X bond and HX becomes less stable to heat. Since H I can decompose when heated with red-hot steel, HSg can therefore decompose when heated with red-hot steel. 14 A (1, 2 and 3) If the third element is O C H O Mass in 100 g 54.5 9.1 36.4 Amt in 100g 4.54 9.1 2.275 Ratio 2 4 1 The empirical formula of the organic compound is C2H4O. Option 1: Correct The molecular formula of CH 3CH2CH2CO2H is C4H8O2 so the empirical formula is C2H4O. Option 2: Correct The molecular formula of CH 2(CHO)CH2CH2OH is C4H8O2 so the empirical formula is C2H4O. Option 3: Correct If the third element is S C H S Mass in 100 g 54.5 9.1 36.4 Amt in 100g 4.54 9.1 1.13 Ratio 4 8 1