2025 Prelim P3 MS (for exchange) H2 Chem HCI

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme 1 HWA CHONG INSTITUTION 2025 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION MARK SCHEME Paper 3 1 (a) (i) Let the solubility of MgCO3 and Mg(OH)2 be x and y respectively, in mol dm–3. For MgCO3 Ksp = [Mg2+][CO32–] [0.5] 1.0 × 10–5 = (x)(x) x = 3.16 × 10–3 [0.5] mol dm–3 For Mg(OH)2 Ksp = [Mg2+][OH–]2 [0.5] 1.1 × 10–11 = (y)(2y)2 y = 1.40 × 10–4 [0.5] mol dm–3 (ii) [Mg2+][OH–]2 = 1.1 × 10–11 [0.5] (3.0 × 10–5)[OH–]2 = 1.1 × 10–11 [OH–] = 6.06 × 10–4 mol dm–3 [0.5] (iii) [0.5] CO32– ions are added first to precipitate Ca2+ as CaCO3 [0.5] such that the filtrate contains mainly Mg2+ / very little Ca2+. [1] This must be controlled to prevent precipitation of Mg2+ as MgCO3 (b) (i) Lattice energy is the heat evolved when 1 mole of solid ionic compound is formed from its constituent gaseous ions. [1] (ii) [3] total for correct Born-Haber cycle Hat (Mg) + 1st + 2nd IE (Mg) + Hf (OH) + EA (OH) x 2 + LE (Mg(OH)2) = Hf (Mg(OH)2) 148 + 736 + 1450 + (6 x 2) – (173 x 2) + LE (Mg(OH)2) = –925 LE (Mg(OH)2) = –2925 (or 2900 or 2930) kJ mol–1 [1] energy / kJ mol–1 Mg(OH)2(s) Mg(s) + O2(g) + H2(g) Mg(g) + O2(g) + H2(g) Mg2+(g) + 2e– + O2(g) + H2(g) Mg2+(g) + 2e– + 2OH(g) Hf(Mg(OH)2) = – 925 Hat(Mg) = + 148 1st + 2nd IE (Mg) = +736 + 1450 2 × ∆Hf (OH) = 2(+6) LE (Mg(OH)2) = ? Mg2+(g) + 2OH–(g) 2 × EA (OH) = 2(–173) 0

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme 2 (iii) [1] correct trend: decreasing magnitude of LE: MgO > Mg(OH)2 > Ca(OH)2 or LE for MgO is larger than for Mg(OH)2 + LE for Mg(OH)2 is larger than for Ca(OH)2 [1] explain why LE of MgO > LE of Mg(OH)2: The cation Mg2+ is the same for both compounds. • O2– anion has larger charge [0.5] • but smaller ionic radius than OH– [0.5] [0.5] explain why LE of Mg(OH)2 > LE of Ca(OH)2: The anion OH– is the same for both compounds and Mg2+ cation has the same charge as Ca2+ but Mg2+ has smaller ionic radius than Ca2+ [0.5] link between ionic radius and charge with magnitude of LE (e.g. quote LE eqn or link to strength of ionic bond / attraction bet. oppositely-charged ion at least once) (c) (i) Al : 1s2 2s2 2p6 3s2 3p1 [0.5] Mg: 1s2 2s2 2p6 3s2 [0.5] [0.5] The 3p subshell of A l is further away from the nucleus than the 3s subshell (or at higher energy level). [0.5] There is weaker attraction between the nucleus and the outermost electron of Al (or less energy is needed to remove the 3p electron) resulting in lower ionisation energy. (ii) MgCl2 dissolves in water to form aqueous ions. MgCl2(s) + 6H2O(l) → [Mg(H2O)6]2+(aq) + 2Cl–(aq) or MgCl2(s) + aq → Mg2+(aq) + 2Cl–(aq) [1] Mg2+ has slightly high charge density . Slight hydrolysis occurs, forming a slightly acidic solution of pH 6.5. [Mg(H2O)6]2+(aq) + H2O(l) ⇌ [Mg(H2O)5(OH)]+(aq) + H3O+(aq) [1] AlCl3 dissolves in water to form aqueous ions. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) or AlCl3(s) + aq → Al3+(aq) + 3Cl–(aq) [1] Al3+ having high charge density, undergoes hydrolysis to a greater extent to give an acidic solution with pH 3. (must compare extent of hydrolysis with Mg2+) Al(H2O)6]3+ (aq) + H2O (l) ⇌ [Al(H2O)5(OH)]2+ (aq) + H3O+ (aq) [1]

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme 3 (d) [1] diagram - label anode, cathode and electrolyte - battery in correct orientation anode: 2Al (s) + 3H2O (l) → Al2O3 (s) + 6H+ (aq) + 6e– [1] cathode: 2H+ (aq) + 2e– → H2 (g) [1] 2 (a) The negative charge on O atom in phenoxide ion is delocalised into the benzene ring, hence negative charge is dispersed, stabilising the phenoxide ion [1]. So phenol is a stronger acid than alcohol. Alkyl group is electron-donating, alkyl group intensifies the negative charge on the alkoxide ion through inductive effect, destabilising the alkoxide ion [1]. (b) (i) step 1: electrophilic substitution [1] step 2: nucleophilic substitution [1] (ii) [1] (iii) Amine in C has one more electron-donating (–CH2C6H5) group [1] that increases the electron density on N, making the lone pair of electrons on N more available for donation to form a dative bond with a Lewis acid [1]. (c) (i) [1] cation structure, minus 0.5 if structure of SO42− was incorrect [1] cation:anion ratio = 2:1 (ii) Salbutamol sulfate forms strong ion-dipole interactions [1] and hydrogen bonding [1] with water molecules. Al object as anode Pt cathode dilute H2SO4 (accept H2SO4(aq))

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme 4 (d) (i) stage equation 2 [0.5] for two arrows [0.5] dipoles 3 [1] for two arrows (ii) Lone pair on O of OCH3 delocalises into C=O in group X (ester) making ester carbon of group X less electron-deficient [1] than aldehyde carbon of group Y, hence ester carbon attracts electron-rich hydride ion less well as compared to aldehyde carbon. Ester carbon of group X is more sterically hindered than aldehyde carbon of group Y for the approach of the hydride ion as OCH3 is larger than H atom [1]. (iii) [1] (iv) In stage 1, the ketone carbon is trigonal planar, there is equal probability for the hydride ion to attack the ketone carbon from the top and bottom [1] of the plane.

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme 5 3 (a) Iron is a d-block element that forms one or more stable ions (or forms a stable Fe 3+ ion) with partially filled d subshells. [1] (b) (i) The phenoxide groups in Cat2− donate their lone pair of electrons on O to the partially-filled 3d orbitals of Fe3+ ion, forming coordinate/dative bonds. [1] (ii) [2] (iv) ligand exchange [1] [Fe(H2O)6]3+ + 3Cat2− → [Fe(Cat)3]3− + 6H2O [1] (c) Fe(OH)3 (s) ⇌ Fe3+(aq) + 3OH−(aq) ---------- (1) [0.5] The catechin in tea leaves could form a complex with Fe3+ ions, hence concentration of Fe3+(aq) decreases [0.5] and shifts the position of equilibrium (1) to the right [0.5] and increases its solubility into the mixture [0.5] to be discarded. (d) (i) [0.5] for each correct shape (no need for axes or labels of orbitals) 𝑑𝑧2orbital 𝑑𝑥2−𝑦2, 𝑑𝑥𝑦, 𝑑𝑥𝑧, 𝑑𝑦𝑧orbitals In an octahedral complex, the ligands approach the central metal ion along the x, y and z axes. Hence the electrons in the dz2 and dx2−y2 orbitals experience stronger repulsion from the direct approach of the ligands, causing the energies of dz2 and dx2−y2 orbitals to be at a higher level than the dxy, dxz and dyz orbitals. This results in d orbital splitting, where the originally degenerate five d orbitals are split into two groups with a small energy gap E between them. The lobes of the dz2 and dx2−y2 orbitals lie along the axes , while those of dxy, dxz and dyz orbitals lie in between the axes. [1] states the 2 different orientations of lobes for the 2 groups of d orbitals [0.5] states how the ligands approach in an octahedral complex [0.5] recognizes 1 group experiences stronger repulsion due to direct approach of ligands [1] states correct group of orbitals in each level (ii) When white light shines on the complex, an electron from the lower energy d orbitals absorbs energy DE and is promoted to vacancies in the higher energy d orbitals. The energy absorbed in this d-d transition corresponds to wavelength of the visible region of the

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme 6 electromagnetic spectrum. Hence, we observe colours complementary to the colours absorbed. [0.5] electron in lower E level absorb energy E to promote to higher level [0.5] vacancies in higher level [0.5] energy absorbed/E corresponds to visible region [0.5] color observed is complementary of that absorbed (e) P undergoes oxidative cl