2024 Prelims HCI H2 Chem P3 (Ans)

2024 HCI C2 H2 Chemistry Prelims / Paper 3 HWA CHONG INSTITUTION 2024 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 3 1 (a) (i) (ii) [0.5] for each point • Down the group, cationic radius increases , and charge remains the same. • Charge density of the cation decreases • Electron cloud of the anion is less polarized and the (N−O) covalent bonds are weakened to a smaller extent going down the group • More energy required to break these bonds hence temperature increases (iii) Number of moles of gaseous products increases as reaction takes place, entropy change is positive. [1] G = H − TS Since H is positive, for G to be negative and reaction to be spontaneous, the magnitude of TS needs to be greater than H. Hence, high temperature is needed. [1] (iv) [3] –602 = + 148+ ½(496) + 736 + 1450 – 142 + 844 + L.E

L.E = –3886 kJ mol–1 = –3890 kJ mol–1 (3 s.f.) [1] (ecf based on energy cycle drawn) (b) MgO dissolves sparingly with water / has limited solubility in water. pH = 9 [0.5] MgO + H2O → Mg(OH)2 Mg(OH)2 ⇌ Mg2+ + 2OH– [0.5] Al2O3 does not react with water [0.5]. pH = 7 [0.5] P4O10 reacts violently with water to give an acidic solution. pH = 2 [0.5] P4O10 + 6H2O → 4H3PO4 [0.5] (c) (i) AlCl3 is electron deficient and can accept a pair of electrons. [1] (ii) [0.5] for each product and (iii) [1] (iv) Chlorine atom is more electronegative than iodine atom, the I-Cl bond is polar and I atom in ICl is more electron deficient as compared to the iodine atom in I2. OR There is a partial positive charge on the I atom. (v) [1] for each step I–Cl + AlCl3 → I+ + AlCl4–

2024 HCI C2 H2 Chemistry Prelims / Paper 3 2 (a) Alkenes do not react/are unreactive with nucleophiles [0.5] as C-H and C=C bonds are non-polar so there is no electron deficient site for nucleophilic attack OR due to their electron rich  cloud/C=C bond which tends to repel nucleophiles [0.5] Carbonyl compounds react with nucleophiles [0.5] due to the presence of polar C=O bond/electronegative O which causes the C=O carbon to be electron deficient. [0.5] (b) (i) A racemic mixture contains equimolar amounts of two enantiomers. [1] Each enantiomer in a racemic mixture rotates plane -polarised light by the same magnitude in opposite directions , [1] hence there is no net rotation of the plane-polarised light. (ii) [1]: Choice of an alkene reaction that produces a carbocation intermediate, e.g. alkene + HCl / HBr / H2O / Cl2 / Br2 OR Choice of a carbonyl compound with a suitable nucleophile, e.g. HCN (in trace KCN) [1]: Explain why racemic mixture is obtained: • 0.5m (trigonal) planar geometry about positively charged carbon of the carbocation intermediate (for alkene reaction) or C=O carbon (for carbonyl reaction); • 0.5m equal probability of attack from top and bottom of the plane by the nucleophile [1]: Correct products drawn as a pair of enantiomers , with tetrahedral geometry about chiral carbon and use of wedge/dash bonds to illustrate 3D structure – mark is lost if there is no chiral carbon at all, or if products are drawn wrongly for the proposed reaction Sample answer: presented in written prose Organic compound and reagents used: , HBr A carbocation intermediate with (trigonal) planar geometry about the positively charged carbon is formed. Br − could attack from top or bottom of the plane with equal probability to produce equal concentrations of both enantiomers / a racemic mixture. Products:

Sample answer: presented using diagram and annotations Organic compound and reagents used: CH3CHO, HCN in trace KCN (iii) Stereoisomers of a drug have different arrangement of atoms in 3D space (different 3D conformation) [0.5], and would hence have different binding properties to chiral binding sites in the body, e.g. enzymes, receptors [0.5] (c) (i) (nucleophilic) substitution [1] (ii) 5-iodopentan-1-ol or 5-iodo-1-pentanol [1] (iii) Orange K2Cr2O7 turns green for B & C [0.5] K2Cr2O7 remains orange for D [0.5] Possible answers for step 2: Reagent [1] Product of B observations [0.5] Product of C observations [0.5] Na2CO3(aq) Effervescence (of CO2(g)) No effervescence Na(s) Effervescence (of H2(g)) No effervescence PCl5(s) White fumes (of HCl(g)) No white fumes (iv) Step 1: Na(s) or NaOH(s) [1] Step 2: ethanoyl chloride (CH3COCl) [1] (d) (i) [1] (ii) Two Br atoms have been incorporated in F [0.5] R group must be at position 4 (or 2) relative to OH group to prevent tri-substitution of phenol when Br2(aq) is added. [0.5]

2024 HCI C2 H2 Chemistry Prelims / Paper 3 (iii) [0.5] × 2 for correct side-chain of F and −CO2H of H [0.5] × 2 for relative positions of Br and side-chain/CO2H in each compound, based on answer in (ii) 3 (a) • In the presence of ligands/ In a ligand field • the degenerate d -orbitals split into two different energy levels with an energy gap, ΔE. • There are vacancies in the higher energy d orbitals/ Partially-filled d orbitals • The promotion of an electron from the lower to higher of these d orbitals • requires absorption of radiation in the visible spectrum corresponding to ΔE. • The colour seen is the complement of the absorbed colour. 6 points [3] 4 to 5 points [2] 2 to 3 points [1] (b) (i) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 [1] (ii) Cr can exhibit variable oxidation states due to the close/similar in energy of the 3d and 4s electrons. Hence, once the 4s electrons are removed (or used for bonding), some or all the 3d electrons may also be removed without requiring much more energy. [1] (c) (i) A: [Cr(H2O)6]2+ (accept Cr2+(aq), 1/2 if Cr2+) B: [Cr(H2O)6]3+ (accept Cr3+(aq), 1/2 if Cr3+) C: Cr(OH)3 / Cr(OH)3(H2O)3 [1 each, total 3 marks] (ii) [Cr(OH)6]3− undergoes oxidation to form CrO42- [1] (d) (i) High ionisation energy needed to remove 6 valence electrons OR Cr6+ has high charge density/ high polarising power and thus undergoes hydrolysis in water to form a polyatomic oxoanion [1] (ii) • Dilution causes the concentration of all aqueous species to decrease (to the same extent). [1/2] • As there are more concentration terms of the right hand side of the equation, position of equilibrium shifts towards the right [1] to produce more CrO42−.

• Hence the (orange) solution becomes yellow. [1/2] (accept ‘less orange’) (to earn this ½ mark, need to mention either the eqm shifts right or more CrO42- is produced) (iii) Ecell = (+1.33) – (+0.17) = +1.16 V [1/2] Since Eocell is positive, the reaction is spontaneous/ r eaction occurs . [1/2] Solution changes colour from orange to green/ Cr2O72- is reduced to Cr3+ Cr2O72- + 2H+ + 3SO2 → 2Cr3+ + 3SO42− + H2O [1] (e) (i) Cr H2O Br Mass in 100 g 13.0 27.0 60 No. of moles 0.25 1.5 0.75 Mole ratio 1 6 3 [1] Formula Cr(H2O)6(Br)3 Working must be shown to earn the mark. (ii) Cream precipitate: AgBr [1] No. of moles of AgBr = 0.188 / (108 + 79.9) = 0.00100 mol No. of moles of E = 0.400 / 399.7 = 0.00100 mol [both amts: 1] No. of moles of Br− anion : No of moles of E 1 : 1 Formula of complex cation in E: [Cr(H2O)4(Br)2]+ [1] (iii) Cr Br Br H2O H2O H2O H2O + [1] (Octahedral; bond should be shown from O to Cr) (e) E / V Fe2+ + 2e ⇌ Fe −0.44 Pb2+ + 2e ⇌ Pb −0.13 Quote E values for Fe and Pb with relevant explanation [1] EoFe2+/Fe is more negative than EoPb2+/Pb, Fe is more easily oxidised by concentrated sulfuric acid than Pb. He