2022 RVHS JC2 H2 CM Prelim P3 (Solutions)

River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination [Turn over Suggested Solutions for H2 Chemistry Prelim Paper 3 1 (a) (i) S(g) → S+(g) + e– [1] (ii) Less energy is required to remove the higher energy 3p electron from Al compared to the 3s electron from Mg. [1] (b) (i) CuCO3 → CuO + CO2 [1] (ii) Ionic radius of Ca2+ = 0.099 nm Ionic radius of Cu2+ = 0.073 nm Decomposition temperature of CuCO3 is expected to be lower. Charge density of Cu2+ is greater than Ca2+ due to the smaller ionic radius of Cu 2+. Cu 2+ ion is able to polarise (the electron cloud of) CO32– ion to a larger extent , hence weakening the C O bond to a larger extent. [2] (c) (i) Na2O(s) + H2O(l) → 2NaOH(aq) pH = 13 [1] (ii) SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq) pH = 2 [1] (d) (i) Na(s) is an electrical conductor due to the presence of delocalised valence electrons which can migrate freely through the metallic structure when a potential difference is applied. Na2O(s) does not conduct electricity because its ions are held in fixed positions in a giant lattice structure / it has no mobile ions. [2] (ii) [1] (iii) Na2O2 has a giant ionic lattice structure while H2O2 has a simple molecular/ covalent structure. More energy is needed to overcome the stronger electrostatic forces of attraction between Na+ and O22– compared to the hydrogen bonds between H2O2 molecules. Hence, Na2O2 has a higher melting point than H2O2 and is a solid at room temperature. [2] (e) (i) 𝐾𝑐 = [Pb2+][Cr2+]2 [Cr3+]2 [1] (ii) Equilibrium [Pb2+(aq)] = ½(2.96  10−4) [2]

2 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination = 1.48  10−4 mol dm–3 𝐾𝑐 = (1.48 × 10−4)(2.96 × 10−4)2 (0.200)2 = 3.24  10−10 mol dm−3 (iii) The concentration of all aqueous species/ ions will be lowered. Since there are more aqueous species/ ions on the right of the equation, the position of equilibrium shifts right. [2] [Total: 17] 2 (a) (i) [2] (ii) 2H2O(l) → O2(g) + 4H+(aq) + 4e− 2Al(s) + 3/2O2(g) → Al2O3(s) [2] (iii) Volume of Al2O3 = 96.2 × 0.03 = 2.886 cm3 Mass of Al2O3 = 3.95 × 2.886 = 11.40 g Amount of Al2O3 = 11.40 2(27.0)+3(16.0) = 0.1118 mol Amount of O2 = 0.1118 × 3 2 = 0.1677 mol Amount of electrons passed = 0.1677 × 4 = 0.6708 mol Q = nF = It 0.6708 × 96500 = 2.0 × t t = 3.24 × 104 s [3] (b) Step 1: 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq) [2]

3 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination [Turn over Step 2: 2Fe2+(aq) + S2O82−(aq) → 2Fe3+(aq) + 2SO42−(aq) (c) (i) BaSO4 [1] (ii) Upon reaction with Br 2, the oxidation state of sulfur increases from +2 in S2O32− to +6 in SO42−. Upon reaction with I2, the oxidation state of sulfur increases from +2 in S2O32− to +2.5 in S4O62−. Therefore, Br2 is a stronger oxidising agent than I2. [2] (d) P does not rotate plane-polarised light.  P does not contain a chiral carbon. P is insoluble in both HC l(aq) and NaOH(aq)/ P does not undergo acid-base reaction.  P is neutral. P undergoes alkaline hydrolysis with hot NaOH(aq).  P contains ester and nitrile groups. 1 mole of R undergoes acid-base reaction with 1 mole of Na2CO3(aq).  R contains 2 −COOH groups. Q undergoes oxidation to give HCOOH and a pale yellow precipitate, CHI3.  Q contains a −CH(OH)CH3 or −COCH3 group. P undergoes reduction with LiAlH4 to form Q and S.  Both Q and S contain a primary −OH group.  S contains a −CH2NH2/ primary amine group. P Q [8] R S [Total: 20]

4 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination 3 (a) (i) Primary amine, primary alcohol [1] (ii) To act as a Lewis base, the lone pair of electrons on N of TRIS is donated into the vacant orbital of H+/ proton from hydrochloric acid, resulting in the formation of a dative bond between N of TRIS and H+. [2] (b) (i) Initial [TRIS] = 121.14 12.0 4 14.0 (16.0 3) 11.0 + +  + = 1.001 mol dm–3 Kb = 14 6 9 10 1.202 108.32 10 − − − = mol dm–3 = 2[TRISH ][OH ] x [TRIS] 1.001 x +− = − Assuming x is very small, [OH–] = x = 6(1.202 10 )(1.001)− = 1.097  10–3 mol dm–3 pOH = –lg(1.097  10–3) = 2.96 pH = 14 – pOH = 11.0 [3] (ii) 7.5 = –lg(8.32  10–9) + lg( [TRIS] [TRISH ]+ ) lg( [TRIS] [TRISH ]+ ) = –0.580 [TRIS] [TRISH ]+ = 10–0.580 = 0.263 Or Ka = 8.32  10–9 = 7.5[TRIS](10 ) [TRISH ] − + Or Kb = 1.202  10–6 = 6.5[TRISH ](10 ) [TRIS] + [1] (iii) [TRIS] [TRISH ]+ = 0.263 = 1.001 x x − [TRISH+] = x = 0.7926 mol dm–3 Amount of HCl required = amount of TRISH+ in 1 dm3 Volume of HCl required = 0.7926/11.0 = 0.0721 dm3 [2]

5 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination [Turn over Or Alternative for finding [TRISH+]: [TRIS] [TRISH ]+ = 0.263 1 TRIS + H+ → TRISH+ Eqm [ ] 0.263 1.0011.263 1 1.0011.263 [TRISH+] = 1 1.0011.263 = 0.7926 mol dm–3 (c) (i) [2] (ii) Formation of cream ppt of AgBr shows that compound B undergoes nucleophilic substitution with NaOH(aq) to release bromide ion for precipitation by AgNO3. Compound B is an alkyl bromide/ halide. [2] (iii) Step 1: CH3CHClCH3, anhydrous AlCl3 Step 2: acidified K2Cr2O7, heat with immediate distillation [2] (iv) Warm with Tollens’ reagent. Silver mirror/ black/ grey ppt formed with cuminaldehyde but no silver mirror/ black/ grey ppt formed with compound D. Or I2 in NaOH(aq), warm. Pale yellow ppt formed with compound D but no ppt formed with cuminaldehyde. Or acidified/H2SO4(aq), K2Cr2O7(aq), heat Orange acidified K2Cr2O7(aq) turned green with cuminaldehyde, but remained orange with D. [2]

6 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination (d) (i) Transition metals have partially filled 3d orbitals. In the presence of ligand field, the 3d orbitals are split into 2 sets of non-degenerate orbitals with small difference in energies. Visible light of the e lectromagnetic spectrum is absorbed for the transfer of an electron from a lower energy d -orbital to an unfilled/ partially filled d orbital of higher energy. The colour of complex observed corresponds to the complement of the absorbed colours. [3] (ii) Complex is violet -red. Since yellow -green light is absorbed for d -d transition, the complementary colour violet-red will be observed. [1] (iii) (M4+ 1s22s22p63s23p6) M 1s22s22p63s23p63d24s2 [1] (iv) Either of the following: • In an octahedral ligand field of F –, the energy gap between the non-degenerate 3d orbitals becomes very large. • Radiation/ light absorbed for d -d transition is not from visible light range. [1] [Total: 23] 4 (a) (i) The CC bond in propanone is formed from the overlap between sp2 and sp3 hybridised carbons while the CC bond in propane is formed from the overlap between sp3 hybridised carbons. sp2 hybridised orbitals have greater s character/ lower p character , are shorter/ smaller and closer to the nucleus, making the CC bond shorter than expected. [2] (ii) [2] (iii) [2]

7 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination [Turn over (iv) S (one of the 2) T (one of the 2) U (one of the 4) [3]

8 River Valley High School 9729/03/PRELIMS/22 2022 Preliminary Examination (b) (i) Cold alkaline KMnO4 [1] (ii) or [1] (c) (i) [CH3COCH3(aq)] remains approximately constant throughout the experiment so rate of reaction is independent of [CH3COCH3(aq)]. Thus experimental res