2019 HCI Prelim H2 Chem P3 ANS

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 Paper 3 1 (a) (i) MgCO3(s)  MgO(s) + CO2(g) [1] (ii) +- +- qqL.E α r +r [1] Both MgCO3 and MgO have the same charges and cationic radius [0.5] Anionic radius for CO32– is bigger than O2– [0.5] Magnitude of L.E. of MgCO3 is smaller than that of MgO. (iii) S is positive as gaseous CO2 is evolved [1]. Hence, –TS term is negative. G = H –TS Since the decomposition of MgCO3 is endothermic, H is positive. So for the decomposition to be spontaneous, for G to be negative, the decomposition should take place at high temperature [1]. (iv) MgCO3 has the lower decomposition temperature. [0.5] Both Mg2+ and Ba2+ have the same charge. The ionic radius of Mg2+ is smaller than Ba2+ (0.5). So Mg2+ has a higher charge density (0.5) and a greater polarizing power and it can distort the electron cloud of the CO32– to a greater extent (0.5 for either point), weakening the C−O covalent bonds in CO32– to a greater extent (0.5), hence less energy is needed for decomposition (0.5). (b) (i) Type of reaction (reaction 1): Nucleophilic Addition [1] Type of reaction (reaction 2): Elimination/Dehydration [1] (ii) [1] (iii) The geometry about the center C is trigonal planar [1]. Hence, there is equal probability for the CN – ion to attack from either side of the plane [1] , giving rise to an equimolar mixture of two stereoisomers. (iv) [1] HWA CHONG INSTITUTION 2019 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

(c) (i) [1] (ii) Reagents and Conditions: Step 1: LiAlH4 in dry ether [1] Step 2: excess concentrated H2SO4, heat [1] Step 3: steam, concentrated H3PO4, high T high P [1] Step 4: [1] J: [1] 2 (a) (i) C3H4O3 [1] (ii) [1] (b) constitutional/structural/functional group isomerism [1] (c) The LF molecule is polar and can form strong/favourable permanent-dipole permanent-dipole interactions/hydrogen bonding/ion -dipole interactions with water molecules. [1] (d) (i) When the onion is cooled, less LF will vaporise and come into contact with the eyes. [1] (ii) Heating the onion can denature the enzyme LF synthase so that LF will not be formed. [1] (e) (i) condensation [1] (ii) C9H16OS3 + 17O2  9CO2 + 8H2O + 3SO3 [1] (iii) n(cepaene) = 5 x 10–3 / 236.3 = 2.116 x 10–5 mol [½] n(O2) = 2.116 x 10–5 x 17 = 3.597 x 10–4 = 3.60 x 10–4 mol [½] ecf from ii volume of O2 mixture at r.t.p. = 3.597 x 10–4 x 24000 = 8.63 cm3 [½] volume of mixture = 8.63 x 100/40 = 21.6 cm3 [1] (iv) Role of copper: reducing agent [1] Cu + ½O2  CuO [1] (v) The blue crystals are CuSO4 [1], which can be formed from the reaction of CuO with SO3. [1]

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 CuO + SO3  CuSO4 (vi) Order of gases released: CO2, then SO2, then water. [1] CO2 is a non-polar molecule, hence intermolecular forces are the weakest – only dispersion forces, so it is released first at a lower temperature. followed by SO2. SO2 is a polar molecule with both intermolecular permanent-dipole permanent- dipole interactions, which are stronger than dispersion forces, and dispersion forces, which are also stronger than those of CO 2 due to SO 2 having more electrons and the larger electron cloud size than CO2. Water has intermolecular hydrogen bonding, which are the strongest intermolecular forces, and most energy required to vapourise it, and hence it is released last. [1] correct types of intermolecular forces for all 3 molecules [1] correct comparison of the strength of the 3 different intermolecular forces and relate to energy required to overcome the intermolecular forces for the gas to escape to the detector. 3 (a) Transition elements are d-block elements that form one or more stable ions with partially filled d-subshell. [1] (b) (i) Precipitation occurs when I.P. = Ksp For Cr3+: Ksp = [Cr3+][OH–]3 1.6 × 10–20 = 1.23×10-2[OH–]3 [OH–] = 1.09×10-6 mol dm–3 [1] For Co2+: Ksp = [Co2+][OH–]2 5.92 × 10–15 = 5.77×10-3[OH–]2 [OH–] = 1.01×10-6 mol dm–3 [1] Since the [OH–] required for IP = Ksp is similar for both precipitates to form, it does not allow for the separation of the two metal ions. [1] (ii) either Cr(OH)3 is sparingly soluble and dissolves to give small concentrations of Cr3+ and OH–. Cr(OH)3  Cr3+ (aq) + 3OH– (aq) (1) [1] When excess OH – is added, complex formation takes place. Cr3+ (aq) + 6OH– (aq)  [Cr(OH)6]3– [1] (accept if [Cr(OH)4−])

The [Cr3+] falls shifting the position of equilibrium of (1) to the right, causing the precipitate to dissolve. [1] or [Cr(H2O)3(OH)3] (s) + 3OH– (aq)  [Cr(OH)6]3– +3H2O (l) [2] ( [1] for [Cr(H2O)3(OH)3] and [Cr(OH)6]3–, [1] for balancing equation) When the [OH–] increases, position of equilibrium shifts forward to offset the increase in the [OH–] concentration. This causes the solid to dissolve. [1] (c) (i) [1] (no need to show axes and ΔE) The 5 d orbitals can be classified into two groups. The dx2 – y2 and dz2 have their lobes along the axis, while the dxz, dxy and dyz have lobes in between the axis. Since the ligands approach the central metal ion along the axis, the repulsion felt by the dxz, dxy and dyz orbitals is less than for the dx2 – y2 and dz2 orbitals. [1] As such the dxz, dxy and dyz orbitals are at the lower energy. (ii) When an electron is promoted from the lower energy d -orbitals, energy is absorbed corresponding to a wavelength in the visible spectrum. [1] The colour observed is the complement of the colours absorbed. [1] (iii) Ligand exchange reaction. [1] (iv) Identify the 1:4 ratio for Co2+: SCN– [1] [Co(SCN)4]2– [1] (d) (i) The energy levels of 3d and 4s electrons in cobalt are similar, hence once the 4s electrons are removed, some or all of the 3d electrons may also be removed without requiring much more energy. [1] However in calcium, once the 4s electrons are removed, the subsequent removal of electrons must come from an inner quant um shell which requires too much energy. [1] (ii) [Co(NH3)6]3+ + e  [Co(NH3)6]2+ E = +0.17 V O2 + 2H2O + 4e  4OH– E = +0.40 V dx2 – y2 dz2 dxy dxz dyz Energy/ kJ mol–1 ΔE

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 E cell = +0.40 – (+0.17) = +0.23V [1] (e) (i) AgCl [1] The number of moles of free Cl– ions/ Cl– counter ions are different in both the complexes. [1] S has one Cl– ion datively bonded to the Co3+ while T has two Cl– ion datively bonded to the Co3+ central metal ion. (ii) complex from S [1] (mark for dative bonds and positive charge) (iii) complex from T [1] (as long as trans structure is shown, ignore dative bonds/charges etc.) (f) Reactant molecules, CO and H2O are adsorbed onto the active sites of the catalyst surface by formation of weak attractive forces. This brings the molecules closer together, weakens the CO and O–H bond, orientating them in the right position for reaction, hence lowering the activation energy. ([2] for all 3 points, [1] for any 2 points) Once the reaction has taken place, the aldehyde formed desorbs and diffuses away from the catalyst surface so that the active sites are exposed for further reaction. [1] adsorb and desorb + active sites

4 (a) (i) Zn(−): Zn + 2OH–  Zn(OH)2 + 2e– [1] Pt(+): O2 + 2H2O + 4e–  4OH– [1] [1] on polarity