2024 Prelims HCI H2 Chem P2 (Ans)

2024 HCI C2 H2 Chemistry Prelims / Paper 2 HWA CHONG INSTITUTION 2024 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 2 1 (a) 3d ___ ___ ___ ___ ___ 3p ___ ___ ___ 3s ___ [1] Correct number of orbitals for each subshell with labels [1] Correct relative energy gap (bigger gap between 3s and 3p than 3p and 3d) (b) (i) First ionisation energy is the energy required to remove one mole of electrons from one mole of free gaseous atoms to form 1 mole of unipositively charged gaseous ions. (b) (ii) 1st IE of Se is lower than As [1/2] Selenium has a lower 1st IE than arsenic because selenium contains a paired electron in one of its p orbitals, thus it experiences inter-electron repulsion. So, less energy is required to remove the outermost electron in selenium compared to arsenic. [1] Explains that inter -electron repulsion in p -orbital leads to lower 1st IE in Se compared to As. 1st IE of Kr is higher than Br [1/2] Krypton has a higher 1st IE than bromine because krypton has a greater effective nuclear charge than bromine. This is inferred from the fact the krypton has a larger proton number / greater nuclear charge , and electrons are added to the same quantum shell / leading to a relatively constant shielding effect. Therefore, the attraction between the nucleus and the outermost electron is stronger, and more energy is required to remove the outermost electron in krypton than bromine. [1] Uses effective nuclear charge to explain the stronger attraction between the outermost electron in Kr and the nucleus compared to that in Br, leading to higher 1st IE. 2 (a) (i) Trigonal planar [1], 120 [1] energy

1 2 (a) (ii) One 2s orbital is mixed with two 2p orbitals to give 3 hybridised sp2 orbitals. [1/2] correct type of atomic orbitals (2s and 2p) [1/2] correct number of atomic orbitals (1 and 2 respectively) (a) (iii) The three sp2 hybridised orbitals are degenerate and will be equally spaced apart in a trigonal planar arrangement to minimise repulsion. [1/2] Indication of number of hybrid orbitals (state ‘three’ or ‘trigonal planar’) [1/2] Applying VSEPR Theory in the context of hybrid orbitals (i.e. hybrid orbitals will be equally spaced apart / minimise repulsion). (b) PTFE has a greater number of electrons / larger electron cloud than F2C=CF2 [1], thus it is more polarisable and has stronger dispersion forces than F2C=CF2. More energy is required to overcome the stronger dispersion forces in PTFE [1], and so it has a higher boiling point, and is a solid at room temperature while F2C=CF2 is a gas. (c) The C–C and C–F bonds in PFAS are very strong [1] and do not break/hydrolyse in typical environmental conditions. (d) (i) RO• + CHCl=CH2 → RO–CHCl–CH2• [1] RO–CHCl–CH2• + CHCl=CH2 → RO–CHCl–CH2–CHCl–CH2• [1] (d) (ii) The carbon atom is chiral because it has a tetrahedral geometry [1/2] and is bonded to four different groups [1/2]. (d) (iii) Type 1 Type 2 Type 3 [1] Type 1 ; [1] Type 2 [1] Type 3 (any variation showing more random arrangement in the stereochemical arrangements)

2024 HCI C2 H2 Chemistry Prelims / Paper 2 3 (a) [1] diagram, taking note of: • axes • labels • shape of graph (starts from intersection of axes), • shaded areas (legend can be embedded in response) [–0.5] for each error At higher temperatures, there is a higher proportion of molecules that have kinetic energies greater than or equal to E a [1] Hence, frequency of effective collisions is higher [0.5], increasing rate of reaction [0.5]. (b) (i) [2] Drawing of mechanism, inclusive of the following: • Partial charges on C−Br atoms • NO indication of slow or fast steps • Balanced equations for the step of the mechanism (including balance of charge) • Correct structure of transition state • Lone pair on O atom of OH− • Correct drawing and placement of curly arrows (b) (ii) When the concentration of 2 -bromopropane increases, the number of 2-bromopropane molecules per unit volume increases / the 2-bromopropane molecules are closer [1]. This increases the frequency of effective collisions which in turn increases the rate of reaction [1]. [-0.5 for not mentioning effective collisions]

2 3 (b) (iii) Circle z [2-bromopropane][OH−]. [1] In the mechanism, both reagents / the 2-bromopropane and OH− are involved in the rate-determining/only/slow step.[1] (b) (iv) rate = y [2-bromopropane] + z [2-bromopropane][OH−] From experiment 1, 7.1 × 10−7 = y(0.1) + z(0.1)(0.1) From experiment 2, 1.2 × 10−6 = y(0.1) + z(0.2)(0.1) Both equations – [1] y = 2.20 × 10−6 units of y = s−1 z = 4.90 × 10−5 units of z = mol−1 dm3 s−1 Each unit – [1] Each value – [0.5] (c) (i) compound observations solubility of residue (if any) in dilute aqueous ammonia in concentrated aqueous ammonia 2-bromopropane cream ppt insoluble soluble/partially soluble [0.5] 2-chloropropane white ppt [0.5] soluble soluble [0.5] iodobenzene no ppt [0.5] - - (c) (ii) When NH3 is added, the Ag+ forms the [Ag(NH3)2]+ complex. [1] Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2]+(aq) The formation of the [Ag(NH3)2]+ complex decreases [Ag+] [1], The ionic product of [Ag+][Cl−] decreases to below Ksp. Hence, AgCl dissolves. [0.5] OR Hence position of equilibrium of AgCl(s) ⇌ Ag+(aq) + Cl−(aq) shifts rightwards to increase [Ag+], allowing AgCl(s) to dissolve. However, as Ksp of AgBr is relatively lower, the addition of dilute NH 3(aq) is insufficient to lower [Ag+] till the [Ag+][Br−] decreases to below that of the Ksp of AgBr. Hence, AgBr does not dissolve. [0.5] 4 (a) Cis-trans isomerism [1] It occurs due to restricted bond rotation caused by the presence of the C=C bond [0.5] and different groups present on each carbon of the C=C. [0.5]

2024 HCI C2 H2 Chemistry Prelims / Paper 2 4 (b) (i) The pH of the buffer is at 6.5 which is higher than pKa,1. Hence the first proton would have completely dissociated at this point. [1] (b) (ii) Ka,2 = [–O2CCHCHCO2–][H+] [HO2CCHCHCO2−] 10−6.22 = (10−6.5) [–O2CCHCHCO2–] [HO2CCHCHCO2−] [–O2CCHCHCO2–] [HO2CCHCHCO2−] = 1.90546  1.91 (3 s.f.) (b) (iii) –O2CCHCHCO2– + H+ → HO2CCHCHCO2– [1] HO2CCHCHCO2− + OH– → –O2CCHCHCO2– [1] (b) (iv) The ratio calculated in (b)(iii) suggests higher concentration of the conjugate base (–O2CCHCHCO2–), hence the buffer removes acids more effectively. [1] (c) (i) s = 6.70 × 10−3 128.1 = 5.230 x 10–5 [1] Ksp = [Ca2+][C2O42–] = s2 = 2.735 x 10–9  2.74 x 10–9 mol2 dm–6 [1] (c) (ii) Ksp = [Ca2+][C2O42–] = [C2O42–](2.20 x 10–3) = 2.735 x 10–9 [C2O42–] = 2.735 x 10–9 ÷ (2.20 x 10–3) = 1.24 x 10–6 mol dm–3 [1] (c) (iii) • Drink more water to reduce the concentration of Ca 2+ and ethanedioate ions in your body. Position of equilibrium 1 will shift to the left and less solid is formed. • Reduce intake of foods high in ethanedioate ions (chocolate, spinach etc.) or Ca2+ions (e.g. milk, dairy product). Position of equilibrium 1 will shift to the left and less solid is formed. • Increase the intake of acidic foods will lower the concentration of ethanedioate ions due to the formation of H 2C2O4. Position of equilibrium 1 will shift left and less solid is formed. (d) (i) step 1: one mole equivalent or limited LiAlH4 (in dry ether) [1] step 2: excess concentrated H2SO4, heat [1] step 3: CH3OH, (a few drops of) conc H2SO4, heat [1] (d) (ii) X: 5 (a) (i) HCl does not decompose even on strong heating. [0.5] Strong heating of HBr yields brown fumes of Br2. [0.5] Violet fumes of I2 are obtained when a red-hot rod is plunged into a jar of HI. [0.5]

3 The thermal stability decreases from HCl to HBr to HI. [0.5] As down Group 17, the H −X bond length increases and bond s