NJC Prelim CHEM P2 Solutions

2015 NJC SH2 H2 Chemistry Prelim Paper 2 Solution with Examiners’ Comments 2015 SH2 H2 Chemistry Prelim Paper 2 Suggested Answers with Examiners’ Comments 1(a) Procedural format Divide the unknown into 5 separate portions - Warm a portion with Fehling’s solution: if brick red ppt forms, unknown is ethanal. - If no ppt, warm another portion with Tollen’s reagent: if silver mirror forms, unknown is benzaldehyde. - If no silver mirror, warm another portion of sample with aq I2 + NaOH(aq): if yellow ppt forms; unknown is 1-iodopentan-2-one. - If no yellow ppt, warm another portion with NaOH(aq) and test gas with moist red litmus paper: if litmus paper turns blue, unknown is propanamide. - If no alkaline gas, add Br2(aq) to last portion: if orange Br2(aq) decolourises & white ppt forms, unknown is phenylamine. Another example of plan: ethanal benzaldehyde unknown potassium manganate(VII)/ sulfuric acid; heat ethanal or benzaldehyde propanamide / 1-iodopentan-3-one / phenylamine purple solution decolourises purple MnO4 – does not decolourise To new sample, aqueous I2/ sodium hydroxide; warm yellow ppt ( CHI3) forms no yellow ppt 1-iodopentan-3-one propanamide or phenylamine To new sample, NaOH(aq); heat yellow ppt ( AgI) forms no yellow ppt Moist red litmus paper turns blue propanamide phenylamine Br2(aq) Orange solution decolourises; white ppt forms To new sample, NaOH(aq); heat; then acidify with aq HNO3; aq AgNO3

2015 NJC SH2 H2 Chemistry Prelim Paper 2 Solution with Examiners’ Comments 1(bi) N NN H NC C H H HH OO O O 1(bii) 1. Suggest suitable quantities of ethanal and 2,4 -DNPH to prepare the solid derivative (hydrazone). 2. Filter solid derivative and then dissolve in minimum volume of hot solvent (non-polar or organic) … use fluted filter paper / pre -heated funnel / Buchner funnel to remove insoluble impurities 3. Cool solution to form crystals …Then filter purified crystals using Buchner funnel. 4. Wash crystals with minimum cold solvent and dry the solid (in desiccator / press between filter papers / low temperature oven). 5. Find melting point of pure and dry crystals (m.p 165 oC). 1(c) As organic compounds are flammable, there should be NO Bunsen flame or naked flame. / Heat or warm reaction mixture in a water bath when conducting experiments. 2(ai) P: 1s2 2s2 2p6 3s2 3p3 S: 1s2 2s2 2p6 3s2 3p4 2(aii) 1st IE of S is lower than that of P: All 3p electrons in P are unpaired. Two of the 3 p electrons are paired in S with inter- electronic repulsion between the paired electrons . Thus less energy is required to remove the 1st paired electron in S 4th IE of S is lower than that of P: 4th e in S is removed from 3p subshell; while 4th e in P is removed from 3s subshell ; 3p subshell is further from the nucleus (at a higher E level) with weaker nuclear attraction. Thus less energy is required to remove this electron in S. 2(bi) p = 23 2 53 )10 x (0.50 (0.687)(1) )5.68x1010 x (0.50 298)(1)(8.31)(    = 2.84 x 106 Pa 2(bii) Using the ideal gas equation, p = nRT/V = )(0.5x10 298)(1)(8.31)( 3 = 4.95 x 106 Pa or

2015 NJC SH2 H2 Chemistry Prelim Paper 2 Solution with Examiners’ Comments pideal = p + 2 2 V an = 2.84 x 106 + 2-3)10 x (0.50 0.687 x 1 = 2.84 x 106 + 2.748 x 106 = 5.59 x 106 Pa 2(biii) Actual pressure exerted by SO 2 is lower than that calculated from ideal gas equation. This is due to significant intermolecular forces of attraction between polar SO 2 and thus less forceful collisions against the walls of container. 2(ci) L Metallic bonding M Ionic bonding N Covalent bonding 2(cii) I: average electronegativity for Ge and O = 2 3.612.01 = 2.81 (accept 2.60 – 3.00) difference in electronegativity for Ge and O = 3.61 – 2.01 = 1.60 (accept 1.40 – 1.80) each + correctly determined and plotted the coordinates for GeO2 The nature of the oxide of germanium is acidic. II: Oxide of germanium has a lower melting point ( giant covalent compound) Ge is below Si in Group IV with a bigger atomic size; overlap of the atomic orbitals of Ge and O will be less effective than those of Si and O; the covalent bonds between Ge and O are weaker. 3(a) Squaric acid dissociates fully in water to form ions which form ion -dipole interactions with water molecules. Energy released in the fo rmation of these interactions can compensate the energy required to break the intermolecular hydrogen bonds in squaric acid and in water. 3(bi) O O OH OH + 2 NaOH O O O-Na+ O-Na+ 2 H2O+ OR C4H2O4 + 2NaOH → Na2C4O4 + 2H2O

2015 NJC SH2 H2 Chemistry Prelim Paper 2 Solution with Examiners’ Comments 3(bii) Amt of excess OH− in 25.0 cm3 = 29.7 1000 x 0.100 = 2.97 x 10−3 mol Amt of excess OH− in 100 cm3 = 2.97 x 10−3 mol x 100 25.0 = 1.188 x 10−2 mol Amt of OH− that reacted with squaric acid in sample = 100 1000 x 0.250 − 1.188 x 10−2 = 1.312 x 10−2 mol Amt of squaric acid = 1.312 x 10−2 /2 = 6.56 x10−3 mol Mass of pure squaric acid = 6.56 x10−3 x 114 = 0.7478 g % purity by mass = 0.7478 4 x 100 = 18.7% 3(c) The dianion can undergo resonance, whereby the lone pair of electrons on O− and  electrons of C=C and C=O can delocalise between the p orbitals of all the C and O to give a symmetrical resonance hybrid with identical C−C bond lengths. OR The lone pair of electrons on O − can delocalise into the  electrons clouds of C=C and C=O to give a symmetrical resonance hybrid with identical C−C bond lengths. 4(ai) Hg(CNO)2 (s) → Hg(l) + N2 (g) + 2CO(g) 4(aii) ΔHr ɵ = Σ mΔHf ɵ(products) − Σ nΔHf ɵ(reactants) = 2 x ΔHf ɵ(CO) − ΔHf ɵ(Hg(CNO)2) = 2 x (−111) − (+386) = −608 kJ mol−1 4(aiii) ΔSɵ is positive as there is an increase in the number of gaseous molecules in the process. ΔGɵ = ΔHɵ − TΔSɵ, since ΔHɵ < 0 and ΔSɵ > 0, ΔGɵ is always < 0; decomposition of Hg(CNO)2 is energetically feasible at all temperatures.

2015 NJC SH2 H2 Chemistry Prelim Paper 2 Solution with Examiners’ Comments 4(bi) By Hess’s law, L.E. = −(+ 64 + ½ (496) + 1007 + 1810 + (−142) + 844) − 91 = −3922 kJ mol−1 = −3920 kJ mol−1 4(bii) Hg2+, same for both compounds, −2 charge of O2− is higher in magnitude than −1 of F−, but r− of O2− bigger than r− of F− (both are isoelectronic, but F has a higher NC) The effect of increase in charge product outweighs that of interionic distance, since |LE| ∝ | q+q− r++ r− |, the magnitude of lattice energy of HgO is likely greater than HgF2. 4(ci) I: acid base reaction or neutralisation N N N O-Na+ O-Na+Na+ -O Points to take note in Born-Haber cycle:  energy of 0 kJ mol−1 for elements in natural state  directions of arrows must correspond to exothermic or endothermic reaction  correctly labelled ΔH symbols or values  balanced equations with state symbols for each stage  formation of cation before anion Hg (l) + ½ O2 (g) Hg2+ (g) + O (g) + 2e +1007 + 1810 Hg2+ (g) + O2− (g) −142+844 HgO (s) −91 Hg (g) + O (g) + 64 + ½ (496) ∆Hlat ɵ / L.E. Energy / kJ mol−1 0

2015 NJC SH2 H2 Chemistry Prelim Paper 2 Solution with Examiners’ Comments II: condensation N N N O O OCH3 O O CH3 CH3 O 4(cii) C=O and N−H bond HN N H NH O OO 5(ai) Ni2+ + 2OH  Ni(OH)2 5(aii) [Ni(NH3)6]2+ Ni(OH)2(s) + aq Ni2+(aq) + 2OH−(aq) [Ni(H2O)6]2+ + 6NH3 [Ni(NH3)6]2+ + 6H2O 5(aiii) Ligand exchange reaction 5(bi) [R] Cu2+ + 2e  Cu Eo red (Cu2+/Cu) = +0.34 V [O] 2S2O3 2  S4O6 2 + 2e Eo red (S4O6 2/ S2O3 2) = +0.09 V Overall equation: Cu2+ + 2S2O3 2  S4O6 2 + Cu Eo cell = +0.25 V > 0 The redox reaction is energetically feasible; blue CuSO4 solution would decolourise and a pink solid of Cu is formed. Alternative accepted answer (in fact, Cu+ is unstable in aq medium) Overall equation: 2Cu2+ + 2S2O3 2  S4O6