YJC H2 Chem ANS Prelim

1 H2 Chemistry 9746 Preliminary Examinations 2009 Suggested Solutions Paper 1: Multiple Choice Questions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 A D B C A D C D C C Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 D D C D B C A D B D Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 B A D B D C D A D A Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 B A B D A B C C B D Paper 2: Structured Questions 1 (a) (i) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (ii) Screening effect on the outermost electron of Cu is more or less the same as that in K, as the electrons in the d-subshell do not contribute significantly to screening effect As nuclear charge of Cu > K, the attraction of the outermost electron by the nucleus of Cu is much stronger than in K ( atomic radius of Cu is smaller than K) (iii) As atomic radius of Cu < K, the copper atoms can pack close together / there will be more copper atoms per unit volume / copper would occupy a smaller volume for the same number of particles As mass of one atom of Cu > mass of one atom of K, hence density (mass / volume) of copper is higher than K. (b) (i) HCl HBr HI Bond Energy (kJmol−1) 431 336 229 (awarded as long as candidate mentions at least two of the bond energies in their answer) ease of breaking the H−X bond: HCl < HBr < HI thermal stability: HCl > HBr > HI (ii) Hydrogen bonding exist between HF molecules (c) 1st I.E.: He = 2370 kJ mol1; Ne = 2080 kJ mol1 Potential difference required = 0 . 242370 2080 = 21.1 V [Total: 11] 9746 / YJC / 2009 / Preliminary Examination / Suggested Solutions

2 2 (a) (i) (ii) CC OO O O 2- 120o trigonal planar with respect to both carbon bond angle = 120 (b) (i) The experiment should be conducted in the fume cupboard to prevent inhalation of the toxic CO(g). (Or any feasible precaution, but must mention the toxic nature of CO. Vague reasons, such as CO is a pollutant, are not accepted.) (ii) amount of CaC2O4 = ) 0 . 16 ( 4 ) 0 . 12 ( 2 1 . 40 128 . 0   = 1.0  103 mol mass of B produced = 028 . 0 128 . 0 = 0.100 g Mr of B = 310 0 . 1 100 . 0  = 100 (iii) CaCO3 (iv) CaC2O4  CaCO3 + CO (v) Any temperature well above 400 C ionic radius of Ba 2+ > ionic radius of Ca2+ or charge density of Ba 2+ < charge density of Ca2+ Ba 2+ ion is less able to polarise / distort electron cloud of the anion. [Total: 12] 3 (a) (i) A is: HC C C HHH H H B is: HC C C H H H OH H H H (ii) Step II: H3PO4, 300 C, 70 atm Step III: KMnO4 / K2Cr2O4, H2SO4(aq), heat / reflux (b) (i) HCN + NaOH  Na+ + CN− + H2O 9746 / YJC / 2009 / Preliminary Examination / Suggested Solutions

3 (ii) (iii) H2SO4(aq) / HCl(aq), heat (c) (i) 335 10 43 . 1 100 . 0 10 05 . 2] [        moldmc K Ha 84 . 2 ) 10 43 . 1 log(3    pH (ii) HAHA NaOHNaOH HA NaOH V c V c n n   1  3 3 0100 . 0200 . 0 10 0 . 20 100 . 0dmc V cV NaOH HAHA NaOH    = 10.0 cm3 (iii) A− + H2O  HA + OH−, pH = 8.5 (any value from 8 to 9) (iv) ) 10 05 . 2 log(5   apK pH = 4.7 (v) pH volume of NaOH added/cm3 (vi) Phenolphthalein. End point pH lies within the pH range of the indicator [Total: 15] C H3 C O CH3 CN 9746 / YJC / 2009 / Preliminary Examination / Suggested Solutions slow C H3 C OH CH3 CN fastC3 CH O CH3 CN + H−CN + CN− 7 0 10 5 4.7 + + −:CN H C3 − CO C H3

4 4 (a) (i) condensation reaction (ii) S will be negative. Forward reaction is accompained by a decrease in number of moles of particles. (b) (i) 1 For nucleophilic substitution of Cl by OH to take place. 2 To prevent the formation of Ag2O (through the reaction of Ag+ and OH−)) (ii) AgCl / silver chloride (iii) AgCl + 2NH3  [Ag(NH3)2]+Cl (must be balanced) (iv) D can be used to detect trichloroethanal. Silver mirror / grey ppt (produced on warming) [Total: 9] 5 (a) (i) OH OH 9746 / YJC / 2009 / Preliminary Examination / Suggested Solutions O H C CH2NHCH3 O H H O H CCH3NHCH2 H O H (ii) O H O H O H C C H H NHCH3 NHCH3 O H C C H H trans−isomer cis −isomer (iii) pKb of benzedrin > pKb of adrenaline the lone pair of electrons on the N atom in adrenaline is more available for donation to H + since it is attached to two electron donating alkyl groups. (b) (i) Bronsted base or base (ii) Gas F: carbon dioxide or CO2 Reagent: water or H2O (iii) 1: CH3NH2 2: C6H5NH2 (iv) Br2(aq) H decolourises Br2(aq) and forms a white ppt and no decolourisation for G. [Total: 13]

5 9746 / YJC / 2009 / Preliminary Examination / Suggested Solutions Paper 3: Free Response Questions 1 (a) Element 1st IE / kJ mol−1 2nd IE / kJ mol−1 3rd IE / kJ mol−1 4th IE / kJ mol−1 Ca 590 1150 4940 6480 V 648 1370 2870 4600 For element such as calcium, there is a significant difference between the 2nd and the 3rd IE (3790 kJmol−1). Hence after losing 2 electrons, further ionisation requires a large energy input. For transition metal such as vanadium, the 4s and 3d subshells have close proximity of the energy levels or the 3s or the 4s orbitals have almost the same energy. Hence vanadium can lose different number of 3d electrons in addition to the 4s electrons when forming stable compounds. (b) (i) (ii) Stable oxidation state = +5 V3+ →VO2+ Eo cell = (+0.98) − ( − 0.50) = +1.48 V, reaction is feasible VO2+ → VO2 + Eo cell = (+0.98) − (+0.16) = +0.82 V, reaction is feasible (iii) Electrode potential values depend on the ligands present in the complex (c) V5+ has very high charge density and hence it will polarise the water molecules and weakens the O−H bonds in the water molecules As a result, water molecules donates protons to form [VO3(H2O)3]− or simply VO3 −. 1.00 mol dm-3 High resistance voltmeter V H2 at 1atm, 25oC VO2+(aq), V3+(aq) and H+(aq) at 1 mol dm_3 Pt Salt Bridge

6 (d) (i) CC H H N N O- O- O- O- CH2 CH2C H H CH 2 CH2 C C O OO C O (ii) Since Kstab [Cd(edta)]2− >> Kstab [Ca(edta)]2− Hence edta can remove the poisonous Cd2+ ions readily through the formation of the more stable [Cd(edta)]2− complex. Due to the close proximity (small difference in magnitude) in Kstab between [Cd(edta)]2− and [Zn(edta)]2−, Zn2+ will also be removed in addition to Cd2+. The unnecessary removal of metal ion, such as Zn2+, can be overcome by taking pills of Zn2+ supplement. Or Use another ligand that has a greater difference between the Kstab of Zn and Cd. (e) (i) Ksp = [Mg2+(aq)] [OH−(aq)]2 : units: mol3dm−9 (ii) [Mg2+(aq)] in the saturated solution = 343 11 10 71 . 14 10 00 . 2     dm mol (iii) [Mg2+(aq)] extracted = 0.0540 − 1.71 x 10−4 = 0.0538 moldm−3 Maximum % of Mg extracted = % 6 . 99 1000540 . 0 0538 . 0   [Total: 20] 2 (a) (i) Mr values are: C3H8O = 122.075, C7H6O2 = 122.038, C9H14= 122.112 Y is C8H10O (its relative molecular mass is closest to mass spectrum value) (ii) Reagent: C2H5OH or ethanol Conditions: heat under reflux with concentrated sulfuric acid (iii) Any two of the following:  a large hydrocarbon chain capable of adopting a ring like a hexagonal ring  a short side-chain containing an electronegative atom such as oxygen  a −CH2O− side chain (iv) CH2CH2OH 9746 / YJC / 2009 / Preliminary Examination / Suggested Solutions

7 (b) (i) 1. Br 2, AlBr3 / FeBr3 / Fe fillings 2. KMnO 4, H2SO4(aq), heat 3. PC l3 / PCl5 / SOCl2, (room temperature) (ii) 1. HCN, KCN / trace of