NYJC H2 CHEM P3 ANS Prelim

1 2011 NYJC Prelim H2 Chemistry 9647/03 Answers 1 (a) (i) I. ½ (CN) 2 + H+ + e HCN E = +0.37 V Cl2 + 2e 2Cl E  = +1.36V E  cell = +1.36 – (+0.37) = +0.99V Overall Equation: Cl 2 + 2HCN (CN)2 + 2H+ + 2Cl-   II. ½ (CN)2 + H+ + e HCN E  = +0.37 V S 4O6 2- + 2e 2S2O3 2- E = +0.09 V E  cell = +0.37 – (+0.09) = +0.28 V Overall Equation: (CN) 2 + 2H+ + S2O3 2-  2 HCN + S4O6 2- III. ½ (CN)2 + H+ + e HCN E  = +0.37 V Cr2O7 2- + 14H+ + 6e 2Cr3+ + 7H2O E = +1.33 V Both are oxidising agents. Hence reaction cannot proceed. (ii) C C N N xx x x x x x xx linear (b) (i) This is because the reaction proceeds with an inversion of configuration. C H (CH2)4CH3 CH3 Cl C H (CH2)4CH3 CH3 NC + Cl -- NC +  To illustrate the inversion in structural formula (3D wedge diagram) (ii) Isomer A: CH3 C (CH2)3CH3 Cl CH2CH3 2

2 Type of reaction: SN1 CH3 C (CH2)3CH3 Cl CH2CH3 + CN - CH3 C + (CH2)3CH3 CH2CH3 + Cl -slow CH3 C + (CH2)3CH3 CH2CH3 CH3 C (CH2)3CH3 CH2CH3 CN   (iii)  Reaction b(i) OR S N2 mechanism is faster.  The rate determining step in b(i) involves NaCN. OR rate = k[NaCN][RX] (c) OH OH O O OH B C K2Cr2O7 , H2SO4 (aq) Heat under reflux HCN NaCN Na in ethanol D heat with concentrated H2SO4 (does not give CO2(g) with Na2CO3) O OH OH CN O OH OH CH2NH2CH3 CH2NH3 + CH3 O O

3 2(a)(i) (There is a plane of symmetry in the molecule or both chiral centres contain the same groups attached to it), the two chiral centres rotate the plane of polarised light to the same extent but in the opposite direction hence cancelling out the optical activity. (ii) Compound A: CH2 CH CH 2 CH2 CH2CH3 CH3 CH3 Compound C: CH3 CC H C H 2 CH3 CH3 CH3 CH3 Compound D: CH3 CC C H 3 CH3 CH3 CH3 CH3 (iii) Isomers A to D are simple molecular compounds with weak van der Waals forces between molecules. However, as we move down the table from A to D, the molecules become increasingly branched, hence they have less surface area of contact with their neighbouring molecules resulting in weaker vdw forces between molecules. Hence less energy is required to break the vdw bonds, resulting in lower b.p. down the table. (iv) 4 molecules + ratio Structural formula mole ratio CH2 CH CH CH 2 CH3CH2 CH3 CH3 Cl 6 3 CH CH CH CH 2 CH3CH3 CH3 CH3Cl 4 2 CH2 CC H C H 2 CH3CH3 CH3 CH3 Cl 2 1

4 CH2 CH CH CH 2 CH3CH3 CH2 CH3 Cl 6 3 (v) CH2 CH C CH 2 CH3CH3 CH3 CH3 CH2CCHCH2CH3 CH3 CH3CH3 OR CHCHCHCH2CH3 CH3 CH3CH3 CH2 CH CH CH CH 3CH3 CH3 CH3 [1st one preferred, because more branched.] (but not CH2 CH CH CH 2 CH2CH3 CH3 CH3 CH2CHCHCH2CH3 CH2 CH3CH3 or any other combination that produces less than 6 branches.) (b) pH = 7.4; pOH = 14-7.4 = 6.6; [OH -] = 10-6.6 = 2.512 × 10-7 moldm-3 K sp of Pb(OH)2 = [Pb2+(aq)][OH-(aq)]2 = 1.43 × 10-20; [Pb 2+(aq)]total = (1.43 × 10-20) / (2.512 × 10-7)2 = 2.26 x 10-7 moldm-3 K sp of PbSO4 = [Pb2+(aq)][SO4 2-(aq)] = 2.53 × 10-8; [SO 4 2-(aq)]total = 2.53 × 10-8/ 2.26 x 10-7 = 0.1116 moldm-3 K sp of BaSO4 = [Ba2+(aq)][SO4 2-(aq)] = 1.08×10-10; [Ba 2+(aq)]total = 1.08×10-10/0.1116 = 9.67 × 10-10 moldm-3 Therefore the barium ion concentration is within safety limits since [Ba2+] is < 1.45 × 10-5 moldm-3 (c)(i) Down the group, the Increase in screening effect and atomic radius outweigh the increase in nuclear charge. Hence, net attraction of outermost electron decreases. Less energy required to remove the valence electron.

5 (ii) Lithium and sodium metals have high boiling points due to their stronger metallic bonding hence can be obtained via electrolysis and (isolated) even in a high temperature environment. [1] Potassium, rubidium and caesium having larger atomic radius, have weaker metallic bonding and vaporises in the high temperature environment. (iii) Lithium ion behaves like Mg 2+. (diagonal relationship). It has a high charge density and is able to polarise the oxygen molecule breaking the O=O double bond[1]. As we come down the group, the other group one ions have larger ionic radii and hence have a lower charge density is unable to polarise the molecule completely, hence giving O 2 2- and O2 - ions. 3(a) N2O3(g) + N2O5(s) 2N2O4(g) 4NO2(g) NO(g) + NO2(g) + O2(g) + N2O3(g) N2O3(g) + N2O5(g) -114.2 2NO(g) + 2NO2(g) +O2(g) -57.2(2) -39.8 -112.5 -54.1 Hrxn ∆Hrxn = ─(─ 39.8) ─(─ 54.1) + 2x(─ 57.2) ─ (─112.5) + (─ 114.2) = ─22.2 kJ (b)(i) graph is a straight line (passing through zero) ∆

6 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0 200 400 600 800 1000 1200 1400 explanation ; since rate is directly proportional to pressure/concentration of N 2O5(g), the reaction is first order wrt N2O5(g) (ii) rate constant = gradient = (any working) = 9 x 10 ─4 sec ─1 (iii) for slow step involving only 1 N 2O5(g) for sum of all steps adding to 2N2O5(g) → 4NO2(g) + O2(g) e.g. I N 2O5(g) → NO2(g) + NO(g) + O2(g) slow II N 2O5(g) +NO(g) → 3NO2(g) fast (c) LiF CsS ZnF SrBr 2 (d) (i) B CO2 C CaCO3 D Ca(HCO3)2 (ii) MO + H 2O → M(OH)2 M(OH) 2 + 2 HCl → MCl2 + 2H2O n HCl for 25.0 cm3 aliquots = 0.0985 x 20.30/1000 = 0.0020 mol n HCl reacting with E = 0.0020 x 250/25 = 0.020 mol nMO = nM(OH)2 = ½ x 0.020 = 0.010 mol Pressure of N2O5(g) remaining /Pa Rate of Reaction / Pa s-1

7 MA = 1.9735 / 0.010 = 197.35 ≈ 197.4 (iii) A is likely to be a carbonate (since CO2 is evolved on heating) Hence formula will be MCO3 Hence Ar of M will be 197.35 ─ 12.0 ─ 3(16.0) = 137.35 M is likely to be Barium. 4(a) Down the group, thermal stability of HX decreases as the H-X bond length increases and it becomes easier to break the H-X bond. (b) The Si-F bond formed is very strong hence the reaction is very exothermic/very stable products are formed. (c)(i) V/T T (ii) Gas X is HF while Gas Y is HCl Work out V/T values. Experiment No. T / K Gas X Gas Y V / dm3 V /T dm3K-1 V / dm3 V /T dm3K-1 1 200 20 000 100 16 500 82.5 2 300 22 500 75 20 000 66.7 3 600 24 000 40 23 500 39.1 The V/T values of Gas X deviates more from a constant value than Gas Y.

8 Hence Gas X behaves less like an ideal gas than Gas Y. Gas X is HF which has stronger hydrogen bonds between molecules while HCl has weaker permanent dipole-permanent dipole interactions. (d)(i) ICl prefers to form stronger permanent dipole-permanent dipole interactions between its polar molecules rather than weak Van der Waal’s forces with non-polar solvent molecules. (ii)   (iii) 2 BrF 3  [BrF2]+ + [BrF4]- or BrF3  [BrF2]+ + F- BrF3 ionizes/forms ions. The mobile ions (in molten state) are able to act as charge carriers. (e)(i) Intermediate X C CH2CH3 CH2 OH Step I: HCN, trace amount of NaOH/NaCN, 10-20oC Step III: LiAlH 4 in dry ether (ii) Step IV. There are 5 isomeric products in all. For reference (not required in ans) C CH2CH3 CH2 OH C CH2CH3 CHOH H C CH CH3 CH2OH H (has cis-trans isomers) (has cis-trans isomers) (iii) (CH 3)2CHMgBr (iv) [NB: Since dimer is planar, iodine should have 4bp and 2lp - square planar]

9 Z C O CH3 CC H 3 O Y C O CH3 OH X CCH3 C CH3 C OH CH3 H H Reaction of CH3MgBr with Aldehdye: Product X contains CH OH H CH3 (deduce from Grignard reaction) Reaction with Na: X contains 1 alcohol group (do not accept phenol or carboxylic acid as these are not possible in C6H12O) Reaction with alkaline aq I2: X contains CH OH H CH3 (do not accept