RI H2 Chem 2012 Prelim P3 Soln

Raffles Institution 2012 Pg 1 Raffles Institution 2012 Year 6 Preliminary Examination H2 Chemistry Paper 3 Suggested Solutions 1(a) (i) The melting point of lithium oxide is higher than that of sodium oxide. As the Li+ ion has a smaller ionic radius than the Na + ion, the Li + ion attracts the O2− ion more strongly (or the lattice energy of Li 2O is more exothermic). Thus, a higher temperature is required to overcome the ionic bonds when lithium oxide is melted. (ii) 6Na 2O + P4O10  4Na3PO4 1(b) (i) Li + + CoO2 + e–  LiCoO2 OR CoO 2 + e–  CoO2  oxidation state of cobalt changes from +4 to +3 (ii) Graphite allows the migration of Li + ions to cobalt oxide. (iii) Lithium reacts with water to form lithium hydroxide. 1(c) Mg reacts with water/steam to produce hydrogen. As hydrogen is (highly) flammable, there will be a risk of explosion. 1(d) Chloride ions are smaller than bromide ions. OR Compared to the electrons of a bromide ion, the electrons of a chloride ion are closer to the nucleus. Compared to an outer shell electron lost by a bromide ion, an outer shell electron lost by a chloride ion is more strongly held by the nucleus. 1(e) (i) A polydentate ligand forms more than one / two or more co-ordinate bonds with the central metal ion. OR A polydentate ligand donates more than one / two or more lone pairs of electrons to the central metal ion. (ii) The number of product particles is greater than the number of reactant particles. OR Disorderliness increases. OR ∆S is positive. −T∆S is negative. ∆G = ∆H − T∆S. As ∆H = 0, ∆G = −T∆S. Thus, ∆G is negative. (iii) 6 (iv) The cyanide ion is strongly bonded to the Co2+ ion (by a co-ordinate bond). 1(f) (i) Q (ii) M

Raffles Institution 2012 Pg 2 (iii) The copper(I) ion, with a co-ordination number of 4, forms a stable complex, Q, with four nitrogen atoms (2 N atoms from L and 2 N atoms from P). This complex allows a molecule of L to be located through the first ring, P, prior to cyclisation with M, to form the two interlocked rings. OR When the four nitrogen atoms form co-ordinate bonds with the Cu + ion to give a stable complex, Q, the two molecules, L and P, are locked in a particular spatial orientation in Q, allowing M to close up the ring afterwards. 2(a) (i) At cathode: CO 2 + 6H+ + 6e–  CH3OH + H2O At anode: 2H 2O  O2 + 4H+ + 4e– Overall: 2CO 2 + 4H2O  2CH3OH + 3O2 (ii) In addition to the electrolysis of water, energy has to be put in to drive the reduction of CO2, a process which is not energetically feasible. (iii) Methanoic acid can be further oxidised by KMnO 4. (iv) CH 3CN + 2H2O + H+  CH3COOH + NH4 + (v) The other organic product is CH3NH3 +. (vi) 2(b) (i) CO 2 + 3H2  CH3OH + H2O H = 2 x 740 + (3 x 436) – (3 x 410) – 360 – (3 x 460) = −182 kJ mol–1 (ii) Methanol is a liqui