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 liquid. As bond energy data is for gaseous species only, the enthalpy change of reaction calculated in (b)(i) did not account for the enthalpy change of vaporisation of methanol. OR Bond energy values are average values. (iii) At high pressure, gas molecules are compressed closer together, and hence, have significant intermolecular forces of attraction. The smaller volume occupied by the gas also means that the actual volume occupied by the molecules becomes more significant in comparison to the volume of the container. (iv) Heterogeneous catalysis The presence of partially filled d orbitals in transition metals and their compounds allows for the ready exchange of electrons to and from reactants, thus facilitating the formation of weak bonds with the reactants. In addition, the increase in surface concentration of the reactants increases the rate of reaction.

Raffles Institution 2012 Pg 3 2(c) (i) Singlet form: 109.5º < θ1< 120º (any one value) Triplet form: 120º < θ2< 180º (any one value) (ii) The triplet form is more stable as electrons tend to occupy orbitals singly first so as to minimise inter-electronic repulsion. (iii) nucleophilic addition (iv) Air could be fed directly into the reaction system as a source of CO2. 3(a) (i) The lone pair of electrons on nitrogen can be delocalised into the two benzene rings, making the lone pair unavailable to form a dative covalent bond with a H + ion. (ii) [H+] = 4.010 0.10  = 3.162 x 10–3 mol dm–3 pH = – lg (3.162 x 10–3) = 2.50 (iii) A buffer solution is one which is able to resist a change in pH (i.e. maintain an almost constant pH) upon the addition of a small amount of acid or base. (iv) Let HA represent diclofenac. HA + NaOH  H2O + NaA NaA  Na+ + A– Amount of A– = [400 / (400 + 600)] x 0.0500 = 0.0200 mol Amount of HA unused = [600 / (400 + 600)] x 0.10 − 0.0200 = 0.0400 mol 3 a 3 0.0200 [A ] (400 600) 10pH pK lg 4.0 lg 3.70 0.0400[HA] (400 600) 10         (v) HA + NaOH  H2O + NaA Amount of HA consumed = 0.0400 − 0.005 = 0.0350 mol Amount of A– produced = 0.0200 + 0.005 = 0.0250 mol 3 a 3 0.0250 [A ] (400 600) 10pH pK lg 4.0 lg 3.85 0.0350[HA] (400 600) 10         Change in pH = 3.85 − 3.70 = 0.15 3(b) (i) The quaternary structure of proteins refers to the spatial arrangement and association of polypeptide subunits. Haemoglobin consists of two  subunits and two  subunits (or 4 subunits). The subunits in haemoglobin interact with one another via ionic interactions, hydrogen bonds, disulfide bonds and van der Waals’ interactions.

Raffles Institution 2012 Pg 4 3(b) (ii) Hb(H2O)4 + 4O2 Hb(O2)4 + 4H2O oxyhaemoglobin One haemoglobin molecule contains 4 subunits. Each subunit can bind to one oxygen molecule. The oxygen molecule acts as a ligand and binds to the Fe 2+ centre reversibly (or use “ ” to show reversibility) through dative covalent bonding. (iii) CO is a stronger ligand than O 2. Hence, CO displaces O 2 in oxyhaemoglobin almost irreversibly to form a more stable complex, carboxyhaemoglobin (HbCO). Consequently, this ligand exchange reaction cuts down the supply of oxygen to the body. Allow the patient to inhale air which has a high concentration of O 2. 3(c) (i) 2 3 c 2 22 [SO ] [SO ] [O ]K  ; units: mol–1 dm3 (ii) 2NaOH + H2SO4  Na2SO4 + 2H2O Amount of H+ in 25.0 cm3 of solution = Amount of OH = 5.00 x (38.5 / 1000) = 0.1925 mol Amount of SO 3, y = amount of H2SO4 = ½ x amount of H+ in 250 cm3 solution = ½ x 0.1925 x (250 / 25.0) = 0.9625 mol 2SO2(g) + O2(g) 2SO3(g) Initial amount/ mol 2 1 0 Change in amount/ mol y ½ y + y Equilibrium amount/ mol 2 y 1 ½ y y Amount of SO 2 = 2 − y = 2 − 0.9625 = 1.0375 = 1.04 mol Amount of O2 = 1 − (y/2) = 1 – (0.9625 / 2) = 0.51875 = 0.519 mol Amount of SO3= y = 0.9625 = 0.963 mol 2 2 3 c 22 22 0.9625 [SO ] 3.60 5.97[SO ] [O ] 1.0375 0.51875 3.60 3.60 K            mol–1 dm3 4(a) Phenylalanine exists predominantly as zwitterions in water. However, due to the close proximity of the two oppositely-charged groups, hydration is not efficient. Together with the hydrophobic phenyl group, phenylalanine is barely soluble in water. In aqueous acids or alkalis, phenylalanine is able to form species which have a non-zero net charge. These charged species form strong ion-dipole interactions with water, so they have higher solubility in acids and alkalis. 4(b) (i) The carboxylic acid group that is nearer to the amino group is more acidic. The conjugate base of that carboxylic acid group is more stabilised due to the negative charge on O being more dispersed by the neighbou