TJC H2 CHEM P3 QN&ANS Prelim

1 9647/03/TJC Prelim 2014 [Turn Over CHEMISTRY (H2) 9647/03 Paper 3 Free Response Tuesday 16th September 2014 2 hours Candidates answer on separate paper. Additional materials: Answer paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name , Civics Group and Index number in the spaces provided on the cover page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers. The number of marks is g iven in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. This question paper consists of 15 printed pages.

2 9647/03/TJC Prelim 2014 [Turn Over Answer any four questions. 1 (a) The table below shows the melt ing point s of aluminium fluoride, aluminium chloride and silicon tetrachloride. Compound Melting point/ °C Aluminium fluoride 1290 Aluminium chloride 190 Silicon tetrachloride -90 (i) Explain the difference in their melting points in terms of their structure and bonding.  AlF3 has a giant ionic structure with strong ionic bonds present between Al3+ and F-. Large amount of energy is required to overcome the strong ionic bonds.  SiCl4 and aluminium chloride have simple molecular structure . Less energy is required to overcome the van der Waals’ forces present between the molecules.  AlCl3 dimerises to form Al2Cl6. Since A l2Cl6 has a larger and more polarisable electron cloud, stronger and more extensive van der Waals’ forces of attraction exist between the molecules. (ii) With the aid of equations, explain the reactions that take place when silicon tetrachloride and aluminium chloride are mixed with excess water.  SiCl4 undergoes hydrolysis with water using the energetically accessible vacant 3d orbitals in Si.  SiCl4(s) + 2H2O(l) → SiO2(s) + 4HCl (g) Or SiCl4(s) + 4H2O(l) → SiO2.2H2O(s) + 4HCl (g) (Note: do not accept Si(OH)4 as product)  AlCl3 undergoes hydration and hydrolysis . The high charge density of aluminium ion polarises water mole cules, weakening the O -H bond and causing it to break.  AlCl3(s) + aq → Al3+(aq) + 3Cl–(aq) [Al(H2O)6]3+ ⇌ Al(H2O)5(OH)2+ + H+ [7] (b) Silver chloride is sparingly soluble in water. AgCl(s) ⇌ Ag+(aq) + Cl–(aq) (i) Briefly describe any difference in t he solubility of silver chloride in water and in aqueous aluminium chloride.  AgCl will be less soluble in aqueous AlCl3 due to common ion effect or higher concentration of chloride ions in the solution. By Le Chatelier’s Principle, position of equilibrium shifts to the left.

3 9647/03/TJC Prelim 2014 [Turn Over (ii) Given that the solubility product of silver chloride is 1 x 10 -10 mol2 dm-6, calculate the maximum [Ag+(aq)] before a precipitate starts to form in 0.0500 mol dm -3 of aqueous aluminium chloride.  [Cl-] = 0.150 mol dm-3 Ionic product = Ksp  [Ag+][Cl-] = 1 x 10-10 [Ag+] = 1 x 10-10 / 0.150 = 6.67 x 10-10 mol dm-3 Silver ions form complexes with ammonia and with amines. Ag+(aq) + 2RNH2(aq) ⇌ [Ag(RNH2)2]+(aq) (iii) Write a KC expression for this reaction and state its units.  KC = [Ag(RNH2)2 +]/[Ag+][RNH2]2 mol–2 dm6 (iv) KC has a numerical value of 1.7 x 10 7 when R = H. Using your expression for K C, calculate the [NH3(aq)] needed to change the [Ag+(aq)] in a 0.010 mol dm–3 solution of silver nitrate to the value that you calculated in (b)(ii). Since the final [Ag+(aq)] is very low, we can assume that most of the Ag+(aq) has gone to the complex and the change in [NH3] is minimal ie [Ag+(aq)] = 6.67 x 10-10 mol dm-3 [Ag(NH3)2 +] = 0.01 mol dm-3  [NH3] = √{[Ag(NH3)2 +]/(Kc[Ag+])} = √{0.01/(1.7 × 107 × 6.67 x 10-10)}  = 0.939 mol dm–3 Other acceptable working: Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2]+(aq) Initial conc/mol dm-3 0.010 x 0 Change/mol dm-3 -0.00999 -2(0.00999) +0.00999 Equilibrium/mol dm-3 6.67x10-10 x – 0.0199 0.00999 = x – 0.020 = 0.01 [NH3] = √{[Ag(NH3)2 +]/(Kc[Ag+])} (x - 0.020) = √{0.01/(1.7 × 107 × 6.67 x 10-10)} x = 0.939 + 0.020 = 0.959 mol dm-3 (v) Explain whether you would expect the K C for the reaction where R = C 2H5 to be greater or less than that for the reaction where R = H. [8]  When R = C2H5, KC is likely to be greater.  The electron releasing ethyl group will cause the lone pair on N to be more available for dative bonding to Ag+ to form the complex.

4 9647/03/TJC Prelim 2014 [Turn Over (c) Compound A is a chloroalkane with the molecul ar formula C 4H9Cl. Compound A rotates plane of polarised light. When compound A is reacted with aqueous sodium hydroxide, an optically inactive product is obtained. (i) Deduce the identity of compound A.  Since compound A rotates the plane of polaris ed light, it has a chiral centre and has no plane of symmetry in the molecule.  Compound A is 2-chlorobutane. (ii) Deduce the type of reaction that has taken place when compound A is reacted with aqueous sodium hydroxide.  A racemic mixture is obtaine d when compound A is reacted with aqueous sodium hydroxide, indicating that there’s an equal probability of nucleophile attacking from top or bottom of the carbocation. Hence compound A has undergone SN1 reaction. (iii) With the aid of Data Booklet, exp lain how the rate of this reaction changes, if any, when the chloroalkane is replaced by an iodoalkane. [5]  Bond energies: C – I = 240 kJ mol-1, C – Cl = 340 kJ mol-1  The C-I bond is weaker, so the rate determining step of nucleophilic substitution reaction i.e. breaking of C – I bond, will be faster for the iodoalkane. The rate of reaction will be faster. [Accept if student quote atomic radius of I (0.133nm) and Cl (0.099nm) and explains that orbital overlap will be less effective for C – I and so weaker.] [Total: 20] 2 This question is about magnesium and its applications. (a) Magnesium does not occur naturally in its elemental form. One method to produce elemental magnesium involves the thermal reduction of magnesium oxide. (i) Predict, with reasoning, whether the melting point of magnesium oxide is higher or lower than that of the Group II oxides further down the group.  MgO has a higher melting point as compared to the Group II oxides down the group.  The strength of ionic bonds is indicated by the lattice energy of the Group II oxides. Down the group, the ionic radii of the Group II cations increase while the charge remains constant. As lattice energy - - rr qq     , lattice energy for the Group II oxides down the group is less exothermic, hence less energy is required to break the ionic bonds in the lattice.

5 9647/03/TJC Prelim 2014 [Turn Over Magnesium oxide can be formed the decomposition