NJC Prelim P2 Solutions

1 NJC SH2 Preliminary Examination 9729 / 02 / 18 [Turn Over NJC SH2 H2 Chemistry P2 Solutions 1 The properties of elements and their compounds show similarities, differences and trends depending on the positions of the elements. (a) The elements in the third period, and their compounds, show trends in their physical and chemical properties. A sketch graph of the first ionisation energies of five successive elements in the third period is shown. (i) Sketch on the graph, the position of the ionisation energy of the two elements that come before Mg in this sequence. Cross shown on first vertical line from the y-axis (group 0/Ne) is clearly higher than all shown. Cross shown on second vertical line from the y-axis (group 1/Na) is clearly lower than all shown. [1] (ii) Explain, with reference to electronic arrangements, the decreases in first ionisation energy between Mg and Al and between P and S. Mg and Al: The most loosely held electron in A l is in the higher energy 3p subshell while that of Mg is in the lower energy 3s subshell. This outweighs the effect of the increase in nuclear charge from Mg to Al. Hence nuclear attraction for the most loosely held electron in Al is weaker, i.e. Al has a lower 1st IE. [1] P and S: The most loosely held electron in S is one of the paired electrons in 3p orbital while that of P is in the singly filled 3p orbital. Inter-electronic repulsion between the paired electrons in the same p orbital outweighs the effect of an increase in nuclear charge. Hence, nuclear attraction for the most loosely held electrons is weaker in S, i.e. S has a lower first IE. [1] [2] (b) The chlorides of the elements in the third period behave in different ways when added to water, depending on their structure and bonding. x x

2 NJC SH2 Preliminary Examination 9729 / 02 / 18 [Turn Over L is a chloride of an element in Period 3. A student investigated L and the results are as given below. • L is a white crystalline solid with a melting point of 987 K. • L dissolves in water to form a weakly acidic solution. • Addition of NaOH(aq) to an aqueous solution of L produces a white precipitate, M . (i) Identify L and M. L: MgCl2 M: Mg(OH)2 [1] (ii) Write an equation to illustrate the formation of the weakly acidic solution. [Mg(H 2O)6]2+(aq) + H2O(l) [Mg(H2O)5(OH)]+(aq) + H3O+(aq) [1] (state symbols not necessary) (c) Some reactions based on the Group 2 metal barium, Ba, are shown below. (i) State the reagent needed for each of reactions 1 and 2. Reaction 1: HNO 3 [1] Reaction 2: H2O [1] (ii) Write an equation for the formation of X. 2Ba + O 2 → 2BaO [1] [Total: 8]

3 NJC SH2 Preliminary Examination 9729 / 02 / 18 [Turn Over 2 The use of Data Booklet is relevant to this question. (a) Copper(II) sulfate, an inorganic compound that has wide uses in organic syntheses and in engraving of zinc plates for inta glio printmaking, can undergo a series of reactions as shown below. (i) Identify D, E, F and G. D: [Cu(NH 3)4(H2O)2]2+ or [Cu(NH3)4]2+ E: CuCr 2O7 F: CuI G: Cu (ii) With the aid of relevant equations, account for the formation of the deep blue solution D from the pale blue precipitate. [Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) ---(1) Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH‒(aq) ---(2) Formation of [Cu(NH3)4]2+ complex lowers [Cu2+]. This causes the ionic product of Cu(OH)2 to become less than Ksp of Cu(OH)2, and hence the pale blue Cu(OH)2 precipitate dissolves to form the deep blue solution. OR Formation of [Cu(NH 3)4]2+ complex lowers [Cu2+]. By Le Chatelier’s Principle, the position of equilibrium (2) shifts to the right to partially increase [Cu2+]. Hence the pale blue Cu(OH)2 precipitate dissolves to form the deep blue solution. [2]

4 NJC SH2 Preliminary Examination 9729 / 02 / 18 [Turn Over (b) (i) By quoting relevant data, account for the trend in the thermal stabilities from HCl to HI. BE(H‒Cl) = 431 kJ mol -1 BE(H‒Br) = 366 kJ mol-1 BE(H‒I) = 299 kJ mol-1 Down the group, less energy is needed to break the weaker H‒X bond, resulting in decreasing thermal stability of HX. [2] (ii) Identify a transition metal cation that can be used to differentiate the oxidising abilities of Br 2 and I2. Explain your answer with appropriate workings. [3] The transition metal cation is Fe2+. From Data Booklet E ϴ / V Br2 + 2e ⇌ 2Br‒ +1.07 I2 + 2e ⇌ 2I‒ +0.54 Fe3+ + e ⇌ Fe2+ +0.77 Br2 + 2Fe2+ → 2Br‒ + 2Fe3+ E ϴcell = +0.30 V > 0 Br 2 can oxidise Fe2+ to Fe3+ since the reaction is spontaneous. I2 + 2Fe2+ → 2I‒ + 2Fe3+ E ϴcell = ‒0.23 V < 0 I2 cannot oxidise Fe2+ to Fe3+ since the reaction is non-spontaneous. (iii) Suggest a series of steps to verify the presence of chloride and iodide ions in a mixture, given the following reagents: • Aqueous silver nitrate • Filter paper • Aqueous ammonia • Filter funnel • Aqueous nitric acid 1. Add excess aqueous silver nitrate to the mixture. Yellowish-white precipitate is formed. 2. Next, add excess aqueous ammonia. Some of the precipitate will dissolve and only a yellow precipitate remains. 3. Filter the mixture with filter paper and filter funnel. Yellow residue indicates the presence of iodide ions. 4. To the colourless filtrate, add excess aqueous nitric acid. White precipitate formed indicates the presence of chloride ions. [3] [Total: 14]

5 NJC SH2 Preliminary Examination 9729 / 02 / 18 [Turn Over 3 Capsaicin is an active component of chili peppers. The reaction scheme involving the formation of a derivative of Capsaicin, C17H25NO2, is shown below. Information on compounds A to D are given on pages 6 and 7. A, C 6H10O4 Capsaicin derivative, C17H25NO2 Compounds A to D react with sodium metal. Compounds A and B react with aqueous sodium carbonate. Compound C, Compound D and the capsaicin derivative reacts with aqueous sodiu m hydroxide but does not react with aqueous sodium carbonate. Compounds B and E also react with cold acidified KMnO4. (a) Name the functional group common to compounds A and B. Carboxylic acid [1] 1,4-dichlorobutane Reaction 2 Reaction 4 H2O 2-methylpropanoic acid Reaction 1 O HO B, C10H18O2 Reaction 3 NH2 OH D, C7H9ON O Cl E, C10H17OCl two-step reaction H2O Cl OH C, C7H7OCl

6 NJC SH2 Preliminary Examination 9729 / 02 / 18 [Turn Over (b) Compound A can be synthesised from 1,4-dichlorobutane in two steps. Suggest re