RI Prelim P2 ANS

Suggested solutions to 2009 RIJC H2 Chemistry Prelim Paper 2 1(a)(i) 1s 2 2s 2 2p 5 (a)(ii) (b)(i) Include IE data O: 1s 22s 22p 4 The 1 st IE increases from O to F as the number of protons increases in F but the shielding effect remains effectively constant . (b)(ii) O +: 1s 22s 22p 3 F +: 1s 22s 22p 4 The 2 nd IE decreases from O to F due to inter–electronic repulsion between the 2p electrons in F (c)(i) 2H +(aq) + Mg(OH) 2(s) → Mg 2+ (aq) + 2H2O( l) (c)(ii) Mg 2+ (aq) + 2F– (aq)  → → →→ MgF 2(s) The Mg 2+ released from the acid–base reaction in the stomac h reacts with the F – present to form a precipitate of MgF 2 which is insoluble and is thus not easily absorbed by the body. (c)(iii) Precipitation occurs when [Mg 2+ ][F – ]2 > K sp [Mg 2+ ] ] 0 . 1 / )0.19 10 1[( 3−× 2 > 5.16 × 10 –11 [Mg 2+ ] > 0.0186 mol dm –3 (vol. of milk of magnesia needed)(1.40) = (0.0186) (vol. of liquid in stomach) i.e. vol. of milk of magnesia needed = 0.0133 dm 3 (i.e. 13.3 cm 3) [1] Data 142 mg of F – in 1 tube of toothpaste Stomach when filled with food about 1 L i.e. 1 dm 3 Milk of Magnesia, 8 g Mg(OH) 2 in 100 mL is (8 / 58.3) ÷ 0.1 = 1.37 mol dm –3 Can also accept p x and p y orbital Zero mark if all three orbitals are drawn x y z

2(a) An ideal gas is one which experiences negligible inter molecular forces of attraction and has negligible volume. OR obeys the ideal gas equation (pV = nRT) perfectly. (b)(i) (b)(ii) At lower pressures, NH 3(g) deviates from ideal behaviour to a greater extent since it experiences stronger Intermolecular forces of interaction (hydrogen bond ing) as compared to CO 2 (instantaneous dipole − induced dipole interaction). (c)(i) Lightning OR forest fires OR volcanoes OR other sensible answers . (but NOT human acts e.g. burning of fuels, combust ion engines etc.) (ii) High temperatures are needed to overcome the strong N≡ N. (d) ]][ [ ][ 22 2 ON NO K c = (e) Let the change in [N 2] be x mol dm −3 Eqm concentrations of N 2, O 2 and NO are 1 – x, 1 – x and 2x mol dm –3 respectively. 2 2 2 2 -14 -7 -3 [NO] [N ][O ] [NO] 6.2×10 = (1-x)(1-x) [NO]=2.49×10 mol dm (assuming x 1) cK = << pV ideal gas p 0 ammonia carbon dioxide [½] – correct shape [½] – CO 2 above NH 3

(f)(i) No change since there are equal numbers of gaseous molecules on either side of the equation. (Note: a change in volume does not affect such a g aseous system.) (ii) Position of equilibrium shifts to the right since there is an increase in concentration of N2. By Le Chatelier’s Principle , the system will respond by favouring the forward reaction and hence the position shifts to the right. 2(g) The system can be treated as the following: N2 + O 2 2NO initial conc. / mol dm –3 (1.0 + 2.0) / 1.5 = 2.0 1.0 / 1.5 = 0.666 – eqm conc. / mol dm –3 2.0 – x 0.666 – x 2x 2 2 2 2 -14 -7 -3 [NO] [N ][O ] [NO] 6.2×10 = (2.0-x)(0.666-x) [NO]=2.873×10 mol dm cK = NO NO -7 -3 -6 p V=nRT p =cRT =(2.873×10 10 )(8.31)(2273) =5.43×10 Pa × 3(a)(i) CaCO 3 CaCO 3 + SO 2 → CaSO 3 + CO 2 (ii) Mass of S in coal = 1 x 10 9 x 2.5 100 = 2.5 x 10 7kg = 2.5 x 10 10 g Amt of S = 2.5 x 10 10 ÷ 32.1 = 7.788 x 10 8 mol Amt of CaCO 3 = Amt of S = 7.788 x 10 8 mol Mass of CaCO 3 = 7.788 x 10 8 x 100.1 = 7.80 x 10 10 g (b) Solution F is calcium chloride solution which is effectively neutral . For BeCl 2, it becomes the aqua complex, [Be(H 2O) 4]Cl 2 upon contact with water. The small, highly polarising Be 2+ weakens the O–H bonds of the water molecules in it s surrounding sphere of coordination and results in the release o f H + (or H 3O+) ions in solution , thus giving rise to an acidic solution. BeC l2(s) + 4H 2O(l) → [Be(H 2O) 4]2+ (aq) + 2C l– (aq) [Be(H 2O) 4]2+(aq) + H 2O(l) [Be(OH)(H 2O) 3]+(aq) + H 3O+(aq) (c)(i) CaCO 3 → CaO + CO 2

Amt of CaO = 1.00/100.1 x 56.1 = 0.560 g (c)(ii) (c)(iii) From Data Booklet ion Mg 2+ Ca 2+ Ionic radius 0.065 0.099 The thermal stability depends on the charge density of the cation. The greater the charge density, the thermally less stable the carbonate. Since charge density and hence polarsing power of Mg 2+ is higher than that of Ca 2+, the distortion of the electron cloud of the carbonate a nion, thus weakening effect of the carbon −− −− oxygen bonds in the magnesium carbonate occurs to a greater extent . Hence MgCO 3 is more unstable than CaCO 3 and should decompose to MgO and CO 2 at a faster rate (since its decomposition temperature is lower than that of CaCO 3.) The mass of MgO obtained is 0.478 g, lower than x (i.e. 0.560 g) 4(a) Structural/positional isomerism (b) Geometric isomerism C C H H CH3 HO OCH3 C C H CH3H HO CH3O cis trans (c) Mr of eugenol = 164 M r of vanillin = 152 time 1.00 0 Mass/ g Decomposition curve of CaCO 3 Decomposition curve of MgCO 3 x

Amt of eugenol = 16.4 ÷ 164 = 0.100 mol The theoretical yield = 15.2 g Percentage yield = 13.0/ 15.2 x 100 % = 85.5 % (d) Reaction with aqueous NaOH. Acid-base reaction O-Na+ OCH3 CH2CHCH2 [Other reagents include Na(s), Br 2 (aq) and acid chlorides.] (e) Add 2,4-DNPH to each compound. Vanillin will give an orange ppt. Or add Tollens’ reagent to each compound and warm [1m for the chemical test] Vanillin will give silver mirror. [Use cold alkaline KMnO 4 or acidified K 2Cr 2O7, heat] [1m for observation] (f) Compound B Reasons why the yield may be low: 1. the cyclic ester has ring strain OR 2. phenol is a weaker nucleophile than its phenoxide. OO OCH3

5(a)(i) (a)(ii) At temperatures above 60 oC, there is sufficient heat energy to break the hyd rogen bonds maintaining the α -helix structure. This causes the α -helix structure to lose its helical shape to become a random coil. (a)(iii) extreme acidic pH (b) Primary structure of gastrin: Glu-Gly-Pro-Gly-Trp-Leu-Glu-Glu-Glu-Glu-Ala-Ala-Tyr-Trp-Met-Asp-Phe (c) Haemoglobin is a transport protein with quaternary structure . It consists of four polypeptide chains/subunits (specifically, two α -subunits and two β -subunits) combined together to form a globular protein. There is considerable amount of R-group interactions between an α -subunit and the neighbouring β -subunit. Each subunit has a haem group bonded to it. The iron( II ) in the haem group can bind to oxygen. (d) Any three of the following: Type of interaction Diagram illustrating the type o f interaction Disulfide bridge/linkage disulphide bridge S S CH2 CH2 b a c k b o n e o f polypeptide chain • N–H ---------:O=C • lone pair on O atom • δ + on H and δ – on N • α -helix

Ionic interaction/ Electrostatic forces of attraction between oppositely charged groups COO- NH3 + CH2 CH2 b a c k b o n e o f polypeptide chain ionic bonding Hydrogen bonding CH2OH O H CH2 δ+ δ− δ− δ+ Or CH2OH δ+ δ− CCH2Ο Ο − van der Waals’ forces/ hydrophobic forces –CH 2CH 2CH 3 + any other group e.g. backbone of polypeptide chain CH2 CH2 CH2CH3 CH2CH3 hydrophobic interactions .