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 favourin