YJC H1 CHEM P1 P2 ANS

Yishun Junior College H1 Chemistry Preliminary Examinations 2015 Paper 1 Answers 1 A 7 A 13 A 19 D 25 B 2 C 8 A 14 B 20 C 26 B 3 B 9 D 15 C 21 C 27 A 4 C 10 D 16 D 22 B 28 C 5 C 11 B 17 D 23 C 29 B 6 D 12 B 18 C 24 A 30 D (A) : 7 (B) : 8 (C) : 9 (D) : 6 Paper 2 Answers 1 (a) Nitrogen has lesser nuclear charge / lesser number of protons and lesser shielding effect/lesser number of electron shells than phosphorus. [1] As the increase in shielding effect outweighs the increase in nuclear charge, the attraction between the nucleus and valence electrons is stronger for nitrogen. [1] Hence, more energy is needed to remove the first electron from nitrogen. [2] (b) Both ammonia and phosgene have simple molecular structure. Lesser energy is needed to overcome the weaker intermolecular permanent dipole-permanent dipole attractions [1] in phosgene compared to the strong intermolecular hydrogen bonds [1] in ammonia. [2] (c) (i) A dative bond is formed when a shared pair of electrons between two atoms is contributed by only one of the atoms. [1] (ii) Dative bond (e− pair from N atom donated to B atom) – [1] Correct dot-and-cross diagram – [1] [2]

(iii) N in ammonia has 3 bond pairs and 1 lone pair of electrons while N in H3NBF3 has 4 bond pairs and no lone pairs of electrons. [1] Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. OR Lone pair of electrons repel more than bond pairs. [1] Hence, the H-N-H bond angle is larger in H3NBF3. [2] [Total:9] 2 (a) alkene [1] and ester [1] [2] (b) [1] (c) [2] (d) (i) The standard enthalpy change of combustion of a substance is the heat evolved when one mole of the substance is completely burnt in oxygen gas at 25oC and 1 atm. [1] (ii) heat released = 100 × 4.18 × (56.5 − 25) = 13200 J [1] (iii) ηA = 2.50 268 = 9.33 × 10−3 mol [1] (iv) ∆HC = − 13.2 9.33 × 10−3 = −1420 𝑘𝐽𝑚𝑜𝑙−1 working – [1] sign + units – [1] [2] [Total:10] O O-Na+ CH3OH [1] [1] O O

3 (a) Test 1: Compound C is a carbonyl compound OR a ketone or aldehyde [1] Test 2: Compound C is an aldehyde or benzaldehyde [1] Test 3: Compound C is not an aliphatic aldehyde OR Compound C is a benzaldehyde [1] [3] (b) [1] (c) OR [1] (d) Each carbon atom has 3 sp2 hybrid orbitals and 1 unhybridised p- orbital. [1] 1 sp2 orbital overlap head-on with the s orbital of H atom to form a C−H single (sigma bonding). The other 2 sp2 orbitals are used to form sigma bonds with the neighbouring carbon atoms (C−C single bonds). [1] The unhybridised p-orbitals for the 6 carbon atoms overlap sideways to form delocalised electron clouds above and below the plane of the carbon ring. OR draw a suitable diagram (as shown above) [1] [3] [Total: 8] 4 (a) oxidation: MnO2 + 4OH−  MnO4 2− + 2H2O + 2e− [1] reduction: ClO3 − + 3H2O + 6e−  Cl− + 6OH− [1] overall: 3MnO2 + ClO3 − + 6OH−  3MnO4 2− + Cl− + 3H2O [1] [3] H3C CHO CH2CHO COCH3

(b) Carbon dioxide reacts with water to form carbonic acid. OR CO2 + H2O  H2CO3 [1] H2CO3 + OH−  HCO3 − + H2O OR H2CO3 + 2OH−  CO3 2− + H2O [1] Carbonic acid react with the hydroxide ions. By Le Chatelier’s principle, the equilibrium position (step 2) shifts to the right so as to replenish the hydroxide ions. Hence, the percentage of MnO4 − increases. [1] [3] (c) (i) Manganese undergoes reduction as its oxidation number decreases from +6 in MnO4 2− to +4 in MnO2 while sulfur undergoes oxidation as its oxidation number increases from +4 in SO3 2− to +6 in SO4 2−. Definition of redox – [1] Oxidation numbers – [1] [2] (ii) η𝐾𝑀𝑛𝑂4 = 0.02 × 26.80 1000 = 5.36 × 10−4𝑚𝑜𝑙 [1] (iii) 𝜂𝑆𝑂32− 𝑖𝑛 25𝑐𝑚3 = 5 2 × 5.36 × 10−4 = 1.34 × 10−3𝑚𝑜𝑙 [1] 𝜂𝑆𝑂32− 𝑖𝑛 250𝑐𝑚3 = 10 × 1.34 × 10−3 = 0.0134 𝑚𝑜𝑙 [1] Mass of pure 𝑁𝑎2𝑆𝑂3 = 0.0134 × 126.1 = 1.69 𝑔 [1] % purity = 1.69 2.00 × 100 = 84.5% [1] [4] [Total:13] 5 (a) There are 2 bond pairs and 1 lone pair of electrons around the S atom in SO2 while there are 3 bond pairs and no lone pairs of electrons around the S atom in SO3. AND To minimise repulsion between the electrons, the electron pairs arrange themselves in a trigonal planar fashion. [1] Shape of SO2 – bent OR v-shaped AND Shape of SO3 – trigonal planar [1] [4] (b) Al2O3 is ionic with significant covalent character and hence is amphoteric in nature. Al2O3 can react with both acids and alkalis. Oxides of S are covalent and hence are acidic in nature. Oxides of S react with alkalis to form a salt and water. [1] sulfur dioxide [1] S O O x x x x x x x x x x x x S O O O x x x x x x x x x x x x x x x x x x sulfur trioxide [1]

Al2O3(s) + 6HCl(aq)  2AlCl3(aq) + H2O(l) [1] Al2O3(s) + 2NaOH(aq) + 3H2O(l)  2NaAl(OH)4(aq) [1] SO2(g) + 2NaOH(aq)  Na2SO3(aq) + H2O(l) OR SO3(g) + 2NaOH(aq)  Na2SO4(aq) + H2O(l) [1] [4] (c) (i) 𝐾𝑐 = [𝑆𝑂3]2 [𝑆𝑂2]2[𝑂2] [1] units: mol-1dm3 [1] [2] (ii) 2SO2 O2 2SO3 Initial amt/mol 5 2 0 Change in amt/mol -3 -1.5 +3 Eqm amt/mol 2 0.5 3 Eqm conc/mol 1 0.25 1.5 [1] 𝐾𝑐 = (1.5)2 (1)2(0.25) = 9.00𝑚𝑜𝑙−1𝑑𝑚3 [1] [2] (iii) By LeChatelier’s Principle, the position of equilibrium will shift to the left. [1] The endothermic backward reaction is favoured so as to absorb the heat added, producing more SO2 and O2. [1] [2] (d) (i) A strong Bronsted acid is a proton donor [1] which undergoes complete dissociation [1] in water. [2] (ii) [𝐻+] = 10−1.2 = 0.0631 𝑚𝑜𝑙𝑑𝑚−3 [1] [𝐻2𝑆𝑂4] = 0.0631 ÷ 2 = 0.0315 𝑚𝑜𝑙𝑑𝑚−3 [1] [2] (e) (i) addition [1] (ii) [1] [Total:20] C H H H C CC H H HH HO H H

6 (a) Product 1: Ethanoic acid Reagents & Conditions: KMnO4, dil. H2SO4, heat under reflux Observations: Purple KMnO4 is decolourised. Equation: CH3CH2OH + 2[O]  CH3COOH + H2O Product 2: Ethanal Reagents & Conditions: K2Cr2O7, dil. H2SO4, immediate distillation Observations: Orange acidified K2Cr2O7 turns green Equation: CH3CH2OH + [O]  CH3CHO + H2O Identify both products – [1] Reagents & Conditions + Observations for each product – [1] x 2 Each equation – [1] x 2 [5] (b) (i) CH3CH2OH + Na  CH3CH2O−Na+ + ½ H2 [1] (ii) Plot volume of gas against time Or Plot (V - Vt) against time with labelled axes [1] All points plotted correctly [1] Best-fit curve [1] Show at least 2 half -lives on graph + half -life  1.5 min + constant half-lives [1] [4]

(iii) Ethanol is in large excess and so the change in the concentration of ethanol in the experiment is negligible. [1] (iv) 𝜂𝑁𝑎 = 2 × η𝐻2= 2 × 60 24000 = 0.00500 𝑚𝑜𝑙 [1] 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎 = 0.005 × 23 = 0.115 𝑔 [1] [2] (v) Sodium is readily oxidised in air to form sodium oxide. [1] Na + ½ O2Na2O [1] [2] (vi) (Correct diagram – [1] ) When temperature increases, the average kinetic energy of the reacting particles is increased. [1] Hence, the proportion of particles with energy equal to or greater than the activation energy increases significantly as seen in the shaded area in the diagram. [1] In addition, when the temperature of the reactants is increased, their average speeds increase and so, the fr