CJC H2 CHEM P1 P2 P3 ANS Prelim

CJC: H2 CHEMISTRY 2009 Preliminary Exam ANSWERS Paper 1 MCQ Answers: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 C B C D A C A C A D C C A C A D A B D B 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 C B C D C C B A B A D B B C B A B A A C Paper 2 1(a) (i) Mg+(g)  Mg2+(g) + e- (ii) The second ionisation involves removal of an electron from the 3p subshell from Si +, and the 3s subshell from A l+. Since the 3p subshell is on average further from the nucleus / at a higher energy level / experiences more shielding than the 3s subshell , the electron removed from Si+ is less tightly held and requires less energy to remove. (b) Neon, being already in the gaseous state, is ionised first and emits light, whereas sodium is in the solid state and takes time to vaporise before it can be ionised. OR Neon has a higher first ionisation energy than sodium. The initial power surge when the light is turned on is able to ionize neon, giving rise to the red light, after which the power drops and is only enough to ionise sodium, giving rise to the orange light. (c) (i) Al 2O3 has a giant ionic lattice structure with strong ionic bonds / high lattice energy / partial covalent character, hence it is insoluble. SiO2 has a giant molecular structure with strong covalent bonds requiring large amounts of energy to break, hence it is insoluble. (ii) AlCl3 dissolves to give [Al(H2O)6]3+(aq) which undergoes hydrolysis due to the high charge density on Al3+: [A l(H2O)6]3+  [Al(H2O)5(OH)]2+ + H+ OR 2AlCl3 + 6H2O ⇌ Al2O3 + 6HCl OR AlCl3 + 3H2O ⇌ Al(OH)3 + 3HCl SiCl4 hydrolyses in water to give HCl, a strong acid / H+ is produced: SiC l4 + 2H2O  SiO2 + 4HCl OR SiCl4 + 4H2O  Si(OH)4 + 4HCl (d) (i) Solubility of Li2O is lower than that of BaO. Therefore, the basicity of Li2O is also lower than that of BaO. (ii) Percentage change in mass = % 100)] 0 . 16 ( 2 ) 9 . 6 ( 2 [ )] 0 . 16 ( 2 ) 9 . 6 ( 2 [ )] 0 . 16 ( 3 0 . 12) 9 . 6 ( 2 [     = 61.1% [Total: 12] 2(a) x NH4NO3(l) + aq  NH4NO3(aq) 71 -50 NH3(g)+ HNO3(g) -31 + 37 NH 3(aq) + HNO3(aq)

CJC: H2 CHEMISTRY 2009 Preliminary Exam ANSWERS From energy cycle, show x = 71 -31 + 37 -50 = +27 kJ mol-1 (b) The reaction is endothermic as the temperature of the reaction mixture drops. S is positive as the number of gaseous molecules increase from 0 to 1. Since reaction is spontaneous, G < 0 . G = H - TS, -TS must be more negative than H. (c) (i) Undergoes substitution (ii) NO2 Br NO 2 Br2, Fe or FeBr3 Conc. HNO3 + conc. H2SO4 50 – 60 oC (i) Sn, conc. HCl, reflux (ii) NaOH(aq) NH2 Br [ T o t a l : 1 0 ] 3(a) Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 or [Ar] 3d5 4s1 Cr 3+: 1s2 2s2 2p6 3s2 3p6 3d3 or [Ar] 3d3 (b) (i) Cr S N Mols 0.298 1.19 2.09 Ratio 1 4 7 x + y + 1 = 7 x: 4 y: 2 (ii) +3 (iii) Cr (III) NH3 NH3 NCS NCSSCN SCN - .. :: :: .. (c)

CJC: H2 CHEMISTRY 2009 Preliminary Exam ANSWERS (i) 2Cr2O7 + 2 HBr  2 KCrO3Br + H2O , remains at +6 (iii) o K (ii) No, oxidation no. of Cr doesn’t change E cell Products sh ould be Cr = +1.33 – (+1.07) = +0.26 V >0 (iv) Reaction didn’t take place at standard condition. (vi) Cr is reduced from +6 (orange solut ion of Cr2O7 2-) to +3 Cr3+ (deep green solution) [Total: 16] H (g) - 0 Equilibrium mol: - 0.045 - 0.055 ota = 0 4 3+(aq) and Br2(l) OR The reaction in step I took place under cooled conditions. (v) A g B r 4 (a) 3 Fe(s) + 4 H2O(g) Fe 3O4(s) + 4 2 Init ial mol: - 0.1 T l mol .0 5 + 0.055 = 0.1 = ( 2Hp 1 . 0 055. 0  1.5 atm = 0.825 atm ) = ( O H2 p 1 . 0 045 . 0) 1.5 atm = 0.675 atm  K p = 4 2 4 2 ) (Hp 4 4 ) 675 . 0 ( ) 825 . 0 ( ) (O Hp = (b) = 2.23 Formation of ammonia is exothermic.  Low temperature is used to obtain hig r y of onhe ield amm ia. Since, low temperature is used, the rate of formation of ammonia is slow . Hence, catalyst is used to increase rate. Formation of nitrogen dioxide is endothermic .  High temperature is used to obtain higher yield of nitrogen dioxide. Since, high temperature is used, the rate of formation of nitrogen dioxide is fast . Hence, catalyst is not required. (c) (i) Step I: conc HNO and conc H SO , 30 oC V tion Step VII: Electrophilic su (iii) P: Q: 3 2 4 Step II: excess Cl2, uv light (ii) Step I : Elimi ater) / Dehydranation (of w bstitution NO2 CH3 N H H O─H │ H─O─C─O─H

CJC: H2 CHEMISTRY 2009 Preliminary Exam ANSWERS ─ │ H O ║ C N CH3 R: S: [Total: 12] 5 (a) (i) (ii) X : Add KMnO 4/H+ (aq) and heat to both samples. For chloroxylenol, purple KMnO4 is decolourised, but not for 4-chlorophenol. (b) (i) reagents & conditions type of reaction organic products NO2 Cl (iii) concentrated D2SO4, heat Dehydration / elimination (of water) OH CH3C H3 Cl OH Cl OO OHO H OH Cl OO OHO H O Cl OO OO Na + Na + Na + + 6[O] KMnO4/H+ heat + 2H2O W + 3NaOH + 3H2O [1] [1] OH Cl OO ClCl [1, ecf if possible] C Cl H CH2 [1] │ C=O NO2

CJC: H2 CHEMISTRY 2009 Preliminary Exam ANSWERS I2, NaOD, D2O C Cl O O Na + and CDI3 [1] [1] (ii) No (orange) ppt is observed. l: 10] Paper 1(a) (i) (ii) Mg(NO )  MgO + 2 NO + ½O Z is not a ca rbonyl compound (is not an aldehyde or ketone) [Tota 3 KNO3  KNO2 + ½O2 3 2 2 2 NO2, which is produced from decomposition of magnesium nitrate, is toxic/a pollutant. OR Decom osition of map gnesium nitrate occurs more readily than potassium nitrate, resulting in urning rata loss in control of b e at the high temperatures reached in fireworks. OR Potassium nitrate releases more mols of O2 per mass than magnesium nitrate because of its (b) I metal be 2 Al + lower Mr. Let the unknown Group I M. 3 2 O2  Al2O3 No. of mols of O = 2 3 41 2 2 8 [39.1 + 14.0 + 3(16.0)] = 2.138 g 01 0 . = 0.01058 3 12 g O + C O2 M ass of KNO3 = 2  0.0105 Mass of MCO = 2.75 – 2.138 = 0.6 MCO3  M No. of mols of CO = 2 24000 M 100 = 4.167  10-3