CJC Prelim H2 Chem 2008 (worked soln for P1 P2 P3) (updated)

CJC H2 Chem Preliminary Exam 2008 Answers Qn No Answer Qn No Answer Qn No Answer Qn No Answer 1 C 11 D 21 B 31 A 2 C 12 D 22 C 32 D 3 C 13 C 23 A 33 C 4 C 14 B 24 D 34 B 5 D 15 B 25 B 35 C 6 A 16 B 26 C 36 B 7 B 17 A 27 C 37 D 8 C 18 C 28 D 38 C 9 A 19 B 29 C 39 D 10 C 20 B 30 B 40 B Paper 1 Paper 2 1 (a) S [Ne] 3s 2 3px 2 3py 1 3pz 1 P [Ne] 3s 2 3px 1 3py 1 3pz 1 S has a lower 1st I.E. than P due to the inter-electron repulsion between paired electrons in a 3p orbital of S. (b) CO 2 exists as simple covalent molecules with weak van der Waals’ forces of attraction between molecules. SiO2 has a giant molecular structure with strong covalent bonds between atoms ∴ SiO2 has a much higher melting point than CO2. (c) 2-nitrophenol is able to form intr a-molecular hydrogen-bonding between the –OH and –NO2 groups. Hence its inter-molecular hydrogen bonding is less extensive than that of 4-nitrophenol. Less energy is required to break the intermolecular forces of attraction and it has a lower melting point. (d) Boiling point is dependent on the intermolecular forces of attraction. Down the group, boiling points of the hydrogen halides generally increase due to stronger van der Waals’ forces of attraction between molecules as the number of electrons increases (with increase in Mr). However, HF has the highest boiling point (an exception) due to the presence of strong hydrogen bonding between molecules. (e) P is more electronegative than As. The bond pair electrons in PCl 3 are closer to the central P atom as compared to that in AsCl3. There is more repulsion between the electrons / greater bond pair-bond pair repulsion and hence the bond angle increases.

2 2 (a) +5 (b) (i) BiO 3 - + 6 H+ + 2 e- → Bi3+ + 3 H2O (ii) 2 Mn 2+ + 5 BiO3 - + 14 H+ → 2 MnO4 - + 5 Bi3+ + 7 H2O (c) Fe 2+ → Fe3+ + e- MnO4 - + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4 H2O ∴MnO4 - ≡ 5 Fe2+ no. of mol of Mn in sample = no. of mol of MnO4 - required = 5 1 × No. of mol of Fe2+ = )1000 00 . 3610 . 0 (5 1 × × = 0.000720 mol ∴ mass of Mn in sample = 0.000720 × Ar of Mn = 0.000720 × 55 = 0.0396 g ∴ % of Mn in sample = 10000 . 1 0396 . 0× = 3.96 % (d) (i) E o cell = E o red - E o oxd Reactants Products Eθ cell / V Can reaction occur? (Y/N) Mn & SO4 2- Mn 2+ & SO2 Eθ cell = +0.17 - (-1.18) = +1.35 Yes Mn2+ & SO4 2- MnO 4 - & SO2 Eθ cell = +0.17 - (+1.52) = -1.35 No Fe2+ & SO4 2- Fe 3+ & SO2 Eθ cell = +0.17 - (+0.77) = -0.60 No (ii) Yes, sulphuric acid is a suitable replacement for nitric acid in this experiment. Reason: H2SO4 can oxidise Mn to Mn2+ but cannot further oxidise Mn2+ to MnO4 -. Also, it does not oxidise Fe2+ to Fe3+. ⇒ H2SO4 does not interfere with the redox reaction. CJC 2008 9746/Answer/prelim/08

3 3 (a) (i) (ii) initial dose = (2 × 500) = 1000 mg 1000 500 250 125 t ½ t ½ t ½ ∴time taken = 3 t ½ = (3 × 2.7) hrs = 8.1 hrs (iii) Curve must show faster reaction (i.e. steeper). (iv) pH of body fluid or body temperature CJC 2008 9746/Answer/prelim/08

4 (b) (i) Cl CHClCH3 (ii) C l2, presence of ultra-violet light (iii) (iv) Only one of the Cl (on the alkyl side-chain) is hydrolysed. Mr of V, C8H8Cl2 = 8(12.0) + 8(1.0) + 2(35.5) = 175 mol of AgCl ppt = mol of V = 175 75 . 1 = 0.010 mol Mr of AgCl = 108 + 35.5 = 143.5 ∴mass of AgCl ppt = (0.010 × 143.5) g = 1.435 g 4 (a) (i) mole ratio C : H = 0 . 12 3 . 92 : 0 . 1 3 . 92 100− = 0 . 12 3 . 92 : 0 . 1 7 . 7 = 7.8 : 7.7 = 1 : 1 ∴ empirical formula of W is CH. Amount of CO2 = ) 0 . 16 ( 2 0 . 12 76 . 1 + = 0 . 44 76 . 1 = 0.0400 mol ∴ mole ratio W : CO2 = 0.005 : 0.0400 = 1 : 8 (or W ≡ 8 CO2) ⇒ 8 C atoms in one molecule of W. CJC 2008 9746/Answer/prelim/08

5 ∴ Molecular formula of W is C8H8. (ii) (iii) sp2 (iv) Geometric isomerism/cis-trans isomerism or optical isomerism (v) or (b) Ni has smaller atomic size due to higher effective nuclear charge. In Ni, electrons are added to d-subshells, which provide poor shielding. However Ni has higher nuclear charge than Ca and so, electrons in Ni are more strongly held by the nucleus / greater electrostatic forces of attraction between electrons and nucleus. Ni has much higher melting point due to stronger metallic bonding. In Ni, both 3d and 4s electrons are delocalised for metallic bonding (since 3d and 4s orbitals have similar energy levels). In Ca, only two valence electrons (4s2) are available for delocalisation in forming metallic bonds. 5 (a) M r of SO2Cl2 = 32.1 + 2(16.0) + 2(35.5) = 135.1 Using ideal gas equation, pV = nRT p = V nRT = 310 1 375 31 . 81 . 135 7 . 6 −× × × × 510 01 . 1 1 × = 1.53 atm CJC 2008 9746/Answer/prelim/08

6 (b) SO2Cl2 (g) ⇌ SO2 (g) + Cl2 (g) initial / atm 1.53 0 0 equil / atm 1.53 - x x x i.e. at equilibrium, = and = 1.53 - 2SOp 2Clp 2 2C SOlp 2SOp ∴ Kp = 2 2 22 C SO CSO l l p p p× = 2 2 SO 2 SO 53 . 1 ) ( p p − Kp = 2 2 SO 2 SO 53 . 1 ) ( p p − 2.4 = 2 2 SO 2 SO 53 . 1 ) ( p p − ⇒ = = 1.06 atm 2SOp 2Clp 2 2C SOlp = 1.53 – 1.06 = 0.47 atm (c) SO2Cl2 (g) ⇌ SO2 (g) + Cl2 (g) initial / atm 1.53 0 1.0 equil / atm 1.53 - x x 1.0 + x i.e. at equilibrium, = + 1.0 2Clp 2SOp and = 1.53 - 2 2C SOlp 2SOp ∴ Kp = 2 2 22 C SO CSO l l p p p× 2.4 = 2 22 SO SOSO 53 . 1 ) 0 . 1 ( p p p − +× ⇒ = 0.86 atm 2SOp 2Clp = 1.86 atm 2 2C SOlp = 1.53 – 0.86 = 0.67 atm CJC 2008 9746/Answer/prelim/08

7 6 (a) Y burns with a sooty flame and reacts with aqueous NaOH but not Na2CO3. ⇒ unsaturated ⇒ acidic but not RCO2H Deduction: Y has a phenol group Y rotates plane polarised light. Deduction: Y has a chiral carbon Y gives a yellow precipitate when warmed with aqueous alkaline I2. Deduction: Y has CH3CH(OH)– or CH3C=O group. Yellow precipitate is CHI3. Y decolourises 2 mol of aqueous Br2 to give steamy white fumes and a white precipitate. ⇒ only 2 position free for rxn ⇒ substitution rxn Deduction: CH3CH(OH)– is at –2 or –4 position with respect to phenol. (Electrophilic substitution takes place, steamy white fumes is HBr.) Y reacts with hot excess conc H2SO4 to give Z, C9H10O, which can exist as stereoisomers. ⇒ dehydration rxn ⇒ cis-trans Deduction: Y undergoes dehydration to form an alkene Z. Y: HO CH2 C H OH C H H H Z: OH C CC H H H HH CJC 2008 9746/Answer/prelim/08

8 (b) (i) By-products (SO 2 and HCl) being gases do not contaminate the product. (ii) ΔS is positive. There are 2 mol of gaseous products formed, which resulted in higher disorder. (iii) ΔG = ΔH - TΔS Since both ΔS and ΔH are positive, ∆G will become more negative / less positive as temperature increases. This reaction is spontaneous (∆G < 0) at high temperatures. [The reaction should be carried out at high a temperature so that ∆G will be < 0, or else “spontaneity” will be lost.] CJC 2008 9746/Answer/prelim/08

9 Paper 3 Essay 1 (a) (i) Electrophilic addition NaCl + H2SO4 → HCl + Na2SO4 Formation of carbocation. slow C=C H CH3CH2 H H Cl H δ+ δ- C H CH3CH2 H C H + H + Cl– The Cl – ion attacks the intermediate carbocation to give the addition product. (ii) The carbocation is planar , hence there is equal probability the nucleophile may attack it from the top or the bottom to give either stereoisomer (optical isomer). (b) (i) CH 3CH2CH(CH3)CH2CH3 CH 3CH2CH(CH3)CHClCH3 2-chloro-3-methylbutane (intermediate) C H 3CH2CH(CH3)CHClCH3 CH 3CH2CH(