VJC H2 CHEM P1 Answers Prelim

2014 JC2 H2 Chemistry Preliminary Examination Paper 1 Detailed Answers 1 D 2 D 3 B 4 C 5 D 6 C 7 C 8 B 9 C 10 A 11 D 12 C 13 D 14 A 15 D 16 C 17 B 18 A 19 D 20 B 21 C 22 B 23 A 24 A 25 A 26 D 27 D 28 C 29 B 30 C 31 B 32 B 33 D 34 A 35 C 36 B 37 A 38 B 39 A 40 C 1 D Molecular formula of compound X is C4H8O2. 2 D [O]: Tl+(aq)  Tl3+(aq) + 2e– [R]: VO3 –(aq) + 6H+(aq) + 3e–  V2+(aq) + 3H2O(l) Redox: 3Tl+(aq) + 2VO3 –(aq) + 12H+(aq)  2V2+(aq) + 3Tl3+(aq) + 6H2O(l) Add 2 mol of VO 3 – to the redox equation will lead to equal number of moles of VO3 –(aq) and V2+(aq): 3Tl+(aq) + 4VO3 –(aq) + 12H+(aq)  2VO3 –(aq) + 2V2+(aq) + 3Tl3+(aq) + 6H2O(l) 3 B Option Valence shell electronic configuration of ion No. of unpaired e– A ns2np2 2 B ns2np5 1 C ns2np3 3 D ns2np2 2 4 C Option Molecule No. of electrons on central atom A 7 B 10 C 8 D 11 5 D For A, SnCl2: ~118o (2 bp, 1 lp), OCl2: 105o (2 bp, 2 lp) For B, both are bent in shape. The electronegativity of O atom is greater than that of S atom. Hence the bond pair of electrons is closer to the nucleus of O atom resulting in stronger bp–bp repulsion in H 2O and a larger bond angle for H2O. For C, both are linear in shape. For D, both are trigonal pyramidal in shape. The electronegativity of F atom is greater than that of C l atom. Hence the bond pair of electrons is further away from the N atom resulting in weaker bp–bp repulsion in NF3 and a smaller bond angle for NF3. 6 C Weak dispersion forces between CH3CH2CH3 molecules, dipole–dipole attractions between CH3CHO molecules, hydrogen bonding between CH3CH2OH molecules and hydrogen bonding between HCO2H molecules. The stronger the attractive forces between the molecules, the greater the deviation from ideal gas behaviour. HCO2H is able to form more extensive hydrogen bonding between its molecules. Hence it has a greater deviation from ideal gas as compared to CH3CH2OH. 7 C The same current flows through all the electrodes. Assuming 6 mol of e - is passed , the reactions at the various electrodes are as shown. E (oxidation): 2Al  2Al3+ + 6e– F (reduction): 6H2O + 6e–  3H2 + 6OH– G (oxidation): 6Ag  6Ag+ + 6e– H (reduction): 3Cu2+ + 6e–  3Cu Electrode F has no change in mass of electrode. Comparing the number of moles of electrode involved in the electrolysis, G has the greatest change in mass of electrode compared to E and H. 8 B Mass of water = 150.00 – 50.00 = 100.00 g Change in temperature of water = 53 – 20 = 33 oC Heat gained by water = mc∆T = (100)(4.18)(33) = 13800 J 9 C For A and B, forward endothermic reaction is favoured at higher temperature, leading to higher Kc and increase in partial pressure of NO2. For C, enthalpy change (∆H) does not change with temperature. For D, a ctivation energy is unchanged unless in the presence of a catalyst. 10 A NH3 + H2O ⇌ NH4 + + OH– For A, BaSO4 or its ions do not react with NH3(aq). For B, AgCl reacts with NH3(aq) to give [Ag(NH3)2]+Cl–. Thus equilibrium is affected. For C, increase in [OH–] will cause equilibrium position to shift left. For D, decrease in [Cu(H 2O)6 2+] due to Cu2+(aq) + 2OH–(aq) ⇌ Cu(OH)2(s) will cause equilibrium position to shift left. 11 D For A, e xcess sodium hydrogencarbonate with hydrochloric acid will produce the buffer consisting of a mixture of H2CO3 and HCO3 –. For B, [H+] = 10–pH = 3.98 x 10–8 mol dm–3 Ka = [H+][HCO3 –] / [H2CO3] [HCO3 –] / [H2CO3] = (7.90 x 10–7) ÷ (3.98 x 10–8) ≈ 20 For C, dilution with water affects the [HCO3 –]/[H2CO3] equally. Hence there is no change in its pH value. For D, pH = p Ka when [HCO3 –] = [H 2CO3], i.e. maximum buffer capacity. Since [HCO3 -] / [H2CO3] ≈ 20, pH ≠ pKa.

12 C Step 2 is the rate –determining step and rate = k’[CHCl3][Cl]. Since C l is an intermediate , it cannot be in the rate equation and [Cl] is dependent on [Cl2] in step 1. Kc = 2 2  [Cl] = (Kc)1/2 [Cl2]1/2 Thus, rate = k’[CHCl3] (Kc)1/2 [Cl2]1/2 = k[CHCl3] [Cl2]1/2 where k = k’(Kc)1/2. 13 D For A, Mg has the highest 1 st IE among the three elements but its melting point is not the lowest. For B, Al has the lowest 1 st IE among the three elements and its melting point is higher than Mg. For C, Si has a lower 1st IE than P and its melting point is the highest. For D, P has the highest 1 st IE among the three elements and its melting point is the lowest. Si has the highest melting point due to strong covalent bonds in a giant molecular structure. S8 has a higher melting point than P4 as it has a greater number of electrons leading to stronger dispersion forces. 14 A From Data Booklet , i onic radi i of Be2+ is 0.031 nm, Mg2+ is 0.065 nm and A l3+ is 0.050 nm. Hence Be 2+ has the highest charge density and Mg 2+ has the lowest charge density. The higher the charge density, the greater the distortion of the electron cloud of nitrate ion leading to lower thermal d ecomposition temperature. Thus, decomposition temperature of Mg(NO3)2 > Al(NO3)3 > Be(NO3)2. 15 D Y2 can undergo displacement reactions with X– and Z–. X2 can undergo displacement reaction with Z– only. Hence on going down Group VII it will be Y, X and Z as oxidising power of halogen decreases down the group. AgY is expected to be the most soluble in aqueous ammonia and AgZ the least soluble in aqueous ammonia. 16 C For A (incorrect), addition of acid (such as H 2SO4(aq)) instead of NaOH(aq) to K 2CrO4(aq) will produce an orange solution of K2Cr2O7. For B (incorrect), the redox reaction between Fe 3+(aq) and I–(aq) is feasible from the Data Booklet: 2Fe3+ + 2I–  2Fe2+ + I2 For C (correct), [Cu(H2O)6]2+ + 4Cl– ⇌ CuCl4 2– + 6H2O, yellow solution of H2[CuCl4] is produced. For D (incorrect), Na2CO3(aq) hydrolyses to form HCO3 –(aq) and OH –(aq), and CrCl3(aq) hydrolyses to form Cr(H2O)5(OH)2+(aq) and H 3O+(aq). Effervescence of CO2 and green ppt of Cr(OH)3 instead of Cr 2(CO3)3 are produced when Na2CO3(aq) is added to CrCl3(aq). 17 B Step 1: redox 2Fe3+ + 2I–  2Fe2+ + I2 Eo cell = +0.77 – (+0.54) > 0 V Step 2: ligand exchange Fe(H2O)6 2+ + 6CN–  Fe(CN)6 4– + 6H2O Step 3: redox Fe(CN)6 4– + ½Cl2  Fe(CN)6 3– + Cl– Eo cell = +1.36 – (+0.36) > 0 V For A, step 3 is incorrect as SO 2 cannot be reduced by Fe(CN)6 4–. For C, step 2 is incorrect as Fe 2+ is oxidised to Fe 3+ while I– cannot be reduced. For D, step 1 is incorrect as reaction is not feasible since Eo cell = +0.77 – (+0.80) < 0 V. Step 2 is incorrect as Fe2+ is oxidised to Fe3+ while I– cannot be reduced. 18 A Compound E contains nitrile, ketone and carboxylic acid. For A, NaBH4 is a weaker reducing agent than LiAlH4 and will reduce nitrile and ketone to sp 3 hybridised carbon atoms , leaving behind the C=O of the carboxylic acid as the only sp2 hybridised carbon atom in the product. For B, LiA lH4 will reduce all the functional groups to sp3 hybridised carbon atoms, leaving no sp2 hybridised carbon atom in the product. For C, nitrile undergoes hydrolysis in the presence of HCl(aq) to give carboxylic acid. There are three sp2 hybridised carbon atoms in the product. For D, carboxylic acid reacts with SOC l2 to give acid chloride. There are two sp2 hybridised carbon atoms in the product. 19 D For A (correct), p ropan–2–ol is a proton acceptor in step 1 and hence it acts as a base. For B (correct), H2SO4 is a catalyst as it is regenerated at the end of step 3. For C (correct), the reactants in step 3 can undergo addition reaction to give CH3CH(OSO3H)CH3. For D (incorrect), the formation of carbocation in step 2 is not likely for primary alcohols as compared to tertiary alcohols since primary carbocation is less stable than tertiary carbocation. 20 B Option Products A CH3CO2H CH3COCH2CH2CO