RI H2 Chemistry TYS Solutions 2013-2019

© Raffles Institution Changes to 2013 H2 Chemistry A Level Question Paper Dear students, The TYS you have purchased is based on the 9647 (old) syllabus. You will be sitting for the 9729 (new) syllabus papers. This document will instruct you on the changes you need to make to the TYS questions. The Planning question for the 9729 syllabus will be in the Paper 4 (Practical). Some concepts are no longer tested in the 9729 syllabus and the values used for some calculations are now different (e.g. molar volume at s.t.p.) which will affect your choice of the answers. You are advised to make the changes on your question papers before you attempt it . Do inform your tutors if you notice any differences which were not highlighted in this document. ---------------------------------------------------------------------------------------------------------------------------------------------- Paper 1 28 Not in syllabus 30 Not in syllabus 34 Amend question “In the reaction below, M represents a Group 2 element.” 37 Option 3 is not in syllabus Paper 2 4(a)(i) Not in syllabus. 4(d)(i) Not in syllabus. Ca(NO3)2(s)  CaO(s) + 2NO2(g) + 1 2 O2(g) 5(a), (c) and (d) - Not in syllabus. Can do 5b, 5e and 5f Paper 3 3a: Change P4O6 to P4O10 4(e)(ii) Bond energy of C=O in CO2 in the Data Booklet (for new syllabus) is the same as that given by the question

© Raffles Institution Q1 3H3PO4  H5P3O10 + 2H2O Ans : B Q2 No. of protons No. of neutrons No. of electrons (A) D3O+ 11 11 10 (B) H3O+ 11 8 10 (C) NH2 9 7 10 (D) OD 9 9 10 Ans: B Q3 Before gaining e After gaining e (A) C+: 1s22s22p1 C: 1s22s22p2 (B) N : 1s22s22p3 N : 1s22s22p4 (C) Si: 1s22s22p63s23p3 Si2: 1s22s22p63s23p4 (D) P+: 1s22s22p63s23p2 P: 1s22s22p63s23p3 Ans: D Q4 660; 1267; 2218; 3313; 7863; 9500 607 951 1095 4550 1637 Since the largest jump is between 4th and 5th ionisation energies, M is a Group 14 element. Ans: C Q5 Substances with giant covalent structure generally do not conduct electricity in all physical states since all its valence electrons are used up for covalent bonding. Ans: A Q6 The behaviour of a gas is most ideal at: (i) low pressure At low pressure, the gas particles are far apart from one another . Hence the volume occupied by gas particles is negligible compared to the volume of the container and the intermolecular forces of attraction are negligible as the gas particles are far apart. (ii) high temperature At high temperature, the gas particles have higher kinetic energy so that the intermolecular forces of attraction between them are negligible. Ans: C Q7 Hrxn = mHfo (pdts)  nHfo (rxts) Hrxn = Hfo (CH3CO2Na.3H2O(s))  3Hfo (H2O(l))  Hfo (CH3CO2Na(aq)) Ans: A Q8 Since a reaction occurs vigorously, G < 0 Reaction is endothermic (H > 0)  temp decreases. G= H  TS  TS is negative since G < 0 and H > 0, thus S is positive. (Eqn of reaction: SOCl2 + Ba(OH)2  BaSO4 + 2HCl) Ans: B Q9 Au3+(aq) + 3e  Au(s) Amt of Au = 6.0 197.0 = 0.03046 mol Amt of electrons = 3 × 0.03046 = 0.09138 mol ne × F = I × t t = 0.09138 96500 0.10 = 8.82 × 104 s Ans: D Suggested Soln for N2013 H2 Chemistry Paper 1 (9647/01) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 B B D C A C A B D C D B D D B D D A C C 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 C D B B B B C C A D A D C B D B A B C A Working for Suggested Solns for N2013 H2 Chemistry Paper 1 (9647/01)

© Raffles Institution Q10 Ionic Prdt of CaCO3 = 0.10×10–9 = 1×1010 < Ksp of CaCO3 (3.8×10–9 )  No ppt of CaCO3 Ionic Prdt of FeCO3 = 0.10×10–9 = 1×1010 > Ksp of FeCO3 (3.2×1011 )  ppt of FeCO3 Ionic Prdt of MnCO3 = 0.10×10–9 = 1×1010 > Ksp of MnCO3 (2.5×1013 )  ppt of MnCO3 Ans: C Q11 (A) [H+] & [OH] are equal at all temperatures since water dissociates to give equal amt of H+ & OH. H2O(l) H+(aq) + OH(aq) (B) At 0 C, Kw is the smallest. Hence, equilibrium position lies furthest to the left at 0oC. (C) Since Kw increases with temperature, equilibrium position shifts right with increasing temperature. Hence, the forward reaction is endothermic. (D) pKw decreases with temperature. Since pH + pOH = pKw and pH = pOH for water, hence pH of water decreases as temperature increases. Ans: D Q12 Nitrogen dioxide from car exhaust fumes can catalyse the oxidation of sulfur dioxide as it is a homogenous catalyst. Step 1: SO2(g) + NO2(g)  SO3(g) + NO(g) Step 2: NO(g) + 1 2 O2(g)  NO2(g) Overall: SO2(g) + 1 2 O2(g)  SO3(g) Ans: B Q13 (A) (B) (C) (D) (D) Across Period 3 (from Na to Cl), effective nuclear charge increases due to increasing nuclear charge (increasing no. of protons) and the relatively constant shielding effect (due to same no. of filled inner principal quantum shells of electrons). With increasing effective nuclear charge, the electrons are more strongly attracted by the nucleus, leading to decreasing atomic radii across Period 3. Ans : D Na Mg Al Si P S Cl Ar 1st I.E. 1st I.E. of Elements

© Raffles Institution Q14 Electrical conductivity depends on amt of electrons in the ‘sea’ of delocalised electrons. Al has 3 valence electrons wh ile Mg has 2 valence electrons, thus el ectrical conductivity of A l is higher than that of Mg. Cu has one unpaired 3d electron & two 4s electron s which can be delocalis ed, while Ca has two 4s electrons, thus electrical conductivity of Cu is higher than that of Ca. Ans : D Q15 5Fe2+ + MnO4 +8H+  4H2O + Mn2+ + 5Fe3+ (green) (purple) (colourless) (yellow) Since KMnO4 is added in excess, the final colour should be purple. The end pt colour is pink due to a slight excess of purple KMnO4 & the yellow Fe3+(aq) present in the conical flask. Ans: B Q16: In QA CH2ClCHICO2H + 2Na  CH2=CHCO2H + NaCl +NaI Upon addition of AgNO3, AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) white ppt AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq) yellow ppt Upon addition of concentrated NH3(aq), Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) White ppt of AgCl dissolves in concentrated NH 3(aq) but not yellow ppt of AgI because Ksp of AgCl is larger than that of AgI. Ans: D Q17 (A) H2S molecule (B) Products of heterolytic fission of Cl -Cl bond are Cl+ and Cl. (C) NH4+ ion (D) Cu2+ in CuO has electronic con figuration of 1s22s22p63s23p63d9. Hence, it has one unpaired 3d electron. Ans: D Q18 From (B), it is clear that X is phosphorus. (A) PCl3(s) + 3H2O(l)  H3PO3(aq) + 3HCl(aq) NaOH(aq) + H3PO3(aq)  Na3PO3(aq) + 3H2O(l) NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)  No precipitates are formed. (B) PCl3 + Cl2  PCl5 (C) Phosphorus (P 4) is a white solid at room temperature and pressure. (D) Oxide of phosphorus gives an acidic solution in water. P4O6 + 6H2O  4H3PO3 P4O10 + 6H2O  4H3PO4 Ans: A Q19 There are 2 ester linkages (circled) and 5 chiral carbon atoms (indicated with asterisks) as shown below. Ans: C Q20 All the carbon atoms in the organic compound are sp2 hybridised, hence all the bon ds indicated involved sp2-sp2 overlap. Ans: C

© Raffles Institution Q21 In the electrophilic addition mechanism , the major product is formed from the more stable s econdary carbocation intermediate while the minor product is formed from the less stable primary carbo cation intermediate: Ans: C Q22 Ans: D Q23 (A) CH3COCl + 2OH  CH3COO + Cl + H2O Amt of Cl = Amt of AgCl = 1 78.5 = 0.0127 mol Mas