RI 2012 A-Level H2 Chemistry Solutions

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 1. O OH H H H O H OH H OH H OH O Cl H H H Cl Cl H Cl H OH Molecular formula when all OH directly bonded to ring replaced with Cl is C6H8O2Cl4, empirical formula is C3H4OCl2 Answer: C 2. [O ]: Fe  Fe2+ + 2 e; [R]: Fe3+ + e  Fe2+ Overall eqn: 2 Fe3+ + Fe  3 Fe2+ To have equal no. of moles of Fe2+ and Fe3+ at the end, must have excess Fe3+, Fe must be limiting, hence A is out since A is in stoichiometric ratio. Fe3+ Fe Fe2+ Initial /mol 5 1 0 Change /mol −2 −1 +3 Final /mol 3 0 3 Answer: C 3. No unpaired s or unpaired p electron : s2, s2p6 or [Ar]3dx4s2 or [Ar]3dx configuration. Look out for species with noble gas configuration or transition metal atoms (exception of Cr and Cu which are [Ar]3d54s1 and [Ar]3d104s1) or transition metal ions with at least +2 charge cos all 4s electrons removed. Sc: [Ar]3d14s2 Answer: D. Suggested Solutions to N2012 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 C C D B C A D C A D D C B A C D D C C C 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 D B B D C A C B B A B D A B B B D A B B Working for Suggested Solutions to N2012 H2 Chemistry Paper 1 (9647/01)

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 4. There are at least 13 electrons, hence this cannot be a period 2 element. Neither is this a group 13 element as the biggest jump is not between the 3rd and 4th electron but between the 2nd and 3rd electron from the first 2 IE. It may be a group 2 element since the big jump is between 2nd and 3rd electron. After the big jump from the 2nd to 3rd electron, the IE of the next 8 electrons removed increases gradually, thus they must be in the same electronic shell. There is a large jump between the 10th and 11th, thus, the 11th to 13th electron must be from the inner electronic shell. Possible configuration: ________ ns2np6 (n+1)s2 It cannot be a d-block element. For a d-block element, a possible configuration is: 1s2 2s2 2p6 3s2 3p6 3dx 4s2. If so, there will not be such a big jump between the 2nd and 3rd electron to be removed as the 3d and 4s orbitals are close in energy. Answer: B 5. Water has 2 lone pairs on O and 2 protonic H. Hence can form 2 H-bonds per molecule. HF has 3 lone pairs on F and 1 protonic H. Hence can only form 1 H-bond per molecule. Water can form more H-bonds on average than HF. This reason is more important than A, where water forms stronger H-bonds than HF as one could also argue that F is more electronegative hence H attached to F is more protonic and HF can form stronger H-bonds than water. Answer: C 6. According to the equation, there is decrease in no. of moles of gas in the forward equation, thus ∆S < 0. Since H-bonds are formed between CO 2 and H2O, ∆H <0. Answer: A 7. Cell 1: 2 Cl−  Cl2 + 2 e Cell 2: 2 H2O  O2 + 4 H++ 4 e Q = I x t and Q = neF Hence, neF = I x t. Since t and F are constant, ne α I. Equating electrons from both eqns: Cell 1: 2 Cl−  Cl2 + 2 e X 2 Cell 1: 4 Cl−  2 Cl2 + 4 e Cell 2: 2 H2O  O2 + 4 H++ 4 e For the same current, Vol of Cl2 will be twice of O2. Hence, for equal vol of Cl2 and O2, current for cell 2 must be 2 I. If O2 must be twice of Cl2 then Cell 2 current must be 4 I. Answer: D 8. 2 H2 CO CH3OH Initial /mol 2 1 0 Change /mol −x − ½ x + ½ x Final /mol 2 – x 1 – ½ x ½ x Answer: C 9. Pyruvic acid is a weak acid, hence upon titration with NaOH, at equivalence point, pH > 7 due to anion hydrolysis. CH3COCOO- CH3COCOOH + H2O + OH- Answer: A

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 10. Aspirin: HA H+ + A−. At pH = 1, there is a high [H+], thus, the above equilibrium position is shifted to the left. [H+] ≠ [A−] since at pH 1, H+ does not come from HA alone but also comes from the stomach acid. Since equilibrium position is shifted to the left, [HA] > [A−]. [H+] > [HA] > [A−] Answer: D 11. HCl and NaOH are mixed: HCl + NaOH  NaCl + H2O. HCl used is in excess. Since there is excess strong acid, the resulting mixture is NOT a buffer. The pH would be due to the excess H+ from the strong acid present. No. of moles of excess HCl = 0.004 − 0.0025 = 0.0015 mol There are 0.0015 mol of H+ in 1 dm3, thus, [H+] = 0.0015 mol dm−3 pH = − lg (0.0015) = 2.82 Answer: D 12. No. of half lives Uranium−238 Lead 0 1 0 1 ½ ½ 2 ¼ ¾ After 2 half lives, ratio of uranium−238: Lead is ¼ : ¾ which is 1:3. Hence after 2 half lives, time taken is 2 x 4.5 x 109 years = 9.0 x 109 years. Answer: C 13. For first order reactions, half life is independent of [H2o2], hence, doubling the concentration from 0.1 mol dm−3 to 0.2 mol dm−3 has no effect. t ½ = ln2/k. Since ln2 and k are constants, t ½ is constant and likewise, time taken to decompose by 10 % will also be constant at 5 minutes. Answer: B 14. Uncatalysed is 1 step, catalyzed reaction will be 2 steps. Catalysed reaction will also have lower E a. However, enthalpy ∆H remains the same since the reactants SO2 + O2 and the product SO3 is the same. Answer: A 15. When conc HCl reacts with Cu2+, the water ligands are replaced with Cl−. No change in oxidation state of Cu2+, hence no.of d-electrons remain the same. However, since there is a change in ligands, the energy gap between the split d-orbitals will change since water ligands and Cl− will cause different degree of repulsion with the electrons in the d-orbitals. [Cu(H2O)6]2+ + 4 HCl [CuCl4]2− + 6 H2O + 4 H+ Answer: C 16. Reactivity of Group 2 elements with water increases from Mg to Ba (down the group) as the valence electrons become further from nucleus, hence more easily lost to reduce water to H2 gas. CaO also readily forms Ca(OH)2 with water. Hence, reaction of Mg with H2O (l) will occur least readily. However, with steam, Mg will react more readily as temperature is higher. NOTE: the product of Mg with steam is MgO as temperature is higher, hence, the Mg(OH)2 would have decomposed to MgO. Thus, MgO is formed instead of Mg(OH)2. Answer: D

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 17. The element has a high density and melting point, hence it must be a transition element. The slightly larger increase from the first to second electron could be due to the fact that the first electron is removed from 4s while second electron is from 3d subshell. Answer: D 18. Not in syllabus With hot sodium hydroxide, chlorine forms sodium chlorate(V), NaClO3. 3Cl2(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l) Answer: C 19. HC CCH 2CH CH 2 There is no 2sp − 2sp2 overlap since the sp and sp2 C are not next to each other. At the C=C and C≡C bonds, the pi bonds are formed by the side on overlap between the 2p orbitals of C. The 1s and 2sp overlap occur between the triple bonded C and the H atom which uses its 1s orbitals for sigma bond with the 2sp hybrid orbital of C. Answer: C 20. X is an alkyl radical. Possible alkyl radicals are: CCH3CH2 CH2 CH3 CH3 C CH3CH CH3 CH3 CH3 CCH2CH2 CH3 CH3 CH3 There are 3 possible X•. (NOTE: the 3 CH3 groups are identical as they are bonded to the same C, thus form the same radical when the H is removed from any of them). Answer: C 21. Reduction only occur at the C=O, hence must use NaBH4. H2 with Ni or Pt will reduce both C=C and C=O of the ketone group. (NOTE: LiAlH4 will reduce both COOH and C=O of aldehydes and ketones). H2O and H+ is used for hydrolysis of CN group to COOH. Answer: D 2sp 2sp