RI 2010 A-Level H2 Chemistry Solutions

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q1 Element Br K Kr Sc Covalent Radius/nm 0.114 0.196 0.117 (in KrF2) 0.144 Atomic radii should fall across the period with increasing effective nuclear charge. This is because across the period, nuclear charge increases but shielding effect does not increase as significantly. Thus K has the largest atomic radius. Ans : B Q2 Amt of CO2= 1.1 𝑥𝑥 107 22.7 = 4.85 × 105 mol since 3 mol of CO2 gives 2 mol of Na2CO3 Amt of Na2CO3 = 4.85 𝑥𝑥 105 3 x 2 = 3.23 × 105 mol Mass of Na 2CO3 = 3.23 x 105 x 106.0 = 3.42 × 107g Ans: B Q3 No of neutrons in 210 84 Po = 210 − 84= 126 He nucleus contains 2 neutrons & 2 protons No of neutrons in X = 126 − 2= 124 No of protons in X = 84 − 2= 82 Ans: B Q4 The interaction between gecko’s toe pads and the glass surface is van der Waals’ forces, depends on surface area (millions of microscopic hairs) and is a reversible process. Ans: D Q5 P H H H 3 bp & 1 lp abt P central atom Bon d angle = 107o Ans: B Q6 MgO has similar ionic lattice structure to NaCl. Each ion is surrounded by six oppositely charge ions (co- ordination number =6), held by electrostatic forces of attraction. Ans: C Q7 NH 3 has the more significant and strongest intermolecular interactions (hydrogen bonding) amongst the 4 gases of the 4 options. The other 3 gases are non-polar covalent simple molecules, thus they have van der Waals’ forces of interactions. Ans: A Q8 (A) Enthalpy change of atomisation, ∆H atm, involves the liberation of gaseous atoms from elements in standard states. This means breaking of bonds between atoms and always be endothermic (B) Enthalpy change of combustion, ∆H C, is the enthalpy change when one mole of substance undergoes complete combustion in excess oxygen. Combustion is usually exothermic, but occasionally endothermic, for example, in N2. (C) Enthalpy change of formation, ∆Hf , is the enthalpy change when one of substance is formed from its constituent elements at their standard states. Whether ∆H f is exothermic or not, it depends on the relative stability of the substance to be formed and its constituent elements in their standard states. (D) ∆ Hsoln =∆Hhyd− L.E. Whether∆Hsoln is exothermic or not, it depends on the relative magnitude of∆Hhyd and L.E. Ans: A Q9 According to Arrhenius eqn, k = A exp ( -Ea/RT), rate constant, k, increases in the presence of catalyst as the Ea is decreased. Energy profile should show lower Ea (energy difference betw reactant & transition state). However, ∆H should be the same since reactants & products are the same, thus product level should be at 5, not lower than 5. Ans: D Q10 When steam condenses at its boiling point of 100°C (373 K), ∆G= 0 since dynamic equilibrium exist when H 2O(g) H2O(l) ∆G= ∆H−T∆S 0 = −44 kJ mol−1 54()18.0 mol − (373 K)(∆S) Suggested Solutions to N2010 H2 Chemistry Paper 1 (9746/01) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 B B B D B C A A D A B D C B C B C C B 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 D A B C C D A B A B C B C A A C Working for Suggested Solutions to N2010 H2 Chemistry Paper 1 (9746/01)

This document is copyrighted, please do not reproduce it without permission © Raffles Institution ∆S = 11132 0.354 354373 kJK JK−−− = −= − Ans: A Q11 Strongest reducing agent has the greatest tendency to be oxidized, i.e. either Fe2+ or Br-. Since reduction potential EFe3+/Fe2+ is less positive than EBr2/Br- Fe2+ is more likely to be oxidized (Fe2+→Fe3+, -0.77V) than Br− (Br−→Br2, -1.07V) , thus Fe2+ is the strongest reducing agent. Ans: B Q12 Overall Eqn: 2MnO 4−+ 5SO2 + 2H2O → 2Mn2+ + 5SO42− + 4H+ Since H+ is produced, [H+] increases, causing pH to decrease generally. Ans: D Q13 N2O4(g) 2 NO2(g) Initial P/atm P ∆ in P/atm −0.5 P + P Eqm P/atm 0.5 P P = 1 3 atm = 2 3 atm Total P at eqm = 1 atm 1.5 P = 1 atm ∴ P = 2 3 atm 22 2 24 2()() 43 1 3() 3 NO P NO atmPK atmP atm = = = Ans: C Q14 Charge Density α q r + + (A) q r + + of Mg2+ = 2 30.80.065nm + = (B) q r + + of Al3+ = 3 60.00.050nm + = (C) q r + + of Fe3+ = 3 46.90.064nm + = (D) SiCl4 hydrolyses completely in water to give Si(OH)4(s) or SiO2.2H2O(s), hence Si4+(aq) will not exist in aqueous solution. Ans : B Q15 (A) Si, not P, has the highest melting pt in Period III. (B) Like P, S can also form 2 acidic oxides, SO2 and SO3. (C) P can form P4 molecule. Other simple molecules in Period III are S8 and Cl2. (D) Like P, Si can also form chloride that hydrolyses completely in H 2O to form acidic soln. SiCl4 + 4H2O → SiO2.2H2O + 4HCl Ans: C Q16 (A) Na2O + H2O → 2NaOH (B) SiO2 is insoluble, with giant molecular structure and all atoms are linked by strong covalent bonds. (C) P 4O10 + 6H2O → 4H3PO4 (D) SO3 + H2O → H2SO4 Ans: B Q17 Examine if halogen can oxidize Fe(II) to Fe(III) Rxn betw Fe2+&Cl2 2Fe2+ + Cl2→ 2Fe3+ + 2Cl− E= +1.36 − (+0.77) = +0.59 V > 0 (feasible) Rxn betw Fe2+&Br2 2Fe2+ + Br2→ 2Fe3+ + 2Br− E= +1.07 − (+0.77) = +0.30 V > 0 (feasible) Rxn betw Fe2+&I2 Fe2+ + I2→ Fe3++ 2I− E= +0.54 − (+0.77) = −0.23 V < 0 (not feasible) Ans: C Q18 A transition element is defined as a d-block element which can form one or more stable ions with a partially filled d sub-shell. The rest of the 3 options are merely properties of transition elements. Ans: C Q19 C1 and C2 are sp1 hybridised and C3 is sp3 hybridised. sp orbitals have higher “s character” than sp2 and sp3 hybrid orbitals. Thus electrons in sp orbitals are more strongly attracted to the nucleus than sp 2 or sp3 orbitals. Hence bonds formed via sp orbitals are stronger than sp2 or sp3 orbitals Ans: B

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q20 * ** * absence of uv light+ 2Br2 Br Br Br Br inert solvent Ans: C Q21 (A) The C=C is unlikely to remain intact upon free radical substitution. (B) C- F bond is extremely strong due to small atomic size of F atom which leads to effective overlapping of orbitals. Thus F− is not readily replaced by NH3 to undergo nucleophilic substitution. (C) CH3 is 2,4-directing, thus NO2+ is unlikely to be substituted in position 3 in the benzene ring. (D) CH3CH2 is 2,4-directing, thus Br+ is likely to be substituted in position 4 in the benzene ring as the major product. Ans: D Q22 (A) H I in aqueous medium implies large excess of H 2O which can be added across C=C bond to form CH3CH2CH(OH)CH3 as major product instead. (B) Sodium ethoxide in alcoholic medium acts as base, favours elimination of HBr to form but-2- ene by Elimination mechanism. (This is the “active species” in the use of alcoholic KOH or NaOH, the detailed mechanism is not in syllabus.) Elimination (E2) mechanism (Not in syllabus) CH3CH2O− C C H H3C H Br H CH3 CH3CH2O C C H H3C H Br H CH3 − CH3CH2OH C C H3C H Br− H CH3 (C) A lCl3 is rapidly hydrolysed by water. The hydrated Al3+ ion will not be able to generate Cl+ electrophile (not electron deficient to function as a Lewis acid to extract Cl− from Cl2). (D) C l2 in aqueous medium cannot undergo homolytic fission to generate Cl radicals. Both Cl2 and cyclohexane must be in same physical state (gaseous state) to undergo free radical substitution. Ans: B Q23 X only undergoes nucleophilic substitution only when heated, thus X must be a halogenoalkane and not acyl halide, as heating is required to break C-X bond. X is Cl as white ppt of AgCl is formed upon addition of AgNO3(aq). Chlorobenzene is not possible as the C-Cl bond has double bond character and cannot undergo nucleophilic substitution. Ans: B Q24 CH2Cl H2, Pt CH2CN CH2CH2NH2 NaCN P Q A