RI 2011 A-Level H2 Chemistry Solutions

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q1 Ti: 22 electrons Ti4+: 22 – 4= 18 electrons Ans: A Q2 First ionisation energy (1 st I.E.) is the energy required to remove 1 mole of electrons from one mole of gaseous atoms in the ground state to form one mole of gaseous singly-charged cations. Ionisation energy (I.E.) of an atom is influenced mainly by two factors: 1. Effective nuclear charge 2. Distance of the valence electron from the nucleus Ne: 1s 22s22p6 F: 1s22s22p5 Both F and Ne are in the same period, hence the shielding effect experienced by their valence electrons is approximately the same. However, Ne has one more proton than F (i.e. greater nuclear charge), hence its valence electrons experience greater effective nuclear than those in F. Ans: D Note: Some students answered B. A complete oct et does not make it harder to remove an electron. Q3 Amt of C in CO 2 = 48 / 24000 = 0.00200 mol Amt of C in (Na2CO3)x.yH2O2 = Amt of C in CO2 x(10/1000 x 0.100) = 0.00200 x = 2 Amt of H2O2 = y(10/1000 x 0.100) = 0.00100y Amt of KMnO4 =24/1000 x 0.0500 = 0.00120 22 4 Amt of H O 0.00100y 2 Amt of KMnO 0.00120 5 = = y = 3 Ans: C Q4 pV = T nR = constant =pV (2p)(V) 273 + 27 T T = 600 K Ans: D Q5 C4H8(g) + 6O2(g) 4CO2(g) + 4H2O(l) 2C2H4(g) + 6O2(g) 2∆Hco [C2H4(g)]= x + y kJ mol−1 Ans: C Q6 + + qqLE rr − − ∝ + The magnitude of lattice energy is proportional to product of charges and inversely proportional to the sum of ionic radii. Ans: C Q7 Sn (white) Sn(grey) ∆Hfo = −2090 – 0 = −2090 J mol−1 ∆Sfo = 44.1 – 51.4 = −7.3 J K−1 mol−1 ∆Gfo = ∆Hfo − T∆Sfo = −2090 − (273+12)(−7.3) = −2090 − (285)(−7.3) Ans: D Q8: Equation in italics is out of syllabus At the anode, chloride is oxidised to chlorine gas. 2Cl− → Cl2 + 2e− At the cathode, water is reduced to hydrogen gas. 2H2O + 2e− → H2 + 2OH− Equation below not in syllabus Chlorine disproportionates in cold NaOH. Cl2 + 2NaOH → NaCl + NaClO + H2O Ans: C Q9 The metal will dissolve in CO2(aq), which is acidic, if Eo(H+/H2) – Eo(Mn+/M) > 0. As Eo(Cu2+/Cu) is the most positive, Cu has the least tendency to be oxidised. Ans: A Q10 As the diagram shows apparatus needed to measure the standard electrode potential of Fe 3+/Fe2+, X has to be an inert electrode, which is Pt , and Y should contain 1 mol dm−3 Fe2+(aq) and 1 mol dm−3 Fe3+(aq). Ans: D Suggested Solutions to N2011 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 A D C D C C D C A D C B B B C C C A C B 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 C B B B C A A D B B D B A B C B D D A B Working for Suggested Solutions to N2011 H2 Chemistry Paper 1 (9647/01) x kJ mol−1 (x + y) kJ mol−1 y kJ mol−1

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q11 A buffer solution is one which is able to resist a change in pH (i.e. maintain an almost constant pH) upon the addition of a small amount of acid or base. Hence statements (A) and (D) are wrong. A mixture of sodium citrate and citric acid functions as an acidic buffer. pH = pKa + lg [citrate] [citric acid] When a small amount of base is added, citric acid will react with the base and form citrate. Hence, [citric acid] decreases while [citrate] increases. Hence, pH of the buffer solution will increase slightly. Ans: C Q12 Let rate = k[NO2]m[SO2]n Comparing expts 2 and 3, where [SO2] = constant: When [NO2] is increased by 1.33 times, the relative rate also increases by 1.33 times. Hence, order of reaction wrt to [NO2] is 1(i.e. m=1). Comparing expts 1 and 2: 1n 1n Expt 2 1.44 [0.006] [0.006]:= Expt 1 1.00 [0.005] [0.005] k k 1.44 = (1.2)(1.2)n n = 1 Ans: B Q13 rate = k’[C 6H5N2+]a, where k’ = k[H2O]b. Rate ∝ (Vgas/t). Since Vgas constant, rate ∝ (1/t). As 1/t ∝ k[C6H5N2+]a, plot [C6H5N2+] vs (1/t). Ans: B Q14 (A) The element that has 4 atoms in its molecule is phosphorus (i.e. P 4), not sulfur. Sulfur exists as S8. (B) Aluminium has the highest electrical conductivity as each A l atom contributes 3 electrons into the ‘sea of delocalised electrons’. (C) In period 3, the element with the highest melting point is silicon (giant molecular structure), not aluminium. (D) The element with the largest anion is phosphorus, not chlorine. Trend in ionic radius is: Cl − < S2− < P3−. Ans : B Q15 Thermal decomposition of HX involves breaking the H−X bond. The strength of the covalent H −X bond is dependent on the effectiveness of the orbital overlap. H− I is weaker than H−Cl bond since the electron cloud of I is more diffuse than that of C l, hence the overlap of orbitals in HI is less effective. Ans: C Q16: Ca(NO3)2 → CaO + 2NO2 + ½O2: Given Amt of Ca(NO3)2 = 8.2 / 164 = 0.050 mol Amt of O2 = 0.5 x 0.050 = 0.025 mol Vol. of O2 = 0.025 x 24000 = 600 cm3 Ans: C Q17 In order to produce oxygen gas, H 2O2 must undergo oxidation and this occurs only in acid solution. In alkaline solution, H2O2 will be reduced to water as it oxidises Cr3+ ions. Ans: C Q18 The enthalpy change of fusion of the period 3 elements is related to melting points of these elements. Trend in enthalpy change of fusion: X > W > Z > Y Si has a giant covalent (or molecular) structure and has the highest melting point. A l, which ha s the 2 nd highest melting point, has a giant metallic structure. S and P have simple molecular structures. The melting point of S is greater than that of P as the electron cloud of S 8 is larger and more polarisable than that of P4. Ans: A Q19 The terminal carbon atoms are sp3 hybridised. (Note: options C and D are mutually exclusive) Ans: C Q20 C xH2x+2 + 6x+2 4 O2 → xCO2 + 2x+2 2 H2O y = 6x+2 4 = 1.5x + 0.5 i.e. straight line with gradient 1.5 Ans: B Q21 CH 2=CHCl is unreactive to nucleophiles for 2 reasons. (i) The electrons on the chlorine atom delocalise into the C–C π bond, giving the C –Cl bond double bond character, thus making it relatively stronger. Thus, the Cl will not be easily substituted. (ii) Nucleophiles are electron- rich, and the carbon attached to the chlorine atom is in a C=C double bond, which is also electron- rich. Hence the carbon attached to the chlorine will not attract the nucleophile. Ans: C

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q22 In an S N1 mechanism (nucleophilic substitution with only 1 species in the rate determining step and is therefore a 2- step reaction), the halogen atom leaves the C, leaving a carbocation as the intermediate. Note: Although this is a primary carbocation, the electrons from the benzene ring delocalise over the positively charged carbon, dispersing the positive charge and stabilising the cation. On the other hand, option (D) shows the S N2 mechanism (2 species in the rate determining step and hence a one- step reaction) . The OHˉ nucleophile and the halogen derivative react in the first step to form a species, at transition state, where the Br atom is just leaving as a bond is just forming between the nucleophile and the carbon atom. Option (A) is an intermediate in a nucleophilic addi tion reaction while option (C) is an intermediate in an electrophilic substitution reaction. Ans: B Q23 Steam with H + (i) can hydrolyse the ester (just as hot, dilute H2SO4 can hydrolyse an ester) (ii) can add water across an alkene (just as cold conc. H 2SO4, followed by warming with water, can add H2O across a