2025 ASRJC Prelim H2Chem P2 MS

9729/02/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2025 JC 2 PRELIMINARY EXAMINATION PAPER 2 SUGGESTED SOLUTIONS NAME:______________________________ ( ) CLASS: 25 / _____ (a) Mild steel is an alloy that contains iron and carbon. A sample of mild steel was analysed, and four different types of atoms were identified; A, B, C and D. Table 1.1 shows information about the four types of atoms found in the sample. Table 1.1 atom relative mass relative % abundance A 12.00 0.238 B 13.00 0.012 C 53.94 5.79 D 55.93 93.96 (i) Calculate the relative atomic mass of carbon in this sample to four significant figures. Show your working. [1] Relative atomic mass of carbon = 12×0.238 + 13×0.012 0.238 + 0.012 = 12.05 [1] (ii) In an experimental set -up, beams of particles travelling at the same speed from different sources are subjected to an electric field as shown in Fig. 1.1. A beam of protons with an angle of deflection of 60° has already been drawn. + protons Fig. 1.1 Under identical conditions, beams of particles 12C2+ and 56Fe2+ were subjected to the same electric field. Calculate the angle of deflection of 12C2+ and 56Fe2+particles under the electric field and sketch the beams of 12C2+ and 56Fe2+ on Fig. 1.1. Label the beams clearly. [2] Angle of 12C2+ = 2 12 x 60o = +10.0o Angle of 56Fe2+ = 2 56 x 60 = +2.14o [1] for both [1] for both beams 12C2+ 56Fe2+

(b) The compound C6H6 has many possible structural isomers. Three suggested structures of C6H6 are shown in Fig. 1.2. Kekulé benzene Dewar benzene Ladenburg benzene Fig. 1.2 (i) Using Fig. 1.2, complete Table 1.2 to predict the number of carbon atoms that have sp, sp 2 and sp 3 hybridisation in Kekulé benzene, Dewar benzene and Ladenburg benzene. Table 1.2 C6H6 structure sp hybridised sp2 hybridised sp3 hybridised Kekulé benzene Dewar benzene Ladenburg benzene [2] [1] row 1 and 2 correct; [1] row 3 correct (ii) Dewar benzene contains both  bonds and  bonds. By reference to the hybridrisation of the carbon atoms and orbital overlap, describe the covalent bonding in Dewar benzene. [2]  bonds: [1] • Head-on overlap of the sp2 orbital of C with sp2 orbital of adjacent C • Head-on overlap of sp3 orbital of C with sp3 orbital of adjacent C • Head-on overlap of sp2 orbital of C with sp3 orbital of adjacent C • Head-on overlap of the sp2 orbital of C with the s orbital of H • Head-on overlap of the sp3 orbital of C with the s orbital of H  bonds: [1] • Side-way overlap of two unhybridised 2p orbital of sp2 C (iv) Suggest why Dewar benzene and Ladenburg benzene are unstable isomers of C6H6. [1] Bond strain or ring strain [1] [Total: 8]

9729/02/H2 [Turn over 2 Aluminum is the most abundant metal in the earth’s crust and has been produced commercially since 1888. It is now the second most used metal in the world after iron. Approximately 75% of aluminum ever produced is still in use today, as it can be recycled endlessly without compromising any of its unique properties or quality. (a) Aluminum objects that have had the aluminum oxide layer removed may be anodised. (i) Complete Table 2.1 to show the relevant half-equations, during the anodisation of an aluminum object. Table 2.1 half-equation anode .…. Al + .….. H2O → ….. Al2O3 + ….. H+ + …...e cathode [2] Table 2.1 half-equation anode 2Al + 3H2O → Al2O3(s) + 6H+ + 6e– [1] cathode 2H+ + 2e– → H2 [1] (ii) During the anodisation of an aluminum object, 3.50 g of a protective layer aluminum oxide is formed in 2 hours. Calculate the value of the current used. [2] Amount of Al2O3 = 3.50 0.034313 mol2 27.0 3 16.0 = +  Amount of electrons passed = 0.034313 x 6 = 0.2059 mol [1] en F I t 0.20588 96500 I (2 60 60) I 2.76A  =   =    = [1]

(b) Fig. 2.1 shows an incomplete energy cycle involving aluminium oxide, Al2O3. Fig. 2.1 (i) Complete line C. Include state symbols. [1] 2Al3+(g) + 3O2−(g) [1] line A: 2Al(s) + O2(g) line D: Al2O3(s) line B: 2Al(g) + 3O(g) line C: …………………………… process 1: enthalpy change of formation of Al2O3(s) process 2: enthalpy change of atomisation of Al and O process 3 process 4: lattice energy of Al2O3(s)

9729/02/H2 [Turn over (ii) Using Fig 2.1, the data in Table 2.2, together with data from the Data Booklet, to calculate the lattice energy of Al2O3(s). Table 2.2 Ho / kJ mol–1 1st electron affinity of oxygen, O(g) + e– → O–(g) –141 2nd electron affinity of oxygen, O–(g) + e– → O2–(g) +790 standard enthalpy change of atomisation of Al(s) +326 standard enthalpy change of formation of Al2O3(s) –1676 [2] ∆Hf = 326 x 2 + 3/2 x 496 + 2(577+1820+2740) + 3 x (–141) + 3 x 790 + LE [1] = –1676 LE = -15293 kJ mol-1[1] (iii) Explain why the first electron affinity of oxygen is exothermic, but the second electron affinity is endothermic. [2] First EA is exothermic as energy is released when an electron is attracted to the neutral oxygen atom by its nucleus. [1] Second EA is endothermic as energy is required to overcome the repulsion between the added electron and the negatively charged O⁻ ion. [1]

(c) When aluminum reacts with dry chlorine, aluminum chloride, AlCl3, is formed. (i) AlCl3 can undergo dimerisation to form Al2Cl6. With the aid of a diagram, name the type of bond form ed during dimerisation and explain why this bond is formed. [2] Al Al Cl Cl Cl Cl Cl Cl [1] In forming the Al2Cl6 dimer, two AlCl3 molecules are joined by dative bonds. Reason: Al is electron deficient and can accept lone pair of electron from chlorine forming octet structure [1] name and reason (ii) When AlCl3 is dissolved in water, a solution of pH 3.0 is formed. Explain with the aid of a balanced equation why the solution has a pH of 3.0. [2] AlCl3(s) dissolves in water to form hydrated [Al(H2O)6]3+(aq). Due to high charge density of Al3+, it is able to polarise the neighbouring water molecule, which further breaks the O -H bonds, thereby producing H + in the solution. Hence, AlCl3 undergoes hydrolysis in water to form an acidic solution. [1] [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) [1] (iii) AlCl3 can be used as a catalyst in the reaction of methylbenzene with chloroethane to form 4-ethylmethylbenzene. Describe the mechanism of this reaction. [3] Electrophilic Substitution Step 1: CH3CH2Cl + AlCl3 CH3CH2+ + AlCl4– CH3 + H CH2CH3 CH3 + CH2CH3 CH3 H++slow step 2 fast step 3 CH3CH2 + Step 4: H+ + AlCl4–