ASRJC 2021 J2Prelims H2Chem P2 soln

1 ASRJC JC2 PRELIMS 2021 9729/02/H2 [Turn over Anderson Serangoon Junior College 2021 JC2 Preliminary Examination H2 Chemistry (9729) Paper 2 Suggested Solutions 1 (a) Describe and explain the trend in the electronegativity of elements across Period 3 from sodium to chlorine. [2] Electronegativity increases across Period 3. • Nuclear charge increases across Period 3. • Shielding effect between nucleus and valence electrons are similar since successive elements in the period have an additional electron in the same valence shell/same no. of inner shells. • There are stronger (electrostatic) forces of attraction between the nucleus and the electron pair in a covalent bond. (b) Some ionic radii are listed in the Data Booklet. (i) Explain the trend in ionic radius down Group 2. [2] Ionic radius increases down the group. • Nuclear charge increase but number of (inner) electronic shells increases. • Outer electrons experienced greater shielding effect and are further from the nucleus. • there are weaker electrostatic forces of attraction between the nucleus and the outer electrons. (ii) Explain the differences between the ionic radii of P3–, Cl– and Ca2+. [1] Ionic radius decreases from P3– to Cl– to Ca2+. • P3–, Cl– and Ca2+ are isoelectronic/same number of electrons and • number of protons / nuclear charge increases from P3– to Ca2+ while the shielding effect remains the same, • resulting in an increasing attraction between the nucleus and outer electrons. (accept there is an increasing attraction between the increasing number of protons and the same number of electrons.) (iii) Hence, suggest a value for the ionic radius of a potassium ion, K+. [1] 0.138 nm (any value between 0.099 and 0.181) K+ is isoelectronic with the ions in (b)(ii) and its nuclear charge is between that of Cl– and Ca2+. Hence, its ionic radius will be between 0.099 and 0.181 nm.

2 ASRJC JC2 PRELIMS 2021 9729/02/H2 (iv) Explain the difference in size between the radius of potassium ion and the radius of a potassium atom. [1] Radius of K+ ion is smaller than that of K atom • nuclear charge remains the same. • the cation has one less electronic shell than the atom. • the outer electrons in the cation are less shielded and closer to the nucleus. • there are stronger electrostatic forces of attraction between the nucleus and the outer electrons in cation. (c) Anodisation of aluminium is a process which coats an oxide layer on aluminium objects. (i) Draw a labelled diagram of the electrolysis cell used to anodise a small piece of aluminium object. Include details of the cathode, anode and electrolyte. [1] (ii) Complete Table 1.1 to show the type of reaction occurring, with the relevant half–equations, during the anodisation of the aluminium object. [2] Table 1.1 type of reaction occurring half–equation(s) anode oxidation 2Al + 3H2O ® Al2O3(s) + 6H+ + 6e– cathode reduction 2H+ + 2e– ® H2 (d) The molecules of alcohol P, C7H16O, are optically active and does not react with hot, acidified Na2Cr2O7(aq). On treatment with Al2O3, P produces a mixture of four different isomeric alkenes with the formula C7H14, only two of which are cis–trans isomers of each other. Suggest the structural formula of compound P and the four alkenes. [4] anode cathode dil. H2SO4 + – Pt Al object e– flow

3 ASRJC JC2 PRELIMS 2021 9729/02/H2 [Turn over P Alkenes formed from P (e) The following equilibrium exists in a sample of aluminium chloride vapour. Al2Cl6(g) ⇌ 2AlCl3(g) (i) Draw a dot–and–cross diagram of the Al2Cl6 molecule, including its co–ordinate bonds. [1] When 1.50 g of aluminium chloride was introduced into an evacuated flask of 250 cm3 capacity and heated to 500 K, the pressure inside the flask rose to 1.16 x 105 Pa. (ii) Assuming the gaseous mixture behaves ideally, calculate the average Mr of the mixture. Give your answer to four significant figures. [2] M = mRTpV M = (1.50)(8.31)(500)(1.16 × 105)(250 × 10–6) = 214.9 g mol–1 Average Mr of mixture = 214.9 (iii) Using the following relationships, calculate the mole fraction of Al2Cl6, x and the mole fraction of AlCl3, y, in the mixture. x + y = 1 average Mr = 267x + 133.5y [1] Given the mole fraction of Al2Cl6 be x and mole fraction of AlCl3 be y. [2(27.0) + 6(35.5)]x + [27.0 + 3(35.5)]y = 214.9 and x + y = 1 267x + 133.5(1 – x) = 214.9 133.5x = 214.9 – 133.5 x = 0.610 y = 0.390 OH Al Cl Cl Cl Cl Al x x x x x x Cl Cl

4 ASRJC JC2 PRELIMS 2021 9729/02/H2 (iv) Hence calculate the partial pressures of Al2Cl6 and AlCl3 in this mixture. [1] partial pressure (p.p.) of Al2Cl6 = 0.610 x 1.16 x 105 = 70729 Pa p.p. of AlCl3 = 0.390 x 1.16 x 105 (or 1.16 x 105 – 70729) = 45240 Pa (v) Write an expression for Kp for the reaction, and calculate its value. Include units in your answer. [2] Kp = pAlCl32pAl2Cl6 = (45240)270729 = 28900 Pa

5 ASRJC JC2 PRELIMS 2021 9729/02/H2 [Turn over 2 (a) When Group 2 iodates(V), M(IO3)2, is heated, it behaves in a similar way to the Group 2 carbonates. Upon heating, it decomposes as shown. 2M(IO3)2(s) ® 2MO(s) + 2I2(g) + 5O2(g) (where M is a Group 2 metal) (i) Using your knowledge of Group 2 carbonates, suggest and explain the trend in thermal stabilities of the Group 2 iodate(V). [2] Down the group, ionic radius of M2+ increases. As a result, the charge density of the M2+ decreases, the M2+ becomes less polarising. The electron cloud of IO3- anion is less distorted. The I–O covalent bond within the IO3- anions is less weakened down the group. Thermal stability of Group 2 iodate increases down the group. X, Y and Z are Group 2 metals (Mg to Ba, not necessarily in that order). X(IO3)2, Y(IO3)2 and Z(IO3)2 are Group 2 iodates(V). The three graphs in Fig. 2.1 show the change in mass when 2.00 g each of X(IO3)2, Y(IO3)2 and Z(IO3)2 were heated separately at a temperature T oC. Fig 2.1 (ii) With reference to the information from Fig. 2.1, show, by calculations, that none of the above iodate(V) samples contains Mg(IO3)2. [2] Assume one iodate(V) given is Mg(IO3)2. Molar mass of Mg(IO3)2 = 374.1 g mol–1 Amount of Mg(IO3)2 = !.##$%&.$ = 5.346 x 10–3 mol Amount of MgO formed = 5.346 x 10–3 mol Molar mass of MgO = 40.3 g mol–1 Mass of MgO formed = (5.346 x 10–3)(40.3) = 0.215 g Since the mass of the oxide calculated does not correspond to that of either XO, YO or ZO, Mg(IO3)2 is not among the given iodate(V). (iii) Hence, suggest the identities of the three iodates(V). [No calculation is required.] [1] iodates(V) X(IO3)2 Y(IO3)2 Z(IO3)2 identity Sr(IO3)2 Ba(IO3)2 Ca(IO3)2

6 ASRJC JC2 PRELIMS 2021 9729/02/H2 (b) When a salt such as a Group 2 sulfate dissolves in water, the lattice energy must be overcome. (i) How will the magnitude of the lattice energy of Group 2 sulfates change from MgSO4 to BaSO4? [1] Mag