RI 2020 Y5 Promo Solutions

1 2020 Y5 Promotional Examinations Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer A B C A D D B D D C C A A B C Question 1 (A) MP: AlCl3 < MgO and MgCl2 AlCl3 is a simple covalent compound while MgO and MgCl2 are ionic compounds. Less energy is required to overcome the weak intermolecular forces of attraction between AlCl3 molecules compared to the strong ionic bonds in MgO and MgCl2. MP: MgCl2 < MgO Melting involves overcoming the ionic bonds between the ions in MgCl2 and in MgO. Since O2– has a higher charge and smaller size than Cl–, the lattice energy of MgO is more exothermic (has a larger magnitude) compare to the lattice energy of MgCl2. Therefore, MgO has stronger ionic bonds and a higher melting point. Question 2 (B) 1 Incorrect. Boiling involves overcoming the hydrogen bonds between H 2O molecules, the id -id interactions between CH 4 molecules and id -id interactions between SiBr4 molecules. It does not involve breaking the covalent bonds within the molecules. 2 Correct. 3 Incorrect. SiBr4 is a simple molecule. Question 3 (C) Lanthanum does not change its oxidation state in this reaction: IO3– → periodate(VII) + I2 + O2 Mole ratio of IO3– : I2 = 5:2: 5IO3– → periodate(VII) + 2I2 + O2 By counting atoms on the LHS and RHS, one periodate(VII) ion contain s one I. To balance the 5(1–) = 5– charge on the LHS, the RHS should have an overall 5– charge. Since I2 and O2 have no charge, the periodate(VII) ion has a 5 – charge i.e. periodate(VII) is IO65–. Question 4 (A) From the Periodic Table, Group 15 Group 16 As Se Sb Te Method 1 The ionization energy (IE) data for As, Sb, Se and Te are not available in the Data Booklet. For group 16 element, ns2 np4 ns2 np3 ns2 np2 ns2np1 ns2 For group 15 element, ns2 np3 ns2 np2 ns2 np1 ns2 ns1 IE4 of a Group 15 element involves the removal of e– from the inner ns subshell, requiring a larger than expected amount of energy compared to IE 3 which involves the removal of e– from the higher energy np subshell. Comparing the values of IE 3 and IE4 in the options, options A and C have large differences between IE3 and IE4 i.e. A and C are group 15 elements. Since antimony is below arsenic, the IE’s of antimony are lower than that of arsenic since IE’s decrease down the Group. Hence, A is Tellurium Antimony. Method 2 Group 15 Group 16 IE decreases down group As Se Sb Te IE increases across period Antimony has the lowest IE out of all the four elements. Look for the option which has the lowest IEs. IE1 IE2 IE3 IE4 IE1 IE2 IE3 IE4

2 Question 5 (D) W, X, Y and Z are consecutive elements. It is important to check whether the elements before and after the element suggested in the options have the correct number of unpaired electrons. A W X Y Z Element Zn Ga Ge As Valence shell electronic configuration 3d10 4s2 4s2 4p1 4s2 4p2 4s2 4p3 No. of unpaired e– 0 1 2 3 B W X Y Z Element Ca Sc Ti V Valence shell electronic configuration 4s2 3d1 4s2 3d2 4s2 3d3 4s2 No. of unpaired e– 0 1 2 3 C W X Y Z Element Zn Ga Ge As Valence shell electronic configuration 3d10 4s2 4s2 4p1 4s2 4p2 4s2 4p3 No. of unpaired e– 0 1 2 3 D W X Y Z Element Cr Mn Fe Co Valence shell electronic configuration 3d5 4s1 3d5 4s2 3d6 4s2 3d7 4s2 No. of unpaired e– 6 5 4 3 Question 6 (D) Equal number of moles of H2 and O2 means that the partial pressure of H2 and O2 are the same. poxygen = phydrogen = ½P Pa. 2H2(g) + O2(g) → 2H2O(l) Since H2 is fully consumed and mole ratio of H2 : O2 = 2:1, poxygen reacted = ½(phydrogen consumed) = ¼P Pa poxygen remaining = ½P – ¼ P = 0.25 P Pa Alternatively 2H2(g) + O2(g) → 2H2O(l) Initial partial pressure / Pa ½ P ½ P -- Since H2 is the limiting reagent and is fully consumed, Change in partial pressure / Pa –½ P –¼P -- Final partial pressure / Pa 0 0.25 P -- Question 7 (B) The definitions of the standard enthalpy change of atomisation of an element and a compound are different. For an element: It is the energy absorbed when 1 mole of gaseous atoms is formed from the element under standard conditions. For a compound: It is the energy absorbed when 1 mole of the compound is converted to gaseous atoms under standard conditions. A Incorrect : O2(g) → 2O(g) [Corrected : ½O2(g) → O(g)] B Hg(l) → Hg(g) C CO2(g) → C(g) + O2(g) [Corrected : CO2(g) → C(g) + 2O(g)] D C2H4(g) → 2C(s) + 4H(g) [Corrected : C2H4(g) → 2C(g) + 4H(g)] Question 8 (D) Heat absorbed by coffee = mcT = (205)(4.18)(42) = 35990 J Since reaction is 80% efficient, heat released by reaction = 35990 / 0.8 = 44990 J 4499064500 0.6975 mol qH n n n  = − − = − = Mass of CaO = 0.6975 x (40.1 + 16.0) = 39.1 g

3 Question 9 (D) 1 Incorrect. The diagram shows an endothermic reaction where the reaction takes in energy from the surroundings. Hence, there is a decrease in the temperature of the surroundings. 2 Correct. Since the products are of higher energy than the reactants, the products are energetically less stable than the reactants. 3 Correct. For the reaction to be spontaneous, G must be less than 0. Since H is positive and G = H – TS, G will be less than 0 only if a negative –TS term outweighs the positive H term. For –TS to be negative, S must be positive. Question 10 (C) We can infer from the information provided that the reaction rate is slow at the beginning but is increased later in the titration. This describes an autocatalytic reaction where the reaction rate is slow initially, but is faster after some reaction has ta ken place because one of the products formed (Mn 2+ in this case) acts a catalyst for the reaction i.e. C is correct. A Incorrect. Since MnO 4– is a titrant, it is consumed immediately after each drop is added to the solution in the conical flask. The concentration of MnO 4– does not “accumulate”, hence there is no concentration of MnO4– to speak of. B Incorrect. The phenomenon described involves the rate of the reaction, not the position of equilibrium. D Incorrect. CO2 gas escaping has no effect on the rate of the reaction. Question 11 (C) Conditions for experiment 1 leading to graph 1 V(Mn2+) / cm3 V(H2O2) / cm3 [H2O2] / mol dm–3 1 25 0.1 From the graph, the reaction for graph 2 • has twice the rate of production of O2 • produces double the O2 compared to that of graph 1. W V(Mn2+) / cm3 V(H2O) / cm3 [H2O2] / mol dm–3 1 25 0.2 Correct. [H2O2] in W is twice that in experiment 1 with the same volume of solutions. Hence, the rate of reaction is double and the rate of production of O 2 is also doubled. This also doubles the amount (and hence volume) of O2 produced. X V(Mn2+) / cm3 V(H2O) / cm3 [H2O2] / mol dm–3 1 50 0.1 Incorrect. X has double the volume of H 2O2 at the same concentration compared to experiment 1. Hence, the amount of H2O2 in X is double that of experiment 1 and will produce double the amount (hence volume) of O2 eventually. X can be split into two containers, each containing the same volume and concentration of H 2O2 compared to experiment 1, but with half the amount of Mn2+ catalyst ⇒ each smaller container produces O2 at half the rate of experiment 1. Combining the effects of the two split containers, the total rate of production of O2 is the same as that of experiment 1. Y V(Mn2+) / cm3 V(H2O) / cm3 [H2O2] / mol dm–3 2 25 0.2 Incorrect. With the same volume of H2O2, • [H2O2] in Y is twice that in experiment 1 ⇒ rate of reaction and hence rate of production of O2 is doubled 1 cm3 Mn2+ 25 cm3 0.1 mol dm–