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