RI 2019 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2019 A-Level H2 C hemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B C B C C A D B D D A B D D D 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B A A D D C A C B B C C B C(B) Q1 (B) 84 38 Sr has 38 protons, 46 neutrons, 38 e– ⇒ W, X and Y have 46 neutrons 84 2 38 Sr + has 38 protons, 46 neutrons, 36 e– ⇒ W, X– and Y2– have 36 e– ⇒ W has 36 e–, X has 35 e– and Y have 34 e– ⇒ W has 36 protons, X has 35 protons and Y have 34 protons Nucleon no. of W = 46 neutrons + 36 protons = 82 Nucleon no. of X = 46 neutrons + 35 protons = 81 Nucleon no. of Y = 46 neutrons + 34 protons = 80 Q2 (C) Electrons, being negatively charged, are deflected towards the (+) plate. Protons, being positively charged, are deflected towards the (–) plate. Electrons, having the same magnitude of charge but lighter than protons, have a higher charge/mass ratio, and are deflected more than protons. Q3 (B) 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 IE 4 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. B and D corresponds to Group 16 elements. Since Tellurium is below Selenium, the IE’s of Tellurium are lower than that of Se since IE’s decrease down the Group. Hence, B is Tellurium. Q4 (C) After forming a single bond between the two oxygen atoms, each oxygen will gain one more e – to achieve octet configuration i.e. peroxide has a 2– charge. Barium, a group 2 element, forms a cation with a 2+ charge. Q5 (C) A All bonds are sigma bonds. B CO2 contains two sigma bonds and two pi bonds. C CH3CHO contains six sigma bonds and one pi bond. D CH2CHCH3 contains eight sigma bonds and 1 pi bond. IE1 IE2 IE3 IE4 IE1 IE2 IE3 IE4

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q6 ( A) r pV nRT masspV RT M = = Since the volume, mass, and temperature for every gas are kept constant, 1 r p M∝ i.e. higher M r ⇒ lower P. Compound CH4 HCHO CH3Cl HCO2H Mr 16.0 30.0 50.5 46 Relative Pressure 1 (highest) 2 4 (lowest) 3 Q7 (D) 2NaN3(s) → 2Na(s) + 3N2(g) Amount of NaN3 reacted = 5.00 23.0 3(14.0)+ = 0.07692 mol Amount of N2 gas produced = (3/2)(0.07692) =0.1154 mol Volume of N2 gas = nRT p