RI 2022 Y5 Promo Solutions

2022 Y5 H2 Chemistry Promotion Examinations Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C B B D D D C C A A B A A C B Question 1 (C) From the question, combustion of CS2 and H2S gives CO2 and SO 2; the hydrogen in H 2S would be converted to H2O, i.e. CS2 + H2S + O2 ⎯→ SO2 + CO2 + H2O Balancing the equation gives CS2 + H2S + 9/2O2 ⎯→ 3SO2 + 1CO2 + H2O Question 2 (B) From period 7 of the Periodic Table, Assuming the ions have a nucleon number of 267, ion no. of protons no. of e– no. of neutrons = 267 – proton no. no. of neutrons – no. of e– Rf2+ 104 104 – 2 = 102 163 61 Db3+ 105 105 – 3 = 102 162 60 Sg4+ 106 106 – 4 = 102 161 59 Bh5+ 107 107 – 5 = 102 160 58 Question 3 (B) A 3rd IE of Na : Na2+(g) ⎯→ Na3+(g) + e– [He]2s22p5 [He]2s22p4 3rd IE of Ne : Ne2+(g) ⎯→ Ne3+(g) + e– [He]2s22p4 [He]2s22p3 Incorrect. Na 2+ and Ne 2+ have the same number of electron shells. Na2+ has a higher IE due to greater nuclear charge and approximately constant shielding compared to Ne2+. B Correct. The following equations show the 4th IE of the five elements. O3+(g) ⎯→ O4+(g) + e– [He]2s22p1 [He]2s2 F3+(g) ⎯→ F4+(g) + e– [He]2s22p2 [He]2s22p1 Ne3+(g) ⎯→ Ne4+(g) + e– [He]2s22p3 [He]2s22p2 Na3+(g) ⎯→ Na4+(g) + e– [He]2s22p4 [He]2s22p3 Mg3+(g) ⎯→ Mg4+(g) + e– [He]2s22p5 [He]2s22p4 The 4th electrons were all removed from the 2p subshell. D Incorrect. Successive IE involve successive removal of electrons from the same element, hence the nuclear charge does not change.

Question 4 (D) molecule shape polarity NF3 4 regions of e– density + 1 lone pair trigonal planar pyramidal polar CH2F2 4 regions of e– density + 0 lone pair tetrahedral non-polar SO2 3 regions of e– density + 1 lone pair linear bent polar PCl5 5 regions of e– density + 0 lone pair trigonal bipyramidal non-polar (The dipole pointing up is cancelled out by the dipole pointing down. The remaining 3 dipoles lie on a trigonal plane and cancel each other out exactly) Question 5 (D) 1 Incorrect. NaHF2 is an ionic compound which dissolves in water to form aqueous ions. These ions form ion-dipole interactions with water. 2 Incorrect. structure no. of lone pairs 4 2 no. of H 2 2 average no. of hydrogen bonds 2 2 H2O2 and N 2H2 form the same average number of hydrogen bonds i.e. they have the same extensiveness of hydrogen bonding. 3 Correct. SF4 is polar which allows it to form permanent dipole – permanent dipole interactions with other polar molecules. 4 Correct. Despite the presence of the –OH groups which allows the molecules to form intermolecular hydrogen bonds, CH3(CH2)17OH has a very long alkyl chain with a very large and