2023 TMJC H2 Chem Prelim P1 (Ans)

2023 H2 Chem Prelim P1 (Ans) Answer: C Let % abundance of 11B be m%. 10.75 = (m)(11)+(100-m)(10) 100 m = 75 2 Use of Data Booklet is relevant to this question. On losing an electron, which particle would have the greatest number of unpaired electrons? A C– B Fe3+ C Ti2+ D P Answer: B C– forms C : [He] 2s22p2 (2 unpaired electrons) Fe3+ forms Fe4+ : [Ar] 3d4 (4 unpaired electrons) Ti2+ forms Ti3+ : [Ar] 3d1 (1 unpaired electrons) P forms P+: [Ne] 3s23p2 (2 unpaired electrons) 3 Particle X has a charge of +1 and a proton number, n. Particle Y has a proton number of (n+1) and is isoelectronic with X. Which of the following statements correctly describes X and Y? A X and Y are isotopes. B The atoms of X and Y have the same electronic configurations. C Y has a charge of +1 and same charge density as that of X. D Y has a charge of +2 and smaller ionic radius than X. 1 A sample of boron contains two naturally occurring isotopes, 10B and 11B. This sample has a relative atomic mass value of 10.75. What is the percentage of the isotope 11B in the sample? A 20 B 25 C 75 D 80

2 2023 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry Answer: D X Y Proton number, n n n + 1 No of electrons n – 1 n – 1 Charge +1 +2 Option A is wrong X and Y have different number of protons. Hence they cannot be isotopes. Option B is wrong X and Y have different number of protons. X atom has n electrons while Y has (n + 1) electrons. Hence X and Y atoms have different full electronic configurations. Option C is wrong Y has charge of +2 while it has same number of electrons as that of X. Hence, Y has smaller ionic radius. This leads to higher charge density. Option D is correct Y has charge of +2 while it has same number of electrons as that of X. Hence, Y has smaller ionic radius. 4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory is used to predict the shapes of molecules or polyatomic ions. Which of the following pairs of species are planar? A F2O and PF5 B SO32– and ClF3 C BrF4– and NO2 D ICl5 and CO2 Answer: C F2O 2 bp 1 lp Bent planar PF5 5 bp 0 lp Trigonal bipyramidal Non-planar SO32– 3 bp 1 lp Trigonal pyramidal Non-planar ClF3 3 bp 2 lp T-shaped Planar BrF4– 4 bp 2 lp Square planar Planar NO2 2 bp 1 lone e - Bent Planar ICl5 5 bp 1 lp Square pyramidal Non-planar CO2 2 bp 0 lp linear Planar

3 [Turn over Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Chemistry 5 Which of these phenomena cannot be explained by hydrogen bonding? 1 NH4Br has a higher boiling point than HBr. 2 Ice is less dense than water. 3 Dimerisation of NO2 to form N2O4. 4 Ethanoic acid forms dimers which dissolved in organic solvents A 1, 3 and 4 only B 1 and 3 only C 2 and 4 only D 1 only Answ