RI 2024 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2024 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 D D C D B B A D C C D A C B A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 D C D D B C D C B D A C B C B Q 1(D) Options A and B are incorrect as X is a cation while both options show anions. No. of e− in Ne atom = 10 protons neutrons electrons 23Na+ 11 12 10 24Mg2+ 12 12 10 T hus, cation X is 24Mg2+ since it has the same number of neutrons as protons. Q2(D) Statement 1 is incorrect as a longer covalent bond generally results in a smaller bond energy. Statement 2 is incorrect as I in H I has a larger nuclear charge than Cl in HC l but HI has a lower bond energy than HCl. Statement 3 is correct as more energy is required to overcome a greater attraction between shared pair of electons and the nuclei of the bonded atoms. Q3(C) Statement 1 is incorrect as H 2S has stronger permanent dipole-permanent dipole forces of attraction than H 2Se and hence a higher boiling point. This is so as the H−S bond is more polar than the H−Se bond due to greater electronegativity difference between H and S than H and Se since S is more electronegative than Se. Statement 2 is correct as H2O has less number of electrons and hence a smaller and less polarizable electron cloud, resulting in weaker instantaneous dipole-induced dipole forces of attraction than H2Se. Statement 3 is correct as the H−O bond is more polar than the H−S bond due to greater electronegativity difference between H and O than H and S since O is more electronegative than S. Q4(D) There are four bond pairs of electrons on N and no lone pair of electrons, hence the bond angle is 109. There are four bond pairs of electrons on C in CH 2 and no lone pair of electrons, hence the bond angle is 109. There are three bond pairs of electrons on N and no lone pair of electrons, hence the bond angle is 120. Q5(B) T otal 10 single unpaired electrons 1 unpaired electron 1 unpaired electron 0 unpaired electron 2 unpaired electrons 3 unpaired electrons 2 unpaired electrons 1 unpaired electron 0 unpaired electron

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q6(B) Option A is incorrect as the highest possible oxidation state of P is +5 and PC l5 contains 85.1% by mass of chlorine. Option B is correct as the highest possible oxidation state of Si is +4 and SiO2 contains 53.2% by mass of oxygen. Option C is incorrect as the highest possible oxidation state of P is +5 and P4O10 contains 56.3% by mass of oxygen. Option D is incorrect as the highest possible oxidation state of S is +6 and SO3 contains 59.9% by mass of oxygen. Q7(A) Ar = 69.3  58 + 26.7  60 + 1.20  61 + 3.80  62 + 1.00  64 69.3 + 26.7 + 1.20 + 3.80 + 1.00 = 58.8 Note: Do not assume that relative abundances are percentages and do not assum