SAJC 2025 Dec Hol Revision Package (SDL) Answers

ANSWERS TO 2024 DECEMBER HOLIDAY REVISION PACKAGE Topic 1: The Mole Concept and Redox Reaction 1. SRJC/PRELIMS 2011/P3/1(a), (b) (a) C H N O % by mass 50.7 7.0 19.8 22.5 Ar 12 1 14 16 Mole Ratio 4.225 7 1.414 1.406 Simplest Ratio 3 5 1 1 Let the molecular formula of acrylamide be C3nH5nNnOn. Mr of acrylamide = 71.0 = n (3 × 12.0 + 5 × 1.0 + 14.0 + 16.0) n = 1 Hence, the molecular formula of acrylamide is C3H5NO. Possible structures of acrylamide: Comments: Knowing the amide structure is the key to derive the possible structures. (b) Maximum mass of acrylamide the woman can take in per day = 0.5 × 10−3 × 50 = 0.025 g Mass of acrylamide in 1 kg of coffee = 1.2 × 10−7 × 71.0 = 8.520 × 10−6 g per kg of coffee Maximum mass of coffee the woman can drink per day = 0.025 ÷ (8.520 × 10−6) = 2934 kg Since coffee has the same density as water, maximum volume of coffee the woman can drink per day = 2934 ÷ 1 = 2934 dm3 C C C O NH2 H H H C O H N H C C H H H

2. RVHS/PRELIMS 2011/P3/4(b) (i) Element K Br O Mass ratio 23.4 47.8 28.8 Mole ratio 23.4 39.1 =0.598 47.8 79.9 =0.598 28.8 16 =1.8 Simplest ratio 1 1 1.8 0.598 =3.01= 3 Formula of white solid is KBrO3 (ii) No of moles of KBrO3= 0.835 167 = 0.00500 mol No of moles of PbCl2 = = 4.17 278 =0150 mol = (iii) 2e + Cl2 → 2Cl– X 3 3H2O + Br– → BrO3– + 6H+ + 6e [acidic medium] 6OH– + + Br– → BrO3– + 3H2O + 6e [alkaline medium] K+ ions are participating ions – Balance accordingly. 3Cl2 + KBr + 6KOH → 6KCl+ KBrO3 + 3H2O 3. H2S → S + 2H+ +2e Amount of BrOx− = (12.00/1000) x 0.300 = 0.0036 mol Amount of S = Amount of H2S = 0.289 / 32.1 = 0.009 mol Mole ratio of BrOx− : H2S = 1 : 2.5 = 2 : 5 1 H2S lost 2 e. 5 H2S lost 10 e. 10 e gained by 2 BrOx−. 1 BrOx− gains 5 e. Final oxidation state of Br in Br2 = 0 Initial oxidation state of Br in BrOx- is +5 5 + (−2)x = −1 x = 3   nKBrO3 nPbCl2 = 0.00500 0.0150   1 3

Criteria of success checklist:  provided clear statements for ALL working  presented intermediate answers to 4 s.f.  presented final answers to 3 s.f.  presented Mr to 1 d.p.  provided units, when applicable Things to note: • Conversion of units for volume: 1 dm3 = 1000 cm3 = 10−3 m3 1 cm3 = 10−3 dm3 = 10−6 m3 1 m3 = 1000 dm3 = 106 cm3 • In the calculation of empirical formula from experimental data, if the following figures are obtained for the number of moles, they are usually multiplied by a factor in order to get the correct simplest ratio. E,g 1.5 (multiply by 2), 1.33 (multiply by 3 ), 1.25 (multiply by 4), 1.2 (multiply by 5) • In balancing half equations, 1. balance the atom that is not O or H first. 2. Balance O by adding H2O 3. Balance H by adding H+ 4. Balance charge by adding electrons Continue with step 5 to 7 for basic medium 5. Add OH- (same number of moles as H+) to both sides of the equation 6. For every mole of H+ + OH- , replace it with 1 mol of H2O 7. Simplify the equation (if necessary) by cancelling out H2O that occurs on both sides of the equation. • For questions to calculate oxidation number in redox reactions, 1. Write out one half equation for the known species 2. Work out the mole ratio of the 2 reactants 3. Using the mole ratio and the half equation written, determine the number of electrons transferred in the redox reaction. The number of electrons given out by the species that is oxidised will be taken in by the species that is reduced. 4. Determine the original/final oxidation number by taking into account the final/original oxidation state and the number of electrons taken in/ given out. • For oxidation state, the sign is written in front of the number e.g +2, -1

Topic 2: Atomic Structure 1(a) (b)(i) Y: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 Z: 1s2 2s2 2p6 3s2 3p6 3d10 (ii) Ionic size of Y is larger than that of Rb +. Y and Rb+ are isoelectronic (same no of electrons) hence the shielding effect is the same. The nuclear charge of Y is lower than Rb+ as it has fewer protons. Therefore effective nuclear charge of Y is lower than Rb+. The valence electrons are thus less strongly attracted towards the nucleus of Y, hence Y has a larger ionic radius than Rb+. (c) X: 1s2 2s2 2p6 3s2 3p4 P: 1s2 2s2 2p6 3s2 3p3 Se: [Ar] 3d10 4s2 4p4 (i) X has lower 1st ionisation energy than of P. This is because the presence of inter- electronic repulsion between the paired electrons in the 3p orbital outweighs the effect of increased nuclear charge from P to X, hence less energy required to remove an electron from the paired 3p electrons in X. (ii) Comparing X and Se, Se has one additional principal quantum shell, thus distance between valence electron and nucleus is greater . The valence electron also experience greater shielding . This leads to weaker attraction between valence electron and nucleus thus less energy is required to remove valence electron from Se. Hence Se has a lower first ionisation energy than X. Things to note: • Criteria of success for electronic configuration include  starting from 1s2 ([Ar] 3d104s24p6 is not acceptable)  superscript to represent number of electron in the subshell (1s2 or 1s 2 are not acceptable) Note that there should not be commas between subshell (1s 2,2s2,2p6 is unacceptable) Particle Electric charge Mass Number Number of Protons Electrons Neutrons X 0 32 16 16 16 Y -1 81 35 36 46 Z +3 70 31 28 39

• For comparison of ionic radius of isoelectronic ions, answers should compare  nuclear charge ( depends on number of protons)  shielding effect (depends on number of inner electrons)  effective nuclear charge  conclude that the species with valence electron more strongly attracted would have smaller ionic radius • Criteria of success for decreasing IE trend down the group, state  Se has a larger number of filled quantum/electronic shells than X, hence shielding/screening effect increases  The distance of the valence electrons from the positively charged nucleus is also greater for Se as compared to X  Hence, the valence electrons in Se are less strongly attracted to the nucleus.  Less energy is needed to remove the 4p (state the subshell) electron in Se compared to 2p (state the subshell) electron in X.  First ionisation energy of Se is lower 2(a) Group 15 1s2 2s2 2p6 3s2 3p3 (b) Either WCl3 Or WCl5 Trigonal pyramidal Trigonal bipyramidal (c) Pair 1: P & X Group 16 Pair 2: Q & Y Group 17

(d) S+ : 1s2 2s2 2p6 T+ : 1s2 2s2 2p6 3s1 Comparing T+ and S+, T+ has a higher nuclear charge but with one additional principal quantum shell, distance between valence electron and nucleus is greater. The valence electron experience greater shielding and hence weaker attraction between valence electron and nucleus, resulting in much lower energy required to remove the 3s electron in T+ than 2p electron in S+. 2nd IE of T is lower than that of S. Things to note: • Criteria of Success for dot-and-cross diagram  Using dot and cross only to represent electrons (other symbols such as circle are not acceptable)  Showing all electrons on the side (Cl) atoms • Criteria of Success for drastic decrease in 2nd IE between S and T  Compare difference in number of filled quantum shells and hence distance between valence electrons and nucleus  Compare difference in shielding experienced by valence electrons  Compare extent of attraction between valence electrons and nucleus  Compare amount of energy required to remove valence electrons.

Topic 3: Chemical Bonding 1. RVHS/PRELIMS 2011/P1/Q5 (a) (b) Dative bond / coordinate bond Comment: PCl3 can donate the lone pair electrons to BF3 and B has empty orbital to ac