2012 JJC H1 Prelim P2 answers

1 Jurong Junior College 8872/02/Prelim 2012 Suggested Answers Suggested MARK SCHEME for 2012 H1 Chemistry Preliminary Examination Paper 2 Section A Minus 1m overall for 3 s.f. mistake General comments: Most students did well for the sketches for the trends of first ionisation energy and ionic radii for the selected elements. However, they still are not careful in their explanation s, often missing out key words in their answers, especially on the reasons for the trend observed in ionic radii from Si4+ to Cl−. 1. (a) (i) Common mistakes: 1. Students plotted the ionic radius of Si4+ as the highest point. (ii) the difference in first ionisation energy between phosphorus and sulfur. the trend in ionic radius from silicon to chlorine. [5] Ionic radius Si P S Cl first ionisation energy Si P S Cl [1m] • Point for P is higher than Si and S but lower than Cl • Point for Si lower than S [1m] • Decreasing trend from P to Cl • Ionic radius for Si4+ is lower than those of P, S and Cl S has a lower first ionisation energy than P due to mutual repulsion between the paired 3p electrons in S which makes it easier to remove one of the paired 3p electrons in S than the unpaired 3p electron in N. [1m] Si4+ has one less quantum shell of electrons than P3−, S2− and Cl−/ the rest of the ions. Hence Si4+ has the smallest ionic radius. [1m] Across the isoelectronic series* (P3−, S2−, Cl−), the ionic radii decreases due to increasing nuclear charge and constant shielding effect by the same number of inner shells’ electrons. [1m]

2 Jurong Junior College 8872/02/Prelim 2012 Suggested Answers 1. (a) (ii) Common mistakes: 1. For 1 st IE: Some students gave the factors resulting in the general increase in first ionisation energies across a period. 2. For 1 st IE: Some did not state the trend (but these students still gained one mark for correct reasoning). 3. For ionic radius: Some students g ave reasons to explain why the ions have different radii from their respective atoms (this is an irrelevant answer). 4. Many students did not quote the identities of the ions in their explanation. For instance, the ionic radius of Si is smaller because it has one less quantum shell. In answers where the subject is not clear, marks were not awarded. 5. Many did not state why shielding effect by inner shell electrons remains relatively constant for P3−, S2− and Cl−. 6. Small of them were uncertain of how to use the term, ‘nuclear charge’. Wrong term like ‘atomic charge’ was seen. (b) (i) (ii) P Cl % by mass 30.4 69.6 Ar 31.0 35.5 Amount/mol 0.981 1.96 Ratio 1 2 Empirical formula is PCl2. [1m] with relevant working Let the molecular formula be (PCl2)n n  31.0 + 2n  35.5 = 200 n  200 102 = 2 Molecular formula is P2Cl 4. [1m] with relevant working involving the use of Mr value of 200 given White fumes of HCl. [1m] PCl3 + 3H2O → H3PO3 + 3HCl [1m], state symbols not required

3 Jurong Junior College 8872/02/Prelim 2012 Suggested Answers 1. (b) (iii) [5] Common mistakes: 1. Some students attempted to draw shape instead of a simple displayed formula for P 2Cl4, making the question difficult for themselves. 2. Some did not appreciate the phrase, ‘typical valencies’, because they drew PP triple bond. 3. A small number of them drew more than four Cl atoms. General comments: Q2 was badly done. Students were required to apply fundamental concepts to novel molecules. Most did not know how to apply their knowledge to new molecules. 2. (a) (i) Common mistakes: 1. Many students failed to recognise that ethanolamine and propylamine have very similar M r and hence, their difference in boiling points cannot be due to difference in strength of VDW forces. 2. Some thought that propylamine molecules form permanent-dipole permanent-dipole interactions. 3. Some thought ethanolamine molecules form pd-pd interactions and propylamine molecules form VDW forces. 4. Some forgot to use the preposition, ‘between’ in their description. P P Cl Cl Cl Cl               [1m], lone pair on each P must be shown More energy is required to overcome the more extensive hydrogen bonding between ethanolamine molecules than the less extensive hydrogen bonding between propylamine molecules. [1m]

4 Jurong Junior College 8872/02/Prelim 2012 Suggested Answers 2. (a) (ii) [2] Common mistakes: 1. Some students drew hydrogen bonding between two ethanolamine molecules. They did not read the question carefully. 2. Some did not write ‘+, −’ as a pair. (b) [1] Common mistakes: 1. Many failed to understand what is meant by a ‘weak Bronsted base in water’. They wrote the following wrong answers: (1) HOCH2CH2NH2 → CH2CH2NH2 + OH− (2) HOCH2CH2NH2 + H2O = −OCH2CH2NH2 + H3O+ (3) −OCH2CH2NH2 + H+ = HOCH2CH2NH2 General comments: For Q3, most students are not familiar with the definition of bond energy. In addition, they made careless mistakes in the substitution of bond energy values when solving for the answer in 3(b)(iii). 3. (a) . [1] Common mistakes: 1. Many did not understand what is meant by ‘orbital overlap’. They label the bonds present with ‘’ and ‘’ symbols. HOCH2CH2NH2 + H2O = HOCH2CH2NH3+ + OH− [1m] C H C H O   bond [1m] [1m] for diagram  illustrating at least one hydrogen bonding between one ethanolamine molecule and a water molecule,  hydrogen bond must be labelled,  +, − must be shown as a pair,  lone pair of electrons on O or N must be shown OCH2CH2N H H − + + ……… O  H H − + + hydrogen bond H  − + O H H − +

5 Jurong Junior College 8872/02/Prelim 2012 Suggested Answers 3. (b) (i) (ii) (iii) [4] Common mistakes: 1. Many students do not know the definition of bond energy. 2. Many students substituted wrong bond energy values such as those of O-O single bond, C-O single bond 3. Some left out the bond energy of oxygen molecule or apply the wrong coefficients. CH2=C=O + 2O2 → 2CO2 + H2O [1m], state symbols not required Bond energy is the energy required to break one mole of covalent bonds in the gas phase/ gaseous state. [1m] Hc(ketene) = E(reactants) − E(products) = 2E(C-H) + E(C=C) + E(C=O) + 2E(O=O)} − {4E(C=O) + 2E(O-H)} = {2(410) + 610 + 740 + 2(496)} – {4(740) + 2(460)} = −718 kJ mol−1 [1m] correct substitution [1m] correct answer + exact value + correct units

6 Jurong Junior College 8872/02/Prelim 2012 Suggested Answers General comments: Q4 was relatively well done. Most students were able to define ‘buffer solution’ and performed the calculations in (a)(ii). Again, they were less able to apply Chemistry knowledge in a given context. Hence, they tend to do badly in 4(a)(iii), (v). For 4(b) , some made mistakes in not being familiar with the different types of reducing agents and the organic functional groups that each type can reduce. 4. (a) (i) (ii) Ka = + 3 23 [H ][HCO ] [H CO ] − [1m] Since pH = 7.40, concentration of H+ = 10−7.40 = 3.98  10−8 mol dm−3. [1m] 3 23 [HCO ] [H CO ] − = a +[H ] K = 7 8 7.90 10 3.98 10 − −   = 19.8 [1m], ecf [H+] calculated, 3 s.f (iii) (iv) Common mistakes: 1. For (a)(iii), some students did not explain the reason, others did not understand the question, hence, they got the answers wrong. 2. There are still students who used equilibrium arrow when writing an equation to show buffering action. 3. For (a)(v) , many students did not make reference to which equilibrium reaction in their explanations. Some did not even explain in terms of equilibrium position shift. Others explain the second part only.