RI 2019 Prelim P2 Ans

-1- Raffles Institution Preliminary Examination 2019 H2 Chemistry Paper 2 (Suggested Answers and Comments) 16 Sep 2019 Q1 (a) Assume 100 g of X. Si N H Mass / g 48.3 48.1 3.6 Amount / mol 48.3 28.1 = 1.72 48.1 14.0 = 3.44 3.6 1.0 = 3.60 Molar ratio 1 2 2 Empirical formula of X is SiN2H2. Comments • Generally well done. Q1 (b) The first ionisation energy of P is higher than that of Si. P has one more proton and hence a higher nuclear charge than Si. Though P also has one more electron than Si, the increase in shielding effect is minimal since this additional electron occupies the outermost shell. Consequently, the valence electrons in P experience a higher effective nuclear charge and are more strongly attracted by the nucleus. Comments • Most students were able to describe the trend across the period well. The key ideas of increased nuclear charge, relatively similar shielding effect and increased effective nuclear charge were all stated. • Students should be more concise. For ionization energy, the key consideration should be simply, i) nuclear charge, ii) shielding effect, hence, iii) effective nuclear charge, and attraction for valence electrons. Q1 (c) Al2O3(s) + 6HCl(aq) ⎯→ 2AlCl3(aq) +3H2O(l Or Al2O3(s) + 6H+(aq) ⎯→ 2Al3+(aq) +3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) ⎯→ 2Na[Al(OH)4](aq) Or Al2O3(s) + 2OH–(aq) + 3H2O(l) ⎯→ 2[Al(OH)4]–(aq) SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l) Or SiO2(s) + 2OH–(aq) → SiO32–(aq) + H2O(l) Comments • Students are reminded to be familiar with all chemical reactions of the Period 3 oxides and chlorides. • This question was poorly answered. Students need to manage the knowledge of inorganic chemistry with understanding and technique, not just rote, as poor recall means inability to answer these standard questions. • Firstly, the acid base properties of oxides across period tends from basic to acidic (correlating to structures: ionic to covalent) . Thus, Al2O3, which has ionic with covalent character, is amphoteric. • Secondly, basic oxides neutralise acids. Acidic oxides neutralise bases. Amphoteric oxide neutralizes both acids and bases. • Thirdly, writing equations involve the following technique: i) The formula of the “salt” needs to be recalled. ii) Balance non-H and non-O atoms first (eg, Na, Al, P, SI)

-2- iii) Balance charges with H+ or OH– if ionic equation is written iv) Always balance the H and O atoms last using H2O Rather than memorise and recall equations (with so many coefficients to note), it is better to know the type of reaction, the expected products, and balance with proper technique. Following steps above i) Identify salts: Al2O3 -> Na[Al(OH)4] ii) Balance Na, Al: Al2O3 + 2NaOH -> 2Na[Al(OH)4] iii) No need. iv) Balance H and O with H2O Al2O3 + 2NaOH + 3H2O -> 2Na[Al(OH)4] Q1 (d) P4O10 has a simple molecular structure with weak instantaneous dipole-induced dipole (id-id) interactions between its molecules. SiO2 has a giant molecular structure with strong Si–O covalent bonds. Much less energy is required to overcome the id -id interactions compared to breaking the stronger Si–O covalent bonds during melting. Hence, P4O10 has a much lower melting point than SiO2. Comments • Be precise in describing structure and bonding, especially of the respective forces between particles – covalent bonds between Si and O atoms, id-id interactions between P4O10 molecules. • When answering questions related to structure and bonding, students must state the type of intermolecular forces between molecules. Q2 (a) xC(g) ⇌ yA(g) + B(g) initial amount/mol 1.80 0.40 0.40 change in amount/mol –0.40 +0.40 +0.20 equilibrium amount/mol 1.40 0.80 0.60 From the change in amount, molar ratio of C : A : B = 2 : 2 : 1 Hence x = 2 and y = 2 Comments • Many students used the equilibrium amounts of A, B and C at time t3. Since some amount of C was added at time t1, the stoichiometric coefficients are valid only if you used the “initial” amounts at time t2. • Other common mistakes include misreading the graph. • Quite a number of students do not realize that the stoichiometric coefficient ratio (the numbers in the equati on) correspond to the ratios of the numbers in the “C” row of the ICE table, not I and not E. C (change) represents the amounts of species reacted and produced in the chemical reaction, according to the stoichiometry of the equation.

-3- (b) Immediately before time t1, Kc = [A]2 [B] [C]2 = (0.80 2 ) 2 (0.60 2 ) ( 1.40 2 )2 = 0.09796 = 0.0980 mol dm-3 Comments • Most students who were able to obtain the correct stoichiometric coefficients from part (a) were able to calculate Kc correctly. • A common mistake was forgetting to convert the amount values to concentration values. (c) G > 0 Comments • Very poorly done! • When K > 1, a reaction is “product -favoured” and hence spontaneous, i.e. G is negative. • Conversely, when K < 1, the reaction is non-spontaneous i.e. G is positive. • Students who prefer mathematical equations may find the following equation useful: G = −RT lnK. However, this equation is not in syllabus and you will not be asked to recall this. (d) The temperature of the system was increased. Comments • A common wrong answer was a change in pressure of the system. Increasing the pressure of the system by decreasing the volume of the container would cause the position of equilibrium to shift left (less moles of gas), increasing the amount of C. (e) Comments • Very poorly done! Adding an inert gas to the system would increase the total pressure of the system, but have no effect on the partial pressures of each gas since the mole fraction of each gas would decrease proportionally. Hence, there will be no change in the position of equilibrium.

-4- Q3 (a) 4 isomers Comments • Students should make use of the given products in Table 3.1 to work out the answer to this question: o There is a chiral carbon in Y . Hence, there are two enantiomeric forms of Y. o There are two different groups bonded to each doubly bonded carbon atom in Z – Z can exist as the cis-isomer or trans-isomer. o X is not considered as one of the mono -brominated products formed since its yield is 0%. (b) Comments • Students are reminded to label the slow/fast steps of the mechanism. • Each step of the mechanism must be balanced. Many students forgot to include Br– as a product of the first step. • Students need to be precise in their arrow pushing e.g. draw arrow originating from the C=C bond and not the C atom. Correct: Incorrect: (c) Carbocation leading to the formation of X Comments • Generally well done (d) (i) (1) Compound X The intermediate is a primary carbocation which is very unstable since there is only one electron -donating alkyl group bonded to the positively charged carbon present to disperse the positive charge. Hence this carbocation does not form and the yield for X is zero at the stated temperature. slow fast + - + Br –

-5- Comments • Students need to answer the question and explain why carbocation X is NOT formed, instead of explaining why carbocation Y is formed. • Students should understand that a carbocation is an ion ic species with a positively charged carbon (e.g. is a carbocation) , a carbocation is not THE positively charged