ASRJC H2 Chem 2022 Prelim P2 Soln

ASRJC JC2 PRELIM 2022 9729/02/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2022 JC 2 PRELIMINARY EXAMINATION SOLUTIONS WITH MARKERS COMMENTS NAME:______________________________ ( ) CLASS: 22 / _____ CHEMISTRY 9729/02 Paper 2 Structured Questions 14 September 2022 2 hours READ THESE INSTRUCTIONS FIRST Please look through the marks scheme and pay attention to the comments from markers. As you read, take note of:  What are your misconceptions?  What are common errors to avoid?  How should you improve next time? Do pay attention as your tutor go through in class and be proactive in asking questions. If you do not ask now, then when? Remember the power of tiny gains and be a little better every day!

1 (a) (i) • Ca has more filled principal quantum shells than Mg. • Nuclear charge and shielding effect of Ca are higher • Outermost electrons are of Ca further from the nucleus. [1] • Less energy is required to overcome the weaker electrostatic forces of attraction between the nucleus and the electron to be removed for Ca. • Second ionisation energies of Ca is lower than that of Mg. [1] Comments: This is a simple comparison of IE down the group. Although it is about the 2nd IE, the electron to be removed is from the same orbital as that of the 1 st IE for both Ca and Mg. Students commonly missed out:  electron to be removed from Ca is further away from the nucleus  electrostatic forces of attraction is between the nucleus and the electron removed Other error:  spelled “principal” wrongly as “principle”  thought that Ca has more valence shells than Mg when there should be only one valence shell for both Ca and Mg.  wrote that the 2nd IE of Ca is higher when the values of ionisation energies can be found in the Data Booklet. (ii) Thermal stability: CaCO3 > ZnCO3 > MgCO3 Zn, Ca and Mg have the same cationic charge of +2 Ionic radius of Ca2+: 0.099 nm Ionic radius of Mg2+: 0.065 nm Ionic radius of Zn2+: 0.074 nm Ionic radius: Ca2+ > Zn2+ > Mg2+ [1] with quoted values Charge density (q/r): Ca2+ < Zn2+ < Mg2+ Hence, polarising power of cation, ease of distortion of anion and weakening of C-O bond increases in the same order. [1] Comments: Many students were not able to use the ionic radius of the 3 cations to relate to the charge density, polarising power etc. Instead, some incorrectly quoted E o values or atomic radius which have NEVER been used to explain thermal stability of carbonates! Several students also missed out key words like the electron cloud of CO 32- was distorted to a greater extent or the C-O bond is weakened to a greater extent etc. Some students spelled “extent” wrongly as “extend” in their answer. (b) (i) [Be(H2O)4]2+ +H2O ⇌ [Be(H2O)3(OH)]+ + H3O+ OR [Be(H2O)4]2+ ⇌ [Be(H2O)3(OH)]+ + H+ [1] Comments: Many students missed out the information from the qns that Be forms a complex ion [Be(H2O)4]2+ and thus couldn’t write the equation for the hydrolysis of this ion. (ii) BeO(s) + 2HCl(aq) BeCl2(aq) + H2O(l) [1] BeO(s) +2NaOH(aq) + H2O(l) Na2[Be(OH)4](aq) [1] [1]

ASRJC JC2 PRELIM 2022 9729/02/H2 [Turn over (Also accept: BeO(s) +NaOH(aq) + H2O(l) Na[Be(OH)3](aq)) Comments: Many students thought beryllium oxide is Be2O3 without realizing that Be is in Group 2 and not Group 13 like Al. They also couldn’t balance the equation. (iii) Be2+ has a high charge size of 2+ and a small ionic radius, giving rise to high charge density and hence high polarising power. It distorts the electron cloud of Cl– anion to a great extent. The extent of sharing of electrons between the two nuclei is so great that BeCl2 exhibits covalent character.[1] Comments: Many students wrote answers like Be is covalent thus not ionic, Be and Cl has similar electronegativity etc that are not accepted. (iv) There is an increase in number of bond pairs around the central Be atoms (increase from 2 to 3) To minimise repulsion (and maximise stability), the 3 bond pairs are arranged as far apart as possible, giving rise to a change in shape from linear to trigonal planar and a smaller (or decrease in) bond angle. Comments: Some students forgot to indicate the number of bond pairs around Be or did not state that the bond pairs are arranged as far apart as possible to minimise repulsion. Students should be specific in their answer and shouldn’t refer to bond pairs only as “electron pairs” as “electron pairs” can mean lone pairs too! Many students incorrectly thought there is a lone pair on Be or thought that the lone pair of electrons on Cl that is used to form the dative bond is considered a lone pair around Be. [Total: 10]

2 (a) (i) CH3 C H C O O O H Comments: Common mistakes include:  Missing dipoles, lone pairs, labels  Drawing of lactic acid instead of lactate  Hydrogen bonding to the wrong H or O (remember that only H bonded to O,N or F can be used for Hydrogen bonding!)  Drew i ntermolecular hydrogen bonding (between two ions) instead of intramolecular hydrogen bonding (within one ion) as required by question  Missing atoms and bonds (ii)  For lactate ion, intramolecular hydrogen bonding formed between the ionised – CO2– group and –OH group which results in the greater stability of the conjugate base/lactate ion compared to the ethanoate ion.  Lactic acid is a stronger acid with larger Ka value. [1] Comments: Common mistakes include:  Explanations involving solubility  Explanations about a second dissociation  Acidity considerations not involving intramolecular H -Bonding despite the qualifier word “Hence”  Answered in terms of pKa  Incorrectly related higher acidity to lower Ka (b) (i) Since n = 2, E = Eo – ( 2 0592.0 ) log10 ]NAD[ ]NADH[  (–0.350) = (–0.320) – ( 2 0592.0 ) log10 ]NAD[ ]NADH[  log10 ]NAD[ ]NADH[  = 1.01 ]NAD[ ]NADH[  = 10.3 [1] (3 s.f.) Comments: Common mistakes include:  Using ln instead of lg  Incomplete calculations + – [1] Hydrogen Bond

ASRJC JC2 PRELIM 2022 9729/02/H2 [Turn over (ii) % of NAD+ = )13.10( 1  x 100% = 8.85 % [1] Comments: Poorly done. Many students cannot properly compute the percentage. (iii) CH3COCO2– + 2H+ + 2e– CH3CH(OH)CO2– [1] Eocell at pH 7 = Eored – Eoox +0.135 = Eored – (–0.320) Eored = –0.185V [1] Comments: Very poorly done. Students are supposed to use th e overall equation to deduce the reduction half-equation for pyruvate. This can be done by subtracting the oxidation half-equation involving NAD+/NADH from the overall reaction equation. overall: CH3COCO2– + NADH + H+  CH3CH(OH)CO2– + NAD+ Ecell o = +0.135 V oxidation: NADH  NAD+ + H+ + 2e– The reduction half -equation is hence as shown below after balancing the particles and charges: CH3COCO2– + 2 H+ + 2 e– CH3CH(OH)CO2– (iv) G at pH 7 = –nFEocell = –2 (96500) (+0.135) = –26 055 J mol–1 = –26 100 J mol–1 (3 s.f.) [1] Comments: Very poorly done. Many students either used the wrong E cell or used the wrong number of mole of electrons transferred or simply misremembered the equation. (c) (i) The amount of heat evolved when one mole of lactic acid in its standard state is completely burned in excess oxygen under standard conditions of 298 K and 1 bar. [1] correct definition with key words/points Comments: Key words such as ‘standard state’, ‘completely’ and ‘excess’ are frequently missing. Students should specify the substance as ‘lactic acid’. (ii) ∆Hco = [3(−393.5) + 3(−285.8)] – (−483.2) = −1554.7 kJ mol−1 = −1550 kJ mol−1 (3 s.f.) [1] (correct value and units) Comments: Students did not take into a