NYJC 2023 H2 Chemistry 9729 P3 Answer

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 1 H2 Chemistry Prelim Exam Answers Paper 3 Answers 1 (a) (i) Electronegativity increases across the period and decreases down the group. Hence, beryllium and aluminium have similar electronegativity and have similar properties. (ii) BeCl2 has simple molecular structure consisting of weaker instantaneous dipole – induced dipole attraction between BeC l2 molecules while BeO has a giant ionic structure consisting of stronger ionic bonds between Be2+ and O2. Less energy is required to overcome the weaker instantaneous dipole –induced dipole attraction than the stronger ionic bonds. Hence, the boiling point of BeCl2 is lower than that of BeO. (iii) BeCl2 undergoes hydration in water to give [Be(H 2O)4]2+ which hydrolyses to give an acidic solution. Due to the high charge density of Be 2+, it can polarise and weaken the O H bond in the coordinated H 2O molecules, releasing H + into the solution. Hence, a weakly acidic solution is formed and causes a beaker of litmus solution to change from purple to red. BeCl2(s) + 4H2O(l)  [Be(H2O)4]2+(aq) + 2Cl–(aq) [Be(H2O)4]2+(aq) + H2O(l) ⇌ [Be(H2O)3OH]+(aq) + H3O+(aq) (b) (i) Since the Ar of Pb is 207.2, compound A can only contain one Pb. % by mass of Pb in A = 207.2 290  100 % = 71.45 % % by mass of F in A = 100 – 71.45 – 3.08 = 25.47 % element Pb Be F % by mass 71.45 3.08 25.47 amount / mol 71.45 207.2 = 0.3448 3.08 9.0 = 0.3422 25.47 19.0 = 1.341 simplest ratio 1 1 4 Hence, the empirical formula of A is PbBeF4 and thus, x = 1 and y = 4. (ii)

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 2 (c) (i) (ii) AlCl3 is acting as a catalyst since it reacted in step 1 and was regenerated in step 4. AlCl3 is also acting as a Lewis acid since it accepted a lone pair of electrons from O atom of phenyl ethanoate via dative bond formation. (iii) (iv) Neutral FeCl3(aq) If 2–hydroxyacetophenone is formed, violet coloration is formed. (v) 18  bonds and 8  electrons (vi) Electron-withdrawing –CO2H reduces the electron density of benzene and makes the benzene ring less electron –rich, causing B to be less reactive toward electrophilic attack. Hence, the rate of Fries rearrangement is slower for B compared to phenyl ethanoate.

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 3 2 (a) (i) Zn and Cu. Density = mass / volume  Zn and Cu have higher relative atomic mass as compared to Mg and Al.  Zn and Cu also have smaller atomic radius. Hence Zn and Cu have higher densities than Mg and Al. (ii) Cu2+ + 2e− ∏ Cu +0.34 V 2H+ + 2e− ∏ H2 0.00 V Ecell = 0.00 – (+0.34) = –0.34 V (<0, not feasible) Ecell calculation & concluding it’s not feasible Cu is the residue. (b) (i) When precipitation occurs, ionic product of the hydroxide = Ksp For precipitation of Cu(OH)2: [Cu2+][OH–]2 = Ksp (1.0) [OH–]2 = 2.20 × 10–20  [OH–] = 1.48 × 10-10 mol dm–3 For precipitation of Al(OH)3: [Al3+][OH–]3 = Ksp (0.0044) [OH–]3 = 4.60 × 10–33  [OH–] = 1.01 × 10-10 mol dm–3  Since the [OH–] required for both precipitates to form is similar, it does not allow for the separation of the two metal ions. (ii)  When NaOH(aq) is added gradually, a white ppt of A l(OH)3 is formed and blue ppt of Cu(OH)2 is formed.  Al3+(aq) + 3OH–(aq) ⇌ Al(OH)3(s) --------- (1)  Cu2+(aq) + 2OH−(aq) ⇌ Cu(OH)2(s)  When excess OH – is added, white ppt of Al(OH)3 dissolves to give a colourless solution due to the formation of complex ion, Al(OH)4 –.  Al3+(aq) + 4OH– (aq) ⇌ Al(OH)4 – -------- (2)  As [A l3+] falls, the position of equilibrium of (1) shifts to the left, causing the white precipitate to dissolve, while the blue ppt of Cu(OH)2 remains. (iii)  In the presence of ligands, the partially filled 3d orbitals of Cu 2+ split into two energy levels, with a small energy gap, ∆E.  An electron in the lower energy d orbital absorbs energy in the visible spectrum corresponding to ∆E and becomes excited to a vacant d orbital at the higher energy level (d-d transition).  Unabsorbed wavelengths are transmitted and the colour observed is complementary to the colour absorbed.

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 4 (c) Ag+ + e− ⇌ Ag (+0.80V) Cu2+ + 2e− ⇌ Cu (+0.34V) Zn2+ + 2e− ⇌ Zn (‒0.76V) The operating voltage is carefully regulated so that it is sufficient for Cu to be oxidised. At anode Cu is preferentially oxidised to Cu 2+ and dissolve in the electrolyte. Zn will be oxidised to Zn 2+ and dissolve in the electrolyte as well since E (Zn2+/Zn) is more negative than E (Cu2+/Cu). Ag will not be oxidised to Ag + as E(Ag+/Ag) is more positive than E(Cu2+/Cu) and will be collected as anode sludge. At cathode Cu2+ is preferentially reduced to Cu and deposited at the cathode as E(Cu2+/Cu) is more positive than E(Zn2+/Zn). Zn2+ will remain dissolved in the electrolyte. (d) (i) ZnCO3(s)  ZnO(s) + CO2(g) (ii)  Ionic radius of the Zn 2+ (0.074nm) is larger than radius of Mg 2+ (0.065nm) but ionic charge remains the same (2+). The charge density of Zn 2+ is lower than charge density of Mg2+, and its polarising power is also lower.  The cation is less able to polarise the electron cloud of the CO32– ion in ZnCO3.  The C–O bonds in CO 3 2– is polarised and weakened to a smaller extent. More energy is needed to break the C–O bond.  Hence thermal decomposition is mor e difficult for ZnCO 3 and a higher temperature is needed before it decompose. (e) Both MgO and Al2O3 react with acids. MgO(s) + 2H+(aq)  Mg2+(aq) + H2O(l) Al2O3(s) + 6H+(aq)  2Al3+(aq) + 3H2O(l) Both P4O10 and Al2O3 react with bases. P4O10(s) + 12NaOH(aq)  4Na3PO4(aq) + 6H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l)  2Na[Al(OH)4](aq) Al2O3 has a giant ionic lattice structure with strong electrostatic forces of attraction between Al3+ and O2- ions. The high charge density of Al3+ allows the electron cloud of O2- to be polarised, resulting in partial covalent character in the ionic bond. Al2O3 is hence amphoteric, exhibiting both acidic and basic properties. (f) (i) Eʅ(X2/X) becomes less positive down the group . This means that halogens are less easily reduced down the group, and so the oxidising power of the halogens decreases down the group. F2 + 2e 2F +2.87 V Cl2 + 2e 2Cl +1.36 V Br2 + 2e 2Br +1.07 V I2 + 2e 2I +0.54 V

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 5 (ii) The relative oxidising ability of bromine and iodine can be illustrated with the reaction with Fe2+ Fe3+ + e Fe2+ +0.77 Br2 + 2e 2Br +1.07 I2 + 2e 2I +0.54 Bromine is a stronger oxidising agent than iodine as bromine is able to oxidise Fe2+ to Fe3+ (as Ecell > 0), but iodine is unable to do so. Br2 + 2Fe2+  2Br + 2Fe3+ Ecell = +1.07 – (+0.77) = +0.30 V > 0 I2 + 2Fe2+  No reaction Ecell =+0.54 – (+0.77) = –0.23 V < 0

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 6 3 (a) (i) Free radical substituition Initiation Br Br uv 2Br Propagation C H H HCH3 Br CH3CH2 + HBr CH3CH2 Br Br CH3CH2Br + Br Termination Br Br Br2 CH3CH2 Br CH3CH2Br CH3CH2 CH3CH2 CH3CH2CH2CH3 [1] for name + initiation [1] for propagation [1] for termination (ii) ∆Hrxn = (+193 + 410) – (+280 + 366) = –43 kJ mol–1 (iii) The brief exposure to bright light produces chlorine radicals that initiate chain reactions of the propagation step to produce more radicals for reaction to proceed. The exothermic reaction results in an increase in temperature, hence rate of reaction increases. (iv) There is a decrease in ON of C1 atom from -1 in CH3CH2 Br to -3 in CH3CH2Li. (v) Cu2+ + 2e- ⇌ Cu +0.34V Li+ +