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