RI 2022 Y6 CT Suggested Solutions

1 © Raffles Institution 2022 9729/J/22 2022 Y6 H2 Chemistry Term 3 Common Test – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D B C B A A A D B C A B C D C MCQ worked solutions 1 IO3– + 5I− + 6H+ ⎯→ 3I2 + 3H2O I2 + 2S2O32− ⎯→ 2I− + S4O62− Mole ratio of IO3– : I2 : S2O32− = 1 : 3 : 6 Amount of S2O32− required = 6  amount of IO3– used = 6  25.0 1000  0.100 = 0.0150 mol [S2O32−] = 0.0150 20.00  1000 = 0.750 mol dm−3 2 Group 2 metals are oxidised when they react with oxygen. The more easily the Group 2 metals are oxidised, the greater their reactivity with oxygen. A Atomic radius of the metal atom (not ionic radius of metal cation) can be used to explain the trend. Down the group, the valence electrons are further away from the nucleus and are more easily lost, resulting in greater ease of oxidation of the metal. B As the reducing power of the Group 2 metals increases down the group, the more likely the Group 2 metals are oxidised and hence the greater their reactivity with oxygen. C Although the electronegativity of the elements decreases down the group, this does not explain why the Group 2 metals oxidise more easily down the group. D Although the magnitude of LE of the metal oxides decreases down the group, this does not explain why the Group 2 metals oxidise more easily down the group.

2 © Raffles Institution 2022 9729/J/22 3 Given their low m.p., elements C to F should be simple molecules with weak intermolecular forces of attraction. Hence, elements C to F should be in Group 15 to 18, and elements A to F should be in the same period. As the variation in m.p. of elements C, D and E corresponds to that of P4, S8 and Cl2, elements A to F must be in Period 3 and elements G to H in Period 4. Based on the melting point trend, the identities of the elements are as follows: Element A B C D E F G H Identity Al Si P S Cl Ar K Ca Type of structure Giant metallic Giant molecular Simple molecular Giant metallic Statement 1 Incorrect. SiO 2 is insoluble in water as a large amount of energy is needed to break the strong Si–O covalent bonds in the giant molecular structure. Statement 2 Correct. Aluminium and chlorine react to form AlCl3, which hydrolyses in water to give an acidic solution of approximately pH 3. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) Statement 3 Correct. Phosphorus and sulfur are solid s at room temperature. Phosphorus can form oxides like P4O6 and P4O10 while sulfur can form oxides like SO2 and SO3. 4 Statement 1 Incorrect. From step 2 (slow step), rate  [B][C]. However, C is an intermediate which does not appear in the rate equation for the overall reaction. From step 1, [C] [A]2. Hence, rate  [A]2[B]. The correct rate equation should be rate = k[A]2[B]. Statement 2 Correct. step 1 2A ⇌ C step 2 B + C → D step 3 B + D → A2B2 overall 2A + 2B → A2B2 Statement 3 Incorrect. Based on the overall equation, rate of formation of A2B2 = 1 2  rate of depletion of B.

3 © Raffles Institution 2022 9729/J/22 5 Statement 1 is correct. ∆Hatomisation = 2353 – (736 + 1450) = +167 kJ mol−1 Statement 2 is incorrect. The enthalpy change of atomisation of an element is the energy absorbed when 1 mole of gaseous atoms is formed from the element. 1 2 O2(g) ⎯→ O(g) ∆Hatomisation = 1 2  BE(O=O) = + 496 2 = +248 kJ mol−1 Statement 3 is incorrect. 2  sum of 1st and 2nd EA = −496 + 1794 = +1298 kJ mol−1 Sum of 1st and 2nd EA = + 1298 2 = +649 kJ mol−1 6 ∆Hvap = 1 3 (–330.5 + 464) = +44.5 kJ mol−1 O2(g) + 4e− 2O(g) + 4e− 2O2−(g) BE(O=O) = +496 kJ mol−1 +1794 kJ mol−1 2  (sum of 1st and 2nd EA) Mg(s) Mg(g) Mg2+(g) + 2e− +2353 kJ mol−1 (736 + 1450) kJ mol−1 ∆Hatomisation

4 © Raffles Institution 2022 9729/J/22 7 X and Y have the same initial pH  both solutions have the same initial [H+]. Upon dilution with the same volume of water, pH of resulting solution of X > Y  the diluted solution of Y contains a higher [H+] than the diluted solution of X.  Y must be a weak acid (e.g. CH3COOH) which undergoes partial dissociation in water. Upon dilution with water, some of the undissociated weak acid undergoes further dissociation to produce more H+ ions.  X must be a strong acid (e.g. HCl) which undergoes complete dissociation in water. Upon dilution with water, there is no more undissociated acid to produce more H+ ions. If both X and Y are strong acids (e.g. HCl, H2SO4), upon dilution with water, the pH of the resulting solution s should be the same as both solutions should contain the same amount/concentration of H+ ions. 8 A pH = −lg(10−4) = 4 B pH = −lg √Ka[HA] = −lg √(10-4.76  10-4) = 4.38 C 1 mol dm−3 of NH4Cl is formed. NH4+ (conjugate acid of the weak base NH3) can undergo hydrolysis in water to form H+ ions. NH4+(aq) ⇌ NH3(aq) + H+(aq) pH = −lg √Ka[HA] = −lg √(10-(14-4.75)  1) = 4.63 D A buffer solution containing 1 mol dm−3 CH3COOH and 0.1 mol dm−3 CH3COONa is formed. pH = 4.76 + log 0.1 1 = 3.76

5 © Raffles Institution 2022 9729/J/22 9 A CaCO3(s) ⇌ Ca2+(aq) + CO32−(aq) ---- (1) 2H+ + CO32− ⎯→ CO2 + H2O With decreasing pH (i.e. increasing [H+]), [CO32−] decreases and the equilibrium position of (1) lies further right. Hence, more CaCO3(s) dissolves and the solubility of CaCO3(s) increases. B Solubility of CaF2 = √( 1 4  4.0  10-11) 3 = 2.15  10−4 mol dm−3 Solubility of CaCO3 = √(3.4  10-9) = 5.83  10−5 mol dm−3 CaF2 has a higher molar solubility than CaCO3. C The value of solubility product is only temperature dependent. D [Ca2+] needed to precipitate CaF2 = (4.0  10−11) / (0.1)2 = 4.0  10−9 mol dm−3 [Ca2+] needed to precipitate CaCO3 = (3.4  10−9) / (0.1) = 3.4  10−8 mol dm−3 CaF2 will precipitate out first. 10 Ag+ + e− ⇌ Ag +0.80 V [cathode] Ni2+ + 2e− ⇌ Ni −0.25 V [anode] Ecell = +0.80 − (−0.25) = +1.05 V When [Ag+] > 1 mol dm−3, E(Ag+/Ag) > +0.80 V When [Ni2+] < 1 mol dm−3, E(Ni2+/Ni) < −0.25 V Hence, Ecell > +1.05 V Electron flows from the anode (Ni electrode) to the cathode (Ag electrode) via the external circuit. 11 A At standard conditions, the pressure of any gas involved is 1 bar. B At standard conditions, the temperature is 298 K (not 293 K). C X should be an inert electrode (e.g. Pt) and Y should be a mixture of 1 mol dm−3 of Fe2+ and 1 mol dm−3 of Fe3+ ions. D At standard conditions, [H+] should be 1 mol dm−3. Since H2SO4 is a dibasic acid, [H2SO4] should be 0.5 mol dm−3.

6 © Raffles Institution 2022 9729/J/22 12 Evidence/Information Deductions P (C6H12) decolourises hot acidified KMnO4 to form Q and R. Strong oxidation/oxidative cleavage has occurred. P is an alkene. C=C in P is cleaved to form Q and R. Q forms a yellow ppt with hot alkaline aq I2. Oxidation reaction has occurred (positive iodoform test). Q is a methyl ketone containing –COCH3 group. (Q cannot be an alcohol with –CH(OH)CH3 group because Q is the oxidation product of alkene P.) R reacts with Na2CO3 to form a gas. Acid-base reaction has occurred. R is a carboxylic acid. CO2 gas is produced. Statement 1 Incorrect. P can be Statement 2 Correct. P is an alkene that can undergo electrophilic addition with aqueous bromine. Statement 3 Incorrect. Q is formed from the oxidation of P by hot acidified KMnO4. Hence, Q cannot be an aldehyde since an aldehyde group would have been oxidised to carboxylic acid in the oxidation reaction. Statement 4 Correct. R, which contains the carboxylic acid group, reacts with PCl5 to form dense white fumes of HCl. 13 Solution Hydrolysis of compound in water P CH3CH2O− + H2O ⇌ CH3CH2OH + OH− Q C6H5O− + H2O