2022 DHS Y6 H2 Prelim Paper 1 Worked Solutions

2022 Y6 Preliminary Examination H2 Chemistry 9729 Paper 1 Suggested Solutions © DHS Chemistry Unit Page 1 of 7 Answer Key 1 2 3 4 5 6 7 8 9 10 B A A B D D C B C D 11 12 13 14 15 16 17* 18 19 20* D C C B A A A C D B 21 22 23* 24 25* 26* 27 28 29 30 D C D B A A C B C D 1 B Given: angle of deflection  q m +15○ = k( +1 1 ) k = +15 –5○ = +15( q m ) q m = – 1 3  A q = 0 m = 1+2 = 3  q m ≠ – 1 3  B q = +3−5 = −2 m = 3+3 = 6  q m = – 1 3  C q = +4−1 = +3 m = 4+5 = 9  q m ≠ – 1 3  D q = +4−3 = +1 m = 4+5 = 9  q m ≠ – 1 3 2 A Interpretation of the graph: C is likely to be in Group 1 since it has the lowest 1 st IE. Since the four elements are consecutive, B is in Group 18 and A is in Group 17. 3 A  1 O3 O O O Bond angle <120  2 PF5 Bond angles 90, 120  3 [PCl4]+ P Cl Cl Cl Cl Bond angle 109.5  4 SF6 Bond angle 90 4 B  A CH3 CH3 H H This hydrocarbon is a non-polar molecule  B CH3 CH3 O The C=O bond has a r elatively large dipole moment (due to the large difference in the electronegativities of carbon and oxygen ) and there are no other significantly polar bonds to offset the dipole moment of the C=O bond Thus, the molecule has the largest overall dipole moment among the four options.  C CH3 CH3 Cl Cl C-Cl bond is polar but the dipole moments cancel out.  D O Cl Cl C=O bond is more polar than C –Cl bond as O is more electronegative than C l but since both bonds are polar, there will be some degree of offset of dipole moments of the bonds, hence this molecule does not have the largest overall dipole moment among the four options. 5 D Both CO 2 and C l2 are non -polar molecules with weaker instantaneous dipole -induced dipole interactions (id-id) as compared to CH3OH and N2H4, which have stronger intermolecular hydrogen bonding. N 2H4 has more extensive hydrogen bonds (an average of 2 H-bonds per molecule) than CH3OH (an average of 1 H -bond per molecule), hence N 2H4 exhibit greatest deviation from ideal gas behaviour.

Dunman High School 2022 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions © DHS Chemistry Unit Page 2 of 7 Per molecule CH3 O HCH3 N N H HH H Total no. of H atoms bonded to O/N atom(s) 1 4 No. of lone pairs on O/N atom(s) 2 2 Average no. of H bonds possible 1 2 6 D Possible identities of X: P4O10(s) + 6H2O(l) → 4H3PO4(aq) SO3(g) + H2O(l) → H2SO4(aq)  Possible identities of Y: Al2O3 is insoluble in water but is amphoteric, however one mole of Al2O3 requires 6 moles of H + for neutralisation, Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l) hence, options A and B are incorrect as they do not match the required mole ratio stated in the question with both given options for X. Na2O(s) + H2O(l) → 2NaOH(aq)  Since, one mol of H2SO4 is completely neutralised by two mol of NaOH, oxide X is SO3 and oxide Y is Na2O. 7 C  A Polarisability pertains to ease of distortion of the anion’s electron cloud and since the anion is the same, i.e. CO32−, in both compounds, the polarisability factor is the same in both.  B Melting is not the same as thermal decomposition/ thermal stability. Melting point involves change in states, not chemical composition.  C Since ionic radius of Ca2+ is smaller than that of Ba2+, the charge density of Ca2+ is higher. charge density  q r + + Ca2+ has a greater polarising power and distorts the electron cloud of CO 32− anion to a greater extent. The C–O covalent bond within the CO32− anions in CaCO 3 is weakened to a greater extent as compared to that in BaCO3. Hence CaCO 3 decomposes at a lower temperature.  D Lattice energy is a measure of the ionic bond strength in metal carbonates and is not a measure of their thermal stability. 8 B Volatility is defined as the tendency of a substance to vapourise. Since the halogens exist as non -polar simple covalent molecules, the volatility of the halogens depends on the strength of the instantaneous dipole -induced dipole (id -id) interactions between molecules.  1 Both bond length and bond strength are about the covalent bond between the atoms in the halogen molecu le. Thus, these factors do not affect volatility. 2  3 Each halogen molecule is comprised of two identical elements from Group 17. Since the elements present in each molecule is identical, the resultant halogen molecule is non -polar. Hence the electronegativity of the halogen atom has no effect on the strength of the id -id interactions.  4 Strength of id -id interactions increases when number of electrons in the molecule increases due to greater ease of distortion of the larger electron cloud. Hence, this statement is correct. 9 C H2SO4 + 2NaOH → Na2SO4 + 2H2O n(H2SO4) = 30 1000 × 0.1 = 0.003 mol n(NaOH) = 40 1000 × 0.2 = 0.008 mol H2SO4 is the limiting reagent. n(H2O) = 2 × n(H2SO4) = 2 × 0.003 = 0.006 mol Heat released from the reaction = n(H2O) × |∆Hneu| = 0.006 × 57.3 × 1000 = 343.8 J = heat absorbed by the solution heat absorbed by the solution = mc|∆T| 343.8 = 70 × 1 × 4.2 × |∆T| |∆T| = 1.2 °C Since heat is absorbed by the solution, ∆T = +1.2 °C. 10 D H S G  A + + − when temperature is high enough such that |TS| > |H|  B − + − at all temperatures not just at low T  C + + − when temperature is high enough such that |TS| > |H|  D − + − at all temperatures

Dunman High School 2022 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions © DHS Chemistry Unit Page 3 of 7 11 D  1 From the Data Booklet, E / V H2O2 + 2H+ + 2e− ⇌ H2O +1.77 O2 + 2H+ + 2e− ⇌ H2O2 +0.68 Ecell = +1.77 – (+0.68) = +1.09 V > 0 Overall equation: 2H2O2 → 2H2O + O2 H2O2 undergoes disproportionation where the oxidation state of O increases from −1 in H2O2 to 0 in O2 and decreases from −1 in H2O2 to −2 in H2O.  2 Both MnO4− and Fe3+ are oxidising agents and will not react. From the Data Booklet, both species are on the left-hand side of the equation and show tendency to undergo reduction. E / V MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O +1.52 Fe3+ + e− ⇌ Fe2+ +0.77  3 From the Data Booklet, E / V I2 + 2e− ⇌ 2I− +0.54 S4O62−+ 2e− ⇌ S2O32− +0.09 Ecell = +0.54 – (+0.09) = +0.45 V > 0 Overall equation: I2 + S2O32− → 2I− + S4O62− I2 oxidises S2O32− to give S 4O62− while S2O32− reduces I2 to give I−.  4 Let the oxidation state of C in HC2O4− be x. (+1) + 2x + 4(−2) = −1 ⇒ x = +3 Note: The oxidation state of both C atoms in HC2O4− can be determined by the calculation above since they are bonded to O atoms in the same way. The oxidation state of each atom is indicated below. O C C O - O O H +3 2 2 +1 2 +3 2 12 C The proposed reaction mechanism must fulfill two criteria: 1. Given that rate = k[NO] 2[H2], the slow step should involve two NO molecules and one H 2 molecule. 2. The overall equation should contain the products N2 and H2O only, as given in the question.  A Does not fulfil both criteria: • Only two NO molecules involved in the slow step, missing one H2 molecule. • Final products are N 2O and H 2O, not N 2 and H2O.  B Does not fulfil criterion (1): • Only one NO molecule and one H 2 molecule involved in slow step, missing one NO molecule.  C Fulfils both criteria.  D Does not fulfil (2): • Products of the reaction are N 2 and H2O2, not N2 and H2O. 13 C Let the orders of reaction with respect to K, L and M be x, y and z respectively. rate = k[K]x[L]y[M]z Comparing experiments 1 and 2, keeping [L] and [M]