RI 2022 Y5 Timed Practice Suggested Solutions

2022 H2 Chemistry Y5 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 C 2&3 D B C A C D C D A B C A Question 1 (D) A Amount of N2 = 2.80 28.0 = 0.100 mol 1 mol of N2 has 2 mol of N atoms Amount of N atoms = 2(0.100) = 0.200 mol No. of atoms = 0.200 x 6.02 x 1023 = 1.204 x 1022 B Amount of Ar = 3.60 39.9 = 0.09023 mol No. of atoms = 0.09023 x 6.02 x 1023 = 5.431 x 1022 C Amount of C2H2 = 950 24000 = 0.03958 mol 1 mol of C2H2 has 4 mol of C and H atoms. Amount of atoms = 4(0.03958) = 0.1583 mol No. of atoms = 0.1583 x 6.02 x 1023 = 9.530 x 1022 D No. of CO2 = 5.10 x 1022 1 mol of CO2 has 3 mol of C and O atoms No. of atoms = 3(5.10 x 1022) = 1.53 x 1023 Question 2 (C) Cl2 + 2e– ⎯→ 2Cl− -- (1) Cl2 ⎯→ Cl–containing product -- (2) The 0.05 mol of Cl− formed in (1) came from ( 0.05 2 ) = 0.025 mol of Cl2. Since 0.03 mol of Cl2 was used, there is (0.03 – 0.025) = 0.005 mol of Cl2 involved in (2). The amount of Cl2 used in equation (1) : equation (2) = 0.025 : 0.005 = 5:1. Therefore, equations (1) and (2) can be written as 5Cl2 + 10e– ⎯→ 10Cl− -- (3) 1Cl2 ⎯→ Cl–containing product -- (4) The 10 mol of e– from equation (3) must be given out in equation (4). So equation (4) becomes 1Cl2 ⎯→ Cl–containing product + 10e– Since 1Cl2 gave 10e–, 1 Cl gives 5 e– to become the chlorine-containing product i.e. the final oxidation state of the Cl in the product is +5. Question 3 (2&3) As 2&3 was not provided as an option, all students were awarded one mark for this question. 1 Incorrect. Electronic configuration of Cr = [Ar] 3d 5 4s1 ⇒ electronic configuration of Cr+ = [Ar] 3d5 2 Correct. All electrons in [Ar] are paired. Cr2+ electronic configuration = [Ar] 3d4 Cr3+ electronic configuration = [Ar] 3d3 For Cr 2+ and Cr 3+, there are no paired electrons. 3 The equation describes the 3 rd IE of Cr which, from the Data Booklet, is +2990 kJ mol–1. Question 4 (D) Since there is a large jump between the 5th and 6th IE for element X, the 6 th electron is removed from an inner shell i.e. X has 5 valence electrons and is from group 15. Since W, X, Y and Z are consecutive elements, W is from group 14, X is from group 15, Y is from group 16 and Z is from group 17. The group 17 element has the highest first IE as it has the highest nuclear charge while having approximately constantly shielding effect as in W, X and Y. This document is copyrighted, please do not reproduce it without permission

Question 5 (B) The cation of J • is polyatomic (made up of many atoms) • has overall charge of 2+ • contains Pt in a +4 oxidation state • dative bonded to 6 species (NH3 or Cl−) If J is made up of 2 Cl and 4 NH3 dative bonded to Pt, then the overall charge of the cation = O.S. of Pt + 2(charge of Cl−) + 4(charge of NH3) = +4 + 2(−1) + 4(0) = 2+ This satisfies the above criteria i.e. cation of J has formula [PtCl2(NH3)4]2+. J has a number of monoatomic anions e.g. Cl−. The 2+ charge of the cation needs to be balanced by a 2– from 2 Cl−. The formula of J should be [PtCl2(NH3)4]Cl2 i.e. Pt(NH3)4Cl4. Question 6 (C) A Incorrect. Ar has 5 regions of electron density (electron pair geometry of trigonal bipyramidal) and 3 lone pairs which gives a molecular shape of linear. B Incorrect. Since F is more electronegative than H, F pulls electron density towards itself causing a − on F and + on H. Hence, HF is polar. C Correct. From the dot-and-cross diagram, Ar in HArF has 10 electrons around itself in its valence shall. D Incorrect. H ydrogen bonds exist between compounds with H –F, H –O and H –N bonds. Since there are no H –F bonds in HArF, there are no hydrogen bonds between HArF. Question 7 (A) 1 Incorrect. Hydrogen bonds exist between compounds with H –F, H –O and H –N bonds. PH3 consists only of 3 P –H and is unable to form hydrogen bonds with other PH 3 molecules. 2 Incorrect. NH3 has a higher bp than CH4 due to the ability of NH 3 to form stronger intermolecular hydrogen bonds (compared to the weaker instantaneous dipole -induced dipole interactions between CH 4 molecules) which require more energy to overcome. 3 Incorrect. NH3 has a higher bp than CH4 due to the ability of NH 3 to form stronger intermolecular hydrogen bonds (compared to the weaker permanent dipole-permanent dipole interactions between PH 3 molecules) which require more energy to overcome. Question 8 (C) A Incorrect. From the data booklet, Ba has a larger ato mic radius compared to K, but this does not relate to the strength of the metallic bonds in Ba and K. B Incorrect. Ba, having a larger molar mass, is heavier than L. However, this does not relate to the strength of the metallic bonds in Ba and K. C Correct. With a higher charge in Ba 2+, there is stronger attraction between the lattice of Ba 2+ cations and the sea of delocalised electrons (compared to that in K +), resulting in stronger metallic bonds in Ba which require more energy to overcome, leading to a higher melting point. D Incorrect. Ba 2+ has more electrons than K +. However, this does not relate to the strength of the metallic bonds in Ba and K. Question 9 (D) Mr of CO2 = 12.0 + 2(16.0) = 44.0 Amount of CO2 = 4.4 44.0 = 0.100 mol pV = nRT p of CO2 = nRT V = (0.100)(8.31)(90+273) 1.00 x 10–3 = 301700 Pa = 3.017 bar (since 1 bar = 105 bar) Final pressure in bottle = pair + pcarbon dioxide = 1 + 3.017 = 4.02 bar Question 10 (C) 1 Correct. This is Avogadro’s law. 2 Correct. For a fixed mass of gas, the no. of moles of gas is also fixed. pV = nRT p = ( nR V )T nR V is a constant since n, R and V are constant in this case. Hence, p is directly proportional to T. 3 Incorrect. p = ( nR V )T p is inversely proportional to V. This document is copyrighted, please do not reproduce it without permission

Question 11 (D) The other information provided do not form a useful part of the energy cycle to determine ∆Hhyd of Br–. Question 12 (A) H2SO4 + 2NaOH ⎯→ Na2SO4 + 2H2O − (1) CH3COOH + NaOH ⎯→ CH3COONa + H2O – (2) If x mol is the amount of water released from (2), then (1) releases 2x mol of water (since the volume and concentrations of H 2SO4 and CH 3COOH are the same i.e. the amounts of H 2SO4 and CH3COOH are the same) Since ∆Hneutralisation = − heat change amount of water , for (1), ∆Hneutralisation(1) = − p 2x for (2), ∆Hneutralisation(2) = − q x (1) involves the reaction of a strong acid with a strong base, while (2) involves the reaction of a weak acid with a strong base. Therefore, |∆Hneutralisation(1)| > |∆Hneutralisation(2)| i.e. p 2x > q x ⇒ p > 2q Question 13 (B) Since the two allotropic forms of sulfur undergo combustion to give a common product, SO 2, the following energy level diagram can be drawn to visualise the difference in energy of -sulfur and -sulfur. A Incorrect. The standard enthalpy change of formation is the energy change when 1 mole of the pure substance in a specified state is formed from its constituent elements in their standard states under standard conditions. The standard state of a substance is its most stable form under standard conditions. From the energy level diagram, the most stable form is -sulfur as it is lower in energy. Hence, ∆Hf of -sulfur = +2 kJ mol−1 since it involves converting -sulfur to -sulfur. B Correct. As mentioned in A, the standard enthalpy change of SO 2 involves SO 2 being formed from its consti