JPJC 2024 H2 Chemistry WA1 Answer

2026 J1 H2 Chemistry (9476) 1 WA1 Revision Paper 1 (Worked Solution) REVISION PAPER for JC1 H2 WEIGHTED ASSESSMENT 1 (REVISION PAPER 1) – 2024 Weighted Assessment 1 Suggested Solutions SECTION A (25 marks) 1 (a) Copper is a transition metal. Write the electronic configuration of a Cu+ ion. Cu atom: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (NOT 3d9 4s2) To form cation, remove electron from outermost shell (4th shell) or subshell. Cu+ ion: 1s2 2s2 2p6 3s2 3p6 3d10 [1m] [1] (b) Copper and ice are solids. (i) Copper has a melting point of 1028 C. Explain this property of copper by referring to its structure and bonding. Copper has a giant metallic structure. A large amount of energy is required to overcome the strong electrostatic attraction (or metallic bonds ) between a lattice of cations and delocalised valence electrons . Thus, copper has a high melting point. [1m] [1] (ii) Draw a labelled diagram to show the significant force of attraction between two water molecules. Include the name of the attraction in your diagram. H O H H O H hydrogen bond [1m] [1m] correct drawing:  dipole (+, ) on the O–H bond of each H2O molecule involved in hydrogen bonding i.e. total two pairs of (+, ) shown  dotted line from lone pair of O atom on one H2O molecule to the H atom of the other H2O molecule [2] [Total: 4] 2 (a) Explain the increase in atomic radius down the group. Down the group, the number of principal quantum shells increases () and shielding effect by inner shell electrons increases (). These effects outweigh the effect of increasing nuclear charge (). Thus, the net nuclear attraction on outermost electrons decreases (). Hence, atomic radius increases across the period. 4(): [2m]; 2-3(): [1m] [2]

2026 J1 H2 Chemistry (9476) 2 WA1 Revision Paper 1 (Worked Solution) (b) The second to fifth ionisation energies of three consecutive elements in the third period of the Periodic Table are given in Table 2.1. The symbols used for reference are not the actual symbols of the elements. Table 2.1 element ionisation energies / kJ mol1 second third fourth fifth W 1450 7740 10 500 13 600 X 1820 2740 11600 14 800 Y 1580 3230 4360 16 100 (i) Write an equation to represent the second ionisation energy of element W. [1] (ii) State and explain the group number of element X. group number ………….. explanation ………………………………………………………………………… ........................................................................................................................ ……………………………………………………………………………………….. [2] (iii) Explain why the second ionisation energy of element Y is less than that of element X. X is from Period 3, grp 13 Y is from Period 3, grp 14 X: [Ne] 3s2 3p1 Y: [Ne] 3s2 3p2 X+: [Ne] 3s2 Y+: [Ne] 3s2 3p1 The 3p electron to be removed from Y + is further away from the nucleus AND has higher energy than 3s electron to be removed from X+. [1m] [1] [Total: 6] W+(g)  W2+(g) + e [1m] state symbols required There is a large increase from 3 rd IE to 4 th IE so the 4th electron is removed from the inner quantum shell. Thus, there are 3 valence electrons. [1m] 13 [1m]

2026 J1 H2 Chemistry (9476) 3 WA1 Revision Paper 1 (Worked Solution) 3 40 cm 3 of gaseous hydrocarbon, C xHy, was exploded with excess oxygen. A contraction in volume of 100 cm 3 occurred. On treating the products with aqueous sodium hydroxide, a further contraction in volume of 120 cm3 occurred. Given that all gas volumes were measured at room temperature and pressure, deduce the molecular formula of the hydrocarbon. Second contraction due to reaction of CO2 with NaOH = = 120 cm3 () First contraction due to combustion = 100    ) x y 2 2 2C H O initial O remaining CO(V V ) (V V = 100  x y 2 2 2C H O initial O remaining COV (V V ) - V = 100 = 100 40 +  120 = 100 Volume of O2 reacted = 180 cm3 () CxHy(g) + (x + 4y ) O2(g)  xCO2(g) + 2y H2O(l) Volume reacted or formed / cm3 40 : 180 : 120 volume ratio = mole ratio 1 : 4.5 : 3 - mole ratio from eqn 1 : (x + 4y ) : X -  x = 3 () no ecf (x + 4 y ) = 4.5  y = 6 () ecf from x & VO2 reacted molecular formula of the hydrocarbon is …………. [2] 5(): [2m]; 2-4(): [1m] [Total: 2] 2COV   x y 2 2C H O reacted COV V V 2O reactedV C3H6 ()

2026 J1 H2 Chemistry (9476) 4 WA1 Revision Paper 1 (Worked Solution) 4 Hydroxylamine, NH 2OH, is commonly used in photography developing solution. Prolonged exposure to NH2OH by photography developers might cause irritation to their respiratory tract. A scientist performed the following experiment to determine the concentration of NH2OH found in a sample of developing solution. 200 cm3 of NH 2OH solution was added to 50.0 cm 3 of 0.120 mol dm 3 acidified iodine solution, which is in excess. The reaction occurs according to the equation given below. 2NH2OH + 2I2  N2O + H2O + 4I– + 4H+ excess (a) Write a half-equation for the conversion of NH2OH to N2O. 2NH2OH  N2O + H2O + 4H+ + 4e [1m] [1] (b) A 25.0 cm 3 sample of the resulting reaction mixture was titrated against 0.0150 mol dm–3 Na2S2O3. I2(aq) + 2S2O32–(aq)  2I(aq) + S4O62–(aq) in 25.0 cm3 The end-point is reached when 22.50 cm3 of Na2S2O3(aq) has been added. (i) Calculate the amount of iodine remaining in the 25.0 cm 3 sample of the resulting reaction mixture. Amount of Na2S2O3 used in titration =  22.500.0150 1000 = 3.375  104 mol S2O32 : I2 2 : 1 Amount of I2 remaining in 25.0 cm3 = 12 3.375  104 = 1.69  104 mol [1m] [1] (ii) Calculate the amount of iodine reacted with NH2OH. Hence, determine the concentration of NH 2OH in 200 cm 3 sample of the photography developing solution. Amount of I2 remaining in (200 + 50) cm3 of resulting mixture =    4(200 + 50) 1.69 10 25.0 = 1.69  103 mol [1m] Amount of I2 reacted with NH2OH =    50.0 0.1201000  1.69 x 103 = 4.31  103 mol [1m] NH2OH : I2 1 : 1 Amount of NH2OH present in 200 cm3 = 4.31  103 mol [NH2OH] =  34.31 10 2001000 = 0.0216 mol dm3 [1m] [3] [Total: 5] After rxn, remaining iodine (I2) is in (200 + 50.0) cm3

2026 J1 H2 Chemistry (9476) 5 WA1 Revision Paper 1 (Worked Solution) 5 The elements sodium to chlorine, in Period 3, all form oxides. (a) Draw a diagram to show the shape of the molecule of each of the oxides, SO3 and Cl2O. Name the shape of each molecule. SO3 3 bond pairs & 0 lone pair around S Cl2O 2 bond pairs & 2 lone pairs around O shape: …………………………………… shape: ………………………………… [3] 2(): [1m] (b) In terms of structure and bonding, explain why (i) the melting point of MgO is higher than that of Na2O. Both MgO and Na2O have giant ionic structures ().  + + q qLE r + r AND Mg2+ has a larger charge and smaller radius than Na+ () More energy (*) is required to overcome the stronger ionic bonds between Mg2+ and O2 ions () in MgO than the ionic bonds between Na+ and O2 ions in Na2O. () Accept “between cations and anions / oppositely charged ions” [2] (ii) the boiling point of SO3 is higher than that of SO2. Both SO2 and SO3 have simple molecular/covalent structures (). SO2 is polar while SO3 is non-polar. Since SO3 has a greater number of electrons per molecule (), more energy (*) is required to overcome the stronger instantaneous dipole-induced dipole attraction between SO3 molecules () than the w