2024 Prelims ASRJC H2 Chem P1 (Ans)

ASRJC JC2 PRELIM 2024 9729/01/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2024 JC2 PRELIMINARY EXAMINATION NAME:______________________________ ( ) CLASS: 24 /____ CHEMISTRY 9729/01 Paper 1 Multiple Choice SUGGESTED SOLUTIONS 1 Use of the Data Booklet is relevant to this question. Which particle has more protons than electrons and more protons than neutrons? (D represents deuterium, 2H) A NH3 B NH3D+ C ND4+ D NHD− Ans: B Approach For a particle to contain more protons than electrons, it is likely to be a cation. Hence you should calculate the number of protons and neutrons for the particles in option B and C. NH3 NH3D+ ND4+ NHD− number of protons 7 + 3(1) = 10 7 + 3(1) + 1 = 11 7 + 4(1) = 11 7 + 1 + 1 = 9 number of neutrons 7 + 0 = 7 7 + 0 + 1 = 8 7 + 4(1) = 11 7 + 0 + 1 = 8 number of electrons 10 10 10 10 From the table above, NH 3D+ has 1 more proton than electron and 3 more protons than neutrons.

2 ASRJC JC2 PRELIM 2024 9729/01/H2 2 The graph shows the second ionisation energies of eight consecutive elements A to H, in Period 2 and 3 of the Periodic Table. [Note that letters A to H are not the atomic symbols of the elements concerned.] Which statements are correct? 1 F and G are in the same period in the Periodic Table. 2 The general increase from A to F is due to increasing atomic radius. 3 The small decrease from C to D is due to repulsion between paired electrons. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 3 only Ans: B Statement 1 is correct. The sharp dip in 2 nd IE from F to G suggest that the second electron in G is removed from an electron shell which is further away / more shielded from the nucleus. The second electron removed from F is from the inner quantum shell. F is in group 1. G is in Group 2. Statement 2 is incorrect. • IE is not affected by atomic radius. IE is affected by nuclear charge, shielding effect and the distance of the valence electron from the nucleus. • From A+ to F+, nuclear charge increases. Number of electron shells remains the same and shielding effect remains approximately constant. Electrostatic attraction between the valence electrons and the nucleus becomes stronger. More energy is required to remove a valence electron from an atom across a period. • Due to the stronger attraction between the valence electrons and nucleus, atomic radius also decreases. Statement 3 is correct. C+: 1s22s22p3 D+: 1s22s22p4 The 2p electron to be removed from D+ is a paired electron while that to be removed from C+ is an unpaired electron. Due to inter -electronic repulsion between paired electrons in the same orbital, less energy is required to remove the 3p electron from D+.

3 ASRJC JC2 PRELIM 2024 9729/01/H2 [Turn over 3 Tyrosine is an amino acid. In which sequence is the bond angles arranged in decreasing order? A x > y > z B y > z > x C z > x > y D z > y > x Ans: D Hence the order is z > y > x 4 bond pairs, 0 lone pair → 109.5o 2 bond pairs, 2 lone pairs → 104.5o 3 bond pairs, 1 lone pair → 107o

4 ASRJC JC2 PRELIM 2024 9729/01/H2 4 The table identifies the shape and polarity of four molecules. Which row is correct? molecule molecular shape polarity A xenon tetrafluoride tetrahedral non-polar B selenium hexafluoride T-shaped non-polar C trichloromethane tetrahedral polar D boron trichloride trigonal planar polar Ans: C XeF4 SeF6 CHCl3 BCl3 No. of bp, lp on central atom 4 bp, 2 lp 6 bp, 0 lp 4 bp, 0 lp 3 bp, 0 lp Molecular shape Square planar octahedral tetrahedral trigonal planar Se F F F F F F C Cl Cl Cl H B Cl Cl Cl polarity non-polar non-polar polar non–polar 5 A vessel is initially filled with argon gas at a particular temperature and pressure. The following changes are made to the vessel. • the gas in the vessel was heated to 1.3 times the original temperature, • the vessel was filled with an inert liquid such that the gas only occupied 1 10 of the volume. What is the number of times the gas pressure is built up in the vessel after the changes were made? A 1.3 B 10 C 13 D 26 Ans: C 𝑃1𝑉1 𝑇1 = 𝑃2𝑉2 𝑇2 Let P1, V1 and T1 be the initial pressure, initial volume and initial temperature respectively and P2 be the final pressure in the vessel after the changes were made. 𝑃1𝑉1 𝑇1 = 𝑃2×0.1𝑉1 1.3𝑇1 𝑃2 𝑃1 = 1.3 0.1 = 13

5 ASRJC JC2 PRELIM 2024 9729/01/H2 [Turn over 6 Use of the Data Booklet is relevant to this question. Which graph correctly describes the behaviour of fixed masses of N 2 and CH 4? The gases behave as ideal gases under the conditions chosen. A B C D Ans: D pV = nRT pV = m M RT Gas N2 CH4 Molar mass 28.0 16.0 Option A: Shape of graph is wrong. Graph should start from origin. Option B: Shape of graph is correct. m M R is the gradient. From above, since N2 has a larger Mr, its gradient should be less steep. Hence graph is wrong. Option C: A graph of pV against V at constant T will give a constant value (i.e. horizontal line). Hence, graph is wrong. Option D: The gas with the smaller molar mass (CH4) will give the plot with the largest pV value while the gas with the larger molar mass (N2) will give the smaller pV value. Graph is correct. pV constant p CH4 N2 T / K pV constant p N2 CH4 V pV constant T N2 CH4 V pV constant T N2 CH4 T / K

6 ASRJC JC2 PRELIM 2024 9729/01/H2 7 The properties of the oxides of four Period 3 elements E, F, G and H are given in the table below. oxide of element melting point electrical conductivity when molten reaction with H2O E 2850 °C good very little F 1720 °C poor none G 1280 °C good exothermic H 580 °C poor exothermic Which of the following is in the correct order of increasing proton number? A E, F, G, H B E, G, F, H C G, F, E, H D G, E, F, H Ans: D E and G forms metal oxides with giant ionic structure since the presence of mobile ions allows these oxides to conduct electricity in the molten state. Since oxides of E has a higher melting point and is only sparingly soluble in H 2O, oxides of E should be MgO. Oxides of G should be NaO which reacts completely and vigorously with water to form a strongly alkaline solution. F and H forms non-metal oxides with giant covalent structure or simple molecular structure. This results in the a bsence of mobile ions or delocalised electrons in the molten oxides and makes them poor electrical conductors. Since oxides of F has high melting point and does not dissolve in water, it has a giant covalent structure with strong and extensive covalent bonds between its atoms, which require a large amount of energy to overcome. Oxides of F is likely to be SiO2. H most likely forms an oxide with simple molecular structure like SO 3 or P4O10. Small amount of energy is needed to overcome the weak instantaneous dipole-induced dipole of attraction between the molecules, thus accounting for its relatively low melting point. It reacts completely and vigorously in water to form a strongly acidic solution. In the order of increasing proton number: Na(G), Mg(E), Si(F), P/S(H)

7 ASRJC JC2 PRELIM 2024 9729/01/H2 [Turn over 8 Theoretical values have been obtained for the two compounds CuCl(s) and TlCl(s) as shown below. Which statements are correct? Compound CuCl(s) TlCl(s) Lattice energy / kJ mol–1 −979 −748 Sum of enthalpies of hydration of separate ions / kJ mol–1 −996 −709 1 Enthalpy change of solution, Hsolution of CuCl is more exothermic than that of TlCl. 2 The strength of interaction with water molecules is greater for the ions of CuCl compared to that of TlCl. 3