2020 HCI H2 Chemistry P1 Answers

2020 HCI C2 H2 Chemistry Prelim Exam / Paper 1 HWA CHONG INSTITUTION 2020 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 D B C B D C B C B D 11 12 13 14 15 16 17 18 19 20 A B A D A B D C A D 21 22 23 24 25 26 27 28 29 30 C C C A A D B A B C 1 D This question tests the writing of electronic configurations. The electronic configuration and the number of unpaired p electrons for each of the species are shown below. A Al2−: 1s2 2s2 2p6 3s2 3p3 (3 unpaired p electrons) O+: 1s2 2s2 2p3 (3 unpaired p electrons) B N: 1s2 2s2 2p3 (3 unpaired p electrons) Cl2+: 1s2 2s2 2p6 3s2 3p3 (3 unpaired p electrons) C C: 1s2 2s2 2p2 (2 unpaired p electrons) Cl+: 1s2 2s2 2p6 3s2 3p4 (2 unpaired p electrons) D B−: 1s2 2s2 2p2 (2 unpaired p electrons) S+: 1s2 2s2 2p6 3s2 3p3 (3 unpaired p electrons) 2 B A simple molecular solid should have a low melting point (due to weak intermolecular forces) and does not conduct electricity (due to lack of mobile charge carriers – ions or free electrons). It may be soluble or insoluble in water depending on the type of intermolecular forces that it can form with water. Since the unknown compound is a solid at room temperature, th e answer cannot be A due to the melting point. 3 C A A  bond involves head-on overlap of the orbitals of two bonded atoms, thus electron density is highest between the two atoms e.g. A is a correct statement.

1 B This is a correct statement. An example of such a compound would be MgCO3 where ionic bonds exist between Mg2+ and CO32− while covalent bonds exists within CO32−. C A triple bond consists of two  bonds and one  bond instead. C is incorrect. D This is a correct statement. This is also part of the reasoning how cis-trans isomerism arises in alkenes. 4 B This question tests your understanding of ideal gas and involves the use of the ideal gas equation PV = nRT. A PV = nRT PV = (mass/Mr)RT Since density = mass / V, P × Mr = density × R × T density = P × Mr/(RT) Thus, the density of an ideal gas at constant temperature is directly proportional to the pressure. B PV = nRT V = 1 mol × RT / P At the same T and P, 1 mol of any ideal gas has the same volume V. This statement is correct. C As seen from the ideal gas equation, V is directly proportional to temperature (in Kelvins), for a given mass (or mol) of the ideal gas at constant P. However, when temperature is increased from 25 C (298K) to 50 C (323K), temperature is not doubled. Thus V is not doubled. D Since the volume of container does not change, the partial pressure of the ideal gas should not change as well. We may also think in terms of: Adding that second gas increases total pressure. However, the mole fraction of the first gas decreases by the same factor, hence its partial pressure does not change. 5 D According to the question, all the carbon atoms in the organic compound are converted into CO2 when burnt completely in excess O2, and the nitrogen atoms are converted into N2.

2020 HCI C2 H2 Chemistry Prelim Exam / Paper 1 Assuming that the formula of the organic compound is C xHyNz, then for every one mole of CxHyNz, x moles of CO 2 will be produced and z/2 moles of N 2 will be produced. The volume ratio of organic compound : CO2 : N2 = 10 cm3 : 30 cm3 : 5 cm3 = 1 : 3 : 0.5 Hence x = 3 and z = 1 which fit the molecular formula of D. 6 C Some general guidelines for solving this kind of oxidation states / electron transfer / mole concept problem:  Work from the reactant whose conversion is given (XeF 2  Xe in this question)  Use the relationship, change in oxidation state for an atom = no. of moles of electrons transferred for 1 mol of that atom  Link the no. of moles of a reactant reacted to the no. of moles of e lectrons transferred XeF2 is converted to Xe, so oxidation state of Xe decreases from +2 to 0. XeF2 is reduced, there is a gain of electrons during reduction. Oxidation state of Xe decreases by 2 units, therefore 2 mol of electrons gained for 1 mol of XeF2. +2 0 XeF2 + 2e–  Xe n(XeF2) reduced = 15.20/1000  25.0/169 = 0.00225 mol n(e–) transferred = 0.00225  2 = 0.00450 mol n(M2+) oxidised = 10.0/1000  0.150 = 0.00150 mol n(e–) transferred for 1 mol of M2+ = 0.00450  0.00150 = 3 Therefore, oxidation state of M increases by 3 units, from +2 to +5  Ans. C 7 B From the melting point trend, it can be deduced that these third period elements are Al, Si, P and S. Property X corresponds to the first ionisation energies of these elements. First ionisation energies generally increase across the period due to increasing effective nuclear charge. However, the first ionisation energy of S is lower than that of P due to the interelectronic repulsion experienced by the electrons in the same p orbital. Students may also check the first IE data from the Data Booklet. Property Y fits well with atomic radii data. Atomic radius decreases across the period due to increasing effective nuclear charge. The outermost electrons are held closer to the nucleus. Students may also check the atomic radii data from the Data Booklet.

2 8 C From the information given, it can be deduced that X is Mg as MgO is commonly used as a refractory lining material due to its high melting point. Y is Ca. When CaCO3 is heated, it forms CaO (Z), which gives Ca(OH)2 when hydrated. Ca(OH)2 can be used in agriculture to treat acidic soils. A Ca is more reactive than Mg. As atomic radii increase, the metal atoms lose their electrons more readily (1st and 2nd ionisation energies decrease) going down the group, so they form M2+ cations more easily. B Reducing power of the group 2 metals increases (tendency to be oxidised increases) down the group. C Mg(NO3)2 decomposes at a lower temperature than Ca(NO3)2. Apply the same reasoning like comparing MgCO3 vs. CaCO3. Mg2+ has smaller radius than Ca2+. Mg2+ has higher charge density and polarising power. Mg2+ polarises and distorts the NO3– electron cloud more, the N–O bond is weakened more and would need less heating to break (the N –O bond). Hence, Mg(NO 3)2 decomposes at a lower temperature than Ca(NO3)2. D L.E.  q+  q– r+ + r– Mg2+ has smaller radius (r+) than Ca 2+. Therefore, the numerical value of the L.E. of MgO is larger than that of CaO.  9 B At is below I in Group 17. 1 As the size of the atoms increases down the Group, the bond length increases and the halogen molecule dissociates more easily due to the decreasing bond strength. As such, it is expected that At2 dissociates more easily than Cl2. 2 Down the group, as atomic radius increases, the bond length of the H–X bond increases and thus bond strength decreases. Hence, less energy is needed to break the H–X bond. Thus, the thermal stability of the hydrogen halides decreases down the group and H–At is expected to be less stable to heat than H–I. 3 Check the Data Booklet QA Notes. AgCl is soluble in aq. NH3, AgBr is partially soluble in aq. NH3, AgI is insoluble in aq. NH3. Based on this trend, AgAt is most likely insoluble in aq. NH3.

2020 HCI C2 H2 Chemistry Prelim Exam / Paper 1 10 D A Increasing volume means there is more space for the particles to move around. There should be more ways to distribute the particles. Hence entropy should increase. B The increase in temperature increases the range of kinetic energies (widens the Boltzman distribution). There is a wider range of energies that can be distributed through the motion of the particles, thus increasing entropy. C When an inert gas is added to a system at constant pressure and constant temperature, the volume of the container will increase given the increase in the total number of moles of gas. Similar to Option A, the entropy will increase.