2022 HCI H2 Chemistry Promo Mark Scheme

2022 HCI C1 H2 Chemistry Promotional Exam / Mark Scheme & Teaching Solutions HWA CHONG INSTITUTION 2022 C1 H2 CHEMISTRY PROMOTIONAL EXAM MARK SCHEME & TEACHING SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 C A B A D A D B D D 11 12 13 14 15 16 17 18 19 20 C D C B C A C A B C Comments 1 C Obtain the proton number based on the identity of the element in the Data Booklet. Calculate the no. of neutrons by taking mass number (indicated as superscript on the element) minus proton number: A No. of protons = 2; no. of neutrons = 2 – 2 = 0 B No. of protons = 4; no. of neutrons = 8 – 4 = 4 C No. of protons = 20; no. of neutrons = 40 – 20 = 20 D No. of protons = 82; no. of neutrons = 210 – 82 = 128 Only 40Ca has both numbers matching the given nuclear magic numbers. 2 A 1 Each nitrogen atom in cyanogen has one lone pair of electrons, hence statement is correct. 2 Both carbon atoms in cyanogen are sp hybridized – there are two electron groups around each carbon atom and the shape around each carbon is linear, hence statement is correct. 3 There should be three σ bonds: C–N, C–N and C–C, hence statement is incorrect. 3 B A The zig-zag chains are flat and unlikely to form regular 3D spaces in the lattice structure like the tetrahedral 3D arrangement for ice, hence statement is incorrect. B Each hydrogen bond in HF is stronger than that in H 2O since F is more electronegative than O, and hence draw away more electron density from H atom, resulting in a greater partial positive charge on H so attraction between the H and the lone pair of another HF molecule is stronger. Statement is correct.

2 2022 HCI C1 H2 Chemistry Promotional Exam / Mark Scheme & Teaching Solutions C The question states that the bond angle around F atom is similar to that around O in ice, hence bond angle around each F atom should be 104.5 o. Statement is incorrect. D Solid HF exists as discrete HF molecules held together by intermolecular hydrogen bonding. It is still a simple molecular structure, not giant covalent. Statement is incorrect. 4 A 1 More gas added into a flask of fixed volume, hence total pressure in the flask will increase. Or consider PV = nRT, where n increases and V is constant, so P will increase. Statement is correct. 2 Mole fraction of a gas in a mixture is “no. of moles of that gas / total no. of moles of gas”, since total moles of gas has increased, the mole fraction of each gas will decrease. Statement is correct. 3 Partial pressure of each gas is defined as the pressure that each gas exerts on the flask as though it is alone in the flask. Since the no. of moles of N2 and O2 did not change and the volume of the flask did not change, their partial pressures will be unchanged. Or consider partial pressure = mole fraction × total pressure = (ngas / ntotal) × Ptotal. Ptotal/ntotal is a constant (use PV = nRT), so if ngas remain the same, partial pressure of that gas remains the same. 5 D The empirical formula of a compound is the simplest formula which shows the ratio of the atoms of the different elements in the compound. The molecular formula of a compound is the formula which shows the actual number of atoms of each element in one molecule of the compound. 1 NaCl is an ionic compound and hence it does not have a molecular formula. Ionic compound formulae are always empirical formula since the actual number of ions is a very large number and it is the simplest ratio of atoms that is reflected in an ionic formula. 2 Water is a simple molecular compound that comprise 2 H atoms and 1 O atom, which means that its formula is both the empirical formula and molecular formula. 3 Propene is a simple molecular compound that comprise 3 C atoms and 6 H atoms, hence its molecular formula is C3H6. The simplest ratio of atom would be CH2 which is the empirical formula.

3 2022 HCI C1 H2 Chemistry Promotional Exam / Mark Scheme 6 A Calculate the number of moles of particles for each option and compare them. The option with the lowest number of moles of particles will give the smallest number of particles since 1 mole of particles contain Avogadro’s number (6.02 × 1023). A No. of moles of atoms in 15 g Fe = (15 ÷ 55.8) = 0.269 mol B No. of moles of ions in 50 g NaCl = (50 ÷ 58.5) × 2 = 1.71 mol C No. of moles of molecules in 15 dm3 of N2 = (15 ÷ 24) = 0.625 mol D No. of moles of molecules in 10 cm3 of C6H14 = (10 × 6.55) ÷ 86.0 = 0.762 mol 7 D 1 The enthalpy change of formation refers to the energy change in forming a substance from its constituent elements (not atoms!) at their standard state. The correct representation of ∆HꝊformation(NH3) is: ½ N2(g) + 3/2 H2(g) → NH3(g) 2 Bond energy refers to the energy required to break one mole of the covalent bond in the gaseous state, the equation is incorrect because it shows the breaking of only half a mole of the Br–Br bond. The correct representation of ∆HꝊbond energy(Br–Br) is: Br2(g) → 2Br(g) 3 Ionisation energy refers to energy required to remove one mole of electrons from one mole of the gaseous atoms. The equation is incorrect because Mg should be in the gaseous state. The correct representation of ∆H Ꝋionisation energy(Mg) is: Mg(g) → Mg+(g) + eˉ 8 B ∆Hsolution = – L.E. + sum of ∆Hhydration of the ions formed = –(–2195) + (–1577) + 2(–363) = –108 kJ mol–1 9 D A At point A, entropy is zero, meaning the particles are at their maximum state of order (no disorder), hence statement is incorrect. B Between points C and D, the liquid water is increasing in temperature. It is not at equilibrium. Dynamic equilibrium between the solid and liquid states occurs at BC, while dynamic equilibrium between the liquid and gaseous states occurs at DE. Hence statement is incorrect. C At BC, the temperature remains constant so average kinetic energy does not change. Hence statement is incorrect. D At DE, the liquid state is converting to gaseous state while at BC the solid state is converting to liquid state. Gas particles have a greater disorder compared to liquid particles, so the increase in entropy in converting from liquid to gas is larger. Statement is correct.

4 2022 HCI C1 H2 Chemistry Promotional Exam / Mark Scheme & Teaching Solutions 10 D A catalyst does not change the initial and final energy levels of the reactants and products, hence the enthalpy change of reaction does not change. This would rule out options B & C immediately. Catalysis offers an alternative reaction pathway that has a lower activation energy. This means that option A is wrong. This leaves option D as the answer as the diagram shows that the enthalpy change of reaction is unchanged while the activation energy is clearly lower for the catalyzed reaction pathway. 11 C 242Cm 4 160 days �⎯⎯⎯⎯� 2 160 days �⎯⎯⎯⎯� 1 160 days �⎯⎯⎯⎯� ½ 160 days �⎯⎯⎯⎯� ¼ 248Bk 1 320 days �⎯⎯⎯⎯� ½ 320 days �⎯⎯⎯⎯� ¼ From the table below, it takes 242Cm 4 half-lives (160 × 4 = 640 days) to go from 4 to ¼, while it takes 248Bk 2 half-lives (320 × 2 = 640 days) to go from 1 to ¼. Hence at the end of 640 days, the ratio of 242Cm : 248Bk = ¼ : ¼ which gives a 1 : 1 ratio. 12 D 1 Comparing Expts 1 & 2, [CH2ICH2I] and [I2] remain constant, but the light intensity increases by 4× while the rate increase by 2 ×. Hence the rate is proportional to �light intensity. 2 Comparing Expts 3 & 4, [CH2ICH2I] and light intensity remains constant but [ I2] increases by 2× while the rate remains constant. Hence the rate is independent of [I2]. 3 Comparing Expts 2 & 3, [CH2ICH2I] increases by 2× and the light intensity and [I2] remain constant, while the rate increases by 2× . Hence the rate is directly proportional to [CH2ICH2I]. 13 C This question tests understanding and application of Le Chatelier’s principle – that a change imposed on a system at equ