NJC Prelim P1 Solutions

1 NJC Preliminary Examination 9729/01/18 [Turn Over + – R S T NJC 2018 SH2 H2 Chemistry Paper 1 Solutions: 1 Use of the Data Booklet is relevant to this question. The following are flight paths of charged particles when accelerated in an electric field. Which correctly identifies S, T and U? S T U A 15O+ 14C+ 14N+ B 15O⎯ 15O+ 28Si+ C 14N⎯ 28Si2+ 14C2+ D 14N⎯ 14C+ 28Si2+ C: Negatively charged ions attracted to positive plate, positive ions to negative plate. particles 14N⎯ 14C2+ 14C+ 28Si2+ Charge /mass 1/14 2/14 = 1/7 1/14 2/28 = 1/14 Since angle of deflection is charge/mass ratio, S and T have roughly the same angle of deflection but different polarity, while U has almost double the angle of deflection as T. 2 In which sequence is the molecules listed in the order of increasing dipole moment? A SO 3, CO2, AlCl3 B H2O, H2S, HBr All non-polar. Dipole moments cancel out due to shape of molecule. H2O and H2S have bent shape. H2O is more polar than H2S as O is more electronegative than S. C CF4, CO, HF D NH3, HF, BeCl2 CF4 is non-polar, dipole moments cancel out due to tetrahedral shape of molecule. CO and HF are linear. HF is more polar than CO, as electronegativity difference is greater between H and F than between C and O. NH 3 and HF are polar but BeCl2 is non- polar, dipole moments cancel out due to linear shape of molecule. S T U

2 NJC Preliminary Examination 9729/01/18 [Turn Over 3 In which row are the molecules arranged in order of increasing bond angle? 1 CH4, AlCl3, XeF2 CH4 (tetrahedral, 109 degrees) AlCl3 (trigonal planar, 120 degrees) XeF2 (linear, 180 degrees) 2 H2S, PH3, NH3, H2S (tetrahedral, 109 degrees) PH3 and NH3 (trigonal pyramidal, around 107 degrees) NH3 has a larger bond angle than PH3 as 1) N is more electronegative than P, N pulls electron density of bond pairs more towards itself, leading to greater bond-pair bond-pair repulsion. 2) N has a smaller lone pair region than P (N is above P in group 15). Lone-pair bond- pair repulsion is smaller, leading to a larger bond angle. 3 NF3 , NCl3, SO3 NF3 , NCl3 (trigonal pyramidal, around 107 degrees) NF3 has a smaller bond angle than NC l3. F is more electronegative than N and N is more electronegative than Cl). F pulls electron density of bond pairs more towards itself/ away from central N atom, leading to smaller bond-pair bond-pair repulsion. SO3 (trigonal planar, 120 degrees) A 1, 2 and 3 B 1 and 2 C 2 and 3 D 1 only 4 Propadiene and propyne both have the same molecular formula, C 3H4. They exist in equilibrium as shown: H2C=C=CH2 CH 3C≡CH propadiene propyne Which bond is present in propadiene but not present in propyne? A a σ bond formed by s – sp overlap B a π bond formed by p – p overlap C a σ bond formed by sp – sp2 overlap D a σ bond formed by sp2 – sp2 overlap propadiene H 2C=C=CH2 hybridisation: sp2 sp sp2 propyne CH3C≡CH hybridisation: sp3 sp sp

3 NJC Preliminary Examination 9729/01/18 [Turn Over A a σ bond formed by s – sp overlap: in propyne CH3C≡C–H, not in propadiene B a π bond formed by p – p overlap: present in both molecules C a σ bond formed by sp – sp2 overlap: in propadiene H2C=C=CH2 , not in propyne D a σ bond formed by sp2 – sp2 overlap :absent in both 5 Use of the Data Booklet is relevant to this question. A reaction scheme regarding manganese compounds is shown below. Which statements are true? 1 Off-white solid S is able to dissolve in excess of NH3(aq). 2 Manganese in brown solid P has an oxidation state of +4. 3 Reagent Q can be acidified [V(H2O)6]3+(aq). 4 Off white solid S turns brown upon standing. A 1, 2 and 4 B 2, 3 and 4 C 2 and 3 D All correct 1 off-white solid S is Mn(OH)2. It is insoluble in excess of NH3(aq). Check Data Booklet. 2 Brown solid P is MnO2. Oxidation state of Mn in MnO2 is +4. KMnO4(aq) oxidises Fe(OH)2 , itself is reduced to MnO2. 3 Reagent Q can be acidified [V(H2O)6]3+ as the Ecell > 0. (Ecell = (+1.52) + (- 0.34) = +1.18V) 4 off-white solid S turns brown upon standing. Check Data Booklet. Mn(OH)2 is further oxidised by air. KMnO4(aq) Fe(OH)2 brown solid P reagent Q [Mn(H2O)6]2+(aq) NaOH(aq) off-white solid S NH3(aq) off-white solid S

4 NJC Preliminary Examination 9729/01/18 [Turn Over 6 In which chemical reaction does the transition metal compound or element behave as the described catalyst? Reaction Catalyst 1 Formation of ethanal from ethanol, using acidified potassium dichromate K2Cr2O7 is an oxidising agent, not a catalyst. Homogeneous 2 Formation of oxygen from hydrogen peroxide, using iron(III) hydroxide Fe(OH)3 is a solid catalyst used in the decomposition of hydrogen peroxide, due to the slow rate of reaction. Heterogeneous 3 Chlorination of benzene, using chlorine and iron(III) chloride FeCl3 is a catalyst as well as a halogen carrier, is regenerated in the last step of the electrophilic substitution. Homogeneous 4 Removal of air pollutants in exhaust systems of cars, using nickel Nickel is a catalyst in the catalytic converter and is in solid phase, a different phase from the gaseous reactant Heterogeneous A 1, 2 and 4 B 2 and 3 C 3 and 4 D 2, 3 and 4 7 Use of the Data Booklet is relevant to this question. Given the following standard enthalpy changes, ∆H /kJ mol−1 C(graphite) + 2H2(g) → CH4(g) −75 a + (2 x436) − (4x 410) = −75 a = +693 kJ mol −1 What is the standard enthalpy change of atomisation of graphite? A +693 kJ mol−1 B +1129 kJ mol−1 C −2151 kJ mol−1 D −2587 kJ mol−1

5 NJC Preliminary Examination 9729/01/18 [Turn Over 8 Ammonia gas and hydrogen chloride gas react to form ammonium chloride as shown in the equation below: NH3(g) + HCl(g) → NH4Cl(s) ∆H o= −176 kJ mol−1 The magnitude of standard entropy change of this reaction is 284 J K−1 mol−1. Which statements are correct? 1 ∆G o = −261 kJ mol−1. No of mol of gas decreases => ∆So = −284 J K−1 mol−1 ∆Go = −176 – 298 (−0.284) = −91.4 kJ mol−1. 2 The reaction becomes non-spontaneous at temperatures higher than 620 K. Crossover temperature occurs when ∆G o = 0 ∆Ho = T∆So T = -176000/ -284 = 620K 3 There is an increase in order as strong hydrogen bonding between NH3 and HCl hold the particles in NH4Cl in fixed positions and close to each other. NH4Cl is ionic lattice with strong ionic bonds between NH 4+ and Cl −, not strong H bonding between the NH3 and HCl molecules. A 1 only B 1 and 2 C 2 only D 2 and 3 9 The graph below shows how the fraction of X, which represents one of the following compounds in the given equilibrium mixture, varies with temperature at pressures of Y Pa and Z Pa. 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) ΔH = −1267 kJ mol−1 Identify X and the correct relative magnitudes of Y and Z. X Pressure A N2 Z > Y B O2 Y > Z C H2O Y > Z D NH3 Z > Y Fraction of X in equilibrium mixture Temperature Y Pa Z Pa

6 NJC Preliminary Examination 9729/01/18 [Turn Over Shape of graph: (i) As temperature increase, fraction of X decrease. (ii) As temperature increase, as forward reaction is exothermic,