2019 TMJC H2 Chem Prelim P1 ANS

1 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 1 Some data of the ions of four unknown elements are given in the table below. A+ B– D2– E3– mass number 21 20 20 21 number of protons – 11 9 – number of electrons 9 – – 12 Which of the following correctly shows pairs of isotopic and isoelectronic species? isotopic isoelectronic A D2–, E3– B–, E3– B B–, D2– A+, B– C B–, D2– B–, E3– D D2–, E3– A+, B– Answer: A A+ B– D2– E3– mass number 21 20 20 21 number of protons 10 11 9 9 D2- and E3- are isotopic number of electrons 9 12 11 12 B- and E3- are isoelectronic 2 The first six ionisation energies of element G are plotted in the graph below. number of electrons removed log10 (ionisation energy) TAMPINES MERIDIAN JUNIOR COLLEGE 2019 JC2 H2 Chemistry Prelim Exam Paper 1

2 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry Which of the following can be deduced from the graph? A There are 2 quantum shells in element G. B Element G is from Group 2 in the Periodic Table. C The fifth and sixth electrons are removed from the same subshell. D The identity of element G is carbon. Answer: C With limited information from the graph, it is unclear if the jump from the 4th to the 5th I.E. is a change in subshell or quantum shell. Hence, options A, B and D cannot be concluded using the graph. Then only conclusion which can be made is that the 5th and 6th electrons are in the same subshell. 3 For which of the following pairs does the first molecule have a higher polarity than the second molecule? 1 CH3OH, CH3I 2 CF4, CCl4 3 cis-CHF=CHF, trans-CHF=CHF A 1 only B 1 and 3 C 2 and 3 D 1, 2 and 3 Answer: B 1 Although both CH3OH and CH3I are polar, CH3OH has higher polarity because O is more electronegative than I, hence the C –O bond is more polar than the C –I bond. 2 Both CF4 and CCl4 are non-polar as both have net zero dipole moment. 3 trans-CHF=CHF is non-polar as both C–F dipole moment cancels out, hence net zero dipole moment. cis-CHF=CHF is polar.

3 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 4 Three substances H, I and J have physical properties shown in the table below. substance melting point/ °C boiling point/ °C electrical conductivity of solid of liquid H 801 1413 poor good I 2852 3600 poor good J 3500 at very high pressure unable to determine poor unable to determine What are the possible identities for substances H, I and J? H I J A NaCl AlCl3 C (diamond) B MgO AlCl3 C (graphite) C NaCl MgO C (graphite) D NaCl MgO C (diamond) Answer: D  Substance H should have a giant ionic structure , where NaCl or MgO are possible identities.  Substance I should also have a giant ionic structure but with larger lattice energy compared to substance H. AlCl3 cannot be substance H because it is a simple molecule. Hence, H must be MgO.  Substance J should have a giant molecular structure, which does not conduct electricity. Hence, J must be diamond.

4 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 5 Which of the following diagrams correctly describe the behaviour of a fixed mass of an ideal gas at constant temperature? [c is concentration] graph 1 graph 2 graph 3 A 2 only B 1 and 2 C 2 and 3 D 1, 2 and 3 Answer: B  1 pV = nRT pV = constant (T is constant) pV plotted on horizontal axis  graph resembles “x = constant” graph.  2 pV = nRT pV = m MRT (where m is mass and M is molar mass) p = m V  constant (T and M are constant) p = c  constant  c = p  constant  graph resembles “y = constant  x” graph.  3 p = c  constant p c = constant  graph resembles “y = constant” graph. V pV c p p c p

5 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 6 In which of the following reactions is the underlined reactant acting as a Lewis acid? A BF3 + F  BF4 B AlCl3 + Cl2  AlCl4 + Cl+ C Ag+ + 2NH3  [Ag(NH3)2]+ D H2O + HSO3  H2SO3 + OH Answer: A Lewis acid: species (molecule or ion) that can accept a pair of electrons  A In the formation of BF4, F donates a pair of electrons for dative bond formation to B in BF3 (B here is electron-deficient, with only 6 electrons around it).  B Cl in Cl2 donates a pair of electrons for dative bond formation to Al in AlCl3 (Al here is electron-deficient, with only 6 electrons around it)  AlCl3 is the Lewis acid.  C NH3 donates a pair of electrons for dative bond formation to Ag+.  D HSO3 donates a pair of electrons for dative bond formation to H+ (from H2O)  H+ is the Lewis acid. 7 Which of the following statements is true of the Period 3 chlorides, NaCl to PCl5? A The melting points of the chlorides show a continuously decreasing trend. B The pH of the solutions formed when the chlorides are added to water range from alkaline to acidic. C Three of the chlorides have simple molecular structures. D The only chlorides which undergo hydrolysis in water are AlCl3 and MgCl2. Answer: C  A (pg 15 of Periodic Table lecture notes)  B The pH of aqueous NaCl is 7, not alkaline.  C NaCl and MgCl2 are both ionic while the other 3 chlorides are covalent. The covalent chlorides have simple molecular structure.  D All the chlorides, except for NaCl undergo slight or complete hydrolysis.

6 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 8 The relative abundances of the isotopes of a sample of titanium are shown in the table below. Given that the relative atomic mass of the sample is 47.862, what is the value of z? A 214.2 B 79.1 C 78.7 D 26.6 Answer: A (46 × 11.2) + (47 × 10.1) + (48 × z) 11.2 + 10.1 + z = 47.862  z = 214.2 9 A metallic salt was found to react exactly with sulfite ions in a 2 : 1 ratio. In this reaction, the sulfite ion is oxidised as follows. SO32(aq) + H2O(l)  SO42(aq) + 2H+(aq) + 2e– What is the new oxidation number of the metal in the salt if its original oxidation number was +3? A +1 B +2 C +4 D +5 Answer: B Since SO32 lost e, the metallic salt gained e. Let the moles of e gained by 1 mole of metallic salt be y Total moles of e gained = Total moles of e lost 2  y = 1  2  y = 1  new ox. state is +2 relative isotopic mass 46 47 48 relative abundance 11.2 10.1 z

7 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 10 The enthalpy changes for the following reactions are given: 2Cr(s) + 3 2 O2(g) → Cr2O3(s) Hf (Cr2O3(s))= 1120 kJ mol1 C(s) + 1 2 O2(g) → CO(g) Hf (CO(g))= 110 kJ mol1 What is the enthalpy change of reaction for the following reaction, in terms of kJ mol1? 3C(s) + Cr2O3(s) → 2Cr(s) + 3CO(g) A  940 B  870 C + 590 D + 790 Answer: D Hrxn = 3(–110) – (–1120) = + 790 kJ mol–1 11 An experiment was conducted by reacting propyne with hydrogen gas in a shock tube. CH3C≡CH(g) + 2H2(g) → CH3CH2CH3(g) The value of G for the reaction is found to be negative at low temperature. What is the sign of H and S for the above reaction? H S A   B +  C  + D + + Answer: A There is a decrease in number of moles of gaseous particles (from 3 moles to 1 mole) after the reaction, thus S is negative. G = H – TS Since G is negative and –TS is positive, H should be negative.

8 2019 H2 Chem Prelim P1 Work Solutions Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 12 The kinetics of the reaction between hydrogen per