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1 Name: ________________________ Student Number: _____________________ Chemistry 1000 Final Exam Version A Fall 2012 1) 2) 3) 4) 5) 6) 7) 8) INSTRUCTIONS Read the exam carefully before beginning. There are 21 questions on pages 2 to 14 followed by 2 pages of “Data Sheet” (including periodic table) and a blank page for any rough work. Please ensure that you have a complete exam. If not, let an invigilator know immediately. All pages must be submitted at the end of the exam. If your work is not legible, it will be given a mark of zero. Marks will be deducted for incorrect information added to an otherwise correct answer. You may use a calculator. Show your work for all calculations. Answers without supporting calculations will not be given full credit. Marks will be deducted for improper use of significant figures and for numerical answers with incorrect/missing units. Do not open the exam until you are told to begin. Beginning prematurely will result in removal of your exam paper and a mark of 0. You have 3 hours to complete this exam. Nobody may leave the exam room during the first hour or the last 15 minutes of the exam. Q Mark Q Mark 1 /4 11 /9 2 /6 12 / 12 3 / 12 13 / 10 4 /4 14 /7 5 /4 15 / 18 6 /5 16 /2 7 / 10 17 /3 8 /2 18 /5 9 /5 19 /3 10 /3 20 /7 21 /1 Total / 132 2 Name: ________________________ 1. Student Number: _____________________ Complete the table below. [4 marks] Name Formula copper(II) hydroxide Cu (OH )2 phosphate ion PO43− 24 12 magnesium-24 ion nitric acid 2. (a) HNO3 Write balanced chemical equations for each of the following processes: 133 Ba decays by electron capture 133 56 Ba + 0−1 e →133 55 Cs (b) Nitrogen reacts with magnesium N 2 ( g ) + 3Mg ( s ) → Mg 3 N 2 ( s ) (c) Potassium reacts with water 2 K ( s ) + 2 H 2 O(l ) → 2 KOH ( aq ) + H 2 ( g ) or 2 K ( s ) + 2 H 2 O(l ) → 2 K (+aq ) + 2OH (−aq ) + H 2 ( g ) (d) Calcium reacts with bromine Ca ( s ) + Br2 (l ) → CaBr2 ( s ) (e) Aqueous sodium chloride is subjected to electrolysis 2 NaCl ( aq ) + 2 H 2 O(l ) → 2 NaOH ( aq ) + H 2 ( g ) + Cl 2 ( g ) or 2Cl (−aq ) + 2 H 2 O(l ) → 2OH (−aq ) + H 2 ( g ) + Cl 2 ( g ) (f) Aluminium oxide is reacted with aqueous sodium hydroxide. Al 2 O3( s ) + 2 NaOH ( aq ) + 3H 2 O( l ) → 2 Na[Al (OH )4 ]( aq ) or Mg 2 + Al 2 O3( s ) + 2OH (−aq ) + 3H 2 O(l ) → 2[Al (OH )4 ]( aq ) − [6 marks] 3 Name: ________________________ 3. Student Number: _____________________ (a) Chlorine, bromine and iodine are all capable of forming several different oxoanions. On the other hand, fluorine can only form one oxoanion, and it is not stable. [12 marks] Draw the Lewis structure of the one oxoanion containing fluorine. [1 mark] (b) [10 marks] Complete the table below for any two of the four oxoanions of iodine. Name of Molecular Oxidation Formula Lewis structure* Geometry State of I Name linear +1 hypoiodite bent +3 iodite trigonal pyramidal +5 iodate tetrahedral +7 periodate (c) Explain why fluorine behaves differently from the other halogens (in this context).[1 mark] F is the most electronegative element on the periodic table and is in period 2. In any other oxoanion of F, it would be necessary to have a positive formal charge on F (which is forbidden) or for F to have more electrons than a complete octet (also forbidden). 4. (a) Lithium-6 and lithium-7 are the only two stable isotopes of lithium. [4 marks] Predict the likely modes of decay for isotopes of lithium that are heavier than the stable isotopes. Briefly explain your reasoning. β– decay [2 marks] Heavier isotopes of an element have more neutrons than the stable isotopes so they will decay in ways that lower N/Z. When a neutron decays to a proton, a β– particle is released. Predict the likely modes of decay for isotopes of lithium that are lighter than the stable isotopes. Briefly explain your reasoning. β+ decay or electron capture [2 marks] Lighter isotopes of an element have fewer neutrons than the stable isotopes so they will decay in ways that raise N/Z. Electron capture converts a proton to a neutron. Positron (β+) emission has the same effect. (b) 4 Name: ________________________ 5. (a) (b) 6. Student Number: _____________________ For each pair of species, list the intermolecular force(s) which can act between the two species. Also, circle the strongest intermolecular force acting between those two species. [4 marks] – I (ion) and chloromethane (polar molecule) (see structure below) ion-dipole ion-induced dipole dipole-induced dipole induced dipole-induced dipole (aka London dispersion forces) H2O (polar molecule) and NH3 (polar molecule) hydrogen bonding dipole-dipole dipole-induced dipole induced dipole-induced dipole (aka London dispersion forces) Briefly describe a general chromatography experiment. Your description should identify the main components of chromatography as well as explain how the technique is used to separate compounds. [5 marks] A chromatography experiment separates compounds based on their relative affinity for two different species: • a stationary phase (bound to a column, packed in a column, etc.) • a mobile phase (travels through a column or similar) This affinity is usually determined by the strength of intermolecular forces. Compounds with a stronger affinity for the stationary phase spend more time adsorbed to it and therefore take longer to travel through the column. Compounds with less affinity for the stationary phase (i.e. more affinity for the mobile phase, relatively speaking) spend less time adsorbed to the stationary phase and travel through the column faster. 5 Name: ________________________ Student Number: _____________________ 7. Define each term and give an example of each (a compound or a specific balanced reaction, as appropriate). [10 marks] For (a) and (b), several good examples have been listed. A Lewis structure in addition to (or instead of) the formula was appreciated. (a) Lewis acid an electron pair acceptor e.g. BF3, CO2, SO3, Fe3+, Cu2+, H+ (b) Brønsted base a proton (H+) acceptor e.g. NH3, OH–, O2–, S2–, CO32– (c) Hydrogen bonding a strong directional intermolecular force in which a lone pair on an atom of N, O or F is partially donated to an atom of H that is bonded to another atom of N, O or F e.g. (d) Ionization energy the energy required to remove an electron from a neutral atom in the gas phase e.g. (e) Enthalpy of electronic attraction (ΔEAH) the energy released when an electron is added to a neutral atom in the gas phase e.g. 6 Name: ________________________ 8. 9. (a) Student Number: _____________________ Choose an element from group 14, group 15 or group 16 that has more than one allotrope. Give two different allotropes that this element can form. You can use either molecular formulae or names. [2 marks] Possible answers carbon: diamond, graphite, graphene, fullerenes, nanotubes, etc. phosphorus: P4 (white phosphorus), red phosphorus, black phosphorus oxygen: O2, O3 (ozone) sulfur: S8, plastic sulfur, S2, S3, S4, S5, S6, S7, etc. [5 marks] Draw the Lewis structures of carbon monoxide and carbon dioxide. Include any nonzero formal charges on the appropriate atoms. [2 marks] (b) What is unusual about the structure of carbon monoxide? [1 mark] It has a negative formal charge on carbon (not a very electronegative element) and a positive formal charge on oxygen (one of the more electronegative elements). (c) Is carbon dioxide a Lewis acid or a Lewis base? Lewis acid (primarily) (d) Is carbon monoxide a Lewis acid or a Lewis base? [1 mark] Lewis base no credit for cutesy answers like “yes” [1 mark] no credit for cutesy answers like “yes” 10. Borane (BH3) does not exist in nature. Rather, it exists as diborane. (a) Draw the Lewis structure of diborane. [3 marks] [1 mark] (b) What is unusual about this structure? [1 mark] It has three-center two-electron bonds (i.e. bonds in which the pair of electrons are shared between three atoms instead of the usual two). This is shown more clearly in the structure on the left. (c) Why do you think that diborane exists while borane does not? [1 mark] The boron atoms obey the octet rule in diborane but they do not in borane (BH3). 7 Name: ________________________ 11. (a) (b) Student Number: _____________________ Draw a full set of 3d orbitals. Each orbital must be drawn on a properly labeled set of axes [9 marks] [5 marks] In an octahedral complex, these orbitals split into two sets of different energies. Indicate which orbitals belong to the higher-energy set and which to the lower-energy set. Explain your choices. [4 marks] The higher-energy set is The lower energy set is and , and . . In an octahedral complex, a metal is surrounded by six atoms, each donating an electron pair to the metal. The six atoms are located along the three axes: There is repulsion between the electron pairs donated by these atoms and the electrons in the d orbitals of the metal, raising the energy of these d orbitals. This effect is stronger for the two d orbitals pointing directly at the electron-donating atoms (i.e. and ). 8 Name: ________________________ 12. (a) Student Number: _____________________ [12 marks] Draw an orbital energy diagram for the valence shell (both occupied and unoccupied orbitals) of a beryllium atom. [2 marks] E (b) or Give one possible complete set of quantum numbers for an electron in the highest energy occupied orbital of beryllium. [2 marks] n = 2, l = 0, ml = 0, ms = +½ n = 2, l = 0, ml = 0, ms = -½ (c) Explain why we can’t measure an enthalpy of electronic attraction for beryllium. [3 marks] The next electron added to Be would start a new subshell (2p). The four electrons already in Be would shield enough of the +4 nuclear charge (from the four protons in the nucleus) that the new electrons would feel as much repulsion from the four electrons as it would feel attraction to the nucleus. Thus, Be– would not form. (d) What is an emission spectrum? [1 mark] An emission spectrum shows the wavelengths of light emitted when excited state atoms relax to the ground state. (e) The lowest energy transition in the emission spectrum of a beryllium atom has a wavelength of 455 nm. To what transition does this line correspond? [1 mark] (f) Calculate the energy difference between the two energy levels involved in the transition described in part (e). [3 marks] The energy difference between the two energy levels is equal to the energy of the photon emitted For light, Therefore 9 Name: ________________________ Student Number: _____________________ 13. Natural nitrogen samples from different sources have different molar masses. Two particular samples of nitrogen have molar masses of 14.00648 g/mol and 14.00711 g/mol respectively. [10 marks] (a) Based on the molar masses given above, what is likely to be the most common isotope of nitrogen? [1 mark] 14 N (nitrogen-14) (b) How many protons and how many neutrons would the nucleus contain for the isotope you named in part (a)? [2 marks] 7 protons and 7 neutrons (c) What experimental method could we use to determine the molar mass? Explain briefly what data we get from this experiment and how a molar mass is calculated from that data. [4 marks] We could use mass spectrometry (MS). This would give the exact mass of each isotope present in the sample and the relative amount of each isotope present in the sample (wchich allows calculation of percent abundance for each isotope). From this information, we use the formula below to calculate molar mass: %n Mn M av = ∑ 100% where n represents each isotope (%n is its percent abundance and Mn is its exact mass) (d) For the two samples described above and the experiment you described in part (c), what data would be the same (within experimental error) and what data would be different? Can you say (at least qualitatively) how they would differ? In other words, what data would be larger/smaller for what sample? [3 marks] The exact mass of each isotope (Mn) is the same. The relative amounts of each isotope will differ, so the percent abundances (%n) will differ. The heavier sample (14.00711 g/mol) will contain a higher percentage of heavier isotopes in comparison to the lighter sample. 10 Name: ________________________ 14. (a) Student Number: _____________________ In proton therapy, a beam of protons is used to destroy tumors inaccessible to normal surgical procedures. Ions (alpha particles, protons, etc.) progressively slow down as they pass through biological tissues, and they do most of their damage just as they are about to stop. By choosing their energy correctly, it is therefore possible to target a tumor at a very specific depth, causing much less damage to intervening tissues. In one particular operation, protons with a kinetic energy of 3.0×10-11 J per proton were used. [7 marks] Calculate the wavelength of these protons. [3 marks] Since protons are particles with mass, E kinetic = 12 mv 2 v= 2 E kinetic = m ( ) kg ⋅ m 2 1 2 2 3.0 ×10 −11 J 1u × × s − 27 (1.007277u ) 1.660539 ×10 kg 1J = 1.9 ×10 8 m s Then use the de Broglie equation to calculate wavelength ( −34 ) kg ⋅ m 2 1 s2 6.626070 ×10 HzJ h 1u = × × = 2.1×10 −15 m λ= mv (1.007277u ) 1.9 ×10 8 ms 1.660539 ×10 − 27 kg 1J ( ) These protons are traveling at high enough speeds that (if we knew how), we would use relativity theory to answer this problem. As it turns out, the answer is approximately the same using classical physics (as done in the solution above) or using relativity. (b) At the stage at which tumors are noticed and treated, they typically have dimensions on the order of centimeters. How does the wavelength of the protons compare? What, if anything, would this imply for their use in proton therapy? [2 marks] The wavelength of the proton is substantially smaller than the dimensions of the tumour (~1013 times smaller) so the protons can be used to precisely target the tumor. (c) A radiotherapist plans to deliver a proton therapy dose of 3.5 Gy to a tumor with an estimated mass of 32 g. (i) How much energy would this dose represent? [1 mark] 1 Gy = 1 J/kg (absorbed dose) 1kg = 0.11J E = 3.5 kgJ × 32 g × 1000 g (ii) How many protons would be needed to deliver this dose? Each proton has a kinetic energy of 3.0×10-11 J E 0.11J # protons = total = = 3.7 ×10 9 protons J E proton 3.0 ×10 −11 proton [1 mark] 11 Name: ________________________ Student Number: _____________________ 15. In power plants, the burning of coal produces SO2 because the coal contains a small amount of sulfur (hence sulfur is also burned). One way to remove SO2 from the flue gases of power plants is to react it with an aqueous solution of H2S. One product of this reaction is sulfur and the other is water. [18 marks] 3 (a) How many moles of SO2 will be produced by burning 1.00 ton (1.00 ×10 kg) of coal containing 3.00% sulfur by mass? [3 marks] Step 1: Calculate the mass of sulfur (S8) that will be burned ( ) 3.00% 3 m S8 = 1.00 ×10 kg = 30.0kg 100% Step 2: Calculate the moles of sulfur that will be burned (molar mass of S8 is 256.528 g/mol) 1000 g 1mol n S8 = 30.0kg × × = 117mol 1kg 256.528 g Step 3: Calculate the moles of SO2 that will be produced (every 1 mole of S8 contains 8 sulfur atoms so will produce 8 moles SO2) 8molSO2 n SO2 = 117molS 8 × = 936molSO2 1molS 8 Note that you should get the same answer for moles of SO2 using any allotrope of sulfur (or even a generic “S”). The only part that will give a different answer for each allotrope is step 2. (b) Write a balanced chemical equation for the reaction between SO2 and H2S. 16 H 2 S + 8SO2 → 3S 8 + 16 H 2 O [1 mark] (c) What are the oxidation states of sulfur in SO2 and H2S? [2 marks] S in SO2: ____+4____ S in H2S: ____-2____ At 25 ˚C (298K) and 100 kPa, how many liters of hydrogen sulfide (H2S) gas would be needed to remove all the SO2 formed in (a)? [3 marks] Step 1: Calculate the moles of H2S(g) required 16molH 2 S n H 2 S = 936molSO 2 × = 1.87 ×10 3 molH 2 S 8molSO 2 (d) Step 2: Calculate the volume of H2S(g) required PV = nRT ( )( ) 3 (298.15K ) 1kPa nRT 1.87 ×10 mol 8.314462 mmol⋅ Pa ⋅K V= = × = 46.4m 3 (100kPa ) P 1000 Pa 3 This volume could be converted into L if you like. That would give 4.64×104 L. 12 Name: ________________________ Student Number: _____________________ 15. (e) continued… If the untreated flue gases of power plants are released into the environment, SO2 will react with another pollutant, NO2, to produce sulfur trioxide and nitrogen monoxide. Draw a formal charge-minimized Lewis structure for SO3. [2 marks] (f) Fill in the following table for SO3. [4 marks] Name of Molecular Geometry Bond Angles Average S-O Bond Order Molecular Polarity? (circle one) trigonal planar 120° 2 polar / nonpolar (g) When SO3 dissolves in atmospheric water, we get acid rain! Write a balanced chemical equation for the reaction of SO3 with water. [1 mark] (h) In the Lewis acid-base reaction described in part (g), is SO3 acting as a Lewis acid or a Lewis base? Support your answer with a reaction showing the movement of electrons using curly arrows. [2 marks] Lewis acid 16. (a) [2 marks] On what quantum number(s) does the orbital energy in a hydrogen atom depend? [1 mark] n Because there is only one electron in hydrogen, n is the only quantum number that affects the energy of the electron. (b) Would your answer be the same for helium? If not, state the quantum number(s) on which the energy of an orbital in a helium atom depends. [1 mark] n and l Helium has more than one electron, so both n and l affect the energy of the orbitals. (e.g. Electrons in 2s and 2p orbitals have different energies in a multi-electron atom.) 13 Name: ________________________ Student Number: _____________________ 17. The Haber-Bosch process is arguably one of the most important industrial reactions. [3 marks] (a) Write a balanced chemical equation for the reaction involved in the Haber-Bosch process. Include all states of matter. [2 marks] (b) What property of nitrogen makes this reaction so challenging? Nitrogen is very unreactive due to the strong N≡N triple bond. [1 mark] [5 marks] 18. Consider the following complex: [Co(OH)6]4– (a) What is the co-ordination number of cobalt in this complex? [1 mark] 6 (b) What is the oxidation state of cobalt in this complex? [1 mark] – +2 six OH ligands; overall charge of -4 so Co must be +2 (c) How many d electrons does cobalt have in this complex? [1 mark] 2+ 7 7 Co electron configuration is [Ar] 3d (d) Would you expect this complex to absorb light with a longer or shorter wavelength than [Co(NH3)6]2+? Explain. [2 marks] longer wavelength (i.e. lower energy light) NH3 is a stronger field ligand than OH–. (see spectrochemical series on data sheet) As such, the energy difference between the two sets of 3d orbitals is larger in [Co(NH3)6]2+. Thus, [Co(NH3)6]2+ will absorb higher energy light and [Co(OH)6]4– will absorb lower energy light. Lower energy light has a longer wavelength. 19. (a) Draw both isomers of [Fe(H2O)3(OH)3]. Your pictures must clearly show the correct geometry at Fe. (b) Clearly label your answers to part (a) as the fac isomer and the mer isomer. [3 marks] [2 marks] [1 mark] 14 Name: ________________________ 20. (a) (b) Student Number: _____________________ The ground-state electron configuration of tantalum (Ta) is [Xe] 6s2 4f14 5d3. [7 marks] The most common oxidation states of tantalum are III, IV and V (i.e. +3, +4 and +5). Write the ground-state electron configuration for each of the following ions: [3 marks] 3+ 14 2 Ta [Xe] 4f 5d Ta4+ [Xe] 4f14 5d1 Ta5+ [Xe] 4f14 Which of these ions would you expect to yield coloured complexes? Explain briefly. Ta3+ and Ta4+ [4 marks] For us to observe colour, the metal in a complex must have a partially filled d subshell so that there are electrons in the lower-energy d orbitals and at least one vacancy in the higher-energy d-orbitals. This is necessary for visible light to be absorbed and excite one electron from a lower-energy to a higher-energy d orbital. When this absorption occurs, we observe the complex as a combination of the colours of light not absorbed by the complex. BONUS: In a special chelating solvent called diglyme, the following reaction can be performed Here, the subscript (sol) indicates a species dissolved in diglyme (the solvent). What is the oxidation state of the tantalum atom in the product complex? [1 mark] -1 that’s correct! a metal with a negative oxidation state! odd… What would the corresponding ground-state electron configuration be? [1 mark] 1 14 5 2 [Xe] 6s 4f 5d would probably be lower energy than [Xe] 6s 4f14 5d4 (similar to electron configuration of Cr: [Ar]4s1 3d5) 21. What was the most useful and/or interesting thing you learned in CHEM 1000? …AND THAT’S ALL FOR CHEM 1000. HAPPY HOLIDAYS! [1 mark] 15 DATA SHEET Fundamental Constants and Conversion Factors Atomic mass unit (u) 1.660 539 × 10-27 kg Avogadro's number 6.022 141 × 1023 mol–1 Bohr radius (a0) 5.291 772 × 10-11 m Electron charge (e) 1.602 177 × 10-19 C Electron mass 5.485 799 × 10-4 u Ideal gas constant (R) 8.314 462 J·mol-1·K-1 8.314 462 m3·Pa·mol-1·K-1 Kelvin temperature scale Planck's constant Proton mass Neutron mass Rydberg Constant (RH) Speed of light in vacuum Standard atmospheric pressure 0 K = -273.15 ºC 6.626 070 × 10-34 J·Hz-1 1.007 277 u 1.008 665 u 2.179 872 x 10-18 J 2.997 925 x 108 m·s-1 1 bar = 100 kPa Formulae or for aqua complexes for oxoacids OpE(OH)q Spectrochemical Series strong field weak field 4CN > ethylenediamine > NH3 > EDTA > H2O > oxalato > OH > F > Cl > Br > I Band of Stability Graph The graph at the right shows the band of stability. Stable isotopes are in black. Isotopes that exist but are not stable are shown in varying shades of gray with the shades of gray corresponding to different half-lives. 16 DATA SHEET 1 Chemistry 1000 Standard Periodic Table 18 1.0079 4.0026 H He 2 13 14 15 16 17 6.941 9.0122 10.811 12.011 14.0067 15.9994 18.9984 Li Be B C N O F Ne 3 22.9898 4 24.3050 5 26.9815 6 28.0855 7 30.9738 8 32.066 9 35.4527 10 39.948 1 2 20.1797 Na Mg 11 39.0983 12 40.078 3 4 5 6 7 8 9 10 11 12 44.9559 47.88 50.9415 51.9961 54.9380 55.847 58.9332 58.693 63.546 65.39 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 19 85.4678 20 87.62 21 88.9059 22 91.224 23 92.9064 24 95.94 26 101.07 27 102.906 28 106.42 29 107.868 30 112.411 31 114.82 32 118.710 33 121.757 34 127.60 35 126.905 36 131.29 Rb Sr 37 132.905 38 137.327 Cs Ba 55 (223) 56 226.025 Fr 87 Ra Y 39 La-Lu Ac-Lr 88 P S Cl Ar 15 74.9216 16 78.96 17 79.904 18 83.80 Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 41 180.948 42 183.85 43 186.207 44 190.2 45 192.22 46 195.08 47 196.967 48 200.59 49 204.383 50 207.19 51 208.980 52 (210) 53 (210) 54 (222) Hf Ta W Re Os Ir Pt Au 72 (261) 73 (262) 74 (263) 75 (262) 76 (265) 77 (266) 78 (281) 79 (283) Rf Db Sg 105 106 138.906 140.115 140.908 144.24 La Ce Pr Nd 57 227.028 58 232.038 59 231.036 60 238.029 Ac Si 14 72.61 40 178.49 104 89 25 (98) Al 13 69.723 Th 90 Pa 91 U 92 Bh 107 Hs Mt Dt Hg Tl Pb Bi Po At 80 81 82 83 84 85 174.967 Rg 108 109 110 111 (145) 150.36 151.965 157.25 158.925 162.50 164.930 167.26 168.934 173.04 Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 61 237.048 62 (240) 63 (243) 64 (247) 65 (247) 66 (251) 67 (252) 68 (257) 69 (258) 70 (259) 71 (260) Np 93 Pu 94 Am 95 Cm 96 Rn 86 Bk 97 Cf 98 Es 99 Fm 100 Md 101 No 102 Lr 103 Developed by Prof. R. T. Boeré