Chem 1000A Final Eamination - Solutions April 15 th, 2003: 9:00 to 12.00 am Your name Instructor: Dr. M. Gerken Student ID Time: 3h No. of pages: 3 + 1 Report all your answers using significant figures. Show all units and their conversions throughout your calculations. Question 1 (5 Marks) Complete the following table. Symbol 140 Ce 120 Sn 2+ Number of electrons 58 48 Number of neutrons 82 70 Number of protons 58 50 Overall charge 0 2+ Question 2 (2 Marks) What is Ψ in Schrödinger s equation? In Schrödinger s equation, Ψ is the symbol for a wavefunction. Question 3 (5 Marks) Draw the five 3d orbitals and label them. y z y z z y d y d z d yz d 2-y2 d z2 Question 4 (4 Marks) What is the maimum number of electrons that can be identified with each of the following sets of quantum numbers? (zero is a possible answer) (a) n = 2, l = 2, m l = -1, m s = ½ zero, l has to be smaller than 2 if n = 2 (b) n = 3, l = 2 10 electrons (in the five 3d orbitals) (c) n = 2 8 electrons (2s and three 2p orbitals) (d) n = 4, l = 3, m l = -½ zero, m l has to be an integer number Question 5 (4 Marks) You dissolve 10.0 mg of NaOH and 10.0 mg of HCl in 100. ml of water. (a) Write the balanced reaction equation. (b) What is the NaCl concentration (neglecting the change in volume during the dissolution and reaction)? (a) NaOH + HCl NaCl + H 2 O 1
(b) M(NaOH) = 39.9971 g mol -1 ; M(HCl) = 36.4606 g mol -1 n(naoh) = 0.0100 g/(39.9971 g mol -1 ) = 2.50 10-4 mol n(hcl) = 0.0100 g/(36.4606 g mol -1 ) = 2.74 10-4 mol The reaction has a 1:1 stoichiometry, therefore, NaOH is the limiting reagent. NaCl produced: n(nacl) = 2.50 10-4 mol [NaCl] = 2.50 10-4 mol/0.100 L = 2.50 10-3 mol L -1 Question 6 (15 Marks) (i) Draw Lewis structures for the following compounds. (Central atom is underlined) (ii) What are the electron-pair geometries and molecular geometries according to the VSEPR model. (iii) Do these molecules have molecular dipole moments. Indicate the dipole moment if applicable. - a) XeF 3 \ - - :F Xe F: :F Xe F: :F: :F: electron-pair geometry: octahedral molecular geometry: T-shaped dipole moment: yes b) SCl 2. Ṡ :Cl Cl :. Ṡ :Cl Cl : electron-pair geometry: tetrahedral molecular geometry: bent dipole moment: yes c) ClO 2 F _ : O : : O : Cl: Cl: : O : O : F: : F: electron-pair geometry: tetrahedral molecular geometry: T-shaped dipole moment: yes d) XeO 2 F 4 F: : F Xe F: : F electron-pair geometry: octahedral molecular geometry: octahedral dipole moment: no _ 2
e) XeO 6 4- O Xe O - : - :. O : : O - - + several resonance structures electron-pair geometry: octahedral molecular geometry: octahedral dipole moment: no Question 7 (5 Marks) For etinguishing fires in rooms with epensive electrical equipment a halon has been used (Halon 1301: CF 3 Br). Unfortunately, this halon is depleting the ozone layer. As a replacement, a new fire-etinguishing agent has been introduced last year. This new fire-etinguishing agent contains 22.8 % C, 7 % F, and 5.1 % O. (a) Determine the empirical formula for this compound. (b) The molar mass of this agent has been found to be 316.046 g mol -1. Determine the molecular formula. In 100 g of sample: 22.8 g C 0 mol 7 g F 3.79 mol 5.1 g O 0.32 mol C : F : O = 0 mol : 3.79 mol : 0.32 mol = 5.96 mol : 1 mol : 0 mol Empirical formula: C 6 F 12 O M(C 6 F 12 O) = 3046 g mol -1 This calculated molar mass is the same than the eperimentally determined one. Molecular mass: C 6 F 12 O Question 8 (8 Marks) Draw the Lewis structures and determine the oidation states for all atoms in the following compounds. a) H 3 C-O-F (connectivity at indicated) H: +I, C: -II, O: 0, F: -I b) IO 2 F 2-5 (central atom:i) I: +VII, O: -II, F: -I c) NF + 4 (central atom: N) N: +V, F: -I d) O 3 S-S-S-SO 2-3 (connectivity as indicated) O -II 3S +V -S 0 -S 0 -S +V O -II 3 2- Question 9 (12 Marks) Balance the following redo reactions in acidic aqueous solutions (add H + if necessary). First, write the balanced half-reactions and combine them. Indicate the (a) electron balance, (b) material balance, and (c) charge balance of the overall reaction equations. a) I +V O -II 3 - + S +IV O -II 3 2- I 0 2 + S +VI O -II 4 2- oidation half-reaction (unbalanced): S +IV O -II 3 2- S +VI O -II 4 2- + 2e - reduction half-reaction (unbalanced): 2I +V O -II 3 - + 10e - I 0 2 electron balance: oidation half-reaction (unbalanced): 5S +IV O -II 3 2-5S +VI O -II 4 2- + 10e - reduction half-reaction (unbalanced): 2I +V O -II 3 - + 10e - I 0 2 material balance: oidation half-reaction (balanced): 5H 2 O + 5S +IV O -II 3 2-5S +VI O -II 4 2- + 10e - + 10 H + reduction half-reaction (balanced): 12H + + 2I +V O -II 3 - + 10e - I 0 2 + 6 H 2 O 3
overall reaction: 2H + - + 2IO 3 + 5SO 2-2- 3 I 2 + 5SO 4 + H 2 O material balance: 2H, 2I, 5S, 21O 2I, 5S, 2H, 21O charge balance: (2+) + (2-) +(10-) = 10-10- correct electron, material, and charge balance! b) H +I Cl +V O -II 3 + Cl -I- Cl +IV O -II 2 + Cl 0 2 oidation half-reaction (unbalanced): 2Cl -I- Cl 0 2+ 2e - reduction half-reaction (unbalanced): H +I Cl +V O -II 3 + e - Cl +IV O -II 2 electron balance: oidation half-reaction (unbalanced): 2Cl -I- Cl 0 2+ 2e - reduction half-reaction (unbalanced): 2H +I Cl +V O -II 3 + 2 e - 2Cl +IV O -II 2 material balance: oidation half-reaction (balanced): 2Cl -I- Cl 0 2 + 2e - reduction half-reaction (balanced): 2H + + 2H +I Cl +V O -II 3 + 2 e - 2Cl +IV O -II 2+ 2H 2 O overall reaction: 2H + + 2HClO 3 + 2Cl - 2ClO 2 + Cl 2 + 2H 2 O material balance: 4H, 4Cl, 6O 4Cl, 4H, 6O charge balance: (2+) + (2-) = 0 0 correct electron, material, and charge balance! Question 10 (6 Marks) Chlorine dioide, ClO 2, (melting point of 11 C) is a reddish-yellow, highly toic gas and has a pungent odor. On gentle heating above 45 C it decomposes by eploding violently into oygen and chlorine. (a) Write the balanced reaction equation for the decomposition. (b) At STP you have 05 g of chlorine dioide in a 500 ml cylinder. Heating the cylinder to 50 C results in the decomposition (nicer word for eplosion) of the chlorine dioide sample. Assuming the cylinder survives the eplosion (which may not be the case in real life so don t try it at home), calculate the partial pressures (in Pa and atm) for oygen and chlorine and the total resulting pressure (in Pa and atm) in the cylinder. (a) 2ClO 2 Cl 2 + 2O 2 (b) M(ClO 2 ) = 67.4515 g mol -1 n(clo 2 ) = 05 g/(67.4515 g mol -1 ) =0.0223 mol n(cl 2 ) = 0.0223 mol ClO 2 (1 Cl 2 produced/2 ClO 2 consumed) = 0.0112 mol Cl 2 n(o 2 ) = 0.0223 mol ClO 2 (1 O 2 produced/1 ClO 2 consumed) = 0.0223 mol O 2 pv = nrt; T = (273.15 + 50) K = 323 K p(o 2 ) = (nrt)/v = 0.0223 mol 8.314 J K -1 mol -1 323 K/0.0005 m 3 = 120000 J m -3 = 120000 kg m 2 s -2 m -3 = 120000 kg m -1 s -2 = 120000 Pa (three sign. Fig.) = 120000 Pa 1 atm/ 101325 Pa = 1.18 atm p(cl 2 ) = ½ 120000 Pa = 60000 Pa = 0.591 atm total pressure = 120000 Pa + 60000 Pa = 180000 Pa = 1.18 atm + 0.591 atm = 7 atm Question 11 (5 Marks) Draw the Lewis structure of chlorine dioide, ClO 2. What class of molecules does ClO 2 belong to? Predict its molecular geometry.. Cl : O O : ClO 2 is a radical, since it has an unpaired electron pair. It has a bent geometry. 4
Question 12 (6 Marks) Write the electron configuration of Europium (Eu) in the orbital bo notation. Eu: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f Question 13 (5 Marks) Write the electron configurations of Cu and Cu 2+ using the noble-gas notation. Cu: [Ar]4s 1 3d 10 Cu 2+ : [Ar]4s 0 3d 9 Both, Cu and Cu 2+ are paramagnetic. Question 14 (4 Marks) (a) What is the mass of 1 mole of H 2 O (in g and in amu)? (b) What is the mass of 1 molecule of H 2 O (in g and amu)? (a) mass of 1 mole H 2 O: 18.0152 g or 85 10 25 g (b) mass of 1 H 2 O molecule: 18.0152 amu or 2.992 10-23 g Question 15 (7 Marks) Describe the bonding situation in protonated cyanic acid (H-C N-H + ) using the valence bond theory. (a) What is the hybridization of the carbon and nitrogen atoms? (b) Draw two energy diagram indicating the formation of the hybrid orbitals on nitrogen starting from the atomic orbitals on N, going to the hybrid orbitals on N. (c) Describe which orbitals are involved in forming bonds between C and H, C and N, and N and H. (a) C: sp hybridization, N: sp hybridization (b) Nitrogen: E 2s 2p p p y sp (c) C-H bond: overlap between the sp hybrid orbital on C and the 1s orbital on H C N bonds: overlap between the sp hybrid orbital on C and the sp hybrid orbital on N, forming the σ bond Overlap between two p orbitals on C with two p orbitals on N, forming the two π bonds N-H: overlap between the sp hybrid orbital on N and the 1s orbital on H 5
Question 16 (4 Marks) Ultraviolet radiation below 290 nm (the ozone cut-off) will be absorbed by the ozone layer and will not reach the earth s surface. (a) What range of frequencies will not reach the earth s surface? (b) Photons of what energy range will be absorbed by the ozone layer? (a) ν = c/λ = 2.998 10 8 ms -1 / (2.90 10-9 m) = 3 10 15 s -1 Frequencies of 3 10 15 Hz and larger will not reach the earth s surface. (b) E=hν = 6.626 10-34 Js 3 10 15 s -1 = 6.85 10-19 J Photons with energies of 6.85 10-19 J and larger will not reach the earth s surface. Question 17 (3 Marks) Assuming an intact ozone layer, can sun light/radiation ecite an electron of a hydrogen atom from the L shell (n = 1) to the M shell (n = 2)? E = R Ryd h c (1/2 2 1/1 2 ) = 97 10 7 m -1 6.626 10-34 Js 2.998 10 8 ms -1 (-0.75) = -34 10-18 J This energy is larger than the energy calculated for 290 nm (6.85 10-19 J). Since energy larger than 6.85 10-19 J is cut off, photons on earth s surface do not have sufficient energy for the ecitation of an electron in a hydrogen atom from the L to the M shell. 1 Electronegativities 18 H He 1 2 13 14 15 16 17 2 Li 3 Na 11 0.9 K 19 0.9 Rb 37 0.8 Cs 55 0.8 Fr 87 Be 4 1.2 Mg 12 3 4 5 6 7 8 9 10 11 12 Ca 20 Sr 38 Ba 56 Ra 88 1.3 Sc 21 1.2 Y 39 1.1 La 57 1.1 Ac 89 1.4 Ti 22 1.3 Zr 40 1.3 Hf 72 V 23 Nb 41 1.4 Ta 73 Cr 24 Mo 42 W 74 Mn 25 Tc 43 Re 75 Fe 26 Ru 44 Os 76 Co 27 Rh 45 Ir 77 Ni 28 Pd 46 Pt 78 Cu 29 Ag 47 Au 79 Zn 30 Cd 48 Hg 80 2.0 B 5 Al 13 Ga 31 In 49 Tl 81 2.5 C 6 Si 14 Ge 32 Sn 50 Pb 82 3.0 N 7 P 15 As 33 Sb 51 Bi 83 3.5 O 8 2.5 S 16 2.4 Se 34 Te 52 Po 84 4.0 F 9 3.0 Cl 17 2.8 Br 35 2.5 I 53 At 85 Ne 10 Ar 18 Kr 36 Xe 54 Rn 86 6
Prefies Pico, p 10-12 ; nano, n 10-9 ; micro, µ 10-6 ; milli, m 10-3 ; centi, c 10-2 ; deci, d 10-1 Fundamental Constants Planck's constant, h 6.626 10-34 J s Rydberg Constant 97 10 7 m -1 Avogadro's number 6.022 10 23 mol -1 Proton mass 7252 10-24 g Elementary charge, e 022 10-19 C Neutron mass 749 10-24 g Electron mass 9.1095 10-28 g Speed of light in vacuum, C 2.998 10 8 m s -1 Gas constant, R 8.314 J K -1 mol -1 = 0.082057 L atm K -1 mol -1 n = m M E = mc m hc ρ = pv = nrt PT = pi pi = Xi PT E = hν = V λ 1 1 E = E E = R hc λ = h mv 2 2 ( ) final initial Ryd 2 n final ninitial Physical quantity Unit Symbol Definition Frequency, f or ν hertz Hz s -1 Energy, W or E joule J kg m 2 s -2 Force, F newton N J m -1 = kg m s -2 Pressure, p pascal Pa N m -2 = kg m -1 s -2 Temperature: 0 K = -273.15 C; 0 C = 273.15 K Pressure: 1 atm = 760 Torr = 760 mmhg = 1325 bar = 101325 Pa; 1 bar = 10 5 Pa Volume: 1 ml = 1 cm 3 ; 1 L = 1000 cm 3 = 1 dm 3 = 0.001 m 3 079 1H i λ = c ν ( mv) > 1 Chem 1000 Standard Periodic Table 18 hydrogen 2 13 14 15 16 17 6.941 3Li lithium 22.9898 11Na sodium 39.0983 19K potassium 85.4678 37Rb rubidium 132.905 55Cs cesium (223) 87Fr francium 9.0122 4Be berrylium 24.3050 12Mg magnesium 3 4 5 6 7 8 9 10 11 12 40.078 20Ca calcium 87.62 38Sr strontium 137.327 56Ba barium 226.025 88Ra radium 44.9559 21Sc scandium 88.9059 39Y yttrium La-Lu Ac-Lr 138.906 57La lanthanum 227.028 89Ac actinium9 47.88 22Ti titanium 91.224 40Zr zirconium 178.49 72Hf hafnnium (261) 104Rf rutherfordium 140.115 58Ce cerium 232.038 90Th thorium 50.9415 23V vanadium 92.9064 41Nb niobium 180.948 73Ta tantalum (262) 105Db dubnium 140.908 59Pr 5961 24Cr chromium 95.94 42Mo molybdenum 183.85 74W tungsten (263) 106Sg seaborgium 54.9380 25Mn manganese (98) 43Tc technetium 186.207 75Re rhenium (262) 107Bh bohrium praesodymium neodymium promethium 2336 91Pa protactinium 144.24 60Nd 238.029 92U uranium (145) 61Pm 237.048 93Np neptunium 55.847 26Fe iron 107 44Ru ruthenium 190.2 76Os osmium (265) 108Hs hassium 150.36 62Sm samarium (240) 94Pu plutonium 58.9332 27Co cobalt 102.906 45Rh rhodium 192.22 77Ir iridium (266) 109Mt meitnerium 1565 63Eu europium (243) 95Am americium 58.693 28Ni nickel 106.42 46Pd palladium 195.08 78Pt platinum 157.25 64Gd gadolinium (247) 96Cm curium 63.546 29Cu copper 107.868 47Ag silver 196.967 79Au gold 158.925 65Tb terbium (247) 97Bk berkelium 65.39 30Zn zinc 112.411 48Cd cadmium 200.59 80Hg mercury 162.50 66Dy dysprosium (251) 98Cf californium 10.811 5B boron 26.9815 13Al aluminum 69.723 31Ga gallium 114.82 49In indium 204.383 81Tl thallium 164.930 67Ho holmium (252) 99Es einsteinium 12.011 6C carbon 28.0855 14Si silicon 72.61 32Ge germanium 118.710 50Sn tin 207.19 82Pb lead 167.26 68Er erbium (257) 100Fm fermium 14.0067 7N nitrogen 30.9738 15P phosphorus 74.9216 33As arsenic 1257 51Sb antimony 208.980 83Bi bismuth 168.934 69Tm thulium (258) 101Md mendelevium h 4π 15.9994 8O oygen 32.066 16S sulfur 78.96 34Se selenium 127.60 52Te tellurium (210) 84Po polonium 173.04 70Yb ytterbium (259) 102No nobelium 18.9984 9F fluorine 35.4527 17Cl chlorine 79.904 35Br bromine 126.905 53I iodine (210) 85At astatine 174.967 71Lu lutetium (260) 103Lr lawrencium 4.0026 2He helium 20.1797 10Ne neon 39.948 18Ar argon 83.80 36Kr krypton 131.29 54Xe enon (222) 88Rn radon 7