Chem 1000A Final Examination - Solutions

Transcription

1 Chem 1000A Final Eamination - Solutions April 15 th, 2003: 9:00 to am Your name Instructor: Dr. M. Gerken Student ID Time: 3h No. of pages: 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 Number of neutrons Number of protons 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

2 (b) M(NaOH) = g mol -1 ; M(HCl) = g mol -1 n(naoh) = g/( g mol -1 ) = mol n(hcl) = g/( g mol -1 ) = mol The reaction has a 1:1 stoichiometry, therefore, NaOH is the limiting reagent. NaCl produced: n(nacl) = mol [NaCl] = mol/0.100 L = 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

3 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 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 S +IV O -II 3 2- I S +VI O -II 4 2- oidation half-reaction (unbalanced): S +IV O -II 3 2- S +VI O -II e - reduction half-reaction (unbalanced): 2I +V O -II e - I 0 2 electron balance: oidation half-reaction (unbalanced): 5S +IV O -II 3 2-5S +VI O -II e - reduction half-reaction (unbalanced): 2I +V O -II e - I 0 2 material balance: oidation half-reaction (balanced): 5H 2 O + 5S +IV O -II 3 2-5S +VI O -II e H + reduction half-reaction (balanced): 12H + + 2I +V O -II e - I H 2 O 3

4 overall reaction: 2H IO 3 + 5SO I 2 + 5SO 4 + H 2 O material balance: 2H, 2I, 5S, 21O 2I, 5S, 2H, 21O charge balance: (2+) + (2-) +(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 e - 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 e - reduction half-reaction (unbalanced): 2H +I Cl +V O -II e - 2Cl +IV O -II 2 material balance: oidation half-reaction (balanced): 2Cl -I- Cl e - reduction half-reaction (balanced): 2H + + 2H +I Cl +V O -II 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 ) = g mol -1 n(clo 2 ) = 05 g/( g mol -1 ) = mol n(cl 2 ) = mol ClO 2 (1 Cl 2 produced/2 ClO 2 consumed) = mol Cl 2 n(o 2 ) = mol ClO 2 (1 O 2 produced/1 ClO 2 consumed) = mol O 2 pv = nrt; T = ( ) K = 323 K p(o 2 ) = (nrt)/v = mol J K -1 mol K/ m 3 = J m -3 = kg m 2 s -2 m -3 = kg m -1 s -2 = Pa (three sign. Fig.) = Pa 1 atm/ Pa = 1.18 atm p(cl 2 ) = ½ Pa = Pa = atm total pressure = Pa Pa = Pa = 1.18 atm 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

5 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: g or g (b) mass of 1 H 2 O molecule: amu or 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

6 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 = ms -1 / ( m) = s -1 Frequencies of Hz and larger will not reach the earth s surface. (b) E=hν = Js s -1 = J Photons with energies of 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 ) = m Js ms -1 (-0.75) = J This energy is larger than the energy calculated for 290 nm ( J). Since energy larger than 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 Li 3 Na K Rb Cs Fr 87 Be Mg Ca 20 Sr 38 Ba 56 Ra Sc Y La Ac Ti Zr Hf 72 V 23 Nb 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 B 5 Al 13 Ga 31 In 49 Tl C 6 Si 14 Ge 32 Sn 50 Pb N 7 P 15 As 33 Sb 51 Bi O S Se 34 Te 52 Po F Cl Br I 53 At 85 Ne 10 Ar 18 Kr 36 Xe 54 Rn 86 6

7 Prefies Pico, p ; 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 J s Rydberg Constant m -1 Avogadro's number mol -1 Proton mass g Elementary charge, e C Neutron mass g Electron mass g Speed of light in vacuum, C m s -1 Gas constant, R J K -1 mol -1 = 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 = C; 0 C = K Pressure: 1 atm = 760 Torr = 760 mmhg = 1325 bar = Pa; 1 bar = 10 5 Pa Volume: 1 ml = 1 cm 3 ; 1 L = 1000 cm 3 = 1 dm 3 = m H i λ = c ν ( mv) > 1 Chem 1000 Standard Periodic Table 18 hydrogen Li lithium Na sodium K potassium Rb rubidium Cs cesium (223) 87Fr francium Be berrylium Mg magnesium Ca calcium Sr strontium Ba barium Ra radium Sc scandium Y yttrium La-Lu Ac-Lr La lanthanum Ac actinium Ti titanium Zr zirconium Hf hafnnium (261) 104Rf rutherfordium Ce cerium Th thorium V vanadium Nb niobium Ta tantalum (262) 105Db dubnium Pr Cr chromium Mo molybdenum W tungsten (263) 106Sg seaborgium Mn manganese (98) 43Tc technetium Re rhenium (262) 107Bh bohrium praesodymium neodymium promethium Pa protactinium Nd U uranium (145) 61Pm Np neptunium Fe iron Ru ruthenium Os osmium (265) 108Hs hassium Sm samarium (240) 94Pu plutonium Co cobalt Rh rhodium Ir iridium (266) 109Mt meitnerium Eu europium (243) 95Am americium Ni nickel Pd palladium Pt platinum Gd gadolinium (247) 96Cm curium Cu copper Ag silver Au gold Tb terbium (247) 97Bk berkelium Zn zinc Cd cadmium Hg mercury Dy dysprosium (251) 98Cf californium B boron Al aluminum Ga gallium In indium Tl thallium Ho holmium (252) 99Es einsteinium C carbon Si silicon Ge germanium Sn tin Pb lead Er erbium (257) 100Fm fermium N nitrogen P phosphorus As arsenic Sb antimony Bi bismuth Tm thulium (258) 101Md mendelevium h 4π O oygen S sulfur Se selenium Te tellurium (210) 84Po polonium Yb ytterbium (259) 102No nobelium F fluorine Cl chlorine Br bromine I iodine (210) 85At astatine Lu lutetium (260) 103Lr lawrencium He helium Ne neon Ar argon Kr krypton Xe enon (222) 88Rn radon 7