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H2AF09FinalKey - Name AQXIMX page 1 ’1 Chem H2A — Final...

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Unformatted text preview: Name AQXIMX page 1 ’1 Chem H2A — Final Exam F\\(\M\ K143 P 12/07/09 -- R. Corn You have two hours to complete the exam, but Q look over all the problems before diving in. Please show all of your work; no credit can be given just for the answer. Some information that you might find useful: Avogadro Constant = NA = 6.022 x 1023 molecules Mass ofan electron = me = 9.109 x 10'31 kg Charge of an electron = e = 1.602 x 10'19 C Speed of light = c = 2.998 x 108 m s-1 Planck's Constant = h = 6.626 x 10-34 J 5 Electron Volt = 1 eV = 1.602 x 10-19 J Atomic Mass Unit = amu = 1.661 x 10'27 kg Gas Constant = R = 8.314 m3 Pa K‘1 mol'1 1 _ (25 points) 7 _ (25 points) 2 _ (25 points) 8 _ (25 points) 3 _ (25 points) 9 _ (25 points) 3 4 _ (25 points) 10 __ (25 points) 1 5 _ (25 points) . 11 _ (25 points) 6 _ (25 points) 12 _ (25 points) Total (3 00 points) Name page 2 1) PIAB (25 points) Consider an electron in a "box" whose energy levels are given by the equation: nzh2 " = 8meL2 where G is a constant equal to 0.6620 eV. E =nzG n= 1,2, 3... a) When placed in a particularly shaped magnetic field, the selection rules for the absorption of photons becomes An = 2, 4 or 6. Please calculate the wavelengths (in mm) that light would be absorbed for an electron in the n=1 ground state, and draw the absorption spectrum you would observe. E9)” 2 (”I") (3 C (66’ ._ 5,26% 6v :7 13H.\ mm 56%! : QS")C' = M5 ‘ ‘55“ ‘V ’7 78.09 nm E : l 4—!) ; ‘WG’ ’— . amp“ 7..“ ‘1 Cr 3| 776 eV $7 30"02 “M \n , we 123,0. 3 7%? E (EV): - 11“,") 3°1 73 73‘! ”km b) We now vary the shape of the magnetic field so that the selection rules change to An = 1, 3 or 5. Please calculate how the absorption spectrum would change. E1,” '2. Cirl)C., : [,0136 e\/ : 52”.“! “M "CW '— Qé-UG ’e M70 ev : 12% nm E42” 2 (76,00 : 23,1?ev =‘ 93-5? mm 53 [1'1 62* Name page 3 2) Atomic Structure and Term Symbols(25 points) a) Write down the electron configurations for the ground state and first excited state of the Helium atom. \ y \31 \S\'Zs‘ l i b) Write down all of the possible the term symbols for these two states. 5:0 $=LO L10 L: O \ ‘30 3s. 3 g 0) Originally, two types of Helium were discovered: ortho-helium and para-helium. Ortho- helium is diamagnetic, but para-helium is paramagnetic. It turned out that there is only one element of Helium (thank goodness!), but para-helium is just a He atom in a particular first excited state. Can you identify which term symbol(s) are para—helium? 33 g d) Due to selection rules, no optical absorption is observed from the ground state to the first excited state of He. However an absorption band IS observed at 58.44 nm to the next highest excited state. What is the electron configuration and Term Symbol(s) of that state(s)? \3‘ 2f $:I,O L=l Name page 4 3) The Photoelectric Effect(25 points) Consider a photoelectric effect experiment where you are shining light of difi'erent wavelengths on a metal surface. a) No electrons are emitted from the surface of the metal until the wavelength is less than 283.7 nm. What is the work function (in J) of this metal surface? \b ME = ' hV’ (1 fl ‘ WV 1 lxc/x ’5 m ’3 - Qw : 7,6071“), 3’ 6.62640 “\Ts 7.379.157” b) For light with a wavelength of 254.3 nm, what velocity (in m/s) do you expect to observe for \b the photoej ected electrons? l/z Me V1: k‘VR-r/e M. v » WfigQ : ‘ 7267191154041? 7-00?"°"75 ) M‘ J 91l07'/0.5’IC5 : \ mmo’m’ _ WW1 1.1070.”th . 31 ,_ qg/S‘ix c) What is the wavelength (in pm) of these ejected electrons? : L1 2! 6 O O M/S ’7‘1 g l= A , 6.62m 3'1 a ? 3.10‘1-l0-3ll()rLl21518 ~ L729“) m = I725 pm Name page 5 4 ) IR and Raman scattering(25 points) Consider the vibrational spectrum of a water molecule isotope with a chemical formula HOD. a) Write down the structure and geometry of this molecule. How many normal mode vibrations does this molecule have? How many are IR—active? g " ‘~ 3 M'b '3 3 magi!) / D ‘ H cm “3‘2 “‘4‘” b) You obtain a Raman spectrum from this molecule using as an excitation source an Argon ion laser with a wavelength of 488.0 nm. You observe three Raman bands at three wavelengths: 523.9 nm ("band A"), 562.9 nm ("band B") and 595.8 nm ("band C"). The 3 bands correspond to the 3 normal mode vibrations. Are these Stokes or Anti-Stokes Raman bands? Please calculate the frequencies of these normal mode vibrations (in cm'l). £19940"... ' mgr: #8846ch .— mm W" \O 5239\“0'7‘W = 9002;? CW" 2:)? C .. 3‘“ "0:7” ‘ 11765 cm" A: Noam" hm} 599-8 - 'o 7m -— (675‘4 W, l5 ‘ 272505 quwdvii, C '— 370?CW‘" Mkjmwlvng c) The normal modes associated with these three Raman bands are an OH stretch, an OD stretch, S and an H—O-D bendpldentify whichgqand (A, B, C) is associated with each normal mode. 2 n BflOD c=> OM g d) If we assume that the OH and OD stretching modes can be modeled by a simple harmonic oscillator with a mass of 1.00 and 2.00 amu respectively, what is the ratio of the two vibration frequencies (Von/V013)? What is the ratio that we observe? T NH ,cQ Cl q/c 5.91 'L 011W) M :K Name page 6 5) Lewis Dot Structures and VSEPR Geometries (25 points) v/For’ffie following molecules, please: i) draw the Lewis Structure; ii) write the VSEPR formula (AXnEm) and name the molecular geometry; iii) draw the molecule with approximate bond angles and the direction of the molecular dipole moment (if there is one) . {SF 13'. S a) cs32- 7%; .i \ H+3~6+L v21 /c\ .. <=> _. F \‘g (:7 ,CK .. ’.'q' 5’. ‘3 «, ‘. / 3: AX3 [120° “low Flow” / V" A7061 3 b) HOCl ’+ ;,q o' " . 6+7 H / O_\gl‘ 4/; «A AXzEz SlicjlnHj less ’l’lAuvx '6750 82an g c) I3+ ' M ."’ ,r +/ t 3'7’l‘20 Ell—‘1"; 49—7 1«::.':,+ (i=2 :l:=‘E'—E , + . . O A1162, F /I\ ) Sllclx‘H‘X (ASS HAOLV‘ I iii/x «‘4 BQVW' l I I: /j\/ +3 —1 s d>AsF3 F .. -." =15“: w 5*3.7: 26 u ‘ > A9“ (AA: gr '; L‘ r .a ’— ‘ 02.5 ”0 :‘lf’- (A? '5': F/ A I- co:\-\\o:v\-\¢ muck) A ’ Pi, F ,w/ le:l O 4/ a use that (0% HlfV‘MA ijflW‘Jw g e) C1F3~ 7+37 :28] 2F" . I F aq— ‘F' [AX’bEz 700 d ’F' shaft” 190 Name page 7 6) X-ray Diffraction (25 points) a) X—ray photons from the Copper Koc line have an energy of 8,027 eV. Calculate the wavelength of these x-ray photons (in picometers (pm)). 8027ev: 1.60240”? .302? e 1.18640”)- ‘0 eV - hi _ 462646335.affideQM/s _ “3.10m 75' E ‘ W ‘ " V” < lSH/5 ”PM m [1 O - 8027: ail—m) ~ O.l‘5"!comm: [swsrm \ S b) In a crystal of gold, planes of gold atoms are separated by a distance of 408.0 pm. Compute the angles for all possible Bragg reflections for these planes if the X-ray photons from the Copper KOL line are used. fl:\ @= [0.0” o “:2 Q; 2225 0 \(\:’3 Q: Dbl.“ «\M 9—. $1.23" “-5 : 711' Name page 8 7) Atomic Unit Cell Calculations (25 points) Niobium is a soft, ductile metal with a bee crystal lattice. The density of Niobium is 8.57 g/cm3. a) How many Niobium atoms are there in the unit cell of the metal? 5 7. \ O b) Calculate the bcc lattice parameter (in pm) of Niobium metal. The Mol. Wt. of Nb is 92.91 g/mol. M «$5 2 - 012m :lmol )0: W = 6.0224023 3 . 6 1‘ 3 10°65 . '2? (A ,: 710 3H 360, [0,21 m3 3 : . . 9,57 34"} 4053.3 '00 (2 m 0t“— 3.30240 m: 3‘30- ? \ O c) Calculate the Nb-Nb distance (in pm) in Niobium metal. h 7;“ l::. 'lal'eei’onk i ,K h: g“ z E Q A l/M J. a a 7.. 7' _L 7, l 'L :5. 09 = z °~ * q a = ‘4 o“ o)” 31, _ E2 33020.44: 286.0?1’“ ’ Ea " 1 l Name page 9 8) Ideal Gases (25 points) A gas mixture being used to simulate the atmosphere of Titan (a moon of Saturn) consists of 500.0 mg. ofnitrogen (MW = 28.01), 42.33 mg ofmethane (MW = 16.04) and 2.250 mg of argon (MW = 39.95). The partial pressUre of Argon in the mixture is 225.6 Pascals at a temperature of 425K. a) Calculate the mole fractions of the three components. mobs PX“ S N2 OISOOSAslo‘i/mo‘ = 0.017% 0357’“ CH4 comm/IbMS/mm = 0.00263? 01qu Ar 01002293wg/m‘ : 5.632469 0,0027% 0.020915 4 b) Calculate the total pressure (in Pa) and volume (in m3) of the sample. \O V’ nRT, 5.63240 vacuums (PM: (Xar' PM. 7’ 22576 c) Calculate the partial pressures (in Pa) of nitrogen and methane in the sample. 5 l7 N1: 0,8688l' 32336 p“ : 7/53‘1 {70k T704“; 04'23‘15’81‘63651 = (0876 P“ c) Calculate the root mean square velocity (in m/s) of the methane molecules in this sample. 5 V = ( (SB—T)”: (3’33” 7rmv 4951c ) 72 my M 00160! 63/4”, \IHM : (650YI70)‘/2 : 813.0 M/S Name page 10 9) MO Theog, Part 1 (25 points) ' Draw the valence MO energy level diagram for the diatomic molecule LiH. (IP of Li is 5.39 eV; IP of H is 13.61 eV). Name the orbitals and fill them with the correct number of electrons. Please write down: a) The bond order of this molecule, b) The number of electrons in the HOMO, c) Whether the molecule is diamagnetic or paramagnetic, d) An estimate of the IP. e) The direction of the dipole moment in this molecule. ” Q \s \ ii 00 yaw» 'V 5 new / M 1 N C) §»iovwajml’l( APB \hoh \aowbm) Q) inwards H Name page 11 10) MO Theory, Part 2 (25 points) Draw the valence MO energy level diagram for the diatomic molecule HF. (IP of F 17.42 eV; IP of H is 13.61 eV). Name the orbitals and fill them with the correct number of electrons. Please write down: a) The bond order of this molecule, b) The number of electrons in the HOMO, c) Whether the molecule is diamagnetic or paramagnetic, d) An estimate of the IP. e) The direction of the dipole moment in this molecule. ”60/ Name page 12 11) Isomers and Hydrogen Bonding(25 points) The molecule nitrophenol (also called hydroxynitrobenzene) has the chemical formula C6H5NO3. As the name suggests, this molecule contains a benzene ring, a hydroxyl group, and a nitro group. This molecule has three isomers. a) Please draw the structures of the three isomers of this molecule. % ® :9 *3 “\70 O l O {,9 0 \g 04"?) b) Two of the isomers exhibit an OH stretching band in their gas phase IR spectrum at a frequency of 3650 cm‘l. The third isomer, called ortho-nitrophenol, has an OH stretching band at 3220 cm'l. The IR spectral evidence shows that there is internal (more properly called INTRAMOLECULAR) hydrogen bonding occurring in one of the isomers. Please draw the structure of this isomer, including the hydrogen bond. Also identify the hydrogen bond donor and acceptors in this molecule. 000 M-..0e/““R[’°' W’N 0/ ’ I I N\ \ O 0 l / c) Ortho-nitrophenol may also form a dimer Via hydrogen bonds. Please draw this dimer % including the hydrogen bonds. [)0 Name page 13 12) Compound Unit Cell Calculations (25 points) Consider the two compounds lead sulfide and gallium phosphide. a) Consider the compound lead sulfide (galena). In the unit cell of a crystal of this compound, the sulfur atoms sit on an fcc lattice, and the lead atoms sit in the octahedral holes in the fcc 5 lattice. What is the chemical formula for lead sulfide, and how many atoms of lead and sulfur are in the unit cell? L{ (Al‘OWI 3 3 "l (moms m H33 b) The unit cell lattice parameter for lead sulfide is 594.6 pm. What is the Pb—S distance (in pm) 8 in this crystal? 5 c) Consider the compound gallium phosphide. In the unit cell of a crystal of this compound, the phosphorus atoms sit on an fcc lattice, and the gallium atoms sit in one half of the tetrahedral holes in the fcc lattice. What is the chemical formula for gallium phosphide, and how many atoms of gallium and phosphorus are in the unit cell? H atoms ? Cap) H atom 5 dc ( O d) The unit cell lattice parameter for gallium phosphide is 545.1 pm. What is the Ga-P distance (in pm) in this crystal? . (ll ’L '7. _Z_ ’L 1 :6 0‘ + (6 E A? A: act : V1615) C& 2 7 3 6 O (9 m k: T22 ‘7‘ k '— £10 page 14 e m N ...
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