ProblemSet6Solutions

ProblemSet6Solutions - Problem 5 1a ® In this problem we...

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Unformatted text preview: Problem 5% 1a ® In this problem, we consider the heteronuclear diatomic molecule CO. The ionization energies ' of an electron from the valence atomic orbitals on the carbon atom and the oxygen atom are listed below. Atom Valence orbital Ionization energy /MJ ~mol‘l 0 2s 3.116 2p 1.524 C 25 1.872 2p 1 .023 Use these data to construct a molecular-orbital energy-level diagram for CO. What are the symmetry designations of the molecular orbitals of CO? What is the electron configuration of the ground state of CO? What is the bond order of CO? Is CO paramagnetic or diamagnetic? O CO C orbitals orbitals orbitals I \ \ \ 2p 2:: 2p ,, ' —ooo-—L'--=:.-_ >1 ‘ ‘$’ 25 S I c If Lu 2: (I I! ——o—l. \~_O___,’ Since C and 0 have similar orbital energies, the bonding will be similar to that in a homonuclear diatomic molecule. In order of increasing energy, the symmetry designations of the molecular , orbitals of C0 are (ignoring the is orbitals) 025, 0‘25, 7Tpr and n2p5,, and finally asz. Note that we do not use the subscripts g and u because CO does not have an inversion center. The electron configuration of CO is (from Example 9—6) KK(0‘25)2(o‘*25)2(7T2px)2(7r2py)2(0‘2pz)2, so the bond order is 3 and CO is diamagnetic because there are no unpaired electrons. The following lines were observed in the microwave absorption spectrum of Hml and D1271 between 60 cm’I and 90 cm". 17/cm‘l I - HmI 64.275 77.130 89.985 D1271 65.070 71.577 78.084 84.591 Use the rigid—rotator approximation to determine the values of B, I , and Rg for each molecule. Do your results for the bond length agree with what you would expect based upon the Born- Oppenheimer approximation? Take the mass of 127I to be 126.904 amu and the mass of D to be 2.014 amu. In the rigid-rotator approximation, the spacing between the lines in a microwave absorption spectrum is 23 (Equation 13.19). The spacing between the lines given for H1271 is 12.855 cm‘I and the spacing between the lines given for DmI is 6.507 cm". Therefore, 9 B“...l = §(12.855 cm") = 6.428 cm" and BDml = CID—l) = CITI—l We now use Equation 13.9 to find I for both molecules: h - 8712c]? I 6.626 x 10-34 J-s 1HI = 2 10 —1 —1 871 (2.998x 10 cm-s )(6.428cm ) I _ 6.626 x 10‘34 J-s 1" ‘ 8212(2998 >< 10'0cm-s-‘)(3.254 em") = 4.355 x 10~47 kg-m2 = 8.604 x 10-“7 kg-mz Now we use the fact that I = ,uRZ for a diatomic molecule to find Re: I=MRZ I l/Z Rio 11 4.355 x 10-47 kg-mz "2 Real-ll : T“ 1.008>< 126.904 [27.912 amu) (1.661 x 10‘27 kg-amu"‘) = 1.619 ><10"O m =161.9pm 8.604 x 10-47 kg-mz "2 Reth = (—‘ 2.014x126.904 128.917 am”)(1-661 X 10—27 kg-amu“) = 1.617 x 10'10 m =161.7 pm These values differ by approximately 0.1%. In the Born-Oppenheimer approximation, the bond length is independent of the isotope of theatoms, in agreement with the above calculations. Calculate the H'Lickel n—electron energies of cyclobutadiene. What do Hund’s rules say about the ground state of cyclobutadiene? Compare the stability of cyclobutadiene with that of two isolated ethylene molecules. The structure of cyclobutadiene is Letting x = (a —- E)/fi, the H'Lickcl determinantal equation is given by xlOl‘ lxlO 01x1 #0 101x Expanding the determinant gives 1 x=0 10 x l l x 01 x xlx O 0 l I x l l l — 0 x — 0 l l l and expanding the above determinants gives x(x3——2x)——(x2—l—l—l)—(l—l—xl—l)=0 x4—4x2=0 x = 2, 0, 0, —2 Because x = (01 — E)/i3, the four n-electron energies of cyclobutadiene are E=ot—2fi E=Ol E=ot E=ot+2fl There are four 7r electrons, and so the two lowest energy levels will be occupied and En =2(oz+2fi)+2oz=4oz+4fi The second energy level (E z or) is doubly degenerate. We need to place two electrons in these orbitals, and according to Hund’s rules, each orbital will contain one electron and these electrons will have the same spin. Therefore, the ground state of cyclobutadiene should be a triplet state. We showed in the text that the energy of the n' orbital in ethene is 2a + 213, so E = E"(cyelobutadiene) — 2E”(ethene) = 0 deloc Cyclobutadiene has the same stability as two isolated ethene molecules. Explain why the doubly degenerate In“ orbitals for a linear XY2 molecule do not remain degenerate when the molecule is bent. Take the z—axis to lie along the bonds of the linear molecule and let the molecule bend in the xz-p'lane. Then one of the 171“ orbitals lies in the plane of the molecule and the other lies in a plane that is perpendicular to the plane of the molecule. For a linear molecule, the overlap between the 2px atomic orbitals (one of the In“ orbitals) and the 2pv atomic orbitals (the other If“ orbital) are identical. Hence these molecular orbitals are degenerate. As the molecule bends, the overlap between the 2 pl orbitals on the three atoms is affected differently than the overlap between the 2 pv atomic orbitals. Therefore, the energy associated with the two 171“ molecular orbitals depends on the bond angle. Explain why the 30“ molecular orbital of a linear XY2 molecule increases in energy as the molecule bends. (Hint: The 30” molecular orbital is a linear combination of the 2pz orbitals from each atom.) The 30“ orbital is a bonding orbital formed from a linear combination of the 2pZ orbitals from each atom. (Recall that we have taken the z—axis to lie along the molecular bond.) When the molecule bends, the overlap between the 2pZ orbitals decreases and the energy associated with this linear combination of atomic orbitals increases. —- a ‘g /¥ 4 , Z ”‘ Vim3+ mare), + cl V} (are) +_ 011 ‘(zt'a air?” (It-(‘5, 3\ \ 7v : D6\, 1" \JCCMTQX A” \r€’\)¢'+~ l (L 9 mm 4: SUV 24pm _ x Z 1' ‘ H“ C chfig/H’mi :3 Away i ’2‘: rxh i _ a x ’> £315; 0% wn/L‘Pa/uma. fl / r~ /—~: a ‘/L\ fix ‘03 b '" ffiim‘s” ->/mexam‘gg \ 2!.8éX/bm’ / ~ w \ r13,” Jib fig? .7, -f —C\ , 1B :1 mm aM J, -; diagnut‘l‘ UNFEFW) J’U M s (35‘) (10> mm :. aflmmm : Max J? M, , 35+\ , _ ,a’WI ‘9’ W ' ’3 ': i =9 Midi; A N\V 12— K“ ) v // 1- r\ ~ \U‘C’i/ 08 /7 Di 0.6-L 0.4— 0.24 ., 13 ,1 A}: :L‘Y‘w "2 (9,1591: x 5345.5) x 8.759316 ha .. ’10 r v—‘a 7— 5. 55 5X10 f :1 2“: EL": Q/M‘ /M j“ N ‘ 7” #fl/ aw Dd m 0:}, min/‘2 be Witt/L M v m .— 6) Ag / 3 I (HAL/x H/UL 15 57-” M K\ A] i 4 f K I N l {:3 ,ULOUACML Mr \5. Lo (5 :9] I ' “31 OWEEMM’L fiHéca ’— mu f Arm $6M! .- r _ F727 ,F JLP f ‘¢ ‘ t, 1 ’4 '- 2 f” ‘\ “‘\ __ 5) w/ / a“ , _ fl ‘3 Lb; .,_ , 7/ f/(A Mfg, ’/ ELL— car/ILL / ‘4 ’ hp... “"5"” a g’ c \3 x ' : Vwm Jib—(fl 7/: i“ n 1 333‘ CXHTE U jim“ 7 f f \x“ ‘ I i" r ‘_ A / I .2” ’“HVB‘CA 7-! ’— |\ X ‘ \V L) /‘\ rs «N A L‘ fl, >L I | / WJLK Kai/k)" J gfl-‘j—I’ .1 / , VFWA \J/UDM r‘f‘a " :1 $.45: Dql‘ _ 93 367:; . , . r f L &‘> ’ ' [1 fl -' 7173/1 W”M”V% In “W “9531” 7141+ #c/ A»? W . i r10. V? l . Fl <5E’IW L g ‘ "K "A' 2 "L3 ‘2 )flw' U , - Iéoma WWII/(ALA! 9% § \ r [M {7/3 me/MJL '1 'L ’ My”; [/VIML'AAID’QL ‘ J Aha CW ’ Mam j, 1”: a!) 139 '17-'55 HES} flLfl G L/L I}. 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We» 1 1;: Slum Wt; ffifl’kacbfmliflx gram 0 [ABA—x1; \Dwfl‘” LOX Hrpm": mTfifl-‘EYmAco am17v'1w‘li‘h'15 rgcs'mfi WE, fimjclfl’m Qangw'mfi Ew'}EC3Ta-\s Saxwe‘x] cathflfrj Sm [curmflc q Sih<fi~flj1§1c 10"” XML] 8 cUth—(C‘k'x’) A?“ h; {39 Bi‘qlam 1 fl "TC lfm <1 1 r H 17* D 511.1(133) bwaIELF—K’JAX ‘3 . x L/ r ._ Mia) “r q 5 L M) 1m a .51 O&_E\ 1F— \G¢L<5 SQ *Mfi 5T)G}CJJETMW ‘E n I -" “LA 05 . imbffiflg . WEEK“: 0.] So irked, '1h’}¢06'1‘1i£;5 $chG—MA EFT-1%? DVQ’Y '3"? Jane thngCAio‘fiS vaolqod 4V? *mefihcmfi 0T @XCLWYJ\@ {of {he “#- lfi‘cfi’l‘fiulrlfi‘fi} ‘H’Iemwr/Ewncjfim figg‘r 10d? "mafia Hm WGCJKQk J 50 41/113 JfT‘con Eriltfxfi’“ (1‘53 W’JQTQ ‘m{¢ng¢- ‘1‘1’Jom JWU: ho§>%l\w>”m mock“ fl: For an np' electron configuration, there are six entries in a table of possible sets of m, and my. We have (using Equations 8.49, 8.50, and 8.52) m, 111‘ ML Ms M] 1 +§ 1 +§ +§ 1 —r 1 —; +r 0. +% 0 +§ +g o —% o a —% —1 +§ —1 +§ —§ r —1 ~r —1 —r —% The M and M s values given here correspond to a 2P state, and the values of M 1 correspond to a value of J of either 1/2 or 3/2. Thus, the term symbols associated with an np‘ electron configuration are 2P3/2 and 2PW. The ground state is determined by using Hund’s rules; by Rule 3, the most stable state (and therefore the ground state) is 2Pl a. An rip5 configuration can be thought of as an np’ configuration because two of the rip orbitals are filled and so Ms and ML are determined by the remaining half-filled p—orbital. Therefore, the term symbols associated with the rip“ configuration will be the same as those for an ripl configuration. From Equation 8.53 we also see that the number of sets of mi, and m” remains the same for an npS configuration as for an np' configuration. K N; (lag)2(la“)2(20g)2(20u)2(171“)2(17ru)2(3ag)' corresponds to |ML| = 0 and |Ms| = g, or} a 2'2 term symbol. The symmetry of the molecule is g. The complete ground-state term symbol of N; is 22:, because the molecular wave function does not change when reflected through a plane containing the two nuclei. N2 (lag)2(lau)2(20g)2(20“)2(1IT")Z(17r”)2(30g)2 corresponds to |ML| = 0 and |M5| = 0, or a 1'2 term symbol. The symmetry of the molecule is g. The complete ground-state term symbol of N2 is '23:, because the molecular wave function does not change when reflected V through a plane containing the two nuclei. o,+ (lag)2(1a“)2(20g)2(20u)2(3ag)2(17r“)2(l7r“)2(ln'g)l corresponds to |ML| = 1 and |MS| = %, or a 2H term symbol. The symmetry of the molecule is g, since the only unfilled molecular orbital has symmetry g, so the complete ground—state term symbol of 0‘; is 211g. 5'2” {Vile U “A? (9’ r551 (‘3‘ 5‘7" 2 4553510133211 ’rc\"‘(1'tr§ (may . 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This note was uploaded on 01/25/2011 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at Berkeley.

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ProblemSet6Solutions - Problem 5 1a ® In this problem we...

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