inal Exam Your ame: Chemistry 639 Thursday, May 9, 00 SS This is your final exam. You can use your notes or a textbook but cannot discuss anything with other students. You have 3 hours to complete the exam. Write ALL your derivations either on these sheets of paper or submit additional pages with the exam if you want to get full credit. The exam consists of 6 questions totalling 101 points. If you choose not to use notes and book, you will get additional 10 points. Using character tables is allowed in both cases. Question 1 (5 points) a) (3points) Arrange in order of increasing energy the terms that may arise from the p 1 3p 1 configuration. L 0,1,; S0,1 3 D 3 P 3 S 1 D 1 P 1 S - in accordance with und s rules #1 and # b) ( points) Identify the lowest energy state in that configuration. J 1,,3 3 D 1 3 D 3 D 3 - in accordance with und s rule #3 Question (5 points) The ammonia molecule is pyramidal with the nitrogen atom lying 0.38 above the plane. a) (10 points) Identify molecular symmerty and explain using Walsh correlation diagram why the molecule is not planar The molecule has C 3v symmetry, and in the correlation diagram it should be compared with D 3h. Since 3 has 8 valence electrons, we need to evaluate energies of the four lowest energy MO which are constracted from valence orbitals of and 3. The three hydrogen s s-orbitals give rise to a 1 and e symmetry adapted linear combinations (SALC) in C 3v, and a 1 and e symmetry species in D 3h. The p z of behaves like a 1 in C 3v, and as a in D 3h. The remainingp x,p y combination transforms according to e symmetry in C 3v, and as e in D 3h. 3 S p z p x,p y C 3v a 1 e a 1 a 1 e D 3h a 1 e a 1 a e The major impact on the shape comes from p z of nitrogen, which is a nonbonding orbital in D 3h but starts contributing significantly to a bonding a 1 in C 3v. inal Exam 000 Page 1
a 1 a'' e e' a 1 a' 1 C 3v D 3h b) (10 points) Identify (label and draw) vibrational modes for this molecule. ow many IR-active and Raman-active fundamental frequencies are there? C 3v E C3 3 v A 1 1 1 1 z x y,z A 1 1?1 R z E?1 0 x,y; R x,r y x? y,xy xz,yz 3 1 0 3A 1 A 4E trarot vibr A 1 A E A 1 E stretch 3 0 1 A 1 E 3 3 0 1 A 1 E three hydrogens s 1 1 1 A 1 s of nitrogen p 1 1 1 A 1 p z of nitrogen p?1 0 E p x,p y of nitrogen There are 3-66 normal modes : 1 a 1 -symmetric stretching, a 1 -symmetric bending, 3 e-asymmetric bending and 4 e - umbrella. All are IR and Raman active. a 1 a 1 e e inal Exam 000 Page
c) (5 points) Sketch microwave spectrum of 3. It is an oblate symmetric top with I m 3m 0.38 4 0.38 0.35675emu.Thus B 16.8576314 m 3m 17 5cm?1 and B94.5cm?1. Since J max kt B? 1/~1, the first line should be the highest at room T 0.35675 cm?1 47. B Question 3 (5 points) Perform vibrational analysis for the planar, 4. Show by arrows (or write in a form of displacement vectors) and classify by symmetry the normal modes, label them and identify their IR and Raman activities. int: treat stretching, bending in plane and bending out of plane modes separately. There should be 3-6 9 normal modes. The molecule has D 4h symmetry. The fastest wave is to introduce - bond stretching, in-plane and out-of-plane bending coordinates. Then: D 4h E C4 C C C i S 4 h v d stretch 4 0 0 0 0 0 4 0 a 1g b 1g e u in?plane 4 0 0? 0 0 0 4? 0 a g e u out?of?plane 4 0 0? 0 0 0?4 0 a u b u e g Streching: 1 a 1g? Raman, 3 b 1g? IR, 6 e u? IR a 1g b 1g e u In-Plane Bending: 4? Raman, 7 e u? IR R z (a g ) e u Out of-plane Bending: a u? IR, 5 b u, inal Exam 000 Page 3
a u b u R x,r y (e g ) Question 4 (10 points) Identify molecular symmetries and sketch microwave spectra and/or IR spectra for sected ones (neglect centrifugal distortion). Show as many details as you wish but explain what you show. B e r 16.8576314 e r e amu cm?1 Molecule Symmetric top? What kind? Microwave Spectrum IR Bond length a) a,b linear flat r 0.74 b) C 3 Cl prolate sym. top c) C 60 a spherical top flat r 1.101 d) CO b linear B1. 93cm?1 ** ~1000cm?1 r CO 1.18 e) C 4 a f) Cl a g) O a h) O a i) CO b spherical top flat r C 1.33 linear *B 10.3cm?1 r Cl 1.75 asymmetric top *B 4.9cm?1 assuming 10 E ) r O 1.193 linear flat r O ~1. linear B 0.39cm?1 *** u a - sketch microwave spectrum b - sketch IR spectrum ~350cm?1 ; u~670cm?1 r CO 1.16 *Microwave spectra have lines spaced by B with intensities reaching maximum at ca. J max ~ kt B? 1 **IR spectrum of CO has only P and R branches made of separated by B lines with J max ~ kt B higher intensity. *** Out of three vibrational frequencies of CO, only two IR actuve: asymmetric stretching at u? 1 and R branch of slightly ~350cm?1 ; and bending at u~670cm?1.the former has only P and R branches separated by B, while the latter also has Q-branch. Question 5 (5 points) a) (4points) Sketch absorption spectrum of a diatomic molecule, for which the energy diagram is shown below. eglect rotational structure. You can draw the spectrum to the right of the energy diagram as a series of sticks with their heights corresponding to intensities and positions - to the transition energies. Pay attention to the relative intensities (qualitatively) or describe them in words. The allowed transitions are vibrational transitions in the ground state X 1 and electronic transitions X 1? A 1. or the latter, the intensities follow ranck-condon factor of overlaping between the vibrational states and the total intensity of absorption become stronger due to the 3 dependence of the Einstein coefficient A. The lowest energy transition, X 1? a 3, is spin forbidden. b) (1point) What would be different if it were a homonuclear diatomic? irst of all, the vibrational transitions in the ground state would not be observed because homonuclear diatomic molecule has not dipole moment. Second, depending on the g-u symmetries of the ground and excited A 1 states, that transition might also become forbidden. inal Exam 000 Page 4
Question 6 (6 points) a) (16 points) Using uckel approximation, estimate energies (i.e. not values but the order of MO is asked) of pyrazine molecule shown below. Label appropriate n and molecular orbitals by their symmetries. Use the table for choosing appropriate and for the two nitrogens. or lone pair electrons on nitrogens take that their enrgies are somewhere under 3 D. as shown on the picture... pc 4 pc 1 pc 3.. z 1 pc y Pyrazine π 6 π 5 π 4 π 3 n n 1 π π 1 uckel parameters for conjugated heteroatoms h k eteroatom h eteroatom bond k C 0 pcpc p 1.0 0.5 pcpp 1.0 1.5 C?? 0.8 inal Exam 000 Page 5
Lone pair electrons on nitrogens: n 1a g 1 1 ;n b 1u 1 1?.Their energies are identical at this lavel of approximation but one can show that mixing with sp orbitals of carbons makes b1u lower energy (less number of nodal planes) our p-orbitals of carbons give rives to the following symmetry adapted lineac combinations (SALC): pc 4 pc 3.. 1 pc 1 pc 1 1 pc 1 pc pc 3 pc 4 E b 1g 1 pc 1 pc? pc 3? pc 4 E 3 1 pc 1? pc? pc 3 pc 4 E? 4a u 1 pc 1? pc pc 3? pc 4 E? Two p-orbitals of nitrogens: 5 1 pn 1 pn E 0.5 6 1 pn 1? pn E 0.5 The carbon based SACL and nitrogen based SACL of the same symmetry can mix but of different symmetry - will not. Such orbitals are only and : The remaining b 1g and a u represent the energies of appropriate orbitals (under uckel approximation). Mixing two constractively and destructively gives following MO structures. Even without calculations, just from analysis of number for nodal planes, we can place orbitals in the appropriate order. b 1g a u b - 3u b - g b + 3u b + g Of course, we can calculate the energies as well. We ll use that pc 1 pn 1 C 1 5 1 pc 1 pc pc 3 pc 4 1 pn 1 pn 1 4 3 6 1 pc 1? pc? pc 3 pc 4 1 pn 1? pn : 1, eigenvalues:? 0.686,. 186, 5 0.5 inal Exam 000 Page 6
: 3?, eigenvalues:? 1. 851, 1. 351 6 0.5 : E? 1. 851 a u : E? : E? 0.686 b 1g : E : E 1. 351 : E. 1861 b) (8 points) rom the electronic configurations identify the ground and the first two excited states of pyrazine. Are both (any) transitions dipole allowed? or allowed transition(s) show the orientation of the transition moment in the molecular frame. π 6 π 5 π 4 a u π 6 π 5 π 4 a u π 3 b 1g π 3 b 1g n n 1 b 1u a g n n 1 b 1u a g π π π 1 π 1
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