Chapter 6 Answers to Problems

Transcription

1 Chapter 6 Answers to Problems 6.1 (a) NH 3 C3v E 2C3 3 v n = 3A 1 A 2 4E trans = A 1 E rot = A 2 E = 2A 2E = 4 frequencies 3n-6 1 Infrared 4 (2A 1 2E) Raman 4 (2A 1 2E) Polarized 2 (2A 1) Coincidences 4 (2A 1 2E) (b) FeCl 3 6 Oh E 8C3 6C2 6C4 3C2 i 6S4 8S6 3 h 6 d

2 3n = A 1g E g T 1g T 2g 3T 1u T2u trans = T 1u rot = T1g = A E T 2T T = 6 frequencies 3n-6 1g g 2g 1u 2u Infrared 2 (2T 1u) Raman 3 (A 1g E g T 2g) Polarized 1 (A 1g) Coincidences 0 Silent modes 1 (T 2u) (c) H2CO C2v E C2 v(xz) v(yz) n = 4A 1 A 2 3B 1 4B2 = A B B = A B B = 3A B 2B = 6 frequencies trans rot n Infrared 6 (3A 1 B 1 2B 2) Raman 6 (3A 1 B 1 2B 2) Polarized 3 (3A 1) Coincidences 6 (3A 1 B 1 2B 2) -87-

3 (d) PF 5 D3h E 2C3 3C2 h 2S3 3 v n = 2A 1' A 2' 4E' 3A 2" 2E" trans = E' A 2" rot = A 2' E" = 2A ' 3E' 2A " E" = 8 frequencies 3n Infrared 5 (3E' 2A 2") Raman Polarized Coincidences 6 (2A ' 3E' E") 1 2 (2A ') 1 3 (3E') (e) C2H 6 (staggered configuration) D3d E 2C3 3C2 i 2S6 3 d

4 3n = 3A 1g A 2g 4E g A 1u 3A 2u 4Eu trans = A 2u E u rot = A 2g Eg = 3A 3E A 2A 3E = 12 frequencies 3n-6 1g g 1u 2u u Infrared 5 (2A 3E ) Raman 6 (3A 3E ) 2u 1g Polarized 3 (3A 1g) Coincidences 0 Silent modes 1 (A 1u) u g (f) H2O2 C2 E C n = 6A 6B trans = A 2B rot = A 2B = 4A 2B = 6 frequencies 3n-6 Infrared 6 (4A 2B) Raman 6 (4A 2B) Polarized 4 (4A) Coincidences 6 (4A 2B) -89-

5 6.2 (a) SeF 5 C4v E 2C4 C2 2 v 2 d n = 4A 1 A 2 2B 1 B 2 5E trans = A 1 E rot = A 2 E = 3A 2B B 3E = 9 frequencies 3n Infrared 6 (3A 1 3E) Raman 9 (3A 1 2B 1 B 2 3E) Polarized 3 (3A 1) Coincidences 6 (3A 1 3E) (b) AsF 4 C2v E C2 v(xz) v(yz) n = 5A 1 2A 2 4B 1 4B2 trans = A 1 B 1 B 2 rot = A 2 B 1 B2 = 4A A 2B 2B = 9 frequencies 3n

6 Infrared 8 (4A 1 2B 1 2B 2) Raman 9 (4A 1 A 2 2B 1 2B 2) Polarized 4 (4A 1) Coincidences 8 (4A 1 2B 1 2B 2) (c) BeF 3 D3h E 2C3 3C2 h 2S3 3 v n = A 1' A 2' 3E' 2A 2" E" trans = E' A 2" rot = A 2' E" = A ' 2E' A " = 4 frequencies 3n Infrared 3 (2E' A 2") Raman Polarized Coincidences 3 (A ' 2E') 1 1 (A ') 1 2 (2E') -91-

7 (d) OSF 4 C2v E C2 v v n = 6A 1 2A 2 5B 1 5B2 trans = A 1 B 1 B 2 rot = A 2 B 1 B 2 = 5A A 3B 3B = 12 frequencies 3n Infrared 11 (5A 1 3B 1 3B 2) Raman 12 (5A 1 A 2 3B 1 3B 2) Polarized 5 (5A 1) Coincidences 11 (5A 1 3B 1 3B 2) (e) trans-fnnf C2h E C2 i h

8 3n = 4A g 2B g 2A u 4Bu trans = A u 2B u rot = A g 2Bg = 3A A 2B = 6 frequencies 3n-6 g u u Infrared 3 (A 2B ) Raman 3 (3A g) Polarized 3 (3A g) Coincidences 0 u u (f) cis-fnnf C2v E C2 v(xz) v(yz) n = 4A 1 2A 2 2B 1 4B2 = A B B = A B B = 3A A 2B = 6 frequencies trans rot n Infrared 5 (3A 1 2B 2) Raman 6 (3A 1 A 2 2B 2) Polarized 3 (3A 1) Coincidences 5 (3A 1 2B 2) -93-

9 (g) S2O 2 3 C3v E 2C3 3 v n = 4A 1 A 2 5E trans = A 1 E rot = A 2 E = 3A 3E = 6 frequencies 3n-6 1 Infrared 6 (3A 1 3E) Raman 6 (3A 1 3E) Polarized 3 (3A 1) Coincidences 6 (3A 1 3E) (h) B2H6 D2h E C 2(z) C 2(y) C 2(x) i (xy) (xz) (yz) i

10 = 4A 2B 3B 3B A 4B 3B 4B trans = B 1u B 2u B 3u rot = B 1g B 2g B3g = 4A B 2B 2B A 3B 2B 3B 3n g 1g 2g 3g u 1u 2u 3u 3n-6 g 1g 2g 3g u 1u 2u 3u Infrared 8 (3B 1u 2B 2u 3B 3u) Raman 9 (4A g B 1g 2B 2g 2B 3g) Polarized 4 (4A g) Coincidences 0 Silent modes 1 (A u) 6.3 C3O 2 as C2h C2h E C2 i h n = 4A g 2B g 3A u 6Bu = A 2B = A 2B = 3A 2A 4B trans u u rot g g 3n-6 g u u Comparing these results with those developed in the text (pp ): C 2h Infrared 6 (2A u 4B u) 4 (2 u 2 u) Raman 3 (3A g) 3 (2 g g) Polarized 3 (3A ) 2 (2 ) Coincidences g D h g -95-

11 6.4 In the correlation diagrams shown below, only the species of T d and D 4h associated with normal modes are shown on the left. The frequency numbering of the undistorted structure has been retained for the distorted structure, with subscripts (a, b, c) for formerly degenerate modes, to show the fate of individual modes on descent in symmetry. (a) Distorted tetrahedron (XY4E) - C2v T d C 2v 1 A1 A1 1, 2a, 3a, 4a i.r., Raman (pol) 2 E A2 2b Raman 3, 4 T2 B1 3b, 4b i.r., Raman B2 3c, 4c i.r., Raman (b) Slightly squashed tetrahedron - D 2d T d D 2d 1 A1 A1 1, 2a Raman (pol) 2 E B1 2b Raman 3, 4 T2 B2 3a, 4a i.r., Raman E, i.r., Raman 3bc 4bc -96-

12 (c) Square pyramid with X at the apex - C 4v D 4h C 4v 1 A1g A1 1, 3 i.r., Raman (pol) 2 B1g B1 2, 5 Raman 4 B2g B2 4 Raman 3 A2u E 6, 7 i.r., Raman 5 B2u 6, 7 Eu 5 2u Note that the formerly silent mode, (B ), of the perfect square plane becomes Raman active as a result of this distortion. -97-

13 (d) Planar MX 4 with two long trans positions - D 2h (C 2' of D 4h retained) D 4h 1 A1g 2 B1g D 2h A1g 1, 2 Raman (pol) 4 B2g B1g 4 Raman 3 A2u 5 B2u 6, 7 Eu B1u 3, 5 i.r. B2u 6a, 7a i.r. B3u 6b, 7b i.r. -98-

14 6.5 Direct correlations from structure I to II to III can be made by using the correlation tables in Appendix B. A correlation from III to IV cannot be made in this way. Although C 2v is a subgroup of C 4v, structure IV is not obtained by retaining pre-existing elements of structure III. Most significantly, the C 2 axis of IV does not exist in III, but rather is newly created. However, it is possible to make a correlation from II (D 4h) to IV (C 2v), using the correlation in Appendix B in which C 2" of D 4h is retained as the C 2 axis of C 2v. If a correlation III (C 4v) IV (C 2v) is attempted, using the correlation tables without realizing the inconsistency of axis orientations in the two structures, the incorrect results for IV will be 6A 1 (R - pol, ir) A 2 (R) 4 B 1 (R, ir) 4B 2 (R, ir). That the results from the correlation II IV shown below are correct can be verified by determining the selection rules for IV de novo. By contrast, a structure whose selection rules could be determined correctly by direct correlation from III would be trans-ma B C. (Structure IV in Fig. 3.1 may be changed to this in the future.) 2 3 In the correlation diagrams shown here (I II III, this page, and II IV, next page) only those symmetry species associated with normal modes are shown. When two or more frequencies occur with the same symmetry, the number of occurrences is indicated in front of the Mulliken symbol. -99-

15 I II Infrared 2 (2T 1u) 5 (2A 2u 3E u) Raman 3 (A 1g E g T 2g) 5 (2A 1g B 1g B 2g E g) Polarized 1 (A ) 2 (A ) 1g Coincidences 0 0 1g Silent modes 1 (T ) 1 (B ) 2u III 2u IV Infrared 8 (4A 1 4E) 13 (6A 1 4B 1 3B 2) Raman 11 (4A 1 2B 1 B 2 4E) 15 (6A 1 2A 2 4B 1 3B 2) Polarized 4 (A 1) 6 (6A 1) Coincidences 8 (4A 1 4E) 13 (6A 1 4B 1 3B 2)

16 6.6 (a) CO 2 [O=C=O] Use D 2h as a working group. D2h E C 2(z) C 2(y) C 2(x) i (xy) (xz) (yz) i n = A g B 2g B 3g 2B 1u 2B 2u 2B3u trans = B 1u B 2u B 3u rot = B 2g B 3g (not B 1g R z) 3n-5 = A g B 1u B 2u B3u In D, = = 3 frequencies h 3n-5 g u u Infrared 2 ( ) u Raman 1 ( g ) Polarized 1 ( g ) Coincidences 0 (b) OCN [O C N] Use C 2v as a working group. C2v E C2 v(xz) v(yz) u 3n = 3A 1 3B 1 3B2 trans = A 1 B 1 B 2 rot = B 1 B 2 (not A 2 R z) 3n-5 = 2A 1 B 1 B2 In C, = 2 = 3 frequencies v 3n

17 Infrared Raman 3 (2 ) 3 (2 ) Polarized 2 (2 ) Coincidences 3 (2 ) (c) H C C H Use D 2h as a working group. D2h E C 2(z) C 2(y) C 2(x) i (xy) (xz) (yz) i n = 2A g 2B 2g 2B 3g 2B 1u 2B 2u 2B3u trans = B 1u B 2u B 3u rot = B 2g B 3g (not B 1g R z) 3n-5 = 2A g B 2g B 3g B 1u B 2u B3u In D, = 2 = 5 frequencies h 3n-5 g g u u Infrared 2 ( ) u Raman 3 (2 ) g Polarized 2 (2 g ) Coincidences 0 u g -102-

18 (d) Cl C C H C2v E C2 v(xz) v(yz) n = 4A 1 4B 1 4B2 trans = A 1 B 1 B 2 rot = B 1 B 2 (not A 2 R z) 3n-5 = 3A 1 2B 1 2B2 In C, = 3 2 = 5 frequencies v 3n-5 Infrared Raman 5 (3 2 ) 5 (3 2 ) Polarized 3 (3 ) Coincidences 5 (3 2 ) (e) H C C C C H Use D 2h as a working group. D2h E C 2(z) C 2(y) C 2(x) i (xy) (xz) (yz) i n = 3A g 3B 2g 3B 3g 3B 1u 3B 2u 3B3u trans = B 1u B 2u B 3u rot = B 2g B 3g (not B 1g R z) 3n-5 = 3A g 2B 2g 2B 3g 2B 1u 2B 2u 2B3u In D, = = 9 frequencies h 3n-5 g g u u -103-

19 Infrared 4 (2 2 ) u Raman 5 (3 2 ) g Polarized 3 (3 g ) Coincidences (a) 3n-6 = A 1' A 2" 2E' = 4 frequencies 1 (A 1'), 2 (A 2"), 3 (E'), 4 (E') The systematic assignment of frequency numbers labels nondegenerate modes before degenerate modes, even though E' is listed before A " in the current D character table. u g 2 3h (b) Association along the C 3 axis of the NO 3 ion causes its symmetry to descend to C 3v. The changes in spectra can be predicted from the following correlation diagram. D 3h R. (pol) 1 A' 1 A1, 1 2 i.r., R. (pol) i.r. 2 A" 2 i.r., R. 3, 4 E E 3, 4 i.r., R. C 3v The descent does not lift any degeneracies, but the spectroscopic activities change significantly. The totally symmetric stretching mode, 1, becomes infrared active. The outof-plane deformation mode,, becomes a Raman-active polarized mode. 2 In-plane association causes the NO 3 ion's symmetry to descend to C 2v. The changes in spectra can be predicted from the following correlation diagram. D 3h R. (pol) 1 A' 1 A1, 1 3a, 4a i.r., R. (pol) i.r. 2 A" 2 B1 2 i.r., R i.r., R. 3, 4 E' B2 3b, 4b i.r., R. C 2v In this case the E' degeneracies of both 3 and 4 are lifted. Thus, both 3 and 4 may split into two frequencies each, with C 2v symmetries A 1 and B 2, giving a total of six frequencies active in both the infrared and Raman spectra

20 The predictions for both modes of association are summarized in the table below. D3h C3v C2v Infrared 3 (A 2" 2E') 4 (2A 1 2E) 6 (3A 1 B 1 2B 2) Raman 3 (A 1' 2E') 4 (2A 1 2E) 6 (3A 1 B 1 2B 2) Polarized 1 (A 1') 2 (2A 1) 3 (3A 1) Coincidences 2 (2E') 4 (2A 1 2E) 6 (3A 1 B 1 2B 2) 0 0 (c) The data are consistent with the predictions of the C 2v model of association. 1 (d) The 830 cm frequency is 2 (A 2"), which is infrared active but Raman inactive for the "free" ion. The first overtone, 2 2, has A 1' symmetry in D 3h, which is a Raman active species. This fundamental becomes Raman active in either the C 2v or C 3v models of association. Regardless of the mode of association, the overtone would be Raman active, since all first overtones of nondegenerate modes are totally symmetric and therefore Raman active. Failure to observe 2 in the Raman spectrum of M Al(NO 3) 3 solutions might seem to suggest that there is no cation-anion or solvent-anion association. However, association cannot be ruled out, because the fundamental, although Raman allowed by association, might be very broad and have too weak an intensity to be observed above the instrumental background. 6.8 (a) True. All even-number direct products of a nondegenerate species give the totally symmetric representation. Thus, all even-number overtones will belong to the totally symmetric representation, which is always Raman allowed. (b) Not always true. Molecules belonging to point groups C 1, C s, C n, and C nv have at least one unit vector transforming as the totally symmetric representation. In these groups, the even-number overtones will be infrared allowed, as well as Raman allowed. Also, in groups with degenerate representations, the direct products will be reducible representations, which may contain an irreducible representation allowing infrared activity. (c) True. All odd-number direct products of any irreducible representation will be or contain the original representation. If a vibrational mode is active as a fundamental, it follows that all its odd-number overtones will be infrared allowed, too. (d) True. The direct product of any irreducible representation with the totally symmetric representation is the non-totally symmetric representation. The symmetry-based selection rules for the non-totally symmetric mode will apply to any combination s ± i, because the symmetry of the combination is the same as. i -105-

21 (e) True. All first overtones belong to or contain the totally symmetric representation. The product of the totally symmetric representation with itself is, of course, totally symmetric. Therefore combinations of the type ± 2 will be Raman allowed. s i 6.9 (a) The 6 frequency arises from three degenerate T 2u modes, which are inactive in both the infrared and Raman spectra. They are silent modes. (b) O h is centrosymmetric, so the rule of mutual exclusion precludes infrared-active modes from being Raman-active, and vice versa. (c) This should occur at approximately 2 x cm = cm. The observed -1 frequency (531 ± 3 cm ) is a little high, but not unreasonably so. The symmetry of this overtone is T 1u x T 1u = A 1g E g T 1g T 2g, making it Raman allowed on the basis of A, E, and T. 1g g 2g This should occur at approximately cm 117 cm = cm, which is -1 good agreement with the observed 380 ± 3 cm. The symmetry of this combination is T 1u x T 2u = A 2g E g T 1g T 2g, making it Raman allowed on the basis of E and T. g 2g If the band at 233 ± 2 cm is assigned as 2 6, then 6 is approximately 117 cm. -1 As shown above, this is consistent with the assignment of the 380 ± 3 cm band to 4 6. The symmetry of the 2 6 overtone is T 2u x T 2u = A 1g E g T 1g T 2g, making it Raman allowed on the basis of A 1g, E g, and T 2g. [Note: The symmetry of T 2u x T 2u is identical to T 1u x T 1u, as shown for 2 4 above.] The observed polarization is consistent with the A component This should occur at approximately 741 cm 739 cm = 1480 cm, which is -1 good agreement with the observed ± 0.5 cm. The symmetry of this combination is A x T = T, making it infrared allowed on the basis of T. 1g 1g 1u 1u 1u This should occur at approximately 652 cm 739 cm = 1391 cm, which is -1 good agreement with the observed ± 0.5 cm. The symmetry of this combination is E x T = T T, making it infrared allowed on the basis of T. g 1u 1u 2u 1u This should occur at approximately 652 cm 263 cm = 915 cm, which is good -1-1 agreement with the observed ± 0.5 cm. ± 0.5 cm. The symmetry of this combination is E x T = T T, making it infrared allowed on the basis of T. g 1u 1u 2u 1u (d) The combination and overtones observed in the Raman spectrum have gerade symmetry, and the combinations observed in the infrared spectrum have ungerade symmetry. Consistent with the rule of mutual exclusion, the Raman-active frequencies cannot be infrared-active, and vice versa

22 (e) In order to be infrared allowed, a combination would have to be or contain T 1u. The possible combinations with 6, their symmetries, and whether infrared allowed are shown below. Combination Symmetries Allowed? 1 6 A 1g x T 2u = T2u No 2 6 E g x T 2u = T 1u T2u Yes 3 6 T 1u x T 2u = A 2g E g T 1g T2g No 4 6 T 1u x T 2u = A 2g E g T 1g T2g No 5 6 T 2g x T 2u = A 1u E u T 1u T2u Yes The predicted frequencies for the allowed combinations are: cm 117 cm = 769 cm cm 117 cm = 434 cm Although allowed, these combinations apparently have too weak intensities to be observed