B F N O. Chemistry 6330 Problem Set 4 Answers. (1) (a) BF 4. tetrahedral (T d )

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1 hemistry 6330 Problem Set 4 Answers (1) (a) B 4 - tetrahedral (T d ) B T d E S 4 6s d G xyz G unmoved atoms G total If we reduce G total we find that: G total = A 1 E T 1 3T 2 Looking at the character table, we can see that: and G rotational = T 1 G translational = T 2 G vibrational = A 1 E 2T 2 By inspection of the character table, the 2T 2 modes are IR active and A 1, E, and 2T 2 modes are Raman active. (b) lno bent ( s ) l N O s E s h G xyz 3 1 G unmoved atoms 3 3 G total 9 3

2 G total = 6A 3A G rotational = A 2A G translational = 2A A G vibrational = 3A IR active: 3A Raman active: 3A (c) XeO 3 trigonal pyramid ( 3v ) O Xe O O 3v E 2 3 3s v G xyz G unmoved atoms G total G total = 3A 1 A 2 4E G rotational = A 2 E G translational = A 2 E G vibrational = 2A 1 2E IR active: 2A 1 2E Raman active: 2A 1 2E (d) l 3 t-shaped ( 2v ) l

3 2v E 2 (z) s v (xz) s v (yz) G xyz G unmoved atoms G total G total = 4A 1 A 2 3B 1 4B 2 G rotational = A 2 B 1 B 2 G translational = A 1 B 1 B 2 G vibrational = 3A 1 B 1 2B 2 IR active: 3A 1 B 1 2B 2 Raman active: 3A 1 B 1 2B 2 (e) S 4 see-saw ( 2v ) S 2v E 2 (z) s v (xz) s v (yz) G xyz G unmoved atoms G total G total = 5A 1 2A 2 4B 1 4B 2 G rotational = A 2 B 1 B 2 G translational = A 1 B 1 B 2 G vibrational = 4A 1 A 2 2B 1 2B 2 IR active: 4A 1 2B 1 2B 2 Raman active: 4A 1 A 2 2B 1 2B 2

4 (2) (a) Axes oriented such that z-axis goes through O-Xe-O bond and x-axis is through Xe, perpendicular to the plane of the molecule D 2h E 2 (z) 2 (y) 2 (x) i s (xy) s (xz) s (yz) G xyz G unmoved atoms G total G total = 2A g B 1g B 2g 2B 3g 3B 1u 3B 2u 3B 3u G rotational = B 1g B 2g B 3g G translational = B 1u B 2u B 3u G vibrational = 2A g B 3g 2B 1u 2B 2u 2B 3u (b) IR active: 2B 1u 2B 2u 2B 3u Raman active: 2A g B 3g (c) To solve this problem we need to look at what happens to A 1, E and T 2 when lowering the symmetry from T d to 2v : T d 2v A 1 A 1 E A 1 A 2 T 2 A 1 B 1 B 2 The normal modes for this molecule are: 4A 1 A 2 2B 1 2B 2 (d) IR active: 4A 1 2B 1 2B 2 Raman active: 4A 1 A 2 2B 1 2B 2 (e) There are several possible answers to this question: (i) Neither structure is correct since neither has the correct number of IR bands. (ii) Structure I, which gives 8 possible IR bands, is correct and one of the stretches is too weak to be seen or lies under another band.

5 (3) N S S N D 2h symmetry Axes oriented such that z-axis goes through the two N-atoms, y-axis goes through the two S-atoms and x axis is perpendicular to the plane of the molecule. D 2h E 2 (z) 2 (y) 2 (x) i s (xy) s (xz) s (yz) G xyz G unmoved atoms G total G total = 2A g B 1g B 2g 2B 3g 2B 1u 2B 2u 2B 3u G rotational = B 1g B 2g B 3g G translational = B 1u B 2u B 3u G vibrational = 2A g B 3g B 1u B 2u B 3u IR active: B 1u B 2u B 3u Raman active: 2A g B 3g S N N S 2v symmetry 2v E 2 (z) s v (xz) s v (yz) G xyz G unmoved atoms G total G total = 4A 1 2A 2 3B 1 3B 2 G rotational = A 2 B 1 B 2 G translational = A 1 B 1 B 2 G vibrational = 3A 1 A 2 B 1 B 2

6 IR active: 3A 1 B 1 B 2 Raman active: 3A 1 A 2 B 1 B 2 If structure A is correct, the IR and Raman spectra should each have only three bands. If either, or both of the spectra contain more than three bands, this would indicate Structure B is correct. (4) Assume the molecule lies in the yz plane, with the z-axis going through the - bond, the x- axis perpendicular to the plane of the molecule and the y-axis pointing down. z x y (a) D 2h E 2 (z) 2 (y) 2 (x) i s (xy) s (xz) s (yz) G xyz G unmoved atoms G total G total = 3A g B 1g 2B 2g 3B 3g A u 3B 1u 3B 2u 2B 3u G rotational = B 1g B 2g B 3g G translational = B 1u B 2u B 3u G vibrational = 3A g B 2g 2B 3g A u 2B 1u 2B 2u B 3u (b) We can separate the bond internal coordinates into one - bond and 4 - bonds. D 2h E 2 (z) 2 (y) 2 (x) i s (xy) s (xz) s (yz) G G G = A g G = A g B 3g B 1u B 2u

7 (c) z Δr 1 Δr 5 Δr 4 Δr 2 Δr 3 x y P " # r & r & P " # r ( r ( r * r r, P -.# r ( r ( r * r r, P - 01 r ( r ( r * r r, P - 21 r ( r ( r * r r, (d) We can use the four -- angles as internal coordinates for in-plane bending modes: θ 4 θ 1 z θ 3 θ 2 x y Each of these four angles can be varied independently of the others;; we therefore expect no spurious modes.

8 D 2h E 2 (z) 2 (y) 2 (x) i s (xy) s (xz) s (yz) G q 1- q G q 1- q 4 = G = A g B 3g B 1u B 2u (e) We can use the results from Part to write the SAL s easily: P " # θ ( θ ( θ * θ θ, P -.# θ ( θ ( θ * θ θ, P - 01 θ ( θ ( θ * θ θ, P - 21 θ ( θ ( θ * θ θ, There are no spurious modes from this choice of in-plane internal coordinates.

9 (f) The choice of internal coordinates for out-of-plane bends is more difficult. Imagine lines through the two atoms, perpendicular to the plane of the molecule. We can use the angles from these lines to the atoms: θ 8 θ 5 x θ 7 θ 6 z y q 9 under q 5 q 10 under q 6 q 11 under q 7 q 12 under q 8 q 5 q 9 = 180 ;; q 6 q 10 = 180 ;; q 7 q 11 = 180 ;; q 8 q 12 = 180 ;; we therefore expect 4 spurious modes. Before we figure G q 5- q 12, let s see what we expect. If we subtract the stretching and inplane bending modes from G vibrational we are left with G out-of-plane = B 2g A u B 3u D 2h E 2 (z) 2 (y) 2 (x) i s (xy) s (xz) s (yz) G q 5- q All of the operations except E move all of the angles. Reducing this rep gives us one irr. rep of each symmetry: G q 5- q 12 = A g B 1g B 2g B 3g A u B 1u B 2u B 3u We cannot increase both q 5 and q 9 at the same time, for this is physically impossible. Likewise, q 6 and q 10, and so forth. Because the supplementary angles are related by s (yz), any irr. rep for which c [s (yz)] = c (E) will lead to a physically impossible mode. Thus, the spurious modes are A g, B 3g, B 1u, and B 2u. (You should try using the projection operators of these irr. reps if you don t follow the logic used above.) or the three remaining modes we expect:

10 P - 2# θ & θ & θ 4 θ 5 θ 6 θ 7 θ (8 θ (( θ (* " " P " 1 θ & θ & θ 4 θ 5 θ 6 θ 7 θ (8 θ (( θ (* " " This mode corresponds to twisting the two 2 fragments relative to one another. P -.1 θ & θ & θ 4 θ 5 θ 6 θ 7 θ (8 θ (( θ (* " " What about the B 1g mode, which has not yet been accounted for? P - 0# θ & θ & θ 4 θ 5 θ 6 θ 7 θ (8 θ (( θ (* " " This is not actually a vibrational mode;; it corresponds to rotation about the - axis (R z ).