5.80 Small-Molecule Spectroscopy and Dynamics

Transcription

1 MIT OpenCourseWare Small-Molecule Spectroscopy and Dynamics Fall 008 For information about citing these materials or our Terms of Use, visit:

2 Fall, 008 Page Lecture #30: What is in a Character Table and How do we use it? Last time matrix representations of symmetry operators representations of group same multiplication table as symmetry operators characters of matrix representations (all we need for most applications) generate representation from convenient set of objects (basis vectors) GOT character table irreducible representations generalization of odd/even notation symmetry label for multi-dimensional integral with several non-commuting symmetry operators GOAL reduction of reducible representations generate and reduce reducible representations how do we get and use the fancy labels to the right of characters (a, b, c) (x, y, z) [conventions for x, y, z, I a I b I c for a, b, c] selection rules: pure rotation and rotation-vibration and Raman. nature of various types of vibration. Example: D 3h totally symmetric E C 3 (z) 3C ( ) σ h (xy) S 3 (z) 3σ v A A E 0 0 A A E 0 0 order of group g = = n ν ν classes: R z belongs to A, z (or T z ) belongs to A (rotational level symmetries and perturbations) Rotations, Translations, IR selection rules, p orbitals electronic selection rules (magnetic dipole) R z (x,y) z (R x,r y ) Polarizability, Raman Selection Rules, d orbitals x + y, z (x y, xy) (xy, yz) (n ν is order of ν-th irreducible representation) equal to number of

3 Fall, 008 Page Use picture to generate representation z T z E C 3 C σ h S 3 σ v y R z + A T z A x R z recall, σ h show with cartoons why R z A from these characters, σ v (x,y) means symmetry operation transforms x into y (must generate D representation using x and y) Selection rules: integrand must contain totally symmetric representation. ψ i O p ψ f dτ 0 Γ( ψ i ) Γ ( O p) must include Γ(ψ f ) because direct product of any irreducible Direct Product: representation with itself contains the totally symmetric representation. χ Γ i Γ j ( χ i (R )χ j ( R ),χ i )χ j (R ( R ), ) Example: E E = ( ) shortcuts (the irreducible representations must all be ) A B = B = g u = u = Decomposition of ( ): a A = [ ] = 0 ae = [4 + ( ) ( ) + 0] = a A = a A = So now we know how to work out all selection rules. Best to work specific example of D 3h molecule BCl 3.

4 Fall, 008 Page 3 Generate 3N dimensional representation. r r r 0 planar E C 3 C ( ) σ h S 3 σ v χ red = + cos π ( ) 4( ) 3 + cos π ( ) χ red = χ A + 3χ E + χ A + χ A + χ E (total of degrees of freedom) 3 translations E (x,y) A z 3 rotations A R z E (R x, R y ) This leaves 6 vibrations χ VIB = χ A + χ E + χ A (total of 6) (four normal modes, two are doubly degenerate) We can go further - to figure out bend vs. stretch or mixed character of the 4 normal modes (especially when there is only mode in a symmetry class) Γ RED from 3 (stretches only) χ red = ( ) = χ A + χ E pure stretch (only A ) mixed bend and stretch (there is another E ) Thus A pure symmetric stretch A pure bend (out of plane because χ(σ h ) = ) E mixed bend and stretch scissors pseudo rotation one out, two in also pseudo rotation

5 Fall, 008 Page 4 (compression of one angle rotates around either clockwise or counterclockwise, but no real rotation) Now we are ready to work out selection rules for vibration-rotation spectra Γ (v,v,v 3,v 4 ) v χ v v 3 v 4 = χ χ 3 χ 4 Γ (0,0,0,0) = A fundamentals Γ ( 0 0 0) A (0 0 0) Γ A Γ (0 0 0) E Γ (0 0 0 ) E E E = ( ) = (A + overtones Γ ( 0 0 0) A Γ (0 0 0) A Γ (0 0 0) A + A + E Γ (0 0 0 ) A + A + E A + E ) (all ) Γ Γ = Selection rules for fundamental bands ( 0 0 0) ( ) A (0 0 0) A (0 0 0) E (0 0 0 ) E in order for transition integral to be nonzero, need Γ x, y, or z = Γ Γ mode # A IR forbidden # A z IR allowed #3 or 4 E (x, y) IR allowed But how will the rotational transitions behave?

6 Fall, 008 Page 5 inertial axes unit vectors Recall M j ( Q) = a ˆ M rotational selection rules j,a +b ˆ M j,b 0 pure rotation spectrum Q e 0 M ( 0 ) + j,a Q i + i Q i ( ) + +c ˆ M ( ) j,c 0 + i i Q i M j,b M j,c Q i e Q i + Q i + e e rotational selection rule in vibration-rotationband vibrational selection rules M j,abc Q i e = ψ(r elect ;Q) Q i ψ * j Γ (abc) a b ψ j dτ c Γ Q i totally symmetric elect be careful if ψ j is degenerate So mode # A M j,abc = 0 Q # A M j,z 0 Q #3 E M j,x or y 0 Q 3 or 4 for BCl 3 an oblate symmetric top z = c x,y = (a,b) mode # fundamental is c type ( ) mode #3,4 fundamentals are a,b type ( )

7 Fall, 008 Page 6 K = 0 weak Q K = ± strong Q General procedure 3N dimensional χ RED find (and classify) all normal mode symmetries (x,y,z) (a,b,c) highest order C n activity and rotational type of each vibrational fundamental Raman Figures from Bernath: Image removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 995.

8 Fall, 008 Page 7 Images removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 995. The E-A energy level diagram is given in Figure 7.5. The energy level structure of an E vibrational state is complicated by the presence of a first order Coriolis interaction between the two components. The selection rules are K = ± and J = 0, ±. Note also that for K = + the transitions connect to the (+ ) stack while for K = they connect with the ( ) stack. The transition can again be represented by a superposition of sub-bands. Notice how the sub-bands do not line up as they do for a parallel transition, but they spread out (Figure 7.5). Each sub-band is separated by approximately [A( ζ) B]. This gives rise to a characteristic pattern of nearly equally spaced Q branches (Figure 7.53).

9 Fall, 008 Page 8 Images removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 995.

10 Fall, 008 Page 9 Images removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 995. What if BCl 3 were not D 3h (planar)? C 3v (like NH 3 ) prolate or oblate, depending on apex angle z C v planar (near oblate) tee

11 z a planar (near prolate) Fall, 008 Page 0 wye z a planar (3 different bond lengths or 3 C s different angles) Truth table point group # of normal modes pure rotational spectrum type # of IR active fundamentals Rotational type of IR fundamentals D 3h 4 3 -c, -a,b C 3v 4 c-oblate 4 -c, -a,b a-prolate C v 6 asymmetric hybrid a,b 6 3-a, -b, -c or b,c Raman active fundamentals For D 3h here will be two polarized (E ) fundamentals There will be one polarized ( A ) fundamental There will be one forbidden (i.e. not observable by IR) fundamental ( A ) (a,b)-type polarized (i.e. K = ±) oblate top vibrational spectrum (the two E modes in BCl 3 ROT E JK = BJ(J +) + (C B)K C B < 0 Strong Q branches. Q branch spikes dominate band profile. K K R Q 0 (J) (C B)[ 0 ] R Q (J) (C B)[ ] (since B is small) P Q (J) (C B)[ 0 ] etc.

12 Fall, 008 Page R Q R Q R Q 0 P Q P Q Red Blue (R and P branches are spread out) headless, subbands overlap strongly ν 0 (c)-type polarized (i.e. K = 0) oblate top vibrational spectrum. (the one A BCl 3 ) weak Q branches, except at J K and high K K = 0 only. All Q Q K (J) tend to pile up as spike at band origin. ν out-of plane bend from P Q R (b,c)-type prolate top bands (x,y,z) (b,c,a) Strong Q branches R Q 0 (A B) R Q (A B) 3 P Q P Q R Q 0 R Q R Q Red Blue ν 0 Looks very similar to type oblate band except that branches are labeled in reverse order and more spread out (less overlap of K sub-bands) because usually A B prolate B C oblate a-type ( ) prolate K = 0, weak Q except at low J K ν Asymmetric tops: Resemble symmetric top when J K More complicated because E JKa K c can t be separated into J-dependent and K- dependent additive terms.

13 Fall, 008 Page Special simplification for linear molecules. zero-point level has only = 0 vibrational angular momentum. There is no K projection of J. Bending mode is π =. type = 0 (anti-symmetric stretch), type = ± (bend) Benzene thru thru h = 4 atoms bonds E C 6 C 3 C 3C 3C i S 3 S 6 σ h 3σ d 3σ v D 6h D 6 A g ( ) = [ 36 ] = 4 A g = ( ) = [ 36 +] = 4 4 B g ( = 36 ) = 0 [ 36 ] = 4 B g = (36 + +) = ( 36 +) = 4 4 E g [ = 7 4] = [ 7] = 6 4 E g [ = 7 + 4] = 4 [ 7] = 6 4 E u = [ 7 + 4] = 4 A u = ( 36 ) = 0 4 A u = (36 + +) = 4

14 Fall, 008 Page 3 symmetries and numbers of normal modes A g 0u 4A g u B 0g u 4B g u 6E g 4u 6E 4g u translation: A u (z), E u (x,y) rotation: A g (R z ), E g (R x, R y )