MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Fall, 2008 Page 1 Lecture #30: What is in a Character Table and How do we use it? Last time matrix representations of symmetry operatorsrepresentations of group — same multiplication table as symmetry operatorscharacters 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-dimensionalGOALintegral with several non-commutingsymmetry operators 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,Ia ≤ Ib ≤ Ic for a, b, c]selection rules: pure rotation and rotation-vibration and Raman.nature of various types of vibration. Example:D3h totallysymmetric E 2C3(z) 3C2(⊥) σh(xy) 2S3(z) 3σv 11 1 1 1 1A′ 1 11 –1 1 1–1 E′ A′ 2 2–1 0 2–10 11 1 –1–1–1A1 ′′ 11 –1 –1–11 E″ A′′ 2 2–1 0 –210 (rotational level symmetries and perturbations) Rotations, Translations, IR selection rules, p–orbitals electronic selection rules (magnetic dipole) Rz (x,y) z (Rx,Ry) Polarizability, RamanSelection Rules,d–orbitals x2 + y2, z2 (x2 – y2, xy) (xy, yz) order of group g = 12 = ∑ν n2 ν(nν is order of ν-th irreducible representation) equal to number of classes: 1 + 2 + 3 + 1 + 2 + 3 Rz “belongs to” A′ 2 , z (or Tz) belongs to A′′ 2 5.80 Lecture #30Fall, 2008 Page 2 Use picture to generate representationz Tz E C3 C2 σh S3 σv Rz11–11 +1–1A′ 2 y Rz Tz1 1–1–1–1 1A″ 2 show with cartoons why Rz ↔ A′ 2 from recall ′, ″↔σh these characters x 1,2↔σv (x,y) means symmetry operation transforms x into y (must generate 2D representation using x and y) Selection rules: integrand must contain totally symmetric representation. ∫ ψiOp ψf dτ ≠ 0 Direct Product: Γ ψi ) ⊗ Γ O ) must include Γ(ψf) because direct product of any irreducible(( prepresentation with itself contains the totally symmetric representation. ≡ χi )χj R1 ),χi )χj R2 ),…χΓi ⊗Γ j ((R1 ( (R2 () Example: E′ ⊗ E″ = (4 1 0 –4 –1 0)shortcuts (the irreducible representations must all be ″) A ⊗ B = B ′⊗″ = ″ g ⊗ u = u 1 ⊗ 2 = 2 Decomposition of (4 1 0 –4 –1 0): a = A2 ′ 121 [4·1·1+1·2·1+ 0 – 4·1·1− 1·2·1+ 0]= 0 1 [4·2·1+ 1·2(−1) + 0 − 4·1·(−2) − 1·2·1+ 0]= 1aE′′ = 12 aA1 ′′ = 1 aA2 ′′ = 1 So now we know how to work out all selection rules. Best to work specific example of D3h molecule BCl3. 5.80 Lecture #30Fall, 2008 Page 3 r r r Generate 3N dimensional representation. 120° planarE C3 C2(⊥) σh S3 σv χred = 12 1 + 2 cos 23 π 2(1–2) 4(2–1) –1+2 cos 23 π 2(2–1) 120 –24 –22 χred = χA1 ′ + 3χE′ + 2χA2 ′′ + χA2 ′ + χE′′(total of 12 degrees of freedom) 3 translations E′↔ (x,y)A′′ 2 ↔ z 3 rotations A′ 2 ↔ Rz E″↔ (Rx, Ry) This leaves 6 vibrations χVIB = χA1 ′ + 2χE′ + χA2 ′′ (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 (especiallywhen there is only 1 mode in a symmetry class) 1 32 ΓRED from (stretches only) χred =(3 0 1 3 0 1)= χA1 ′ + χE′ pure stretchmixed bend and stretch (only A′ 1) (there is another E′) Thus A′ 1 pure symmetric stretch A′′2 pure bend (out of plane – because χ(σh) = –1)2E′ mixed bend and stretch one out, two in pseudo rotation scissors also pseudo rotation 5.80 Lecture #30Fall, 2008 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(v1,v2,v3,v4 ) ⎡⎣χ1⎤⎦ v1 ⎡⎣ ⎤⎦ v2 ⎡⎣ ⎤⎦ v3 ⎡⎣ ⎤⎦ v4 = ⊗ χ2 ⊗ χ3 ⊗ χ4 Γ(0,0,0,0)= A′ 1 fundamentals Γ(1 0 0 0) A′ 1 Γ(0 1 0 0) A′′ 2 Γ(0 0 1 0) E′ Γ(0 0 0 1) E′ E′⊗ E′ =(4 1 0 4 1 0)=(A′ 1 + Selection rules for fundamental bands overtones Γ(2 0 0 0) A′ 1 Γ(0 2 0 0) A′ 1 Γ(0 0 2 0) A′ 1 + A′ 2 + E′ Γ(0 0 0 2) A1 ′ + A′ 2 + E′ A′ 2 + E′) (all ′) Γ′⊗Γ″ = (1 0 0 0) ← (0 0 0 0) A′ 1 (0 1 0 0) A′′ 2 (0 0 1 0) E′ (0 0 0 1) E′ in order for transition integral to be nonzero, need Γx, y, or z = Γ′⊗Γ″ mode #1 A′ 1 IR forbidden #2 A′′ 2 z IR allowed #3 or 4 E′ (x, y) IR allowed But how will the rotational transitions behave? 5.80 Lecture #30Fall, 2008 Page 5 Qe ≡ 0inertial axes unit vectors ( ) + 0 Qi +… e Qi +… ⎤⎥⎦ ⎤⎥⎦ ⎟⎠⎞ ⎛ ∑ Q( ) = a Mˆ j,a ⎡⎢⎣ Recall Mj ∂Mj,a ⎤ ⎥⎦ ∂Qi eQi +… ⎟⎠ rotational selection rules pure rotationrotational selection rule vibrational spectrum selection rulesin vibration-rotation-band Γ(abc) ΓQi totally symmetric elect be careful if ψj is degenerate So mode#1 A′ 1 ∂Mj,abc = 0 ∂Q1 #2 A′′ 2 ∂Mj,z ≠ 0 ∂Q2 #3 E′ ∂Mj,x or y ≠ 0 ∂Q3 or 4 for BCl3 an oblate symmetric top z = c x,y = (a,b)mode #2 fundamental is c type (||)mode #3,4 fundamentals are a,b type (⊥) ⎞ ⎟ ⎟⎟⎠ ⎤ ⎥ ⎥ ⎥⎦ ⎡ ⎢ ⎢ ⎢⎣ ⎜⎝ i e ⎞⎟⎠ ⎛ ⎜⎝ ∑ ( ) ++b Mˆ j,b 0⎡⎢⎣ i ⎞ ψ*j ⎛ ⎜ ⎜⎜⎝ ⎛ ⎜⎝ ∂ ∑ i ( ) ++c Mˆ j,c 0⎡ ⎢⎣ ∫ ⎞ ⎛ ⎟⎠ ⎜⎝ ψj dτ a b c ∂Qi ∂M j,b ∂Qi ∂Mj,c ψ(relect;Q) ∂Qi = e ∂M j,abc ∂Qi 5.80 Lecture #30Fall, 2008 Page 6 || ∆K = 0 weak Q ⊥ ∆K = ±1 strong Q General procedure3N dimensional χRED find (and classify) all normal mode symmetries(x,y,z) ↔ (a,b,c) ↑ highest order Cn 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 Atomsand Molecules. New York, NY: OxfordUniversity Press, 1995.5.80 Lecture #30Fall, 2008 Page 7 The E-A1 energy level diagram is given in Figure 7.51. The energy level structure of an E vibrationalstate is complicated by the presence of a first order Coriolis interaction between the two