Lecture #30: What Is in a Character Table and How Do We Use It?

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 notation generalization of odd/even 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, Ia ≤ Ib ≤ Ic for a, b, c] selection rules: pure rotation and rotation-vibration and Raman. nature of various types of vibration. Example: D3h totally symmetric A′1 A′2 E′ A1′′ A′′2 E″ E 1 1 2 1 1 2 2C3(z) 1 1 –1 1 1 –1 3C2(⊥) σh(xy) 2S3(z) 3σv 1 1 1 1 1 1 –1 –1 0 2 –1 0 1 –1 –1 –1 –1 –1 –1 1 0 –2 1 0 (rotational level symmetries and perturbations) Rotations, Translations, IR selection rules, p–orbitals electronic selection rules (magnetic dipole) Polarizability, Raman Selection Rules, d–orbitals x2 + y2, z2 Rz (x,y) (x2 – y2, xy) z (Rx,Ry) (xy, yz) order of group g = 12 = ∑ n 2ν (nν is order of ν-th irreducible representation) equal to number of ν classes: 1 + 2 + 3 + 1 + 2 + 3 Rz “belongs to” A′2 , z (or Tz) belongs to A′′2 Use picture to generate representation z Tz E 1 1 Rz Tz y C3 C2 1 –1 1 –1 σh 1 –1 S3 +1 –1 σv –1 A′2 1 A″2 show with cartoons why R z ↔ A′2 from these characters Rz recall ′, ″↔σh 1,2↔σv (x,y) means symmetry operation transforms x into y (must generate 2D representation using x and y) x Selection rules: integrand must contain totally symmetric representation. p ψ dτ ≠ 0 ∫ ψi O f p ) must include Γ(ψ ) because direct product of any irreducible Γ ( ψi ) ⊗ Γ ( O Direct Product: f representation with itself contains the totally symmetric representation. χ Γ i ⊗Γ j ( ) ≡ χ ( R1 ) χ ( R1 ) , χ ( R 2 ) χ ( R 2 ) ,… i j Example: E′ ⊗ E″ = (4 1 0 –4 –1 0) (the irreducible representations must all be ″) Decomposition of (4 1 0 –4 –1 0): 1 a A2′ = [ 4·1·1 +1·2·1 + 0 – 4·1·1 −1·2·1 + 0 ] = 0 12 1 a E′′ = [ 4·2·1 +1·2(−1) + 0 − 4·1·(−2) −1·2·1 + 0 ] = 1 12 a A1′′ = 1 a A2′′ = 1 So now we know how to work out all selection rules. Best to work specific example of D3h molecule BCl3. i j shortcuts A⊗B=B g⊗u=u ′⊗″=″ 1⊗2=2 r r r 120° planar Generate 3N dimensional representation. E C3 S3 C2(⊥) σh χ = red χ red 12 1 + 2 cos 2π 3 2(1–2) 12 0 –2 E′ A2′′ A2′ = χ + 3χ + 2χ + χ + χ E′′ A1′ 4(2–1) –1+2 cos 2π 3 σv 2(2–1) 4 –2 2 (total of 12 degrees of freedom) 3 translations E′ ↔ (x,y) A′′2 ↔ z A′2 ↔ R z E″ ↔ (Rx, Ry) 3 rotations This leaves 6 vibrations (total of 6) χVIB = χ A1′ + 2χ E′ + χ A2′′ (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 1 mode in a symmetry class) 1 2 ΓRED from 3 (stretches only) χ red = ( 3 0 1 3 0 1) = χ A1′ + χ E′ pure stretch (only A′1) Thus A′1 A′′2 2E′ pure symmetric stretch pure bend (out of plane – because χ(σh) = –1) mixed bend and stretch scissors pseudo rotation one out, two in also pseudo rotation mixed bend and stretch (there is another E′) (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 ⎤⎦ ⊗ ⎡⎣χ 2 ⎤⎦ ⊗ ⎡⎣χ 3 ⎤⎦ ⊗ ⎡⎣χ 4 ⎤⎦ v1 v2 v3 v4 Γ (0,0,0,0) = A′1 A′1 overtones Γ(1 0 0 0) (0 1 0 0) A′′2 Γ (0 0 1 0) Γ E′ (0 0 0 1) Γ E′ E′ ⊗ E′ = ( 4 1 0 4 1 0 ) = ( A′1 + A′2 + E′) fundamentals Γ(2 0 0 0) Γ(0 2 0 0) Γ(0 0 2 0) Γ(0 0 0 2) (all ′) A′1 A′1 A′1 + A′2 + E′ A1′ + A′2 + E′ Γ′ ⊗ Γ″ = Selection rules for fundamental bands A′1 (1 0 0 0) ← (0 0 0 0) (0 1 0 0) A′′2 (0 0 1 0) E′ (0 0 0 1) E′ x, y, or z in order for transition integral to be nonzero, need Γ = Γ′ ⊗ Γ″ mode #1 #2 #3 or 4 A′1 A′′2 E′ IR forbidden z IR allowed (x, y) IR allowed But how will the rotational transitions behave? Qe ≡ 0 inertial axes unit vectors   ⎡ ⎤ ⎛ ∂M j,a ⎞ ( ) ˆ Recall M j (Q ) = a ⎢M j,a 0 + ∑ ⎜ ⎟ Q i + …⎥ ∂Q ⎝  ⎣ ⎦ i ⎠e i ⎡ ˆ +b ⎢M j,b (0 ) + ∑ ⎣ i ⎡ ˆ +c ⎢M j,c (0 ) + ∑ ⎣ i ⎤ ⎛ ∂M j,b ⎞ ⎟ Q i +…⎥ ⎜ ⎝ ∂Q i ⎠e ⎦ ⎤ ⎛ ∂M j,c ⎞ ⎜ ⎟ Q i + …⎥ ∂Q ⎝ ⎦ i ⎠e rotational selection pure rotation rules spectrum rotational selection rule in vibration-rotation band ψ (relect ;Q) ⎛ ∂M j,abc ⎞ ⎜ ⎟ = ⎝ ∂Q i ⎠e ∫ vibrational selection rules Γ(abc) ⎛ ⎡a ⎤ ⎞ ∂ ⎜ *⎢ ⎥ ⎟ ⎜ ψ j ⎢b⎥ ψ j ⎟ dτ ∂Q i ⎜ ⎟ ⎥ ⎢ ⎝ ⎣c⎦ ⎠ Γ Qi totally symmetric be careful if ψelect j is degenerate So for mode BCl3 #1 A′1 #2 A′′2 #3 E′ ∂M j,abc =0 ∂Q1 ∂M j,z ≠0 ∂Q 2 ∂M j,x or y ≠0 ∂Q 3 or 4 an oblate symmetric top z=c x,y = (a,b) mode #2 fundamental is c type (||) mode #3,4 fundamentals are a,b type (⊥) || ⊥ ∆K = 0 ∆K = ±1 weak Q strong Q General procedure 3N 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 Atoms and Molecules. New York, NY: Oxford University Press, 1995. Images removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 1995. The E-A1 energy level diagram is given in Figure 7.51. 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 = ±1 and ∆J = 0, ±1. Note also that for ∆K = +1 the transitions connect to the (+) stack while for ∆K = –1 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 7.52). Each sub-band is separated by approximately 2[A(1–ζ)–B]. This gives rise to a characteristic pattern of nearly equally spaced Q branches (Figure 7.53). Images removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 1995. Images removed due to copyright restrictions. Please see: Bernath, P. F. Spectra of Atoms and Molecules. New York, NY: Oxford University Press, 1995. What if BCl3 were not D3h (planar)? C3v (like NH3) z C2v prolate or oblate, depending on apex angle planar (near oblate) c tee z ↑a planar (near prolate) wye z ↑a Cs planar (3 different bond lengths or 3 different angles) Truth table point group D3h C3v # of normal modes 4 4 C2v 6 pure rotational spectrum type — c-oblate a-prolate asymmetric hybrid a,b or b,c # of IR active fundamentals 3 4 Rotational type of IR fundamentals 1-c, 2-a,b 2-c, 2-a,b 6 3-a, 2-b, 1-c Raman active fundamentals For D3h here will be two ⊥ polarized (E′) fundamentals There will be one || polarized ( A′′2 ) fundamental There will be one forbidden (i.e. not observable by IR) fundamental ( A′1 ) (a,b)-type ⊥ polarized (i.e. ∆K = ±1) oblate top vibrational spectrum (the two E′ modes in BCl3 2 E ROT JK = BJ(J +1) + (C − B)K C−B< 0 Strong Q branches. Q branch “spikes” dominate band profile. K′ − K′′ R Q 0 (J) ≈ (C − B) [12 − 0 2 ] R Q1 (J) ≈ (C − B) [ 2 −1 ] P 2 2 Q1 (J) ≈ (C − B) [ 0 2 −12 ] etc. (since ∆ B is small) R Q2 R R Q1 Q0 P P Q2 Q1 Red Blue (R and P branches are spread out) headless, subbands overlap strongly ν0 ν (c)-type || polarized (i.e. ∆K = 0) oblate top vibrational spectrum. (the one A′′2 out-of plane bend from BCl3) weak Q branches, except at J ≈ K and high K ∆K = 0 only. All QQK(J) tend to pile up as spike at band origin. Q P R (b,c)-type ⊥ prolate top bands (x,y,z)↔(b,c,a) Strong Q branches R Q0 (A – B) 1 R Q1 (A – B) 3 P Q2 P Q1 R Q0 R R Q2 Q1 Red Blue ν0 ν 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 Asymmetric tops: ∆K = 0, weak Q except at low J ≈ K Resemble symmetric top when J ≈ K More complicated because E JKa Kc can’t be separated into J-dependent and Kdependent additive terms. 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 ↔  = 1. || type ∆ = 0 (anti-symmetric stretch), ⊥ type ∆ = ±1 (bend) Benzene E 36 2C6 0 2C3 0 C2 0 D6h 1 (−12 +12 +12 + 36) = 2 24 1 A 2g = ( 36 +12 +12 −12 ) = 2 24 1 ( 36 −12 −12 −12 ) = 0 B1g = 24 1 B2g = ( 36 +12 −12 +12 ) = 2 24 1 [ E1g = 72 − 24 ] = 2 24 1 [ E 2g = 72 + 24 ] = 4 24 1 [ E 2u = 72 + −24 ] = 2 24 1 ( 36 −12 −12 −12 ) = 0 A1u = 24 1 A 2u = ( 36 +12 −12 +12 ) = 2 24 A1g thru atoms ↓ 3C′2 –4 thru bonds ↓ 3C′′2 0 D6 1[ 36 −12 ] = 2 12 1[ 36 +12 ] = 4 12 1[ 36 −12 ] = 2 12 1( 36 +12 ) = 4 12 1[ ] 72 = 6 12 1[ ] 72 = 6 12 h = 24 i 0 2S3 0 2S6 0 σh 12 3σd 0 3σv +4 symmetries and numbers of normal modes 2A1 2g – 1 0u 4A2 2g 2u – 1 2B1 0g 2u 4B2 2g 2u 6E1 2g – 1 4u – 1 6E2 4g 2u 18 18 –3 –3 15 15 translation: A2u(z), E1u(x,y) rotation: A2g(Rz), E1g (Rx, Ry)