Molecular Spectra and Molecular Structure - Lec 36

Page 1 of 4 pages Lecture #36: CNPI Group Theory (Final Lecture — used transparencies) What do we do when a molecule is not rigid large amplitude vibrational motions? electronic transitions between states belonging to 2 different point groups. C2v C2h E, C2(z), σv(xz), σv(yz) E, C2(z), σh, i What is a rigorous symmetry? permutations and inversion (lab, not body!) e.g. G8 E, P12, Pab, H1CaCbH2 * P12ab, E*, p12 , p*ab , Cb2v E Cb2 E c 2 C c 2h C σv(xz) σh * p12 ab σv(yz) i Need to keep track of nuclear permutation symmetry (s, a) and total parity (+, –). ↑ Γ* is s– electric dipole transitions are s– too much information in character table * feasible and nonfeasible permutation-inversion operations * irreducible representation labels with zero statistical weight (those that are odd with respect to Pab because 12C has I = 0) Page 2 of 4 pages i –state of C2H2 Experiments relevant to E * 1 photon VUV ordinary absorption ⎧no rotation-vibration cooling difficult because ⎨broad lines due to predissociation ⎩ most useful rotational structure is unavailable FC favors transitions to linear ungerade Rydberg states * < TOF PES favors transitions to linear Rydberg states because they go to linear ion. valence states are bent and predissociate rapidly REMPI (through E or A intermediate state) Logic - Summary I. II. III. IV V. S1 all 3 are electric dipole → not centro symmetric a-type from A Ka = 1 → one electronic symmetry possibility each for cis and trans b-type from Ka = 0,2 non-planar near cis 1A2- with tunneling through cis barrier (explains hot bands via ν′′4 + ν′′5 ) nonzero inertial defect → non-planar tunnel doublets (below barrier) → regular spacing with respect to barrier we claim non-planar near-cis Wilkinson claims non-planar near trans i ←i E X only hot bands v4 = 1, 2, 3, 4, 5 in C2D2 v4 = 1, 2, 3 also hot combinations C2D2 1ν4 + 1ν5, 2ν4 + 1ν5, 3ν4 + 1ν5, 2ν4 + 2ν5 in transitions to origin level Herzberg - linear Herman and Colin planar trans-bent 3 + 1 REMPI to origin fluorescence dip 1φ VUV ← 1+1+1 MPI ← UVUVDR favors linear u states favors linear g states C2H2 sign of asymmetry splitting Page 3 of 4 pages 1. 2. 3. linked cycle of 3 elect. dipole transitions involving centro symmetric i A states → implies X and i ~ E state is non-centro symmetric. rotation structure - large negative inertial defect → non-planar rotation and vibration structure → group theory [ Cb2 symmetry] C2h, C2v point groups are not useful → need true symmetry labels. (+,–) total parity (s, a) H(D) permutation rigorou s all levels of i A are either (s–) or (a+) so final states are all (s+) or (a–) used J = 1, Ka = 1 (a+) v′3 = 2 rQ0(1) possibilities D∞h C2h Cb2v Ca2v non-planar trans cis vinylidene C2c and C2b N N near trans near cis C2D2 expect narrower bands than C2H2 use of || vs ⊥ polarization effects to prove electric dipole transition ~ a -type (∆Ka = 0, ∆Kc = ±1) A Ka = 1 → E one member of each Ka = 1 asymmetry doublet is s, other a (nuclear permutation) energy order dependent on vibronic symmetry → since rotational pattern is universal get Beff values for s vs. a levels a s Ka = 1 (odd J) (opposite for even J) compatible with trans 1Bg or cis 1A2 B A Ka = 0 and 2 (vs. Ka = 1 above) Contrast i see ⊥ selection rules! → b type for sign of asymmetry splitting got A, B, C → 1 φg alternative assignment ruled out * non-centro symmetric * would be OK if different CH’s bond lengths → break (a,s) symmetry and FC activity in antisymmetric CH stretch Page 4 of 4 pages * see all, not alternate v′′4 hot bands * isotope shifts proves that observed level is v = 0 level so = 0 and Ka = 1 (can’t see both K = 1 and 3) OK for cis-bent planar because can have s±, a± (but only 2 in even Ka, other 2 only in odd-Ka) i ←i but would require E X to be A2←A1 (forbidden) find θHCC = 110°, τHCCH = 75° (0° is cis) 1 1 1 − − ≤0 C B A accidental symmetric top → expected for τHCCH = 45° – 135° Ic ≤ Ib + Ia so