Molecular Spectra and Molecular Structure - Lec 31

Page 1 of 11 pages Lecture #31: Electronic Spectra of Polyatomic Molecules Electronic spectra are more complex than vibration-rotation spectra. Polyatomic molecule spectra are more complex than diatomic molecule spectra * Many vibrational properties (such as Franck-Condon factors) can’t be calculated because we know too little about V ′ Q′ and V ′′ Q′′ . ( ) ( ) QUALITATIVE IDEAS LIKE NODE COUNT STILL WORK! * Inter-surface interactions are more subtle in 3N–6 dimensions than in 1 dimension (avoided crossings along high symmetry directions). * Non-radiative processes (other than Predissociation and autoionization) cause spectral lines to broaden and fluorescence quantum yields to approach zero. E.g. Bixon and Jortner J. Chem. Phys. 48, 715 (1968). This subject is so complex that only a case-by-case approach seems feasible. Some crucial background information: 1. Electronic States ← electronic configurations ← MO’s. symmetry labels effect of each orbital on favored molecular shape (i.e. RAB’s and kAB’s and θABC’s) Qualitative MO Theory e.g. Walsh diagrams localization of orbitals onto “chromophores” concept of chromophores is an enormous simplification because: * its orbitals and electronic properties are known from other molecules **** large changes in molecular shape are confined to the region of the chromophore. 2. Electronic Transitions A single orbital promotion, e.g. π*←n, gives direction of transition moment relative to body fixed â , b̂ , ĉ axes ↔ rotational selection rules direct product Γ φi ⊗ Γ φf Often mixed transition type when chromophore does not lie || on ⊥ to a principal axis. 2 i 2 A′ − i E.g. HCO B X A′ ( µ in plane of molecule) AB Often axis-switching effects when â , b̂ , ĉ are not oriented the same way in both electronic states because of a large change in geometry e.g. 1 i A Au 1 i X ∑ +g π*←π in HCCH Page 2 of 11 pages 3. Vibrational Structure Symmetry selection rules for vibrational bands in an allowed electronic transition ΓV ′ ⊗ ΓV ′′ ⊂ Γ totally symmetric   allowed to see: all v of totally symmetric mode alternate (even) v’s of non totally symmetric modes except in combination with other non-totally symmetric modes FRANCK-CONDON propensity rules ∆v = 0 for all normal modes that are unchanged by φ′←φ″ electronic transition. Long progressions in modes exhibiting large change in ω and Qe - just as for diatomic molecules. Only modes localized on the chromophore are expected to exhibit Franck-Condon activity. Massive Simplification! Crucial diagnostic! Symmetry Forbidden Electronic Transitions Γ e′ ⊗ ΓV ′ ⊗ Γ e′′ ⊗ ΓV ′′ ⊂ Γ totally symmetric ( ) ( ) “ vibronic ” symmetry a non-totally symmetric normal mode distorts the molecular framework and causes to be mixed in some character of another electronic state that would give rise to a symmetry allowed electronic transition in the lower symmetry point group. In this case the vibrational intensity depends strongly on the number of quanta in the non-totally symmetric “promoting mode”. 1 1 A A2 − i X A1 transition. See Vibronic coupling handout and the H2CO i 4. Rotational Structure a, b, or c-type selection rules determined by Γe′ ⊗ Γe″ or electronically allowed Γev′ ⊗ Γev″ vibronically allowed crucial clue to symmetry assignment. Large changes in A, B, C rotational constants: * spectra look much more complex than a vibration-rotation band because Q branches are not line-like and do not fall into regular patterns of sub-bands; * crucial key to configurational assignment because the A, B, and C rotational constants tell us about gross changes in equilibrium geometry. Page 3 of 11 pages Two classic examples: 1. The trans-bent π*←π excited state of HCCH [see K. K. Innes, J. Chem. Phys. 22, 863 (1954) and G. W. King and C. K. Ingold, Nature 169, 1101 (1952)] 2. The non-planar π*←n excited state of H2CO Acetylene has a linear ground state HOMO LUMO 14 total e– 1 X ∑ +g 1σ 2g 1σu2 C 1s 2σ2g 2σu2 3σ2g net of 2 C– H bonds H1s, C2s C–C 2 CC σ-bond π-bonds C 2p 1πu4 1πg0 3σ 0u π* some sp polarization Numbering within symmetry type is in order of energy and number of nodal surfaces. Listed in increasing energy order. What are the plausibilities for the upper state of a π*←π transition? linear upper state D∞h cis-bent C2v four possibilities hole in π (π–1) single e– in π* (π–1) in plane ⊥ to plane in plane ⊥ to plane trans-bent Again, 4 possibilities for π–1 and π* orientations. C2h Page 4 of 11 pages π orbital OR vinylidene C2v necessarily in plane of C 1 1 H H or πσ diborane bridged structure D2h H C C H see Bogey Phys. Rev. Lett. 66 413 H (1991) for non-planar Si Si H near trans near cis Non-planar (like HOOH) Unequal CH bond lengths planar non-planar C2 Cs C1 Linear possibility unlikely because π*←π gives ∆u, 3∆u, 1 ∑+u , 3 ∑+u , 1 ∑−u , 3 ∑−u states all of which except 1 ∑+u are forbidden from 1 ∑+g . 1 In CO and N2 the analogous 1 ∑+u state is very high lying and above the first dissociation limit. 1 ∑+u ← 1 ∑+g is expected to have || type rotational selection rules (strong R, P, weak Q) and the observed bands obviously contain strong Q lines. In addition, we do not expect the π*←π states to be linear because a trans-bend distortion mixes the C–H bonding 2σu MO into the π* MO, thereby strengthening the two CH bonds AND DIMINISHING the anti-bonding interaction between the C’s OR a cis-bend distortion mixes the CH bonding 2σg MO into the π* MO, thereby strengthening the two CH bonds. So we expect the and possibilities to be most likely. Page 5 of 11 pages Electronic Spectra of Polyatomic Molecules Last time: Survey of relevant factors ⎧electronic ⎪ * symmetry selection rules ⎨vibrational ⎪ ⎩rotational * propensity rules localization onto chromophores Franck-Condon factors * expected shapes of molecules predictions based on qualitative MO theory proof of one vs. other structure based mostly on group theory The S1 ← S0 transition in HCCH (first to be rotationally analyzed) rules out linear excited state. Today: 1. Why would the HCCH π*←π excited states want to be bent? QMOT (Qualitative Molecular Orbital Theory) 2. rule out 3. support for assignment assignment How do various MOs respond to distortion of the molecular shape: e.g. HCCH? For QMOT, look at books by John Lowe, Gimarc, and papers by Walsh. The following diagram shows how MOs in acetylene change energy as the molecule is distorted from linear to cis-bent (left side of diagram) or linear to trans-bent (right side of diagram). Page 6 of 11 pages C2v cis D∞L linear C2h trans Bu B1 3σ∗u +(–) –(+) A2 B1 and Bg 1π∗g +(–) +(–) ⊥ Ag || and Au B2 1š u A1 Bu A1 Ag 3σg (mostly C–C bonding) B1 2σu (mostly CH bonding) bend ⊥ P no extra CH bonding bend || P extra CH bonding Bu Ag 2σg (mostly CH bonding) Anti-bonding interactions are always slightly more unfavorable energetically than corresponding bonding interactions. A1 b a C2v A1 E 1 C2(z) 1 σv(xz) 1 σv(yz) 1 z A2 1 1 –1 –1 Rz 1 –1 1 –1 Tx, Ry 1 –1 –1 1 Ty, Rx B1 B B B2 B B (a, b, c) = (x, z, y) B Page 7 of 11 pages π* symmetric with respect to σv(xz) anti-symmetric with respect to σ(yz). in plane B1 ⊥ plane A2 π–1 in plane A1 ⊥ plane B2 favors bent π* (B1) neutral π–1 B1 A2 A2 (A1) → B1 a type B2 A1 B2 → → → A2 A2 B1 forbidden forbidden a type slightly bent because π* favors bend slightly more than π did strongly bent linear linear neutral excitation favors linear The only allowed transition is B1 ← A1 (= ∑+g ) which is a-type ∆Ka = 0, weak Q, which does not agree with spectrum. Moreover, the B1 excited states are expected to be either linear or only slightly bent. b B a B Note the tilt (exaggerated as drawn)!!!! C2h Ag E 1 C2(z) 1 σh(xy) 1 i 1 Rz Au 1 1 –1 –1 Tz 1 –1 –1 1 Rx, Ry 1 –1 1 –1 Tx, Ty Bg B B Bu B B (a, b, c) = (x, z, y) π* π–1 in plane Ag ← favors bent ⊥ plane Bg ← neutral in plane ⊥ plane Bu ← favors bent Au ← neutral B favors bent excitation favors Bu linear Bu Au neutral Au neutral × × × × Ag Bg Ag Bg → → → → Bu Au Au Bu x,y = a,b z=c c a,b next to lowest E weakly ben t highest E linear lowest E strongly bent next to highest E linear Page 8 of 11 pages expect to find either c-type or a,b-type 1 Au ← 1 ∑+g strongly bent 1 Bu ← 1 ∑+g weakly bent B c-type ∆Ka = ±1, ∆Kc = even, strong Q a-type ∆Ka = 0, ∆Kc = odd, weak Q b-type ∆Ka = ±1, ∆Kc = odd, strong Q X1 ∑+g linear ground state is special because it has only = 0 (which acts like Ka) in zero-point level. This means that cold bands will sample only limited number of K′a values (0 and/or 1). THIS IS WHY HCCH WAS 1ST POLYATOMIC MOLECULE FOR ROTATIONAL ANALYSIS OF AN ELECTRONIC TRANSITION. Observed bands are probably c-type because they consist of strong K′ – ″ = ±1 sub-bands and contain a strong Q branch. This can be checked because, for a near prolate asymmetric top, the prolate-oblate correlation diagram requires that the higher energy member of each asymmetry doublet have the lower Kc value JKa Kc = J – K a Kc = J – K a + 1 and the magnitude of the asymmetry splitting is controlled mostly by B-C. So we can figure out which rotational branches are naively * allowed for c vs. a,b-hybrid transitions. * Neglecting nuclear spin and rovibronic symmetry requirements. Page 9 of 11 pages Supplement c-type Ka = 1 b-type Ka = 1 a-type Ka = 0 J + 1 1,J J + 1 1,J + 1 J1,J – 1 J1,J ² ′J J+10,J + 1 J – 11,J – 2 J – 11,J–1 1A J0,J J–1 0,J – 1 ² Ka = odd ² Kc = even ² Ka = odd ² Kc = odd “Combination defects” ← a-type R(J) − Q(J) = 2(J +1)B′ + ² ′J b-type R(J) − Q(J) = 2(J +1)B′ − ² ′J Q(J) − P(J) = 2JB′ − ² ′J Q(J) − P(J) = 2JB′ + ² ′J ² Ka = even ² Kc = even no Q What else do we expect? Activity in the C-C stretch (expect frequency near that of ethylene ground state) because π*←π should weaken kCC and lengthen rCC expect ωCC - 1623 cm–1 X -State ν2 in Activity in the trans-bend because the bend angle decreases from 180° (sp hybridization) and the bend frequency nearly doubles. Page 10 of 11 pages sp2 hybridization 120° B B Small activity in symmetric CH stretch because CH might get slightly shorter and stronger (expect ~3100 cm–1). No activity in torsion (< 1000 cm–1), antisymmetric (~ 3100) CH stretch, and antisymmetric in-plane bend (< 1000 cm–1). B All Confirmed! Inertial defect B x IA + IB – IC = 0 is useful criterion for planar molecule. All atoms in xz plane. z Ix = ∑ m i z 2i mi i x Iz = ∑ m i x 2i i C.M. z [x2 + z2]1/2 Iy = ∑ m i (x 2i + z 2i ) i Page 11 of 11 pages HCCH 1 1 i X ∑ +g i A Au trans-bend C—C ν3 ν2 1047.70 cm–1 1389 A0 B0 C0 12.94 1.1243 1.0297 B B rCH rCC θCCH 1 1 1 + − = −0.0044 cm A B C 1.08Å 1.388Å 120° ν4 608.26 ν2 1973.5 B 1.1766 1.058 1.208 180°