Molecular Spectra and Molecular Structure - Lec 32

Page 1 of 8 pages ~ ~ Lecture #32: The H2CO A1A2 ← X1A1 Transition H2CO was 1st (asymmetric top) polyatomic electronic transition to be rotationally analyzed. G. H. Dieke and G. B. Kistiakowsky Phys. Rev. 45, 4 (1934). It is more complicated than linear HCCH because many values of the Ka rotational quantum number can have significant thermal population in the V′′ = 0 level. For a linear molecule in a ∑-state, Ka = 0, and = 0 in V =0. The S1 ← S0 transition in H2CO * is electronically forbidden in C2v * the excited state appears to be non planar - hence C2v may not be relevant. * “quasi-planar” molecule – inversion barrier is low resulting in staggering of bending levels ~ ~ * 3 distinct transition mechanisms, each with own selection rules, contribute to A —X system. Outline: i. ii. iii. iv. v. classification of orbitals and normal modes what do we expect “vibronically” rather than electronically allowed system – false origin, promoter mode surprise in hot band spectrum – peculiar spacings in upper state out-of-plane bend low barrier to inversion through planarity Next time: Vibronic Coupling Body fixed axis system: 0 b y a inertial axes x ** z (b, c, a) = (y, x, z) specific corresponden ce point group axis labels Molecualr orbitals from atomic orbitals: IP CO 14.014eV IP H 13.595 ∴ H’s are above CO HOMO IP C 11.264 IP O 13.614 ∴ σ2s, π2p are polarized toward O C2v A1 A2 B1 B2 E 1 1 1 1 C2(z) 1 1 –1 –1 σv(xz) 1 –1 1 –1 σv(yz) 1 –1 –1 1 B Tz Rz Tx, Ry Ty, Rx B *** see p. 2002 *** “Report on Notation” z is axis of symmetry x ⊥ to plane NB ↑ Page 2 of 8 pages What happens to orbitals in a plane? Figure of Correlation of the orbitals of planar H2 XY to those of the united molecule Y2 and to those of 2H + XY removed due to copyright restrictions. σ → ⎯z⎯ + + – π + ⎧ …… ⎪ + H’s ⎨ ⎪ ⎩ π orbitals ⊥ to plane of paper paper is yz plane + …… → ⎯⎯ A1 → ⎯⎯ → ⎯⎯ A1 → ⎯⎯ A1 ⎯⎯ → B2 → ⎯⎯ B1 B2 B B – – + H B C H B O B Page 3 of 8 pages B H ° C O B H π is necessarily ⊥ to plane B The two “lone pair” or “non-bonding” (n) orbitals on O are and + – + – A1 more stable because avoids e– in C—O region B2 HOMO B B The antibonding orbital, viewed edge-on, is LUMO B HH C O B paper is xz plane HOMO → LUMO is B1 ← B2 π* ← n B1 ⊗ B2 = A2 which is electric dipole forbidden (because A2 does not transform as Tx, Ty, or Tz). B What do we know about ground state? Figure out symmetries of vibrations from ΓRED = [12, – 2, 2, 4] = 4Γ A1 + 1Γ A2 + 3Γ B1 + 4Γ B2 B Vibrations: 3 × A1, 0A2, 1B1, 2B2 (after removing 3 translations and 3 rotations) 1 2e– in each orbital i X A1 A0 9.4053 cm–1 B0 1.2954 C0 1.1343 B 1 A 1 1 + B – C = –0.0033 cm (small inertial defect implies planarity 2766.4 cm–1 ω1(a1) 1746.1 ω2(a1) 1500.6 ω3(a1) 1167.3 ω4(b1) 2843.4 ω5(b2) 1251.2 ω6(b2) rCO = 1.210Å RCH = 1.102Å θHCH = 121.1° sym CH stretch CO stretch scissors out of plane anti-sym CH in plane wag Numbering of vibrational modes, by convention in order of frequency within each symmetry type, symmetries in order A1, A2, B1, B2. ____________________ What do we expect for π* ← n electronic transition, provided that it is observable? Page 4 of 8 pages * lengthen C—O bond * change hybridization on C to sp3 to minimize antibonding interaction! ⎪⎧θHCH → 109° This could cause ⎨ ⎪⎩nonplanar distortion If the molecule becomes non-planar, we can no longer work in C2v point group. The vibrations no longer divide into 3 symmetry species. In Cs we have 4A′ + 2A″ (old modes 1-4) (5,6) mode 4 could become symmetry active in all quanta, not merely even quanta! So whether we see odd and even quanta of ν′4 seems to be a key question. Strongly F–C active modes: definitely CO possibly scissors possibly out-of-plane ω2 ω3 ω4 Observe long progression in 1182 cm–1 must be C—O stretch. (Note that π*←n decreases bond order from 2.0 to 1.5.) Each strong band is observed with a short progression built on it: in 824 cm–1 out -of-plane and 1322 cm–1 scissors and 2872 cm–1 symmetric CH But is 824 cm–1 ω′4 or 2ω′4 ? ⎧ A = 8.69 B = 1.156 C = 1.041 for excited state ⎪ HINT: ⎨ 1 1 1 ~ � ⎪⎩ + − = 0.0195 cm larger inertial defect than in X state. A B C ~1 � suggests much less planar than X A1 state (so if nonplanar → Cs point group → 824 cm–1 is 1 ω′4 not 2 ω′4 . But in C2v the 1A2 ← 1A1 transition is electronically forbidden but it might be vibronically allowed Γ(ev)′ ⊗ Γ(ev)″ = Γa or Γb or Γc start with cold bands Γ (ev )′′ = Γ A1 . Expect that should only find strong transitions built on odd quanta of non-totally symmetric modes: ν′4 (b1 ) ⇒ Γ (ev)′ = Γ B2 = Γ b (b − type)⎤ ⎥ rotational selection rules (a − type)⎦⎥ ν′5 or ν′6 (b 2 ) ⇒ Γ (ev )′ = Γ A1 = Γ a Page 5 of 8 pages a “false origin” ~ (0, 0, 0, 0, 0, 0) level. � So expect to see V′ (0 or 1, v 2, 0 or 1, 1, 0, 0) b-type band s from X Never expect to see V′ = (0, 0, 0, 0, 0, 0) i.e. the 0 00 band. Modes 5 and 6 are not expected to be F–C active. ~� � ~ − X spectrum are observed to be b-type (∆Ka = ±1, Kc = odd). All strong cold bands in A So all looks clear. But is 824 cm–1 ν′4 or ν′4 ? However hot bands and emission bands contain some surprises! Level diagram - for hot bands ~ � -state. ν′′4 is lowest frequency mode in X v′4 = 1 very strange! –1 113 cm V′ = 0 A – 1280 + 1167 cm–1 = A – 113 A X A – 1280 cm–1 observed v′′4 = 1 1167 cm–1 V ′′ = 0 0 These vibrational bands have some contribution from the rotational constants A′ or A″ for K′ = 1 or K″ = 1 included. Page 6 of 8 pages Observed Bending Level Diagram ~1 � A A2 43 B2 407 A2 2 4 Why is this so irregular? 417 41 40 ~1 � X A1 B2 A2 124 43 B1 42 A1 41 B1 1167 cm –1 A1 40 vibronic symmetry ~� 1 Bending potential for A A″ State symmetric double minimum potential “Quasi-non-planar” slightly small large very small non-planar planar Cs C2v get a situation where it is appropriate to work in the higher symmetry group. J. C. D. Brand, J. Chem. Soc. p. 858 (1956) tunneling doubling inversion Page 7 of 8 pages ~ � deduced that zero point energy in inversion potential for A –state is 400 cm–1, top of barrier is 650 cm–1, thus v4 = 1 level lies just below top of barrier. ~1 ~1 � � What mechanisms could make the A A2 − X A1 transition observable? 1. ⎧alternate quanta of ν′4 or ν′′4 Vibronic coupling via the Q4, out-of-plane vibration: ⎨ ⎩b-type rotational selection rules z-component of magnetic dipole 2. Magnetic dipole (A1 ⊗ A2) = A2 = Rz a a-type rotational selection rules [actually observed as very weak a-type bands into opposite inversion doublet component as main bands] 3. ⎧built on ν′5 = 1 or ν′6 = 1 Vibronic coupling via the Q5 or Q6 in plane vibrations ⎨ ⎩c-type bands weak bands of this type also observed. ~ ~ � � − X transition does derive its intensity from 3 mechanisms: The A 1. Vibronic coupling to a 1B1 excited electronic state via the ν4 (b1) promoting mode (this is the only B1 overall vibrational level) 2. Vibronic coupling to a 1B2 excited electronic state via ν5 and ν6 (b2) promoting modes (only B2 overall vibrational levels) 3. magnetic dipole transitions (only totally symmetric vibrations). All 3 symmetric modes are governed by Franck-Condon like intensity factors. In the vibronic coupling cases, we have to worry about the vibrational structure of the state from which intensity is borrowed, times a factor controlling the strength of the vibronic coupling which increases monotonically with the number of odd quanta in the promoting mode. Page 8 of 8 pages A2 has χ(σv) = –1 (odd parity) total parity of A1 vs. A2 states A1 — K=0 3 A2 + A1 ± + 2 – ∓ – 1 + ± + J=0 – J0J J2 J1 J0 prolate J1J–1 JJ–1 J1J J0J JJ oblate K=1 ∓ 211 212 110 111 ± ∓