Q1(o). The incorrect statement on electronic spectra
Identify the INCORRECT statement and correct it briefly: (i) Electronic transitions occur much faster than vibrational motion. (ii) Franck–Condon principle explains the intensity distribution of vibrational structure of electronic spectra. (iii) Fluorescence always originates from the lowest vibrational level of the excited singlet state. (iv) Phosphorescence involves a spin-forbidden transition.
Answer in brief
(iii) is incorrect. Fluorescence generally comes from the lowest vibrational level of S₁ (Kasha's rule), after vibrational relaxation, but not always: azulene fluoresces from S₂ and in a low-pressure gas emission can come from higher levels.
Key points
- (i) Correct: electronic transition 10⁻¹⁵ s, nuclear vibration 10⁻¹³ s (the Born–Oppenheimer / Franck–Condon basis: vertical transitions).
- (ii) Correct: the intensity is proportional to the Franck–Condon factor | v'|v”|².
- (iii) Incorrect because of “always”. Corrected form: Fluorescence usually originates from v=0 of S₁ (Kasha's rule), since vibrational relaxation (10⁻¹² s) is faster than emission (10⁻⁹ s).
- (iv) Correct: T₁→ S₀ changes the spin multiplicity (Δ S≠0).
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
- (i) Correct: electronic transition 10⁻¹⁵ s, nuclear vibration 10⁻¹³ s (the Born–Oppenheimer / Franck–Condon basis: vertical transitions).
- (ii) Correct: the intensity is proportional to the Franck–Condon factor | v'|v”|².
- (iii) Incorrect because of “always”. Corrected form: Fluorescence usually originates from v=0 of S₁ (Kasha's rule), since vibrational relaxation (10⁻¹² s) is faster than emission (10⁻⁹ s).
- (iv) Correct: T₁→ S₀ changes the spin multiplicity (Δ S≠0).
Coverselectronic spectraThe incorrect statement on electronic spectra
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