CHEM106: Assignment 12
Electronic Transitions
1. Which molecular processes correspond to absorption of microwave, infrared,
ultraviolet-visible photons?
Rotational excitation  microwave
Vibrational excitation  IR
Electronic excitation  UV-Vis
2. Electronic transitions of a molecule can be probed by both absorption and emission
spectroscopy.
A. Draw a diagram to illustrate the basic setup of the experimental apparatus for the
electronic absorption spectroscopy and fluorescence spectroscopy, respectively.
a. Absorption Spectroscopy
b. Fluorescence Spectroscopy
B. Explain why the fluorescence spectroscopy is more sensitive than the absorption
spectroscopy.
The key difference is that fluorescence spectroscopy measures the intensity of
fluorescence perpendicular to the direction of incident beam, whereas absorption
spectroscopy measures the ratio of the intensities of two beams (incident and
passed beams). The fluorescence signal is directly proportional to the intensity of
the excitation source. Increase in the intensity of the excitation beam leads to a
subsequent increase in the fluorescence signal, resulting in greater sensitivity.
On the other hand, absorption depends on the ratio of incident (I0) to passed light
(I). So simply increasing I0 also increases I (see above figure).
3. Define the FranckCondon principle, and illustrate it with a diagram.
According to the Franck–Condon principle, an electronic transition is most likely to
occur without changes in the positions of the nuclei in the molecular entity and its
environment. Thus, the electronic transition is a vertical transition. The quantum
mechanical formulation of this principle is that the intensity of a vibronic transition is
proportional to the square of the overlap integral between the vibrational
wavefunctions of the two states that are involved in the transition. As illustrated in
the following diagram, the FranckCondon principle dictates that the transition from
v’’ = 0 to v’ = 2 is the most intense for the absorption spectrum, and the transition
from v’ = 0 to v’’ = 2 is the most intense for the emission spectrum.
4. The carbonfluorine bond in tetrafluoromethane (CF4) is one of the strongest in
organic chemistry, with a bond dissociation energy of 113.0 kcal/mol. Calculate the
maximum wavelength of light capable of breaking the CF bond.
First, we convert E from kcal/mol to J/molecule.
E
113.0 kcal 4184 J
mol


 7.857  10 19 J / molecule .
mol
kcal 6.02  10 23 molecules
According to the Einstein equation, E  h  hc /  . Solve for , and we have
  hc / E 
(6.626  10 34 Js )( 3  108 m / s )
 2.531  10 7 m  253.1 nm .
7.857  10 19 J
Thus, it takes a photon in the UV region to break a CF bond.
5. What is the difference between a singlet (S) state and a triplet (T) state? What is the
selection rule that governs the electronic transitions among S states and T states?
A singlet state is a many-electron state in which all electron spins are paired. Total
spin angular momentum for the singlet state is zero (S = 0), yielding a spin
multiplicity of 2S + 1 = 1. This is commonly the multiplicity of neutral molecules in the
ground state.
A triplet state is a many-electron state in which two electron spins are parallel. Total
spin angular momentum for the triplet state is 1 (S = 1), yielding a spin multiplicity of
2S +1 = 3.
According to the spin selection rule for electronic transitions, S  0 . This means
that singlet–singlet transition is spin allowed, and singlet–triplet transition is spin
forbidden.
6. Consider a molecule that can fluoresce from the S1 state and phosphoresce from the
T1 state. Draw an energy diagram representing absorbance, fluorescence, and
phosphorescence.