1.
Which of the following molecular parameters are likely to change when a molecule is
electronically excited?
(a) ωe (b) ωexe (c) μ (d) De (e) k
All of them, except the (c) μ.
When a molecule is excited, the bonding between the atoms is changed. This will alter every
property of the molecule that depends on the bonding. It will change the potential energy curve. For
example, if the bonding is weaker in the excited state:
(a)
The harmonic frequency, ωe, will be lower. It depends on the force constant (see (e)).
Weaker bonds have lower vibrational frequencies.
(b)
The anharmonicity, ωexe, is likely to be larger as the weaker bond will dissociate more
easily.
(c)
The mass is not changed by strength of the bonding.
(d)
The dissociation energy, De, will be lower in the excited state due to the weaker bonding.
(e)
The force constant, k, depends on the ‘stiffness’ or strength of the bond. This depends on the
bonding.
2.
Consider the four sketches below, each depicting an electronic transition in a diatomic
molecule. Note that more than one answer may be possible.
(a)
Which depicts a transition to a dissociative state?
Figure (d) represents a dissociate state. There is no energy minimum on the top curve.
(b)
Which depicts a transition in a molecule that has a larger bond length in the excited
state?
Figures (b) and (c) show excited states with longer bond lengths than in the ground state. The bond
length is given by the position of the minimum in the well.
The minimum for the potential energy curves is at the same distance in the ground and excited
states for (a). There is no minimum in the potential energy curve for the excited state in (d) as it is
unbound. Only in (b) and (c) are the minima in the potential energy curves for the excited states
displaced to the right (longer bond length) compared to the ground state.
(c)
Which would show the largest intensity in the 0-0 transition?
Figure (a) would show the largest intensity in the 0-0 transition.
The largest intensity corresponds to the where the vertical transition hits the excited states potential
energy curve. Only for (a) does the arrow hit the 0 level in the excited state.
(d)
Which represents molecules that can dissociate after electronic excitation?
Figure (d) and possibly (c) represent molecules that can dissociate after electronic excitation.
In (a) and (b), the electronic excitation is to a low vibrational level in the excited state, well below
the dissociation energy. In (c), excitation is to a higher vibrational level which looks similar to the
dissociation limit. In (d), electronic excitation leads to an unbound state as there is no minimum in
the potential energy curve for the excited state and so dissociation occurs.
(e)
Which represents the states of a molecule for which the v”=0 → v’=3 transition is
strongest?
In both (b) and (c), the vertical arrow starts at v’’ = 0 and hits the excited state as v’ = 3. The
vertical transition gives the strongest absorption.
In (a), the v’’ = 0 to v’ = 0 transition is the strongest, see part (c). In (d), there are no vibrational
levels in the excited state and only a broad and diffuse spectrum is observed.