Part 11. Photophysics and Molecular Spectroscopy - W.H. Freeman

Photophysics and Molecular
Spectroscopy
11
EXPERIMENT 33
Objective
Introduction
Excited State Properties of 2-Naphthol
Part I: Excited State Acidity Constant
To determine the acidity constants of the ground and lowest electronically excited
states of 2-naphthol in aqueous solution.
The electronic structure of a molecule determines such physical and chemical properties
as its charge distribution, geometry (therefore dipole moment), ionization
potential, electron affinity, and general chemical reactivity. If the electronic structure
of a molecule is changed, therefore, we would expect its physical and chemical
properties to be altered. Such a rearrangement in electronic structure can, in
fact, be brought about (and very rapidly, 10 13 s) if the molecule is raised to
an electronically excited state by the absorption of a quantum of light (a photon)
whose energy matches the gap between the molecular ground and excited-state
energy levels.
For most organic molecules that contain an even number of electrons, the
ground state is characterized by the pairing of all electron spins; the net spin angular
momentum is zero, and this arrangement is called a singlet state. When considered
in terms of molecular orbitals (MO), electornic excitation involves the
promotion of an electron from a filled MO to a higher, vacant MO. This new
orbital configuration, which characterizes the electronically excited state, may be
one in which the two electrons in the singly occupied MOs have opposite spins.
Accordingly, this electronically excited state is also a singlet. The ground and lowest
electronically excited singlet states are often denoted S 0 and S 1 , respectively.
Higher excited singlet states are referred to as S 2 , S 3 , . . . , S n . This experiment
deals with excited singlet states.
Although measurements of the physical and chemical properties of a molecule
in its ground state can be carried out at leisure, more or less (assuming that the
molecule is thermally stable), the examination of these properties in its excited
states is severely hampered because these states are very short-lived. For most
molecules, S 1 states have lifetimes ranging from 10. 6 to 10 11 s. Excited states
are metastable; they undergo decay processes that dissipate the energy they possess
relative to more stable products. For example, the excited state of a molecule
may in general spontaneously return to the ground state via photon emission
(fluorescence), convert electronic excitation into ground-state vibrational
energy (heat), or undergo bond dissociation or rearrangement or a change in elec-

11-2 PART 11 Photophysics and Molecular Spectroscopy
tron spin multiplicity. Because spontaneous emission from an excited state (that
is, fluorescence) often takes place very rapidly, fluorescence can be used as a probe,
or measurement, of excited-state concentration (for example, fluorescence assay).
In addition, fluorescence studies can provide information about the physical and
chemical properties of these short-lived singlet states. This field of experimentation
is called photophysics.
In this experiment, you will determine some ground- and excited-state properties
of the organic molecule 2-naphthol (ArOH).
OH
2-Naphthol (ArOH)
In aqueous solution, ArOH behaves as a weak acid, forming the hydronium ion
and its conjugate base and naphthoxy ion ArO .
ArOH H 2 O 7 ArO H 3 O
It is instructive to measure the acidity constant of ArOH in its lowest excited
electronic state, denoted as K a *, and to compare this value with that of the ground
state K a . This information indicates how the change in electronic structure alters
the charge density at the oxygen atom. The experimental method is best introduced
in terms of the energy-level diagram shown in Figure 1.
Figure 1. Schematic
diagram of the ground
and first excited singlet
state energies of free
naphthol and its
conjugate base, the
naphthoxy ion, in
aqueous solution.
S 1
ArO – *
∆H*
K* a
S 1
ArOH*
h ArO –
h ArOH
S 0
∆H
ArO – + H +
K a
S 0
ArOH
The relative energies of the free acid and its conjugate base (the naphthoxy
ion) are indicated for both the electronic ground (S 0 ) and (lowest) excited (S 1 )
states in aqueous solution. Each anion is elevated with respect to its free acid by
an energy, H and H*, respectively. These are the enthalpies of deprotonation.
Both the ground-state acid and its conjugate base can be transformed to their respective
excited states via the absorption of photons of energy hv ArOH and h ArO .
For simplicity, these absorptive transitions are shown to be equal to the fluorescence
from the excited to the ground states of the acid and conjugate base. (The
ground- and excited-state vibrational levels involved in the transitions are not
indicated.)

We can express the free energy of deprotonation of ArOH in terms of the enthalpy
and entropy of deprotonation and the equilibrium (ionization) constants
and
Experiment 33
G H T S RT ln K a (1)
G* H* T S* RT ln K a * (2)
for the S 0 and S 1 states, respectively. If we make the assumption that the entropies
of dissociation of ArOH and (ArOH)* are equal, it follows that
Ka
11-3
K*
H H* RT ln
a
, (3)
and thus from Figure 1, it can be deduced that
H N A h ArO
H* NA h ArOH , (4)
where h is Planck’s constant. Avogadro’s number N A has been included to put each
energy term on a molar basis. Combining equations (4) and (3) and rearranging,
K*
ln
a
N A h( ArO
ArOH )
. (5)
Ka RT
Thus knowledge of the energy gap between the ground and first excited states
for both the free acid and its conjugate base leads to an estimate of K a *, if K a is
known. The analysis presented here, which accounts for the observed thermodynamic
and spectroscopic energy differences, was first developed by Th. Förster
(1949, 1950). This approach is thus often referred to as a Förster cycle. Equation
(5) can be recast into a more convenient form:
N
pK* a pK a A hc
[˜ArOH ˜ArO ], (6)
2.303RT
where c, the speed of light, has been incorporated in the expression to express
the transition energies of ArOH and ArO into wavenumbers (cm 1 ), a common
spectroscopic energy unit, rather than Hz (s 1 ). The acidity constants are expressed
as pK values. The 2.303 is needed in the denominator because the pK’s
are defined as log 10 K.
The question now is how to obtain the spectroscopic energy difference (˜ArO
˜ArOH ), or ˜, pertinent to the Förster cycle in 2-naphthol. Three approaches can
be considered. The first is to base the measurement on the absorption maxima of
the free acid and conjugate base. The second is to use the fluorescence maxima
of the two species, and the third is to use what is called the 0-0 energies of ArOH
and ArO . The first two methods are somewhat more straightforward. Obtaining
˜ from absorption data has the advantage that the energy difference that is
obtained does not require an instrumental wavelength correction. Although obtaining
˜ from the fluorescence maxima seems simple enough, fluorimeters usually
produce spectra that are distorted by the wavelength response of the monochromator/photomultiplier
combination. Thus, unless these spectra are corrected
for this sensitivity distortion, the recorded fluorescence maxima will depend (al-

11-4 PART 11 Photophysics and Molecular Spectroscopy
though perhaps subtly) on the particular instrument used and thus will not represent
the “true” or absolute spectroscopic properties of the ArOH/ArO system.
The third approach provides the value of ˜ that best represents the energy
difference between S 1 and S 0 implied in the Förster cycle, ˜0-0 . Unfortunately
˜0-0 cannot be determined directly in all cases (such as ArOH), but it can be estimated
from an analysis of both the absorption and fluorescence spectra. It is
shown schematically in Figure 2 for a single species (ArOH or ArO ).
˜
max (fluor)
˜
max (abs)
Figure 2. Schematic
diagram of the
absorption and
fluorescence spectra of a
molecule. It is assumed
that these transitions are
between the same two
electronic states, that is,
S 0 i S 1 .
Fluorescence
˜
00
˜
absorption
Mirror image of fluorescence
Safety Precautions
The absorption and (preferably instrument-corrected) fluorescence spectra are
both plotted on a common energy axis. Furthermore, the spectra are presented
so that each has the same peak maximum value. The point of intersection of these
spectra can be approimated as the 0-0 energy gap between S 0 and S 1 . This approach
does not apply to cases in which the S 0 –S 1 transition is distorted by a
nearby (and especially more intensely absorbing) S 0 to S 2 trnsition.
Whichever method is used to determine ˜, it should be used consistently with
both species ArOH and ArO .
◆
Always wear safety goggles in the laboratory. Ultraviolet light-absorbing
eye protection is required. Some plastic safety goggles do not completely
absorb ultraviolet radiation.
◆
◆
2-Naphthol is an irritant. If you are to prepare the solutions from solid
material, you must wear gloves; if possible, work in a fume hood.
Be sure you have been instructed to use the proper pipetting techniques
when handling 2-naphthol solutions. Never pipet using suction by mouth.
Procedure
◆
If you are to obtain fluorescence spectra, be sure that any ozone produced
by the ultraviolet source is vented. Ozone is a noxious, dangerous gas that
has an acrid odor. If you detect this gas, leave the vicinity, inform your
instructor, and increase air circulation at once.
In this experiment, you will determine K a from the pH dependence of the absorption
spectrum of aqueous 2-naphthol.
1. Obtain the absorption spectra of two solutions of ArOH (each at
2 10 4 M; the actual concentration must be accurately known). In one

Experiment 33
11-5
solution, the free acid must predominate (low pH), and in the other, the
conjugate base must be the major naphthol component (high pH). Use
stock solutions of HCl and NaOH (for example, 0.10 M) to create [H]
and [OH ] of 0.02 M, respectively. It is important that the ArOH
concentration be the same in each case. Label and save these solutions.
Record or display these spectra on a common wavelength axis so that the
spectra overlap.
2. Now obtain absorption spectra of ArOH (also 2 10 4 M) at
intermediate pH values. Adjust pH levels using ammonium chloride buffer
solutions (NH 4 OH/NH 4 Cl), for example, 0.1 M/0.1 M, 0.1/0.2, or 0.2/0.1.
These solutions can be conveniently prepared from 1.00 M stock solutions
of NH 4 OH and NH 4 Cl. Obtain at least three spectra and scan the entire
wavelength range as in step 1. These spectra will indicate both free acid
and conjugate base absorption. If necessary, choose appropriate ratios of
the NH 4 OH and NH 4 Cl stock solutions to produce a satisfactory series of
ArOH/ArO spectra. It is essential that the 2-naphthol concentrations be
identical and accurately known in each case. Immediately after recording
each spectrum, measure the actual pH of the solution using a properly
calibrated pH meter. Label and save these solutions. It is instructive to
display or overlap each spectrum on a common axis. Make sure that the
temperature of the samples is constant (to within 1°C) or otherwise
controlled throughout the experiment. If the bulk ArOH concentration, for
example, [ArOH] [ArO ], is invariant, the spectra, when properly
overlapped, should intersect at a common wavelength called the isosbestic
(equal absorption) point. The presence of an isosbestic point indicates that
there is a closed system (as a function of the variable, pH) consisting of
two species in equilibrium.
3. Using the same solutions in step 1, obtain the fluorescence spectra of
naphthol and its conjugate base. If possible, correct the spectra for the
monochromator/photomultiplier response of the fluorimeter.
4. If possible, display the fluorescence spectra of the intermediate pH solutions
on the same wavelength axis so that the spectra overlap. Again, you should
observe a common wavelength where the spectra overlap. This is sometimes
referred to as an isostilbic (equal brightness) point. Likewise, the presence
of an isostilbic point indicates that there are two excited-state species that,
aside from emitting light, interconvert only with each other.
Calculations
Because the bulk naphthol concentrations [ArOH] 0 are identical in each of the
solutions studied, the following material balance applies:
[ArOH] 0 [ArOH] [ArO ]. (7)
Under the condtioins that (a) both the free acid and conjugate base absorb at
max ArOH, the wavelength of maximum absorption for ArOH, and (b) Beer’s
law holds,
A( max ArOH) ArOH [ArOH] ArO
[ArO ], (8)

11-6 PART 11 Photophysics and Molecular Spectroscopy
where A is the absorbance (for a 1-cm path length) and ArOH and ArO
are the
molar absorptivity coefficients of the free acid and conjugate base at max (ArOH),
respectively. Combining equations (7) and (8) produces
A( max ArOH) { ArOH ArO
}[ArOH] {ArO }[ArOH]0 . (9)
This relation allows [ArOH] to be determined under equilibrium conditions. Once
it is known, the value of [ArO ] at the same pH can be obtained from equation
(7). Because the acidity constant is
[H
K a 3 O ][ArO ]
(10)
[ArOH]
(assuming that the activity coefficient ratio is unity in these dilute solutions), you
can obtain the pK a value for ArOH from a linear regression of pH vs.
log([ArO ]/[ArOH]):
[ArO
pH pK a log ]
. (11)
[ArOH]
You can obtain the values of [ArOH] and [ArO ] for each solution from equations
(9) and (7), respectively. Report the standard deviation in pK a . Compare
your results with literature values.
Calculate pK a * for (ArOH)* from equation (6) using the pK a value you obtained
and whatever method(s) you choose to determine (˜ArOH ˜ArO
). Compare
these values and perform an error analysis. Discuss the errors that the primary
measurements have on the derived value of pK a *.
Other molecules that you can study using this procedure are 2-naphthoic acid,
acridine, and quinoline. Their solubilities in water are limited but sufficient to allow
you to prepare the dilute solutions needed for the experiments.
O
C
OH
2-Naphthoic acid
N
Acridine
N
Quinoline
Questions and Further Thoughts
1. In comparing the values of K a and K*, a what can you deduce about the change in electron
density at the O atom in 2-naphthol in the electronically excited state relative to the
ground state If you have access to a molecular orbital program, such as HyperChem,
MOPAC, or Gaussian, perform a suitable MO calculation on the ground and first excited
singlet states of 2-naphthol. The AM1 Hamiltonian is appropriate, and a configuration
interaction calculation involving at least the highest-filled and lowest-unfilled
molecular orbitals is required for the excited-state property. These calculations will provide
the atomic electronic charges.
2. If pK* a pK a (the excited-state species is a stronger acid), can you comment on the relative
(absolute) magnitudes of the enthalpies of deprotonation See Figure 1.

Experiment 33
11-7
Further Readings
3. On what basis can we justify the assumption that the entropies of deprotonation of
the ground- and excited-state 2-naphthol molecules are equal [see equations (1)–(3)]. Can
you think of another possibility in which this assumption is a poor one
4. Ohter molecules that can be studied using this technique are 2-naphthoic acid, acridine,
and quinoline (see the last paragraph in the Calculations section). Indicate the protolytic
reactions for these molecules, i.e., write the aqueous acid-base reactions.
5. Can you predict before doing an experiment whether the excited state of a molecule
is a stronger or weaker acid than its respective ground state What information would
you need to perform such an assessment
General
P. Atkins and J. de Paula, Physical Chemistry, 8th ed., p. , W. H. Freeman (New York),
2006.
I. N. Levine, Physical Chemistry, 5th ed., pp. 777–780, McGraw-Hill (New York), 2002.
R. J. Silbey, R. A. Alberty, and M. G. Bawendi, Physical Chemistry, 4th ed., pp. 519–522,
Wiley (Hoboken, NJ), 2005.
Förster Cycle and Excited-State Acidity
Th. Förster, Naturwiss., 36:186 (1949).
Th. Förster, Z. Electrochem., 54:531 (1950).
C. Parker, Photoluminescence of Solutions, pp. 328–341, Elsevier (Amsterdam), 1968.
J. Van Stam and J. E. Loefroth, J. Chem. Educ., 63:181 (1986).
A. Weller, in G. Porter, ed., Progress in Reaction Kinetics, vol. 1, pp. 189–214, Pergamon
(New York), 1961.
Fluorescence and Photophysics
J. B. Birks, Photophysics of Aromatic Molecules, Wiley-Interscience (London), 1970.
J. R. Lakowicz, Principles of Fluorescence Spectroscopy, Plenum Press (New York), 1983.
N. J. Turro, Modern Molecular Photochemistry, Benjamin/Cummings (Menlo Park, CA),
1978.

EXPERIMENT 34
Objective
Introduction
Excited-State Properties of 2-Naphthol
Part II: Deprotonation and Protonation Rate Constants
To determine the deprotonation and protonation rate constants of 2-naphthol in
its lowest excited singlet state in aqueous solution.
In the previous experiment, we determined acidity constants for aqueous
2-naphthol (ArOH) for both the ground and lowest excited (singlet) states. These
constants pertain to the equilibrium
k d
ArOH H2 O 0ˆ ˆ9 ArO H 3 O , (1)
k p
where the rate constants for the forward (deprotonation) and reverse (protonation)
reactions are indicated as k p and k d , respectively. A similar equilibrium can
be written for the electronically excited-state species, which is produced via photon
absorption:
k d
ArOH* H 2 O 0ˆ ˆ9 ArO * H 3 O , (2)
k p
in which the values of the forward- and reverse-rate constants may be different
from those in the ground state because of differences in the properties of the
2-naphthol in these two states (for example, different K a values).
We can express the ratio of the concentrations of free acid and the conjugate
base as a function of the pH:
[ArOH]
[ArO ]
log
pK a pH, (3a)
where we use molar concentrations to approximate activities. An analogous
equation,
[ArOH*]
[ArO *]
log
pK a* pH (3b)
applies to excited state species. Equations (3a) and (3b) show that if the pH of
the solution is less than pK a of 2-naphthol (in either electronic state), the free acid
form will predominate over that of conjugate base: [ArOH] [ArO ]. Likewise,
if pH pK a , then [ArO ] [ArOH].
Suppose that by using a suitable buffer, the pH of the medium is established
to be less than pK a but greater than pK a *. The ground state of the system will
then consist primarily of ArOH. Electronic excitation via light absorption will instantaneously
(10 13 s) transform ArOH into ArOH*. We may assume that in
this experiment, the buffer holds the pH of the medium constant during and after
electronic excitation. This is a valid assumption because the number of photons
absorbed per unit volume is much less than the ground-state concentration

Experiment 34
11-9
of ArOH. Thus [ArOH*] [ArOH]; moreover, ArOH* will spontaneously dissociate
to form ArO * in order to establish new equilibrium conditions. Under
these circumstances, [ArO *] must be greater than [ArOH*] because pH pK a *
[see equation (3a)]. In fact, most of the fluorescence observed from ArO * takes
place from species that were formed via ArOH* deprotonation after elecronic excitation.
As excited-state equilibrium is approached, the ArOH* concentration
decreases while that of ArO * increases. The strategy of this experiment in determining
k d and k p is to measure the dependence of [ArOH*] on the pH of the
solution. [ArOH*] is monitored through its fluorescence intensity I f , assuming
that it is proportional to [ArOH*] (see the next section). The pH is established
(and varied) using an ammonium acetate buffer.
Kinetic Analysis
Because in this experiment pH pK a , we consider only light absorption by the
protonated form ArOH:
ArOH h abs ArOH*.
(absorption)
The ArOH* thus produced is, like any excited state, metastable, and when it relaxes
(dissipates the excitation energy) it can undergo a number of different decay
processes, such as
k r
1. ArOH* l ArOH h fluor (fluorescence) rate k r [ArOH*].
k nr
2. ArOH* l ArOH heat (nonradiative decay) rate k nr [ArOH*]
k d
3. ArOH* l ArO * H (aq) (deprotonation) rate k d [ArOH*]
k Ac
4. ArOH* Ac l ArO * HAc (deprotonation via Ac ) rate k Ac
[ArOH*][Ac ]
In addition to radiative (fluorescence) decay (1) and nonradiative relaxation
(2), ArOH* can undergo “unassisted” (3) and “acetate-assisted” (4) deprotonation.
This distinction is significant because the rate of deprotonation will be enhanced
in the presence of the acetate ion in the bimolecular step indicated in
process 4. Undoubtedly, solvent plays a role in the unassisted deprotonation (3),
but this step can be considered pseudo first order in ArOH* because the concentration
of “solvent” is so much larger than [ArOH*]. The reverse steps of the
deprotonation processes, which are bimolecular and proportional to [H 3 O ] and
[HAc] (see steps 3 and 4, respectively), are ignored because under these experimental
conditions, [H 3 O ] and [HAc] are very small.
If the pH of the solution is much lower than pK a * (for example, in the presence
of sulfuric acid), deprotonation by either process is suppressed and fluorescence
from ArOH* predominates. In this case, the fluorescence intensity I f 0 , which
is proportional to the ratio of the rate of radiative decay to the total ArOH* decay
rate, can be expressed
I 0 Ck
f r [ArOH*]
, (4)
k r [ArOH*] k nr [ArOH*]

11-10 PART 11 Photophysics and Molecular Spectroscopy
or, canceling [ArOH*],
I 0 Ck
f , r
, (5)
kr k nr
where C is an instrumental constant.
In contrast, when pH pK a * (but less than pK a ) under NH 4 Ac buffer conditions,
the deprotonation steps become kinetically important, and thus the denominator
of equation (4) will contain the aditional terms k d [ArOH*] and
k Ac
[ArOH*][Ac ]. Therefore, the ArOH* fluorescence intensity, now denoted
as I f , become [see equation (5)]
Ck
I f
r
. (6)
k r k nr k d k Ac[Ac ]
The deprotonation of ArOH* causes a diminution in its fluorescence intensity;
thus I f I f 0 . Assuming that [ArOH*] is identical in all the solutions studied, the
ratio of ArOH* fluorescence intensity in a solution containing sulfuric acid (low
pH) to that containing ammonium acetate buffer (higher pH) is obtained by dividing
equation (5) by (6):
0
I k
r k nr k d k Ac
[Ac f ]
.
If
kr k nr
(7)
0
I f
1 k
d
If kr k nr kr k nr
(8)
We can rearrange equation (7) to a more convenient form,
or
k Ac
[Ac ]
0
I f
If
1 0k d 0 k Ac
[Ac ],
where 0 1/(k r k nr ) and is the “lifetime” of ArOH* in the absence of significant
deprotonation. A plot of ( I 0 f /I f 1) versus [Ac ] should be linear with slope
0 k Ac
and intercept 0 k d . To determine k d , one must obtain 0 from a separate
experiment.
The information provided by this type of study, which involves timeindependent,
or steady-state, measurements could also be obtained directly using
a transient, or kinetic, approach by monitoring the fluorescence decay of ArOH*.
After photoexcitation by a very short pulse of light (10 9 s), the ArOH* fluorescence
intensity follows the first-order decay law
I f (t) [ArOH*] 0 exp{(k r k nr k d k Ac
[Ac ])t}, (9)
where [ArOH*] 0 is the concentration of photoexcited ArOH produced immediately
after excitation (at t 0) and t is the time after excitation. Note again that
equation (9) represents the proportionality between fluorescence intensity and excited-state
concentration. The coefficient of t in equation (9) is the reciprocal of
the fluorescence lifetime of ArOH*, (1/ Ac
), in the presence of the NH4 Ac buffer.

Experiment 34 11-11
The information provided by the steady-state Stern-Volmer-like plot—equation
(8)—could be directly obtained from a plot of 1/ Ac
versus [Ac ]. These experiments
can be carried out using a nanosecond (10 9 s) fluorescence decay
spectrometer.
In this experiment, you will obtain k d using steady-state approach previously
described. You can obtain the value of 0 from the provided data for the time dependence
of ArOH* fluorescence intensity. This information comes from a fluorescence
decay experiment (see the Appendix to this experiment). We emphasize
again that in deriving equations (4) to (8), it is assumed that throughout the series
of fluorescence intensity measurements, first in sulfuric acid and then in different
NH 4 Ac buffer solutions, the concentration of ArOH* is invariant. This
condition requires that the amount of light absorbed by ArOH per unit time be
constant; thus, not only must the formal concentration of 2-naphthol be identical
in each of the samples but also the intensity of the excitation source must not
fluctuate. Satisfying these conditions is crucial for the success of the experiment.
Once you obtain a value of k d from the analysis of the data as discussed, you
can determine the value of k p (the protonation rate constant for the excited state
naphthoxy ion) from K* a because
k
K* a d
(10)
kp
for this set of elementary reactions.
Safety Precautions
◆
◆
◆
◆
Always wear safety goggles; these glasses should block ultraviolet light.
Ordinary plastic safety goggles or glasses may not be effective in absorbing
all the ultraviolet radiation. Check with your instructor.
2-Naphthol is an irritant. If you are to prepare the solutions from solid
material, you must wear gloves; if possible, work in a fume hood.
Be sure you have been instructed to use proper pipetting techniques when
handling 2-naphthol. Never pipet by mouth.
When using the fluorimeter, be sure that any ozone produced by the
ultraviolet source is vented. Ozone is an irritating, dangerous gas that has a
sharp, pungent odor. If you notice this kind of odor during the experiment,
inform your instructor at once. The laboratory must be immediately
ventilated and, if necessary, evacuated.
Procedure Using Scanning Fluorimeter
1. Prepare a series of aqueous solutions in 25-mL volumetric flasks, each
having the same ArOH concentration (about 4.0 10 4 M). One of them
will be 0.1 M in H 2 SO 4 (for I f 0 ), another 0.1 M in NaOH (in order
to observe predominantly ArO * fluorescence). The remainder of the
solutions will have varying NH 4 Ac concentrations between 0.01 M and
0.10 M. Study at least five NH 4 Ac-containing ArOH solutions in this
range. Stock solutions of H 2 SO 4 , NaOH, ArOH, and NH 4 Ac will be
provided. Label each solution and record the ambient temperature.
2. You will be shown how to operate the fluorimeter. Using an excitation
wavelength of 320 nm, obtain the fluorescence spectra of the ArOH

11-12 PART 11 Photophysics and Molecular Spectroscopy
solutions starting with the solution containing H 2 SO 4 . Next, obtain the
fluorescence spectrum of the NaOH-containing solution. Then proceed with
the NH 4 Ac-containing solutions in order of increasing NH 4 Ac
concentraion. It is recommended that you display these spectra on a
common wavelength scale. If the excitation source remains steady and the
solutions have the same bulk ArOH concentrations, a distinct isostilbic
(equal brightness) point should be produced. This is the common
wavelength point of all the ArOH fluorescence spectra. See Experiment 34
for a further discussion of the isostilbic point.
Data Analysis
3. If a fixed-wavelength fluorimeter is used or if time permits, perform a
second determination of data at the emission wavelength of the fluorimeter
corresponding to the maximum for the protonated ArOH* (360 nm).
First, place the H 2 SO 4 -containing sample in the cavity and establish the
“maximum” intensity setting on the chart paper by moving the pen close to
its full displacement (or just read and record the signal strength). Carefully
and systematically obtain values of the fluorescence intensity of the
buffered solutions in order of increasing NH 4 Ac concentrations. If there is
any doubt about the constancy of the instrument, remeasure I f 0 and
compare it with the original value. If the agreement is unsatisfactory, you
will have to repeat the series of measurements.
1. Using the time-dependent fluorescence intensity data provided in the
Appendix for ArOH in 0.10 M H 2 SO 4 , determine 0 . The fluorescence
quantum efficiency of ArOH under these conditions has been determined
to be 0.18; this quantity is equal to k r 0 . Report values of both k r and k nr
for ArOH.
2. Tabulate I f 0 and the I f and corresponding [Ac ] values for the samples, and
construct a Stern-Volmer-like plot indicated by equation (8). Determine
values of k d and k Ac
and their standard deviations using linear regression.
Consider the error in 0 [obtained in step (1)] in your error analysis of the
rate constants.
3. Determine k p from your previously obtained value of K a * [see equation (10)].
4. Tabulate all the rate constants determined and include estimates of their
respective uncertainties.
5. Using the Stokes-Einstein-Smolouchowski (SES) equation for a diffusioncontrolled
rate constant (see Experiment 22), calculate k diff :
8000RT
k diff (dm 3 mol 1 s 1 ),
3
where is the solvent viscosity. The indicated units for k diff are obtained if
R and are in SI units, that is, 8.314 J K 1 , and in N m 2 s, respectively
(1 cP 10 3 N m 2 s). Compare your values of the second-order rate
constants k d and k Ac
with kdiff . You can use the viscosity of water at the
appropriate temperature for this comparison. The SES equation applies to
the reaction between two neutral species (or a neutral and a charged
reaction pair). For anion–cation pairs (each of single charge), the rate
constant is about an order of magnitude larger than k diff .

Experiment 34
11-13
Questions and Further Thoughts
Further Readings
1. The pH of the aqueous medium is assumed to be unchanged as a result of electronic
excitation of 2-naphthol. Why is this justified
2. Can you predict how the value of the deprotonation rate constant of ArOD* (ArOD)
would differ from that of ArOH* (ArOH) What would you need to know to answer
this question
3. Can you suggest an experiment in which the deprotonation rate constant of ArOH
(ground state) could be determined
4. Ozone is sometimes produced by ultraviolet light sources. What is the mechanism by
which ozone is produced Can you suggest a way in which the ozone production by a
given ultraviolet source can be eliminated
5. As the source of acetate ion you used ammonium acetate. Why is this compound chosen
What would be the advantage or disadvantage of using sodium acetate
2-Naphthol Protolysis
R. Boyer, G. Deckey, C. Marzzacco, M. Mulvaney, C. Schwab, and A. M. Halpern, J.
Chem. Educ., 62:630, 1985.
J. Van Stam and J. E. Loefroth, J. Chem. Educ., 63:181 (1986).
Appendix
Fluorescence and Photophysics
J. R. Lakowicz, Principles of Fluorescence Spectroscopy, Plenum Press (New York), 1983.
N. J. Turro, Modern Molecular Photochemistry, Benjamin/Cummings (Menlo Park, CA),
1978.
Fluorescence Decay Data for 2-Naphthol
in 0.10 M H 2 SO 4 at 25°C
Time (ns)
Intensity*
0.00 21753
1.00 18907
2.00 16380
3.00 14171
4.00 12432
5.00 10757
6.00 9288
7.00 8138
8.00 7083
9.00 6014
10.00 5350
*Photons emitted per unit time.

EXPERIMENT 35
Objective
Introduction
Enthalpy and Entropy of Excimer Formation
To determine the enthalpy and entropy of formation of the dimer between ground
state and electronically excited pyrene molecules in solution.
We know that the physical and chemical properties of a molecule are dictated by
the nature of its electronic structure. It is not surprising, therefore, that such properties
as acidity, dipole moment, and chemical reactivity may change considerably
on electronic excitation, in which there is a very rapid change in the characteristics
of a species’ electronic wavefunction. In this experiment, you will
investigate the pairwise interaction between pyrene molecules. We are interested
in determining how the interaction potential between two ground-state molecules
differs from that between an electronically excited molecule and its ground-state
counterpart.
Consider, then, the approach of two pyrene molecules along an axis perpendicular
to their molecular planes. If both molecules are in the ground electronic
state, there will be a very weak van der Waals attraction at moderate distances.
As they become closer, of course, significant intermolecular repulsion occurs. If,
however, one of the molecules is in its electronically excited state (as a result of
light absorption) and is allowed to approach a ground-state species with the appropriate
orientation, a stable dinner will form. This dimer, whose structure is
proposed to have a “sandwich” configuration, is formed reversibly; that is, it can
thermally dissociate back into an electronically excited pyrene molecule (excited
monomer) and a ground-state species. 1 This unusual dimeric species is stable as
long as it possesses electronic excitation but dissociates as soon as this electronic
excitation is dissipated and the system returns to the ground state. Such a species
is called an excimer (from excited dimer).
Pyrene
Pyrene crystal dimer
A bound state that arises from the interaction between two dissimilar species,
one in its electronically excited state, is sometimes called an exciplex (excited complex).
Like an excimer, an exciplex is dissociative in the electronic ground state.
Because the excimer forms only when an electronically excited molecule and a

Experiment 35
11-15
ground state molecule come into specific and close contact with each other, excimer
fluorescence is often used as a probe of molecular interactions. For example,
pyrene end-capped polymers are used to study the conformational structure
and dynamics of alkane chains. Excimer (or exciplex) emission is also used as a
probe of the intercalation or binding of aromatic molecules into the crevices of
certain macromolecules.
Figure 1 illustrates the potential energy diagram for electronic transitions and
excimer formation. At large intermolecular separation (which pertains to low concentration),
electronic transitions involving the excited monomer (absorption and
fluorescence) are shown. The involvement of molecule vibrations in these transitions
is also indicated. Although the monomer absorption and fluorescence spectra
show vibrational structure (usually involving a skeletal mode), excimer emission
is structureless. This is characteristic of electronic transitions (absorption or
emission) between bound and unbound states.
B
Figure 1. Potential
energy of two pyrene
molecules as a function
of intermolecular
separation. The
coordinate r describes
the approach of the
molecules having the
equilibrium orientation
of the excimer. The
right-hand transitions
represent the isolated
monomer (M).
V(r)
h D M00
h M
R
r
The maximum in the excimer emission spectrum corresponds to transitions
between the minimum of the excimer potential well (which exists at the equilibrium
separation between the excimer components) and the unbound, ground-state
dimer having the same intermolecular separation as the excimer. Hence, excimer
formation involves the creation of a bound species subsequent to photon absorption.
Conceptually, this is the reverse of photodissociation, in which light absorption
results in bond breakage. Moreover, photon emission in the excimer (fluorescence)
brings about molecular dissociation, since the repulsive ground-state
pair rapidly dissociates.
The potential energy diagram in Figure 1 can also be used to compare the
emission energies of monomer and excimer in terms of the excimer binding energy
B and the ground-state repulsion energy R. Thus,
h M h D B R, (1)
where h M represents the monomer fluorescence energy (called the 0-0 energy because
the transition takes place between excited- and ground-state monomers pos-

11-16 PART 11 Photophysics and Molecular Spectroscopy
sessing zero quanta of vibrational energy), and h D is the energy of the excimer
fluorescence maximum. The binding energy of the excimer is just the negative of
its enthalpy of formation, that is, B H (the system is studied at constant
pressure).
Kinetic Scheme
Excimer formation (called photoassociation) and reversible dissociation can be
represented by the following set of elementary processes, or mechanism.
Process Equation Rate Constant
Electronic excitation M h abs M* (rate I abs )
Photoassociation M* M D* k DM
Excimer dissociation D* M M* k MD
Excited monomer decay M* M h M k fM
M* M k iM
Excimer decay D* (M 2 )h D k fD
D* 2M k iD
M
hv abs
k fM
M k DM
M*
D*
kiM k MD M k iD k fD
M hv M M 2M 2M hv D
In this scheme, I abs is the number of moles of photons (einsteins) absorbed by groundstate
monomer per cubic decimeter per second. For convenience, k M and k D are defined
as the total intrinsic (intramolecular) decay rate constants of monomer and excimer,
respectively, independent of photoassociation or dissociation:
k M k fM k iM and k D k fD k iD . (2)
k DM is the second-order rate constant for excimer formation from M and M*
(DM denotes “dimer from monomer”), and k MD refers to the first-order rate constant
for the dissociation of the excimer into M* and M (MD denotes “monomer
from dimer”). The formation rate of excited monomer (in mol dm 3 s 1 ) is
d[M*]
dt
I abs (k M k DM [M])[M*] k MD [D*], (3)
while that for the excimer is
d[D*]
dt
k DM [M][M*] (k D k MD )[D*]. (4)
Note that k DM [M] is the pseudo first-order rate of formation of excimer from excited-
and ground-state monomers. Also, it is implied that [M*] [M]; that is,

Experiment 35 11-17
a very small fraction of monomer is electronically excited. In experiments using
conventional light sources (arc lamps, not focused lasers) this is indeed the case.
Under conditions of steady-state illumination, the rates in both equations (3)
and (4) are equal to zero (photostationary conditions). The ratio of [D*] to [M*]
can be thus obtained:
[D*] kD
M[ M]
. (5)
[M*] k D kMD
It stands to reason that since the objective of this experiment is the determination
of thermodynamic quantities, we must determine the temperature dependence
of the excimer formation equilibrium constant. Therefore, the first thing
we need to do is to obtain the equilibrium concentrations of M, M*, and D*.
First, we will assume that the equilibrium ground-state concentration [M] is equal
to the bulk pyrene concentration, consistent with the inequality [M*] [M] discussed.
The second crucial (and fundamental) assumption is that the fluor (or fluorescent
species) concentration ([M*] or [D*]) is proportional to its respective
fluorescence intensity. This is a basic tenet of spectrofluorimetry and is perhaps
analogous to Beer’s law in absorption spectrophotometry.
This experiment is based on the possibility of obtaining the ratio of excimer
to excited monomer concentrations from the measured fluorescence intensities,
each measured at the appropriate wavelength. Both the fluor concentrations are
assumed to be proportional to the respective integrated fluorescence intensities
(areas under the monomer and excimer fluorescence spectra),
I fM k fM [M*] and I fD k FD [D*], (6)
where the proportionality constants are the radiative rate constants. The dimensions
of I f are einsteins dm 3 s 1 (an einstein is 1 mol of photons). Combining
equations (5) and (6),
I k
fD k DM [M]
fD
. (7)
IfM kfM (k D k MD )
Equation (7) is central to the application of photostationary techniques to the
study of photoassociation. (The distinction between the integrated and “instantaneous”
fluorescence intensities will be discussed later.)
Because we seek thermodynamic data, we examine the temperature dependence
of equation (7). First, we make the general observation that both the formation
and dissociation rate constants are temperature-dependent and can be expressed
in Arrhenius form:
E
k DM A DM exp R
DM
T
and k MD A MD exp E
MD
R
T ,
(8)
where the A’s and E’s are the preexponential factors and activation energies, respectively.
Because intermolecular excimer formation is restricted only by molecular
transport, or diffusion, the activation energy E DM is related to the activation to
viscous flow of the solvent medium. E MD , however, reflects the intrinsic strength of
the “excimer bond” as well as the energetics of molecular diffusion. This is because
E MD represents the activation energy for the dissociation ofthe bound excimer
into separated and individually solvated excited- and ground-state constituents.

11-18 PART 11 Photophysics and Molecular Spectroscopy
If we assume that the temperature dependence of k MD , the rate constant for
thermally activated excimer dissociation, is much larger than that for k D , the intrinsic
decay rate of excimer (indeed, for many systems, k D is nearly temperatureindependent),
k MD k D in the limit of high temperature. Applying this result
to equation (7),
I fD
IfM
k
fD k DM [M]
(high temperature). (9)
kfM k MD
If we furthermore make the general assumption (which has been confirmed by
measurements of several monomer/excimer systems) that both radiative rate constants
(k fM and k fD ) are temperature-independent, an Arrhenius plot of I fD /I fM
yields, in the limit of high temperature, a slope
I
d ln
f D
I f M (E
DM E MD )
(high temperature). (10)
1
d R
T
See equations (7) and (8). Because E MD E DM , the desired binding energy (the
negative of the enthalpy of formation) of the excimer can be determined as the
slope of such a plot.
At low temperatures, where excimer dissociation is slow compared with its intrinsic
decay rate (k MD k D ), we can write equation (7) as
I fD k fD k DM [M]
(low temperature). (11)
IfM
kf Mk D
An Arrhenius plot of the left-hand side of (7) will have a slope approaching
I
d ln
f D
I f M
1
d
T
E DM
(low temperature). (12)
R
E DM has been found to be closely related to the activation energy associated with
molecular diffusion (hence bulk viscous flow) in the solvent medium. This is one
of the reasons that excimer formation is interpreted as a diffusion-controlled
process.
An Arrhenius plot of equation (7) illustrates the behavior of the monomer and
excimer fluorescence spectra over a wide temperature range (Figure 2). At low
temperature, the ratio of excimer to monomer emission is small because excimer
formation is impeded by the high viscosity of the solvent; indeed, in a glass or
rigid medium, excimer formation does not take place because of the inability of
excited-monomer and ground-state species to diffuse. (It is possible, however, that
weak van der Waals dimers may be stable at low temperatures and that emission
from these directly photoexcited dimers may be observed. Strictly speaking, this
is not excimer emission.) At very high temperature, excimer emission is also suppressed
because of the high dissociation rate of the excimer. Thus there is an intermediate
temperature at which there is a maximum excimer emission intensity

Figure 2. Arrhenius plot
of the ratio of excimer
to monomer
fluorescence intensities.
The low-temperature
regime (right) reflects
the activation energy of
excimer formation,
whereas the hightemperature
regime
(left) indicates the
“equilibrium” enthalpy
of excimer formation.
In
l fD
l fM
Experiment 35
k k fM I fD
K eq DM
.
kMD kfD I fM [M]
(13)
11-19
High T
1/T
Low T
k DM [M] k M and k MD k D ,
relative to the excited monomer. This temperature depends not only on the nature
of the excimer system (its binding energy and entropy of formation) but also
on the solvent characteristics, such as its viscosity and activation to viscous flow.
Another very important condition, which is pertinent to the high-temperature
limit and is crucial to this experiment, concerns the fact that if interconversion
between photoexcited monomer and excimer is rapid relative to the intrinsic decay
rates of M* and D*, these species are in dynamic equilibrium. Thus with
K eq [D*] eq /[M*] eq [M] eq , and we can express the true equilibrium constant (we
assume unit activity coefficients) for photoassociation as the ratio of the formation
and dissociation rate constants. Using this result and rearranging equation
(9),
The high-temperature or dynamic equilibrium regime can be depicted by an
analogy to two leaky containers connected to each other by two tubes, each containing
a pump (Figure 3). The containers are filled with a liquid. If the leak rate
A
B
A
A
B
B
Figure 3. Hydrodynamic
model of reversible
reactions between two
metastable
(electronically excited)
species.

11-20 PART 11 Photophysics and Molecular Spectroscopy
in container A is small compared with the rate of transport of liquid from container
A to container B and the leak rate of container B is small relative to the
flow rate from B to A, the amounts of liquid present in each container depend
only on the A l B and B l A flow rates. This situation thus represents the dynamic
equilibrium state; the volumes of liquid in A and B are intrinsic to the
plumbing between them. On the other hand, if one of the leak rates is too large,
the volume of liquid in that container will be depleted and will not reflect the
“equilibrium” condition.
The link between K eq and the enthalpy and entropy of excimer formation is
RT ln K eq G° T S°, (14)
and if we combine the expressions in equations (13) and (14), we can represent
the temperature dependence of I fD and I fM as
I fD
IfM
k fM
S°
R
H°
RT
ln
ln . (15)
kfD [M]
Assuming that k fM and k fD are temperature-independent, a plot of ln (I fD /I fM )
versus 1/T should be linear, having a slope equal to H°/R. The intercept can
also provide a value of S° if we know the ratio of the radiative rate constants
of the excimer and monomer. While this information can be obtained from fluorescence
lifetime and efficiency studies, Stevens and Ban have described a photostationary
fluorimetric technique for obtaining S° and H°. 2
High-temperature conditions lead to another interesting consequence. Although
the distribution between D* and M* changes with temperature in this
regime, the total concentration of excited states (D* M*) remains constant. In
other words, M* and D* form a “closed system.” Hence,
[M*] T [D*] T constant [M*] 0 , (16)
where the temperature dependence of the excited monomer and excimer concentrations
is explicit. The spectrometric significance is that there is a specific wavelength
between the maxima of the monomer and excimer emission spectra at
which the emission intensity is independent of temperature. This position, called
the isostilbic (equal brightness) point, is analogous to the isosbestic point (equal
extinction) seen in absorption spectra when there is linear relationship between
the concentrations of two absorbing species.
Proceed with the fluorimetric analysis, we must consider the important distinction
between the total molecular fluorescence intensity I fM (area under the
spectrum) and the fluorescence intensity at a specific wavelength, f M . The latter
is a function of energy or wavenumbers [f M (˜) represents the monomer emission
spectrum], whereas the former is a consatnt at a given temperature and monomer
concentration. This distinction is important, because in this experiment, we measure
fluorescence intensities of the monomer and excimer emission spectra at specific
wavelengths (for example, the respective maxima) rather than the total areas
under the spectra. The relationships beween the total intensity and the
instantaneous intensity are
I fM C f M (˜)d˜ and I fD C f D (˜)d˜. (17)

Experiment 35
11-21
C and C are instrumental constants that link the integrals of instantaneous intensities
with the molecular fluorescence strengths.
One problem with the analysis presented thus far is that we need the ratio of
the excimer and monomer radiative rate constants k fD /k fM [see equations (7), (9),
and (11)]. Although the information can be obtained independently from fluorescence
lifetime and (absolute) quantum efficiency measurements, Stevens and
Ban have described an approach that gets around this problem. Observe in equation
(15) that the value of k fD /k fM is needed only to determine the entropy of
photoassociation (that is, an absolute intercept).
Although the method for obtaining this information from fluorimetric measurements
is straightforward, the development of the result is presented in the
Appendix to this experiment. The approach is as follows.
Let R D ° and R M ° be the recorded intensities of excimer and monomer at their
respective maxima. In the high-temperature regime, a plot of R D ° against R M ° (for
differrent temperatures) is expected to be linear with a negative slope (a/b).
Thus, as the temperature increases, the excimer intensity decreases with a concomitant
increase in the monomer intensity. The relation used to obtain the entropy
and enthalpy of excimer formation is
R D °
RM °
b
a
S°
R
H°
RT
ln
ln [M] . (18)
Safety Precautions
◆
◆
◆
◆
◆
◆
◆
Safety glasses that block ultraviolet light must be worn during this
experiment.
Wear protective gloves when working with pyrene. Do not allow the
material or its solutions to come in contact with the skin.
Make sure you are familiar with proper pipetting techniques. Never pipet
by mouth.
If solid pyrene or its solutions come in contact with the skin, immediately
wash the affected area with soap and water.
A cylinder containing N 2 at high pressure is used. Be sure it is securely
attached to a firm foundation. A reducing valve is used to deliver the N 2 at
a pressure slightly above ambient (5 psig). Do not change this pressure.
The fluorimeter may produce ozone. Make sure the ultraviolet source is
vented. If you notice a sharp, pungent odor, inform your instructor
immediately.
The experiment must be performed in an open, well-ventilated laboratory.
Procedure
Prepare 25 mL of a 5.0 10 3 M solution of pyrene in methylcyclohexane. The
pyrene should be of high purity (and should be sublimed or recrystallized before
use if necessary). The solvent must have a negligible fluorescence background and
preferably should be of spectrometric quality.
Add sufficient solution to a fluorescence cell that is equipped with a gas-tight,
PTFE (Teflon) stopper. Deaerate by bubbling a fine, gentle stream of pure dry N 2

11-22 PART 11 Photophysics and Molecular Spectroscopy
through the solution for 4–6 min. Control the gas flow so that solution is not expelled
from the cell.
(Deaerating with N 2 is essential because it displaces dissolved oxygen, which
significantly quenches the pyrene fluorescence. This procedure is an application
of Henry’s law; the air above the liquid is replaced by an atmosphere of nitrogen,
and thus the solubility of oxygen approaches a very low value commensurate
with the residual oxygen composition in the nitrogen used. Bubbling of the
nitrogen through the liquid hastens gas-liquid equilibrium.)
Immediately after deaerating, firmly stopper the fluorescence cell. Wipe the
four windows (cleaning with ethanol if necessary) using lens paper or the equivalent;
be sure not to scratch the cell. Place the cell in the fluorimeter cavity.
Your instructor will show you how to obtain emission spectra using the particular
fluorimeter and how to vary and measure the temperature of the sample.
Use an excitation wavelength of 370–380 nm. When the sample is exposed to
the ultraviolet radiation, you should be able to see the bright blue-violet fluorescence.
Scan the emission between 370 and about 500 nm. The number of spectra
you obtain will depend on the available time, but you should acquire at least
five spectra, taken between 60 and 100°C. The temperature sholuld remain constant
(to within 0.5°C) during a given scan.
If possible, display these spectra on a common wavelength axis. Preferably,
and if time permits, run spectra over a wider temperature range, e.g., 0 or 20 to
100°C. Note that even if the excitation lamp fluctuates or drifts during the experiment,
the data are still valid as long as this change does not occur during a
particular scan.
Data Analysis
Tabulate in a spreadsheet the rcorded fluorescence intensities of the monomer and
excimer (R M ° and R D °) at each temperature; choose convenient wavelengths (for
example, at or near the respective maxima).
Plot R M ° against R D ° and determine b/a from the linear portion (corresponding
to the high-temperature regime).
From equation (18) determine S° and H° for the pyrene excimer formation
in methylcyclohexane using the values of b/a and [M].
Determine the binding energy of the pyrene excimer as well as the repulsion
energy of the ground state pyrene pair having the excimer geometry [equation
(1)].
If you achieved low enough temperatures in the experiment, estimate the activation
energy to excimer formation and compare it with the activation energy
of viscosity of the solvent (or similar hydrocarbon).
Estimate the errors in S°, H°, B, and R.
Questions and Further Thoughts
1. What effect would solvent polarity (for example, acetonitrile versus cyclohexane) have
on the transition energies of the monomer and excimer and the binding energy of the excimer
Consider the same effect(s) on an exciplex, for example, pyrene (or anthracene)
and N,N-dimethylaniline.
2. The “excimer” laser is based on transitions between the bound state formed with rare
gas and halogen atoms [for example, (ArF)*, XeCl)*] and the dissociative ground state
of these species. One of the advantages of an excimer laser is that after emission takes
place from the excited complex, the ground state is rapidly removed because it is dissociative.
This allows a significant population inversion to be achieved, which enhances lasting
efficiency.

Notes
Further Readings
Appendix
Experiment 35
3. Does the following statement make sense The ground state of an excimer is an excited
state.
4. The probability that the ends of a linear chain oligomer (or polymer) come into close
proximity can be probed by attaching pyrene probes to the ends of the molecule (endcapped
polymers). How does such a technique work What type of experiment is needed
to gather information about the end-to-end interactions
5. The fluorescence of some excimers can readily be observed only at moderately low
temperatures (and high concentrations), yet at very low temperatures, where the solvent
becomes glassy or crystalline, excimer fluorescence nearly vanishes. How can you explain
these observations
6. Suppose the pyrene excimer were formed in the gas phase, and each ground-state pyrene
molecule formed as a result of excimer fluorescence had a recoil energy. Estimate the
speed of the recoiling molecules. Describe the trajectory.
7. Reducing the dissolved O 2 concentration by bubbling the solution with N 2 (deaeration)
is an application of Henry’s law. Explain how deaeration works in the context of
Henry’s law.
8. Suggest some ideas that account for the fluorescence quenching effect by dissolved O 2 .
1. J. B. Birks, Photophyscis of Aromatic Molecules, pp. 301–371, Wiley-Interscience (London),
1970.
2. B. Stevens and M. I. Ban, Trans. Faraday Soc., 60:1515 (1964).
f i
m
C
d
I fM C I fD . (A3)
11-23
J. B. Birks, D. J. Dyson, and I. H. Munro, Proc. Roy. Soc. A, 275:575 (1963).
B. Stevens, “Photoassociation in Aromatic Systems,” in Advances in Photochemistry, ed.
J. N. Pitts, Jr., G. S. Hammond, and W. A. Noyes, Jr., vol. 8, pp. 161–226, Wiley-
Interscience (New York), 1971.
The fluorescence intensity at the isostilbic point (at wavenumber I) is composed of the
separate contributions of monomer and excimer emission:
f i f iM f iD .
(A1)
Each emission is, in turn, proportional to its respective integrated fluorescence spectrum:
f iM m f M (˜)d˜ and f iD d f d (˜)d˜). (A2)
After combining equations (17), (A1), and (A2), we have for the isostilbic intensity,
Using equation (6), which expresses the absolute fluorescence intensities in terms of radiative
rate constants and excited-state concentrations, and equation (16), which states
the “closed system” constraint of the high-temperature limit, equation (A3) becomes
which can be factored to give
m
d
f i k fM [M*] C k fD {[M*] 0 [M*]}, (A4)
C
d
C
m
C
f i k fD [M*] 0 [M*] k fM k fD
. (A5)
d
C

11-24 PART 11 Photophysics and Molecular Spectroscopy
Becauase both f i and the first term of the right-hand side of equation (A5) are independent
of temperature (in the high-temperature regime) and [M*] is temperature-dependent,
the bracketed term in equation (A5) must be zero. Thus, d/m k fM /k fD , and combining
this result with equations (A2) and (17) furnishes the relation
f iD
f iM
dI k fM I fD
fD
,
mIfM ffD /I fM
or
(A6)
I fD
IfM
.
The importance of equation (A6) is that we can now express the enthalpy and entropy
of photoassociation in terms of the experimentally determined contributions of
monomer and excimer fluorescence to the isostilbic intensity. Combining equations (A6)
and (15), we have
f
iD
f iM [M]
k
fD
kfM
S°
R
H°
R
ln
. (A7)
The problem is to find a way of decomposing f i into f iM and f iD . We again follow
the Stevens and Ban procedure. To understand this approach, we must make the distinction
between the absolute fluorescence intensity (f iM , referred to earlier), and the observed
instrumental response, or signal strength (R iM ). Because the combination of the analyzing
monochromator and photomultiplier tube in the particular fluorimeter used has a
unique, nonuniform wavelength dependence, a given emission spectrum will be distorted
by an instrumental sensitivity function S(), which can be expressed
f iD
f iM
R() [S()][I()],
(A8)
where R() is the actual wavelength distribution of the instrument response to a radiant
source of known wavelength distribution I().
Various experimental techniques can be used to determine the sensitivity function
S(). For example, a standard emission source having a known output spectrum [absolute
power density as a function of wavelength, I()] may be used to calibrate the analyzing
system. These sources are often a quartz halogen tungsten lamp—a black-body radiator—
for the near-infrared and visible region and a deuterium lamp for the ultraviolet range.
The observed emission profile of the lamp is obtained using the particular fluorimeter
R(), and the sensitivity function is then obtained by performing a point-by-point division
of R() by I(). This calculation can easily be achieved if the R and S data are acquired
digitally. Many fluorimeter manufacturers furnish these calibration data with the
particular instrument.
At the isostilbic point, the observed emission intensity R i is composed of contributions
from excited monomer and excimer at that wavelength:
R i R iD R iM .
(A9)
Because the sensitivity function S i is the same for both species at a common wavelength
(for example, the isostilbic point), the ratio of the response contributions will be equal to
that of the absolute intensities:
f R
iD
iD
. (A10)
f iM RiM
Now, we relate these response values to those at different reference wavelengths at
which it is assumed only excimer and excited monomer, respectively, emit (these reference
wavelengths may be taken to be the maxima of the monomer and excimer emission
spectra; see Figure 1). Referring to the observed instrumental responses at these reference

wavelengths as R D ° and R M ° , respectively, we can assume the following proportion between
R iD and R D °, and so on:
R iD a R D ° and R iM b R M °, (A11)
where the constants a and b are temperature-independent. Combining equations (A10)
and (A11) gives
f a R D °
iD
. (A12)
b RM °
The ratio a/b can be obtained from a plot of R M ° vs. R D ° in which the set of {R M ° , R D ° } is
obtained over the temperature range for which k MD k D and k DM [M] k M (the hightemperature
regime). This is justified by using equations (A9) and (A11):
Differentiation of R M ° from R D ° gives
R i constant a R D ° b R M °.
d R M °
d RD °
a
(high temperature). (A13)
b
Hence, a/b can be obtained by plotting R M ° vs. R D ° in this way, the needed fluorescence
intensity ratio can be obtained from the observed instrumental response data. See equation
(A12).
The final working result in the Stevens-Ban method, obtained by combining equations
(A7) and (A12), is equation (18):
R D °
RM °
f iM
11-25
b
a
S°
R
H°
RT
ln
ln [M] .
Experiment 35

EXPERIMENT 36
Objective
Introduction
Rotational-Vibrational Spectrum of HCl
To obtain the rotationally resolved vibrational absorption spectrum of vapor
phase HCl and to determine its molecular constants.
In this experiment you will examine the energetics of vibrational and rotational
motion in the diatomic molecule HCl. The experiment involves detecting transitions
between different molecular vibrational and rotational levels brought about
by the absorption of quanta of electromagnetic radiation (photons) in the infrared
and near-infrared regions of the spectrum. The introductory discussion consists
of three parts: (1) vibrational and rotational motion and energy quantization, (2)
the influence of molecular rotation on vibrational energy levels (and vice versa),
and (3) the intensities of rotational transitions.
Vibrational Motion
Consider how the potential energy of a diatomic molecule AB changes as a function
of internuclear distance. A simple way of expressing this change mathematically
is
V(x) 1 ⁄2 kx 2 , (1)
where the coordinate x represents the change in internuclear separation relative
to the equilibrium position; thus, x 0 represents the equilibrium configuration
(the position with the lowest potential energy). For x 0 the two atoms are closer
to each other, and for x 0 the two atoms are farther apart. k is called the force
constant and its value reflects the strength of the force holding the two masses to
each other. Equation (1) is a consequence of Hooke’s law, which states that the
force between two linked masses is proportional to the displacement of the two
masses from their equilibrium position: F(x) kx. The minus sign indicates that
the restoring force is opposite in direction to the displacement.
Equation (1) is referred to as the harmonic potential, and the application of
Newtonian (or classical) mechanics to this problem results in a description of the
behavior of the masses called simple harmonic (or periodic) motion. The system
is commonly referred to as the harmonic oscillator. The harmonic potential, which
is parabolic, is shown in Figure 1. According to this model, the steepness of the
potential is determined by the force constant k. This simple picture of molecular
energetics predicts that the potential energy of a diatomic molecule increases symmetrically
with respect to bond elongation or compression and does so without
limit. Chemical intuition, however, suggests that this description cannot be entirely
true. It is correct that as the nuclei are brought very close to each other,
strong repulsive forces are brought to bear and these forces increase sharply with
decreasing x. In contrast, as the nuclei are pulled apart, the chemical bond is
weakened and eventually broken, and the potential energy therefore becomes constant
as the two “free” (unassociated) atoms are produced. Thus the harmonic
potential is unrealistic for large internuclear displacements. This inadequacy of
the harmonic potential is referred to as anharmonicity and is discussed in the next
section.

Experiment 36
11-27
Figure 1. Potential
energy curve for a
harmonic oscillator,
where x represents
displacements from the
equilibrium position. The
horizontal lines indicate
the eigenvalues (total
energies).
v(x)
0
x
The application of quantum mechanics to the harmonic oscillator reveals that
the total vibrational energy of the system is constrained to certain values; it is
quantized. The expression for these energy levels (or eigenvalues) is
E v (v 1 ⁄2)h e (joules), (2)
where v 0,1,2,3, . . . is the vibrational quantum number, h is Planck’s constant
(6.62608 10 34 J s), and e is the harmonic frequency (s 1 ), which is equal to
( 1 ⁄2)(k/)1 ⁄ 2. k is the force constant [see equation (1)], and is the reduced mass
of the oscillator, i.e., m A m B /(m A m B ). Notice that the lowest eigenvalue
(v 0) is nonzero: E 0 ( 1 ⁄2)h e . This is the zero point energy of the oscillator. It
represents the residual vibrational energy possessed by a harmonic oscillator at
0 K; it is a “quantum mechanical effect.” The eigenvalues E v , which represent the
levels of total energy (kinetic and potential) of the harmonic oscillator, are superimposed
on the potential energy function in Figure 1. As indicated by equation
(2), these levels are equally spaced with a separation of h e .
It is convenient to express the harmonic frequency in a unit called the wavenumber
˜ (nu tilde) or reciprocal centimeter cm 1 . The wavenumber, which is the quotient
of the frequency, (s 1 ) and the speed of light c (in cm s 1 ) widely used in
atomic and molecular spectroscopy. Using e ˜ec, we can express equation (2)
as
E v (v 1 ⁄2)hc˜e (joules). (3)
For diatomic molecules, ˜e is typically on the order of hundreds of thousands of
wavenumbers. For example, for I 2 and H 2 , ˜e values (which roughly represent)
the extremes of the vibrational energy spectrum for diatomic molecules) are 215
and 4403 cm 1 , respectively.
Next, let us introduce an important and necessary complication to the harmonic
oscillator: anharmonicity. Rather than describing the potential energy function
as parabolic [equation (1)], we will use a more realistic potential, the Morse
function, which takes into account molecular dissociation at large values of x.
The Morse potential is represented analytically as
V(x) D e [1 exp(ax)] 2 . (4)
A plot of this function is shown in Figure 2. In equation (4), D e is the potential
energy (relative to the bottom of the well) at infinite A-B separation (x ), and
a is a constant that, like k in equation (1), determines the curvature of the po-

11-28 PART 11 Photophysics and Molecular Spectroscopy
Figure 2. Potential
energy curve for an
anharmonic oscillator
(the Morse function).
The vibrational levels
become more closely
spaced at higher
energies. A harmonic
potential with its energy
levels is also shown for
comparison.
v(x)
0
x
tential well and hence reflects the vibrational frequency; in fact a k/(2D e ) 1/2 .
The use of the Morse potential instead of the harmonic potential results in the
following expression for vibrational eigenvalues:
E v (v 1 ⁄2)hc˜e (v 1 ⁄2) 2 hc˜e e (joules), (5)
where ˜e is the harmonic frequency (as above) and ˜e e is called the a nharmonicity
constant. It is a molecular constant that, for the Morse oscillator, is equal to
ha 2 (4). Because the anharmonicity term in the eigenvalue expression (5) is multiplied
by (v 1 ⁄2) 2 , the spacing between eigenvalues rapidly becomes smaller
for higher v (and ˜e e 0). As the total energy approaches the dissociation limit,
the levels become very closely bunched together. The harmonic potential and its
energy levels are also shown in Figure 2 so that these characteristics can be compared
with the Morse oscillator. For relatively low excitations (only a few vibrational
quanta), the Morse potential provides a good description of molecular vibrations.
Rotational Motion
Consider the rotation of the molecule AB. In the absence of external electric or
magnetic fields, the potential energy is invariant with respect to the rotational coordinates;
rotational motion is isotropic (independent of spatial orientation). If
the AB bond length is assumed to be constant, or independent of rotational energy,
then AB is called a rigid rotator. A quantum mechanical treatment of this
simple (but not unreasonable) model gives the eigenvalue expression
E J J(J 1)hcB e (joules), (6)
where J is the rotational quantum number (J 0,1,2, . . . ) and B e is the rotational
constant (cm 1 units), which is equal to
I e is a molecular constant called the moment of inertia, which for a diatomic molecule
is I e r e 2 , and and r e are the reduced mass m A m B /(m A m B ) and the
equilibrium bond length of the molecule AB, respectively. Note that r e is the inh
B e . (7)
8 2 cI e

Experiment 36 11-29
ternuclear separation for which x 0 in equations (1) and (4) (that is, the value
of x at the bottom of the potential well). For the general case in which r is considered
a variable (a nonrigid rotor) I r 2 , and thus the rotational constant is
a function of r:
h
B . (8)
8 2 cr 2
Notice that the rotational constant contains structural information about the molecule.
In fact, the determination of molecular structure can be achieved with high
precision from studies of rotational transitions, usually in the gas phase (microwave
spectroscopy).
One complication that must be considered is that molecules are, in fact, not
perfectly rigid. A chemical bond may be weak enough so that the nuclei respond
to increasing rotational energy by moving farther apart. In this case, the molecule
experiences centrifugal elongation: r e increases with increasing J. The rotational
eigenvalues for the nonrigid rotator are approximated by
E J J(J 1)B e J 2 (J 1) 2 D, (9)
where D is called the centrifugal distortion constant. From equation (9) we can
see that the effect of nonrigidity is to bring the rotational levels closer together
for higher energy states. Although in principle this complication is expected to alter
the energy levels implied by the rigid rotator model, the effect is often very
small and can sometimes be neglected, and as a result the spacing between rotational
levels increases with J. For example, D is typically several orders of magnitude
smaller than B e , and therefore centrifugal distortion is significant only for
highly rotationally excited states (large J). D can be expressed in terms of B e and
˜e:
4B e
3
D (cm 1 ) (10)
˜ 2
e
Rotational and Vibrational Motion
If rotational and vibrational motion were completely separable, that is, if molecular
vibrations had no effect on rotational states and vice versa, the total energy
of a rotating, vibrating, nonrigid diatomic molecule would be expressed as the
sum of equations (5) and (9):
E v, J (v 1 ⁄2)hc˜e (v 1 ⁄2) 2 hc˜e e J(J 1)hcB e
J 2 (J 1) 2 hcD (joules). (11)
However, the complete separation of rotational and vibrational motion is not
justifiable. Because the mean internuclear separation depends on the degree of vibrational
excitation (v), the rotational constant depends on 1/r 2 [see equation
(8)], and therefore the “effective” rotational constant for a vibrating molecule will
vary with the mean value of 1/r 2 for the vth eigenstate, 1/r 2 v . Taking this effect
into account, we can approximate the rotational constant as
B v B e e (v 1 ⁄2) (cm 1 ), (12)

11-30 PART 11 Photophysics and Molecular Spectroscopy
where B v is the rotational constant that accounts for vibrational excitation and
e is defined as the rotational-vibrational coupling constant. It turns out that for
an anharmonic potential (for example, the Morse potential), e 0, whereas for
the harmonic potential, e 0. This is evident in the expression for e :
6B e
2
e
1/2 1 (cm 1 ). (13)
˜e
˜e e
Be
Notice that for an anharmonic oscillator, B v decreases as v increases. Also observe
that according to equation (13), B v B e even for the case of no vibrational
excitation (v 0). Thus B e , r e , and ˜e are intrinsic molecular parameters that are
referenced to the bottom of the potential well (see Figures 1 and 2); they are independent
of vibrational energy. Finally, in applying this treatment of coupled
rotational-vibrational motion, we replace B e in equation (11) by B v [equation 12)]
and write
E v,J (v 1 ⁄2)hc˜e (v 1 ⁄2) 2 hc˜e e J(J 1)hcB e
(14)
(v 1 ⁄2)J(2J 1)hc e J 2 (J 1) 2 hcD
(joules).
Equation (14) shows that the total energy contains a “pure” vibrational contribution
(containing only v), a “pure” rotational term (containing only J), and
vibrational-rotational coupling containing both v and J, as well as the centrifugal
distortion constant. We emphasize again that equation (14) represents the energy
of a vibrating, rotating molecule; furthermore, the molecular constants ˜e,
˜e e , B e , e , and D are all expressed in cm 1 . We will now express the energy of
the coupled rotational vibrational state in cm 1 . To do this we divide both sides
of equation (14) by hc, and replace E v, J /hc by the quantity T v, J , called the term
value, which expresses the rotational-vibrational state energy expressed in
wavenumbers.
E v, J
T v,J (v 1 ⁄2)˜e (v 1 ⁄2) 2 ˜e e J(J 1)B e hc
(15)
(v 1 ⁄2)J(J 1) e J 2 (J 1) 2 D (cm 1 ).
Equation (15) is the key expression you will use to interpret and analyze the experimental
data.
Experimental Method
Rotational energy levels are much more closely spaced than vibrational levels.
For example, for HCl the spacing between the lowest two rotational energy levels
(J 0 and J 1) is about 20 cm 1 , whereas the gap between the lowest vibrational
level (v 0, ground state) and the next highest one ( 1, first vibrational
excited state) is about 2900 cm 1 . The arrangement of rotational and
vibrational levels is shown schematically in Figure 3.
Because we wish to examine transitions between different vibrationalrotational
levels, the spectrometer must be able to resolve the rotational “fine
structure” in the vibrational spectrum. The absorption spectrum involving the
lowest energy vibrational transition, v 0 to v 1, called the fundamental, is
observed in the infrared region at about 2900 cm 1 , or 3400 nm. [The wave-

Experiment 36
11-31
v
2
J
0
3
2
1
1
0
3
0
2
1
0
Figure 3. Schematic representation of rotational and
vibrational (rovibronic) eigenstates and eigenvalues
denoted by J and v, respectively.
length of a transition is the reciprocal of its energy in cm 1 ; the nm (10 9 m) is
a common wavelength unit.] Infrared spectrophotometers that are based on dispersive
optics, that is, a grating or a prism used to separate the radiation into
narrow wavelength “bands,” generally do not possess the required resolution for
this experiment. Fourier transform (FT) instruments, however, which operate on
a different principle, 1 are usually capable of providing resolution of 0.5 cm 1 ,
which is satisfactory for this experiment.
An acceptable and interesting alternative to using an FT infrared spectrophotometer
is to use a near-infrared instrument that has the desired resolution
(1 cm 1 ). Such an instrument is an NIR-VIS-UV spectrophotometer (nearinfrared,
visible, ultraviolet) that uses dispersive optics. In this case, we examine
not the fundamental transition but the next higher one: v 0 (ground state) to
v 2 (second excited vibrational state). This transition is called the first overtone.
This nomenclature is in analogy with sound waves, or a vibrating string, in
which the first overtone occurs at half the wavelength (twice the frequency) of
the fundamental, that is, longest-wavelength vibration. The energy range in which
this transition occurs in HCl lies just below the visible region of the spectrum
from about 5300 cm 1 (1900 nm) to 5900 cm 1 (1700 nm); this is in the nearinfrared
(NIR) region.
One complication that arises in examining the overtone transition is that it is
much weaker than the fundamental. It has a much smaller molar absorptivity coefficient;
in fact, the overtone transition is formally forbidden and is made weakly
allowed because of anharmonicity. For HCl, the absorption intensity of the overtone
transition is less than 2% that of the fundamental. For this reason, a higher
concentration and/or longer optical path length of HCl vapor is needed to make
the experiment practical. In this case, HCl at 1 atm pressure in a 10-cm pathlength
cell is used. In studying the fundamental transition in the “ordinary” infrared
region, a pressure of a few tens of torr in a 10-cm cell is needed.
NOTE: The discussion that follows treats both the fundamental and the first
overtone transitions. Equations that pertain to these transitions are identified by
the letters a and b, respectively.
The fundamental relation in spectroscopic transitions is derived by equating
the photon energy with the difference in state energies involved in the transition.
For the case where radiation of frequency is absorbed, the energy of the transition
is
E h hc˜ E v, J E v, J ,

11-32 PART 11 Photophysics and Molecular Spectroscopy
where ˜ is the photon frequency in cm 1 and v,J and v,J are, respectively, the
vibrational/rotational quantum numbers of the upper and lower energy states involved
in the transition. (The use of single and double primes to denote the upper
and lower states, respectively, is traditional notation in spectroscopy.) Expressed
in wavenumber units, the preceding equation is
˜ T 1, J T 0, J (fundamental), (16a)
˜ T 2, J T 0, J (overtone), (16b)
in which the fundamental and overtone transitions are made explicit (v1;
v0) and (v2; v0), respectively.
Equations (16a) and (16b) imply that the frequencies of the fundamental and
overtone transitions are not precisely equal to the vibration-only energy gap,
vv, but are instead altered by the simultaneous changes in rotational energy.
Logically, we can anticipate three generael possibilities: an increase, a decrease,
or no change in rotational quantum number. In proceeding further, we introduce
an important (and simplifying) constraint in the way in which J changes as a result
of photon absorption. This restriction is called a selection rule, and for pure
rotational transitions it states that
J 0
J 1
is forbidden,
is allowed,
and all other combinations of J values are forbidden. For vibrational transitions,
the selection rule requires that v 1; this rule is rigorous only for the harmonic
oscillator. Anharmonicity causes the vibrational selection rule to break
down, and thus the v0 v2 transition is weakly allowed.
Returning to the rotational-vibrational spectrum of HCl, we can group the
transitions into two classes. In the first, there is a decrease in rotational quantum
number (J 1, or JJ1), and in the second case, there is an increase in
rotational energy (J 1, or JJ1). The allowed transitions are depicted
schematically in Figure 4. The J 0 transitions are shown by dashed lines. Each
vertical line in Figure 4 is expected to give rise to an observable rotational band
2
1
J = 0 v = 1
Figure 4. Rotationalvibrational
transitions
between v 0 and v 1
(fundamental). Dipole
allowed transitions are
for J 1 (R) and J
1 (P). J 0 forbidden
transitions (Q) are
indicated by dashed
lines.
2
1
J = 0
P Q R
v = 0
J 1 J 0 J 1

Experiment 36 11-33
in the “high” resolution vibrational absorption spectrum of the molecule. The
lines for which J 1 are all positioned at lower energy, and they are called
the P-branch. The lines corresponding to J 1 are situated at higher energy
and are referred to as the R-branch. If J 0 transitions were allowed (they are
under some circumstances and are called the Q-branch), they would “pile up” in
a narrow grouping of lines between the P and R branches. The Q-branch would
appear as a single line only if the rotational constants for the upper and lower
vibrational states were identical.
The rotationally resolved overtone and fundamental transitions can be expressed
as a function of rotational quantum numbers J (ground state, v 0) and
J (upper state, v 1 or v 2). This is done by combining equations (16a), (16b)
and (15). After grouping terms containing B e , e , and D e ,
T 1, J T 0, J ˜ 2˜e e B e [JJ 1) J(J1)]
(17a)
e
[3J(J1) J(J1)] D[J2 (J1) 2 J 2 (J1) 2 ]
2
T 2, J T 0, J2˜e 6˜e e B e [J(J 1) J(J1)]
(17b)
e
2
[5J(J1) J(J1)] D[J2 (J1) 2 J 2 (J1) 2 ].
These relations are simplified under the constraint of the selection rule for rotational
transitions J 1; therefore, we can write the transition energy in terms
of a common rotational quantum number, which we choose to be J.
For the P-branch, JJ1, where J1,2,3, . . . , and equations (17a) and
(17b) become
˜P(J) (˜e 2˜e e ) (2B e 2 e )J e (J) 2 4D(J) 3 ,
˜P(J) (2˜e 6˜e e ) (2B e 3 e )J2 e (J) 2 4D(J) 3 .
(18a)
(18b)
Equations (18a) and (18b) show that the energies of the P-branch lines decrease
nonlinearly with increasing J.
For the R-branch, JJ1, where J0,1,2, . . . and equations (17a) and
(17b) read
˜R(J) (˜e 2˜e e 2B e 3 e 4D)
(2B e 4 e 12D)J( e 12D)(J) 2 4D(J) 3 ,
(19a)
and
˜R(J) (2˜e 6˜e e 2B e 5 e 4D)
(2B e 7 e 12D)J(2 e 12D)(J) 2 4D(J) 3 .
(19b)
Note that the transition energies of the R-branch members increase (also nonlinearly)
with increasing J. Above a certain J value, the R-lines will begin to decrease
in energy because of the negative coefficient of the (J) 2 - and D-containing
terms. For HCl, this does not occur until J 27.

11-34 PART 11 Photophysics and Molecular Spectroscopy
If the Q-branch (JJ) were observable in this experiment, its components
would appear at
˜(J) ˜e 2˜e e e [J(J) 2 ]
for the fundamental
and
˜(J) 2˜e 6˜e e 2 e [J(J) 2 ]
for the overtone
and would be displaced to lower energies. In the absence of rotational vibrational
coupling ( e 0), the Q-branch would appear as a single line at an energy equal
to the gap in the vibrational levels, v 0, v 1 (or v 0, v 2). For convenience,
this gap is defined as
˜0a ˜e 2˜e e for the fundamental, (20a)
˜0b 2˜e 6˜e e for the overtone. (20b)
Determination of the Molecular Constants
We see from equations (18a) and (18b) and (19a) and (19b) that the P- and
R-branch lines (abbreviated P1, P2, P3, . . . and R0, R1, R2, . . . , respectively)
are cubic in J. Thus, in principle, if we were to fit the observed lines to these
equations, we could obtain the four constants ˜0, B e , e , and D. Note that we
cannot directly determine the anharmonicity constant ˜e e unless both the fundamental
and first overtone spectra are acquired. To obtain a reasonably reliable
value of D, you must obtain data that range to high enough J values in the P-
and R-branches to fully characterize the cubic nature of equations (18a,b) and
(19a,b). (You should be able to make assignments from about P12 to R12.)
It would appear that equations (18a,b) and (19a,b) are redundant because they
contain the same constants. However, it turns out that if you were to analyze the
P- and R-branch data separately you would obtain slightly different values for
the molecular constants. Furthermore, when analyzed separately, the P- and
R-branch data do not usually provide reliable values of D because their maximum
J values are not usually large enough.
It is possible, however, to pool the P- and R-branch information and analyze
the data globally, that is, simultaneously. Doing so requires us to combine equations
(18a,b) and (19a,b) into one relationship. To accomplish this, we define the
index N as a new variable such that
and
N J for the P-branch, (21)
N J1 for the R-branch. (22)
By substituting J from either equation (21) or (22) into equations (18a,b) or
(19a,b) for the P- and R-branch expressions, respectively, we obtain
˜ ˜0a 2(B e e )N e N 2 4DN 3 ,
˜ ˜0b 2(B e 3 e )N 2 e N 2 4DN 3 .
(23a)
(23b)

Experiment 36 11-35
You can use equations (23a) and (23b) to analzye your experimental data. It
is therefore important to understand how to assign N values to your P- and
R-branch assignments. Suppose, for example, that the lowest-energy P-branch line
is P12 and the highest-energy R-branch line is R12. The P12 feature is assigned
the index N 12, as indicated in equation (21), and N is increased by 1 for
each subsequent P-branch line. Note that N is not assigned a value of 0. For the
lowest-energy R-branch line, R0, N has a value of 1 and increases by 1 until
N 13 for R12.
Notice that although you cannot determine ˜e or ˜e e from the data, you could
obtain these values by analyzing both the fundamental and overtone transitions.
In this case you would use equations (23a) and (23b) to obtain ˜0a and ˜0b and
then solve for ˜e and ˜e e from equations (20a) and (20b). The Solver utility in
Microsoft Excel is capable of finding the regression values of all parameters from
a global analysis of the P- and R-branch assignments.
Transition Intensities
We now examine the intensity dependence of the P- and R-branches. We can
write the integrated absorption strength I 0, J ;1, J of a particular rotationalvibratinal
transition (fundamental) or I0 , J ;2,J (overtone) as
I J , J ˜ N J M J , J 2 , (24)
where ˜ is the frequency (photon energy) of the transition, N J is the (relative)
population of molecules in the Jth rotational eigenstate, and M J , J is called the
transition moment integral. In equation (24), only the initial and final rotational
quantum numbers J and J are indicated because we are interested in the intensity
pattern of rotational structure within a given vibrational transition (overtone
or fundamental). In the range of the overtone (or fundamental) transition ˜ varies
very slightly (for example, from 2600 to 3100 cm 1 for the fundamental and
6000 to 6500 cm 1 for the overtone transition). Notice that, all things being
equal, the transition intensity depends directly on the frequency of the transition.
Thus as a rule of thumb, pure rotational transitions are intrinsically much weaker
than vibrational transitions, which, in turn, are weaker than electronic (visible
and ultraviolet) transitions. The population factor in equation (24) indicates the
fraction of molecules that exists in the absorbing rotational energy level denoted
by J at thermal equilibrium. Specifically,
E J
N J g J exp, (25)
kT
where g J , called the degeneracy factor, is equal to (2J1); E J is the energy of
the Jth rotational quantum level E J J(J1)hcB e ; and k is the Boltzmann
constant. The exponential term in equation (25) is called the Boltzmann factor.
The transition moment integral in equation (24) reflects the degree to which
the electromagnetic radiation field interacts with the initial and final rotational
quantum states J and J. There is a slight but important J dependence to the
square of this integral:
M J , J 2 JJ1
. (26)
2J1

11-36 PART 11 Photophysics and Molecular Spectroscopy
This information can be used to predict the shapes of the P- and R-branch
structure in the rotational-vibrational spectrum of HCl. Combining equations
(25), (26), and (24),
and
J(J1)hcB
e
kT
I J ˜ J exp for the P-branch, (27)
J(J1)hcB e
kT
I J ˜(J1)exp
for the R-branch, (28)
Safety Precautions
Procedure
Note that in equation (27), J1,2,3, . . . whereas in equation (28) J0,1,2,3,
... . These equations predict that the P- and R-branches have somewhat different
line profiles.
◆
◆
◆
◆
◆
Safety glasses must be worn during this experiment.
Fill the cell in a fume hood. If you are using a pressurized HCl cylinder
and a gas-handling vacuum rack, make sure you understand the filling
procedure.
If the HCl supply develops a leak or the vacuum system breaks or
malfunctions, notify your instructor immediately.
Do not use or handle liquid nitrogen until your instructor has explained its
hazards and proper use.
If the cell filled with HCl breaks outside the fume hood, notify your
instructor immediately.
1. Your instructor will describe the procedure to follow in filling the
absorption cell and will also tell you the type of spectrum to be obtained
(fundamental or first overtone). The procedure will determine the nature of
the cell [Infrasil or NaCl (or KBr) windows], the HCl pressure used to fill
the cell, and the type of spectrophotometer (infrared or near-infrared).
2. If you are to use the vacuum rack, the instructor will review its design and
use. Make sure there is sufficient liquid nitrogen in the trap (dry ice should
not be used). A 10-cm absorption cell will be filled with about 40 torr of
HCl vapor for the fundamental spectrum and about 1 atm for the overtone
spectrum. Lightly grease the cell ball joint and attach it to the rack.
Evacuate the cell and pump it for several minutes to remove adsorbed
water; if necessary, dry the cell using a heat gun (CAUTION!). Evacuate
the gas line connecting the HCl cylinder to the manifold. After isolating the
manifold, fill the system with HCl vapor until the pressure gauge reads the
appropriate value. Carefully but firmly, close the cell stopcock. Close the
HCl cylinder valve(s). Pump out the manifold, gas-filling line, and the
station to which the cell is attached. Shut off the gas inlet and cell station
stopcocks and remove the gas inlet tubing and the absorption cell; wipe off
the stopcock grease from the joints.

Experiment 36
11-37
3a. Obtain an absorption spectrum of HCl vapor under optimal instrumental
resolution. Your instructor will explain these operational details. If
possible, record the spectrum as absorbance versus cm 1 to simplify
subsequent data reduction. For the fundamental transition, the region
2500–3200 cm 1 should be displayed.
3b. An NIR spectrophotometer operating in the near-infrared can be used to
obtain the overtone spectrum. Howeer, the resolution of this
spectrophotometer should be less than 2 cm 1 . Acquire the spectrum
between about 1700 nm to about 1900 nm. If you obtain a hardcopy
spectrum, display it on an expanded wavelength scale so that you can easily
see and characterize the positions of the structure.
3c. If you use a computer-interfaced instrument, you can save your data onto a
removable disk. You can also obtain your spectrum in hardcopy form using
a plotter or printer. However, this document is useful for presentational
purposes only. Some instruments are also capable of analyzing the acquired
spectrum and creating a “peak table,” a summary of the positions and
intensities of all features that exceed a user-specified threshold. You should
use this feature, if available, because a properly created peak table can save
considerable time in reading and organizing the data. Otherwise, the band
peak locations must be displayed individually on the monitor. You can, if
necessary, hand-enter these wavenumber values in your notebook.
Data Analysis
4. After obtaining an acceptable spectrum, attach the cell to the vacuum rack
and evacuate the cell by pumping on it for several minutes. Close the cell
stopcock, and remove it from the rack. If the cell is fitted with KBr (or
NaCl) windows, store it in a dessicator. Remove the Dewar flask with the
liquid nitrogen cryogen and shut down the rack. Wipe all ground-glass
joints clean. Place the outer jacket of the trap in a fume hood to vent the
HCl.
1. If you used high enough resolution in acquiring your data, you will notice
that each rovibrational “line” shows two components. The two components
result from the presence of two Cl isotopes; thus, the sample is a mixture
of H 35 Cl (the major component) and H 37 Cl, presumably representing the
natural abundance of these isotopes. These two isotopomers have different
reduced masses and hence different B e values [see equation (8).] You should
at this time determine which isotope corresponds to the higher-energy
component of the split features.
2. Tabulate (in cm 1 ) the P- and R-branch components corresponding to
H 35 Cl. Assign these features as P1, P2, P3, . . . R0, R1, R2, . . . , and give
each the appropriate value of N. (If you acquire the spectrum on recorder
paper that is linear in wavelength, determine the wavelength of each feature
as precisely as possible.)
3. Enter the ˜,N data into a scientific spreadsheet. From the tabulated data,
determine the spectroscopic constants ˜0, B e , e , and D (and their standard
deviations) from a third-order regression analysis of the appropriate
equation [(23a) or (23b)]. From B e determine I, the moment of inertia of
1 H 35 Cl [equation (8)], and from the definition of I, calculate r e (m 1H

11-38 PART 11 Photophysics and Molecular Spectroscopy
Questions and Further Thoughts
1.0078 units; for M 35Cl , see step 4). Note: you may not be able to obtain a
reliable value of D from your data unless the range of N is sufficiently
large. In this case a second-order analysis of equations (23a) and (23b) will
yield values of ˜0, B e , and e .
4. If the two isotopic HCl rovibrational features are reasonably well resolved,
determine (from the mean line heights for several features) the relative
abundance of the tweo Cl isotopes. Estimate the atomic weight of natural
abundance Cl assuming that 35 Cl and 37 Cl have masses of 34.969 and
36.966 units, respectively. Compare your estimate with the reported atomic
weight of Cl.
5. Using your values of B e and D, calculate ˜e, the harmonic frequency, from
equation (10) and then find the anharmonicity constant, ˜e e , from your
value of ˜0 [see equations (20a) and (20b)]. Compare your values of ˜e, B e ,
e , ˜e e , and D with those reported in Herzberg’s book. 2
6. (Optional) Compare the intensities of the P- and R-branch lines of your
spectrum with those predicted from equations (27) and (28). To do this,
calculate relative intensities for the P- and R-branch lines using the rigid
rotator model [equations (27) and (28)]. The integrated line intensities
should be used in this calculation; as an alternative, however, the peak
intensities can be taken. Scale these intensities to your data using the
observed band heights of the R0 and P1 features and compare the
predicted “sticks” with the observed bands. In other words, divide the peak
heights of all the tabulated R-branch bands by the height of R0 and do
likewise with the P-branch bands and the height of P1. Comment on the
quality of this comparison.
1. Why must the cryogenic medium used with the trap on the vacuum line be liquid nitrogen
rather than dry ice
2. The force constant for H 35 Cl is 5.1669 10 5 dynes cm 1 (ergs cm 2 ). What are the
expected values ˜e and the anharmonicity constant
3. Use the result of question 2 with your value of B e in equation (13) to calculate the
value of e . Compare it with the value determined in this experiment.
4. What additional data would be required to determine the anharmonicity constant. Outline
the qeuations and calculations required.
5. A high-power infrared laser, a very narrow-band light source that can deliver substantial
energy in a very small time, can be used to achieve separation of the 35 Cl and
37 Cl isotopes from a sample of gas phase, natural abundance HCl. At what (single) frequency
should such a laser be “tuned” How do you think this process works How
would the chlorine isotopes be collected
6. Explain in detail how the rovibrational spectrum of a thermally equilibrated sample of
HCl vapor can be used to determine its temperature. Try this method on the data from
your spectrum!
Notes
1. P. W. Atkins, Physical Chemistry, 8th ed., pp. -, W. H. Freeman (New York), 2006.
2. G. Herzberg, Molecular Spectra and Molecular Structure, I: Spectra of Diatomic Molecules,
2d ed., p. 534, Van Nostrand Reinhold (New York), 1950.

Experiment 36
11-39
Further Readings
R. A. Alberty, R. J. Silbey, and M. G. Bawendi, Physical Chemistry, 4th ed., pp. 475–491,
Wiley (New York), 2005.
P. W. Atkins, Molecular Quantum Mechanics, 2d ed., pp. 298–303, Oxford University
Press (New York), 1983.
P. W. Atkins, Physical Chemistry, 8th ed., pp. -, W. H. Freeman (New York), 2006.
E. D. Glendening and J. M. Kansanaho, J. Chem. Educ., 78:824 (2001).
G. Herzberg, Molecular Spectra and Molecular Structure, I: Spectra of Diatomic Molecules,
2d ed., chap. 3, pp. 66–82, 90–97, 103–115, 121–130, Van Nostrand Reinhold
(New York), 1950.
M. Karplus and R. N. Porter, Atoms and Molecules, pp. 458–484, W. A. Benjamin (New
York), 1970.
I. N. Levine, Physical Chemistry, 5th ed., pp. 748–759, McGraw-Hill (New York), 2002.
C. H. Townes and A. L. Schawlow, Microwave Spectroscopy, pp. 18–24, Dover (New
York), 1975.
D. L. Williams, P. R. Minarik, and J. H. Nibler, J. Chem. Educ., 73:608 (1996).