Guide to Second Four Experiments
Please read this guide carefully, as it contains useful information on how to carry out the data analysis for the second four experiments in the class, and also indicates particular things that should appear in the lab report.
For three of the four experiments a formal lab report is required. For the fourth experiment (the MO calculation) an informal report will be turned in. There is a separate handout and worksheet for the MO experiment.
Dye molecule spectrum experiment
(Experiment 34 – Absorption Spectrum of a Conjugated Dye)
Some of the analysis you will do for this experiment is based on material in the lab manual, but I am also asking you to do additional analysis in the lab report. Please follow the guidelines given below rather than those given in the lab manual for all parts of the report.
The three dye molecules you worked with are given on the next page, along with a table of their molecular masses. Note that each structure gives one of two equivalent resonance structures for the molecule. The second resonance structure interchanges both the single and double bonds in the molecule and the location of the nitrogen atom with the positive charge. The actual structure of the molecule is, to a first approximation, a combination of the two resonance structures.
In this experiment you will measure the absorption spectrum of each of the dye molecules in methyl alcohol. The particle in a box model will be used to predict the location of the lowest energy electronic transition for the dye molecules (this is the most intense and longest wavelength transition observed in the spectrum). The oscillator strength for the transitions will also be calculated.
Particle in a box model (see Atkins and de Paula, pp. 318-322; Garland et al., pp. 380-385)
The pi electrons in the conjugated bonds between the nitrogen atoms of the dye molecules can be (crudely) modeled as a one dimensional particle in a box, where the box is the length of the region containing the pi electrons. This model ignores several important features of the real molecules, including the fact that the potential does not go to infinity at the ends of the box, that the potential does not have a constant value inside the box, that there is electron-electron repulsion between the pi electrons, and that the box is three dimensional rather than one dimensional. Because of this, we expect the model to be at best qualitatively correct.
Figure 1. Structure of dye molecules.
Figure taken from Truman College, “Absorption Spectra of Conjugated Dyes”
Table 1. Molecular masses for dye molecules.
Dye / Molecular mass (g/mol)1,1'-2,2'-cyanine iodide / 454.36
1,1'-diethyl-2,2'-carbocyanine chloride
(pinacyanol) / 388.94
1,1'-diethyl-2,2'-dicarbocyanine iodide / 506.43
To use the particle in a box model we need to know the following:
Length of the box (L) – The “box” is defined as the region between the two nitrogen atoms in the dye molecules. The length of the box is taken as (p+3), where is the average bond length for a carbon-carbon or carbon-nitrogen bond, p is the number of carbon atoms between the two nitrogen atoms, and an additional length equal to one bond on each side of the terminal nitrogen atoms is added to account for the fact that the potential does not go to infinity at these points. Since there are p+1 covalent bonds between the nitrogen atoms, this gives L = (p+1+2) = (p+3) as the length of the box. Since a C-C single bond has a typical bond length of 0.134 nm, and a C=C double bond has a typical bond length of 0.154 nm, we will use the average value, = 0.144 nm, in the calculations (this is slightly larger than the value used in Garland et al.).
Number of electrons in the box – The carbon atoms in the carbon chain between the two nitrogen atoms all have sp2 hybridization. These atoms each have a pz orbital perpendicular to the chain, containing one electron per carbon atom. The terminal nitrogen atoms also have sp2 hybridization and also have a pz orbital perpendicular to the carbon chain. The neutral nitrogen atom has two electrons in the pz orbital, while the positively charged nitrogen atom has a single electron in the pz orbital.
If we combine the above, the number of pi electrons in the box is equal to the number of carbon atoms in the carbon chain, plus the number of nitrogen atoms, plus one (since one of the nitrogen atoms contributes two electrons to the box). We can write the number of electrons in the box in terms of p, previously defined.
# pi electrons = p+3 (1)
As a check, for 1,1'-diethyl-2,2'-cyanine iodide the length of the box is L = 6 = 0.864 nm and # pi electrons = 6.
The energy levels for the particle in a box are given by the expression
En = n2E0 n = 1, 2, 3, … (2)
where
E0 = h2 (3)
8meL2
where me is the mass of an electron (me = 9.109 x 10-31 kg) and h is the Planck constant (h = 6.626 x 10-34 Js).
The ground state (lowest energy) electron configuration for the pi electrons in the dye molecule can be found by use of the Aufbau principle (electrons add to the lowest energy available orbital) and Pauli principle (a maximum of two electrons per energy level). The lowest energy excited electronic state can be found by moving one electron from the highest occupied energy level to the lowest unoccupied energy level. The difference in energy between these two electronic states corresponds to the lowest energy required to electronically excite the molecule. This energy, (E), can be set to the energy of the photon needed to carry out the excitation
E = hc (4)
where c is the speed of light (c = 2.998 x 108 m/s) and the wavelength of the absorbed photon.
For each of the above dye molecules use the particle in a box model to predict the wavelength at which the first electronic transition should occur. Give sufficient details in your lab report to make it clear how you calculated the values for wavelength. You should compare your calculated values for wavelength to the values observed experimentally in two ways: qualitatively (is the trend in the values for wavelength predicted to occur as the size of the carbon chain increases the same as what is observed experimentally) and quantitatively (how well to the predicted values of wavelength compare to the values observed experimentally). Based on this, you should be able to make general comments on how well the data are modeled by the particle in a box.
Oscillator strength
In addition to fitting your longest wavelength absorption peaks to the particle in a box and Huckel MO models you can use your spectra and the concentration of dye molecule to find the oscillator strength for the lowest energy electronic transition. The oscillator strength, f, is defined as
f = 40mec2(ln10) () d = K () d (5)
NAe2
K = 40mec2(ln10) = 4.320 x 10-9 molcm2/L
NAe2
In the above expression 0 is the permittivity of free space, me is the mass of the electron, c is the speed of light, e is the fundamental electrical charge, and NA is Avogadro’s number. If we substitute the values for the constants into eq 5 we get K = 4.320 x 10-9 molcm2/L. The integral in eq 5 is in terms of frequency and is carried out over the absorption band being studied. Note that the concentration of dye molecule will be used to convert from A (absorbance) to (extinction coefficient), as discussed below. Equation 5 is derived from Einstein’s model for light absorption and emission (Atkins and de Paula, pp. 477-478; Hollas, pp. 28-32).
The significance of the oscillator strength comes from the fact that the value for f is different for allowed and forbidden transitions. For an allowed electronic transition f 1, while for a forbidden electronic transition f is typically around 10-3.
In the present experiment we will calculate a value for f for the lowest energy electronic transition for each dye molecule. To find the value for the integral in eq 5, we can approximate the first peak in the spectrum by a triangle.
() d (1/2) max (6)
where
max = Amax (units will be L/molcm)(7)
c
where Amax is the experimental absorbance of the dye at the peak of the spectrum, c is the concentration of dye molecule in solution (in units of mol/L) and is the pathlength of the cuvette (1.00 cm for a standard cuvette). , the width of spectrum (in units of cm-1) is found by converting the wavelengths corresponding to the base of the triangle used to approximate the absorption band as a triangle, as shown in the figure on the next page. Since the base of the triangle runs from 715. nm (13990. cm-1) to 810. nm (12350. cm-1), then = (13990. – 12350.) = 1640. cm-1.
Figure 2. Example for finding .
While the above method gives a rather crude approximation for the integral, and therefore the oscillator strength, it is more than sufficient for determining whether we are observing allowed or forbidden transitions in the dye molecules.
For each of the dye molecules find the value for oscillator strength. Based on the values, determine whether the transitions we observe are allowed or forbidden.
Your final lab report should include the results of your modeling of the dye spectra using the particle in a box model, and the values for oscillator strength for the lowest energy electronic transition for each dye molecule. You should also address the points mentioned in the discussion above.
References
Atkins, P. W. and J. de Paula Physical Chemistry, 10th Edition. New York: W. H. Freeman, 2014.
Garland, C. W., J. W. Nibler, D. P. Shoemaker Experiments in Physical Chemistry, 8th Edition. New York: McGraw-Hill, 2009.
J. M. Hollas Modern Spectroscopy, 2nd Edition. New York: Wiley & Sons, 1992.
F. L. Pilar Elementary Quantum Mechanics. New York: McGraw-Hill, 1968.
Truman College. CHEM326Labs/ ConjugatedDyes.htm Retrieved 3/3/2014.
I2 vapor spectrum experiment
(Experiment 39 – Absorption and Emission Spectrum of I2)
Details concerning the experiment are given in the lab manual. In this discussion the main focus is on understanding what is observed experimentally and obtaining spectroscopic constants for the upper electronic state of I2.
Figure 1 shows the vapor spectrum for I2. The spectrum indicates the regions where peaks are observed for v"=0 v', v"=1 v', and v"=2 v', where v" is the vibrational quantum number in the ground electronic state (the X1g+ state) and v' is the vibrational quantum number in the excited electronic state (the B30u+ state). Here, and in the remainder of this handout, we follow the usual convention of using a single prime to label constants for the upper (B30u+) electronic state and a double prime to indicate constants in the lower (X1g+) electronic state.
Figure 1. Absorption spectrum of I2 vapor. The sample was I2 vapor above solid iodine, in a capped 1 cm cuvette. T = 50. C, SBW = 0.20 nm, averaging time = 1. s.
The observed electronic transition in the I2 molecule is spin forbidden (since it is a singlet triplet transition) but the selection rules for electronic transitions tend to be less closely followed for diatomic molecules containing heavy atoms, such as is the case for the I2 molecule. Because the vibrational constant in the ground electronic state of the molecule is smaller than that for most diatomic molecules, and because we work at an elevated temperature (to increase the vapor pressure above the solid iodine crystals) transitions are observed from the v" = 0, 1, and 2 states. The intensity of the peaks in the spectrum is determined by the relative population of molecules in each of these vibrational states and the Franck-Condon factors (the overlap between the vibrational wavefunction in the ground state and that in the final state). The Franck-Condon factors change in a regular manner, first increasing in size and then decreasing in size, and so the peaks observed for a particular value for v" also increase and decrease continuously.
Figure 1 indicates the regions where v" = 0, 1, and 2 peaks are observed in the spectrum. Note that when you assign your spectrum, the peak assignments are given as (v, v), where v is the vibrational quantum number in the higher energy electronic state, and v is the vibrational quantum number in the lower electronic state. So, for example, the transition v"=1 v'=18 would be labeled (18,1).
Examination of Figure 1 indicates that there is some overlap for the transitions from v = 0, v = 1, and v = 2. In the regions of overlap we can observe double peaks, or sometimes shoulders on the side of a peak. In these regions it is difficult to identify which peaks (or shoulders) belong to which value for v". One guide to doing this is that the spacing between adjacent peaks is slowly changing and so approximately constant. Therefore, if you know the location of two peaks in the spectrum, you can use this information to predict the approximate location of peaks at longer or shorter wavelengths, and so extend the assignment of the spectrum. In the region corresponding to wavelengths less than 550. nm, all peaks can be attributed to a progression v"=0 v', since in this region overlap with peaks from v" = 1 or 2 is weak and so not observed. For wavelengths greater than 550 nm it is more difficult to assign peaks to the (v, 0) progression, but careful analysis should make it possible to identify at least a few additional peaks out to about 560. nm.
The one remaining difficulty is assigning peaks in the spectrum of I2 to specific transitions between vibrational levels in the ground and excited electronic states. In general this is not an easy thing to do. We will therefore give the assignment for two transitions, and use that as a starting point to assign the other peaks in the spectrum that will be used in the analysis. The peak appearing at approximately 551. nm is due to the transition v"=0 v'=24 (labeled 24,0). The next peak, at approximately 548. nm, is due to the transition v"=0 v'=25 (labeled 25,0). Successive peaks at higher energies (shorter wavelengths) are in order, and so to v' = 26, 27, 28, ... , and successive peaks at lower energies (longer wavelengths) are also in order, and so v = 23, 22, 21, …and so forth.
Once you have assigned as many peaks from v = 0 as possible, the data can be analyzed to find out information about the upper electronic state of the molecule. The starting point in doing that is the equation for the vibrational energy for a diatomic molecule
E(v) = e(v+1/2) - exe(v+1/2)2 (1)
where e and exe are vibrational constants. Using this expression, the energy at which transitions occur between the v" vibrational state in the lower electronic state and the v' vibrational state in the upper electronic state is
E = Te' + E(v') - E(v") (2)
= Te' + [ e'(v'+1/2) - exe'(v'+1/2)2 ] – [ e"(v"+1/2) - exe"(v"+1/2)2 ] (3)
where Te' is the energy at the minimum of the upper electronic state potential relative to an energy E = 0 for the energy at the minimum in the ground electronic state potential (see Fig. 2 below). If we limit ourselves to transitions where v" = 0, and use the known values for e" and exe" for the vibrational constants in the lower electronic state (obtained using Raman spectroscopy, and available at the NIST website), then eq 3 can be rearranged to give
E = [T'e – ("e/2) + (exe"/4)] + e'(v'+1/2) - exe'(v'+1/2)2 (4)
Note that eq 4 is quadratic in (v'+1/2). If we define x = v'+1/2, we can then plot E vs x. If we fit the plot to a quadratic equation of the form
E = a0 + a1x + a2x2 (5)
then by comparing eq 5 to eq 4 it follows that
a0 = [T'e – ("e/2) + (exe"/4)] (6)
a1 = e' (7)
a2 = - exe'(8)
We can find the vibrational constants for the upper electronic state directly using the best fit values for a1 and a2, and obtain the value for Te' using the best fit value for a0 and the known values for e" and exe".
Figure 2. Potential energy curves for the X1g+ and B30u+ states of I2.
The data analysis should be done as follows:
1) Assign the peaks for the v"=0 v'=24 and v"=0 v'=25 transitions, as discussed above.
2) Use the above peak assignments to assign other peaks in the spectrum at shorter wavelengths. You need only assign peaks originating in the v = 0 vibrational state in the ground electronic state of the molecule.
3) Convert peak locations for all assigned peaks from wavelength to energy. Give the energies for the peak locations in cm-1. Recall that energy in wavenumber can be found by the expression
E = 1/ (9)
where is given in units of cm. So, for example, if = 400. nm then E = 20000. cm-1.
4) Fit your peak locations to a quadratic function using eq 5. Recall that y = E, and x = v' + ½, where v' is the vibrational quantum number in the upper electronic state of I2.
5) Use eq 6, 7, and 8 to find experimental values for Te', e', and exe', the energy of the upper electronic state (relative to the minimum in the ground state potential energy curve), and the vibrational constants for the upper electronic state. Note that you will need to go to the NIST website (given below) to find the values for e" and exe" required to use eq 6 to find Te'.
6) Compare your results for the upper electronic state constants to those given at the NIST website for I2. ( ).
Your lab report should give the results of your data analysis and a comparison of your molecular constants for the B30u+ state to those found in the literature. This is a formal lab report, but the introduction and experimental sections can be short (no more than 1-2 pages total). You do not have to find confidence limits for the values you obtain for the molecular constants.
HCl spectrum
(Experiment 37 – Vibrational-Rotational Spectra of HCl and DCl)
Details concerning how the relationships used in the data analysis are derived are given in the lab manual. In a few cases I have used a slightly different (and more common) notation than the lab manual uses.
A sample HCl spectrum is given below, indicating the P-branch and R-branch transitions and giving an assignment for the first few peaks in each branch. You can use this as a guide for carrying out your own assignment of the HCl spectrum.
Figure 1. Infrared spectrum of HCl
In the first part of the data analysis you need to make two tables. Each table will have peak assignments and peak locations - one table for the 1H35Cl molecules and the other table for the 1H37Cl molecules. Because the reduced masses of these two isotopic forms of the HCl molecule are close to one another, you will observe pairs of peaks in your spectrum. For each pair of peaks, the higher intensity (and higher energy) peak belongs to the 1H35Cl molecules (35Cl natural abundance is ~ 75%) and the lower intensity (and lower energy) peak belongs to the 1H37Cl molecules (37Cl natural abundance is ~ 25%). Please list the peaks in each table in order from lowest energy to highest energy