Lab Reports Second Three Experiments 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 below, 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. 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 388.94 (pinacyanol) 1,1'-diethyl-2,2'-dicarbocyanine iodide 506.43 In this experiment you measure the absorption spectrum of each of the above dye molecules in methyl alcohol. Two models 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. 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 sp 2 hybridization. These atoms each have a pz orbital perpendicular to the chain, containing one electron per carbon atom. The terminal nitrogen atoms also have sp 2 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 = n 2 E0 n = 1, 2, 3, (2)
where E0 = h 2 (3) 8meL 2 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 J s). 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 10 8 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. Huckel Molecular Orbital (MO) Approximation (see Atkins and de Paula, pp. 427-431; Pilar, pp. 584-596) An alternative to the particle in a box model for the conjugated pi electrons in the dye molecules is the Huckel molecular orbital approximation. This method uses the pz atomic orbitals to construct multicentered pi bonding and antibonding molecular orbitals. It is a simple example of a semiquantitative LCAO-MO (Linear Combination of Atomic Orbitals to form Molecular Orbitals) approach. For the case of a linear chain of atoms, the energy levels for the pi electrons are given by the expression (Pilar, p. 595) Ej = - 2 cos[j /(p+3)] j = 1, 2, 3,, (p+2) (5)
where p is as previously defined in the particle in a box model, and and are constants, called the Coulomb integral and resonance integral, respectively. and are given by integrals of a particular type. Note that since we begin with p+2 atomic orbitals (p atomic pz orbitals from the carbon atoms, and an additional two pz atomic orbitals from the terminal nitrogen atoms) we end up with p+2 molecular orbitals. In principle, and can be calculated, but is usually found from experimental data (the value for drops out when expressions for energy differences are found, and so is not needed). Based on the form of the resonance integral, it is almost always the case that < 0. As a consequence, the bonding (lower energy) orbitals will be the ones where the value of cos[j /(p+3)] is larger than zero, and the antibonding (higher energy) orbitals will be the ones where cos[j /(p+3)] is less than zero. As was the case for the particle in a box model, the ground state (lowest energy) electron configuration and the lowest energy excited electronic state electron configuration can be found using the Aufbau and Pauli principles. From this, values for E can be found for each dye molecule, and set equal to the energy of the longest wavelength photon capable of electronically exciting the molecule. Your expressions for E and wavelength will be in terms of. Choose by matching the observed and calculated value for wavelength for pinacyanol (the middle dye), and then use the same value for to calculate the wavelength for the other two dye molecules. As with the particle in a box model, you should compare your calculated results to the experimental results both qualitatively and quantitatively (keeping in mind that because of the way you use the model you will automatically get exact agreement for the wavelength for pinacyanol). Also comment on how well the Huckel MO approximation does compared to the particle in a box model. 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 = 4 0mec 2 (ln10) ( ) d = K ( ) d ; K = 4 0mec 2 (ln10) (6) NAe 2 NAe 2 where 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 6 we get K = 4.320 x 10-9 mol cm 2 /L. The integral in eq 6 is in terms of frequency and is carried out over the absorption band being studied. Equation 6 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 6, we can approximate the first peak in the spectrum by a triangle. where ( ) d ( 1 /2) max (7) max = Amax (units will be L/mol cm) (8) 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 example below. 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 and Huckel MO models, 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, 10 th Edition. New York: W. H. Freeman, 2014. Garland, C. W., J. W. Nibler, D. P. Shoemaker Experiments in Physical Chemistry, 8 th Edition. New York: McGraw-Hill, 2009. J. M. Hollas Modern Spectroscopy, 2 nd Edition. New York: Wiley & Sons, 1992. F. L. Pilar Elementary Quantum Mechanics. New York: McGraw-Hill, 1968. Truman College. http://chemlab.truman.edu/chemlab_backup/pchemlabs/ CHEM326Labs/ ConjugatedDyes.htm Retrieved 3/3/2014.
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 1 H 35 Cl molecules and the other table for the 1 H 37 Cl 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 1 H 35 Cl molecules ( 35 Cl natural abundance is ~ 75%) and the lower intensity (and lower energy) peak belongs to the 1 H 37 Cl molecules ( 37 Cl natural abundance is ~ 25%). Please list the peaks in each table in order from lowest energy to highest energy It is useful to recall that the set of peaks at lower energies than the central gap in the spectrum are the P-branch transitions ( J = - 1 peaks) and the set of peaks at higher energies than the central gap are the R-branch transitions ( J = + 1 peaks). The two peaks on either side of the central gap are the P1 peak (lower energy side) and R0 peak (higher energy). Moving away from the central gap, the peak numbering increases by one as you move from peak to peak (P1, P2, P3, as you move to lower energies than the central gap, and R0, R1, R2, as you move to higher energies than the central gap). Your tables should have include the values for m for each peak. Note that m is defined as follows (Garland et al., p. 418)
m = - J" (P-branch) (1) m = J"+ 1 (R-branch) (2) So, for example, m = -3 for the P3 transition, and m = 5 for the R4 transition. By defining this new label for the peaks, this allows us to do a single fit to all of the peaks, both the P- branch and the R-branch transitions. The equation to which the peak locations will be fit to is the following: E = 0 + (2Be - 2 e)m - em 2 (3) where E is the peak location in units of cm -1. Note that 0, Be, and e will also have units of cm -1. Equation 3 is a quadratic equation in m. We can find the values for the vibrational and rotational constants for each isotopic form of the HCl molecule by fitting our data to a second order polynomial. By comparing the coefficients of our fit to eq 3 we can find our best fit values for 0, Be, and e. For the 1 H 35 Cl molecules you should compare your values for 0, Be, and e to the values listed at the NIST website ( http://webbook.nist.gov/chemistry/form-ser.html ). You will have to calculate a value for 0 using the relationship 0 = e - 2 exe (4) where e and exe are taken from the NIST website. You should also calculate predicted values for the 1 H 37 Cl constants 0 and Be using the relationships (Garland et al., p. 419) and ( 0*/ 0) = ( / *) 1/2 (5) (Be*/Be) = ( / *) (6) where the * indicates values for the 1 H 37 Cl molecules, and, the reduced mass, is given by the expression AB = mamb (7) ma+mb Values for the masses of 1 H, 35 Cl, and 37 Cl are given in Atkins and de Paula (Appendix, Table 0.2, pp. 968-969). Note that eq 5 really applies to e, but it should be approximately correct for 0.
Your report should have a discussion of the degree of agreement between the constants you find for 1 H 35 Cl and those given at the NIST website, and how well the values for 0 and Be predicted for 1 H 37 Cl compare to those found experimentally. You do not need to give confidence limits for your experimental values for 0, Be, or e. References Atkins, P. W. and J. de Paula Physical Chemistry, 10 th Edition. New York: W. H. Freeman, 2014. Garland, C. W., J. W. Nibler, D. P. Shoemaker Experiments in Physical Chemistry, 8 th Edition. New York: McGraw-Hill, 2009.