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EXPERIMENT 13. ELECTRONIC ABSORPTION SPECTRA

Organic compounds which are colored (i.e. absorb in the visible) generally have some weakly bound or delocalized electrons such as the odd electron in a �ee radi-cal, or ! electrons in a conjugated organic molecule. In this experiment we are con-cerned with the determination of the visible absorption spectrum of several symmetrical cyanine dyes and with the interpretation of these spectra using the “�ee-electron” model (particle in a box) and Hückel molecular orbital theory.

Introduction UANTUM MECHANICS IS VERY useful for interpreting spectroscopic transitions. In this ex-periment you will use quantum mechanics to model the electronic transition energy of a

molecule between its ground state and its �rst excited state. �e absorption arising from the electronic excitation of colored compounds, such as polymethine or cyanine based dyes, occurs in the visible region of the electromagnetic spectrum (from 800 nm to 400 nm). In this experi-ment, you will obtain the absorption spectrum of several dye molecules. �e wavelength maxi-mum of the absorption is used to determine the energy di�erence between the ground and excited states. You will compare your data with two theoretical models,

•! �e particle in a one-dimensional box (the free-electron model) •! �e Hückel molecular orbital theory.

Both models treat only the energy levels associated with the 2p electrons delocalized along a conjugated chain. Although they are crude models, they capture the essential features of the absorption spectra in these molecules.

�e general molecular structure of the cyanine based dyes you will study is shown in Figure 1. �e dyes will di�er in the length of the conjugated chain. �ese dyes are:

•! Dye(0): 3-3!-diethylthiacyanine iodide x = 0 •! Dye(1): 3-3!-diethylthiacarbocyanine iodide x = 1 •! Dye(2): 3-3!-diethylthiadicarbocyanine iodide x = 2 •! Dye(3): 3-3!-diethylthiatricarbocyanine iodide x = 3

You will measure the bridge length dependence of the energy gap between the ground and �rst excited electronic state and compare this dependence with the predictions of the two models mentioned above.

Q


Figure 1 General structure of cyanine based dyes.

Theory Here we provide a brief description of the two theoretical models. Please refer to your textbook for more detail.

One-dimensional box model

�e quantum mechanical expression for the energy levels of an electron in a one-dimensional box is

E =n2h2

8mL2"

# $

%

& ' (1)

where m is the mass of the electron, h is Planck’s constant, n is the quantum number (1, 2, 3, …) and L is the length of the box. For these molecules, you should consider the box size to be the distance between nitrogen atoms plus one bond length beyond each nitrogen. More precisely, let L = (p + 3)l, where p is the number of carbon atoms in the chain (including the terminal ones in the 5-membered ring systems) and l is the average bond length of the chain.

Figure 2. Dye(2) showing the “box” in which the electrons in the conjugated system are allowed to move.

Figure 3. Dye (0). Circled atoms represent the 1-D box the electrons are moving in. Note, p = 3 for Dye(0).

N

S

Et

N

S

Et

x

N

S

Et

N

S

Et


Figure 4. �e p-orbitals for Dye(0) that are being used to form the 1-D box and the "-molecular orbitals in the Hückel molecular-orbital method.

�e number of " electrons1 in the chain, which we will denote as N, is equal to p + 3. Because each �lled energy level consists of two spin-paired electrons, the highest occupied molecular orbital (HOMO) will have the quantum number n given by n = N ⁄ 2.

Correspondingly, the lowest unoccupied molecular orbital (LUMO) will have the quantum number n! given by n! = (N ⁄ 2) + 1. To promote an electron from the HOMO to the LUMO re-quires a photon whose energy is equal to this energy di�erence.

"E = E LUMO( ) # E HOMO( ) =h2 N +1( )8mL2

$

% &

'

( ) (2)

If we let l = 1.39 Å (a typical C–C distance in a conjugated system) the energy of the transition is given in electron volts (eV) by

"E eV( ) =19.46p + 4( )p + 3( )2

#

$ % %

&

' ( ( (3)

Because this model has only one parameter, you can calculate directly the amount of energy needed for the transition simply from your knowledge of the dye’s molecular structure (i.e. the value for p can be determined from the structure of each of the four dyes).

In order to compare the model prediction with the spectroscopic data for the dyes, the energy in equation (3) must be converted to a wavelength. �is conversion uses the relation #E = h$ = hc ⁄ %. Substitution of this relation, converting units and rearranging yields

" nm( ) = 63.7p + 3( )2

p + 4( )

#

$ % %

&

' ( ( (4)

Note that for chains having easily polarized groups at the ends, the length of the “box” is e�ec-tively increased. �is e�ect can be included in the model by adding an empirical parameter, &, to equation (4) to yield

1 One electron from each carbon atom, two electrons due to the lone pair on each nitrogen, and the removal of one

electron due to the positive charge.


! nm( ) = 63.7p+3+"( )2

p+ 4( )

!

"##

$

%&& (5)

�is parameter is adjusted for an homologous series of dyes in order to obtain a best �t to the date; i.e. it is a “fudge factor”. Note that & should increase with increasing polarizability of the end groups. In your analysis, you must perform the calculations both with the & parameter and without the & parameter.

Hückel molecular orbital (HMO) model

�e Hückel theory is a speci�c case of the more general molecular orbital theory. It is typical in the molecular orbital theory to write the molecular wavefunction (for each electron) as a linear combination of atomic orbitals (LCAO), such that

" = ci# ii$ (6)

where the 'i are normalized atomic orbitals (e.g., 1s; 2s; 2p; …) and the ci are weighting coe�-cients which must be determined.

We can write the Schrödinger equation for the molecule as

ˆ H " E( )# = 0 (7) Where

ˆ H is the electronic Hamiltonian, ( is the molecule’s electronic wavefunction and E is the electronic energy. Using the LCAO method, we can write

ciˆ H " E( )# i = 0

i$ (8)

where the sum is from i = 1 to N, (equal to the number of atomic orbitals we choose to com-bine). Generally speaking, we would now minimize E and seek the coe�cients that lead to the minimum energy. Using some basic linear algebra techniques we can solve equation (8) by form-ing the secular determinant. �is leads to an Nth order polynomial in E. Each value of E from the polynomial equation can be substituted back in to equation (8) and along with the require-ment for ( to be normalized (i.e. ∫(2d) = 1), the set of coe�cients (and hence the LCAO mo-lecular orbital) corresponding to each E can be determined. �e secular determinant is written below.

H11 " ES11 H12 " ES12 ... H1N " ES1NH21 " ES21 H22 " ES22 ... H2N " ES2N

. . . .HN1 " ESN1 HN 2 " ESN 2 ... HNN " ESNN

= 0 (9)

See your textbook for details. In these equations

Hii = " i* ˆ H # " id$ = i ˆ H i (10)

and represents the energy of an electron in an atomic orbital 'i. �e o� diagonal Hamiltonian terms are de�ned


Hij = " i* ˆ H # " jd$ = i ˆ H j (11)

and represent the energy of interaction (stabilization or destabilization) between two atomic orbitals. �e overlap term is

Sij = " i*# " jd$ = i j (12)

To obtain the energy for the molecular orbitals, we would need to solve this N by N determi-nant.

The Hückel approximation

In the Hückel approximation, we assume that the electrons of the atomic inner cores and of the localized sigma bonds merely provide (along with the nuclei) an e�ective �eld in which the re-maining pi electrons move. �is allows us to write the full electronic wavefunction as a product: (electronic = ("!(*-framework. �e form and energies of the "-molecular orbitals can be calculated by solving a reduced secular determinant, only involving the atomic orbitals that can combine to form "-molecular orbitals (the unhybridized pz atomic orbitals on carbon and the lone-pairs on nitrogen in our cyanine dye molecules).

Further drastic assumptions are made within the Hückel approximation. We assume that unless the two orbitals i and j are on the same atom, that Sij = 0. Otherwise, Sii is equal to 1. �e cou-lomb integral of an orbital i, Hii, represents the self-energy of an electron in that orbital and is nonzero (and negative). �e energy of interaction (resonance energy) between orbitals i and j is also ignored, unless they are on adjacent atoms.

Hii is assumed to be identical for all carbon atoms and is assigned a value of +. Hij is assigned a value of , for pz-orbitals on adjacent carbon atoms. Hij is referred to as the resonance integral and is a negative quantity.

Consider the methylene cyclopropene molecule:

�e secular determinant corresponding to the " energies and form of the molecular orbitals is given by:

" # E $ 0 0$ " # E $ $

0 $ " # E $

0 $ $ " # E

= 0

Dividing each row of the determinant through by ,,2 followed by a substitution of x = (+ – E) ⁄ ,, leads to a secular determinant that is much easier to work with:

2 �e determinant is thus reduced by a factor of ,4, from 0 to 0 ⁄ ,4, which of course still equals zero!

2

3 4

1


x 1 0 01 x 1 10 1 x 10 1 1 x

= 0

Expanding out the determinant, we are le� with a polynomial in x.

x4 – 4x2 – 2x + 1 = 0

which has roots of x = 1.48, 1.00, –0.31 and –2.17. Using our de�nition of x, we �nd that the energy levels are given by

En = + – x,

So our four energy levels are: + + 2.17, (lowest energy), + + 0.31, (homo), + – , (lumo), and + – 1.48, (highest energy).

Notice that since , is a negative quantity, + + 2.17, is lower in energy than + – 1.48,. If we assume that these energy levels are unchanged upon promotion of an electron, then the HOMO-LUMO energy gap, #E = (+ – ,) – (+ + 0.31,) = –1.31 ,. By comparing #E from measuring the wavelength at which the dye absorbs, with #E predicted using Hückel theory, then it should be possible to determine the value of ,.

�e same technique can be applied to larger, more complex molecules such as the dyes used in this experiment. Because of the large number of atoms involved, and thus the much larger secu-lar determinant, a computer program will be used to solve these equations.

Rather than model the whole dye molecule, we will only model the conjugated system in the center of the dye. For example, we will model Dye(2) as:

Energy

! + 2.17"

! – 1.48"

! – "

! + 0.31"

N NR'

R

R'

R'' R''

R


An introduction to SHMO—A HÜCKEL MOLECULAR ORBITAL CALCULATOR �e program we will use to calculate the Hückel molecular orbitals and energies resides on the website: http://chem4431.ssuchemistry.com. Look for the SHMO link on the le� toolbar. It was written in Java by Richard Cannings at the University of Calgary. �e homepage for the program is located at the following URL: http://www.ucalgary.ca/rauk/shmo Although the program may be run from the Calgary website, the �rst mentioned website has an expanded drawing canvas that will make it easier to sketch the conjugated cyanine based dyes that we are using.

It is possible that the program will not run without a few tweaks. Under Microso� Windows, try the following if SHMO will not run:

•! Click on the Windows “Start” menu in the lower-le� corner of the screen ! “Control Panel” ! “Programs” ! “Java”

•! In the “Java Control Panel” window that appears, click the Security tab

•! In the “Exception Site List” section in this window, click on the "Edit Site List..." button ! Click the “Add” button ! Copy and paste the web server location for the Java ap-plet “http://chem4431.ssuchemistry.com/SHMO/SHMO.html” ! Click the OK button ! Click the OK button.

•! Close ALL of the open windows. You must completely QUIT the web browser or this workaround will fail

•! Re-launch the web browser

•! Go to the SHMO page

•! When the Security Warning window asks “Do you want to run this application?” ap-pears, if there is a “I accept the risk and want to run this app.” option, checkmark it ON �rst then ! Click the “Run” button.

If this does not work, please let Dr. Napper know ASAP!

�e program is fairly easy to use, and the following procedure will allow you to become ac-quainted with its use (as well as some of its quirks). We will demonstrate how this program can be used to calculate the HOMO and LUMO of pyrrole, an important heterocyclic molecule.

Pyrrole has the formula C4H5N, and the structure

We will use the SHMO program to calculate the energy of the pi-electron MOs.

Start o� by clicking on the Clear button to erase any previous drawings. Click anywhere on the canvas to place a carbon atom. Place �ve atoms on the canvas in the shape of a pentagon.

N

H


To add bonds, click on the �rst atom in the bond, and then draw the bond over to the second atom. Since the Hückel MO method only deals with the pi-electrons, you only need to draw the bonds that make up the conjugated system. If you make a mistake, then click on the Erase button and delete the o�ending atom(s). Continue drawing the bonds by clicking on the Add button and proceeding as before.

To convert the bottom carbon atom into a nitrogen atom, then click on the Change button, and select the atom.


�ere are two choices for nitrogen: N2 and N3. �e number a�er the element symbol N refers to the number of atoms that the nitrogen atom is bonded to. In pyrrole, the nitrogen atom is bonded to three atoms, so we should select N3. In pyridine, the nitrogen atom is bonded to two atoms, so you should select N2 for this atom. Click OK a�er making your selection.

Due to a bug (oops) in the program, you will need to do the following to properly calculate the orbital energies! (�e following procedure is only required when a heteroatom is present.)

• Click on the on-the-�y-calculation check box (turn it o� ) • Click on the verbose button (turn it on) • Click on the show-orbitals button • Click on the verbose button (turn it o� ) • Click on the on-the-�y-calculation check box (turn it on)

You can now click on the (properly) calculated energy levels and display the energies and mo-lecular-orbital shapes. Click on the highest-occupied energy-level, and the program will display the shape of the orbital, and the energy of the orbital. For pyrrole, the HOMO energy should be displayed as “alpha – 0.618 |beta|”. (Note, since beta is a negative quantity, this energy is actually equal to alpha + 0.618 beta! Confusing, eh?)

Electrons can be added or removed from the structure (to create anions or cations respectively) by clicking on the + or – button in the bottom le� corner of the program’s window.

N

pyridine

N

H

pyrrole


Since the resonance integral, beta, is a negative quantity in HMO theory, this corresponds to an energy for the HOMO of EHOMO= + + 0.618 ,.

Clicking on the lowest unoccupied energy level, you should be able to display the wavefunction corresponding to the LUMO, and verify the energy of ELUMO = + – 1.112,. (Be careful with the sign convention!)

If we assume that absorption of a photon of light corresponds to excitation of an electron be-tween the HOMO and the LUMO, then we can equate h$ to #E, where #E = ELUMO – EHOMO = (+ – 1.112,) – (+ + 0.618 ,) = –1.730 ,.


If we take %max to be equal to 290 nm, then , would be equal to –3.96 x 10–19 J (–238 kJ-mol-1 or ca. –2.5 eV). �is is consistent with many estimates of ,.3

Laboratory procedure �e goal of this experiment is to obtain electronic spectra for a series of conjugated dyes and compare the results with theoretical models. Stock solutions of the cyanine based dyes dissolved in methanol (ca. 1 x 10–5 M concentration) are available from your laboratory instructor.

NOTE: THESE DYES ARE TOXIC AND MAY BE CARCINOGENIC IN

SMALL AMOUNTS. HANDLE THE DYES WITH GLOVES ONLY AND

AVOID CONTACT WITH MOUTH AND EYES. IF POSSIBLE, PLACE

SOLUTIONS IN DARK AREAS TO AVOID DECOMPOSITION.

•! Read Appendix A. Using the HP 8453 UV-VIS diode array spectrophotometer in order to familiarize yourself with the operation of the diode array UV-VIS spectrophotometer.

•! Use the quartz 1-cm cuvette for all your measurements. •! Collect a background (blank) spectrum using spectrophotometric grade methanol. •! Collect a sample spectrum using each of the four dye solutions. Make sure that you

thoroughly rinse out the quartz cuvette between samples. �e washings should be dis-posed of in the relevant chemical waste bottle. Scan each of the dyes from 400 nm to 800 nm.

•! If the absorbance of the dye solutions exceeds 1.0 at any wavelength, then dilute the sample (in the cuvette) with methanol and record a new spectrum.

•! Record the wavelength at which maximum absorption occurs for each of the dye solu-tions.

•! Note the general color of each of the four dye solutions. •! Print a copy of the UV-VIS spectrum for each of your four dyes. Attach to your �nal re-

port in a properly labeled appendix.

3 Heilbronner, E.; Murrell, J. N.; J. Chem. Soc., 1962, 2611.


Data analysis Using equations (3) through (5),

(1) Calculate the transition energies and wavelengths predicted by the particle in a box model without the & parameter

(2) Using your experimental data for the transition energies, obtain an average value for &.

(3) Calculate the transition energies and wavelengths predicted by the particle in a box model with the & parameter.

Using the Hückel program’s output, �nd the di�erence in units of , between the HOMO and LUMO energies for each dye molecule. �is di�erence (#coe�) is the energy gap (HOMO to LUMO) in units of ,. Remember that since , is negative, the more stable bonding orbitals will have posi-tive , coe�cients.

•! Using #E = hc ⁄ %, convert your experimental wavelengths to energy in eV. •! Since #Eexp = ,#coe�, you can plot #Eexp vs. #coe� and the slope of the line4 will equal ,.

�e units of , will be eV. •! To �nd the calculated HMO excitation energy, obtain the product of the #coe� and , val-

ues.

The laboratory report Your report should include a table of your experimental data, sample calculations, and the com-puted transition energies using the particle-in-a-box model (with & and without &) and the Hückel model. You should report the percent error for each of the calculated energies versus your experimental energies. You should also report the values of any parameters (e.g. ,) ob-tained by these comparisons. Lastly, you should answers questions 1 – 3 given below.

Questions 1. Which method of calculating the excitation energy is more accurate, Hückel theory or

the particle-in-a-box model without &? Which of these two theories do you expect to give more accurate results and why?

2. Given the information in question 1 about the particle-in-a-box theory, why does a par-ticle-in-a-box model with & give good results?

3. What is the color of each dye solution? Using your measured absorption spectra, explain why the di�erent dye solutions have the colors you observe.

4 Hint: #Eexp = ,#coe� not #Eexp = ,#coe� + constant. Be sure you curve-�t to a y = mx expression, not y = mx + b!


Pre-Lab Questions 1.! Using the particle in a 1-D box model, estimate %max for 1,3-butadiene.

2.! Given the experimental %max for 1,3-butadiene is 217 nm, use the 1-D PIB model to cal-culate what the box length is. How does this compare to the actual length you used in Q1?

3.! Explain why a solution that absorbs green light does not appear green.

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