1 1 diethyl 2 2 dicarbocyanine iodide color

CHEM 4525: Physical Chemistry Laboratory I Experiment 2

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Experiment 2 – Spectrophotometry of Conjugated Dyes

1. Objectives

1) Establish the connection between the fundamental principles of quantum mechanics - energy

quantization – via the simple model of the particle-in-a-box and real world absorption spectroscopy.

2) Practice the application of the quantum mechanical formalism in the context of the particle-in-a-box

model to predict experimental observables and compare with experiment.

3) Test the applicability and limitations of the simple free electron particle-in-a-box model for the

description of the electronic states of conjugated systems.

2. Background

2.1. Spectroscopy in the visible region

Absorption bands in the visible region of the spectrum (350 - 700 nm) correspond to transitions

from the ground state of a molecule to an excited electronic state which is 160 to 280 kJ above the ground

state. In many substances, the lowest excited electronic state is more than 280 kJ above the ground state

and no visible spectrum is observed. Those 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 free radical or the

 electrons in a conjugated organic molecule. In most other molecules the energy difference between the

ground and excited states is much higher. As consequence, they do not absorb visible light but a higher

energy (shorter wavelength) ultraviolet (UV) light and are not colored. Molecules that are colored, i.e. those

that do absorb visible light, are often called dyes.

2.2. Conjugated systems

Conjugated systems are those with alternating single and double bonds. This includes linear as well

as cyclic (aromatic) compounds. The electronic strucutre of conjugated molecules is unique in that it has

delocalized π-orbitals, which are formed from sp2 hybridized p atomic orbitals of carbon above and below

the plane of the molecules (conjugated molecules are planar – just in case you didn’t know that). This is

illustrated in Figure 1 for a butadiene.

Figure 1: Formation of delocalized π-electronic orbitals in a conjugated linear hydrocarbon.

The visible absorption of the conjugated molecules comes from these delocalized -states, which are very different from all the other electronic states and orbitals in the molecule. The fact that an electon in one of

CHEM 4525: Physical Chemistry Laboratory I Experiment 2

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these is effectively delocalized over the entire molecule suggests a simple model to approximate the

behavior of the conjugated systems: a free electron that can move within the confines of the molecule (or

its conjugated part). In other words, it can be assumed that each electron moves in a constant potential,

which rises sharply to infinity at the ends. This is effectively the particle-in-the-box that you know from

your physical chemistry class.

2.3. Theory – interpretation of absorption wavelengths.

The quantum mechanical behavior of one particle in a one-dimensional box of length L is found by

solving Schrodinger's equation with appropriate boundary conditions [1]. The solutions reveal that the

particle can occupy any one of an infinite number of discrete levels, the energy of the n-th level En is given

by:

...,3,2,1 8 2

22

 n mL

nh En (1)

where h is the Planck constant, and m is the mass of the electron. The corresponding wavefunctions are:

  ...,3,2,1sin 2

 

  

  n

L

xn

L xn

  (2)

for 0 ≤ x≤ L, and 0 otherwise. The Pauli exclusion principle limits the number of electrons in any given energy level to two and

these two electrons must have opposite spins: +1/2 and -1/2. The ground state of a molecule with N electrons

will have the N/2 lowest energy levels filled and all higher levels empty. When the molecule absorbs light,

one electron jumps from the highest filled level (n1 = N/2) to the lowest empty level (n2 = N/2 + 1). The

energy change for this transition is:

    2

2

2

2

1

2

2

2

8

1

8 mL

Nh

mL

nnh E

 

  (3)

The frequency of wavelength of the abrorbed light must satisfy the resonance condition:



 hc

hE  (4)

Combining (3) and (4) leads to:

 1 8 2

 

Nhc

mL  (5)

which directly relates the observed absorption wavelength to the size of the box (L), mass of the electron

(m) and the number of the electrons in the system.

2.4. Interpretation of absorption intensities

The intensities of electronic transitions have always proved extremely difficult to calculate

accurately from molecular wavefunctions. The reason for this is that the calculated values are quite sensitive

to the exact details of the wavefunctions, so that approximate wavefunctions tend to give poor agreement

with experiment. However, different types of transitions observed in molecules have intensities that vary

over a range of about eleven orders of magnitude, so that even approximate calculations of intensities can

be very valuable in identifying observed spectral bands.

CHEM 4525: Physical Chemistry Laboratory I Experiment 2

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The intensity of an electronic transition, σ is related to the absolute value square of the integral of

the electric dipole moment operator, ̂ , and the wavefunctions of the two states involved:

2

21 ˆ  dnn (6)

For the one-dimensional particle in the box, the dipole moment operator is

xe ˆ.ˆ  (7)

where e is the electron charge. The intensity can be expressed in a number of ways. The most convenient

one for an absorption experiment is the molar extinction coefficient  which for the particle is [2]:

2

0 21

23

6909

)(8 )( 

L

nn A dxx

hc

eN 

  (8)

The x in the integral above is the variable x (not a multiplication sign) that comes from the dipole operator

(7), λ is the mean transition wavelength in nm (from your calculated ΔE) and ρ(λ) represents the shape of

the absorption band, which is normalized so that ∫ ρ(λ)dλ = 1. We will be interested at the extinction

coefficient at the maximum (λ = λmax). Since we do not want to integrate the band, the value of ρ(λmax) will

be estimated by assuming it has the shape of a symmetric triangle. Then, we can write for ρ(λmax):

  FWHM

 

 1

max (9)

where FWHM is the Full-Width-at-Half-Maximum of the absorption band in nm. Substituting to the

formula (8) along with the expressions for the wavefunctions (eqn. 2) and the relationship between the

quantum numbers n1 and n2 and the number of electrons N, we get:

 

2

0

max

23

max

12 sin

2 sin

6909

32 )(  



  

   

  



 

L

FWHM

A dx L

xN x

L

xN

hcL

eN 



  (10)

The extinction coefficient calculated according to (10) is the same extinction coefficient we find

in Beer-Lambert law for absorbance:

Cl I

I A 



   

 

0

10log (11)

where A is the absorbance, I0 and I are intensities of the light entering and exiting the sample, respectively,

C is molar concentration and l is the optical pathlength (Figure 2, problem 9). Since A is measured directly

using a UV-Vis spectrophotometer, C and l are known, ε is easily found experimentally and, thanks to eqn.

(10), can be directly compared with the theoretical prediction using the particle-in-a-box model.

CHEM 4525: Physical Chemistry Laboratory I Experiment 2

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Figure 2: Absorption experiment: I0 is the intensity of the incident light, I is the intensity of light

transmitted by the sample, l the optical path, C molar concentration and α the extinction coefficient.

2.5. Application to cyanine dyes

What remains for practical applications of the above formular is to determine the number of

electrons – N and the length of the “box” – L.

The dyes investigated in this experiment are 1,1’-diethyl-2,2’-cyanine iodide, 1,1’-diethyl-2,2’-

carbocyanine iodide and 1,1’-diethyl-2,2’-dicarbocyanine iodide (Figure 3). Ignoring the benzene rings,

which are also conjugated but separately from the rest, the conjugated hydrocarbon chain in these molecules

corresponds to the carbons between the two nitrogens in the cation [3]. Each contributes an electron to the

π system. In addition, the two nitrogens together contribute three more electrons. If there are p such carbons

in the molecule, the number of the π electrons in the system is N = p + 3.

Figure 3: Cyanine dyes. Top to bottom: 1,1’-diethyl-2,2’-cyanine iodide, 1,1’-diethyl-2,2’-carbocyanine

iodide and 1,1’-diethyl-2,2’-dicarbocyanine iodide.

CHEM 4525: Physical Chemistry Laboratory I Experiment 2

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For the length, we will take the number of bonds between the nitrogen atoms plus one extra on each

side, yielding L = (p + 3)l, where p is again the number of carbons and l is the length of a C-C bond in a

conjugated system l = 1.39 Å.

3. Experimental

1. Using analytical balance and volumometric glassware, prepare 25 mL of exactly 10 M solutions of each of the three dyes in methanol. You will do the necessary calculations as the part of your pre-lab quiz.

2. The instructor will demonstrate how to use the UV-vis spectrophotometer.

3. Using a standard 1 cm cuvette, run a scan of the cuvette filled with just the solvent (methanol) over the

350-800 nm range. This will be your blank.

4. Measure the UV-Vis absorbance spectra of the three dyes. Export the data in the text (ASCII) format for

plotting and analysis. Bring a USB memory stick if you have one (if not, hopefully someone from your

group does, or the instructor can provide one).

5. Properly dispose of the solvent and sample solutions. There will be a waste container set up for you –

don’t wash your solution down the sink. Clean the glassware after yourself.

4. Data Analysis

1. Find the wavelengths of the absorption maxima and, using Beers law (eqn. 11), extinction coefficients

at these maximum wavelengths.

2. Using the particle-in-a-box model, calculate the absorption wavelengths for the three dyes (eqn. 5).

Compare to the experimental values. Discus the performance of the particle-in-a-box model with respect to

the observed qualitative trends and quantitative correspondence with the experimental values.

3. Calculate the extinction coefficients predicted by the particle-in-a-box model (eqn. 10) and compare with

experiment. Again discuss the validity of the model as far as qualitative trends and quantitative

correspondence.

References:

1. Any Physical Chemistry/Quantum Chemistry Text, e.g. Peter W. Atkins, Julio de Paula and Ronald

Friedman: “Quanta, Matter and Change: A Molecular Approach to Physical Chemistry”, W. H. Freeman

and Co., New York, 2009.

2. John A. Schellman, “Circular Dichroism and Optical Rotation”, Chem. Rev. 75 (1975), 323-331.

3. David. P. Shoemaker, Carl W. Garland and Joseph B. Nibler, Experiments in Physical Chemistry, 5th

edition, McGraw-Hill, New York, 1989.

Prelab quiz questions:

1. Why most simple molecules do not absorb in the visible region and therefore are not colored?

2. Which types of molecules do absorb visible light?

3. What color would a substance have if it absorbs light between the following wevelengths:

a) 450 – 480 nm

b) 540 – 570nm

CHEM 4525: Physical Chemistry Laboratory I Experiment 2

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c) 400 – 450 nm

d) 660 – 700 nm

e) 620 – 660 nm

4. What is a conjugated molecule? Why are conjugated molecules planar?

5. What types of orbital hybridizations for a carbon atom can you name? What are the corresponding

shapes and examples (one each) of compounds that undergo these types of hybridization?

6. For a quantum mechanical particle-in-a box, the energy spacing between the successive levels

a) increases proportionally to n

b) decreases proportionally to n

c) increases proportionally to n2

d) decreases proportionally to n2

c) stays the same

as the quantum number n increases.

7. For a quantum mechanical particle-in-a box, the energy spacing between the same levels (i.e. with the

same quantum numbers)

a) increases

b) decreases

c) stays the same

as the size of the box (L) increases.

8. Based on your answer to the previous two questions, would you expect the absorption maximum to

shift to:

a) longer wavelengths

b) shorter wavelengths

c) not shift at all

as the size of the conjugated dye increases? Justify your answer.

9. Show that when a solution with the molar concentration of absorbing species C is irradiated with light

of intensity I0, the intensity of light emerging from the sample I (see Fig. 2) is given by:

CleII  0 (12)

10. The most common form the Beer-Lambert law (eqn. 11) is written as:

ClII  100 (11’)

rather than the form just obtained in the previous exercise (eqn. 12). Derive the mathematical relationship

between the extinction coefficient  from eqn. (12) and  from eqn. (11).

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11. What is the “blank” for the absorption experiment and why is it needed?

12. In a laboratory class, students are supposed to measure an absorption spectrum of a certain solute

dissolved in a common solvent.

Student A approaches it the following way: she fills the cuvette with the solvent, measures the spectrum

and saves it as a blank. Then she fills the cuvette (after cleaning and drying it, of course) with the

sample solution and measures the spectrum. She takes the spectrum of the sample with the solvent used

as a blank as the spectrum of the solute.

Student B, on the other hand, measures the spectrum of an empty instrument with nothing in it but air,

and saves that as a blank. Then he measures the spectrum of just the solvent and saves it and, after again

properly cleaning and drying the cuvette, measures the spectrum of the sample solution. Finally, he

subtracts the spectrum of the solvent from that of his sample solution. He takes the difference as the

desired spectrum of the solute.

Show mathematically, using the Beer-Lambert law (eqn. 11), that both approaches are equivalent.

Which one would you take?

13. Look up the molecular masses of the three dyes to be studied (Figure 3). Calculate how much of each

you have to weigh out to make 25 mL of 10M solution. Write it in your notebook (in addition to the

answers that you’ll turn in at the beginning of the lab) and use these numbers to prepare your samples.

14. Calculate the number of the electrons N that have to be considered for each of the three conjugated

dyes in Figure 2. What is the highest occupied level (i.e. the quantum number n) for each of the three

dyes?

15. Calculate the length of the “box” L for each of the three studied dyes (Figure 3).

16. True/False

a) When finished with my experiment, I will wash the solutions down the sink, so that I get a lousy

grade for my lab conduct.

b) When finished with my experiment, I will dispose of my solutions properly by pouring them into the

provided waste container and then I will thoroughly clean the glassware.

c) When finished with my experiment, I will leave the lab as quickly as possible, leaving all the mess

behind, which will likely result in a reduced number of points for my lab conduct.