CHEMISTRY 373

PHYSICAL CHEMISTRY

QUANTUM MECHANICS AND SYMMETRY

LABORATORY MANUAL

FALL, 2005

DEPARTMENT OF CHEMISTRY

THE UNIVERSITY OF CALGARY

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General Information

This manual provides the information required for the Chemistry 373 wet and dry laboratory program. If at any time you are in doubt as to what is required, ask your instructor. The nature of the laboratory work in this class varies from one part of the course to another, from week to week. Because of the nature of the course material, the laboratory component is divided into two main groups of exercises. We begin the laboratory with a few weeks of “wet labs,” in which students will be exposed to some basic principles of quantum mechanics. These wet labs will be followed by several weeks of dry labs on molecular symmetry and point groups. During the last week, a combination of dry and wet lab procedures will be conducted. For the wet labs, most of the exercises are of the “confirmatory” type in which some law or principle is demonstrated by measurements on a system that, hopefully, follows the rules closely. (Bear in mind that every laboratory must deal with a specific example - but that it is the general concept rather than the particular case that is of concern.) The wet laboratories involve practical spectroscopy. Spectroscopy is one of the most fundamental proving grounds for quantum mechanics and hence the knowledge of one supports the other. Further, this laboratory exposure is required for the first part of Chemistry 471, which continues an exploration of the various spectroscopic methods used. Laboratory Safety Some specific precautions are noted at the appropriate places in the instructions. All chemicals and apparatus can be hazardous if mishandled. All chemicals should be assumed to be dangerous. The following rules apply. 1. Students are not allowed to work in the laboratory unless an instructor is present. 2. Laboratory coats (full length) and safety glasses must be worn at all times when working in the

laboratory. The safety glasses issued fit over most modern prescription glasses. 3. Smoking, eating and drinking are not permitted in the laboratory. 4. Laboratory work must be attended continuously. 5. Flammable solvents should be stored only in small quantities away from flames or other

possible sources of ignition. They should be dispensed in the fume hoods and capped immediately after use. Similarly, any chemical that produces toxic vapors must be used in the fume hoods.

6. A pipette bulb must be used whenever a liquid is being transferred by pipette.

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In addition, the following guidelines will provide you and those around you with a safer working environment. 1. Determine the location and means of operation of the fire extinguishers, safety showers, eyewash

facilities, and other emergency equipment. 2. Dress appropriately – a layer of clothing between your skin and any chemical that may be

spilled may considerably reduce injury. Long hair should be tied back so that it does not become entangled in apparatus or dangle in bunsen burner flames. Rings should be removed during the laboratory class. Sandals do not provide good footing. Contact lens wearers run the risk of a more serious injury to their eye in the event of a foreign body or chemical contaminating the eye.

3. Keep the work area, sink, and fume hoods clean and uncluttered. 4. A spatula or other appropriate apparatus should be used to minimize skin contact when handling

chemicals. Washing your hands immediately after handling and prior to leaving the laboratory for the day will remove any chemicals you have contacted.

5. Beware of burns from forgotten, still-lit burners and from hot glassware. 6. Ask for instructions when disposing of used chemicals. 7. Never hurry when performing experiments. Safety always has the highest priority. 8. Last but not least, think before acting. Common sense is the most fundamental rule of all, as a

perusal of the above will confirm. If unsure of what to do or how to do it, ASK THE INSTRUCTOR!

All injuries, including minor cuts, etc. treated initially in the laboratory, should be attended to by the University Health Services (MacEwan Hall, Room 370, 220-5765, 09:00 - 16:30, M - F). In the event of a serious accident, call Campus Security immediately using a red phone, or 220-5333 from any campus phone. The technician (SA 201) is available for assistance at any time. WHMIS and Chemical Laboratory Safety Course All students taking chemistry laboratory work are required to complete the Chemistry Laboratory Safety Course offered by the Department of Chemistry Safety Office. This is a federal requirement. The course is CD-ROM based. Contact your TA if you have not completed this course.

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Laboratory Reports (Wet Labs) As was noted above, the nature of the laboratory work varies considerably from week to week. It is therefore impossible to give a hard and fast set of rules as to the content and format of the report. Your instructor will have more to say about specific cases at the appropriate time. Some particulars:

The title page must give the name of the exercise; the author's name and the name(s) of any partner(s); and finally, the instructor's name as well as the laboratory section number and day and date on which you completed the experiment.

The next page should contain just your brief Abstract summarizing the whole laboratory. Although it appears first, this is the last part you should write. The abstract should say what you measured, how you measured it, and what your result was (including the expected or literature value for comparison). In general, you will only need one sentence for each of these three aspects.

Your Introduction should provide a short statement about the objectives of the experiment, and then comment on the basis of (a) the method of measurement; and (b) the theory as it is applicable to the specific laboratory. Much of the theory is available in this Manual, but you must paraphrase and extend or condense as is appropriate. The introduction should contain the background information necessary to understand the discussion of the results.

The experimental Data and any other observations made in the course of the experiment should be tabulated with appropriate table title, headings and units. If you use data acquired by other individuals, its source must be acknowledged but the data itself need not be presented. Figures should be titled; use Figure numbers to refer to figures in your report.

The Calculations and Results follow, expressed to an appropriate precision and in the correct units. A sample of each type of calculation must be shown in detail (to demonstrate what is being done in the spreadsheet or other tables), but the rest should be tabulated in summary form. In some cases, it may be more sensible to combine the data and calculations sections in one large table, i.e. in a spreadsheet as long as the sample calculations are clear to the reader in this format.

The final major section is the Discussion and Conclusions, which deals with the significance of the results, any unexpected outcomes, comparison with literature values, etc. The final part of the discussion should refer to the objectives and conclude whether or not they have been achieved.

It is sometimes difficult to decide what material belongs in the introduction and what in the discussion. If the item is independent of the measurements actually made, it is introductory material. Conversely, no new concepts should be introduced in the discussion - it should involve the interpretation or significance, in light of the measurements just made, of the concepts that were introduced earlier.

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The final item in the report is a list of the References Cited, reported in the correct bibliographic notation. In particular, note that each source must be cited at the appropriate place(s) in the report. No material from any source should be copied; a brief résumé and/or reference is sufficient. Except for the calculations, laboratory reports must be done in ink or on the computer. The Abstract, Data, and References Cited are fundamental expectations for any laboratory report and therefore are not awarded any marks. Instead, their absence or inadequate completion will result in the loss of up to two marks each (to a maximum of six marks per report, but no total mark of less than zero will be awarded). The Introduction, Calculations and Results, and Discussion and Conclusion, on the other hand, will be earn marks in a positive sense, to a maximum of five marks for each of the three. The maximum total mark is thus fifteen marks. Dry Lab Assignments Specific instructions will either accompany the description of the dry lab or will be given to you by your laboratory instructor. Normally, each dry lab will be accompanied by practice problems and a specific set of problems which are to be completed by each student during the laboratory period and submitted to the laboratory instructor for grading. Preliminary Questions and Exercises These must be completed and available for inspection by the instructor at the beginning of the laboratory period. It is not really part of the report but is included on a separate sheet at the end. The Sources of Laboratory Marks The laboratory mark is determined by the reports and assignments for both of the wet and dry labs. Laboratories 1, 2 and 6 will be marked out of 30. The maximum scores for each dry lab assignment will be given with the returned graded assignments.

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Chemistry 373 Lab Schedule Week of Monday Title Location 2005-09-12 NO LABS THIS WEEK!! 2005-09-19 NO LABS THIS WEEK!! 2005-09-26 (A) 1A. The Absorption of Linear Polyene Dyes SA 206 (particle in a box) 2005-10-03 (B) 1B. The Absorption of Linear Polyene Dyes SA 206 (particle in a box) 2005-10-10 (A) 2A. The Uncertainty Principle SA 206 (diffraction) 2005-10-17 (B) 2B. The Uncertainty Principle SA 206 (diffraction) 2005-10-24 NO LABS THIS WEEK!! 2005-10-31 (A, B) 3. Molecular Symmetry and Point Groups B01, B02, B04 in SA015 (Dry Lab) B02, B05 in SB149 2005-11-7 NO LABS THIS WEEK!! READING WEEK 2005-11-14 (A, B) 4. Character Tables B01, B02, B04 in SA015 (Dry Lab) B02, B05 in SB149 2005-11-21 (A, B) 5. Huckel, part I All sections in first year (Dry Lab) CAL, SA 204 2005-11-28 (A, B) 6. Huckel, part II SA 206 Polycyclic Aromatic Hydrocarbons (1.5hr each section A and B) 2005-12-05 HAND IN ALL MATERIALS

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Table of Contents

General Information iii Laboratory Safety iii Laboratory Reports (Wet Labs) v Dry Lab Assignments vi Schedule vii Table of Contents ix Safety Compliance Form xi Laboratory Exercises

1 The Absorption of Linear Polyene Dyes 12 The Uncertainty Principle 113 Molecular Symmetry 174 Character Tables 335 Hückel, Part I 536 Hückel, Part II 61

Appendices

Appendix I Character Tables Appendix II Fundamental Physical Constants Appendix III Laboratory Survey

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Safety Compliance Form THIS FORM MUST BE COMPLETED, SIGNED, AND SUBMITTED TO THE LABORATORY INSTRUCTOR BEFORE ANY LABORATORY WORK IS BEGUN. Name (please print) ID Laboratory section Day Time Instructor I certify that I have read and understood the safety regulations that appear in this manual, recognize that it is my responsibility to observe them, and agree to abide by them throughout this course. Date Signature (To be completed by the Laboratory Instructor) The above-named student has indicated to me that s/he has read and understood the safety regulations that appear in the Chemistry 373 laboratory manual, and agrees to abide by them. Date Signature I certify that I have successfully completed the University of Calgary Department of Chemistry/Safety Office Chemistry Laboratory Safety Course. Date Signature The above named student has indicated to me that s/he has successfully completed the University of Calgary Department of Chemistry/Safety Office Chemistry Laboratory Safety Course. Date Signature This information is being collected under the authority of the Freedom of Information and Protection of Privacy Act. The information you provide is intended for use by the Department Head to assess instructors, by instructors to assist them in improving instruction, and by the Department of Chemistry to improve the quality of the course. Your constructive comments will be read and taken into consideration for future decisions. These forms will be held until after your instructor has submitted the final course grades to the Registrar. If you have any questions about the collection or use of this information, contact the Office of the Head, Department of Chemistry, Science A 105 (403) 220-5340.

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1. The Absorption of Linear Polyene Dyes NOTE: There is a pre-lab exercise to submit for this Lab. Your TA will check this before you start your lab. You will also need a disk on which to save your data. Introduction Consider a simple system of alternate double and single bonds in a hydrocarbon such as butadiene,

CH2

CH2 In theory we draw the double bonds at specified locations signifying localized electrons, but in practice the double bonds are not localized—the π electrons are free to move from one end of the molecule to the other. These electrons, however, cannot move beyond the last atom without ionization occurring—in other words, without a large increase in the potential energy of the system. The movement of the π electrons is similar to that of a particle in a box, and so this model can be used to predict the behavior of conjugated π electron systems. The length of the box is the length of the molecule, or that fraction of it over which the electrons can move. The equations of quantum mechanics can predict that the energy put into a system will be sharply absorbed as soon as the energy of the incident radiation exactly matches the energy of an electronic transition, En → En+1. This results in a series of lines where we can write, hν = E1 – E2, hν = E2 – E3, hν = E3 – E4, etc. These lines are the eigenvalues, or solutions to the Schrödinger equation, HΨ = EΨ. The eigenfunction is,

)cos()sin()( kxDkxCxk +=Ψ=Ψ [1] which can be simplified to,

⎟⎠⎞

⎜⎝⎛=Ψ

LxnCxn

πsin)( [2]

where L is the length of the box, C is a coefficient equal to (2/L)1/2 (D is another coefficient), k is the wave vector equivalent to the quantum number, n (n = 1, 2, 3, etc.). This becomes the eigenfunction equation we recognize,

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=Ψ

Lxn

Lnπsin2 2/1

[3]

2

The eigenvalue to solve the Schrödinger equation is then,

mLhnE

8

22

= [4]

where n is the quantum number of the electronic transitions which produce multiple eigenvalues for a system, h is Planck’s constant, m is the mass of the electron, and L is the length of the box. From first year chemistry, you are familiar with the Balmer series for the hydrogen atom, which is an example of this in reverse, i.e. emission not absorption (Figure 1).

n = 1

n = 2

n = 3

n = 4n = 5

red

blue -g re e n

blue

violet

n = 6

Figure 1: Balmer series

The Balmer series lines show transitions from a higher energy state to a lower one, E3 → E2, E4 → E2, etc. To observe these lines the initial energy put into the system must be at least equal to that energy which is released. If this does not occur, i.e. the energies do not match, we do not observe the lines for a specific transition. If lines are observable for the transitions of the released energy, as we see in the Balmer (and other) series, then we would expect to see corresponding lines for that energy put into the system, the transitions, E1 → E3, E2 → E10, etc. This, however, is not the case for the following two reasons.

(1) The eigenvalues that are solutions to the Schrödinger equation as shown above (Eq. 4) correspond to the energy of a system that is a rigid box of length L. These equations are used to determine the energy of the transitions, but they fall short because, in reality, the molecule has a nuclear framework that is constantly vibrating and rotating. Both these motions produce some variation in the length of the box and so the box cannot be considered truly rigid. The rotational motion alters the box length by virtue of the fact that the electrical field of the incident radiation (energy being put into the system) “sees” different lengths as the molecule rotates. The result

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is that each line of the expected spectrum spans a larger frequency spread; i.e. it is broadened into a band spectrum rather than being a simple line spectrum (Figure 2).

ε

hν = E1 - E2

ν

hν = E3 - E2

(a)

ε

ν(b)Resonance

frequency

Δν1/2 - bandwidth at half height

Figure 2. (a) Line spectrum. (b) Band Spectrum Integrated absorption (ε) refers to the area under each peak of this band spectrum and measures the total strength (intensity) of the transition. The absorption at a single frequency may not give an indication of the true intensity of a transition because this single absorption does not reflect the total energy of the system. The single line spectrum (Figure 2a) omits some of the energy input that the wider band includes (Figure 2b). The integrated curve, ε(ν), may effectively be used to give a relative concentration of a sample according to the equation which we know as the Beer-Lambert Law: ∫ == CdA lεννε )( [5] where A is the measured absorbance, ε(ν) is the absorption coefficient as a function of frequency (ν), ε is the coefficient itself, ℓ is the pathlength of the cell in the instrument, and C is the concentration of the sample.

(2) According to quantum mechanics, when absorption occurs the correct amount of energy must be provided and the wavefunction of the system must change. The wavefunctions corresponding to the energies observed for the transitions are given above (Eq. 3). The energies (eigenvalues) that correspond to these wavefunctions can be written in terms of the box length, L, (Eq. 4)

mLhnE

8

22

= [4]

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which can then be written in a form describing a one dimensional particle,

( )

2

2

81

mLhNE +

= [6]

where N is the number of π electrons in the system. This gives an experimental value for L, as the energy is based on the measured wavelength, λ. The measured energy of a particular transition is expressed,

λ

ν hchE == [7]

For example, in the transition E2 → E10, quantum mechanics predicts the scenario illustrated in Figure 3.

ψ π10

2 101

2

= ⎛⎝⎜

⎞⎠⎟L

xL

sin

ψ π2

2 21

2

= ⎛⎝⎜

⎞⎠⎟L

xL

sin

ψ π1

21

2

= ⎛⎝⎜

⎞⎠⎟L

xL

sin

E

E1

E2

E10

hν = E2 - E10

Figure 3. The E2 E10 transition.

This transition can occur only if all conservation laws, including energy, are obeyed. If any conserved quantity is changed as a result of the photon absorption, then the two wavefunctions (Ψ2 and Ψ10) depicting energy absorbed and energy released must balance this change, otherwise, the transition is forbidden. In this situation only the ψn ψn±1 are allowed transitions. This means that only transitions for + 1 or – 1 energy increments are allowed, such as, E1 → E2 or E3 → E2. Transitions of more than n ± 1 (n ± 2, for example) are forbidden and their lines will not be observed. This is called a selection rule and will be covered in more depth in advanced courses.

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Prelaboratory Exercise Show that the energy change associated with the spectral transitions observed is

( )2

2

81

mLhNE

EEE HOMOLUMO

+=Δ

−=Δ

where N is the number of π electrons in the box. Hint: some of the references listed will be helpful. Experimental Procedure The samples for this experiment are cyanine dyes. These three dyes differ in color, but only differ structurally in the length of the aliphatic connecting link, which is what we will consider to be the length of the box, L (Figure 4). I. 1,1'-diethyl 2,2'-cyanine iodide II. 1,1'-diethyl 2,2'-carbocyanine chloride III. 1,1'-diethyl 2,2'-dicarbocyanine chloride

NC=CH-(CH=CH)

Et

R CN

Et

+

X

Figure 4. The three dyes - I, r = 0; II, r = 1; and III, r = 2.

..

Use the methanolic stock solutions provided and note the concentrations. Record the spectrum of each dye between 350 and 800nm in 10mm cells with methanol in the reference cell. We will be using the Cary 100Bio UV-visible spectrometer. Your TA will help with the set-up and running of the instrument.

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Instructions for using the Cary UV-vis: 1) Turn computer and instrument on 15 min before running samples 2) Go to PChem students, look for Cary WinUV icon and click on this; the

programme is now open. You will see No graph selected at the bottom if the page and a traffic light and stop sign at the top. The traffic light will show a green light when the instrument is warmed up. (Your TA may have done these steps for you.)

3) Fill one cuvette with methanol and the other with your sample. Be sure your cuvettes are clean, use gloves when handling them and wipe with a kimwipe to ensure this.

4) Place the reference cuvette in the back compartment and the sample cuvette in the front compartment of the instrument. The reference cuvette may be left in the instrument for the duration of this experiment.

5) Go to Setup and note the scan can be changed from the default start and stop wavelengths. You will want the change the stop wavelength to 350nm, then click OK.

6) Click on the traffic light to begin the analysis. The programme will ask you to save each spectrum before you run it, so find the 2005 Chem 373 folder under

My Documents and save to your lab section folder (i.e. Mon 2pm). Create a file name (dye I-Mike) and a sample name (dye I) for each run and click OK—the spectrum will now be running. Remember your file name as you will need to save it later to your disk (step 9).

7) You will be able to view the spectrum on the screen as the instrument is running. When it is done you can look at various options:

a. Peak labels allows you to label peaks – make sure you choose a peak threshold of 0.010 and X and Y labels under label type to ensure all relevant peaks will be labeled

b. Axes Scales allows you to change the axes of your spectrum. We are only concentrating on the peaks between 350 and 800nm so you may wish to set the x-axis accordingly if you did not do so in step 5. Make sure the peaks are not greater than 1.0 Absorbance units. If this is the case, a lower concentration will be necessary.

8) Go to Print to print your spectrum and choose the Lexmark T630 P53 and number of copies for your lab report. ALL lab members will require a copy of EACH spectrum for calculating the length of the box. ONLY one spectrum will be saved later to a disk for each group to analyse the absorption coefficient.

9) Go to File, then Save data as and choose 3 1/2 Floppy (A:). Name your file for your disk and choose Spreadsheet Ascii (*.CSV) as the file type, then Save. This will save your data on your disk such that it will be readable by Excel and you will have all spectral information (Absorbance and wavelength values) which you can plot later.

10) Go to Graph and choose Remove graph, this will leave a blank screen ready for the next sample.

11) Remove your sample from the instrument and wash the cuvette out well with methanol.

12) Repeat steps 4 through 10 for each group.

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13) When all groups have run their spectra you can exit the programme. Go to My Documents and look for 2005 Chem 373, then your lab section folder. You will see all the files you saved for each run. If you did not save your data to a disk in step 9, you can do this now (as described in step 9).

14) Make sure you exit all windows, then shut down the computer and the instrument.

Calculations Integrated Absorption Coefficient, ε The integrated absorption coefficient has been shown to relate to the familiar equation of the Beer-Lambert Law (Eq. 5) CA lε= [5] The experimental spectrum gives A as a function of λ. Include a copy of each experimental spectrum in your report. Your calculations will use wavenumbers. Note that your observed peaks are not symmetrical but appear to have one or more smaller peaks on one side (shoulders). Your TA will assign a dye to your group. For your analysis, use Excel to convert your wavelength data into wavenumber data for one of the dyes. Plot this spectrum using data points from 12500/cm (800nm) to about 10 points past the maximum (this will differ depending on the dye you are working with). Do not include data points around the shoulder of the peak. This curve can be manipulated in Excel to show a half peak. If the main peak is symmetrical, doubling the area of the half peak will give the area of the whole peak. You can estimate this most simply by using the height of the peak multiplied by half its width, Amax × Δν 1

2. This

area is your theoretical area. There are other methods for determining the area under the curve: (1) by counting squares printed on the graph, provided the squares are sufficiently small, (2) by cutting out and weighing the peak, and (3) by use of numerical integration using graphing software. Your TA will show you how using Origin. You should determine the area by means of method 3 and compare this experimental area with the theoretical value. Convert each of your areas to the integrated absorption coefficient by dividing by the cell length and the concentration using the Beer-Lambert Law (Eq. 5).

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Length of the Box Tabulate the absorbance maximum from each of the three spectra and use this data to calculate the empirical value of L for each molecule considering the electrons as one-dimensional particles Eq. [6]. Note that the nitrogen on the left in Figure 3 contributes two π electrons, while each polymethine carbon and the nitrogen on the right contributes one electron. The observed transition is between the highest occupied π orbital and the lowest unoccupied one. Calculate the length of the box using basic trigonometry. You should make the following assumptions:

the molecule is completely planar both C-N-C and C-C-C angles are exactly 120° the length of a C-C bond with bond order 1.5 (as in benzene) is 0.139 nm and the

length of a N-C bond is 0.145 nm the box extends from one bond length on either side of the nitrogens. Thus in (I) 1,1'-

diethyl 2,2'-cyanine iodide, the relevant segment of the molecule is C-N-C-C-C-N-C. Compare your trigonometric values with those obtained from the spectra measurements and rationalize any systematic differences. Hints—units are VERY important! Make sure you write out equations appropriately with all

units to cancel at the end of the calculation. Include a sample of these careful calculations in your report.

Discussion questions 1) Why use MeOH as the reference? 2) Why do we want the spectra to fall between 0.2 and 1.2 absorbance units? 3) Compare the two methods to determine areas for calculating the extinction coefficient 4) What are the lengths of the added carbon chains in dyes II and III (theoretical and

experimental 2C and 4C links) and what assumptions were made to determine these? 5) Show how dye II and III differ in length from dye I for both theoretical and experimental

lengths. Comment on how these compare. *** Attach ALL spectra with proper labels

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References Atkins, P. W. Physical Chemistry, 7th ed. Freeman: New York, 1998, Chapter 12. Bocarsly, J. R., David, C. W. Chem. Educator. 1997, 2, 4. (The Chemical Educator internet access only: http://chemeducator.org.ezproxy.lib.ucalgary.ca/tocs.htm) Kuhn, H. J. J. Chem. Phys. 1949, 17, 1198. Moog, R. S. J. Chem. Ed. 1991, 68, 506. Shoemaker, D. P., et al. Experiments in Physical Chemistry, 5th ed. McGraw-Hill: New

York, 1989, pp 440 – 446 (dyes), 482-489 (Balmer series). Wilson, J. M., et al. Experiments in Physical Chemistry, 2nd ed. Oxford: Pergamon, 1968, pp. 171 – 173 (Balmer series), 57 (Beer-Lambert Law).

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2. The Uncertainty Principle Introduction In 1801, Thomas Young conducted an experiment to show the wave nature of light. His setup consisted of a point source of light shining through two slits and a screen at a distance from those slits. The results showed interference patterns of dark and light bands on the screen. The dark bands represented the points at which the light waves were destructively interfering, i.e. canceling each other out so no light was seen. The light bands observed represented the points at which the light waves were interfering constructively, amplifiying each other to create very bright spots. Were light to consist solely of particles, this would not have been the result; rather, there would have been no breaks in the observed band of light.

light source

light source

constructiveinterference

destructiveinterference

Figure 1: Schematic of Young’s experiment

Young’s experiment served to show the wave-like nature of light. Physical phenomena of light such as reflection, refraction, diffraction, and interference can all be explained by considering light as a wave. To investigate the uncertainty principle in this version of the double slit experiment we will make certain assumptions about the variables involved. The following diagram shows these variables:

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ΔWz

y

z

xs

hν Lθ

Figure 2: Schematic of the experiment showing variables involved

The laser light is represented by hν that passes through the slit of width s. The pattern appears on the wall at a distance L from the laser beam and the distance between the closest two lines of destructive interference that give the bright central spot is designated the size ΔW. The angle θ represents the z component of the beam’s momentum p (sometimes referred to as pz). This component of the beam is restricted in the experiment. The beam travels through the slit parallel to the xy plane. The component pz relates proportionately to ΔW, i.e. as pz gets bigger, so does ΔW. It is assumed that the central spot of ΔW contains the majority of the photons from the laser beam and serves to define the uncertainty in the z direction momentum, pz. The equation used to express the uncertainty principle can be written, hzpz ≥Δ×Δ [1] where Δpz is the z component of the uncertainty in the momentum and Δz is the uncertainty in the position of the photons (i.e. the restriction of the beam). The angle θ relates the variables ΔW and L using the small angle approximation:

LW

L

W

22

1tansin Δ

=Δ

≈≈ θθ [2]

If we consider and zz pp 2≈Δ θsinppz ≈ , then we can write,

LWp

LWpp

ppp

z

zz

Δ≈

Δ≈Δ

≈≈Δ

22

sin22 θ [3]

The total momentum, p, relates to the wavelength, λ, as such, λhp = and so we can write,

LWh

LWppz λ

Δ≈

Δ≈Δ [4]

Since , we make the substitution into the uncertainty equation to obtain our final equation,

sz =Δ

1≥Δ

→≥Δ

→≥Δ×ΔLWsh

LWhshpz z λλ

[5]

All the variables, s, ΔW, λ, and L, are measurable in the lab and in using them in this equation can show that the uncertainty principle holds. This will be the basis of our investigation.

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Prelaboratory Exercise In his paper, Pedro Muiño presents a thought experiment regarding Heisenberg’s uncertainty principle. Write a short paragraph on what the uncertainty principle means and what issues Muiño’s thought experiment raises. Muiño, P. L. J. Chem. Ed. 2000, 77, 1025 is available online through the University of Calgary Library website. Procedure

WARNING: You will be working with lasers in this lab. Although they are in the form of laser pointers, they still pose a danger. Do not under any circumstances look directly into the beam. Be careful when taking your measurements that you do not look into the laser. Consult the website listed in the References for more safety information.

The wavelength of a common red light laser pointer is 650nm ± 10nm. Each group will work with a set of two slit slides. One slide contains four single slits labeled A through D with the listed widths. The other slide consists of two sets of double slits with set widths and spacings listed.

single slit double slit Figure 3: Schematic of slit slides

Set up a ring stand so that you can clamp a single slit slide in such a way so that the laser beam will shine through the slit (Figure 4). You will need one clamp for the laser and three clamps for the slit slide.

paper screen

slit slide (s)

laser pointer

ring stand

L

Wλ

Figure 4: Diagram of apparatus set-up

Do not turn the laser on until you feel you have a good set-up and your TA has checked it.

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This apparatus should be pointing to a wall or door. It can be set on a table or the floor. Turn the laser on briefly and find where the resulting pattern appears on the wall. Tape a piece of paper to the wall so that you can mark the pattern you see. You will have to move the paper periodically, so that the patterns and your markings of them do not overlap. Measure the distance between the laser and the wall, this is your L distance. Turn the laser on and direct the beam through the largest single slit. You will see a faint pattern on the wall where your paper is. Be patient - you may have to move the slide into place before you see a clear pattern. You will also have to turn the lights off in the lab to see a brighter pattern. Mark where you see the central band so that it can be measured later and label this (slit width, s, and distance, L). Using the same slit move the ring stand to a different distance and mark the pattern you observe. Try to measure at least three distances. Repeat this for the next smallest slit and use the same distances you did for the larger slit. You may want to try to see the patterns of the two very small slits, but these are more difficult to detect. Go back to using the largest slit and change the orientation of the slide (90° turn in the clamp) and examine the resulting pattern. Describe how this is different from your initial measurement and come up with an explanation. Keep in mind what Muiño mentions in his paper, i.e. that it is erroneous to consider light behaving like a fluid that must squeeze and expand through the slit. Now insert the double slit slide into the clamp and shine the laser through the smallest slit spacing. Note the pattern you see and mark this on a piece of paper. Count the number of small bands you see in the centre band. This centre band corresponds to the single slit central band you saw earlier. Do this for as many of the slit spacings you can and use the same distances you did with the single slit experiment. Convert all measurements to meters before doing your calculations to evaluate the equation given above. Discussion and Report Guidelines

1) Describe the 90° reorientation of the slide and the effect on the pattern observed. 2) Tabulate your data and discuss any trend you see between the small and large slits.

Show examples of your calculations. 3) Discuss the pattern seen with the double slits. What is the explanation for this? 4) Discuss errors in this experiment. Can any of these be eliminated? 5) How would the use of a green light laser (λ = 532nm) in this investigation affect the

values of the calculation?

15

References Marcella, T. V. European Journal of Physics. 2002, 23, 615. (multi-slit diagrams) McEvoy, J. P., Zarate, O. Introducing Quantum Theory. UK: Icon Books, Ltd., 1996. Muiño, P. L. J. Chem. Ed. 2000, 77, 1025. Nariz, O., Arndt, M., Zelinger, A. Am. J. Phys. 2003, 71, 319. (slit diagrams) Rioux, F. J. Chem. Ed. 2005, 82, 1210. Rioux, F., Johnson, B. J. Chem. Educator. 2004, 9, 12. (The Chemical Educator internet access only: http://chemeducator.org.ezproxy.lib.ucalgary.ca/tocs.htm) (diagrams of slit experiments) Helpful Websites http://zebu.uoregon.edu/~imamura/208/jan27/hup.html http://plato.stanford.edu/entries/qt-uncertainty/ http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c1 Laser safety information http://www.hpa.org.uk/radiation/understand/information_sheets/laser_pointers.htm

16

17

3. Molecular Symmetry & Point Groups NOTE: There is a pre-lab exercise to submit for this Dry Lab. Your TA will check this before you start your lab. A more comprehensive understanding of symmetry will be achieved if a molecular model kit is used to complete the exercises in this lab. Your TA will be able to supply an octahedral centre if you do not already have one. For further practice of symmetry principles consult the list of websites at the end of this lab.

I Introduction Finite symmetries are useful in the study of molecules. They are used in the classification of molecules, simplifying quantum mechanical calculations on molecules, determining the presence of certain molecular properties such as molecular polarity and chirality. It will become apparent how important symmetry considerations are for molecules and orbitals in physical processes such as NMR spectroscopy and x-ray crystallography in molecular structure determination. In the next two dry labs, we will focus on finite symmetries and groups of finite symmetry operators. We will use P.W. Atkins, Physical Chemistry, Chapter 15, as our principle reference. Another book by F.A. Cotton, entitled Chemical Applications of Group Theory, either the first or second edition, is an excellent source book too. In Chemistry 331/333, you have had some exposure to finding symmetry elements and symmetry operations. This Dry Lab is meant to be either a practical review or an introduction to symmetry elements and symmetry operations. The set of all symmetry operations for a molecule form a mathematical structure called a group. Here, we will look at group structure, classes of symmetry operations, and the naming of molecular point groups. We will examine how to use the group structure to predict when a molecule is polar or chiral. You will apply these ideas to several molecules or molecular ions. II Symmetry elements and operations A symmetry operation transforms a molecule into itself so that the transformed molecule is indistinguishable from the original structure. Also, at least one point in the molecule is always left undisturbed by the transformation. This is the origin of the term, molecular point group. Often, two or more atoms are permuted during the course of the molecular transformation. Since atoms of the same type are indistinguishable, the transformed molecule is indistinguishable from the starting molecule. So, a molecular point group consists of all those symmetry operations that leave a point in the molecule invariant and permute identical atoms.

18

Symmetry operations come in several flavors: 1) no operation, 2) rotation, 3) reflection, 4) inversion, 5) improper rotation. The operation is denoted by symbols in plain text (E, Cn, etc.), whereas the actual axis or plane referred to in a molecule is denoted by symbols in italics (E, Cn, etc.). 1) E The identity transformation This operation does nothing and is a symmetry element of all molecules. The symbol E is used from the German Einheit, meaning unity.

H1

O

H2 H1

O

H2

E 2) Cn The rotation operation This takes place about a rotation axis. The molecule is rotated by an angle of 2π/n radians (n is an integer). For a water molecule the rotation is 2π/2 = π radians or 180˚:

H1

O

H2 H2

O

H1

C2

C2

By convention, a rotation by a positive angle indicates a counterclockwise direction(as above) and a rotation through a negative angle is performed in a clockwise direction. Each rotation of the molecule into itself is an n-fold rotation along the axis. Some molecules have more than one axis, as in the example below (C4 and C2 axes exist):

MBB

BB

A

A

C4

C2

In these cases, the convention dictates that the axis with the higher value of n is called the principal axis (C4 above). Some molecules contain more than one operation along an axis. Consider the square planar structure XeF4 – it has a C4 axis through the Xe atom that is perpendicular to the plane of the molecule. Carrying out two consecutive 4-fold rotations, i.e. C4

2 = C4C4, about a C4 axis is equivalent to conducting a C2 rotation about the same axis. Therefore, the C2 axis is

19

coincident with the C4 axis. In XeF4, the C4 axis is the principal axis of rotation with a C2 axis coincident to it. In H2O, the C2 axis is the principal axis of rotation with no other axes coincident to it.

XeF3F4

F2F1

Xe

F2F3

F1F4

C4 XeF3F4

F2F1

Xe

F1F2

F4F3

C2

For a diatomic molecule (HCl), a rotation by any arbitrary angle can be performed about the internuclear axis. Such an axis is called a C∞ axis and is the principal axis of rotation.

H ClC∞

3) σ Reflection through a plane A reflection takes place in a plane of symmetry, sometimes called a mirror plane (speigal is German for mirror). Reflection planes can either contain rotation axes or be perpendicular to an axis of rotation. When a plane is perpendicular to a principal axis of rotation it is designated by the symbol σh and sometimes called the horizontal plane.* When a plane contains a principal axis of rotation it is usually denoted by σv. That is to say the plane and axis are parallel to one another. This is usually called the vertical plane.*

XeF3F4

F2F1

σh

H1

O

H2

σv

A special case of plane is the dihedral plane, denoted σd. In the BF3 molecule below there are three reflection planes, each containing BFn (only the plane through B-F2 is shown). Each of these planes bisects the angle between the remaining two fluorines in the rotation of the secondary C2 axis (the principal axis here is the C3 axis through the central boron with the fluorines perpendicular to that axis; only one of the C2 axes is shown – the one through B-F2 permuting F1 and F3).

* Keep in mind that not all σh planes are ‘horizontal,’ nor are all σv planes ‘vertical.’

20

B F2

F3

F1

σd

C2C3

Note that often other criteria are needed to distinguish between σv planes and σd planes. For example, water has two reflection planes, but both of them are σv planes—one as shown above, the other which is perpendicular to it. These σv planes are both parallel to the principal axis, which is C2. There are no other C2 axes perpendicular to this principal axis which can be bisected by a plane, so neither of these planes can be σd. In XeF4, the reflection plane perpendicular to the principal C4 axis is a σh plane. There are four planes which bisect the FnXeFm angles. There are also secondary and tertiary C2 axes lying along Fn-Xe-Fm and bisecting the FnXeFm angle respectively. Those planes corresponding to the bisecting C2 axis are those that are σd planes. Take a look at benzene for dihedral planes. Your TA can help you to see these. 4) i Inversion through the origin The inversion operation, denoted i or I, occurs through the centre of inversion, i. A point in the molecule (x, y, z) is taken and transformed to another point (-x, -y, -z). The chemical environments are identical. Water does not posses an inversion centre, whereas XeF4 does.

M

BA

AB

(-x, -y, -z)

(x, y, z)

Xe

F3F4

F2F1

(-x, -y, -z)

(x, y, z)

H1

O

H2

(-x, -y, -z)

(x, y, z)

5) Sn Improper rotation Improper rotation is composed of two successive transformations: 1) an n-fold rotation about an axis (using S notation instead of C notation to denote the axis—Sn); 2) a reflection in a plane that is perpendicular to that Sn axis.

CC C

H1

H2

H4

H3

CC C

H2

H1

H4

H3

CC C

H4

H3

H2

H1

90° rotation

step 1 step 2

reflectionIn the 1,3-disubstituted allene above, the n-fold rotation does not need to match an actual n-fold rotation axis in the molecule (the n-fold rotation here is by 90˚ whereas a rotation axis

21

should be 180˚ for complete symmetry). This particular rotation through the three-carbon chain does not yield a representation of the structure that is indistinguishable from the original molecule (step 1). A reflection through a plane through the central carbon must be carried out to achieve this (step 2). By the same reasoning, the reflection plane perpendicular to the Sn axis need not be an actual reflection plane. In the complex ion [Co(NCS)4]+2, the reflection plane that is being considered for step 2 (after the rotation of step 1) is not a true mirror plane. However, upon the reflection step we see the original structure.

90° rotation reflection

step 1 step 2Co NCSSCN

NCS

Co NCSSCN

NCS

Co NCSSCN

NCS

NCSNCS NCS•

Water does not possess an improper axis of rotation because it does not have a plane of symmetry that is perpendicular to the axis of rotation (it does have vertical reflection planes, but these have no bearing on the Sn operation). The CH4 molecule, however, has four S4 axes (one through each CH bond), but no C4 axis.

90° rotation reflectionC

HH

H H

C

H H

HH

C

HH

H H

Improper rotation can simply be denoted by the Sn notation,

S4C

HH

H H

C

HH

H H

but it is easier to write out both the rotation and reflection steps as we have done previously. We can know the n value for the Sn operation by the angle used for the rotation step. In CH4 this is 90˚, which is rotation by 2π/n or 360º/4, so n = 4 and the molecule has S4 improper rotation. This is also true for [Co(NCS)4]+2

which requires a 90˚ rotation, so it also has an improper rotation S4. If a molecule had a 60˚ rotation this would mean it had S6 improper rotation (360º/6 = 60˚).

22

A convenient way to determine point groups for molecules is to use a flow chart (Figure 1). Consider ammonia as an example. Is the molecule linear? The answer is no. It has only one axis of rotation, a C3 axis that is the principle axis of rotation. Are there any C2 axes perpendicular to the principal axis? The answer is no. Is there a reflection plane perpendicular to the principal axis? Again, the answer is no. Are there three σv planes containing the C3 axis? The answer is yes. Therefore NH3 must have C3v symmetry. Pre lab exercise: Assign point groups to five common objects such as candlesticks, wine glasses, and traffic signs (stop sign with and without the word STOP) using the Schoenflies flow chart provided. (For more assistance, refer to McKay and Boone, 2001.)

Exercise 1: Determine the molecular point groups for the species given by your TA in your laboratory session. Use the flow chart. It will help to build some of these with a molecular model kit.

Your TA will provide you with about 10 sample molecules and/or molecular ions to work on.

23

Start Here

molecule

linear?

molecule has high symmetry;more than 2 Cn axes with n>2

i?

Td

yes

no

no

no

yesC4 or C5?

i?

Cnv

no

σh?

Cnh

Oh

Ih

C4

D∞h

no

yesyes

Cn?are there nC2 axesperpendicular to Cnaxes? Dnh

Dnd

Dn

σ?σh?

i?

σd?

Cnvσv?

S2nSn?

Cs

yes

yes

yes

no

no

no

no

yes

no

yesyes

no

yes

Ci

C1

no

no

no

yes

yes

C5

Cs

Figure 1: Schoenflies flow chart for determining point groups of molecules

24

III Symmetry Groups We use the phrases symmetry groups and molecular point groups synonymously. A mathematical group, G = {G,·}, consists of a set of elements1 G = {E, A,B,C,D, ...} and a binary relation, called group multiplication or group product or simply multiplication or product, denoted by ‘·’, which satisfies the following properties:

(a) The product of any two elements A and B in the group is another element in the group, i.e., we write A·B ∈ G. (b) If A, B, C are any three elements in the group then (A·B)·C = A·(B·C). Therefore, group multiplication is associative, and frequently, we omit the brackets.

(c) There is a unique element E in G such that E·A=A·E=A, for every element A in G. The element E is called the identity element.

(c) For every element A in G, there is a unique element X in G, such that X·A = A·X = E. The element X is referred to as the inverse of A and is denoted A-1. The identity is its own inverse.

The number of elements in a group is called the order of the group. Frequently, this is denoted by the symbol h. It is common to omit the symbol ‘·’ when no confusion will arise Also, when there can be no confusion, we will use the symbol G for the group rather than G. We can think of the group elements as symmetry operations of a molecule. This means that we can “translate” A·B to: “first we perform the symmetry operation B on the molecule followed by symmetry operation A.” The net result of such consecutive action on a molecule is another symmetry operation. Take dichloromethane as our example. The molecular structure and Cartesian axis system are shown below.

C

ClCl

x

y

z

HH

A B

A B In this figure, you can think of the origin of the Cartesian axes as centred on the C atom with the z-axis bisecting the HCH and the ClCCl angles. The x-axis lies in the HCH plane, while the y-axis is in the ClCCl plane.

1 Do not confuse symmetry elements discussed in section 2 with group elements. The set of symmetry elements does not form a group, only the symmetry operations form a group. The term element used in this definition is standard usage in set theory.

25

The identity operation, E, leaves the molecule unchanged. The C2 axis lies along the z-axis. The C2 operation transforms the dichloromethane molecule as A B

A B

B

AB

A

H

C

Cl Cl

HH

C2

ClCl

H

C

Carrying out two consecutive C2 operations is equivalent to the identity transformation. There are two reflection planes in the molecule; both contain the rotation axis. One plane is the plane of the page containing the ClCCl plane. We will label this plane σ′ (yz). The second plane is perpendicular to the plane of the page; we will label it σ(xz).

The action of σ′ (yz) is to give the arrangement of atoms shown , where the two hydrogen atoms have been interchanged, while the two chlorine atoms and carbon are unchanged.

C

ClClA B

A BHH

C

ClCl B

AB HH

A

σ(yz)

The σ(xz) permutes the chlorine atoms, but leaves carbon and the two hydrogen atoms fixed

A B

A B

B

B

A

C

ClCl

HH

C

ClCl

HH

σ(xz)A

Applying the plane σ(yz) twice, i.e., (σ′ (yz))2 = σ′ (yz)σ ′ (yz) = E, we get the identity. This means that σ′ (yz) is its own inverse (see (d) above). Similarly, we find that (σ(xz))2 = σ(xz)σ(xz) = E, and σ(xz) is its own inverse.

Now, if we carry out a σ(xz) reflection first and follow it by a σ′ (yz) reflection, we get

A B

A B

B

B

A

C

ClCl

HH

C

ClCl

HH A

σ(yz)σ(xz)

Comparing this diagram to that of a C2 rotation we see that the result is identical. Therefore, we say that σ′ (yz)σ(xz) = C2. You can show that performing the reflections in reverse order yields the same result. Note that the symmetry elements remain fixed and are not transformed to new positions when the atoms in the molecule move to new positions.

What about carrying out a C2 rotation followed by the reflection σ′ (yz)? Performing these symmetry operations yields

C

ClClA B

A B

ClCl B

B

A

AHH

σ(yz)C2

H

C

H

This is equivalent to a σ(xz) operation. You can show that carrying out these operations in reverse order affords the same result.

Next, we compute the product σ(xz)C2

A B

A B

B

B

A

AHH

σ(xz)C2

H

C

H

C

ClCl ClCl

This is identical with a σ′ (yz) operation. Again, check that the reverse sequence of operations yields the same result.

26

Using the definition of the group and the products of symmetry operations that we have just uncovered, we can construct a group multiplication table:

C2v E C2 σ(xz) σ’(yz) E E C2 σ(xz) σ’(yz) C2 C2 E σ’(yz) σ(xz) σ(xz) σ(xz) σ’(yz) E C2

σ’(yz) σ’(yz) σ(xz) C2 E This table contains all the information about the group and its structure. The name of this molecular point group is C2v. There are some observations to make about this table. (1) Notice the inner four-by-four box. In each row and each column, each operation

appears once and only once. In other words, each row and each column is a permutation of the others. This is a feature possessed by all group multiplication tables.

(2) We can identify smaller groups within the larger one. For example, {E,C2} is a group. There are two others; what are they? These smaller groups are called subgroups of C2v.

(3) In this particular table, we observe that the group product is commutative. This is not necessarily true for other groups.

In the ammonia molecule, the nitrogen atom is the fixed point:

NH H

HA

B

C12

3

The molecule has a C3 axis of rotation. Note that the both the C3 and C3

2 operations occur about the same C3 axis. There are three reflection planes; each plane contains an NH bond and bisects the opposing HNH angle. Each plane can be labeled by the number on the hydrogen atom it contains; thus σ1 is the plane containing HA

1. This reflection interchanges atoms HB and HC, leaving HA fixed

AB

C12

3

AB

C

12

3N

HH

HN

H

σ1H H

27

Note that the numbers remain fixed to their original positions. The symmetry elements must not shift with the atoms when they are transformed to new positions. In the previous example, we used labels for the reflection planes that were expressed in terms of the fixed axes external to the molecule. In this example, it is not obvious how to do that. So keep in mind here that the numbers stay fixed and the letters move. You will want to use the double labels whenever it is inconvenient to label axes with references outside the molecule. It is irrelevant whether the numbers stay fixed and the letters move or vice versa. Just be consistent within a given application. Since σ1

2 = σ22 = σ3

2 = E, each reflection is its own inverse. Since C33 = C3

2C3 = C3C32 = E,

C32 is the inverse of a C3 rotation (or C3 is the inverse of C3

2). Recall that a C3 rotation is a 120° rotation in the counterclockwise direction about the rotation axis, while C3

2 is a 240° counterclockwise rotation. Also, we can interpret a C3

2 rotation as a -120° rotation (clockwise). Given these definitions and considerations, here is the complete group multiplication table:

C3v E C3 C32 σ1 σ2 σ3

E E C3 C32 σ1 σ2 σ3

C3 C3 C32 E σ3 σ1 σ2

C32 C3

2 E C3 σ2 σ3 σ1

σ1 σ1 σ2 σ3 E C3 C32

σ2 σ2 σ3 σ1 C32 E C3

σ3 σ3 σ1 σ2 C3 C32 E

Exercise 2: Confirm the results in the table for C3v using ammonia, NH3, as the sample molecule. If you use another molecule or different labeling for the symmetry elements, the table will be basically the same except some rows and columns may be interchanged. Follow the examples given below and draw your results out in the same manner using the same numbering sequence for the hydrogen atoms. Rules to remember:

1. the C3 axis rotates counterclockwise 2. in a multiplication the row across is the first term and the column down is the

second term 3. each row (column) is a permutation of another row (column) 4. no symmetry element can occur more than once in a row or column 5. labels for reflection planes always correspond to those in the original molecule,

despite the numbering system after an operation (see examples B and C below)

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Example A: rotation, C3 × C3 = C32, for ammonia

N

H1 H2

H3 C3N

H3 H1

H2 C3N

H2 H3

H1

N

H1 H2

H3N

H2 H3

H1C3

2

Example B: reflection, σ3 × σ2 = C3, for ammonia

N

H3

H1 H2

σ3

NH1

H3

H2

σ2

N

H2

H3 H1

N

H1 H2

H3N

H3 H1

H2C3

Example C: both operations combined, C3 × σ1 = σ2, for ammonia

N

H1 H2

H3N

H3 H2

H1σ2

N

H1 H2

H3N

H3 H2

H1C3 N

H2

H3

H1

σ1

29

Exercise 3: The point group of the BrF5 molecule is C4v. There are eight symmetry operations in the group. Construct the group multiplication table using the template below. You may use the textbook, models, or other resources to help you. Use the given molecule and numbering system as well as the rules from Exercise 1 and the following hints. Draw out your results as you did before. Hints for constructing the table:

1) start with the identity row and column 2) do the rotational axes × rotational axes 3) remember σ planes × themselves = E 4) look for patterns in the table and confirm these by drawing out structures

rotations for BF5

eflection planes for BF5

Br

F

F1 F3

F4

F2

C4C42C4

3

r

σ2

Br

F

F1 F3

F4

F2

Br

F

F1 F3

F4

F2

σ4

σ1

σ3

30

C4v E C4 C4

2 C43 σ1 σ2 σ3 σ4

E C4 C4

2 C4

3 σ1 σ2 σ3 σ4

Happily it is not necessary to derive a group multiplication table each time you want to find the point group of a molecule or molecular ion. It is sufficient to determine the presence of only a particular subset of symmetry elements (Cn or σx). Once you have discovered the molecular point group, the character table for the point group will give you the remaining group operations. See the Appendix for printed character tables.

IV Physical implications of symmetry Chemists use symmetry in a wide variety of ways. In quantum chemistry, symmetry allows us to classify molecular orbitals and state wave functions. With this classification, we can simplify calculations of certain matrix elements or expectation values. Often, it is possible to decide when matrix elements or expectation values are zero just on the basis or symmetry alone. In vibrational spectroscopy, frequently, the vibrational modes of a molecule are classified according to their behavior under the symmetry operations of the molecular point group. We shall explore some of these ideas in a future dry lab and later you will see them in action in higher level physical chemistry courses dealing with spectroscopy. You are already familiar with polarity and chirality from earlier chemistry courses. Now we will examine these concepts in greater depth with respect to the specific symmetry of a molecule. Relating these concepts to symmetry will help in future chemistry courses. Polarity: The idea is very simple; a polar molecule has a permanent electric dipole moment, μ, which is a vector quantity. The dipole moment has a specific orientation in the molecule. If a molecule has a rotation axis, then the dipole moment must lie along the rotation axis, since no dipole moment can change under a rotation. Consider water, its dipole moment lies along its C2 axis:

H1

O

H2

μ

31

If a molecule possesses a secondary axis or a plane of symmetry perpendicular to the axis of rotation, then it can have no dipole moment. This would mean that μ would have to lie along all such rotation axes (both principal and secondary) and all mirror planes. This cannot happen and for such molecules μ = 0. Look at the example of XeF4, it has both principal and secondary axes which are perpendicular to each other. The individual dipole moments along each Xe-F bond cancel each other out and the result is no dipole moment, μ = 0.

XeFF

FF C4 C2

Along similar lines of reasoning, if a molecule has a centre of inversion, then it can have no permanent dipole moment. A dipole moment would change sign under inversion and that is not physically possible. Another way of looking at this is that with a centre of inversion, the dipole moment would have to be a point, and that is not possible for a vector quantity. Note that molecules from the point groups Td, Oh, and Ih will not have dipoles. Exercise 4: Look back at the examples from Exercise 1. Using symmetry, identify the molecules with permanent electric dipole moments. Chirality: If a molecule is chiral, it is optically active. A molecule is chiral when mirror images of the molecule cannot be superimposed on one another. For a molecule to exhibit chirality, it must have no improper axes of rotation. Note that S2 = i (an inversion) and that S1 = σ (a reflection). Thus, if a molecule possesses reflection planes or an inversion centre, it cannot be chiral. Molecules from the point groups Dnd, Dnh, Td, and Oh are not chiral. Exercise 5: From the set of molecules your TA provides, identify the chiral species. Hint: You may want to build models of these to help visualize the symmetry elements.

32

Useful Websites http://radchem.nevada.edu/chem431/lecture_4__molecular_symmetry.htm (lecture notes) Dr. Arvi Rauk: http://www.chem.ucalgary.ca/SHMO/ (SHMO, orbital symmetry) http://www.reciprocalnet.org/common/index.html (go to alphabetical listing)

References Atkins, P., de Paula, J. Physical Chemistry, 7th ed. New York: W. H. Freeman and Company, 2002. Bishop, D. M. Group Theory and Chemistry. Oxford: Clarendon Press, 1973. Blinder, S. M. Introduction to Quantum Mechanicis. New York: Elsevier Academic Press, 2004. Cotton, F. A. Chemical Applications of Group Theory, 3rd ed. New York: Wiley, 1990. Pilar, Frank L. Elementary Quantum Chemistry, 2nd ed. New York: Dover Publications, Inc., 1990, pp. 553ff. Vincent, Alan. Molecular Symmetry and Group Theory. England: John Wiley & Sons, Ltd., 1977.

More help with symmetry Lloyd-William, P., Giralt, E. J. Chem. Ed. 2003, 80, 1178 (Td symmetry) McKay, S. E., Boone, S. R. J. Chem. Ed. 2001, 78, 1487 (symmetry with common objects)

4. Applications of Group Theory—Character Tables We will use P.W. Atkins, Physical Chemistry, 7th ed., Chapter 15 as our primary reference for this Dry Lab. I Introduction When a molecule possesses symmetry, we can use that symmetry to classify the atomic and molecular orbitals. In calculating molecular properties, such as dipole moments or evaluating selection rules for electronic transitions, we can use symmetry to simplify the calculations. Molecular symmetry is a valuable tool in classifying vibrational motion in the molecule and is useful in predicting which vibrational transitions will occur. Thus symmetry operations have many implications where orbitals in a molecule are concerned and can be approached in the same way as atoms in a molecule such as we saw in the previous symmetry lab. For example, atomic orbitals move with the operations in the same way as atoms:

C2

s s

+ + C2

no sign changesign change

p p

Note the sign changes for the different orbitals under the C2 operation, the s orbital does not change whereas the p orbital does. Bonding can be described in terms of electron distribution within an orbital of a molecule. This is represented with molecular orbitals (MO) which are the result of a linear combination of atomic orbitals (LCAO). An energy diagram of this is drawn as shown:

H2 H2+

HA HAHB HB

σ∗

σ σ

σ∗

Notice the differences in the MO diagrams for the diatomic molecule H2 and its molecular ion, H2

+. The electron density for each of these is described by writing the wave equation as a linear combination of two 1s atomic orbitals, one on each of the two hydrogen atoms.

33

Labeling the hydrogens as HA and HB, the bonding σ combination can be written as ( )ψ b S A Bs s= +

+1

2 11 1

( ),

while the antibonding σ* combination can be expressed as ( )ψ a S A Bs s= −

−1

2 11 1

( ),

where the 1 and 1 orbitals are centred on hydrogen atoms A and B, respectively. In the bonding molecular orbital, σ, electron density is built up in internuclear region (as seen in the diagram where the electrons have been placed). In the antibonding orbital, σ*, electron density is removed from the internuclear region.

sA sB

This approach to describing the electron distribution in molecules is called molecular orbital (MO) theory. Expressing the MO as a linear combination of atomic orbitals is called the MO-LCAO method. For all complex molecules or molecular ions, MO-LCAO theory provides a good model for understanding bonding in molecules. We will endeavor to show how symmetry can lead to deriving and reading character tables as well as explaining and understanding orbitals and nodes in a molecule. II Transformations In the lab on Molecular Symmetry we saw how the atoms in a molecule were permuted when the molecule was subjected to symmetry operations. Now, we shall associate with each atom some object or collection of objects and see what happens to these objects under a symmetry transformation. Examples of these objects include coordinate systems and atomic orbitals. These are the most common objects to be associated with atoms in molecules. We will work with SO2, the same example as in Atkins (Ch 15); however, our approach will differ slightly. Let us attach a coordinate system to each atom in SO2 as shown,

S

OBOA

z

y

x where xS is the x-axis associated with sulfur, xA the x-axis associated with oxygen atom A, and xB is the x-axis associated with oxygen atom B and similarly for the y-axes and z-axes. The entire molecule lies in the yz-plane. Recall from your molecular symmetry lab, the SO2 molecule belongs to the C2v molecular point group and the symmetry operations are, E, C2, σv, σv′: E identity operation C2 180˚ rotation along z-axis, in this case, through the centre of the S atom bisecting the O-S-O angle σv reflection plane along the C2 axis in the xy plane σv′ reflection plane perpendicular to the C2 axis in the yz plane

34

Consider what happens to the axes attached to each atom under each of the group operations. Of course, the identity operation E leaves each axis system unchanged. Observe the axes on each atom through the σv′(yz) operation (we start with this operation rather than C2 for simplicity as we will see later on):

S

OBOA

σv'(yz)z

y

x x axis changes sign for any atom through this reflection plane; this is because the plane bisects the axis (think of a traditional Cartesian coordinate system

with zero at the centre of each atom and positive and negative values on each side of the axis)

y axis does NOT change sign z axis does NOT change sign These observations can be written out as such:

σv′(yz) xS = -xS σv′(yz) yS = yS σv′(yz) zS = zS σv′(yz) xS = -xS σv′(yz) yA = yB σv′(yz) zA = zB σv′(yz) xS = -xS σv′(yz) yB = yA σv′(yz) zB = zA

We could represent all the coordinates as a 9-component vector and each transformation as a 9×9 matrix. That is cumbersome and by noticing that x-axes always transform into x-axes, y-axes into y-axes, and z-axes into z-axes, it is possible to represent the transformation by three 3×3 matrices. Thus we write for he σv′(yz) reflection:

σv'(yz) (xS, xA, xB) = (xS, xA, xB)

σv'(yz) (yS, yA, yB) = (yS, yA, yB)

σv'(yz) (zS, zA, zB) = (zS, zA, zB)

-1 0 0 0 -1 0 0 0 -1

1 0 0 0 1 0 0 0 1

1 0 0 0 1 0 0 0 1

Each line in the matrix represents the value of the sign as well as the position in space of the axis. Consider the C2 rotation operation:

35

S

OBOA

C2 z

y

x

x axis changes sign for any atom through this rotation because the axis now points in the opposite direction relative to the C2 axis y axis also changes sign because the axis points in the opposite direction relative to the C2 axis z axis does NOT change sign; axis on any atom is unchanged by this operation Results of the C2 operation on all the axes is written out as such:

C2xS = -xS C2yS = -yS C2zS = zS C2xA = -xB C2yA = -yB C2zA = zB C2xB = -xA C2yB = -yA C2zB = zA

Results for the matrices corresponding to these axes:

C2 (xS, xA, xB) = (xS, xB, xA)

C2 (yS, yA, yB) = (yS, yB, yA)

C2 (zS, zA, zB) = (zS, zB, zA)

-1 0 0 0 0 -1 0 -1 0

1 0 0 0 0 1 0 1 0

-1 0 0 0 0 -1 0 -1 0

All lines of the matrix show negative values for the x- and y- axes through the rotation; however, the second and third lines show that xA and xB change positions. The oxygen A atom rotates into the position of the oxygen B atom and vice versa. Thus the values in a line of the matrix show the position (+ or -) of the axis as well as the position of the atom in space (A or B).

36

We can write out the results for the other reflection plane, σv(xz) in the same way:

S

OBOA

σv(xz) z

y

x

σv (xz) (xS, xA, xB) = (xS, xB, xA)

σv (xz) (yS, yA, yB) = (yS, yB, yA)

σv (xz) (zS, zA, zB) = (zS, zB, zA)

-1 0 0 0 0 -1 0 -1 0

1 0 0 0 0 1 0 1 0

1 0 0 0 0 1 0 1 0

The exchange of position of the oxygen atoms occurs for the reflection operation through this plane as it did under the C2 operation. The identity transformation, E, is simply represented by the identity matrix for each axis (we will not write this out). Considering all the transformations from the symmetry operations for the x-axis (xS, xA, xB), we have the four matrices representing all four operations which are written out as follows:

D(E) =1 0 00 1 00 0 1

D(σv'(yz)) =-1 0 0 0 -1 0 0 0 -1

D(C2) =-1 0 0 0 0 -1 0 -1 0

1 0 00 0 10 1 0

D(σv(xz)) =

These four distinct matrices are said to form a (reducible) representation of the group C2v. In fact, the matrices mimic the group operations; taking products of these matrices would allow us to reconstruct the group multiplication table (as seen in the previous lab). The four matrices representing the transformations of the y-axis (yS, yA, yB) and the z-axis (zS, zA, zB) also form (reducible) representations. For (yS, yA, yB), we have

D(E) =1 0 00 1 00 0 1

D(σv'(yz)) =D(C2) =-1 0 0 0 0 -1 0 -1 0

-1 0 0 0 0 -1 0 -1 0

D(σv(xz)) =1 0 00 1 00 0 1

In this case, notice that not all of the matrices are distinct compared to the matrices for the x-axis. D(E) is the same as D(σv’(yz)) and D(C2) is the same as D(σv(xz)) for the x-axis. The x-axis representation is said to be faithful, the y-axis is not. Observe that the representation constructed from (zS, zA, zB) is not faithful either:

D(E) =1 0 00 1 00 0 1

D(σv'(yz)) =D(C2) = D(σv(xz)) =1 0 00 1 00 0 1

1 0 00 0 10 1 0

1 0 00 0 10 1 0

The dimensions of these representations are equal to the dimension of the matrices (3×3) and also to the number of objects that are used to construct the representation (three axes, x, y, and z). Every set of objects on the atoms (coordinate system or orbital) produces a matrix representation. If we were to place atomic p orbital functions on each atom, we would construct a set of matrices like those computed in Atkins (Chapter 15), with σv (his) = σv(xz) (ours). His matrix for the transformation of the three px orbitals is the same as our matrix for the transformation of (xS, xA, xB). Think about why this is the case.

37

Exercise 1: Find the matrix representations for the py and pz orbitals on SO2. Write these as shown in the example above for the x-, y-, and z-axes (the results above are the matrix representations for the px orbital on SO2) . You will have 12 matrices for each orbital (py, pz) with x-, y-, and z-axes for a total of 24 matrices. Label the sets of matrices as faithful or not faithful.

S

OBOA

S

OBOA

py pz

It is possible to have many different matrix representations; the larger the group the larger the number of matrices in each representation. It is often easier to work with a more invariant quantity—the trace of the matrix. The trace is the sum of the diagonal elements of a matrix, D:

χ(R) = Tr RD(R) ==∑D jjj

d( ),

1

where d is the dimension of the representation and is the (jj)-th element of the matrix D(R). The symbol χ is frequently used in two senses. First, when written as χ(R), it means the trace of the matrix representing the transformation corresponding to the symmetry operation R (χ(C

D jj ( )R

2) for example). Secondly, when it is written alone as χ, it means the set of all traces for all symmetry operations in the group. It is often called the character of the representation when used in the latter sense. Even so, people will speak of χ(R) as the character of the individual operation R. For example, for the representation defined by (xS, xA, xB), we have Table 1: Reducible representation for x-axis C2v E C2 σv (xz) σv′(yz) Γ 3 -1 1 -3

where Γ is a symbol denoting the (reducible) representation. It is represented by the set of traces for each symmetry operation, i.e., the character. So, we would write:

χ(E) = Tr χ(C2) = Tr= 3 = -11 0 00 1 00 0 1

-1 0 0 0 0 -1 0 -1 0

χ(σv'(yz)) = Tr-1 0 0 0 -1 0 0 0 -1

1 0 00 0 10 1 0

χ(σv(xz)) = Tr = -3= 1

Another advantage of working with characters is that the traces of matrices representing elements in the same class are identical. This is no advantage here since each element is in a class by itself (each symmetry operation is distinct). However, for larger groups, this can simplify calculations to a considerable extent. Look at ammonia from the previous lab for example, symmetry operations include C3 and C3

2. These are considered to belong to the same class and the same applies to all three σ operations as well. We need only to find the trace for one element of each of these classes—two traces total as opposed to five.

38

Looking again at the matrix representation for (xS, xA, xB), we observe that there is still some structure that we have not made use of. The matrices are of block diagonal form:

D(E) =1 0 00 1 00 0 1

Since no transformation in the group interchanges xS with either xA or xB, we can actually look at xS separately (the upper left hand box). The matrix representation for xS is a 1×1 matrix representation: D(E) = 1, D(C2) = -1, D(σv(xz)) = 1, D(σv′(yz)) = -1 This is labeled Γ1 and leaves us with xA and xB, the matrices for which can be written in 2×2 form (lower right hand box):

D(E) = D(C2) = 0 -1-1 0 D(σv(xz)) = 0 1

1 0 D(σv'(xz)) = -1 0 0 -1

1 00 1

trace (Γ2) = 2 = 0 = 0 = -2 These are the traces for xA and xB and are called Γ2. We say that we have reduced the 3×3 matrix representation to a 1-dimensional representation and a 2-dimensional matrix representation. We refer to the 3×3 matrix representation as a reducible representation. The 1-dimensional representation cannot be reduced further and it is said to be an irreducible representation. Note that the traces for the 2×2 matrices can be further reduced; this is then a reducible representation for xA and xB. The original 3×3 matrix can be reduced down or decomposed into 1×1 and 2×2 matrices. Table 2: xS, xA, xB

C2v E C2 σv (xz) σv’(xz) Γ1 1 -1 1 -1 Γ2 2 0 0 -2

Exercise 2: The above example just showed a decomposition of the matrices to obtain the representations for the x-axis. Using the templates below, repeat this analysis for the representations based on (yS, yA, yB) and (zS, zA, zB) to obtain the “character tables” analogous to Tables 1 and 2 (hint – use your results for the py and pz orbitals from Exercise 1 and start by finding the trace of each matrix and work from there). You will need to write out the 2×2 matrices and traces on a separate sheet of paper.

39

Table 3: yS, yA, yB

C2v E C2 σv(xz) σv’(yz)Γ

Γ1 Γ2

Table 4: zS, zA, zB C2v E C2 σv(xz) σv’(yz)Γ

Γ1 Γ2

We can find all the irreducible representations of a point group in this way. The number of irreducible representations is equal to the number of classes. For C2v, there are four classes, one for each symmetry operation, so there must be four irreducible representations. The characters for each of these irreducible representations can also be determined. These have been found for all molecular point groups and have been compiled into tables called character tables. III Character Tables Atkins provides the character tables for most of the important molecular point groups in his Appendix. The five tables for the point groups Cn (n = 2, 3, 4, 5, 6) are listed there and in many other texts (see references). We shall examine the character tables for the two point groups C2v and C3v and our comments easily generalize to other point groups. This is the character table for the C2v point group:

C2v E C2 σv(xz) σv'(yz) h = 4 A1 1 1 1 1 z x2,y2,z2

A2 1 1 -1 -1 Rz xy BB1 1 -1 1 -1 x, Ry xz BB2 1 -1 -1 1 y, Rx yz

40

The top row in the table gives the name of the point group, followed by the symmetry operations arranged according to classes. For C2v, each symmetry operation is in a class by itself. The next four rows and the first five columns give the name and character for the irreducible representations. There are four irreducible representations (equal to the number of classes of symmetry operations)—A1, A2, B1, B2. The A and B notation is used to signify the character under the principal rotation (in this case C2), A if the character is +1, B if it is –1. Each irreducible representation is 1-dimensional since the trace of the matrix for the identity operation is 1. (If the trace were 2, then the representation would be 2-dimensional, if 3, then the representation would be 3-dimensional; 2- and 3-dimensional representations are not irreducible.) The sum of the squares of the values of the irreducible representations for each class must be equal to the order of the group (h), in this case, 12+12+12+12 = 4 for the class represented by E. In the last two columns, functions at the fixed point are listed according to which irreducible representation they transform. For example, x transforms as the B1 irreducible representation, as does a px orbital. The product xy transforms as A2 and therefore so does a dxy orbital. The terms Rx, Ry, and Rz refer to operators such as angular momentum operators. (You can think of as an example of R$Lx x.) If we compare Γ1 (also χ1) for the transformation of xS in Table 2 with the irreducible representations in the character table for C2v, we see that it corresponds to B1. That is because the S atom is at the fixed point of the point group. We say that xS forms a basis for the representation B1. The same is not true for xA and xB; they form a basis for a 2-dimensional reducible representation (recall the 2×2 matrices above). The character, Γ2 (or χ2), for this representation is given in Table 2 as well. Given the character of a reducible representation, there is a formula that connects the number of occurrences of an irreducible representation in the reducible representation. Let Γ denote a reducible representation and Γ′ an irreducible representation of a point group. If is the number of times the irreducible representation, Γ′, appears in the reducible representation, Γ, then

cΓ '

c g C ChΓΓ

'

'( ) ( ) ( )( )= ∑1 χ χ C [1]

where h is the number of symmetry operations in the group (4 in this case), g(C) is the number of symmetry operations in the class of operations C (here this is 1, but for ammonia it would be either 2 or 3 because there are two symmetry operations in the C3 class and three in the class of σ). The term, χ(Г´)(C), is the character of the irreducible representation, Γ′ (from the character table). Finally, χ(C) is the character for the reducible representation for the class of operations C (this is Γ2, the actual values in Table 2). Recall that all symmetry operations in a class (Cn or σ) have the same character. Considering the class of symmetry operations and referring to the character table for C2v for the character of each irreducible representation, and Table 2 for the character of each reducible representation, we can write out the equations: E C2 σv (xz) σv’(yz) cA1 = ¼[(1)(1)(2) + (1)(1)(0) + (1)(1)(0) + (1)(1)(-2)] = 0 cA2 = ¼[(1)(1)(2) + (1)(1)(0) + (1)(-1)(0) + (1)(-1)(-2)] = 1 cB1 = ¼[(1)(1)(2) + (1)(-1)(0) + (1)(1)(0) + (1)(-1)(-2)] = 1 cA2 = ¼[(1)(1)(2) + (1)(-1)(0) + (1)(-1)(0) + (1)(1)(-2)] = 0

41

We have shown that the reducible representation Γ2 decomposes into the irreducible representations A2 and B1. Frequently, we denote this fact by writing, . Also, we can decompose the original 3-dimensional reducible representation into, .

Γ2 1= ⊕A2 BA B2 1⊕2

We will use the character table for C3v as an example of how to determine the basis functions for these irreducible representations:

C3v E 2C3 3σv h = 6 A1 1 1 1 z x2+y2, z2

A2 1 1 -1 Rz E 2 -1 0 (x,y)(Rx, Ry) (x2-y2, xy) (xz, yz)

We can see upon examination of the character table that there are three irreducible representations and three classes. Two of the irreducible representations, A1 and A2, are 1-dimensional (both +1 as they do not change sign under the rotation operation) and the third, E, is 2-dimensional. (Do not confuse the identity operation, E, and the label, E, which designates 2-dimensional representations. Other dimensions can exist and are labeled T for 3-dimensional and F for 4-dimensional, but we will not investigate these here.) The sum of the squares of the values (dimensions) of the three irreducible representations is 6 (E—[12 + 12 + 22], or 2C3—2[12 + 12 + (-1)2], or 3σv—3[12 + (-1)2 + (0)2]). This is the order of the group, h. Suppose that we have been given the following reducible representation:

C3v E 2C3 3σv

Γ 5 2 -1 Then, decomposing the reducible representation into irreducible representations according to Eq. [1], we get: E 2C3 3σv cA1 = 1/6[(1)(1)(5) + (2)(1)(2) + (3)(1)(-1)] = 1 cA2 = 1/6[(1)(1)(5) + (2)(1)(2) + (3)(-1)(-1)] = 2 cE = 1/6[(1)(2)(5) + (2)(-1)(2) + (3)(0)(-1)] = 1 and so we would write for the decomposition. Γ = ⊕ ⊕A A1 22 E Exercise 3: Decompose the reducible representations for px and py obtained in Exercise 2 (Γ2 from Tables 3 and 4) into their irreducible components by using equation [1] as was shown in the above ammonia C3v decomposition. Filling out the templates given will help keep track of your values. From the decomposition determine the basis functions for these irreducible representations of C2v (i.e. A1, A2, B1, or B2) and then draw the py and pz orbitals on the oxygen atoms which correspond to these bases. (Note that we have ignored the s orbitals on both the S and the two O atoms. A full treatment would include them.)

42

py decomposition E C2 σv (xz) σv´(yz) cA1 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = cA2 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = cB1 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = cB2 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = pz decomposition E C2 σv (xz) σv´(yz) cA1 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = cA2 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = cB1 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = cB2 = ¼[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] = IV Molecular Orbitals We want to be able to write down the basis for the reducible representations in terms of the basis for the irreducible representations. For example, in the C2v case, we have been working with the reducible representation Γ2 which has the basis (xA, xB). This representation decomposes into A2 ⊕ B1 as we have seen (Γ2 = A2 + B1). What are the bases for A2 and B1 in terms of xA and xB? Atkins gives a description of the procedure in his chapter on character tables and we shall use the method that he describes. The technique is called building symmetry-adapted linear combinations (SALC) of atomic orbitals. It works quite well for 1-dimensional representations but care must be exercised when applying it to 2- and 3-dimensional representations. (Atkins illustrates this in his discussion) The steps given in Atkins are repeated here with corresponding results: 1. Construct a table showing the effect of each symmetry operation on each orbital (or

any function for that matter) of the original basis (see Table 3a below). 2. To generate the combination of a specified symmetry species, take each column in

turn and: (i) Multiply each member of the column by the character of the corresponding operation (Table 3b, 3c).

(ii) Add together all the orbitals in each column with the factors as determined in (i) (iii) Divide the sum by the order of the group (total number of symmetry operations, h)

Atkins treats ammonia (C3v) as an example. As our example, we shall work with Γ2 for a C2v molecule, where we work with px orbitals rather than the x-axis (recall that these transform equivalently). Following the instructions in step 1, we set up the table:

43

Table 3a: Transformations for px orbitals Original Basis

Operation pxA pxB

E pxA pxBC2 -pxB -pxA

σv(xz) pxB pxA σv’(yz) -pxA -pxB

Notice the operation of C2 on pxA yields -pxB and the same is true for this operation on pxB (the result is -pxA) because the atoms in these points swap positions.

Step 2 follows by using the A2 irreducible representation (1, 1, -1, -1) to obtain results for the first column, pxA (2(i) through 2(iii)): Table 3b: Transformations for px orbitals

Original Basis (step 2 i) Operation pxA (A2) pxB (A2)

E pxA (+1) pxB (+1) C2 -pxB (+1) -pxA (+1)

σv(xz) pxB (-1) pxA (-1) σv’(yz) -pxA (-1) -pxB (-1)

¼ [(pxA)(+1) + (-pxB)(+1) + (pxB)(-1) + (-pxA)(-1)] = ( ) ( )1

412p p p p p pxA xB xB xA xA xB− − + = − ,

The process is repeated for the second column (pxB) still using A2 irreducible representation, ¼ [(pxB)(+1) + (-pxA)(+1) + (pxA)(-1) + (-pxB)(-1)] = ( ) ( )1

412p p p p p pxB xA xA xB xB xA− − + = − .

These two functions differ by a minus sign which is unimportant. If we apply each operation in turn to this linear combination, we generate the character for the A2 irreducible representation. The same process can be applied to B1 for each column resulting in the following, Table 3c: Transformations for px orbitals

Original Basis (step 2 i) Operation pxA (B1) pxB (B1)

E pxA (+1) pxB (+1) C2 -pxB (-1) -pxA (-1)

σv(xz) pxB (+1) pxA (+1) σv’(yz) -pxA (-1) -pxB (-1)

44

For pxA, ¼ [(1)(pxA) + (-1)(-pxB) + (1)(pxB) + (-1)(-pxA)] = ( ) ( )1

412p p p p p pxA xB xB xA xA xB+ + + = + ,

For pxB, ¼ [(1)(pxB) + (-1)(-pxA) + (1)(pxA) + (-1)(-pxB)] = ( ) ( )1

412p p p p p pxB xA xA xB xB xA+ + + = + .

These results are identical. Again, applying each symmetry operation in turn to this function, we obtain the character for the B1 irreducible representation. The significance of this process is founded on the following principle: Only linear combinations of atomic orbitals transforming according to the same irreducible representation can have nonzero overlap. In the above example, only pxS and ( )1

2 p pxA xB+ orbitals transform according to B1. They can be combined in two possible ways ( )a bp p pxS xA xB+ + , [2] ( )a b' ' ,p p pxS xA xB− + [3] where and b are constants. Absolute magnitudes are shown so that the signs may be indicated explicitly. The first combination represents a bonding π molecular orbital, the second, an antibonding π* molecular orbital. The bonding orbital combination in Eq. [2] is shown in Figure 1. There is one nodal plane, the plane of the molecule (yz), and delocalization of electrons is possible throughout the π-network. The entire molecular orbital will transform as the B

a

1 representation. The antibonding combination in Eq. [3] is portrayed in Figure 2. There are now three nodal planes, one in the molecular plane (yz) and one between each SO bond. This orbital also transforms according to the B1 representation.

SO

O

xz

y

SO

O

xz

y

Figure 1: Bonding π combination (B B1) Figure 2: Antibonding π* combination (A2)

In the SO2 molecule, all molecular orbitals can be classified in terms of the irreducible representations of the point group C2v. We have discovered the symmetry of the π MO and π* MO only. The actual numerical values of the coefficients, a and b (from Eqs. [2] and [3]), must be determined by solving the Schrodinger equation within some approximate scheme. Once the MOs have been obtained, an energy diagram can be constructed and electrons placed in each MO according to the Pauli exclusion principle. Not all of the MOs are used in this process. The last MO to be filled is called the highest occupied MO or HOMO. The next orbital in the energy sequence, which is left unfilled, is called the lowest unoccupied MO or LUMO. Usually, in simple molecules, if a π-orbital is present, it corresponds to the HOMO and the π* to the LUMO.

45

Exercise 4: The point group of ethene, C2H4, is D2h. The coordinate axes and the molecular structure are shown in Figure 3. Assume all bond angles are 120°. We intend to construct symmetry adapted linear combinations (SALCs) of atomic valence orbitals to describe the bonding in ethene. NOTE: there are three parts to this exercise, a, b, and c.

C1 C2

HB

HA

HC

HD

x

z

y

Figure 3: Ethene structure and coordinate system

Table 4: Character table for D2h group

D2h E C2 (z) C2 (y) C2 (x) I σ(xy) σ(xz) σ(yz) Ag 1 1 1 1 1 1 1 1 BB1g 1 1 -1 -1 1 1 -1 -1 BB2g 1 -1 1 -1 1 -1 1 -1 BB3g 1 -1 -1 1 1 -1 -1 1 Au 1 1 1 1 -1 -1 -1 -1 BB1u 1 1 -1 -1 -1 -1 1 1 z BB2u 1 -1 1 -1 -1 1 -1 1 y BB3u 1 -1 -1 1 -1 1 1 -1 x

The point group D2h consists of eight symmetry operations: the identity operation, E, a 2-fold rotation about the z-axis, C2(z), a 2-fold rotation about the y-axis, C2(y), a 2-fold rotation about the x-axis, C2(x), a centre of inversion, i, a reflection in the xy plane, σ(xy), a reflection in the xz plane, σ(xz), and a reflection in the yz plane, σ(yz). The basis on each hydrogen is a 1s orbital and these are denoted (sA, sB, sC, sD). On each carbon there is one 2s and three 2p orbitals denoted (s1, s2) and (px1, px2), (py1, py2), and (pz1, pz2), respectively. (a) Calculate the character for the reducible representations for each type of orbital on ethene (remember that reducible representations will have values greater than 1), then decompose these reducible representations into their irreducible components. To do this apply each operation to the orbital listed (HsA for example) and fill in the template below. Then use Eq. [1] as you did in Exercise 3 for SO2 to obtain the irreducible representations. This method of determining reducible representations differs from the previous method as it does not involve matrices. Use the labeling system in Figure 3! Other labeling systems will give different (but not necessarily incorrect) results. Hint—look at the number of atoms for a particular orbital. For example, there are four H atoms in total, so perform the operation on one and then multipy by 4 (ΓHs for the operation E is 1 × 4 = 4, so place the value 4 in the table). Also, there are two C atoms, so the character for these orbitals is multiplied by 2 (Γ2sC for the operation E is 1 × 2 = 2). The same strategy is used for the 2 p orbitals for coordinates x, y, and z.

46

Table 5: Reducible representations for ethane

E C2 (z) C2 (y) C2 (x) i σ (xy) σ (xz) σ (yz) ΓHs Γ2sC Γ2px Γ2py Γ2pz Use the following set up for all basis sets from the character table for determining irreducible representations: E C2 (z) C2 (y) C2 (x) i σ (xy) σ (xz) σ (yz) cAg = 1/8[( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )] =

(b) Building SALCs: step 1 from Atkins rules above has been done for you in Tables 6 through 10. Table 6: Transformations for Hydrogen 1s orbitals

Original basis operation sA sB sC sD

E sA sB sC sDC2 (z) sB sA sD sCC2 (y) sD sC sB sAC2 (x) sC sD sA sBi sC sD sA sBσ(xy) sD sC sB sAσ(xz) sB sA sD sCσ(yz) sA sB sC sD

Table 7: Transformation of carbon 2s orbitals

Original basis operation s1 s2

E s1 s2C2 (z) s1 s2C2 (y) s2 s1C2 (x) s2 s1i s2 s1σ(xy) s2 s1σ(xz) s1 s2σ(yz) s1 s2

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Table 8: Transformation of carbon 2px orbitals Original basis

operation px1 px2

E px1 px2C2 (z) -px1 -px2C2 (y) -px2 -px1C2 (x) px2 px1i -px2 -px1σ(xy) px2 px1σ(xz) px1 px2σ(yz) -px1 -px2

Table 9: Transformation of carbon 2py orbitals

Original basis operation py1 py2

E py1 py2C2 (z) -py1 -py2C2 (y) py2 py1C2 (x) -py2 -py1i -py2 -py1σ(xy) py2 py1σ(xz) -py1 -py2σ(yz) py1 py2

Table 10: Transformation of carbon 2pz orbitals

Original basis operation pz1 pz2

E pz1 pz2C2 (z) pz1 pz2C2 (y) -pz2 -pz1C2 (x) -pz2 -pz1i -pz2 -pz1σ(xy) -pz2 -pz1σ(xz) pz1 pz2σ(yz) pz1 pz2

Using these tables choose the one which corresponds to a particular set of orbitals and determine the bases for each irreducible representation for that set of orbitals. For example, in part (a) you found ГHs had a certain number of irreducible representations; one of the tables corresponds to the Hs orbitals. Apply the characters from the D2h character table which corresponds to the irreducible components you obtained in (a) and follow step 2 from Atkins rules. This will show the orbital symmetry for each basis set. Draw out the orbital symmetries which match the irreducible representations.

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(c) Create a table of symmetry types and their bases. Place the orbitals you drew out in (b) into the appropriate boxes of the table below. Label the drawings as σ, σ*, π, and π*. When appropriate calculations are performed, the π MO corresponds to the HOMO and the π* to the LUMO. Label the HOMO and LUMO of the π to π* transition. Ag 1sH orbital looks something like this:

Ag 2sC orbital looks something like this (σ):

Note that not all the boxes will be filled.

1sH 2sC 2px 2py 2pz

Ag

BB1g

BB2g

BB3g

Au

BB1u

BB2u

BB3u

49

V Expectation Values and Matrix Elements One application of group theory that is important is in the calculation of matrix elements for selection rules and expectation values. Direct product representations We denote the direct product of two irreducible (reducible) representations by Γ × Γ‘. The character of the direct product representation is obtained by multiplying the characters of each irreducible representation. For example, for the group C2v, if we take the direct product of the A2 and B1 representations we get the following: C2v E C2 σv (xz) σv’(yz) A2 1 1 -1 -1 BB1 1 -1 1 -1 A2 × B1 1 -1 -1 1 Comparing the character of this 1-dimensional representation to the others in the C2v character table we see that A2 × B1 = B2. This is general; the direct product of two or more 1-dimensional representations always yields a 1-dimensional representation. The direct product of two or more 2- or 3-dimensional representations usually yields a reducible representation which can then be decomposed into a sum of irreducible representations according to methods we detailed above. Molecular states When we determine the molecular state for a system we take the direct product of the irreducible representations of each MO that each electron occupies. For closed shells (no unpaired electrons), the resulting state is usually the totally symmetric state, A1, in the case of C2v. The full designation is 1A1 since for a closed subshell, the ground state is always a spin singlet. Excited states are usually obtained by exciting one or more electrons into empty (virtual) orbitals. For example, in our SO2 example, suppose an electron is excited from the π-orbital (B1) to the π* orbital (B1). Then, the state of the molecule would be B1 × B1 = A1, the totally symmetric representation. Both the ground and excited states would then be A1. The resulting excited state may be either a singlet or triplet spin state, 1A1 or 3A1. Exercise 5: The ground state of ethene is 1Ag. Given the symmetries of the π and π* orbitals of ethene in Exercise 4(c), find the symmetry species of the excited state formed by exciting an electron from the π orbital to the π* orbital. Will the state be a singlet or a triplet or are both possible? (Hint—use direct product) Note that the direct product of any representation with itself always contains the totally symmetric representation. As used here, the term “contains” means that the direct product is either equal to the totally symmetric representation or it decomposes into a sum of representations at least one of which is the totally symmetric representation. This property is important for what comes next.

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Matrix elements and expectation values Let O be some operator, then a matrix element involving the operator O is written [4] d OτΨ Φ*∫ .

Then, when , the matrix element is equivalent to an expectation value, if the state Ψ Φ= Ψ is normalized. Since they represent physically observable quantities, for matrix elements to be nonzero, they must remain unchanged after any symmetry operation. Let the transform as the irreducible representation, O as the irreducible representation, and as the Γ irreducible representation. Now, compute the direct product representation

ΨΓΨ ΓO Φ Φ

[5] Γ Γ ΓΨ ΦO .If this new representation contains the totally symmetric representation, then the integral will be nonzero. As an example, we shall consider the dipole moment operator, written as

v vμ = −er , [6] where vr xe ye ze= + +~ ~ ~

1 2 3 . [7] This operator appears not only in expectation values but also in matrix elements that determine the probability of whether a particular transition between states will occur. Let us show that the dipole moment of SO2 can be nonzero. Let the ground state wave function be Ψ, and it transforms as A1 of C2v. We have [μ τ τ τ= − + +∫ ∫e d x d y d zΨ Ψ Ψ Ψ Ψ Ψ* * * ]∫

Ψ

. [8] For the dipole moment to be nonzero, at least one of these integrals must be nonzero. Referring to the C2v character table, we see that x, y, and z, transform as B1, B2, and A1, respectively. Since ( ), we have for the first integral, ΓΨ = A1 Γ ΓΨ* =Γ Γ ΓΨ Ψx = =A B A B1 1 1 1 , which is not the totally symmetric representation. Hence, the first integral must be zero. Also, we can show that the second integral is zero since it transforms as BB2, ΓΨΓyΓΨ = A1BB2A1 = B2. For the third integral, we see that Γ Γ ΓΨ Ψz = =A A A A1 1 1 1, and hence it may be nonzero. This is consistent with the dipole moment having to have a nonzero component in the z-direction since that corresponds to the C2 rotation axis. Let Ψ be the ground state function and Φ be the wave function for some excited state and suppose that the spin multiplicity of both the ground and excited states are the same (ΔS=0). Then, a transition between these states has nonzero probability if at least one of the matrix elements, ,*,*,* ∫ ∫ ΦΨΦΨΦΨ∫ zdydxd τττ [9] is nonzero. This is because the band intensity is proportional to the square of the modulus of these three integrals. For the integrals to be nonzero, the direct product representation in Eq. [5] must contain the totally symmetric representation. For SO2, we have already discussed the case of forming an excited state 1A1 from the π and π* orbitals. Is a transition from the ground state to this excited state allowed? The answer is yes, since Γ Γ contains the AΓΨ z Φ 1 representation.

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Exercise 6: Using the result from Exercise 5, show whether a transition from the π to π* orbital is allowed or forbidden in ethene. Set up a calculation according to Eq. [5] and pay attention to x, y, and z components from the D2h character table. Helpful Websites See previous Molecular Symmetry lab Collection of character tables online: 1) http://www.mpip-mainz.mpg.de/~gelessus/group.html2) http://mathworld.wolfram.com/CharacterTable.html Slides laying out examples of setting up decompositions: http://mutuslab.cs.uwindsor.ca/macdonald/250-LectureNotes/Fall2002/Bonding-Notes5.pdf References and helpful materials Atkins, P. W. Physical Chemistry, 7th ed. Freeman: New York, 1998, Chapter 15. Bishop, David M. Group Theory and Chemistry. Oxford: Clarendon Press, 1973. Blinder, S. M. Introduction to Quantum Mechanics in Chemistry, Materials Science, and Biology. Boston: Elsevier Academic Press, 2004. Cotton, F. A. Chemical Applications of Group Theory, 3rd ed. New York: John Wiley & Sons,1990. Kettle, Sidney F. A. J.Chem.Ed. 1999, 76, 675 (sym, gr th and nodes, understand char tables) Levine, Ira N. Physical Chemistry, 5th ed. Boston: McGraw Hill, 2002, pp. 804-16. Pilar, Frank L. Elementary Quantum Chemistry, 2nd ed. New York: Dover Publications, Inc., 1990, pp. 553ff. Vincent, Alan. Molecular Symmetry and Group Theory. England: John Wiley & Sons, Ltd., 1977

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5. Hückel, Part I Introduction We have seen the importance of symmetry in determining polarity and chirality in a structure. Symmetry is also important in spectroscopy. We can use symmetry along with calculations to predict the likely energies for transitions in a spectrum and how the transitions that produce this spectrum are polarized i.e. in what direction, x, y, or z, the incident energy needs to be to induce an excitation. Spectroscopy There are various types of transitions from one state to another in a molecule or system with a gain or loss of quantum energy. These can be observed spectroscopically and are dictated by certain rules, called selection rules. The energy that is either absorbed or emitted by the transition, if it is an allowed transition, is dependent on ν (E = hν). Each transition has a distinct E, or more appropriately, a distinct frequency, ν. This ν is what we will measure in this lab using UV-visible spectroscopy. The allowed transition is properly characterized by an electric dipole within the structure. The radiation vector from the excitation source (in our case, the instrument) interacts with the electromagnetic field in the molecule that serves to polarize the transition in a particular direction, x, y, or z. The transition is then labeled as being polarized in x, y, or z. You observed something of this nature in the last lab where you determined the direct product of the ground state and the excited state characters with x, y, and z components to find where the allowed π → π* transition occurred for ethene. The direct product of states (characters) that gave a non-zero result (A1 symmetry) showed the allowed transition and its polarization. Some molecules will have such high symmetry that the coordinates, x and y, or x, y, and z, are jointly included in an irreducible representation in the character table. You can see this in benzene where x and y share the E1u character. If the result of the direct product using this irreducible representation gives the non-zero result, then such molecules are termed ‘xy’ polarized and the radiation vector will cause a transition in this plane. Hückel Molecular Orbital Theory The aim of Hückel molecular orbital calculations is to find the energies of the electron pairs in a system. Hückel theory is successful in analyzing planar conjugated systems such as butadiene and benzene. A matrix of equations is built, the roots of which describe the π orbital energy levels. From this data, we can then know the delocalization energy of a conjugated system (resonance stabilization) and the wavelength of the lowest energy electronic transition.

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Hückel calculations are somewhat complicated, but can be simplified. These simplified calculations are called SHMO and are characterized by using four basic rules:

1. overlap integrals, S, are set to zero 2. resonance integrals, Hij, between non-neighboring atoms are set to zero 3. resonance integrals, Hij, between neighboring atoms are set to β 4. Coulomb integrals, Hii, are set to α

The α term is the core energy of an electron localized to the 2p atomic orbital of a C atom. The β term is the energy associated with the interaction of the two carbon 2p orbitals overlapping in a π fashion, hence the π → π* energy is always given in terms of β. In setting up a SHMO calculation, we can determine the coefficient for β which can then be used to calculate the wavelength of the electronic transition. In Laboratory Tutorial Molecular orbital diagrams and energy levels Recall that the point group of benzene is D6h. Consult the Appendix for the character table. We will be using the SHMO programme for this lab and you will copy and draw information off the screen by hand; you can go to the site directly:

http://www.chem.ucalgary.ca/SHMO/ This site can also be accessed through the department’s website, look for Dr. Arvi Rauk under Emeritus Professors and go to his site, you will see a link called Interactive SHMO. Click on this and go to SHMO. The page will look like this (but you will not see the structure drawing yet!):

Figure 1: Screen shot from SHMO programme

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To draw the benzene molecule as shown above, click Add and then click on the blank screen. Drag and click to add all 6 carbons in a ring. Go to Minimize (this will orient all the bonds into a proper ring). You may need to rotate the structure so that it is upright as in the above diagram (though this is not necessary for further calculations and observations). It is a good idea to Minimize after each change you make to your structure. This ensures good energy level data. Click on Show Orbitals on the lower right hand corner. You will see the orbital configuration for the ground state energy level. Click Up (Down) to view the other orbitals or click directly on the energy level of the desired orbital. When the following coordinate system is assigned, these orbitals can be considered the px orbitals. You will not be able to see the full px orbital using SHMO, but you can determine the + or – value by consistently assigning one of the colors (red or blue) to represent the positive lobe of the orbital in all the diagrams.

z

yx y

xz

Figure 2: Coordinate axis for benzene showing three of six px orbitals

Below is a diagram of what you see in the programme. Notice the orbitals and the number of nodes at each energy level for the benzene molecule.

Energy

Figure 3: Energy level diagram for benzene showing corresponding orbitals

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Tutorial Question 1 Write out the energies calculated by the SHMO programme that correspond to the different levels in the diagram above, then indicate the HOMO and LUMO. What do the HOMO and LUMO represent? (The SHMO programme gives coefficients for the α term, ignore these and consider all values as 1α.)

Determining the character of each energy level Each energy level in the MO diagram has a corresponding irreducible representation. We determine this through applying the symmetry operations to the px orbitals. A shortcut to drawing out all the orbitals can be used in the case of benzene. Look at the MO diagram, there are six MOs, but four of these show degeneracy. These four orbitals will not have the characters A or B because they are doubly degenerate. Consulting the character table shows us that these MOs must be labeled E, the character for double degeneracy. We can also take a shortcut in terms of the u and g designations. Look at the orientation of the orbitals with regard to the operation of the inversion centre. Two of the orbitals below represent the ground state (Figure 4A). The shaded area of one orbital does not invert through the inversion centre and is designated with a negative sign (-1) in the character table. The inversion is therefore unsymmetrical or ungerade and labeled u. The other two orbitals are from another symmetry configuration and show a positive inversion, which is labeled g (gerade) in the character and +1 in the character table (Figure 4B).

y

xz

i i

A B Figure 4: Orbital diagram for benzene showing

u (ungerade) inversion (A) and g(gerade) inversion (B) Tutorial Question 2

Assign u and g designations to the orbitals in Figure 3 above. Your TA can help you and will check your answers.

To properly determine the characters for the lowest and highest energy states we must consult the character table for benzene. We know these states are not E because they are not degenerate, so they must be either A or B. Recall from your lab on character tables that A and B notation signifies the character of the principal rotation—A if the operation is +1, B if it is –1.

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Tutorial Question 3 Consult the character table for benzene and assign the character for the remaining energy levels according to the principle rotation.

Normally we would follow this procedure by performing the direct product of the π and π* states to find the character for the excited state. The benzene molecule has a difficult character table to implement this due to the presence of degenerate energy levels. You will use the techniques you learned in the character table lab to manipulate orbital symmetry in an analysis of the molecules naphthalene, quinoline, and phenanthrene. Performing the Hückel calculation A traditional Hückel calculation for benzene would consist of six equations set up in a matrix set equal to zero:

H11 - ES11 H12 - ES12 H13 - ES13 H14 - ES14 H15 - ES15 H16 - ES16

= 0

H23 - ES23 H25 - ES25H22 - ES22 H24 - ES24 H26 - ES26H21 - ES21H33 - ES33 H35 - ES35H32 - ES32 H34 - ES34 H36 - ES36H31 - ES31H43 - ES43 H45 - ES45H42 - ES42 H44 - ES44 H46 - ES46H41 - ES41

H54 - ES54 H55 - ES55 H56 - ES56H51 - ES51 H52 - ES52 H53 - ES53

H61 - ES61 H62 - ES62 H63 - ES63 H64 - ES64 H65 - ES65 H66 - ES66 These equations follow the numbering system for the benzene molecule where atom 1 has neighboring atoms 2 and 6 and so on.

1

2

3

4

5

6

Figure 5: Benzene molecule with labeled carbon atoms

This matrix can be simplified according to the above four rules to achieve the following matrix:

α - E β 0 0 0 β

0 β α - E β 0 0 0 0 β α - E β 0

0 0 0 β α - E β

β 0 0 0 β α - E

β α - E β 0 0 0

= 0

This can further be simplified by dividing all quantities by β and letting

βα Ex −

=

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We then get the following matrix:

x 1 0 0 0 11 x 1 0 0 0

0 0 1 x 1 00 0 0 1 x 1

0 1 x 1 0 0

1 0 0 0 1 x

= 0

Expansion of the determinant gives us a polynomial equation which has six roots, x = ±1, ±1, and ±2. These roots correspond to the six energy levels (recall that two of these were degenerate) you saw in the SHMO programme, E = α ± β and E = α ± 2β. There are various sources that give different values for β based on what property we are trying to examine: bond strength, resonance or delocalization energy, electronic excitation energy in a spectrum. For example, Rauk cites that the π bond strength in the ethylene molecule can be determined using a value for β of 33kcal/mol. Platt shows that there is a β value that relates to the spectroscopic signal of a molecule and can be used to estimate the differences in energies between MOs, he uses -23,000/cm for this value. (Cotton, p. 440, Platt, p. 1168.) Another value that has been successfully used for spectroscopic data is given by Schwenz; this is 75kcal/mol. Hückel calculations of naphthalene and quinoline In applying symmetry operations to the px orbitals you must use the coordinate systems given here or your answers will not be consistent. Use this numbering system for all your calculations.

1

2

3

45

6

7

8

910z

yx N

1

2

3

45

6

7

8

910

naphthalene quinoline Figure 6: Naphthalene and quinoline with coordinate axes and numbered atoms

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Exercise 1 Follow the instructions provided above for benzene and draw naphthalene using the SHMO programme. From your results, sketch the molecular orbital diagram showing the energy levels and their corresponding orbitals, labeling the HOMO and LUMO. What is the point group of naphthalene? (Some of the lobes will be different sizes, this reflects their relative contributions and is not important here in our analysis. You may draw your diagrams with same size lobes. Again, ignore the coefficients for α in the SHMO programme.) You will not need to draw a new structure to analyze the orbitals and energies for quinoline. Simply click on Change then click on the atom to be changed (see Fig 1, benzene screen). A box will show a number of changes that are possible, pick N2 and click ok. There should now be a blue nitrogen atom in place of the carbon at position 4. The orbitals will be slightly different from naphthalene as well as the energies. Click Minimize to ensure the lowest energy conformation of the molecule has been achieved. Exercise 2 Draw out the MO diagram for quinoline including energy levels and corresponding orbitals. Label the HOMO and LUMO as well. What is the point group of quinoline? Exercise 3 Assign characters to each MO for both naphthalene and quinoline following the instructions for the benzene example. You will need to consult the character tables in the Appendix to do this. Be sure to use the axes given (Figure 6). Exercise 4 Indicate the π → π* transition for each molecule on their respective MO diagrams. What is the energy value in terms of β for each transition? Exercise 5 Determine whether these transitions are allowed and with what polarization. To do this you may want to review your lab on character tables. First find the ground state of the molecule (the totally symmetric state) and then find the excited state by direct product of the characters of the two states, HOMO and LUMO. Next, set up the direct product equation for the transition using Eq. 5 from the character table lab. Consult the character table to see where each axis component is listed (x, y, z). Exercise 6 Set up the Hückel matrix using the numbering sequence for naphthalene (Figure 6). If you were to solve this matrix, what would be the roots? Exercise 7 Review the possible values given above for β (Platt and Schwenz). Convert these values to energy units (Joules) and determine the likely frequency range for the naphthalene excitation spectrum.

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Discussion Compare the results of your investigation of naphthalene and quinoline. How does the presence of a non-carbon atom in the structure change the energies and the orbitals? Discuss how symmetry plays a role in this with respect to the different characters that were assigned to the energy levels. Your report should show all of your work from the Exercises: MO diagrams with energy labels and characters, results from direct products, and Hückel matrices. References Blinder, S. M. Introduction to Quantum Mechanics in Chemistry, Materials Science, and Biology. New York: Elsevier, 2004, (Ch 12). Cotton, F. A. Chemical Applications of Group Theory, 3rd ed. New York: John Wiley & Sons, 1990, (Ch 5, Ch 7, Appx III, Appx IX). Platt, J. R. J. Chem. Phys. 1950, 18, 1168. Rauk, A. Orbital Interaction Theory of Organic Chemistry, 2nd ed. New York: John Wiley & Sons, 2001, (Ch 5). Rauk, A. http://www.chem.ucalgary.ca/SHMO/ Schwenz, R. W., Moore, R. J. Physical Chemistry Developing a Dynamic Curriculum. Washington, DC: American Chemical Society, 1993, (pp. 280-291).

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6. Hückel, Part II PAH spectroscopy lab

Introduction Now that you have learned about some Hückel calculations, we can see how these can be made useful in the lab. Molecular spectroscopy studies the absorption of light by a sample and gives information about the sample’s energy levels. This information is in the form of a spectrum of lines that are due to quantized electronic transitions as well as rotational and vibrational motion. We are interested in the electronic transitions in this laboratory and will concentrate on the π → π* transition. Quantum calculations can lead to predictions of the frequencies of these lines seen in the spectrum. You have seen in the last exercise that it is possible to calculate the energy of a one-electron jump, the parameter β in SHMO. This transition or change in the energy state of a molecule corresponds to a frequency which can be seen in a spectrum. Polycyclic aromatic hydrocarbons (PAH) are found in many materials upon their incomplete combustion including diesel fuels, wood, grilled food, and cigarette smoke [Wingen, et al., J.Chem.Ed. 75, 1599 (1998)]. There is some environmental concern as some PAH are considered carcinogenic to humans. PAH compounds have a very recognizable shape—they are all combinations of benzene rings, either in chains (naphthalene, anthracene) or in groups (phenanthrene, pyrene). We will concentrate on two PAH compounds in this lab, naphthalene and phenanthrene and examine how their symmetry and the Hückel calculations relate to their respective spectra. Prelaboratory Exercise Draw your phenathrene molecule in SHMO as shown and use the numbering system given below for all your calculations.

1

2

3 4

5 6

7 8

9

1011

1213

14z

yx

phenanthrene Figure 1: Phenanthrene molecule with axis label and numbered carbons

Perform the same Hückel evaluation sequence on your phenanthrene molecule as you did for naphthalene and quinoline. Start by drawing the MO diagram and finish with writing out a Hückel matrix.

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Be sure to cover the following points (your TA will check this when you come to the lab):

1. Indicate the point group for phenanthrene 2. Draw the MOs and assign characters to them. Add electrons to the diagram and label

the HOMO and LUMO (ignore SHMO values for α) 3. Determine the coefficient of β for the π → π* transition 4. Find the characters corresponding to the ground state and the excited state by direct

product and determine whether there is an allowed transition for a π → π* excitation and with what polarization (x, y, or z)

5. Set up a Hückel matrix using x and 1. What are the roots that result from solving this matrix? (Hint: you will have a 14×14 matrix and you do not have to solve the matrix to find the roots.)

6. Use the coefficient of β you obtain from the MO diagram to predict the frequency range where you will see the peaks for the electronic transition in the UV-visible spectroscopy experiment next week. You will work with the two values for β from the previous lab, use these to obtain two predictions, one will fit the naphthalene spectrum and one will fit phenanthrene.

Experimental Procedure We will be using the Cary 100Bio UV-visible spectrometer. Review the instructions given in Lab 1, The Absorption of Linear Polyene Dyes. Stock solutions of naphthalene and phenanthrene are available to you. These compounds have been dissolved in methanol (about 0.1g in 1L; 10-4M). Note the concentrations. Run the spectra for naphthalene and phenanthrene from the stock solutions using MeOH in the reference cell. You will use the default parameters of 800-200nm. When you see your spectrum, you may need to manipulate the scales (x and y axes) to better observe the peaks. At the concentrations of the stock solutions, all the naphthalene peaks can be seen, but not all the phenanthrene peaks are apparent. Given the concentration of the stock solution, it will be necessary to do a dilution to achieve a concentration on the order of 10-5M (accuracy is not important, around a 10 fold dilution is acceptable). Each group will do a dilution using the 100ml volumetric flasks provided and calculate their final concentration. Run the diluted phenanthrene solution against the MeOH reference and observe the spectrum comparing the ranges of the peaks that appear. You will collect a total of three spectra—one naphthalene, one concentrated phenanthrene, one dilute phenanthrene.

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Discussion and Report Guidelines 1. Compare your spectral data with the literature (Silverstein or others, do some digging or check given references. Note that you may find discrepancies due to different solvents used; this is acceptable and is known as ‘solvent shift’ that you will cover in later advanced courses). 2. Show your calculations for values of β and assign the best value for both your samples, given the ranges you see in your spectra. 3. Discuss whether the SHMO programme gave a good prediction of where peaks would appear in the spectra of both naphthalene and phenanthrene. Be sure to attach all your spectra with appropriate and informative titles. References Blinder, S. M. Introduction to Quantum Mechanics in Chemistry, Materials Science, and Biology. New York: Elsevier, 2004, (Ch 12). Cotton, F. A. Chemical Applications of Group Theory, 3rd ed. New York: John Wiley & Sons, 1990, (Ch 5, Ch 7, Appx III, Appx IX). Platt, J. R. J. Chem. Phys. 1950, 18, 1168. Rauk, A. Orbital Interaction Theory of Organic Chemistry, 2nd ed. New York: John Wiley & Sons, 2001, (Ch 5). Rauk, A. http://www.chem.ucalgary.ca/SHMO/ Schwenz, R. W., Moore, R. J. Physical Chemistry Developing a Dynamic Curriculum. Washington, DC: American Chemical Society, 1993, (pp. 132-137). Silverstein, R. M., Webster, F. X., Kiemle, D. J. Spectrometric Identification of Organic Compounds, 7th ed. New York: John Wiley & Sons, Inc., 2005. Interesting Websites http://www.chem.ucl.ac.uk/cosmicdust/pah.htm#what http://chrom.tutms.tut.ac.jp/JINNO/DATABASE/00alphabet.html http://www.dhfs.state.wi.us/eh/ChemFS/fs/PAH.htm

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Appendix I Character Tables

Cnv Groups

C2v E C2 σ (yz) σ'(yz) A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 -1 Rz xy BB1 1 -1 1 -1 x, Ry xz BB2 1 -1 -1 1 y, Rx yz

C3v E 2C3 3σv A1 1 1 1 z x2 + y2, z2

A2 1 1 -1 Rz E 2 -1 0 (x,y) (Rx, Ry) (x2 - y2,xy) (xz,yz)

Nonaxial Groups

C1 E A 1

Cs E σh A' 1 1 x, y, Rz x2, y2, z2, xy A" 1 -1 z, Rx, Ry yx, xz

Ci E i Ag 1 1 Rx, Ry, Rz x2, y2, z2, xy, xz, yz Au 1 -1 x, y, z

Dnh Groups

D2h E C2(z) C2(y) C2(x) i σ(xy) σ(xz) σ(yz) Ag 1 1 1 1 1 1 1 1 BB1g 1 1 -1 -1 1 1 -1 -1 RzBB2g 1 -1 1 -1 1 -1 1 -1 RyBB3g 1 -1 -1 1 1 -1 -1 1 RxAu 1 1 1 1 -1 -1 -1 -1 BB1u 1 1 -1 -1 -1 -1 1 1 z BB2u 1 -1 1 -1 -1 1 -1 1 y BB3u 1 -1 -1 1 -1 1 1 -1 x

D6h E 2C6 2C3 C2 3C2' 3C2" i 2S3 2S6 σh 3σd 3σv A1g 1 1 1 1 1 1 1 1 1 1 1 1 A2g 1 1 1 1 -1 -1 1 1 1 1 -1 -1 RzBB1g 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 BB2g 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 E1g 2 1 -1 -2 0 0 2 1 -1 -2 0 0 (Rx, Ry) E2g 2 -1 -1 2 0 0 2 -1 -1 2 0 0 A1u 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 A2u 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 z BB1u 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 BB2u 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 E1u 2 1 -1 -2 0 0 -2 -1 1 2 0 0 (x, y) E2u 2 -1 -1 2 0 0 -2 1 1 -2 0 0

Appendix II Physical Constants and Conversions

Quantity symbol value Planck’s constant h 6.626 076 × 10-34 J·s ħ = h/2π 1.054 572 7 × 10-34 J·s speed of light, vacuum c 2.997 924 6 × 108 m/s Avagodro’s constant NA 6.022 136 7 × 1023 /mol mass of an electron me 9.109 390 × 10-31 kg elementary charge e 1.602 177 3 × 10-19

C conversions: 4184J/kcal

Chemistry 373 Laboratory Survey

Since feedback is an essential first step in any realistic attempt to improve the quality of a course, your co-operation is requested in responding to the items below. Your comments will be regarded as confidential; further, they will not be read until after the laboratory grades have been submitted.

The success or failure of a laboratory program depends on the content of the exercises and the quality of the instruction. It would be naive to think that these two factors are completely independent, but please try to separate them as much as possible.

Please use this page of the form for your personal comments. For responding to the objective questions on the reverse side, please use the standard opscan answer sheet which you will obtain from your laboratory instructor during the last laboratory period.

Laboratory section _______ Day _______ Time _____ Instructor __________________________________

In your opinion, what was the most positive aspect of the laboratory program? ........................................................ ....................................................................................................................................................................................... ....................................................................................................................................................................................... What was the most negative? ...................................................................................................................................... ....................................................................................................................................................................................... ....................................................................................................................................................................................... Have you any additional comments about any aspect of the laboratories? .................................................................

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(Survey continued on reverse side)

For answering the following questions, please use a soft pencil only to blacken the corresponding space on the opscan sheet. Please indicate the extent of your agreement with each of the following statements:

A - strongly agree; B - agree; C - neutral; D - disagree; E - strongly disagree

INSTRUCTION 1. The instructor was punctual. A B C D E

2. The instructor was prepared. A B C D E

3. The instructor was knowledgeable about the subject matter. A B C D E

4. The instructor provided clear directions. A B C D E

5. The instructor conducted laboratory sessions in an organized, A B C D E well-planned manner.

6. The instructor answered students’ questions clearly and A B C D E thoroughly.

7. The instructor displayed enthusiasm in conducting laboratory A B C D E sessions.

8. The instructor provided helpful comments and feedback A B C D E where appropriate.

9. The instructor treated students professionally and respectfully. A B C D E

10. Laboratory reports were returned promptly (e.g. in the next A B C D E laboratory period).

11. The grading of reports was fair. A B C D E

12. The comments on the reports were useful. A B C D E

13. The instructor was available during office hours. A B C D E

CONTENT

Please rate each laboratory exercise by assigning it a letter grade. 14. Exp. #1: Particle in a Box A B C D E

15. Exp. #2: The Uncertainty Principle A B C D E

16. Exp. #3: Molecular Symmetry A B C D E

17. Exp. #4: Character Tables A B C D E

18. Exp. #5: Hückel, Part I A B C D E

19. Exp. #6: Hückel, Part II A B C D E

This information is being collected under the authority of the Freedom of Information and Protection of Privacy Act. The information you provide is intended for use by the Department Head to assess instructors, by instructors to assist them in improving instruction, and by the Department of Chemistry to improve the quality of the course. Your constructive comments will be read and taken into consideration for future decisions. These forms will be held until after your instructor has submitted the final course grades to the Registrar. If you have any questions about the collection or use of this information, contact the Office of the Head, Department of Chemistry, Science A 105 (403) 220-5340.