+ All Categories
Transcript

PHYSICAL CHEMISTRY II

CHEM 3720

Syllabus & Lecture Notes

Prepared by Dr. Titus V. Albu

Department of Chemistry

University of Tennessee at Chattanooga

Spring 2017

12

HH

e

34

56

78

910

LiB

eB

CN

OF

Ne

1112

1314

1516

1718

Na

Mg

AlSi

PS

Cl

Ar

1920

2122

2324

2526

2728

2930

3132

3334

3536

KC

aSc

TiV

Cr

Mn

FeC

oN

iC

uZn

Ga

Ge

AsSe

Br

Kr

3738

3940

4142

4344

4546

4748

4950

5152

5354

Rb

SrY

ZrN

bM

oTc

Ru

Rh

PdAg

Cd

InSn

SbTe

IXe

5556

57-7

172

7374

7576

7778

7980

8182

8384

8586

Cs

Ba

Hf

TaW

Re

Os

IrPt

AuH

gTl

PbB

iPo

AtR

n

8788

89-1

0310

410

510

610

710

810

911

011

111

211

311

411

511

611

711

8

FrR

aR

fD

bSg

Bh

Hs

Mt

Ds

Rg

Cn

Uut

FlU

upLv

Uus

Uuo

5758

5960

6162

6364

6566

6768

6970

71

Lant

hani

des

LaC

ePr

Nd

PmSm

EuG

dTb

Dy

Ho

ErTm

YbLu

8990

9192

9394

9596

9798

9910

010

110

210

3

Actin

ides

AcTh

PaU

Np

PuAm

Cm

Bk

Cf

EsFm

Md

No

Lr8A

Perio

dic

Tabl

e of

the

Elem

ents

1A

1.00

794

2A3A

4A5A

6A7A

4.00

2602

6.94

19.

0121

8210

.811

12.0

107

14.0

067

15.9

994

18.9

9840

3220

.179

7

30.9

7376

232

.065

35.4

5339

.948

39.0

983

40.0

7844

.955

912

47.8

6750

.941

551

.996

1

7B┌

──

──

─ 8

B ─

──

──

┐1B

2B26

.981

5386

28.0

855

22.9

8976

924

.305

03B

4B5B

6B

78.9

679

.904

83.7

9854

.938

045

55.8

4558

.933

195

58.6

934

63.5

4665

.38

85.4

678

87.6

288

.905

8591

.224

92.9

0638

95.9

6

69.7

2372

.64

74.9

2160

114.

818

118.

710

121.

760

127.

6012

6.90

447

131.

293

[98]

101.

0710

2.90

550

106.

4210

7.86

8211

2.41

1

[209

][2

10]

[222

]18

6.20

719

0.23

192.

217

195.

084

196.

9665

6920

0.59

[223

][2

26]

Actin

ides

[267

][2

68]

[271

]

204.

3833

207.

220

8.98

040

132.

9054

519

137.

327

Lant

hani

des

178.

4918

0.94

788

183.

84

[284

][2

89]

[288

][2

93]

[294

][2

94]

[272

][2

70]

[276

][2

81]

[280

][2

85]

[227

]23

2.03

806

231.

0358

823

8.02

891

[237

][2

44]

[243

]

151.

964

157.

2513

8.90

547

140.

116

140.

9076

514

4.24

2[1

45]

150.

36

[259

][2

62]

[247

][2

47]

[251

][2

52]

[257

][2

58]

168.

9342

117

3.05

417

4.96

6815

8.92

535

162.

500

164.

9303

216

7.25

9

The University of Tennessee at Chattanooga

i

Physical Chemistry II

Spring 2018

CHEM 3720, CRN 20987, 4 credit hours (including CHEM 3720L lab)

Instructor: Dr. Titus V. Albu

Phone and Email: 423-425-4143; [email protected]

Office Hours and Location: Monday 8:30-11:00 am & Wednesday 8:30-11:00 am or by

appointment; Grote 314

Course Meeting Days, Time, and Location: Monday/Wednesday/Friday 1:00 pm-1:50 pm;

Grote 411

Course Catalog Description: Continuation of 3710 with primary emphasis on kinetics, quantum

mechanics, and spectroscopy. Spring semester. Lecture 3 hours, laboratory 3 hours. Prerequisite:

CHEM 3710 with a minimum grade of C and 3710L with a minimum grade of C; or department

head approval. Corequisite: CHEM 3720L or department head approval. Laboratory/studio

course fee will be assessed.

Course Pre/Co Requisites: see Course Catalog Description above

Course Student Learning Outcomes: Through classroom lectures, assigned textbook reading

and homework, and the laboratory work the students are expected to advance the ability to

interpret and reason with physical chemistry concepts, laws, and theories. Having completed the

class, a student is expected to be able to: Understand and use basic physical chemistry language,

Identify, discuss and analyze factors influencing molecular properties and chemical kinetics;

Apply physical chemistry principles and laws to problems or issues of a chemical nature;

Critically interpret and reason physical chemistry data.

Course Fees: Laboratory/studio course fee will be assessed.

Course Materials/Resources:

Textbook and Topics: Physical Chemistry 10th edition by Atkins and de Paula (W. H.

Freeman and Company, New York, 2014, ISBN: 978-1-4292-9019-7) is the textbook of

record in this class. We will be covering topics that are presented in the textbook as

follows:

Quantum Chemistry (Chapters 7-11)

Spectroscopy (Chapters 12-13)

Statistical Thermodynamics (Chapter 15)

Reaction Dynamics (Chapter 21)

ii

The topics will be covered using lecture notes organized as units, which have a slightly

different arrangement than the textbook. In addition, selected topics of nanomaterials

will be presented and discussed toward the end of the semester.

Lecture Notes: The lecture notes are available on UTC Learn, should be printed and

brought to the class. I strongly recommend that you print and have bound the entire

course package containing the syllabus and the lecture notes. In addition to the lecture

notes, I may also refer you to or have you find additional information from various online

sites.

Technology Requirements for Course:

Computer: You need access to a computer with a reliable internet connection for this

course. Test your computer set up and browser for compatibility with UTC Learn at

http://www.utc.edu/learn/getting-help/system-requirements.php. Although not required,

the computer might need to have speakers or headphones. You should also have an

updated version of Adobe Acrobat Reader, available free from

https://get.adobe.com/reader/.

UTC Learn: Access this class by selecting “SP18.CHEM.3720.20987: Physical

Chemistry II” course on UTC Learn (http://www.utc.edu/learn/). Log in using your

utcID and password (the same as for your UTC email). In this class, UTC Learn will be

used for: (1) Course announcements; (2) Syllabus; (3) Course Materials: Lecture notes,

old/practice exams, and homework assignments; and (4) Individual grades.

Technology Skills Required for Course: You will need to have basic computer skills including

using the learning management system (UTC Learn), using MOCSNet email, completing online

homework, and downloading and printing pdf files.

Technology Support: If you have problems with your UTC email account or with UTC Learn,

contact IT Solutions Center at 423-425-4000 or email [email protected].

Course Assessments and Requirements: Your overall course grade will be computed based on:

In-class exams 45%

Final exam 20%

Homework assignments 10%

Laboratory 25%

Exams: There will be 3 one-hour exams during the class period. The lowest-scored

exam grade (or a missed exam grade) can be replaced by the final exam grade, if the final

exam grade is higher. A grade of 0 will be assigned for any exam that is not taken, and

one grade of 0 can be replaced by the final exam grade. The typical average exam score

is around 60-65. Exams are based on class lecture notes, textbook, and homework. You

should bring a working calculator and two pencils to exams. You may not share a

calculator during exams. No other paper, notes, books or stored information is to be used

except what will be provided to you. After the first person leaves an exam, no one else

can come late and start the exam. No cell phone use, texting, or checking phone in class

iii

at any time. The tentative exam dates are given below. NO MAKEUP exams will be

given.

Final Exam: The final exam is a standardized ACS exam that is scheduled for Friday,

April 27, 2018: 10:30 am-12:30 pm. Final exam contains 60 questions: 40 questions

covering Quantum Theory and Spectroscopy, 10 questions covering Statistical

Thermodynamics, and 10 questions covering Dynamics. The final exam grade will be

determined by dividing the number of questions answered correctly to 0.50 and will not

be adjusted any further. The typical average on this test (after adjusting) is expected to be

around 55-60. There is NO MAKEUP for the final exam.

Homework Assignments: Homework assignments will be assigned regularly and will

primarily contain exercises and problems from the textbook except the last assignment

which is a class presentation. All homework assignments will be posted in your UTC

Learn course and will be collected during the lecture period. Assignments containing

problems will be graded with scores of 1/0.5/0 point per problem. Each problem should

be worked out on one page (or more separate pages), and spread them out showing all

steps. Do not put more than one problem per page unless otherwise noted in the

instructions. If problems are not done in this format, points will be deducted. It is

assumed you are working problems in a timely manner. Late homework assignments will

be accepted as long I did not return the graded ones back but a 25% deduction per day

will be enforced. Some homework problems are more difficult and are for you to

struggle with and be satisfied with your answer. You must try to work all problems by

yourself with help only to guide you and not to replace working or thinking about the

problem. If I will be asked questions about homework problems before the due date, I

may be able to point you in the right direction, but all the details of the work are up to

you. In addition to reading the textbook and studying the lecture notes, working the

assigned homework problems as we discuss each chapter is a good way to prepare for the

exams. You can also use the worked examples in the textbook or old exams for practice.

Remember that learning chemistry requires thinking and doing, and not just listening and

reading.

Laboratory: Laboratory grades will be provided by the CHEM 3720L instructor (Dr.

Han Park), who has full responsibility for these grades. The typical class lab average is

around 90.

iv

Course Grading

Course Grading Policy: The unadjusted overall course score can be adjusted (up or

down) by the instructor, by up to 3%, based on (but not limited to) class attendance and

participation, homework effort, involvement in the lab experiments, general interest in

the presented material, the score/average of a particular exam, etc. Your letter grade in

the class will be determined based on the adjusted overall course score, and it is expected

to be determined according to the following scale:

F < 50% < D < 60% < C < 70% < B < 83% < A

Instructor Grading and Feedback Response Time: I will try my best to grade all

assignments before the next class period, and provide written feedback when necessary.

Course and Institutional Policies:

Late/Missing Work Policy: Late homework assignments will be accepted as long I did

not return the graded ones back but a 25% deduction per day will be enforced. There is

no makeup for in-class exams. As presented above, the lowest-scored exam grade (or a

missed exam grade) can be replaced by the final exam grade, if the final exam grade is

higher. There is no makeup for the final exam.

Student Conduct Policy: UTC’s Academic Integrity Policy is stated in the Student

Handbook. A violation of the honor code could result in appearing in honor court and

receiving a course grade of F. Instructor will not tolerate academic dishonesty.

Specifics: (1) Any attempt (successful or unsuccessful) to cheat during any of the exams

will automatically result in an “F” grade in the course; (2) Presenting a homework

assignment that resembles too closely the assignment of another student will result in a

grade of 0 for that assignment (for both students) for the first infraction, and an overall

grade of 0 for the homework (10% of the overall grade) for the second infraction.

Honor Code Pledge: I pledge that I will neither give nor receive unauthorized aid on any

test or assignment. I understand that plagiarism constitutes a serious instance of

unauthorized aid. I further pledge that I exert every effort to ensure that the Honor Code

is upheld by others and that I will actively support the establishment and continuance of a

campus-wide climate of honor and integrity.

Course Attendance Policy: Students are expected to attend every lecture, be punctual,

and be respectful of others in the class. Classroom behavior such as talking to your

neighbor during lecture, reading, dozing, or checking cell phone, might interfere with my

ability to teach effectively and others ability to learn. I might require you to meet with

me before you are allowed to take the next exam so I can explain more clearly why your

activities are a problem. I might also ask you to leave the classroom. Laptop computers

can be a big distraction in class so no laptops may be used at any time during class.

Similarly, there should be no cell phone use of any kind during class. You are

responsible for everything covered in the lecture. Information or points missed during

unexcused absences cannot be reclaimed from me so check with a fellow student who is

able to share notes and go over items you missed. The only acceptable (but not

necessarily accepted) excuses are the ones received from The Dean of Students Office.

v

During lecture period, there might be some additional assignments, class pop-quizzes, or

attendance quizzes that might be added as bonus points to the next exam grade or

considered in the homework grade. Negative points will be assessed for missing class

during attendance quizzes. Class will include lecture and discussion with assumption that

you have read and studied textbook and lectures notes ahead of where we are in class and

you can discuss topic, ask relevant questions, and/or respond to questions. Always bring

your printed lecture notes to class.

Communication: Class announcements will be made through UTC Learn and email. UTC email

is the official means of communication between instructor and student at UTC. Please check

your UTC email and UTC Learn on a regular basis. (i.e., daily). I will try to answer emails from

students with questions/comments/concerns within 24 hours (Monday through Friday) although

occasionally it might take longer. I might not answer student emails if they require repeating

information already mentioned in a class that the student missed.

Course Participation/Contribution: The course contains several learning objectives that are

critical to building a solid foundation in physical chemistry. For this reason, several methods will

be employed, including (but not limited to): lecture, group study, pre-class reading, and post-

class work. To be successful in this course, I recommend that you engage in all methods.

Course Learning Evaluation: Course evaluations are an important part of our efforts to

continuously improve the learning experience at UTC. Toward the end of the semester, you will

receive a link to evaluations and are expected to complete them. We value your feedback and

appreciate you taking time to complete the anonymous evaluations.

Syllabus Changes: Although unlikely, some things on this syllabus are subject to change at the

discretion of the instructor. Every attempt will be made to follow this syllabus, however, if

changes are made, they will be announced in class, by email, and/or on UTC Learn, and it is the

responsibility of the student to keep up with the changes.

Course Calendar/Schedule: The tentative exam schedule below is based on the assumption that

no classes will be cancelled (due to weather or other emergencies).

Exam 1: February 7, 2018 (covering Units 2-5)

Exam 2: March 7, 2018 (covering Units 6-9)

Exam 3: April 16, 2018 (covering Units 11-13)

Final Exam: April 27, 2018 10:30 am – 12:30 pm

CHEM 3720

1

Unit I

Introduction

A. Introduction to Physical Chemistry

1. Physical Chemistry is the part of chemistry dealing with application of

physical methods to investigate chemistry.

2. Physical Chemistry main subdivisions are:

a. Quantum Mechanics

□ deals with structure and properties of molecules

b. Spectroscopy

□ deals with the interaction between light and matter

c. Computational Chemistry

□ deals with modeling chemical properties of reactions using

computers

d. Statistical Mechanics

□ deals with how knowledge about molecular energy levels (or

microscopic world) transforms into properties of the bulk (or

macroscopic world)

e. Thermodynamics

□ deals with properties of systems and their temperature dependence

and with energetics of chemical reactions

f. Electrochemistry

□ deals with processes in which electrons are either a reactant or a

product of a reaction

g. Chemical Kinetics

□ deals with the rates of chemical reactions or physical processes

CHEM 3720

2

B. Classical Physics Review

1. Classical Physics was introduced in the 17th century by Isaac Newton.

2. At the end of 19th century, classical physics (mechanics, thermodynamics,

kinetic theory, electromagnetic theory) was fully developed and was

divided into:

a. the corpuscular side or particle domain (the matter)

b. the undulatory side or wave domain (the light)

3. Some useful classical physics equations:

a. Total energy E:

VKE

○ K is the kinetic energy (or energy arising from motion)

○ V is the potential energy (or energy arising from position)

b. Kinetic energy K:

m

pmK

2v

2

1 22

○ m is the mass

○ v is the velocity (or speed)

○ p is the momentum

c. Frequency (Greek letter nu):

2

~ cc

○ is the wavelength (Greek letter lambda)

○ c is the speed of light

○ ~ is the wavenumber (read “nu tilde”)

○ is the angular frequency (Greek letter omega)

4. Classical mechanics was successful in explaining the motion of everyday

objects but fails when applied to very small particles. These failures led to

the development of Quantum Mechanics.

CHEM 3720

3

C. The Classical Wave Equation

1. It is a prelude to Quantum Mechanics because it introduces (or reminds

you) concepts that are similar to the ones in Quantum Mechanics.

2. The classical wave equation describes various wave phenomena:

a. a vibrating string

b. a vibrating drum head

c. ocean waves

d. acoustic waves

3. The classical (nondispersive) wave equation for a 1-dimensional wave:

2

2

22

2 ),(

v

1),(

t

txu

x

txu

□ ),( txu is the displacement of the string from the horizontal

position

□ v is the velocity or the speed that the disturbance moves

□ t is the time

a. The classical wave equation is a partial differential equation (a linear

partial differential equation) because ),( txu and its derivatives appear

only to the first power, and there are no cross terms.

b. The x and t are independent variables.

c. The ),( txu is a dependent variable.

4. Example: A 1-dimensional wave describing the motion of a vibrating string

a. The displacement ),( txu must satisfy certain physical conditions: the

amplitude should be zero at the end of the string.

□ 0),0( tu

□ 0),( tlu

CHEM 3720

4

b. These conditions are called boundary conditions because they satisfy

the behavior at the boundaries.

c. To solve the differential equation, we assume that ),( txu factors into a

function of x times a function of t:

)()(),( tTxXtxu

d. This technique (or method) is called the separation of variables.

e. Solving further the equation: – Substituting ),( txu in the equation above:

2

2

22

2 )()(

v

1)()(

dt

tTdxX

dx

xXdtT

– Dividing by )()(),( tTxXtxu :

Kdt

tTd

tTdx

xXd

xX

2

2

22

2 )(

)(

1

v

1)(

)(

1

– In order for this equation to be true for every x and t, each side should be equal to a

constant K called the separation constant.

– The problem of finding ),( txu transformed into two problems of finding X(x) and T(t) by

solving the following linear differential equations with constant coefficient (they are

ordinary differential equations):

0)()(

2

2

xKXdx

xXd

0)(v)( 2

2

2

tTKdt

tTd

f. Solving for X(x): l

xnBxX

sin)(

– Trivial solution is obtained (that is X(x) = 0) if 0K .

– If K 0, set 2K ( is real):

0)()( 2

2

2

xXdx

xXd

– The general solution for this equation is: xixi ececxX 21)(

– Considering Euler equation ( xixe ix sincos ):

xBxAxX sincos)(

– This solution of X(x) should verify the boundary conditions:

0)0( X A = 0

0)( lX 0sin lB nl where n = 1, 2, ...

CHEM 3720

5

g. Look more closely to the solutions:

Number of

wavelength

that fits in 2l:

n = 1

n = 2

n = 3

n = 4

Number of

wavelength

that fits in 2l:

n = 1

n = 2

n = 3

n = 4

□ By generalization: n

ln

2

○ This is called the eigenvalue condition.

□ The solutions are a set a functions called eigenfunctions or

characteristic functions.

xBxl

nBxX

nnnn

2sinsin)(

□ Also, angular frequencies 0vv2

2

n

l

n

nnn (where

l

v0

) are called eigenvalues or characteristic values.

1

2

3

4

CHEM 3720

6

h. Solving for T(t) but keeping in mind that l

n

0)(v)( 22

2

2

tTdt

tTd

– Similar to above, the solution is:

tEtDtT nn sincos)( where: l

nn

vv

i. Coming back to ),( txu :

)()(),( tTxXtxu

l

xntGtFtxu nn

sin)sincos(),( ; n = 1, 2,…

– There is a ),( txu function for each n so a better notation would be:

l

xntGtFtxu nnnnn

sin)sincos(),( ; n = 1, 2,…

– The sum of all ),( txun solutions is also a solution of the equation (This is called the

principle of superposition.) The general solution is:

1

sin)sincos(),(

n

nnnnl

xntGtFtxu

– Make the transformation: )cos(sincos tAtGtF where (Greek letter phi)

is the phase angle and A is the amplitude of the wave.

– Rewrite the general equation as:

11

),(sin)cos(),(n

nn

nnn txul

xntAtxu

□ Each ),( txun is called:

○ a normal mode

○ a standing wave

○ a stationary wave

○ an eigenfunction of this problem

j. The time dependence of each mode represents a harmonic motion of

frequency: n

nn

l

n

v

2

v

2 where the angular frequency is:

nnn

l

nv

v2v2v .

CHEM 3720

7

k. Solutions:

□ First term is l

xt

lA

sin)

vcos( 11

○ First term is called fundamental mode or first harmonic.

○ The frequency is: l2/v1

□ Second term is l

xt

lA

2sin)

v2cos( 22

○ Second term is called first overtone or second harmonic.

○ The frequency is: l/v2

○ The midpoint has a zero displacement at all times, and it is

called a node.

□ Third term is l

xt

lA

3sin)

v3cos( 33

○ Third term is called second overtone or third harmonic.

○ The frequency is: l2/v33

○ This term has two nodes.

□ Fourth term is l

xt

lA

4sin)

v4cos( 44

CHEM 3720

8

l. Let’s consider now the case of:

l

xt

l

xttxu

2sin)

2cos(

2

1sin)cos(),( 21

4

t1

2

4

30

□ This is an example of a sum of standing waves yielding a traveling

wave.

m. Thinking backwards, any general wave function can be decomposed

into a sum or superposition of normal modes.

n. The number of allowed standing waves on a string of length l:

□ increases as the wavelength decreases the possible high-

frequency oscillations outnumber the low-frequency ones.

n

ln

2

– Consider that l so we can approximate the set of integers n by a continuous

function )(n .

dl

dnl

nn

2

22

– The negative sign indicates that the number of standing waves decreases as

increases.

o. The number of standing waves in an enclosure of volume V (use c not

v for the speed):

d

Vdn

4

4 but

v

c and

cv ;

d

cdv

2 dv

cd

2

dvvc

Vdv

c

Vdn 2

3

2

4

4))(

4(

CHEM 3720

9

5. Example: A 2-dimensional wave equation = the equation of a vibrating

membrane:

2

2

22

2

2

2

v

1

t

u

y

u

x

u

where ),,( tyxu

a. Solving this equation:

– Similar to the one-dimensional problem, use separation of variables:

)(),(),,( tTyxFtyxu

2

2

2

2

2

2

2

2 ),(

1

)(v

1

y

F

x

F

yxFdt

Td

tT

– Use separation of variables for ),( yxF :

)()(),( yYxXyxF

– Divide by ),( yxF : 0)(

)(

1)(

)(

1 2

2

2

2

2

dy

yYd

yYdx

xXd

xX

– Solve two equations:

2

2

2 )(

)(

1p

dx

xXd

xX and 2

2

2 )(

)(

1q

dy

yYd

yY

where 222 qp

– Solutions for )(xX and )(yY are:

a

xnBxX

sin)( (n = 1,2,…) and

b

xmDyY

sin)( (m = 1,2,…)

where 2

2

2

2

b

m

a

nnm

– Solution for )(tT :

)cos(sincos)( nmnmnmnmnmnmnmnm tGtFtEtT

where

2

2

2

2

vvb

m

a

nnmnm

b. The general solution for ),,( tyxu :

1 1

1 1

sinsin)cos(

),,(),,(

n mnmnmnm

n mnm

b

ym

a

xntA

tyxutyxu

x

y

a

b

x

y

a

b

CHEM 3720

10

c. Again, the general function is a superposition of normal modes

),,( tyxunm but in this case one obtains nodal lines (lines where the

amplitude is 0) instead of nodes.

d. Examples:

m =

n =

m =

n =

m =

n =

m =

n =

1

1

2

1

1

2

2

2

e. The case of a square membrane ( ba ), the frequencies of the normal

modes are given by:

22vmn

anm

□ For the cases of n = 1, m = 2 and n = 2, m = 1 one can see that:

a

v52112

although ),,(),,( 2112 tyxutyxu

f. This is an example of a degeneracy.

□ The frequency 2112 is double degenerate or two-fold

degenerate.

□ This phenomenon appears because of the symmetry ( ba ).

CHEM 3720

11

D. Unit Review

1. Important Terminology

frequency

wavelength

wavenumber

angular frequency

independent variables

boundary conditions

separation of variables

eigenfunctions

eigenvalues

stationary wave

traveling wave

node

degeneracy

CHEM 3720

12

2. Important Formulas

VKE

m

pmK

2v

2

1 22

2

~ cc

CHEM 3720

13

Unit II

The Development of Quantum Mechanics

A. Introduction

1. Failures of classical physics

a. The classical physics predicts the precise trajectory of a particle and

allows the translational, rotational, and vibrational modes of motion to

be excited to any energy by controlling the applied force.

b. These observations are found in everyday life in macroscopic world

but do not extend to individual atoms.

c. Classical mechanics fails when applies to transfers of very small

quantities of energy and to objects of very small mass.

2. Historic prospective on Quantum Mechanics (QM)

a. 1887 Hertz The discovery of photoelectric effect

b. 1895 Roentgen The discovery of x-rays

c. 1896 Becquerel The discovery of radioactivity

d. 1897 J. J. Thomson The discovery of the electron

e. 1900 Plank The quantum hypothesis of blackbody radiation

f. 1905 Einstein The quantum hypothesis of photoelectric effect

g. 1907 Thomson Model of atom

h. 1909 Rutherford Scattering experiment with particles

i. 1911 Rutherford The nuclear model of atom

j. 1913 Bohr The quantum hypothesis applied to the atom

k. 1924 de Broglie The prediction of the wave nature of the matter

l. 1925 Heisenberg QM in a form of matrix mechanics

m. 1926 Heisenberg Uncertainty principle

n. 1926 Schrödinger QM in a form of wave mechanics

CHEM 3720

14

B. Blackbody Radiation

1. Background

a. All objects are absorbing and emitting radiation and their properties as

absorbers or emitters may be extremely diverse.

b. It is possible to conceive the existence of objects that are perfect

absorbers of radiation, and they are called blackbodies.

c. A blackbody is an ideal body, which absorbs and emits all frequencies.

d. Blackbody radiation is the radiation emitted by the blackbody.

e. Ideal blackbodies do not exist. Most of substances absorb (or emit) all

frequencies only in a limited range of frequencies.

f. The best lab blackbody is not a body but a cavity

that is constructed with insulating walls, and in

one of which a small orifice is made.

g. When the cavity is heated, the radiation from the

orifice is a good sample of the equilibrium

radiation within the heated enclosure, which is

practically ideal blackbody radiation.

2. Experimental observations

a. Materials at the same temperature T have the

same blackbody radiation spectrum.

□ “Materials look the same”.

b. The brightness increases as T increases.

c. As the temperature increases, the maximum

shifts toward higher frequencies (or toward

lower wavelength).

d. There is a simple relationship between the wavelength at the maximum

intensity and the temperature, relationship known as the Wien

displacement law: Km108979.2const 3max T

CHEM 3720

15

3. Analogy to classical systems

a. There is a similarity between the behavior of radiation within such a

cavity and that of gas molecules in a box.

b. Both the molecules and the radiation are characterized by a density,

and both exert pressure on the confining walls.

c. One difference is that the gas density is a function of V and T, whereas

radiation density is a function of temperature alone.

4. The classical explanation of the blackbody radiation

a. Rayleigh and Jeans try to explain the observed blackbody radiation

based on the laws of classical physics.

b. Assumptions:

□ Blackbody radiation is coming from standing electromagnetic

waves in the cavity that are at equilibrium with the vibrating atoms

(or electrons) in the walls.

□ The waves that are leaked out are observed.

□ The atoms in the blackbody are assumed to vibrate like harmonic

oscillators (these harmonic oscillators may be seen as

electromagnetic oscillators), and to be in thermodynamic

equilibrium with the radiation in the cavity.

□ According to the principle of equipartition energy, an oscillator in

thermal equilibrium with its environment should have an average

energy equal to TkB (that is TkB21 for kinetic energy and TkB2

1

for the potential energy).

□ We already found out that 32 /4 cdVdn for electromagnetic

waves from classical physics but we should multiply this result by

2 because for the electromagnetic radiation, both electric and

magnetic fields are oscillating

dc

Vdn 2

3

8

CHEM 3720

16

c. According to Rayleigh-Jeans law, the radiant energy density is the

product of the density of states with the average energy of the state:

dc

Tk

V

dnTkdTTd 2

3B

B8

)(),(

where dT is the radiant energy density between the

frequencies and + d.

d. Rayleigh-Jeans law reproduces the experiment at low frequencies but

diverges at high frequencies (at low wavelength) as the radiation enters

the ultraviolet region.

e. Because of that, this divergence was called the ultraviolet catastrophe.

f. This was the first failure of classical physics in explaining theoretically

naturally occurring phenomena that could be explained by quantum

ideas.

5. The quantum explanation of the blackbody radiation

a. Proposed by Planck in 1900.

b. Assumptions:

□ The vibrating atoms in the walls have quantized energies or there

is a collection of N oscillations with fundamental frequency:

nhEn

where En is the energy of an oscillation, is the frequency, h is

a constant, and n = 0,1,2,…

○ Another way to say it is that oscillators take up energy in

increments h.

○ These increments (discrete units) are called quanta.

□ All frequencies are present.

c. Planck distribution law for blackbody radiation in terms of frequency:

– The number of oscillators having energy nh is given by the Boltzmann distribution: Tknh

n eNN B/0

where 0N is the number of oscillations in the lowest energy state (n = 0)

CHEM 3720

17

– Total number of oscillators:

0

/0

/20

/00

BBB ......

n

TknhTkhTkheNeNeNNN

– Total energy of oscillators is the product of the number of oscillators and their energy:

0

/0

/20

/00

BBB .....20

n

TknhTkhTkhneNheNheNhNE

– The average energy of an oscillator becomes:

11 BB

B

/

0

0

0

/

0

/

Tkhx

n

nx

n

nx

n

Tknh

n

Tknh

e

h

e

h

e

ne

h

e

neh

N

E

where Tk

hx

B

1

8

1)(),(

BB /

3

3/

TkhTkh

e

d

c

h

V

dn

e

hdTTd

where dT )( is the radiant energy density between the

frequencies and + d.

d. Planck distribution law for blackbody radiation in terms of wavelength:

1

8

1)(),(

BB /5/

TkhcTkh

e

dhc

V

dn

e

hdTTd

where dT )( is the radiant energy density between the

wavelength and + d.

e. Successes of Planck’s distribution law

□ It reproduces experimental data for all frequencies and

temperatures within the experimental error if Js10626.6 34h .

○ Units of energy·time = action.

○ The constant h is now known as the Planck constant.

□ It explains the constant in the Wien distribution law:

Bmax

965.4 k

hcT

□ It introduces the idea of energy quantization: an oscillator acquires

energy only in discrete units called quanta.

CHEM 3720

18

C. Photoelectric Effect

1. Experimental observations:

a. Hertz (1887) a spark would jump a gap more readily when the gap

electrodes were illuminated by light from another spark gap than when

they were kept in the dark.

b. The phenomenon was due to the emission of electrons from the surface

of solids upon incidence of light having suitable wavelengths. These

emitted electrons were called photoelectrons.

c. Whether or not electrons are emitted from the surface (plate of

electrodes) depends only on the frequency of the light and not at all on

the intensity of the beam.

d. The number of electrons emitted is proportional

to the intensity of the light.

e. There is no time delay between the light beam

striking the surface and the emission of the

electrons.

f. Lenard (1902) determined that the maximum kinetic energy of the

emitted electrons depends on the frequency of the incident light, and

below a certain frequency called threshold frequency 0 no electrons

were ejected. Above 0, the kinetic energy of the electrons varies

linearly with the frequency .

2. The classical interpretation of the photoelectric effect

a. Electromagnetic radiation is an electric field oscillating perpendicular

to its direction of propagation, and the intensity of radiation is

proportional to the square of the amplitude of the electric field.

b. Increasing intensity of the light, the electrons oscillate more violently

and break away from the surface.

c. KE will depend on the amplitude (intensity) of the field.

CHEM 3720

19

d. Photoelectric effect should occur for any frequency of light as long as

the intensity is sufficiently high.

e. For weak intensities and reasonable values of the frequency a long time

should intervene before any electron would soak up enough energy to

be emitted from the metal.

f. None of these predictions were verified experimentally. Classical

physics failed badly.

3. The quantum interpretation of the photoelectric effect

a. Einstein extends Plank’s idea of quantized oscillators comprising the

blackbody radiation by suggesting that radiation itself is quantized.

b. He suggests that the radiation itself exists as small packets (quanta) of

energy known as photons:

hE

c. Entire quantum of energy is accepted by a single electron and cannot

be divided among all the electrons present.

d. Kinetic energy of the ejected electrons is the difference between the

energy of the incident photons h and the minimum energy required to

remove an electron from the surface of a particular metal called the

work function of the metal and denoted .

hm 2v2

1KE

e. One can write as: = h0

where 0 is called the threshold frequency.

)(KE 00 hhh

f. The work function is usually

expressed in eV:

J10602.1eV1 19

where V1C1J1

CHEM 3720

20

D. Atomic Spectrum of Hydrogen Atom; The Bohr Model

1. Experimental facts known at the beginning of 20th century

a. Structure of the atom

□ The nuclear model for the atom had been proposed and accepted.

□ The model has a positively charged nucleus but the model was

unstable according to classical electromagnetic theory.

b. Existence and properties of atomic spectra

□ It was known that the emission spectra of atoms consist of certain

discrete frequencies called line spectra. The simplest such spectra

was the spectrum of hydrogen atom.

□ For the hydrogen atom the line spectra is composed from a number

of series.

□ One of these series is in the visible range of the radiation and is

called Balmer series in honor of Balmer who showed (1885) that

the emission lines could be described by the equation:

Hz4

1102202.82

14

n where n = 3,4,5,…

○ Using wavenumbers (c

~ ) instead of frequency:

122

cm1

2

1109680~

n where n = 3,4,5,…

○ This is the Balmer formula and it describes the lines in the

hydrogen spectrum occurring in VIS and near UV regions.

○ As n increases the lines bunch up toward the series limit.

□ Other series have been discovered in UV and IR regions and

Balmer’s formula had been generalized by Rydberg and Ritz:

22

21

111~

nnRH

where n1 < n2

CHEM 3720

21

○ This is called Rydberg formula.

○ 1-H cm109677.57R is the Rydberg constant.

□ These series were named as:

Lyman 11 n ,...3,22 n UV

Balmer 21 n ,...4,32 n VIS

Pashen 31 n ,...5,42 n Near IR

Bracket 41 n ,...6,52 n IR

○ Series limits are obtained for 2n .

□ Ritz (1908) showed that:

21~ TT

where 21

1n

RT H and

22

2n

RT H are called terms.

○ This is called the Ritz combination rule.

□ Conway (1907) proposed that a single atom produces a single

spectral line at a time, and the emission is due to a single electron

in an “abnormal state”.

2. The Bohr atomic model (1911)

a. Assumptions:

□ The spectral lines are produced by atoms one at a time.

□ A single electron is responsible for each line.

□ The Rutherford nuclear atom is the correct model.

CHEM 3720

22

□ The quantum laws apply to jumps between different states

characterized by discrete values of angular momentum and, Bohr

added, energy.

□ The angular momentum L of the electron is given by the

Ehrenfest’s rule, i.e., the angular momentum L is given by:

nh

nrmL 2

ve where n is an integer.

□ Two different states of the electron in the atom are involved. They

are called allowed stationary states, and the spectral terms of Ritz

correspond to these states.

□ Planck-Einstein equation E = h holds for the emission and

absorption. If the electron makes a transition between two states

with energy Em and En, the frequency of the spectral line is given

by h = Em – En.

□ Bohr said: “We must renounce all attempts to visualize or to

explain classically the behavior of the active electron during a

transition of the atom from one stationary state to another.”

b. The picture of the atom: a massive nucleus (proton) considered fixed

with the electron revolving around it.

c. The forces between the proton and the electron:

□ Attraction given by the Coulomb’s law:

20

2

4 r

ef

□ Repulsion given by the centrifugal force:

r

mf

2ev

CHEM 3720

23

d. Bohr assumed the existence of stationary states so the electron is not

accelerated toward nucleus.

□ The radius of the stationary orbit is given by:

r

m

r

e2

e2

0

2 v

4

2e0

2

v4 m

er

For hydrogenic atoms (He+, Li2+, etc): 2

e0

2

v4 m

Zer

e. The angular momentum of the electrons is quantized:

nh

nrm 2

ve with n = 1,2,… rm

nh

e2v

f. The allowed radius must satisfy the condition:

nrem

n

em

hnr

2e

220

2e

220 4

g. The smallest radius, obtained for n = 1, is called the first Bohr radius.

bohr1A5292.0m10292.5 110 a

h. The energies of the stationary states (or allowed orbits):

□ The total energy is given by:

r

e

r

emVKE

0

2

0

22

e84

v2

1

□ By introducing the expression of the allowed orbit radius:

2220

4e 1

8 nh

emEn

with n = 1,2,…

□ When n = 1, one gets the ground state and the ground state energy: 1

1 cm109690kJ/mol1312eV6.13 E

□ When n = 2,3,…, one gets excited states and excited state energies: 1

2 cm74202kJ/mol328eV4.3 E

13 cm12187kJ/mol134eV5.1 E

CHEM 3720

24

i. The lines in the spectrum were a result of two allowed stationary states

(corresponding to the spectral terms of Ritz) and Plank-Einstein

equation E = h holds with the energy being equal to energy difference

between two states n1 and n2:

EEEh 12

□ This is the Bohr frequency condition.

j. The observed spectrum is due to transitions from one allowed energy

state (also called stationary state or orbit) to another one.

□ The energy difference is being given to a photon:

hnnh

emE

22

21

220

4e 11

8

□ In terms of wavenumbers:

22

21

320

4e 11

8

~

nnch

em

c

where 32

0

4

8 ch

eme

is the Rydberg constant

1cm32.109737 R

○ R∞ is only 0.005% off the RH.

Energy Level

Diagram:

CHEM 3720

25

E. The wavelike properties of the matter

1. Wave-particle duality

a. In the beginning of the 1920’s it was well established that the light

behaves as a wave in some experiments and as a stream of photons in

others.

b. This is known as the wave-particle duality of the light.

c. For the light:

chhE and 2mcE

2mcc

h

p

h

mc

h

where p is the photon momentum

d. In 1923-24 Louis de Broglie proposed the idea that the matter might

also display wavelike properties under certain conditions. Under those

conditions, similar equations should hold for the matter.

e. A particle of mass m moving with the speed v will exhibit a de Broglie

wavelength given by:

vm

h

p

h

f. For particles with big mass the de Broglie wavelength is so small that it

is completely undetectable and of no practical consequences.

g. Examples:

Baseball: m102.1m/s40mph90v

kg0.14oz5.0 34

m

Electron: m1043.2m/s102.998v

kg109.109 106

31

m

h. The wave property of the electron (as well as other particles like the

neutron or the hydrogen atom) has been observed experimentally.

i. The electron diffraction is actually used in the electron microscopy.

CHEM 3720

26

2. The de Broglie interpretation of Bohr radius

a. The Bohr condition that says that angular momentum of the electron

should be a multiple of :

nh

nrm 2

ve

b. This is equivalent with saying that an integral number of complete

wavelength must fit around the circumference of the orbit:

nr 2

c. Substituting vem

h

p

h one obtains:

v2

em

hnr n

hnrm

2ve

d. This is equivalent of saying that the de Broglie waves of the orbiting

electron must “match” or be in phase as the electron makes one

complete revolution.

e. Without the matching the amplitude of the wave gets cancelled during

each revolution and the wave will disappear.

CHEM 3720

27

F. Heisenberg’s Uncertainty Principle (Principle of indeterminacy)

1. Heisenberg’s Uncertainty Principle in terms of position and momentum

a. In classical mechanics it was possible to be able to determine both the

position and the momentum of a particle. What happens if the particle

has wave properties?

b. Look at the example of an electron: during the measurement of the

position of an electron, the radiation used to do that changes its

momentum.

c. If we wish to locate an electron within a region x there will be an

uncertainty in the momentum of the electron (denoted p).

d. Heisenberg showed that:

4

hpx

□ This is the Heisenberg Uncertainty Principle.

2. Heisenberg’s Uncertainty Principle in terms of energy and time

a. A similar expression can be deduced for the uncertainty in the energy

E and the uncertainty in the time t:

m

pE x

2

2

; xxxxx p

t

xpp

m

pE

v

xptE x

4

htE

b. The uncertainty in terms of E and t is used in spectroscopy:

□ stable states sharp lines (large t small E)

□ unstable states diffuse lines (small t large E)

c. To measure the energy of a system with accuracy E the measurement

must be extended over a period of time of order of magnitude h/t.

CHEM 3720

28

G. Unit Review

1. Important Terminology

blackbody

blackbody radiation

Wien displacement law

ultraviolet catastrophe

quanta

Planck constant

photoelectrons

photon

threshold frequency

work function

line spectra

Lyman, Balmer, Pashen, Bracket series

series limit

CHEM 3720

29

Balmer formula

Rydberg formula

Rydberg constant

term

angular momentum

stationary state or orbit

first Bohr radius

ground state

excited states

Bohr frequency condition

wave-particle duality

de Broglie wavelength

Heisenberg Uncertainty Principle

CHEM 3720

30

2. Important Formulas

nhEn or hE

1

8)(

B/

3

3

Tkh

e

d

c

hdT

1

8)(

B/5

Tkhc

e

dhcdT

)(v2

1KE 00

2 hhhhm

22

21

111~

nnRH

nh

nrm 2

ve

2e

2204

em

nrn

2220

4e 1

8 nh

emEn

EEEh 12

2

221

11~

nnR

vm

h

p

h

4

hpx

4

htE

CHEM 3720

31

Unit III

Postulates and Principles of Quantum Mechanics

A. The Schrödinger Equation

1. Introduction

a. It is the fundamental equation of quantum mechanics.

b. The solutions of the time-independent Schrödinger equation are called

stationary-state wave functions.

2. Time-independent Schrödinger equation

a. Schrödinger equation is the equation for finding the wave function of a

particle and come up based on idea that if the matter possesses

wavelike properties there must be a wave equation that governs them.

b. Schrödinger equation cannot be demonstrated (it can be seen as a

fundamental postulate) but it can be understood starting from classical

mechanics wave equation:

– Classical wave equation:

2

2

22

2

v

1

t

u

x

u

– The solution is: txtxu cos)(),(

– The t dependence appear as cost or T(t) or exp(2ivt).

– The spatial amplitude of the wave, , is obtained from the equation:

0)(v

)(

2

2

2

2

xdx

xd

where

v22 v

0)(4)(

2

2

2

2

xdx

xd

– Rearrange the equation considering:

Vm

pVKE

2

2

)(2 VEmp

)(2 VEm

h

p

h

0)()]([2)(

22

2

xxVEm

dx

xd

CHEM 3720

32

)()()()(

2 2

22

xExxVdx

xd

m

c. This is the one-dimensional time-independent Schrödinger equation.

d. The solutions (wave functions) of this equation are called stationary-

state wave functions.

e. Schrödinger equation for three dimensions:

□ Rewrite the equation and generalize to three dimensions:

),,(),,(),,(

),,(2 2

2

2

2

2

22

zyxEzyxzyxV

zyxzyxm

□ Rewrite including the notation for the Laplacian operator:

○ 2

2

2

2

2

2

2

zyx is the Laplacian operator.

○ An operator is a symbol that tells you to do something (a

mathematical operation) to whatever (function, number, etc)

follows the symbol.

EzyxVm

),,(

2

22

□ Rewrite including the notation for the Hamiltonian operator:

○ HzyxVm

ˆ),,(2

22

is the Hamiltonian operator.

EH ˆ

This is the simple form of the Schrödinger equation.

CHEM 3720

33

3. Eigenvalue-eigenfunction relation in quantum mechanics

a. The eigenvalue-eigenfunction relation in quantum mechanics is written

in the form:

)()(ˆ xxA

where A is an operator

(x) is an eigenfunction or characteristic function

is an eigenvalue or characteristic value

b. The wave functions in Quantum Mechanics are eigenfunctions of the

Hamiltonian operator and the total energy is the eigenvalue.

4. Interpretation of the wave function :

a. The *(x)(x)dx is the probability that the particle to be located

between x and x + dx (the one-dimensional case).

b. The function * is the complex conjugate of the wavefunction .

□ The complex conjugate * is obtained by replacing i with –i in the

expression of the wavefunction .

□ The * product becomes real.

CHEM 3720

34

B. Postulates of Quantum Mechanics

1. Enunciations

a. Postulate 1: The state of a quantum-mechanical system is completely

specified by a function (x) that depends upon the coordinate of the

particle. All possible information about the system can be derived

from (x). This function, called the wave function or the state

function, has the important property that *(x)(x)dx is the probability

that the particle lies in the interval dx, located at the position x.

Note: This principle can be stated using the time-dependent wave

function (x,t) instead of the time-independent wave function (x).

b. Postulate 2: To every observable in classical physics there corresponds

a linear, Hermitian operator in quantum mechanics.

c. Postulate 3: In any measurements of the observable associated with the

operator A , the only values that will ever be observed are the

eigenvalues an, which satisfy the eigenvalue equation: nnn aA ˆ

d. Postulate 4: If a system is in a state described by a normalized wave

function , then the average value of the observable corresponding to

A is given by:

space all

ˆ dxAa

e. Postulate 5: The wave function, or state function, of a system evolves

in time according to the time-dependent Schrödinger equation:

t

txitxH

),(),(ˆ

f. Postulate 6: All electronic wave functions must be antisymmetric under

the exchange of any two electrons.

g. Note: The principles above are stated for a one-dimensional space. To

generalize to three dimensions one can replace (x) with (x,y,z) and

(x,t) with (x,y,z,t).

CHEM 3720

35

2. First postulate of quantum mechanics

a. Classical mechanics deals with quantities called dynamical variables:

position, momentum, angular momentum, and energy. A measurable

dynamical variable is called an observable.

b. The movement of a particle can be described completely in classical

mechanics. This is not possible in quantum mechanics because the

uncertainty principle tells one cannot specify or determine the position

and the momentum of a particle simultaneously to any desirable

precision. This leads to the first postulate of quantum mechanics.

c. Postulate 1: The state of a quantum-mechanical system is completely

specified by a function (x) that depends upon the coordinate of the

particle. All possible information about the system can be derived

from (x). This function, called the wave function or the state

function, has the important property that *(x)(x)dx is the probability

that the particle lies in the interval dx, located at the position x.

d. This principle can also be stated using the time-dependent wave

function (x,t) instead of the time-independent wave function (x).

e. The one-dimensional probability will be replaced by the three-

dimensional probability:

dxdydzzyxzyx ),,(),,(*

f. The case of two particles:

212121* ),(),( dxdxxxxx

g. The total probability of finding a particle somewhere must be unity:

1)()(

spaceall

* dxxx

□ The wave functions that satisfy this condition are said to be

normalized.

CHEM 3720

36

□ The functions are called normalizable if:

1)()(

spaceall

* Adxxx

because they can be normalized once they are divided by A .

h. Other conditions for the wave function(x),

its first derivative dx

xd )(, and its second

derivative 2

2 )(

dx

xd should be:

○ single-valued

○ continuous

○ finite

□ If these conditions are met, the wave

functions is said to be “well behaved”.

3. Second postulate of quantum mechanics

a. Postulate 2: To every observable in classical physics there corresponds

a linear, Hermitian operator in quantum mechanics.

b. An operator is a symbol that tells to do a mathematical operation to

whatever follows the symbol.

□ The operators are usually denoted by a capital letter with a little hat

over it called a carat (like in H ).

□ What follows the operator is called operand.

c. An operator A is linear if:

xfAcxfAcxfcxfcA 22112211ˆ)(ˆ)()(ˆ

CHEM 3720

37

d. An operator A is Hermitian if it has the property of being linear and if:

dxAdxA

spaceall12

spaceall2

*1 )ˆ(ˆ

or dxxfAxgdxxgAxf

spaceallspaceall

)](ˆ)[()(ˆ)(

for any pair of functions 1 and 2 (or f and g) representing a physical

state of a particle.

□ These expression are true in derivative form as well.

e. All the quantum operators can be written starting from the operators in

the table below using classical physics formulas:

Classical

Variable

QM

Operator

Expression for operator

x X x

xP xP x

ixi

t T (or t ) t

E E t

iti

or zyxVm

,,2

22

□ Examples:

○ Kinetic energy: m

pK

2

2

1-dimensional: 2

222

22

ˆˆ

xmm

pK

3-dimensional: 22

2

2

2

2

2

22

22ˆ

mzyxmK

○ Potential energy: VV ˆ

CHEM 3720

38

○ Total energy: VKE

The operator for the total energy is the Hamiltonian H :

Vm

Vzyxm

VKH

2

2

2

2

2

2

2

22

22ˆˆˆ

○ The angular momentum along the x axis:

yzx pzpyL

)(ˆy

zz

yiLx

f. Commutation

□ It is a property of operators in which )(ˆˆ xfBA is compared to

)(ˆˆ xfAB .

○ )](ˆ[ˆ)(ˆˆ xfBAxfBA

○ )](ˆ[ˆ)(ˆˆ xfABxfAB

□ Operators usually do not commute: )(ˆˆ)(ˆˆ xfABxfBA

□ When )(ˆˆ)(ˆˆ xfABxfBA for every compatible f(x) the operators A

and B are said to commute.

□ Example: dx

dA ˆ and 2ˆ xB

dx

xdfxxxfxfBA 22ˆˆ

dx

xdfxxfAB 2ˆˆ

□ Rewrite and drop f(x).

□ Define the commutator (i.e., the commutator of A and B ) as:

ABBABA ˆˆˆˆ]ˆ,ˆ[

□ The commutator of commuting operators is the zero operator:

0ˆˆˆˆ ABBA

where 0 is the zero operator.

CHEM 3720

39

g. A special property of linear operators (for example, the operator A) is

that a linear combination of two eigenfunctions of the operator with the

same eigenvalue is also an eigenfunction of the operator:

□ Consider that two eigenfunctions have the same eigenvalue:

11ˆ aA and 22

ˆ aA

(This is a two-fold degeneracy.)

□ Then any linear combination of 1 and 2 is also an eigenfunction

of A :

)(ˆˆ)(ˆ 221122112211 ccaAcAcccA

□ Example:

– Solving )()( 2

2

2

xfmdx

xfd gives the eigenfunctions imxexf )(1 and

imxexf )(2 .

– A linear combination of 1f and 2f , )()()( 21112 xfcxfcxf x , is also an

eigenfunction:

)()(

)()()(

122

212

22

212

122

xfmececm

emcemcdx

xfd

imximx

imximx

4. Third postulate of quantum mechanics

a. Postulate 3: In any measurements of the observable associated with the

operator A , the only values that will ever be observed are the

eigenvalues an, which satisfy the eigenvalue equation:

nnn aA ˆ

b. For an experiment designed to measure the observable associated to A ,

we will find only the values naaa ,...,, 21 corresponding to the states

n ,...,, 21 and no other values will be observed.

c. Example: If HA ˆˆ nnn EH ˆ (Schrödinger equation) and only

the nE energies (eigenvalues) will be experimentally observed.

CHEM 3720

40

5. Fourth postulate of quantum mechanics

a. Postulate 4: If a system is in a state described by a normalized wave

function , then the average value of the observable corresponding to

A is given by:

space all

dxÂa

□ a is the symbol that represents the average value.

b. Determine the variance, 2, of experiments:

□ Variance is a statistical mechanics quantity.

□ Assume we have: )()(ˆ xaxA nnn

nnn adxxAxa

)(ˆ)(*

□ Also: )()(ˆˆ)(ˆ 22 xaxAAxA nnnn

22*2 )( nnnn adxxaxa

□ Variance of the experiments become:

022222 nna aaaa

□ The standard deviation a is zero so the only values observed are

the an values.

CHEM 3720

41

6. Fifth postulate of quantum mechanics

a. Postulate 5: The wave function, or state function, of a system evolves

in time according to the time-dependent Schrödinger equation:

t

txitxH

),(),(ˆ

b. Time dependence of ),( tx :

– Assume H does not depend on time.

– Separation of variable: )()(),( tfxtx

Edt

tf

tf

ixH

x

)(2

)()(ˆ

)(

1

(where E is a constant)

)()(ˆ xExH (time-independent Schrödinger equation)

)()(

tEfi

dt

tdf

– The solution is: tiiEt eetf /)( where hvE

/

)(),(tiEnextx

(assuming a set of solutions nE )

c. The probability density and the average values are independent of time

(the function of time cancels out):

dxxxdxexexdxtxtxtiEtiE nn )()()()(),(),( *//**

)(xn are called stationary-state wave functions

CHEM 3720

42

C. More Properties

1. Properties of the Eigenfunctions of Quantum Mechanical Operators

a. The eigenfunctions of QM operators are orthogonal.

b. The eigenfunctions of QM operators are functions satisfying the

equation: nnn aA ˆ where na are real (although A and n can be

complex) when the operator is Hermitian.

c. If the eigenvalues na are real then eigenfunctions have the property:

0)()(*

dxxx nm

□ If a set of wave functions satisfies this condition then the set is said

to be orthogonal or that the wave functions are orthogonal to each

other.

□ If they are also normalized, 1)()(*

dxxx nn , than the set is

said to be orthonormal and wave functions are orthonormal.

d. Generalizing:

ijji dx

*

where

ji

jiij

if0

if1 is the Kroenecker delta symbol.

e. Example: sin and cos functions are

orthogonal in 0–2 (or 0–) interval.

f. An even function is always orthogonal to an

odd function over an interval centered at 0.

□ An even function is a function for which f(x) = f(–x).

□ An odd function is a function for which f(x) = –f(–x).

g. The eigenfunctions of QM operators form a complete set.

CHEM 3720

43

2. Uncertainty Principle written based on operators

a. The uncertainty in the measurements of a and b ( a and b ) are

related by:

dxxBAxba )(]ˆ,ˆ[)(2

1 *

where ABBABA ˆˆˆˆ]ˆ,ˆ[ is the commutator.

□ If A and B commute then 0]ˆ,ˆ[ BA then 0ba a and b

can be measured simultaneously to any precision.

□ If A and B do not commute then a and b cannot be simultaneously

determined to arbitrary precision.

b. Example:

□ For x the operator is xX ˆ

□ For xp the operator is dx

diPx ˆ

□ The commutator of X and xP is IiPXXPXP xxxˆˆˆˆˆ]ˆ,ˆ[

where I is the identity operator (multiplication to 1).

dxxIix xxp )()ˆ)((2

1 *

22

1 ixp

where 22 xxx and 22 ppp

CHEM 3720

44

D. Unit Review

1. Important Terminology

Schrödinger equation

wavefunctions

operator

operand

Laplacian operator

Hamiltonian operator

eigenvalue-eigenfunction relation

complex conjugate

QM postulates

wavefunction interpretation

Hermitian operator

observable

average value

CHEM 3720

45

normalized/normalizable

commutation

commutator

linear operators

variance/standard deviation

orthogonal

orthonormal

even function

odd function

complete set

CHEM 3720

46

2. Important Formulas

)()()()(

2 2

22

xExxVdx

xd

m

2

2

2

2

2

2

2

zyx

EzyxVm

),,(

2

22

HzyxVm

ˆ),,(2

22

)()(ˆ xxA

space all

ˆ dxAa

xi

xiPx

ˆ

1)()(

spaceall

* dxxx

0)()(*

dxxx nm

ijji dx

*

CHEM 3720

47

Unit IV

Applications of Quantum Theory

A. The Particle in a Box

1. Introduction

a. The particle-in-a-box refers to the quantum mechanical treatment of

translational motion.

b. One tries to solve the Schrödinger equation to obtain the wavefunctions

and the allowed energies for the particle.

c. The time-independent three-dimensional Schrödinger equation:

),,(),,(),,(2 2

2

2

2

2

22

zyxEzyxzyxVzyxm

2. Particle in a one-dimensional box (or one-dimensional motion)

a. Consider only a one-dimensional motion

(along x coordinate) drop y and z in the

equation above.

b. Consider that the particle experiences no

potential energy between the position 0

and a V(x) = 0 (for 0 x a)

c. The Schrödinger equation for this problem:

Edx

d

m

2

22

2

d. The mathematical solution of this equation is:

kxBkxAx sincos)(

where h

mEmEk

22)2( 2/1

m

khE

2

22

8

aa

CHEM 3720

48

e. Apply the boundary conditions and the normalization condition to

determine A and B:

0)0( 0A

0)( a 0sin kaB nka where n = 1,2,…

f. The allowed energy levels are:

2

22

8ma

nhEn where n = 1,2,…

where n is called quantum number.

□ The energy is quantized.

□ The energy levels increase as the quantum

number increases.

□ The energy separation between the allowed

energy levels increases as the quantum

number increases.

□ The energy levels and the energy separation

between the energy levels increases as the

size of the box or the mass decreases.

g. Quantum numbers appear naturally when the boundary conditions are

put in the Schrödinger equation (like in the string problem) and are not

introduced ad hoc like in Plank model for blackbody radiation or Bohr

model of hydrogen atom.

h. Determining the wavefunctions:

□ Find B by setting the condition that the function (x) is normalized

1)()(

0

* dxxxa

1sin

0

22

a

a

xnB

1

2

2 a

B a

B2

○ B is called the normalization constant.

a2)

a2)

CHEM 3720

49

□ The normalized eigenfunctions n(x) are given by:

a

xn

axn

sin

22/1

where 0 x a and n = 1,2,…

i. Solving the Schrödinger equation for the particle in a box problem

gives a set of allowed energies (or eigenvalues) and a set of wave

functions (or eigenfunctions).

j. Representations including the energies and the wave functions (a) and

the probability densities (b,c) for the first few levels for the particle in a

box:

CHEM 3720

50

k. The particle-in-a-box model can be applied to electrons moving freely

in a molecule (also called free-electron model).

l. Example:

□ Butadiene has an adsorption band at 4.61104 cm–1. As a simple

approximation, consider butadiene as being a one-dimensional box

of length 5.78 A = 578 pm and consider the four electrons to

occupy the levels calculated using the particle in a box model.

○ The electronic excitation is given by )23(8

22

2e

2

am

hE

○ The calculated excitation energy

14 cm1054.4~

hc

E

compares very well with the

experimental value.

○ This simple free-electron model can be quite successful.

m. The probability of finding the particle between x1 and x2 is given by:

2

1

)()(*x

xnn dxxx

□ If x1 = 0 and x2 = a/2 then 2/1)2/0(Prob ax for all n.

□ For n = 1: )2/)4/(Prob)4/0(Prob axaax

□ As n increases (for example n = 20) these 2 probabilities become

equal.

n. Generalizing, the probability density becomes uniform as n increases.

o. This is an illustration of the Correspondence Principle that says that

quantum mechanics results and classical mechanics results tend to

agree in the limit of large quantum numbers.

□ The large-quantum-number limit is called the classical limit.

CHEM 3720

51

p. The eigenfunctions are orthogonal to each other:

0)()(

0

* a

nm dxxx (where m n)

□ Example: look at 1(x) and 3(x):

03

sinsin2

)()(

003

*1

aa

dxa

x

a

x

adxxx

q. Average values and variances for the position and the momentum of

the particle:

– Average value of position: 2

sin2

)()(

0

2

0

* adx

a

xnx

adxxxxx

aa

nn

– Average value of position square: 22

22

0

2*2

23)()(

n

aadxxxxx

a

nn

– Variance in position:

2

32

222222 n

n

axxx

– Standard deviation in position:

2/122

22 232

n

n

axxx

– Average value of momentum:

a

dxa

xn

a

xn

a

nip

02

0cossin2

– Also: 2

222

2

2222

222

a

n

ma

nmEmp

;

2

2222

a

np

– Finally: 2

232

2/122

npx (Heisenberg Uncertainty Principle)

3. Particle in a two-dimensional box (or two-dimensional motion)

a. The Schrödinger equation for this problem:

yxEyxm

,2 2

2

2

22

b. Allowed energies (or eigenvalues)

2

2

2

22

8 b

n

a

n

m

hE

yxnn yx

b

a

b

a

CHEM 3720

52

b. The wavefunctions (or eigenfunctions)

b

yn

a

xn

bayYxXyx

yx

sinsin22

)()(),(

4. Particle in a three-dimensional box (or three-dimensional motion)

a. The Schrödinger equation:

zyxEzyxm

,,2 2

2

2

2

2

22

b. Solutions:

c

zn

b

yn

a

xn

cbazyx zyx

nnn zyx

sinsinsin

222),,(

2

2

2

2

2

22

8 c

n

b

n

a

n

m

hE zyx

nnn zyx

c. Average values of the position and the momentum of the particle:

kjiRr222

),,(ˆ),,(*

0 0 0

cbazyxzyxdzdydx

a b c

where kjiR ZYX ˆˆˆˆ

0),,(ˆ),,(*

0 0 0

zyxzyxdzdydx

a b c

Pp where

zyxi kjiP ˆ

d. The case of a cubic box (a = b = c):

□ Three sets of quantum numbers

give same energy:

2

2

1121212118

6

ma

hEEE

CHEM 3720

53

□ The energy level 2

2

8

6

ma

hE is degenerate.

□ The energy level 2

2

1118

3

ma

hE is nondegenerate.

□ Degeneracy is equal to the number of states with same energy.

○ Once the symmetry is destroyed then degeneracy is lifted.

5. Separation of variables

a. The Hamiltonian for the particle in a 3-dimensional box:

zyx HHHzyxm

H ˆˆˆ2

ˆ2

2

2

2

2

22

□ It can be written as a sum of terms where:

2

22

xmH x

, 2

22

ymH y

, 2

22

zmH z

○ The operator is said to be separable.

b. The eigenfunctions zyx nnn are written as a product of eigenfunctions

of each operator xH , yH , and zH , and the eigenvalues zyx nnnE are

written as a sum of the eigenvalues of each of the operator xH , yH ,

and zH .

c. This is a general property in quantum mechanics: If the Hamiltonian

(or an operator in general) can be written as a sum of terms involving

different coordinates (i.e., the Hamiltonian is separable) then the

eigenfunctions of H is a product of the eigenfunctions of each

operator constituting the sum and the eigenvalues of H is a sum of

eigenvalues of each operator constituting the sum.

d. Example: )(ˆ)(ˆˆ

21 wHsHH )()(),( wswsnm and mnnm EEE

where )()()(ˆ1 sEssH nnn and )()()(ˆ

2 wEwwH mmm

CHEM 3720

54

B. The Harmonic Oscillator

1. Introduction

a. The harmonic oscillator refers to the quantum mechanical treatment of

vibrational motion.

b. The harmonic oscillator in classical mechanics: – Force: kxkf )( 0 (Hooke’s Law) where k is the force constant.

– Hooke’s Law combined with Newton’s equation: 02

2

kxdt

xdm

– The solution of this equation: tAtx cos)( where m

k

– The potential energy of the oscillator: tkAxk

xV 222 cos2

1

2)(

– The kinetic energy of the oscillator: tkAdt

dxmK 22

2

sin2

1

2

1

– The total energy of the oscillator: 2

2

1kAKVE

The total energy is conserved; it is transferred between K and V.

c. The harmonic motion in a diatomic molecule:

– Consider the movement of atoms of masses m1 and m2.

)( 01221

2

1 xxkdt

xdm

)( 01222

2

2 xxkdt

xdm

– By summing the two equation above:

02

2

dt

XdM where 21 mmM and

M

xmxmX 2211

where X is the center the mass coordinate.

– Subtracting equation 2 divided by m2 from equation 1 divided m1:

02

2

kxdt

xd where 012 xxx and

21

21

mm

mm

where is the reduced mass of the system and x is the relative coordinate.

□ The movement of a two-body system can be reduced to the

movement of a one-body system with a mass equal to the reduced

mass of the two-body system.

CHEM 3720

55

2. The quantum-mechanical harmonic oscillator

a. The Schrödinger equation for quantum-mechanical harmonic

oscillator:

)()()(2 2

22

xExxVdx

d

where 2

2

1)( kxxV

0)(2

12)( 2

22

2

xkxE

dx

xd

b. The eigenvalues are:

2

1

2

1

2

1nhnn

kEn

where ,...2,1,0n ;

k

;

k

2

1

c. The wave functions (eigenfunctions) are:

2/2/1 2)()( x

nnn exHAx where 2

k

□ The normalization constant is given by

4/1

!2

1

n

An

n

□ The wave functions form an orthonormal set.

□ The )( 2/1 xHn are polynomial functions called Hermite

polynomials where )(nH is a nth degree polynomial in .

○ Here are the first few Hermite polynomials: 1)(0 ξH 2)(1 ξH

24)( 22 ξH 128)( 3

3 ξH

124816)( 244 ξH 12016032)( 35

5 ξH

even polynomials: f(x) = f(–x) odd polynomials: f(x) = –f(–x)

□ Continuous odd functions properties: 0)0( f and

A

A

dxxf 0)(

CHEM 3720

56

d. The normalized wave functions and the probability density:

e. The existence of the zero-point energy

□ The minimum energy (the ground-state energy) is not zero even for

n = 0.

□ This energy is called the zero point energy (ZPE):

h2

1ZPE

□ It is a result in concordance with the Uncertainty Principle that

says one cannot determine exactly both the position (for example x

= 0) and the momentum (for example p = 0) at the same time.

f. The quantum-mechanical harmonic oscillator model accounts for the

IR spectrum of a diatomic molecule.

□ It is a model for vibrations in diatomic molecules.

□ The transitions between various levels in harmonic oscillator

model follow the selection rule:

1n

□ The quantum-mechanical harmonic oscillator predicts the

existence of only one frequency in the spectrum of a diatomic, the

frequency called fundamental vibrational frequency:

nn EEhE 1obs

CHEM 3720

57

khnn

kE

22

1

2

1)1(

k

2

1obs and

k

c2

1~obs

○ The quantity x from above is, in this case, the difference

between the interatomic distance during the vibration and the

“equilibrium” distance.

0llx

○ If the fundamental vibrational frequency is known, one can

determine the force constant as:

2obs

2obs 2~2 ck

g. Typical values of obs~ are in the 100 – 4000 cm–1 range.

h. The average value of position and momentum for the harmonic

oscillator:

dxxxxx )()(* 0x

dxxdx

dixp

)()(* 0 p

i. Note that vibrational quantum number n is usually labeled as v in other

textbooks.

j. The probability density is different than 0 (bigger than 0) even in

regions where E < V.

□ This is equivalent to a negative kinetic energy and is an example of

quantum mechanical tunneling, a property of quantum mechanical

particles that is nonexistent in classical mechanics.

□ The quantum mechanical particles have the property of non-zero

probability in regions forbidden by classical mechanics.

CHEM 3720

58

C. The Rigid Rotator

1. Introduction

a. The rigid rotator (or rigid rotor) refers to the quantum mechanical

treatment of rotational motion.

b. The quantum mechanical rigid rotator is a model for a rotating

diatomic molecule.

c. Treatment of rotation in diatomic molecules – For the center of mass: 2211 rmrm

– Velocities: 1rot11 2 rrv ;

2rot22 2 rrv where rot2 is the

angular speed.

– Kinetic energy: I

LImvmvK

22

1

2

1

2

1 222

221

– Moment of inertia: 2222

211 rrmrmI where 21 rrr and

21

21

mm

mm

– Similar to moment of inertia for a single rotating particle ( 2mrI ).

□ The movement of a two-body system can be replaced by the

movement of a one-body system where the mass is replaced with

the reduced mass.

2. Quantum-mechanical rigid-rotator model

a. The Schrödinger equation:

0ˆ VV ; 2

2ˆˆ

KH

,

2

2

22,

2,

2

2

2

2

2

2

2

22

sin

1sin

sin

11

rr rrrr

rr

zyx

b. The special case when r is constant (i.e., the rotator is rigid):

),(),(ˆ EYYH

where ),( Y are the rigid-rotator wave functions and are also

known as spherical harmonic functions.

CHEM 3720

59

,,sin

1sin

sin

1

2 2

2

2

2

EYYI

– Further rearrange:

0)sin(sinsin 2

2

2

Y

YY

where

2

2

2

22

ErIE

– The rigid rotator wave functions ),( Y will be given later.

c. By solving the equation above, one can determine that must satisfy:

)1( JJ

d. The discrete set of allowed rotational energy levels is given by:

)1(2

2

JJI

EJ

where ,...2,1,0J

○ J is the rotational quantum number.

e. Each level has a degeneracy given by:

12 JgJ

f. The selection rule for the rigid-rotator model:

1J

○ Only transitions between adjacent states are allowed.

g. The energy and frequency for transition between levels:

hJI

hEEE JJ )1(

4 2

2

1

)1(4 2

JI

h

h. The frequency of these transitions is about Hz1010 1110 which is in

the microwave range of electromagnetic radiation microwave

spectroscopy.

i. Usually, write the frequency in terms of the rotational constant B:

)1(2 JB where I

hB

28

□ Typical values are 1010 – 1011 Hz.

CHEM 3720

60

j. In terms of wavenumbers:

)1(~

2~ JB where cI

hB

28

~

□ Typical values are 0.1 – 20 cm–1.

k. From experimental B~

one can determine the moment of inertia I then

the bond distance r in a diatomic molecule ( 2rI ).

l. Note that rotational quantum number J is sometimes (i.e., in our

textbook) labeled as l.

□ The quantum number l is usually used for the orbital angular

momentum (of an electron orbiting around a nucleus), and the

quantum number J is usually used for rotating molecules.

CHEM 3720

61

D. Tunneling

1. Tunneling is the quantum mechanics property of the wavefunction to be

non-zero in classically forbidden zones.

2. Examples:

a. Penetration through an energy barrier of energy V:

□ The wavefunction and its derivatives should be continuous.

c. The transmission probability T is higher as:

□ the barrier is narrower

□ the incident energy is closer to the height

□ the mass of the particle is smaller

Measure of barrier

width height mass

Incident energy/Barrier height E/V

d. Also true is that the transmission

probability T < 1 even when E > V.

e. Chemical reaction can occur

exclusively through tunneling even if

involves heavier atom (i.e., carbon)

tunneling.

CHEM 3720

62

E. Appendix

1. Cartesian and spherical coordinates

a. The position of a point can be

specified by using Cartesian

coordinates (x,y,z) or by using

spherical coordinates (r,,).

b. Relations between Cartesian and spherical coordinates:

cos

sinsin

cossin

rz

ry

rx

and

zy

zyx

z

zyxr

/tan

cos222

222

c. Other relations:

ddrdrdxdydzdV sin2

0 0

2

0

2 sin

ddrdrdxdydz

0 0

2

0

2 ,,sin,,

rFdddrrdxdydzzyxF

,

2

2

22,

2,

22

2

2

2

2

2

22

sin

1sin

sin

11

rr rrrr

rr

zyx

rsin

z

x

y

r

(x, y, z)

(r, , )rsin

z

x

y

r

(x, y, z)

(r, , )

CHEM 3720

63

2. Comparison between linear and circular motion

v

v

r

Linear Motion Angular Motion

Speed: Angular speed:

v (m/s) r

v (rad/s)

Mass: Moment of inertia:

m I = mr2

Linear momentum: Angular momentum:

p = mv L = Iω = mvr

Kinetic energy: (Rotational) kinetic energy:

m

pmK

2v

2

1 22

I

LIK

22

1 22

Momentum (as a vector): Angular momentum (as a vector):

p = mv L = Iω

L = r p

Components:

xyz

zxy

yzx

ypxpL

xpzpL

zpypL

CHEM 3720

64

F. Unit Review

1. Important Terminology

particle-in-a-box

quantum number

normalization constant

probability density

free-electron model

Correspondence Principle

classical limit

separable Hamiltonian

harmonic oscillator

force constant

reduced mass

Hermite polynomials

zero-point energy

CHEM 3720

65

selection rule

IR spectrum

vibration

fundamental vibrational frequency

tunneling

rigid rotator

moment of inertia

spherical harmonic functions

degeneracy

microwave spectroscopy

rotational constant

spherical coordinates

angular momentum

CHEM 3720

66

2. Important Formulas

2

22

8ma

nhEn

a

xn

axn

sin

22/1

21

21

mm

mm

2

2

1)( kxxV

2

1nhEn

k

2

1

2/2/1 2)()( x

nnn exHAx ; 2

k ;

4/1

!2

1

n

An

n

h2

1ZPE

1n ; nn EEhE 1obs

k

2

1obs ;

k

c2

1~obs

2222

211 rrmrmI

)1(2

2

JJI

EJ

12 JgJ

1J

)1(4 2

JI

h

)1(2 JB ; I

hB

28 ; )1(

~2~ JB ;

cI

hB

28

~

CHEM 3720

67

Unit V

Quantum Mechanical Treatment of the Hydrogen Atom

A. Schrödinger Equation for the Hydrogen Atom

1. Introduction

a. The hydrogen atom will serve as prototype for the treatment of more

complex atoms.

b. By solving the Schrödinger equation for the hydrogen atom, one

obtains the allowed energies (eigenvalues) and the wavefunctions of

the electron (eigenfunctions), which gives the atomic orbitals.

2. Schrödinger Equation

a. Consider the proton fixed at the origin and the electron (of mass me)

interacting with it.

b. The potential energy (interaction energy) is given by the Coulombic

potential:

r

eV

0

2

4

c. Hamiltonian operator becomes:

r

e

mH

0

22

e

2

42ˆ

d. Schrödinger equation:

Er

e

m

0

22

e

2

42

□ Rewrite the Laplacian operator in terms of spherical coordinates:

,

2

2

22,

2,

2

2

2

2

2

2

2

22

sin

1sin

sin

11

rr rrrr

rr

zyx

CHEM 3720

68

e. Schrödinger equation in terms of the spherical coordinates:

Er

e

rm

rmrr

rrm

re

ree

0

2

,2

2

22

2

,2

2

,

2

2

2

4sin

1

2

sinsin

1

2

1

2

where ),,( r will be called the H atom orbitals.

f. Simplifying Schrödinger equation:

– Multiply by 2mer2.

– Separate the terms that depend on r from the ones that don’t.

□ Make a separation of variables: ),()(),,( YrRr

– Divide the equation by ),()( YrR .

0sin

1sin

sin

1

),(

)(4

2

)(

2

2

2

2

0

2

2

2e2

2

YY

Y

rREr

erm

dr

dRr

dr

d

rR

□ Each of the two terms in the equation above should be constant:

2

0

2

2

2e2

2

)(4

2

)(

rRE

r

erm

dr

dRr

dr

d

rR

2

2

2

2

2

sin

1sin

sin

1

),(

YY

Y

3. Solving for Y(,)

a. The equation that gives Y(,) is the same as the one for the rigid

rotator model.

b. The solutions Y(,) are called therefore rigid-rotator wave functions

or spherical harmonics.

c. Simplifying the equation:

– Rearrange the equation in the form:

0)sin(sinsin 2

2

2

Y

YY

CHEM 3720

69

□ Make a separation of variables: )()(),( Y

22sinsinsin)(

1m

d

d

d

d

and 2

2

2

)(

1m

d

d

d. Solve for ():

2

2

2

)(

1m

d

d

□ The solution is: immm eA )(

○ where ,...2,1,0 m

○ 2/12 mA is the normalization constant.

im

m e2

1)(

e. Solve for ():

22sinsinsin)(

1m

d

d

d

d

□ Change the variables: xcos and )()( xP

0)(1

2)1(2

2

2

22

xP

x

m

dx

dPx

dx

Pdx

○ where ,...2,1,0 m (as determined above)

□ This is called Legendre equation.

□ Certain conditions should be met:

○ = l(l+1); l = 0,1,2,…

○ |m| l where |m| is the magnitude of m

0)(1

12)1(2

2

2

22

xP

x

mll

dx

dPx

dx

Pdx

○ where l = 0,1,2,… and lm ,...,2,1,0

□ The solution P(x) depends on two integers l and m.

CHEM 3720

70

□ The case of m = 0:

)13(2

1)(

)(

1)(

22

1

0

xxP

xxP

xP

are called Legendre polynomials

○ These polynomials are orthogonal to each other:

1

1

0)()( dxxPxP nl

□ The case of m 0:

)()1()( 22 xPdx

dxxP lm

mmm

l

○ These are called associated Legendre polynomials.

○ Examples of )(xPm

l:

sin)1(

cos

1

2/1211

01

00

xxP

xxP

xP

2222

2/1212

2202

sin3)1(3

sincos3)1(3

)1cos3(2

1)13(

2

1

xxP

xxxP

xxP

○ The associated Legendre functions are normalizable:

nl

mn

m

lm

nm

l

ml

ml

l

dPPdxxPxP

)!(

)!(

)12(

2

sin)(cos)(cos)()(

0

1

1

○ The normalization constant is: 2/1

)!(

)!(

2

)12(

ml

mllNlm

CHEM 3720

71

f. Looking back at Y(,) = ()():

imm

lm

leP

ml

mllY cos

)!(

)!(

4

)12(),(

2/1

□ The ),( ml

Y functions form an

orthonormal set with respect to

ddsin , i.e., orthonormal over the

surface of a sphere therefore are called

spherical harmonics.

0

2

0

* ),(),(sin kmnlkn

ml

YYdd

g. The spherical harmonics are also the eigenfunctions of the square of

the angular momentum operator:

2

2

2

22

sin

1sin

sin

L

),()1(),(ˆ 22 ml

ml

YllYL

□ The eigenvalues are: )1(22 llL

□ The magnitude of the angular momentum: )1( llL

h. For the rigid rotator:

I

LH

2

ˆˆ

2

),(2

)1(),(ˆ

2

ml

ml

YI

llYH

so the energies for the rigid rotator are I

llE

2

)1(2

i. Angular momentum and its components:

coscotsinˆˆˆ iPzPyL yzx

sincotcosˆˆˆ iPxPzL zxy

iPyPxL xyz

ˆˆˆ

CHEM 3720

72

– The function ime is an eigenfunction of zL : imimz emeL )(ˆ so the spherical

harmonics are also eigenfunctions of :ˆzL

),(),(ˆ ml

mlz mYYL

– Lz depends only on so part of the wavefunction will be like a constant.

□ The measured values of zL are integral multiples of .

○ The is a fundamental measure of the angular momentum in

quantum mechanics.

□ The zL and 2L operators commute so 2L and zL can be

determined simultaneously and precisely. – Spherical harmonics are not eigenfunctions of xL or yL .

– The xL (or yL ) operator and 2L do not commute so 2L and xL (or yL ) cannot be

measured simultaneously with any precision.

– Also, the xL and zL (or yL and zL ) are not commuting so the 3 components of angular

momentum cannot be measured simultaneously with any precision.

– Because the orientation is chosen arbitrarily, one can say that one of the xL , yL , or zL

commute with 2L but do not commute among themselves.

– Show that lm :

,,ˆ 222 ml

mlz YmYL and ,1,ˆ 22 m

l

m

l YllYL

But 2222 LLLL zyx 22222 1 llmLL yx

lmllm 12

For a value of l, the m can take (2l +1) values:

lm ,......2,1,0 .

– The case 1l , 1,0 m :

22 1 llL ; 2L ; ,0,zL

The maximum value of zL is less than L ,

which implies that L and zL cannot point in the

same direction.

□ As a conclusion: It is possible to observe precise value of L2 and

one of the components of angular momentum simultaneously but it

is not possible to observe precisely the other two components. We

will know nothing about the component.

CHEM 3720

73

4. Solving for R(r)

a. This is the radial part of the wavefunction (r,,).

b. Including all results and conditions from above:

2

0

2

2

2e2

2

)(4

2

)(

rRE

r

erm

dr

dRr

dr

d

rR

0)(

42

1

2 0

2

2e

22

2e

2

rRE

r

e

rm

ll

dr

dRr

dr

d

rm

The term

r

e

rm

ll

0

2

2e

2

42

1

(the Coulombic

attraction and the angular momentum = centrifugal

repulsion term) forms an effective potential, Veff.

□ This is an ordinary differential equation in r.

□ It gives the radial dependence of hydrogen atomic orbitals (in the

form of function R) so it is called the radial equation.

c. The allowed energies:

□ Solving the equation one see that the energy is quantized:

220

4e

8 n

emEn

where ,...2,1n

□ By introducing the Bohr radius: 2

e

20

0em

a

2

00

2

8 na

eEn

; ,...2,1n

□ The same energies were obtained from Bohr

model of the hydrogen atom so the explanation

of the electronic transitions responsible for the

hydrogen atom spectrum is same as before.

CHEM 3720

74

□ Difference compared with the Bohr model: the electron is not

restricted to the orbits of Bohr but is described by its wave function

(r,,).

d. Solving for R(r), one can determine a condition that must be satisfied:

1 ln 10 nl where ,...2,1n

e. The solutions of the equation above:

0

12/2/3

0

2/1

3

22

!2

!1)( 0

na

rLer

nalnn

lnrR l

lnnarl

l

nl

□ )(rRnl depends on two quantum numbers: n and l.

0

12 2

na

rL l

ln are associated Laguerre polynomials.

!7)(

)6(!6)(

)2

1510(!5)(

)6

1264(!4)(

!5)(

)4(!4)(

)2

133(!3)(

!3)(

)2(!2)(

1)(

77

56

235

3214

55

34

213

33

12

11

xL

xxL

xxxL

xxxxL

xL

xxL

xxxL

xL

xxL

xL

f. The )(rRnl functions are normalized with respect to the integration

over r as:

1)()(

0

2*

drrrRrR nlnl

where drr 2 is the volume element.

□ This is the radial part of the spherical coordinate volume element:

ddrdr sin2.

CHEM 3720

75

B. Hydrogen Atom Orbitals

1. The complete hydrogen atom wavefunctions

a. The wavefunctions are called orbitals and depends on 3 quantum

numbers:

),()(),,( mlnlnlm YrRr

b. The wavefunctions are orthonormal because they are:

□ normalized

0 0

2

0

*2 1),,(),,(sin

rrdddrr nlmnlm

□ orthogonal

0 0'''

2

0

*'''

2 ),,(),,(sin

mmllnnnlmmln rrdddrr

c. Examples:

□ 0/ azr and z is the charge of nucleus

n l m

e

a

z2/3

0100

1 1 0 0

2/2/3

0200 2

32

1

e

a

z 2 0 0

cos32

1 2/2/3

0210

e

a

z 2 1 0

ieea

z

sin

64

1 2/2/3

0121 2 1 1

ieea

cossin

1

81

1 3/22/3

0

3 2 –1

CHEM 3720

76

2. Representing the wavefunctions

a. The radial and angular parts are considered separately.

b. The probability of the electron to be located in the spherical volume

between a sphere of radius r and one of r + dr, drrrRnl22)( or

)(22 rRrnl

, is usually plotted.

c. More appropriate is to represent d* which is the probability of the

electron to be located in element of volume d .

d. What is typically represented is the boundary surface of the orbital,

which is the surface (of equal electron density) that contains 90% of

electron density.

e. The term “orbital” is used loosely to represent the wavefunction

depending on 3 quantum numbers (n, l, and m), the probability density,

or the boundary surface.

CHEM 3720

77

f. The number of nodes in the radial part is given by (n – l – 1) and these

nodes are called radial nodes.

□ The r = 0 is not considered a node.

3. The 1s orbital

a. The radial function for 1s orbital (l = 0): 0/

2/30

12

)(ar

s ea

rR

□ This function is normalized:

0

221

1drrRs

b. These results are different than the incorrect Bohr’s proposal of

restricted movement in a fixed, well-defined orbit.

c. Probability for the 1s orbital:

drera

Rar

s0/22

30

21

4

323.00 Prob 0 ar

d. Average value of r is 1s orbital: 012

3ar s

e. The most probable value of r (the maximum probability): 0mp ar

□ The first Bohr radius is equal to the most probable radius (or

distance) for the 1s orbital.

f. The angular part for 1s orbital:

4

1,0

0Y .

□ This function is normalized with respect to

integration over a spherical surface:

0

2

0

00

*00

1,,sin YYdd

g. The 1s orbital: 0/2/130

00101 )(),,(

ars eaYRr

□ This orbital is also normalized.

CHEM 3720

78

4. The 2s orbital

a. Expression for the 2s orbital:

0/

0

2/3

0

00202 2

1

32

1),(),,(

ars e

a

r

aYRr

b. Average value of r is 2s orbital: 02 6ar s

□ The 2s electron is, on average, farther from nucleus than the 1s

electron.

5. Orbitals with l = 1

a. They are possible only for n ≥ 2.

b. The radial part remains the same so look at the angular part only.

cos4

3),(

2/10

1

Y

ieY sin8

3),(

2/111

ieY

sin

8

3),(

2/11

1

where

22

11

211 sin

3

8),(),( YY

c. Any linear combination of 11

Y and 11Y is also an eigenfunctions and

the following combinations are preferred because they are real:

cossin4

3

2

1 11

11 YYpx

sinsin4

3

2

1 11

11 YY

ipy

d. The number of nodes in the angular part is given by l and these nodes

are called angular nodes (or nodal plane).

e. The usual representation is in form of tangent spheres but this

representation is not an accurate one.

CHEM 3720

79

6. Orbitals with l = 2

a. They are possible only for n ≥ 3.

b. The possible values of m are: 2,1,0 m .

c. For 2,1m take linear combinations of ml

Y to get the orbitals:

022 Yd

z

)(2

1 12

12

YYdxz ; )(2

1 12

12

YYi

d yz

)(2

1 22

2222

YYdyx

; )(2

1 22

22

YYi

d xy

d. This way the orbitals (wave functions) are real functions and are

preferred by chemists.

e. These combinations are not unique but are used because the wave

functions are real and have orientation consistent with molecular

structure.

CHEM 3720

80

C. Quantum Numbers and Other Properties

1. The number of quantum numbers

a. The hydrogen atomic wave functions depend on three quantum

numbers.

b. In general, an wavefunction (or an orbital) is described by three

quantum numbers: n, l, and m.

c. A subshell is described by two quantum

numbers: n and l.

d. A shell is described by one quantum

numbers: n.

e. Depending on the quantum numbers,

orbitals are organized in shells (given by

the value of n) and subshells (given by the

value of l).

f. Later, we will see that the state of an electron (in an atom) is described

by four quantum numbers.

2. The principal quantum number n:

a. Can take values of n = 1,2,…

b. It describes the shell or level.

c. The energy of hydrogen atom depends only upon n:

200

2

8 na

eEn

3. The angular momentum quantum number l:

a. It is also called azimuthal or secondary or orbital quantum number.

b. Can take values between 0 and n–1: l = 0,1,…,n–1

c. It describes the type of subshell (or shape of orbital), and, in addition to

n, the actual subshell or sublevel.

d. The n and l quantum numbers determine the radial wave function.

CHEM 3720

81

e. Quantum number l completely determine the angular momentum of the

electron about the proton (nucleus):

1 llL

f. The value of l is customarily denoted by a letter as follows:

l = 0 s (sharp)

l = 1 p (principal)

l = 2 d (diffuse)

l = 3 f (fundamental)

l = 4 g

4. The magnetic quantum number m:

a. More specifically denoted ml as we will see later.

b. It can take (2l + 1) values: lm ,...,2,1,0

□ Same as the degeneracy of the l sublevel or subshell.

c. It describes the orientation (or type) of orbitals, and, in addition to n

and l, the actual orbital.

d. Quantum number m completely determine the z component of the

angular momentum (Lz):

mLz

e. The quantum number m is called magnetic because the energy of

hydrogen atom in a magnetic field depends on m.

□ Each energy level has a degeneracy of (2l + 1) which is removed in

magnetic field.

□ This is knows as the Zeeman effect.

CHEM 3720

82

5. The degeneracy of each of the hydrogen atomic energy levels

a. The energy is equal for different sets of quantum numbers because it

depends only on n.

b. The degeneracy of each level is given by the number of levels

(different l and m) with the same energy (same n):

□ For each n: l = 0,…,n –1:

□ 2

1

0 2

1212 nn

nnl

n

l

6. The number of nodes in an orbital (in an wavefunction):

a. The number of radial nodes is n – l – 1.

b. The number of angular nodes is l.

c. The total number of nodes is n – 1.

7. The average value of r and r2 in various orbitals

a. For an ns orbital: 20

2

3nar ns

□ All s orbitals are spherically symmetric.

b. More general:

2

02

)1(1

2

11

n

ll

Z

anr nl

22

20

42 2/11

12

31

n

ll

Z

anr nl

20

1

na

Z

r nl

2/1

132

0

2

2

lna

Z

r nl

CHEM 3720

83

D. Unit Review

1. Important Terminology

atomic orbitals

Legendre polynomials

associated Legendre polynomials

angular momentum magnitude

associated Laguerre polynomials

boundary surface

radial node

angular node

nodal plane

most probable value of r

principal quantum number

azimuthal/secondary/orbital quantum number

magnetic quantum number

CHEM 3720

84

2. Important Formulas

r

eV

0

2

4

r

e

mH

0

22

e

2

42ˆ

),()(),,( YrRr ; )()(),( Y

im

m e2

1)(

)13(2

1)(

)(

1)(

22

1

0

xxP

xxP

xP

)()1()( 22 xPdx

dxxP lm

mmm

l

imm

lm

leP

ml

mllY cos

)!(

)!(

4

)12(),(

2/1

),()1(),(ˆ 22 ml

ml

YllYL

)1(22 llL

),(),(ˆ ml

mlz mYYL

220

4e

8 n

emEn

0

12/2/3

0

2/1

3

22

!2

!1)( 0

na

rLer

nalnn

lnrR l

lnnarl

l

nl

),()(),,( mlnlnlm YrRr

drrrRnl22)(

CHEM 3720

85

Unit VI

Approximation Methods in Quantum Mechanics

A. Introduction

1. Atomic units

a. Are used in atomic and molecular calculations.

b. An advantage in their use is that the values expressed in atomic units

are not affected by refinements in various constants (me, e, h, etc).

c. Unit of angular momentum: ħ

d. Unit of length: 2

e

20

04

ema

named bohr and denoted as a0.

□ A529177.01 0 a

e. Unit of energy: 22

02

4e

16

emE named hartree and denoted as Eh.

□ eV2116.27kJ/mol5.26251 h E

2. Quantum mechanical treatment of the He atom

a. The hydrogen atom serves as prototype for the treatment of more

complex atoms.

b. By solving the Schrödinger equation for the hydrogen atom:

),,(),,(42 0

22

e

2

rErr

e

m

one obtains the allowed energies and the wavefunctions of the electron.

c. He atom is constituted from the nucleus and two electrons.

d. The wavefunction of the system will depend on:

□ the position of the helium nucleus R

□ the positions of the two electrons r1 and r2

CHEM 3720

86

e. Schrödinger equation for the He atom:

),,(

),,(44

2

4

2

),,(222

21

21210

2

20

2

10

2

2122

e

221

e

22

2

rrR

rrRrrrRrR

rrR

E

eee

mmM

□ He atom is a three-body system and solving its Schrödinger

equation is more complicated. (It is impossible to be solved

exactly.)

f. Solving Schrödinger equation for the He atom

□ Consider the nucleus to be fixed at the origin:

),(

),(4

),(11

4

2

),(2

21

21210

2

21210

2

2122

21

e

2

rr

rrrr

rr

rr

E

e

rr

e

m

□ This equation cannot be solved exactly.

□ This is due to the existence of the interelectronic repulsion term.

□ Without it, one can apply the separation of variables technique.

g. Schrödinger equation for the He atom represents a system that cannot

be solved exactly. Solving these types of equations requires

approximate methods:

□ Variational method

□ Perturbation theory

CHEM 3720

87

B. Variational Method

1. Variational principle (or Variation principle)

a. Consider an arbitrary system for which the ground-state wavefunction

and ground-state energy satisfy 000ˆ EH .

d

dHE

0*0

0*0

0

ˆ

where d represents the volume element

b. If one substitute 0 with any other function and calculate:

d

dHE

*

* ˆ

then, according to variational principle,

0EE

where 0EE only if 0 .

c. Variational Principle: If an arbitrary wavefunction is used to calculate

the energy, then the calculated value is never less than the true ground-

state energy E0.

□ Variational principle says that one can calculate an upper bond to

E0 by using any other function.

2. Trial functions

a. One can choose the function called trial function that depends of

some arbitrary parameters , , ,… called variational parameters.

b. The energy calculated based on this trial function will also depend on

these parameters:

0,, EE

c. Optimize the variational parameters (, , ,…) to get the lowest

ground-state energy therefore obtaining the best trial wavefunction.

CHEM 3720

88

d. Example of the ground state of the hydrogen atom

□ The Hamiltonian is r

e

dr

dr

dr

d

rmH

0

22

2e

2

42

ˆ

□ Choose the trial function: 2

)( rer where is a variational

parameter.

□ Determine E():

2/3

02/1

22/1

e 22

3)(

e

mE

where

420

3

42e

18

em

□ Optimize the expression of E() with respect to :

220

2

4e

min16

424.0

emE

○ compare to

220

2

4e

016

500.0

emE

□ The trial function with the optimal value of :

20

2 /)9/8(

2/1

30

2/3

1

3

8 are

ar

○ compare to s1

3. Variational method for the He atom

a. Rewrite the Hamiltonian as

120

2 1

4)2(ˆ)1(ˆˆ

r

eHHH HH

□ j

jHr

e

mjH

1

4

2

2)(ˆ

0

22

e

2

b. Neglect 120

2 1

4 r

e

term so separation of variables is possible.

CHEM 3720

89

c. One has: ),,(),,()(ˆjjjHjjjjHH rErjH get and Ej.

d. Use a trial function )()(),( 2111210 rrrr ss

e. Calculate E(z):

h22

220

2

4e

008

27

8

27

16

ˆ)( Ezzzzem

dzHzE

□ hE is the atomic unit of energy called hartree.

f. Minimizing E(z):

16

27min z hmin 8477.2 EE

□ The value of minz can be interpreted as an effective nuclear charge

each electron partially screens the nucleus from the other

( 2z )

□ Compare with accurate calculated value: h9037.2 E

□ Experimental value: h9033.2 E

4. Variational method for a trial function obtained as a linear combination of

functions

a. Consider a more complex trial function:

...22111

fcfcfcN

nnn

□ c1, c2, … coefficients are variational parameters

b. Example for 2N

2211 fcfc

c. The energy:

d

dHccE

*

*

21

ˆ),(

22222121122111

21

ˆ HcHccHccHcdH

2222122111

21

2 ScSccScd

□ Hij and Sij are called matrix elements.

CHEM 3720

90

□ dfHfH jiijˆ is called Coulomb integral.

□ dffS jiij is called resonance integral or overlap integral.

□ jiij HH if H is Hermitian.

2222122111

21

2222122111

211

212

2),(

ScSccSc

HcHccHcccE

d. Minimize the energy with respect to the variational parameters:

01

c

E 01212211111 ESHcESHc

02

c

E 02222212121 ESHcESHc

□ Coefficients 1c and 2c are nonzero if and only if:

022221212

12121111

ESHESH

ESHESH

○ This is called a secular determinant.

□ Solving the secular determinant:

02121222221111 ESHESHESH

0212

2212

212

2122211

2112222112211 SESEHHSSESHSHEHH

0212221111222211

212

212

2122211

2 HHHSHSHSHESSSE

○ This is a second-order equation called secular equation.

○ The smaller-value solution is the variational approximation for

the ground-state energy.

e. For the case of larger N than the determinant in N order:

0

...

............

......

...

2211

22221212

1112121111

NNNNNNNN

NN

ESHESHESH

ESHESH

ESHESHESH

f. Once E is determined, one can go back in determining coefficients ic .

g. A trial function that depends linearly on the variational parameters

leads to a secular determinant.

CHEM 3720

91

5. Slater orbitals

a. A more complex function can be used for trial functions:

N

jjj fc

1

2rj

jef

where function jf is function of few coefficients as well.

□ Solving for the wavefunction becomes more demanding but

algorithms are available.

b. Slater filled the need for general and suitable trial functions that are not

necessarily same as the hydrogen wavefunctions by introducing a set of

orbitals, called Slater orbitals.

c. Slater orbital are defined as:

,,, 1 ml

rnnnlm YerNrS

where )!2(

)2( 2/1

nN

n

n

is a normalization constant

,ml

Y are the spherical harmonics

d. Properties of Slater orbitals:

□ (zeta) is arbitrary and is not necessarily equal to Z/n as in

hydrogen-like orbitals.

□ The radial part of the Slater orbitals does not have nodes.

□ ),,( rSnlm is not orthogonal to ),,( rS lmn .

□ n can be also considered as a variational parameter and is

optimized to get the lowest energy.

CHEM 3720

92

C. Perturbation Theory

1. Description

a. Suppose one is unable to solve the Schrödinger equation EH ˆ

for the system of interest but one can solve it for a similar system: )0()0()0()0(ˆ EH

where )1()0( ˆˆˆ HHH

□ )0(H is called unperturbed Hamiltonian operator.

□ )1(H is called perturbation.

b. If the perturbation is small, the solutions of H will be similar to those

of )0(H .

c. Anharmonic oscillator example

...246

1

2

1

2ˆ 432

2

22

xb

xkxdx

dH

□ 22

22)0(

2

1

2ˆ kx

dx

dH

is the harmonic oscillator operator.

□ The solutions are known: xn)0( ; hnEn 2/1)0(

□ ...246

1ˆ 43)1( xb

xH is the perturbation.

d. Perturbation theory says that the wavefunction and the energy of the

unperturbed system can be successively corrected:

...)2()1()0(

...)2()1()0( EEEE

□ )0( is the wavefunction of the unperturbed system.

□ )0(E is the energy of the unperturbed system.

□ )1( , )2( ,… are successive corrections to )0(

□ )1(E , )2(E ,… are successive corrections to )0(E

□ A basic assumption is that those successive corrections become

smaller. Expressions are available for these terms.

CHEM 3720

93

e. We will only work with )1(E which is the first-order correction to )0(E .

dHE )0()1()0()1( ˆ

f. )1()0( EEE is the energy through first-order perturbation theory.

g. )1()0( is the wavefunction through first-order perturbation

theory.

h. )2()1()0( EEEE is the energy through second-order perturbation

theory.

2. Application to Helium atom

a. Schrödinger equation for He atom

),(),(4

),(11

4

2),(

2

2121210

2

21210

2

2122

21

e

2

rrrrrr

rrrr

Ee

rr

e

m

b. Unperturbed system

2ˆ1ˆˆHH

)0( HHH

where i

ie r

e

miH

0

22

2

H4

2

22211111)0( ,,,, rr ss

22

220

2

42

21

220

2

42)0(

3232 n

emz

n

emzE ee

c. The perturbation is 120

2)1(

r

eH

d. The first-order correction to the energy

h220

2

4)1(

8

5

168

5zE

emzE e

CHEM 3720

94

e. The energy (in hartree) through first-order perturbation theory

zzzzzEEE8

5

8

5

2

1

2

1 222)1()0(

□ when 2z h750.2 EE

□ variational result: h8477.2 EE

□ experimental value: h9033.2 EE

□ The results do not look as good as variational method but second

order perturbation theory gives: h9077.2 EE

□ Considering that kJ/mol5.26251 h E , h05.0 E is a substantial

value so higher level of corrections are needed for very accurate

results.

f. Overall both variational method and perturbation theory can give

reasonable results.

CHEM 3720

95

D. Unit Review

1. Important Terminology

atomic units

interelectronic repulsion

variational principle

trial function

variational parameters

effective nuclear charge

matrix elements

Coulomb integral

resonance integral

secular determinant

secular equation

Slater orbitals

Perturbation theory

CHEM 3720

96

unperturbed Hamiltonian operator

perturbation

first-order correction

energy through first-order perturbation theory

wavefunction through first-order perturbation theory

2. Important Formulas

d

dHE

*

* ˆ

0EE ; 0,, EE

...22111

fcfcfcN

nnn

dfHfH jiijˆ ; dffS jiij

022221212

12121111

ESHESH

ESHESH

0

...

............

......

...

2211

22221212

1112121111

NNNNNNNN

NN

ESHESHESH

ESHESH

ESHESHESH

,,, 1 ml

rnnlnlm YerNrS

)1()0( ˆˆˆ HHH

...)2()1()0( ; ...)2()1()0( EEEE

dHE )0()1()0()1( ˆ

CHEM 3720

97

Unit VII

Multielectron Atoms

A. Electron Spin

1. Introduction

a. The idea was introduced initially when it was observed that some lines

in Na spectrum that are predicted to be a singlet actually appear as a

doublet.

b. The electron behaves like a spinning top with a z component of ±ħ/2.

c. This classical picture is not accurate; the spin is strictly a nonclassical

(quantum mechanical) concept.

d. Spin is an intrinsic (built-in) angular momentum possessed by

elementary particles.

2. Spin quantum numbers

a. New spin-angular-momentum quantum numbers are introduced, s and

ms, that are analogues to the orbital-angular-momentum quantum

numbers l and m (or more specific ml).

b. The spin quantum number ms determines the z component of the

electron spin angular momentum just as ml determines the z component

of the electron orbital angular momentum in hydrogen atom.

□ The idea of spin is though more general than just the case of

electrons in atoms.

□ Similar to ml = –l,…,+l, ms can take 2s + 1 values: ms = –s,…,+s.

□ The analogy is not complete because a given species of elementary

particle can have only one value for s.

c. The value of s may be half-integral (21 ,

23 ,…) as well as integral

(0,1,…).

CHEM 3720

98

□ Examples:

○ electrons, protons, neutrons have 21s

21

21 ,sm

○ 35Cl nuclei have 23s

23

21

21

23 ,,, sm

○ 12C nuclei have s = 0 0sm

○ photons have s = 1 1,0,1 sm

d. The particles with half integer spin are called fermions, and the

particles with integer spin are called bosons.

e. Because 21s only and cannot have big values, the spin cannot be

classical (cannot have classical behavior).

3. Spin eigenfunctions

a. There are two operators associated with the spin (similar to the angular

momentum): 2S and zS .

b. For the electron (21s ;

21sm ), there are two eigenfunctions

associated with the spin (similar to the spherical harmonics) and

with the properties:

222

222

12

1

2

11ˆ

12

1

2

11ˆ

ssS

ssS

○ similar to ),()1(),(ˆ 22 ml

ml

YllYL

2

2

sz

sz

mS

mS

○ similar to ),(),(ˆ ml

mlz YmYL

□ Spin eigenfunction looks like 2/12/1

Y , and looks like 2/1

2/1Y .

CHEM 3720

99

c. Similar to 122 llL , we have that 122 ssS where 2S is

the square of the spin angular momentum.

d. The spin eigenfunctions and are orthonormal:

1** dd

0** dd

where is the spin variable (and it has no classical analog).

4. Sixth postulate of quantum mechanics

a. Postulate 6: All electronic wavefunctions must be antisymmetric under

the exchange of any two electrons.

b. A more familiar statement is the Pauli Exclusion Principle:

□ Two electrons in an atom cannot have the same values for the four

quantum numbers n, l, ml, ms.

c. The Pauli Exclusion Principle is a special case of a general statement

called the Pauli Principle:

□ When the labels of any two identical fermions are exchanged, the

total wavefunction changes sign ( wavefunction is

antisymmetric).

□ When the labels of any two identical bosons are exchanged, the

total wavefunction retains the same sign ( wavefunction is

symmetric).

5. Total wavefunction for an electron

a. “Total wavefunction” means the entire wavefunction, including the

spin of the particle.

b. Total wavefunction is given by a product of the spatial part and the

spin part because those parts are independent of each other.

c. For the case of an electron:

)(),,(),,,( zyxzyx or )(),,( zyx

CHEM 3720

100

d. The one-electron wavefunctions are called spin orbitals.

□ Spin orbitals are normalized.

e. The case of He atom:

21

21 100100

)2(1)1(1)2,1(

ss

□ But the electrons are indistinguishable (our label cannot make a

distinction between the electrons) so the wavefunction

)1(1)2(1)1,2( ss is equivalent.

□ Linear combinations of the two wavefunctions are also possible:

)1(1)2(1)2(1)1(1)1,2()2,1(1 ssss

)1(1)2(1)2(1)1(1)1,2()2,1(2 ssss

□ Out of the two possible linear combinations, only 2 is

antisymmetric:

)2,1()2,1()1,2()1,2( 22

f. Slater determinants

□ Antisymmetric wavefunctions are represented by Slater

determinants:

)2(1)2(1

)1(1)1(1

2

1)2,1(

ss

ss

○ This is called determinantal wavefunction.

□ More general (the case of N electrons):

)(.....)()(

....................

)2(....)2()2(

)1(....)1()1(

!

1),....2,1(

21

21

21

NuNuNu

uuu

uuu

NN

N

N

N

where u are orthonormal spin orbitals.

□ Properties of the wavefunction written is determinantal form:

○ = 0 if two columns are the same.

○ changes sign if columns are interchanged.

CHEM 3720

101

B. Hartree-Fock Method

1. The case of helium atom

a. Determining ground-state energy using variational method and a trial

function of the form )()(),( 2121 rrrr the minimum-energy limit

that one can obtain will be h8617.2 EE .

□ The result is too high compared to the experimental value of

h9033.2 EE .

b. This is the best value of the energy that can be obtained using a trial

function of the form of a product of one-electron wavefunction (i.e.,

the Orbital Approximation).

□ This is called the Hartree-Fock limit.

c. The idea of one-electron orbitals (wavefunctions) is preserved in the

Hartree-Fock approximation and is abandoned in more accurate

methods.

d. To improve the variational results one should include explicitly the

interelectronic distance:

□ Example: 121221 1),,( 21 creerrrZrZr

.

e. Variational method with a trial function with 1078 parameters gives

very good results.

2. Description of the Hartree-Fock method

a. Hartree-Fock equations are solved using self-consistent field (SCF)

method known also as HF.

b. Example for He:

□ Write the two-electron wavefunction as a product of one-electron

wavefunctions (orbitals) and assume the same wavefunction for

both electrons:

)()(),( 2121 rrrr

CHEM 3720

102

□ Electron 1 at r1 experiences a potential energy from electron 2

given by:

2212

2*

1eff

1

1rrr d

rrV

□ Define an effective one-electron Hamiltonian operator (in a.u.):

1eff

11

211

eff1

2

2

1ˆ rVr

H r

□ Write Schrödinger equation for electron 1 (same for the second

electron):

1111eff1

ˆ rrr H

○ This is a Hartree-Fock equation.

c. The method to solve this equation is called self-consistent field method

and consists in the following steps:

□ Guess the form , which is the same for (r2)

□ Determine )( 1eff

1rV

□ Define an effective one-electron Hamiltonian operator.

□ Write eff1

H and solve for a new , which is the same for (r2)

□ Compare the old with the new .

○ The initial and final are usually different.

□ Continue the process until is self-consistent (the new with the

old ) are the same

d. Rewrite the Hartree-Fock equation taking in consideration that the

wavefunction should be a spin orbital:

iiiiF ˆ

□ iF is the effective Hamiltonian operator or Fock operator.

□ The expression for the Fock operator is more complicated then

given above because the wavefunction is written in a form of a full

Slater determinant.

CHEM 3720

103

e. The final wavefunctions i are self-consistent orbitals and are called

Hartree-Fock orbitals.

f. In practice, the trial function is (or can be) a sum of Slater orbitals

and what are optimized are the linear coefficients (and maybe the

exponent in the e–r term).

g. The energies i are called orbital energies.

□ Koopman’s theorem given an interpretation for i:

○ It is the ionization energy of the electron from the ith orbital.

□ Example: Argon configuration: 62622 33221 pspss

Ionization Process Koopman’s

theorem

HF

calculation experimental

62622 33221 pspss 6262 33221 pspss + e 311.35 308.25 309.32

62622 33221 pspss 6262 33221 pspss + e 32.35 31.33

62622 33221 pspss 62522 33221 pspss + e 25.12 24.01 23.97

62622 33221 pspss 6622 33221 pspss + e 3.36 3.20 2.82

62622 33221 pspss 52622 33221 pspss + e 1.65 1.43 1.52

Experimental values are determined using photoelectron spectroscopy.

The values are in MJ/mol

3. Correlation energy

a. When the wavefunction for a two-electron system is written as a

product of two one-electron wavefunctions )()(),( 2121 rrrr , the 2

electrons are considered to be independent of each other (or just to

have an average potential created by the other one) and the electrons

are said to be uncorrelated.

b. Define the correlation energy:

HFexactCE EE

□ Example: kJ/mol110CE for the He atom.

□ The Hartree-Fock energy ( HFE ) is about 99% exact but is missing

the correlation energy.

CHEM 3720

104

C. Electron Configurations and Atomic Term Symbols

1. Electron configurations

a. The electron configuration gives the subshells (or the orbitals) that are

occupied and how many electrons are in which orbital.

□ Example: C atom 1s22s22p2

b. Relative energies of the atomic orbitals determine their occupation, and

the order in which they are occupied is from bottom to top (i.e., from

the most stable to the least stable).

□ The order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, …

c. The electron configuration doesn’t say anything about the four

quantum numbers of each electron.

d. For each configuration there are a number of such possibilities.

e. The energies are different for each of these possibilities so there is a

need for a more detailed description of the electronic states of atoms

than the one provided by electronic configurations.

2. Russell-Sounders coupling

a. Russell-Sounders coupling (called RS coupling) is a method of

providing more detailed information on the electronic state in an atom

in a form of an atomic term symbol.

□ An atomic term symbol is also called a spectroscopic term symbol.

b. The general idea is:

□ Determine the total orbital angular momentum L

□ Determine the total spin angular momentum S

□ Vectorially add L and S to get total angular momentum J

c. Define atomic term symbols as:

JS L12

□ J is the total angular momentum quantum number

CHEM 3720

105

d. For the total orbital angular momentum quantum number L, use a

similar notation to the one used for orbitals:

...

,...3,2,1,0

FDPS

L

e. The value of 2S + 1 (where S is the total spin angular momentum

quantum number) gives the spin multiplicity:

0S 112 S singlet

21S 212 S doublet

1S 312 S triplet

f. The total orbital angular momentum L and its z component:

i

ilL

Li i

liziz MmlL

g. The total spin angular momentum S and its z component:

i

isS

Si i

siziz MmsS

h. The total angular momentum J and its z component:

SLJ

JSLzzz MMMSLJ

i. The allowed J values are:

||,...,1, SLSLSLJ where

LSLS

SLSLSL

if

if||

j. The number of values for the z component:

□ The z component of L can have 2L + 1 values: LLM L ,...,

□ The z component of S can have 2S + 1 values: SSM S ,...,

□ The z component of J can have 2J + 1 values: JJM J ,...,

○ The maximum value of ML will be determined by L.

CHEM 3720

106

k. Define a microstate as a set of ml and ms values for each electron in the

atom.

l. The degeneracy (i.e., the number of microstates) of a term:

□ For a term written as LS 12 : )12)(12( LS

□ For a term written as JS L12 : 12 J

m. Recall that the number of distinct ways to assign N electrons into G

spin orbitals belonging to the same subshell (equivalent orbitals) is

given by:

)!(!

!

NGN

G

n. The term symbols are read as spin multiplicity (as a word describing a

number) followed by L (as a letter) followed by J (as a number).

□ 2/52D is read as “doublet D five halves” (

25

21 ,2, JLS )

□ 11P is read as “singlet P one” ( 1,1,0 JLS )

□ 03P is read as “triplet P zero” ( 0,1,1 JLS )

□ 2/34S is read as “quartet S three halves” (

23

23 ,0, JLS )

o. The term symbol of the ground state electronic state is given by Hund’s

rules:

□ The state with the largest value of S is the most stable (has the

lowest energy), and stability decreases with decreasing S.

□ For states with the same value of S, the state with the largest value

of L is the most stable.

□ If few states have the same value of L and S.

○ for a subshell that is less than half filled, the state with the

smallest value of J is the most stable.

○ for a subshell that is more than half filled, the state with the

largest value of J is the most stable.

CHEM 3720

107

p. Examples of possible terms and the term symbol of the ground state for

a certain configuration:

□ The ns2 configuration:

0

0

0

2

1

L

l

l

M

m

m

0

2/1

2/1

2

1

S

s

s

M

m

m

L = 0, S = 0 J = 0

○ The only possible term symbol and therefore the terms symbol

of the ground state is 01S .

□ Same result is obtained for the np6 configurations and in every

situation when a subshell is completely filled.

□ The 1s12s1 configuration:

○ The total number of microstates (possibilities of putting two

electrons in two spin orbitals): 2 2 = 4

○ A microstate is represented by a number representing ml and a

superscript + or – representing the sign of ms.

○ For the case of only 2 electrons, one possibility to determine

the possible terms is to make a table to show all possible

microstates:

LM

1 SM

0

–1

0 0+0+ 0+0– ; 0–0+ 0–0–

○ The two 0+0– microstates are not equivalent because the two

orbitals are not equivalent.

L = 0, S = 1 13S term 3 microstates

L = 0, S = 0 01S term 1 microstate

Total 4 microstates

○ The ground state is 13S because of its highest spin multiplicity.

CHEM 3720

108

□ The np2 configuration:

○ The number of microstates: 15!2!4

!6

○ Maximum L = 2l = largest LM

○ Maximum S = 1 s = largest SM

LM

1 SM

0

–1

2 1+1+ 1+1– 1–1–

1 0+1+ 1+0–; 1–0+ 0–1–

0 0+0+; 1+–1+ 1+–1–; –1+1–; 0+0– 1––1–; 0–0–

–1 0+–1+ 0+–1–; 0––1+ 0––1–

–2 –1+–1+ –1+–1– –1––1–

○ Microstates forbidden by the Pauli exclusion principle: 1+1+,

1–1–, –1+–1+, –1––1–, 0+0+, and 0–0–.

Largest 2LM ( 0SM ) 21D term 5 microstates

Next largest 1LM ( 1SM ) 23P term 5 microstates

13P term 3 microstates

03P term 1 microstate

Next largest 0LM ( 0SM ) 01S term 1 microstate

Total 15 microstates

○ All these five terms will have different energies.

○ The ground state of a np2 configuration is: 03P

□ The p2 and p4 configurations give the same terms symbols but the

ground state will be different.

□ Same terms are obtained also for the nd x and nd10–x configurations.

CHEM 3720

109

D. Atomic Spectra

1. The use of atomic term symbols to describe atomic spectra

a. The energy of the hydrogen atom depends only on principal quantum

number n.

b. It was experimentally observed that various levels show small splitting.

□ The reason for this fine splitting is the spin-orbit coupling.

c. Hamiltonian should include a small term (i.e., a perturbation) that

define the spin-orbit interaction which represents the interaction of the

magnetic moment associated with the spin of the e– with the magnetic

field generated by the electric current produced by the electron’s own

orbital motion.

d. The increased spectral complexity caused by the spin-orbit coupling is

called fine structure.

e. Any state of the atom, and any spectral transition, can be specified

using term symbols.

□ By convention, in writing the expression for the transition, the

upper term precedes the lower term.

f. Transitions in atomic spectra are restricted by some selection rules:

□ 0S (no change of the overall spin the light does not affect

the spin)

□ 1,0 L

□ 1l

□ 1,0 J (except 0J 0J transition)

g. The selection rules show that a transition is accompanied by a change

in the angular momentum of an individual electron.

CHEM 3720

110

2. The spectrum of the hydrogen atom

a. The n = 2 n = 1 in Bohr model appear at 82258.19 cm–1.

Configuration Term Symbol Energy / cm–1

1s 2/12S1s 0.00

2p 2/12P2 p 82258.917

2s 2/12S2s 82258.942

2p 2/32P2 p 82259.272

□ Allowed transitions:

○ 2/12

2/32 S1P2 sp

○ 2/12

2/12 S1P2 sp

□ The two allowed transitions make the first line in Lyman series to

appear as a doublet.

b. The 3d 2p (n = 3 n = 2) transition shows a fine structure:

□ The line also appear as a doublet.

c. The term symbols and their energies for various atoms are tabulated.

CHEM 3720

111

D. Unit Review

1. Important Terminology

spin

spin quantum numbers

spin operator

spin eigenfunctions

spin variable

fermion

boson

sixth postulate of QM

Pauli exclusion principle

Pauli principle

spin orbital

Slater determinant

determinantal wavefunction

CHEM 3720

112

Hartree-Fock method

Hartree-Fock approximation

self-consistent field method

effective one-electron Hamiltonian operator

Hartree-Fock equation

Fock operator

Hartree-Fock orbitals

Koopman’s theorem

correlation energy

electron configuration

Russell-Sounders coupling

total angular momentum quantum number

total orbital angular momentum quantum number

total spin angular momentum quantum number

CHEM 3720

113

atomic term symbol

spin multiplicity

Hund’s rules

microstate

fine splitting

spin-orbit coupling

fine structure

CHEM 3720

114

2. Important Formulas

222

222

12

1

2

11ˆ

12

1

2

11ˆ

sS

ssS

2

2

sz

sz

mS

mS

)2(1)2(1

)1(1)1(1

2

1)2,1(

ss

ss

)(.....)()(

....................

)2(....)2()2(

)1(....)1()1(

!

1),....2,1(

21

21

21

NuNuNu

uuu

uuu

NN

N

N

N

2212

2*

1eff

1

1rrr d

rrV

1eff

11

211

eff1

2

2

1ˆ rVr

H r

iiiiF ˆ

HFexactCE EE

JS L12

i

ilL ; Li i

liziz MmlL

i

isS ; Si i

siziz MmsS

SLJ ; JSLzzz MMMSLJ

)12)(12( LS ; 12 J

0S ; 1,0 L ; 1l ; 1,0 J

CHEM 3720

115

Unit VIII

Diatomic Molecules

A. Schrödinger Equation for Diatomic Molecules

1. Introduction

a. Solving Schrödinger equation for (diatomic) molecules represents the

quantum mechanical treatment of chemical bonding.

b. Hamiltonian for H2 molecule:

R

e

r

e

r

e

r

e

r

e

r

e

mMH

0

2

120

2

B20

2

A20

2

B10

2

A10

2

22

21

e

22B

2A

2

44

4444

22ˆ

2. The Born-Oppenheimer approximation

a. Due to larger mass of the nuclei compared with the electrons, this

approximation considers the nuclei fixed in positions relative to the

motion of the electrons (i.e. neglecting the nuclear motion).

b. This results in dropping two kinetic energy terms in the expression for

the Hamiltonian.

c. The Hamiltonian becomes:

R

e

r

e

r

e

r

e

r

e

r

e

mH

0

2

120

2

B20

2

A20

2

B10

2

A10

222

21

e

2

4444

442ˆ

□ In atomic units:

Rrrrrr

H111111

2

12B2A2B1A1

22

21

d. By making the Born-Oppenheimer approximation, the nuclear motion

can be treated separately on a potential energy curve (or surface) that

includes the electronic energy and the internuclear repulsion energy.

CHEM 3720

116

B. Molecular Orbital Theory

1. Introduction

a. It is a method to describe the bonding properties of molecules in a form

of orbitals (or electron wavefunctions) distributed over the whole

molecule called molecular orbitals.

2. The case of H2+ ion

a. This is a one-electron system.

b. Avoids the interelectronic repulsion term that makes the equation not

to be solved exactly (using elliptic coordinates).

c. The Schrödinger equation can be solved exactly for H2+, within the

Born-Oppenheimer approximation.

d. H2+ looks more like hydrogen atom (one-electron system) while H2

looks more like helium atom (two-electron system).

e. Hamiltonian: Rrr

H111

2

BA

2

f. Schrödinger equation: );,();,(ˆBABA RrrERrrH

□ (rA, rB; R) show that equation is solved at fixed R distance.

g. Although an exact solution can be found, we will choose to present the

problem using the variational method (approximate but easier to

understand). This method is called molecular orbital theory.

h. Choose trial functions as B2A1 11 scsc

□ This is a molecular orbital formed as a linear

combination of atomic orbitals (LCAO).

i. Solve first for )11(11 BAB2A1 sscscsc :

2)(

CHEM 3720

117

RERH ,,ˆ rr

*

* ˆ

r

r

d

HdE

22

B*BA

*BB

*AA

*A

2

BA*B

*A

*

)22()11(

11111111

1111

cSSSc

ssdssdssdssdc

sscssdrcd

rrrr

r

○ B*A11 ssdS r is called an overlap integral.

○ S contains a product of orbitals located on different atoms.

○ Normalization condition: 1)22( 2 cS S

c22

1

31

2RReRS R for 1s orbitals of hydrogen.

○ S is bigger when R is smaller.

KJSEc

ssRrr

ssdc

Hd

s 2212

11111

2

111

ˆ

12

BABA

2*B

*A

2

*

r

r

A

BA s

RrsdJ 1

111 *

r is called Coulomb integral.

A

BB s

RrsdK 1

111 *

r is called exchange integral.

ReJ R 1

12

○ ReR

SK R 1

S

KJEE s

11

CHEM 3720

118

○ The exchange integral (also called resonance integral) is

responsible for the existence of the chemical bond in H2+.

□ The energy of H2+ with respect to the dissociated species (H and

H+):

S

K

S

J

S

KJEEE s

1111

j. Solve for B2A1 11 scsc :

□ The energy with respect to the dissociated species (H and H+):

S

KJEEE s

11

k. The wavefunctions (or orbitals):

BAb 11

12

1ss

S

○ This is a bonding orbital.

BAa 11

12

1ss

S

○ This is an antibonding orbital.

□ We obtain only 2 molecular orbitals because we start with only 2

atomic orbitals.

○ Starting with more atomic orbitals, one will get more

molecular orbitals.

2)(

CHEM 3720

119

3. The case of the H2 molecule

a. Orbitals for H2 are similar with the one of helium atom.

)1()2()2()1(2

1)2()1(

)2()2(

)1()1(

!2

1bb

bb

bb

b. This method on construction molecular orbitals is called LCAO-MO

(linear combination of atomic orbitals-molecular orbitals).

c. To calculate (approximately) the energy, abandon the spin part from

the wave function (and the determinantal form of the wavefunction).

□ Total wavefunction, MO, is a product of molecular orbitals.

21211111

12

1BABAMO ssss

S

)2,1(ˆ)2,1( MO*MO21MO HddE rr

sEEE 1MO 2

1binding molkJ260

EE at pm8561.1 0e aRR

d. Molecular energy level diagram:

4. Labeling molecular orbitals

a. The label includes the symmetry about the internuclear axis:

□ - cylindrically symmetric with respect to the internuclear axis

○ There are no nodal planes containing the internuclear axis.

□ - have 1 nodal plane that contains internuclear axis

□ - have 2 nodal planes that contain internuclear axis

b. The label includes the atomic orbital from which the molecular orbital

is formed:

□ Example: 1s or 2s or 2px or 2py or 2pz …

CHEM 3720

120

c. The label includes the type of molecular orbital:

□ bonding

□ antibonding (*)

d. The label includes (for homonuclear

diatomics) the sign under the inversion

of the orbital through the midpoint:

□ u = ungerade (odd)

□ g = gerade (even)

5. Representing molecular orbitals

a. The interaction between 1s orbitals:

g1s (or 1s)

u1s (or *1s)

b. The interaction between 2s orbitals (larger and with radial nodes):

g2s

u2s

CHEM 3720

121

□ The energies of these MO: g1s < u1s < g2s < u2s

c. The interaction between 2p orbitals (higher in energy than 2s for atoms

other than H):

□ Consider the internuclear axis to be the z axis.

□ Interaction along the internuclear

axis:

Bonding orbital: g2pz

Antibonding orbital: u2pz

○ The combination of 2pz atomic orbitals is cylindrically

symmetric about the internuclear axis therefore orbital.

□ Interaction off the internuclear axis:

Bonding orbital: u2py

Bonding orbital: u2px

○ The combination of 2px (or 2py) AO is NOT cylindrically

symmetric about the internuclear axis orbital.

CHEM 3720

122

Antibonding orbital: g2py

Antibonding orbital: g2px

○ These orbitals have a nodal plane halfway between the nuclei.

○ These orbitals are degenerate orbitals and so are the bonding

counterparts.

6. Molecular orbital theory for homonuclear diatomics

a. A molecular orbital energy diagram shows the relative energy and the

occupancy of molecular orbitals.

Li2 Be2 B2 C2 N2 O2 F2

u2pz

g2px , g2py

g2pz

u2px , u2py

u2s

g2s

Li2 Be2 B2 C2 N2 O2 F2

u2pz

g2px , g2py

g2pz

u2px , u2py

u2s

g2s

CHEM 3720

123

b. The molecular orbital energy diagram is filled

out according to Pauli exclusion principle.

c. Determine the electronic configuration by

naming each orbital that contains electrons

(similar to electronic configurations for atoms).

□ B2: 112222222211 yuxuugug ppssss

□ N2: KK 2222222222 zgyuxuug pppss

d. Molecular Orbital (MO) theory the magnetic

properties of the molecules:

□ Molecules that have only paired electrons are called diamagnetic.

○ Examples: F2, N2, C2, O22+, etc.

□ Molecules that have unpaired electrons are called paramagnetic.

□ MO theory predicts that O2 is paramagnetic:

O2: KK 2222222222 zgyuxuug pppss 11

22 ygxg pp

e. Frontier orbitals

□ The highest occupied molecular orbital or HOMO

□ The lowest unoccupied molecular orbital or LUMO

○ For N2, HOMO is g2pz.

○ For N2, LUMO is g2px or g2py.

f. Define the bond order as:

)orbitals gantibondinin #orbitals bondingin (#2

1BO ee

□ Examples:

Li2 Be2 B2 C2 N2 O2 F2

BO 1 0 1 2 3 2 1

and

H2 He2 H2+ He2

+

BO 1 0 1/2 1/2

CHEM 3720

124

□ When BO = 0, the molecule is

unstable and does not exist.

□ A high bond order is equivalent

to a small internuclear distance

(i.e., bond distance) and a high

bond energy.

□ The variations between B2 to F2

of bond order, bond energy, and

bond distance:

□ Ne2 is a molecule with a very

long distance (3A) and a very

weak “bond energy” (< 1kJ/mol).

7. Photoelectron spectroscopy

a. Photoelectron spectroscopy supports the existence of molecular

orbitals.

b. The ionization energy is the energy required to eject an electron from a

molecule (and is a direct measurement of how strongly electrons are

bound within a molecule).

N2N2

CO CO

8. Molecular orbital theory for heteronuclear diatomics

a. Heteronuclear diatomics are molecules in which the two nuclei are

different.

□ Examples: NO, CO, CN, CN–, etc

CHEM 3720

125

b. Procedure for creating and using a MO energy level diagram

□ Start the diagram by sketching the atomic orbitals (AO) for the two

atoms constituting the molecule.

□ The energies of the AO on the two atoms will be different.

□ The more electronegative atom will have the atomic orbitals lower

in energy.

□ Write the molecular orbitals as a combination of atomic orbitals

considering that, to “interact and create” a molecular orbital, the

atomic orbitals should have:

○ similar energies

○ proper overlap or interaction or orientation

2s + 2pz: 2s + 2py:

○ Example: a 2s orbital will not interact with a 2px orbital

○ Example: a 2py orbital will not interact with a 2px or 2pz orbital

□ Be sure that the total number of molecular orbitals equals that or

atomic orbitals.

□ Label (and draw) all orbitals in the diagram.

○ For heteronuclear diatomics, the u and g description of the

molecular orbitals disappear because atoms are different.

□ The atomic orbital contributions to the MO are not equal: the more

stable atomic orbital will contribute more to the more stable MO.

□ Fill in the MOs with the appropriate number of electrons.

○ This number is the sum of electron from each atom in the

atomic orbitals that are considered in the diagram.

□ Write the electronic configuration.

□ Determine the bond order.

CHEM 3720

126

c. Example: CN– ion

x* y

*

x y

*

*

2p (N)

2s (N)

2p (C)

2s (C)

□ The electronic configuration is: 2222*22*2 )2()2()2()2()2()1()1( zyx pppssss

□ Bond order: BO = 6/2 = 3

d. Example: HF molecule

2p (F)

2s (F)

1s (H)

CHEM 3720

127

□ The molecular orbital for HF: zH pcsc 21 21

□ The 2s, 2px, and 2py orbitals on F (2sF, 2pxF, 2pyF) are nonbonding

orbitals.

□ The 1s orbital on H does not have the right orientation to interact

with 2px or 2py (net overlap between these orbitals is zero).

□ Configuration: 22222 2221 yFxFbFF ppss

□ Bond order: BO = 1

○ 2sF, 2pxF, 2pyF orbitals are nonbonding so they are not

influencing the bond order.

e. More advanced treatment

□ A more complicated way to create a molecular orbital is as a linear

combinations of atomic orbitals whose coefficients are determined

self-consistently.

....222211 BA3BA2BA1 zz ppcsscssc

□ Solve self consistently for coefficients to get SCF-LCAO-MO

wavefunctions by Hartree-Fock-Roothaan method.

□ Because the molecular orbitals are combinations of more than just

one type of atomic orbitals, drop the atomic orbital name (1s, 2s,

etc) from the molecular orbital notation.

○ g (notg2s); u (not u2px)

○ add an index of the molecular orbital

CHEM 3720

128

C. Molecular Term Symbols

1. Introduction

a. Designates the electronic state of a molecule (the symmetry properties

of molecular wavefunction).

b. Similar as for atomic term symbols.

c. The same electronic state of a molecule applies to a whole potential

energy curve (or a potential energy surface) not just one geometry (i.e.,

nuclear arrangement).

2. Molecular term symbols for homonuclear diatomic molecules (or ions)

//

12ugL

SM

a. ML = sum of orbital angular momentum of the e– occupying the

molecular orbitals

□ liL mM

○ ml = 0 for orbitals

○ ml = 1 for orbitals

○ ml = 2 for orbitals

□ The capital Greek letter are used to represent a certain value of ML:

| ML | 0 1 2 3

Letter used

b. MS = sum of the spin angular momentum of the e– occupying the

molecular orbitals

□ sS mM

c. Use g or u to define if the total wave function is odd (u) or even (g).

□ This is a product of the parity of each electron.

□ g g = g; u u = g; u g = g u = u

d. Use + or – to define if the wave function is symmetric (+) or

antisymmetric (–) with respect to a plane containing internuclear axis.

CHEM 3720

129

e. Use Hund’s rules to figure out the ground state.

f. The molecular terms are used to describe electronic ground state and

excited states.

g. Examples:

□ H2: 2)1( g g

1

□ O2: 112222222 )1()1()1()1()3()2()2()1()1( gguugugug

g3

□ O2+: 12222222 )1()1()1()3()2()2()1()1( guugugug

g2

□ B2: 112222 )1()1()2()2()1()1( uuugug g

3

□ C2: 222222 )1()1()2()2()1()1( uuugug g

1

CHEM 3720

130

D. Unit Review

1. Important Terminology

Born-Oppenheimer approximation

potential energy curve (or surface)

Molecular Orbital Theory

molecular orbitals

linear combination of atomic orbitals (LCAO)

overlap integral

Coulomb integral

exchange integral

bonding orbital

antibonding orbital

Molecular energy level diagram

orbital, orbital, orbital

u (ungerade) and g (gerade)

CHEM 3720

131

homonuclear/heteronuclear diatomic

diamagnetic

paramagnetic

highest occupied molecular orbital (HOMO)

lowest unoccupied molecular orbital (LUMO)

bond order

photoelectron spectroscopy

nonbonding orbital

molecular term symbol

overlap

CHEM 3720

132

2. Important Formulas

Rrrrrr

H111111

2

12B2A2B1A1

22

21

B2A1 11 scsc ;

*

* ˆ

r

r

d

HdE

)22(2* Scd r

KJSEcHd s 2212ˆ1

2* r

B*A11 ssdS r ;

31

2RReRS R

A

BA s

RrsdJ 1

111 *

r ;

ReJ R 1

12

A

BB s

RrsdK 1

111 *

r ; ReR

SK R 1

S

KJEEE s

11 ;

S

KJEEE s

11

BAb 11

12

1ss

S

BAa 11

12

1ss

S

)orbitals gantibondinin #orbitals bondingin (#2

1BO ee

//

12ugL

SM

liL mM ; sS mM

CHEM 3720

133

Unit IX

Bonding in Polyatomic Molecules

A. Valence Bond Theory and Hybrid Orbitals

1. Introduction

a. The concept of hybrid (atomic) orbitals was introduced to interpret the

molecular shape.

b. The concept of hybrid orbital was also introduced in general chemistry

to explain why C is tetravalent.

□ Recall that carbon atom with the configuration 1122 2221 yx ppss

may be expected to form only 2 bonds.

□ An electron is promoted from the 2s orbital into the 2pz orbital.

□ The four singly occupied orbitals combines to form four equivalent

sp3 hybrid orbitals.

2. The sp hybridization

a. Example: BeH2 molecule:

□ two Be–H bonds that are equivalent

□ H–Be–H angle is 180º

b. Beryllium electronic configuration: 1s22s2 (term symbol: 1S0)

□ The 2s and 2pz orbitals combine to form two sp hybrid orbitals

(atomic orbitals):

zsp ps 222

1

CHEM 3720

134

c. The chemical bonds are formed as a bonding between one sp hybrid

orbital of Be and the s orbitals of one H atom:

)1H(3)2Be(2)2Be(1HBe sps cccz

3. The sp2 hybridization

a. Example: BH3 molecule:

□ three B–H bonds that are equivalent

□ H–B–H angle is 120º

b. The appropriate hybrid orbitals (i.e., orbitals with proper orientation)

are constructed by combining the 2s orbital and two 2p orbitals

resulting in three sp2 hybrid orbitals.

c. The chemical bonds in BH3 are formed as a linear combination

between one hybrid orbital on B and the 1s orbitals of one H atom.

d. The orthonormal (normalized and orthogonal) sp2 hybrid orbitals are:

zps 23

22

3

11

xz pps 22

12

6

12

3

12

xz pps 22

12

6

12

3

13

CHEM 3720

135

4. The sp3 hybridization

a. Example: CH4 molecule:

□ four C–H bonds that are equivalent

□ H–C–H angle is 109.5º

b. Four sp3 hybrid orbitals are obtained combining the 2s orbital and three

2p orbitals of C. The orthonormal sp3 hybrid orbitals are:

zyx ppps 22222

11

zyx ppps 22222

12

zyx ppps 22222

13

zyx ppps 22222

14

5. The sp3d2 hybridization

a. Example: SF6 molecule:

□ Six S–F bonds that are equivalent

□ F–S–F angles are 90º and 180º

□ The geometry is octahedral.

b. The ground-state electron configuration of S is [Ne]3s23p4.

c. For S, combine 3s orbital with the three 3p orbitals and two 3d orbitals

to create 6 equivalent sp3d2 hybrid orbitals.

SF

F

F

F

F FS

F

F

F

F

F F

d. An d2sp3 hybridization gives similar results.

CHEM 3720

136

6. Other hybridizations

a. Example: H2O molecule:

□ Two O–H bonds that are equivalent

□ H–O–H angle is 104.5º

b. This geometry is described by the hybrid orbitals that have 104.5º

between them.

zy pps 255.0271.0245.01

zy pps 255.0271.0245.02

O

H H

y

z

O

H H

y

z

□ These hybrid orbitals are intermediate between the pure p and the

sp3 hybrid orbitals.

□ There are two other hybrid orbitals that contain the lone pairs of

the oxygen.

□ All these hybrid orbitals are orthonormal.

7. The s and p character of hybrid orbitals

a. For a normalized hybrid orbital whose wavefunction is a combination

of s and p atomic orbitals written as:

znpynpxnpns npcnpcnpcnsczyx

○ The s character of the orbital is (cns)2.

○ The p character of the orbital is (cnpx)2 + (cnpy)

2 + (cnpz)2.

□ To understand this result, look at the electron density associated

with this orbital:

122222 zyx npnpnpns ccccd

○ (cns)2 is the “amount” of the electron that is in the s atomic

orbital giving therefore the s character.

CHEM 3720

137

□ Example: the hybrid orbitals in BH3 have:

○ 33.03

12

s character

○ 67.02

1

6

122

p character

□ Example: the hybrid orbitals in H2O have:

○ 0.4522 = 0.20 s character

○ 0.5522 + 0.7122 = 0.80 p character

b. A hybrid orbital that is x% s character and (100 – x)% p character can

be said to be an sp(100 – x)/x hybrid orbital.

□ Example: the bonding hybrid orbitals in H2O are sp4 hybrid

orbitals.

c. An spy hybrid orbital is an orbital that is 100/(1 + y)% s character and

100y/(1 + y)% p character.

□ Example: the sp3 hybrid orbitals in CH4 have:

○ 100/(1 + 3) = 25% s character

○ 100·3/(1 + 3) = 75% p character

CHEM 3720

138

B. Molecular Orbital Theory for Triatomic Molecules

1. Molecular orbital energy diagrams

a. Molecular orbital theory can explain molecular geometry of some

triatomic molecules.

□ BeH2 molecule is linear.

□ H2O molecule is bent.

b. The number of valence electrons on the central atom is important.

□ Be has 2 electrons and O has 6 electrons.

c. Obtain the molecular orbitals as a combinations of atomic orbitals on

Be (or O) and H (using an LCAO-MO procedure).

d. Molecular orbital energy diagram for BeH2:

Be HH–Be–H

1sB1sA

2s

2p

Be HH–Be–HBe HH–Be–H

1sB1sA

2s

2p

□ The 1g MO (same as the 1s atomic orbital of Be) is not shown.

□ The 2px and 2py orbitals of Be do not have the proper symmetry (or

orientation) to interact with H atomic orbitals so they are

nonbonding orbitals.

□ The configuration is (2g)2(1u)2.

CHEM 3720

139

e. Molecular orbital energy diagram for H2O:

O HOH H

2s

2p

1sB1sA

O HOH H

2s

2p

1sB1sA

□ The 2py orbital of O start interacting with the atomic orbitals of H.

□ Only the 2px orbital of O in nonbonding.

□ The configuration is (2a1)2(1b2)2(3a1)2(1b1)2.

□ The labels for a bent molecule

(i.e., a1, a2, b1, b 2) reflect the

symmetry of the molecule.

2. Walsh correlation diagrams

a. A Walsh correlation diagram is a plot

of the energy of a molecular orbital

as a function of a change in

molecular geometry.

b. The Walsh correlation diagram for an

AH2 triatomic molecule:

CHEM 3720

140

c. The molecular geometry (if a molecule is linear or bent) depends on

which energy is lower.

d. Bending destabilizes 2g, 1u, 3g, and 2u orbitals.

e. Bending stabilizes 1 u orbital.

f. The 1 u orbital get stabilized more than the 2g and 1u gets

destabilized at angles close to 180º.

g. The case of BeH2: (2g)2(1u)2 configuration is more stable (lower in

energy) than (2a1)2(1b2)2 configuration.

□ The BeH2 molecule is linear.

h. The case of H2O: (2g)2(1u)2(1u)4 configuration is less stable (higher

in energy) than (2a1)2(1b2)2(3a1)2(1b1)2 configuration.

□ The H2O molecule is bent.

i. The exact angle of bending (for example 104.5º for H2O) can be

determined using more complicated computational techniques.

j. A Walsh correlation diagram for a

XY2 triatomic (X and Y are second

row elements) can be used to

determine if CO2, NO2, NO2+, CF2,

etc are linear or bent.

k. Molecules with 8, 10, 12, 14, or 16

valence electrons are predicted to be

linear.

l. Molecules with 18 or 20 valence

electrons are predicted to be bent.

CHEM 3720

141

C. Hückel Molecular Orbital Theory

1. Introduction

a. It is a method applied mainly to conjugated and aromatic

hydrocarbons.

b. It uses the -electron approximation.

□ The sp2 hybrid orbitals of C and s orbitals of H create a -bond

framework.

□ The -bond framework is in xy plane.

□ The pz orbitals are perpendicular to the plane and form bonds.

□ One can say that electrons are moving in a fixed, effective

electrostatic potential due to the electrons in the framework.

○ This is the -electron approximation.

c. The problem of electrons (which are delocalized MO occupied by

electrons) can be treated separately from the problem of electrons.

2. Application to ethene (CH2=CH2)

a. The wave function for orbital in ethene:

B2A1 22 zz pcpc

b. Solving for c1 and c2, using variational method, results in the secular

determinant:

022221212

12121111

ESHESH

ESHESH

□ Hij are integrals involving an effective Hamiltonian operator (that

includes interaction of electrons with electrons)

□ Sij are overlap integrals between 2pz atomic orbitals.

c. The H11 and H22 are Coulomb integrals.

□ H11 = H22 in ethene.

d. The H12 integral is called a resonance integral (or exchange integral).

CHEM 3720

142

e. Hückel assumptions:

ji

jiS ijij

if0

if1

□ Hii are assumed to be same and denoted .

ji

jiHij

toboundnot is if0

tobound is if.

○ Use the same value for each pair of neighbors.

f. Rewrite the secular determinant:

0

E

E

E

g. To get the energy one must have and .

□ Instead of calculating those values, one can determine their values

from comparison with experiment.

□ By doing so = –75 kJ/mol.

h. Represent the energies of the orbitals considering as reference ( is

the energy of an electron into a 2pz orbital):

i. The energy of the electrons (total electronic energy):

222 E

□ This is the energy of a double bond in Hückel theory.

j. Solve for c1 and c2 to get the orthonormal wavefunctions.

□ Conditions are: c12 = c2

2 and c12 + c2

2 = 1.

B,A,1 22

12

2

1zz pp

B,A,2 22

12

2

1zz pp

CHEM 3720

143

3. Application to butadiene (CH2=CH–CH=CH2)

a. Consider that the molecule is linear (although in reality it is not).

1 2 3 4

b. The wavefunctions for orbitals:

4

1,2

jjziji pc

c. These MOs created as a LCAO leads to the secular determinant:

0

4444343424241414

3434333323231313

2424232322221212

1414131312121111

ESHESHESHESH

ESHESHESHESH

ESHESHESHESH

ESHESHESHESH

ijijijii SHH

;0

;But

0

00

0

0

00

E

E

E

E

d. Rearrange by factoring and writing xE

.

0

100

110

011

001

x

x

x

x

e. This determinant leads to a fourth-order equation.

013 24 xx

f. There are 4 solutions: x = 1.618 and x = 0.618.

g. Using xE

xE , one obtains the four energies

associated with the four molecular orbitals.

CHEM 3720

144

618.1

618.0

618.1

618.0

618.1

618.0

618.1

618.0

h. The energy of the electrons (also called the total electronic energy):

472.44618.02618.12 E

i. The delocalization energy:

□ This is the difference in energy between the total electronic

energy and the energy of localized double bonds (or radicals).

□ The delocalization energy gives the relative stabilization of the

molecule.

kJ/mol35472.0)HC(2)HC( 4284deloc EEE

j. Solve for the coefficients of the atomic orbitals in the molecular

orbitals (cij) to get the orthonormal wavefunctions:

43211 23717.026015.026015.023717.0 zzzz pppp

43212 26015.023717.023717.026015.0 zzzz pppp

43213 26015.023717.023717.026015.0 zzzz pppp

43214 23717.026015.026015.023717.0 zzzz pppp

□ The MOs are delocalized over the whole molecule.

CHEM 3720

145

4. Application to benzene (C6H6)

a. The bonds between carbon atoms in the

ring and between the carbon atoms and the

hydrogen atoms form the framework of

benzene.

b. Write the secular determinant using

the notation

Ex

:

1

2

3

4

5

6

c. This determinant leads to a sixth-order equation.

0496 246 xxx

d. There are 6 solutions: x = 1, 1, 2.

e. The MO energy diagram is given below.

f. The energy of the electrons:

86422 E

g. The delocalization energy:

kJ/mol1502)HC(3)HC( 4266deloc EEE

h. The wavefunctions for benzene (top view):

2

2

2

2

0

10001

11000

01100

00110

00011

10001

x

x

x

x

x

x

CHEM 3720

146

D. Photoelectron spectroscopy

1. It is a method of probing the molecular energy (i.e., the energy of

molecular orbitals).

2. Molecular orbital energy diagram for CH4

a. The small circles represent the number of atomic or molecular orbitals.

3. Molecular orbital energy diagram for C2H4

a. This diagram explains the UV absorption of 58500 cm–1.

CHEM 3720

147

E. Unit Review

1. Important Terminology

hybrid orbitals

sp, sp2, sp3, d2sp3 hybridizations

s and p character of hybrid orbitals

Walsh correlation diagrams

Hückel theory

-electron approximation

-bond framework

Hückel assumptions

total electronic energy

delocalization energy

localized energy

CHEM 3720

148

2. Important Formulas

znpynpxnpns npcnpcnpcnsczyx

(cns)2 = s character

(cnpx)2 + (cnpy)

2 + (cnpz)2 = p character

x% s character = (100 – x)% p character = sp(100 – x)/x hybrid orbital

spy hybrid orbital = y1

100% s character =

y

y

1

100% p character

4

1,2

jjziji pc

ji

jiS ijij

if0

if1

Hii =

ji

jiHij

toboundnot is if0

tobound is if

= –75 kcal/mol

222 E

xE

xE

CHEM 3720

149

Unit X

Computational Chemistry

A. Molecular Modeling

1. Introduction

a. Computational chemistry is a field of chemistry that is been developing

in the past two decades.

b. Computers have become powerful enough that some molecular

properties can be calculated with an accuracy that approaches (or

sometimes exceeds) the accuracy of the experimental data.

c. Molecular modeling, a part of computational chemistry, deals with:

□ developing theories and models of molecules that can be used to

predict molecular properties

□ applying these models and theories

2. Quantum Mechanics

a. The wave treatment of the matter is necessary for light particles like

electrons, atoms and/or molecules.

b. The properties and behavior of a particle are specified by the

wavefunction, which is obtained by solving the Schrödinger equation:

0)(8

2

2

2

2

2

2

2

2

VE

h

M

zyx

○ M is the mass of particle.

○ E is the total energy.

○ V is the potential energy.

c. Wave equation can be solved exactly for the H atom (and a few other

systems) but approximations are needed for multielectron systems and

especially for polyatomic molecules.

CHEM 3720

150

d. For polyatomic molecules, Born-Oppenheimer approximation allows

the separation of the total wavefunction ),(molecular Rr into an

electronic wavefunction );(e Rr and a nuclear wavefunction )(N R .

)();(),( Nemolecular RRrRr

e. The electronic wavefunction is obtained by solving the Schrödinger

equation at a fixed nuclear arrangement.

f. Electrons create a potential in which nuclei move, and this potential is

called a potential energy surface for polyatomic molecule or a potential

energy curve for a diatomic molecule.

g. Examples:

□ Torsional potential of ethane (or propane):

-1.0

0.0

1.0

2.0

3.0

0 60 120 180 240 300 360Torsion angle (degree)

E (

kca

l/m

ol)

2.9 kcal/mol

for ethane

2.9 kcal/mol

for ethane

□ Torsional potential of butane:

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 60 120 180 240 300 360Torsion angle (degree)

E (

kca

l/m

ol)

3.4kcalmol

0.8kcalmol

3.4kcalmol

3.4kcalmolkcalmol

0.8kcalmol

0.8kcalmolkcalmol

H

CH3

HC

H H

H3

H

CH3

H

H

H CH3

H

CH3

HH

H CH3

H

CH3

H

CH3

H H

H

CH3

HH

H3C H

H

CH3

H

H

CH3 H

H

CH3

HC

H H

H3

H

H

H

H H (CH3)

H

H

H (CH3)

HH

H

H

CHEM 3720

151

3. Molecular Mechanics

a. Molecular mechanics is a classical mechanical model that represents a

molecule as a group of atoms held together by elastic bonds.

b. Molecular mechanics looks at the bonds as springs that can be

stretched, compressed, bent at the bond angles, and twisted in torsional

angles.

□ Interactions between nonbonded atoms also are considered.

c. The sum of all these forces is called the force field of the molecule.

d. A molecular mechanics force field is constructed and parameterized by

comparison with a number of molecules, for instance a group of

alkanes.

e. This force field then can be used for other molecules similar to those

for which it was parameterized.

f. Various force fields are available: MM2, MM3, AMBER, etc.

g. To make a molecular mechanics calculation, a force field is chosen and

suitable molecular structure values (natural bond lengths, angles, etc.)

are set.

h. The structure then is optimized by changing the structure incrementally

to minimize the strain energy (by spread it over the entire molecule).

i. This minimization is orders of magnitude faster than a quantum

mechanical calculation on an equivalent molecule.

j. Molecular mechanics does have two weaknesses:

□ Force fields are based on the properties of known, similar

molecules, and if one is interested in the properties of a new type

of molecule, an appropriate force field is probably not available for

that molecule.

CHEM 3720

152

□ Because molecular mechanics models look at molecules as sets of

springs, they cannot be used to predict electronic properties of

molecules, such as dipole moments and spectroscopy.

○ To make predictions about the electronic properties of a

molecule, you must use quantum mechanical models.

k. Molecular mechanics is very useful in investigating very large systems

of few thousands atoms (enzymes, nanotubes, etc.) for which the

explicit treatment of electrons is prohibited.

4. Electronic Structure Theory

a. Electronic structure theory includes a multitude of methods in which

electrons are treated explicitly.

b. Some of these methods invoke the independent electron

approximation: the electronic wavefunction can be written as a product

of one-electron functions.

c. The electronic wavefunction, which must be antisymmetric, can be

represented by a Slater determinant:

)(.....)()(

....................

)2(....)2()2(

)1(....)1()1(

!

1),....2,1(

21

21

21

NuNuNu

uuu

uuu

NN

N

N

N

○ u are orthonormal spin orbitals obtained as a product of the

spatial part and the spin part.

d. For molecules, one-electron wavefunctions are called molecular

orbitals, and they are usually obtained as a linear combination of

atomic orbitals (LCAO).

e. The energy is found by variationally optimizing the coefficients in the

LCAO-MO, this process leading to a secular determinant:

CHEM 3720

153

0

...

............

......

...

2211

22221212

1112121111

NNNNNNNN

NN

ESHESHESH

ESHESH

ESHESHESH

○ The elements of this determinant are called matrix elements

and involve calculation of integrals (between atomic orbitals).

f. The electronic structure theory methods can be classified as:

□ semiempirical

○ These methods ignore, approximate some of the integrals, or

introduce some parameters from experimental data.

○ Examples: Hückel, Extended Hückel, ZINDO, AM1, PM3,

etc.

□ ab initio

○ These methods solve all of these integrals without

approximation.

○ The simplest such method is the Hartree-Fock (HF) method

which is a self-consistent field (SCF) method that has the

shortcoming of not including the correlation energy.

○ More advanced methods, called post-HF, include some or

most aspect of electron correlation.

○ Examples: MP2 (Møller-Plesset second order perturbation

theory), MP4 (Møller-Plesset fourth order perturbation

theory), CI (configuration interaction) and derivations, CC

(coupled cluster) are derivations, etc.

□ density functional theory

○ In density functional theory (DFT) methods, the energy is

obtained from the electron density through a functional (a

function of a function) that exists but is not known explicitly.

CHEM 3720

154

○ The electronic energy is a sum of the kinetic energy term, the

nuclear-electron attraction (and internuclear repulsion) term,

electron-electron (Coulomb) repulsion term, and the exchange-

correlation term including the remaining electron-electron

interaction.

○ The exchange-correlation term is usually a sum of the

exchange part and a correlation part.

○ There are many density functional theory methods available

based on the functional form of the exchange and/or

correlation part.

○ DFT methods are typically inaccurate for some molecular

properties.

□ hybrid density functional theory

○ In hybrid density functional theory (HDFT) methods, certain

exchange contributions obtained in HF method are included in

the exchange-correlation energy.

○ The extent of HF contributions is typically parameterized.

○ The new methods are slightly more computationally expensive

but much more accurate than the “pure” DFT methods for

investigating molecular properties.

○ The name of the HDFT method typically include the acronym

for the exchange functional used followed by a number

indicating the number of parameters used followed by the

acronym for the correlation functional used.

○ For example, the mPW1PW91 method (used in the lab

experiment) is a one-parameter method using the modified

Perdew-Wang exchange functional and the Perdew-Wang

correlation functional.

CHEM 3720

155

5. Basis sets

a. The set of atomic functions used to construct LCAO-MOs is called the

basis set.

b. Slater orbitals

□ Slater orbitals were used in computations for polyatomic

molecules.

□ Slater orbitals are defined as:

,,, 1 ml

rnnnlm YerNrS

)!2(

)2( 2/1

nN

n

n

is a normalization constant

,ml

Y are the spherical harmonics

□ Properties of Slater orbitals:

○ (zeta) is arbitrary and is not necessarily equal to Z/n as in

hydrogen-like orbitals.

○ The radial part of the Slater orbitals does not have nodes.

○ Snlm(r,,) is not orthogonal to Snlm(r,,).

○ n is also considered as a variational parameter.

□ (zeta) and n are obtained based on some rules.

□ The disadvantage of Slater orbitals is that the multicenter integrals

are difficult to calculate.

c. Gaussian-type orbitals

□ Use Gaussian functions also called Gaussian-type orbitals instead

of Slater orbitals.

□ Gaussian-type orbitals has the form:

,,,21 m

lrn

nnlm YerNrG

□ The Slater orbitals and Gaussian orbitals have different behavior

for small values of r.

CHEM 3720

156

□ Using more Gaussian functions, with different values, improve

the description the fitting of Gaussian orbitals to a Slater orbital.

□ The STO-3G basis set is a basis set in which each atomic orbital is

described by a sum of three Gaussian functions that try to mimic a

Slater orbital.

),()( GF3

1,

STOinlm

iinlmnlm

rdr

d. Double-zeta basis sets

□ These basis sets are generated as a sum of two Slater orbitals with

different orbital exponents (which differ in the value of the

exponent).

),(),()( 2STO

1STO rdrr

nlmnlmnlm

CHEM 3720

157

e. Split-valence basis sets

□ These basis sets have the inner-shell electrons described by a

single Slater-type orbital and the valence orbitals expressed by a

double-zeta (triple-zeta or higher) representation.

□ Example: 6-31G basis set has the inner-shell electrons described

by a Slater orbital obtained as a sum of 6 Gaussian functions and

the valence electrons by a sum of two Slater orbitals obtained as a

sum of 3 and 1, respectively, Gaussian function.

f. Polarization functions

□ Improved basis sets are obtained by adding a function of higher

angular momentum, l, to the mathematical expression of a given

angular momentum.

□ For example, for a 2p orbital, add a 3d–type function.

□ Example: 6-31G* basis set has a polarized function (3d) added to

the 6-31G basis set for atoms of the second row while 6-31G** has

(in addition to the polarized function from above) a polarized

function (2p) for the atoms of the first row (i.e., H atom).

g. Diffuse functions

□ Better basis sets are obtained by adding a diffuse function (a

function of low exponent) that describes electron density further

away from nucleus.

□ These basis sets are very useful for investigating anions.

□ Example: 6-31+G** basis set will have a diffuse function added to

the 6-31G** basis set for atoms of the second row while

6-31++G** has (in addition to the diffuse function from above) a

diffused function for the atoms of the first row (i.e., H atom).

CHEM 3720

158

B. Unit Review

1. Important Terminology

molecular mechanics

force field

electronic structure theory

independent electron approximation

semiempirical methods

ab initio methods

MP2, MP4, CI, and CC

density functional theory methods

exchange-correlation term

hybrid density functional theory methods

basis set

double-zeta, split-valence basis sets

polarization and diffuse functions

CHEM 3720

159

Unit XI

Molecular Symmetry and Group Theory

A. Introduction to Molecular Symmetry

1. The molecular symmetry properties can be used to:

a. Reduce the high-order secular determinants in Hückel method

b. Determine the IR or Raman activity of vibrational normal modes

c. Label and designate molecular orbitals

d. Derive selections rules for spectroscopic transitions

2. Symmetry elements and symmetry operations

a. The symmetry of a molecule is described by its symmetry elements.

b. Each symmetry element has (one or more) symmetry operations

associated with them.

Symmetry elements Symmetry operations Examples

Description Symbol Symbol Description

Identity E E No change

n-Fold axis of symmetry nC nC Rotation about the axis by

360/n degrees

C2, C3

C4, C6

Plane of symmetry

(mirror plane)

d

h

v

d

h

v

ˆ

ˆ

ˆ

ˆ Reflection through a plane

Center of symmetry i i Reflection through

the center of symmetry

n-Fold

rotation reflection

axis of symmetry

(improper rotation)

nS nS

Rotation about the axis by

360/n degrees followed by

reflection through a plane

perpendicular to the axis

CHEM 3720

160

c. The axis with the highest value of n is called the principal axis.

d. The planes of symmetry can be:

□ v: the plane of symmetry is parallel to a unique axis or to a

principal axis

□ h: the plane of symmetry is perpendicular to a unique axis or to a

principal axis

□ d: the plane of symmetry bisects the angle between C2 axes that

are perpendicular to a principal axis

○ d is a special type of a v plane.

e. A symmetry element may have more than one symmetry operation

associated with it.

□ The 3-fold axis of symmetry (C3) has two symmetry operations

associated with it.

○ 3C (rotation with 1203

360 degrees)

○ 3323

ˆˆˆ CCC (rotation with 240 degrees)

□ The 4-fold axis of symmetry (C4) has three symmetry operations:

○ 4C (rotation with 904

360 degrees)

○ 4424

ˆˆˆ CCC (rotation with 180 degrees)

○ 44434

ˆˆˆˆ CCCC (rotation with 270 degrees)

3. Point groups

a. A group (or set) of symmetry operations constitutes a point group.

b. Each point group consists of a number of symmetry elements.

c. The total number of symmetry operations is called the order of the

point group.

□ The total number of symmetry operations can be greater than the

total number of symmetry elements.

CHEM 3720

161

d. Common point groups of interest to chemists.

Point group Symmetry elements Molecular examples

C2v E, C2, 2v H2O, CH2Cl2, C6H5Cl

C3v E, C3, 3v NH3, CH3Cl

C2h E, C2, i, h trans-ClHC=CHCl

D2h E, 3C2, i, 3v C2H4, naphthalene

D3h E, C3, 3C2, h, S3, 3v SO3, BF3

D4h E, C4, 4C2, i, S4, h, 2v, 2d XeF4

D6h E, C6, 3C2, i, S6, h, 3v, 3d C6H6

D2d E, S4, 3C2, 2d H2C=C=CH2

Td E, 4C3, 3C2, 3S4, 6d CH4

□ The number in front of a symmetry element is the number of times

such a symmetry element occurs.

e. Some point groups have some common features.

□ Cnv n-fold axis and n v mirror planes

□ Cnh n-fold axis and a mirror plane perpendicular to the n-fold

axis

□ Dn n-fold axis and n 2-fold axes perpendicular to n-fold axis

□ Dnh n-fold axis and n 2-fold axes perpendicular to n-fold axis

plus a plane perpendicular to n-fold axis

□ Dnd same as Dn plus n-dihedral mirror planes

CHEM 3720

162

f. A summary of the shapes corresponding to different point groups.

CHEM 3720

163

g. Molecule examples: identify all the symmetry elements and the point

group that each molecule belongs to.

H2O XeF4 C6H5Cl H2C=C=CH2

Xe

F

FF

F

Cl

E,C2,2v E,C4,4C2,i,S4,h,2v,2d E,C2,2v E,S4,3C2,2d

C2v D4h C2v D2d

SO3 CH2Cl2 C6H6 trans-ClHC=CHCl

S

O

O

O

C

H H

Cl Cl

E,C3,3C2,h,S3,3v E,C2,2v E,C6,3C2,i,S6,h,3v,3d E,C2,i,h

D3h C2v D6h C2h

C2H4 BF3 CH3Cl CH4

Cl

HHH

E,3C2,i,3v E,C3,3C2,h,S3,3v E,C3,3v E,4C3,3C2,3S4,6d

D2h D3h C3v Td

cis-ClHC=CHCl NH3 naphthalene meta-C6H4Cl2

C2v C3v D2h C2v

CHEM 3720

164

h. A flow diagram for determining the point group of a molecule.

□ Start at the top and answer the question posed in each diamond

(Y = yes, N = no).

□ Example: benzene

CHEM 3720

165

B. Group Theory

1. Introduction

a. The set of symmetry operations of a molecule form a point group.

b. A group is a set of entities (A, B, C…) that satisfy certain requirements:

□ Combining (i.e., multiplying) any 2 members of the group gives a

member of the group.

○ “A group must be closed under multiplication.”

□ The multiplication must be associative:

A (B C) = (A B) C

□ The set of entities (i.e., the members of the group) contains an

identity element E such that:

E A = A E (and E B = B E)

□ For every entity in the group (for example A) there is an inverse

(A–1) that is also a member of the group so that:

A A–1 = A–1 A = E (and B B–1 = B–1 B = E)

2. Group multiplication table

a. Example of a group of symmetry operations: the case of water

□ There are four symmetry operations: E , 2C , v , 'ˆv

□ These four symmetry operations form the C2v point group.

□ Consider an arbitrary vector (u) and investigate how each

combination of symmetry operations change the vector.

First Operation

Second Operation E 2C v 'ˆ v

E E 2C v 'ˆ v

2C 2C E 'ˆ v v

v v 'ˆ v E 2C

'ˆ v 'ˆ v v 2C E

b. The table above is called the group multiplication table of the C2v point

group.

OHH

u

CHEM 3720

166

c. The four symmetry operations of C2v satisfy the conditions of being a

group and are collectively referred to as the point group C2v.

d. Example of a group of symmetry operations: the case of ammonia

□ There are six symmetry operations: E , 3C , 23

C , v , 'ˆv ,

''ˆv

□ These six symmetry operations form the C3v point group.

First Operation

Second Operation E 3C 23C v 'ˆ v ''ˆ v

E E 3C 23C v 'ˆ v ''ˆ v

3C 3C 23C E

23C 2

3C E 3C

v v E 'ˆ v 'ˆ v E ''ˆ v ''ˆ v E

□ Fill in the rest of the table.

3. The character table of a point group

a. Symmetry operations can be represented by matrices.

□ Example of H2O or the case of the C2v point group.

○ Consider a vector: u = uxi + uyj + uzk

○ For the 180 degree rotation along z axis (Ĉ2):

Ĉ2ux= –ux; Ĉ2uy= –uy; Ĉ2uz= uz

○ One can write:

z

y

x

z

y

x

u

u

u

C

u

u

u

C 22ˆ where

100

010

001

2C

○ Similarly:

100

010

001

E ;

100

010

001

v ;

100

010

001'v

CHEM 3720

167

b. A set of matrices that multiply together in the same manner as a group

multiplication table is said to be a representation of that group.

c. These four matrices form a representation of the C2v point group or,

more specific, a three-dimensional representation because it consists of

33 matrices.

d. There are an infinite number of such representations, but some of them,

called irreducible representations, can be used to express all the others

that are called reducible representations.

□ Finding the irreducible representations has been already done for

all point groups.

e. The case of C2v point group

□ The irreducible representations are denoted A1, A2, B1, and B2.

○ Use notation A if the representation is symmetric with respect

to principal axis (C2).

○ Use notation B if the representation is antisymmetric with

respect to principal axis (C2).

□ A1 is the totally symmetric one-dimensional irreducible

representation.

□ The irreducible representations of the C2v point group:

E 2C v 'ˆ v

A1 (1) (1) (1) (1)

A2 (1) (1) (–1) (–1)

B1 (1) (–1) (1) (–1)

B2 (1) (–1) (–1) (1)

CHEM 3720

168

f. The case of C3v point group

□ The irreducible representations of the C3v point group:

E 3C 23C v 'ˆ v ''ˆ v

A1 (1) (1) (1) (1) (1) (1)

A2 (1) (1) (1) (–1) (–1) (–1)

E

10

01

21

2

3

2

3

21

21

2

3

2

3

21

10

01

21

2

3

2

3

21

21

2

3

2

3

21

□ Two-dimensional irreducible representations are designated by E

(not the same as the symmetry operation E).

○ It can be obtained by applying the symmetry operations to a

vector in x-y plane.

○ Because x and y transform together, the result of a given

operation is written as a linear combination of x and y.

□ The x and y are said to form a basis for E or to belong to E.

g. Three-dimensional irreducible representations are designated by T.

h. For almost all applications of group theory one do not uses the

complete matrices, only the sum of the diagonal elements called its

trace or, in group theory, its character.

i. The characters of the irreducible representations of a point group are

displayed in a table called character table.

j. Certain symmetry operations (for example 3C and 23

C or v , 'ˆv , and

''ˆv in the C3v point group) are essentially equivalent (have the same

characters) and are said to belong to the same class.

k. The number of classes is equal to the number of irreducible

representations ( the character tables are squared).

l. For the point groups that has a center of symmetry i, the irreducible

representations are labeled as g or u to describe whether they are

symmetric or antisymmetric under the inversion.

CHEM 3720

169

m. Examples of character table for some useful point groups.

v3C E 2 3C 3 v

1A 1 1 1 z 22 yx , 2z

2A 1 1 –1 zR

E 2 –1 0 ),( yx ),( yx RR ),( 22 xyyx ),( yzxz

v2C E 2C v v ˆ

1A 1 1 1 1 z 2x , 2y , 2z

2A 1 1 –1 –1 zR xy

1B 1 –1 1 –1 x, yR xz

2B 1 –1 –1 1 y, xR yz

h2C E 2C i h

gA 1 1 1 1 zR 2x , 2y , 2z , xy

gB 1 –1 1 –1 xR , yR xz , yz

uA 1 1 –1 –1 z

uB 1 –1 –1 1 x, y

h3D E 2 3C 3 2C h 2 3S 3 v

1A 1 1 1 1 1 1 22 yx , 2z

2A 1 1 –1 1 1 –1 zR

E 2 –1 0 2 –1 0 ),( yx ),( 22 xyyx

1A 1 1 1 –1 –1 –1

2A 1 1 –1 –1 –1 1 z

E 2 –1 0 –2 1 0 ),( yx RR ),( yzxz

CHEM 3720

170

n. Description of character tables

□ The second to last column of the character table lists how the three

axis (x, y, and z) (or the translation along the three axis) and how

the rotation along the three axis (Rx, Ry, and Rz) transform in that

particular point group.

○ It can also be said that x (or y or z) form a basis of a certain

irreducible representation.

○ For the C2v point group, x forms a basis for the B1

representation, y forms a basis for the B2 representation, and z

forms a basis for the A1 representation.

○ For the C3v point group, x and y form jointly a basis for the two

dimensional representation E.

○ Example of rotation around the z axis in the C3v point group:

– Depict the rotation around an axis by vectors.

top viewtop view

– The rotation along the z axis transforms as the A2

representation.

□ The last column lists how combinations of the axis (also

components of the molecular polarizability) (x2, y2, xy, etc)

transform in that particular point group.

○ When there is no axis that transform as a two-dimensional

irreducible representation, the combination of the axis is just

the product between the particular axis.

zzv

zzv

zzv

zz

zz

zz

RR

RR

RR

RRC

RRC

RRE

)(ˆ

)(ˆ

)(ˆ

)(ˆ

)(ˆ

)(ˆ

23

3

CHEM 3720

171

4. Mathematical properties derived from character tables

a. Notations that will be used in the following mathematical relations

involving the characters of irreducible representations:

□ R is an arbitrary symmetry operation

□ )ˆ(R is the character of the matrix representation of R

□ )ˆ(Rj is the character of the jth irreducible representation of R

b. The order of a point group can be written as:

N

jjdh

1

2

□ h = order of the group (= number of symmetry operations)

□ dj = dimension of the jth irreducible representation

□ N = number of irreducible representation

c. Considering that the dimension of the jth irreducible representation

equal to the character of the jth irreducible representation for the

identity operation:

jj dE ˆ 2

1

)ˆ(

N

jj Eh

d. The irreducible representations are orthogonal.

0)ˆ()ˆ(ˆ

RR jR

i

classes

0)ˆ()ˆ()ˆ( RRRn ji

○ where n is the number of symmetry operations in the class

containing R

□ Mathematically: the product of two vectors is given by

n

kkk vu

1

vu

CHEM 3720

172

□ Example: A1 and B2 representations of group C2v are orthogonal:

011)1(1)1(111

)ˆ()ˆ()ˆ()ˆ(

)ˆ()ˆ()ˆ()ˆ(

2121

2121

BABA

2B2ABA

vvvv

CCEE

e. If i for i in equations above is the totally symmetric irreducible

representation (A1):

1)ˆ( Ri

0)ˆ()ˆ()ˆ(classesˆ

RRnR jR

j if 1Aj

f. For an irreducible representation:

Rjj hRRnR

ˆ classes

22)ˆ()ˆ()ˆ(

□ Looking at a n-dimensional irreducible representation like a vector

for which one define the length of the vector as:

n

kk

v1

22)length(vv

□ Example: A2 representation of C3v point group:

6131211)ˆ()ˆ( 222

classes

2 RRn j

g. Combining the equations from d. and f.:

ijjijR

i hRRRnRR classesˆ

)ˆ()ˆ()ˆ()ˆ()ˆ(

h. Reducing a reducible representation as a sum of irreducible

representations:

□ Assume that R is the character of the symmetry operator in

reducible representation .

□ Write this R as a combination of the characters of irreducible

representations: j

jj RaR ˆˆ

CHEM 3720

173

□ Find the coefficients aj, multiply by )ˆ(Ri and sum over R :

ji

jR

ij

jR

i RRaRR

ifonly0

ˆˆ)ˆ()ˆ()ˆ()ˆ(

and considering ijjR

i hRR )ˆ()ˆ(ˆ

R

ii RRhaˆ

)ˆ()ˆ(

classesˆ

)ˆ()ˆ()ˆ(1

)ˆ()ˆ(1

RRRnh

RRh

a iR

ii

i. Example: Reduce the reducible representation = 4 2 0 2 as a sum of

irreducible representations belonging to C2v group.

4)ˆ( E ; 2)ˆ( 2 C ; 0)ˆ( v ; 2)ˆ( v

2121012144

11A a

1121012144

12A a

0121012144

11B a

1121012144

12B a

2212 BAA

□ Verification:

2A1 21 21 21 21

A2 1 1 –1 –1

B2 1 –1 –1 1

4 2 0 2

CHEM 3720

174

C. Applications of Molecular Symmetry

1. Hückel theory for benzene

a. When using the pz orbitals on various C atoms, Hückel theory leads to:

0

10001

11000

01100

00110

00011

10001

x

x

x

x

x

x

b. Instead of using atomic orbitals, use “symmetry orbitals” that are

obtained as a linear combination of the pz orbitals:

)(6

16543211

)(6

16543212

)22(12

16543213

)22(12

16543214

)22(12

16543215

)22(12

16543216

c. Using these orbitals leads to the following determinant:

0

12

10000

2

110000

0012

100

002

1100

000020

000002

xx

xx

xx

xx

x

x

CHEM 3720

175

□ This determinant is easier to solve and the solutions are

2,1,1 x .

□ The construction of those “symmetry orbitals” takes advantage of

the symmetry of the system.

2. The use of symmetry to predict the elements of the secular determinant that

are zero.

a. Look at dHH jiijˆ* and dS jiij

*

b. These integrals should be independent on the molecule orientation (the

value should not change upon applying a symmetry operation):

ijjiij SdRRSR ˆˆˆ *

ijjiij HdRHRRHR )(ˆ)ˆ(ˆ)(ˆˆ *

c. The overlap integrals

□ Consider i* is a basis for one irreducible representation a.

□ Consider j is a basis for one irreducible representation b.

** )ˆ(ˆiai RR and jbj RR )ˆ(ˆ

ijbajibaij SRRdRRS )ˆ()ˆ()ˆ()ˆ( *

○ So the )ˆ()ˆ( RR ba product should be equal to 1 for every

symmetry operation. If the )ˆ()ˆ( RR ba is equal to –1 Sij

should be zero for the equation above to be true.

□ Sij 0 only if i and j are bases of the same irreducible

representation.

□ Sij = 0 only if i and j are bases of different irreducible

representations.

d. The exchange integrals

□ The molecular Hamiltonian operator is symmetric under all

symmetry operations.

ijijbAajiij HHRRRdRHRRHR )ˆ()ˆ()ˆ()(ˆ)ˆ(ˆ)(ˆˆ1

*

CHEM 3720

176

○ So )ˆ()ˆ()ˆ(1A RRR ba should be equal to 1 for all symmetry

operations R in order for the Hij terms not to be zero. The

condition is similar as the one obtained for Sij.

□ Only the elements of the determinant that are bases of the same

irreducible representation are not zero.

e. Example: H2O

□ The 2px orbital of O transform as B1.

□ The 2py orbital of O transform as B2.

□ The 2pz orbital of O transform as A1.

□ Consider a linear combination of 1s orbitals (BA HH 11 ss ) that

transform as A1.

□ So the overlap integral of 2px orbital of oxygen with

BA HH 11 ss is zero because they transform differently.

3. The generating operator

a. The symmetry orbitals are obtained using the generating operator

defined as:

RRh

dP

Rj

jj

ˆ)ˆ(ˆ

ˆ

b. The generating operators are used to find linear combination of atomic

orbitals that are bases for irreducible representations.

CHEM 3720

177

D. Unit Review

1. Important Terminology

symmetry element

symmetry operation

n-fold axis of symmetry

plane of symmetry

center of symmetry

n-fold rotation reflection axis of symmetry

principal axis

point group

the order of the point group

Cs, C2v, C3v, C2h, D2h, D3h, D4h, D6h, D2d, Td

group multiplication table

irreducible representation

reducible representation

CHEM 3720

178

the character table

one, two, and three-dimensional irreducible representations

symmetry orbitals

use of symmetry to predict the elements of the secular determinant

the generating operator

2. Important Formulas

N

jjdh

1

2

ijjijR

i hRRRnRR classesˆ

)ˆ()ˆ()ˆ()ˆ()ˆ(

classesˆ

)ˆ()ˆ()ˆ(1

)ˆ()ˆ(1

RRRnh

RRh

a iR

ii

ijbajibaij SRRdRRS )ˆ()ˆ()ˆ()ˆ( *

ijijbAajiij HHRRRdRHRRHR )ˆ()ˆ()ˆ()(ˆ)ˆ(ˆ)(ˆˆ1

*

RRh

dP

Rj

jj

ˆ)ˆ(ˆ

ˆ

CHEM 3720

179

Unit XII

Molecular Spectroscopy

A. Introduction to Molecular Spectroscopy

1. Energy levels in molecules

a. Let’s consider a system with

electrons

nuclei

n

N

b. This system will be (completely) described by a wavefunction (r, R)

depending on the coordinates of all the nuclei and all the electrons.

c. According to Born-Oppenheimer approximation, the wavefunction is

separated: )();(),( RRrRr nuel

□ Solve for electronic wavefunction (el) considering a fixed

position of the nuclei (i.e., at one nuclear configuration).

□ Electrons create a potential in which nuclei move, and this

potential is called a potential energy surface for polyatomic

molecule or a potential energy curve for a diatomic molecule.

□ The nuclear motion (which is also quantized) can be separated in

vibrational, rotational and translation motions (or degrees of

freedom).

1 nucleus

(atom)

2 nuclei

(diatomic)

N (>2) nuclei

(polyatomic)

el depends on

3n coordinates

el depends on

3n coordinates

el depends on

3n coordinates

nu depends on

3N = 6 coordinates

nu depends on

3N coordinates

Linear Nonlinear

3 translational 3 translational 3 translational

2 rotational 2 rotational 3 rotational

1 vibrational 3N – 5 vibrational 3N – 6 vibrational

CHEM 3720

180

2. Introduction to spectroscopy

a. Spectroscopy is dealing with the study of the interaction of

electromagnetic radiation with atoms and molecules.

b. Molecular spectroscopy deals with the radiation interacting with

molecules.

c. The electromagnetic radiation is usually divided into different energy

(frequency of wavelength) regions.

CHEM 3720

181

d. The frequency or the energy of the radiation corresponds (is

associated) with different types of molecular transitions.

)cm(~ 1 Region of

Electromagnetic Radiation

Molecular Process

0.033-3.3 Microwave Rotation of polyatomic

3.3-330 Far infrared Rotation of small molecules

330-14500 Infrared Vibrations of flexible bonds

14500-50000 Visible and UV Electronic transitions

e. The energy difference between two energy levels is related to the

frequency of radiation by:

hEEE lu

f. Sometimes it is preferred the use of wavenumbers defined as:

1~ c

CHEM 3720

182

B. Diatomic Molecules

1. Pure rotational spectra

a. The simplest model for the rotation in a diatomic molecule is the rigid

rotator model.

b. The energy levels predicted by the rigid rotator model:

)1(2

2

JJI

EJ

□ I is the moment of inertia ( 2eRI ).

□ Rotational quantum number J = 0,1,2,…

c. The degeneracy of Jth level is gJ = 2J + 1.

d. The rotational energy of a molecule (in wavenumbers) is denoted by

F(J) and is called rotational term.

e. The energies (in wavenumbers) predicted by the rigid rotator model:

)1(~

)( JJBJF

□ The rotational constant cI

hB

28

~

f. The selection rule is J = 1 (plus the molecule should have a

permanent dipole moment).

g. Frequency of absorption is given by a transition from J to J + 1

(J J + 1) and is given, within rigid rotator model, by:

)1(~

2)()1(~ JBJFJFv where J = 0,1,2,…

h. Within rigid rotator model, the rotational spectrum of diatomics is

predicted to be a series of lines equally spaced (separated) in

microwave region of electromagnetic radiation.

i. Experimentally the lines are not equally spaced, they get close to each

other as the quantum number increases.

j. In reality, the energy levels are slightly different than predicted by rigid

rotator model.

CHEM 3720

183

k. The energies predicted by the nonrigid

rotator model: 22 )1(

~)1(

~)( JJDJJBJF

□ B~

is the rotational constant.

□ D~

is the centrifugal distortion constant.

l. Frequency of absorption is given by a

transition from J to J + 1 (J J + 1): 3)1(

~4)1(

~2)()1(~ JDJBJFJFv where J = 0,1,2,…

□ The B~

and D~

constants are obtained by

fitting this equation to the experimental data.

m. Rotational transitions also accompany vibrational transitions.

2. Vibrational spectra

a. The simplest model for the vibration in a diatomic molecule is the

harmonic oscillator model.

b. The energy levels predicted by the harmonic oscillator model:

hnEn

2

1

2/1

2

1

k is called fundamental frequency and

BA

BA

mm

mm

.

□ Vibrational quantum number: n = 0,1,2,…

c. The vibrational energy of a molecule (in wavenumbers) is denoted by

G(n) and is called vibrational term.

d. The energies (in wavenumbers) predicted by the harmonic oscillator

model:

~2

1)(

n

hc

EnG n where

k

c2

1~ and n = 0,1,2,…

e. The selection rule is n = 1.

CHEM 3720

184

f. Within harmonic oscillator model, only one line in the spectrum (called

the fundamental) should be observed, in the IR region, at:

k

2

1obs or

k

c2

1~obs

g. Experimentally it can be observed a series of less intense lines, almost

integral multiples of the fundamental, called overtones due to

transitions from n = 0 to n = 2,3,…

□ The first overtone: n = 0 n = 2.

□ The second overtone: n = 0 n = 3.

h. Harmonic-oscillator model does not represent accurately the potential

energy far from the equilibrium distance.

– Potential energy V(R) may be expanded in a Taylor series about Re:

....2462

....!3

1

!2

1)()(

44332

3e3

32

e2

2

e

ee

xxxk

RRdR

VdRR

dR

VdRVRV

RRRR

where j

j

jdR

Vd

– Harmonic–oscillator approximation (model) consists in keeping only quadratic term.

i. Once the anharmonic terms (or contributions) are included, the

vibrational term becomes:

...~~~)(2

21

ee21

e nxnnG where n = 0,1,2,…

□ e~x is the anharmonicity constant.

□ The anharmonic corrections are much smaller than the harmonic

term (i.e., e~x << 1).

j. The energy levels are not evenly spaced in anharmonic oscillator

model, and the energy difference gets smaller as quantum number n

increases.

)1(~~~)0()(~eeeobs nnxnGnG where n = 1,2,…

CHEM 3720

185

□ Selection rule: n = 1, 2, 3,…

□ Considering anharmonic terms improves agreement with

experiment.

3. Vibration-rotation spectra

a. Rotational transitions accompany vibrational transitions.

b. The energies predicted by the rigid rotator-harmonic oscillator

approximation:

)1(~~)()(

~21

, JJBnJFnGE Jn

c. The absorption in IR region of spectrum:

1

1

J

n

□ The J = +1 case:

1~2~~~

)1(~,1,1obs JBEEJ JnJn

○ This is called the R branch.

□ The J = –1 case:

JBJ~

2~1~obs

○ This is called the P branch.

□ The case of J = 0 gives the Q branch.

d. The example of the HCl system:

e. The spectrum is predicted to be a series of lines equally spaced (with

one line missing) in the IR region of electromagnetic radiation.

CHEM 3720

186

f. Labeling the lines in vibration–rotation spectrum is based on the

branch, and on the value of J in the lower state (J):

....

)1(

)0(

)( R

R

JR and

....

)2(

)1(

)( P

P

JP

g. Experimentally the lines are not equally spaced:

□ On the P branch they get farther apart from each other as the

quantum number increases.

□ On the R branch they get closer to each other as the quantum

number increases.

h. Vibration-rotation interaction explains the unequal spacing between

lines in vibration-rotation spectrum.

i. Because Re vary (increases) slightly with n, consider that rotational

constant B is dependent on the vibrational state n:

)1(~~~

21

, JJBnE nJn

j. The dependence of B~

on n is called vibration-rotation interaction.

k. The frequencies for the lines in the vibration-rotation spectrum

considering the vibration-rotation interaction: 2

01011,01,1 )~~

()~~

3(~

2~~ JBBJBBBEE JJR

20101,01,1 )

~~()

~~(~~ JBBJBBEE JJP

l. The dependence of nB~

on n is given by:

21

ee~~~

nBBn

m. One can see that if 01~~BB :

□ The spacing between P branch lines increase with J.

□ The spacing between R branch lines decrease with J.

CHEM 3720

187

n. The energies predicted by the nonrigid rotator-anharmonic oscillator

model including the vibration-rotation interaction:

)1(~)1(~

)1(~

~~~~

21

e22

e

2

21

ee21

e,

JJnJJDJJB

nxnE Jn

□ The absorption in IR region of spectrum

1

3,2,1

J

n.

□ Neglecting the 22 )1(~

JJD term:

2eeeee0

,01,1obs

~)~4~

2()~3~

2(~

~~)1(~

JJBB

EEJ JJ

2eee0

,01,1obs

~)~2~

2(~

~~)1(~

JJB

EEJ JJ

where eee0~~2~~ x

□ Introducing a new quantity m where m = J + 1 for the R branch and

m = –J for the P branch. 2

eee0~)~2

~2(~)(~ mmBm

□ Considering the 22 )1(~

JJD term:

32eee0 4~)~2

~2(~)(~ DmmmBm

4. Electronic spectra

a. Molecules can undergo transitions

between electronic levels and the

electronic transitions are accompanied

by both rotational and vibrational

transitions.

b. According to BO approximation, the

electronic energy is independent of the

vibrational-rotational energy.

CHEM 3720

188

c. Using anharmonic oscillator-nonrigid rotator approximation but

excluding translational energy, the total energy is:

222

21

ee21

eel

eltotal

1~

1~~~~~

)()(~

JJDJJBnxn

JFnGE

□ el~ is the energy (in wavenumbers) at the minimum of the

electronic potential energy curve.

d. Ignore the rotational levels from the discussion (and formula) because

their energies are much smaller.

e. All vibrational transitions in electronic spectra (vibronic transitions)

are allowed.

f. Usually the transitions originate from the ground state electronic state

with n = 0 vibrational state.

1~~~~~~~~~~~eeee4

1e2

1ee4

1e2

1eobs nnxnxxT e

□ e~T is the difference in minimum of the two electronic potential

energy curves.

□ notation is used for upper electronic energy state, and notation

is used for lower electronic energy state.

g. Look at a case of two electronic states and

use Morse curves to represent these curves:

2)(e

e1)(RR

eDRV

h. Define dissociation energies:

□ De is the difference between the

minimum of the potential energy

curve and the dissociated atoms.

□ D0 is the difference between the

ground vibrational level and the

dissociated atoms.

CHEM 3720

189

i. For a harmonic oscillator: hDD21

0e

j. For an anharmonic oscillator: )( ee21

e21

0e xhDD

k. Define 0 0 vibronic transition as the transition between the zero-

point energies of upper and lower electronic states.

ee41

e21

ee41

e21

0,0~~~~~~~~ xxTe

)1(~~~~~eee0,0obs nnxn where ...2,1,0n

l. The spectrum look like a bunch of lines called n progression.

m. It can be used to get information about vibrational parameters of

excited electronic states.

n. The example of I2 molecule:

CHEM 3720

190

5. Franck-Condon principle

a. Franck-Condon principle says that the electronic transitions “are

vertical”.

□ Because the motion of the electrons

is almost instantaneous relative to

the motion of nuclei, when the

molecule makes a transition from

one electronic state to another by an

electron transition, the nuclei do not

move appreciably during the

transition.

b. The Franck-Condon principle allows

estimation of relative intensity of vibronic

transitions.

□ The intensity is proportional to the overlap between the wave

function in the upper and lower vibronic states.

c. Sometimes if the excited state potential curve is displaced to sufficient

larger internuclear distance than at the ground state potential curve,

then the absorption spectra may not contain a line corresponding to

0,0~ .

d. A detailed analysis of the intensities of such vibronic transitions yields

more information about the potential energy curves (that cannot be

found in vibrational or rotational spectra).

CHEM 3720

191

C. Polyatomic Molecules

1. Rotational spectra

a. A rigid body is characterized by its principal moments of inertia

defined as follows (the moments of inertia are defined about 3 arbitrary

axes):

N

jjjjxx zzyymI

1

2cm

2cm )()(

N

jjjjyy zzxxmI

1

2cm

2cm )()(

N

jjjjzz yyxxmI

1

2cm

2cm )()(

□ cm stands for the center of mass of the molecule (body)

□ xj, yj, and zj are the coordinates of atom j

□ N is the number of atoms in the molecule

b. There will be other terms of form

M

jjjjxy yyxxmI

1cmcm ))((

c. Choose 3 axis (X, Y, Z) called principal axes so the Ixy, Ixz, and Iyz terms

vanish.

d. In this case, Ixx, Iyy, and Izz are called principal moments of inertia.

e. Principal axes are easy to find for molecules with some degree of

symmetry.

□ For planar molecules, one principal axis is perpendicular per

molecular plane.

□ For CH3Cl, one principal axis is the C3 axis.

CHEM 3720

192

f. We do not need to know how to find those axes (they are tabulated) but

we will work with the rotational constants (in cm–1):

A28

~

cI

hA

,

B28

~

cI

hB

, and

C28

~

cI

hC

□ CBA~~~

because CBA III

g. Based on the relative magnitudes of the three principal moments of

inertia of a rigid body, they can be divided in:

□ linear rotor or top (one moment of inertia is zero)

○ Examples: CO2, C2H2

□ spherical rotor or top (all three

principal moments of inertia are

equal)

○ Examples: CH4, SF6

□ symmetric rotor or top (two

principal moments of inertia are

equal)

○ Examples: NH3, C6H6

○ Molecules with Cn (n 3) are at least a symmetric top.

□ asymmetric rotor or top (none of the principal moments of inertia

are equal)

○ Examples: H2O

h. Spherical top

□ The problem can be solved exactly and the energy levels are:

)1(~

)( JJBJF where J = 0,1,2,…

□ The degeneracy of J level is: 2)1(2 JgJ

i. Oblate symmetric top

□ The unique moment of inertia is larger than the other two.

zzyyxx III ( xxII A , yyII B , zzII C )

CHEM 3720

193

□ Examples: BCl3, C6H6

□ Energy levels are given by 2)~~

()1(~

),( KBCJJBKJF

□ Degeneracy is 12 JgJK , ,...2,1,0J , ,...2,1,0 K

□ The quantum number J is a measure of total rotational angular

momentum and K is a measure of the component of rotational

angular momentum along the unique axis of symmetry.

j. Prolate symmetric top

□ The unique moment of inertia is smaller than the other two.

zzyyxx III ( xxII C , yyII B , zzII A )

□ Examples: CH3I

□ Energy levels are given by 2)~~

()1(~

),( KBAJJBKJF

□ Degeneracy is 12 JgJK , ,...2,1,0J , ,...2,1,0 K

k. Selection rules:

0for0and1

0for0and1,0

KKJ

KKJ

□ If J = +1 (absorption) and K = 0 )1(~

2~ JB

l. Polyatomic molecules are less rigid than diatomics so centrifugal

distortion effects are more pronounced.

□ One needs to apply nonrigid rotator model.

□ The formulas become more complicated.

m. Asymmetric top

□ The formulas are more complicated.

CHEM 3720

194

2. Vibrational spectra

a. A N-atom molecule has 3N degrees of freedom (3 for each nucleus).

Linear Nonlinear

Translational 3 3

Rotational 2 3

Vibrational 3N – 5 3N – 6

Total 3N 3N

b. Examples:

□ The number of vibrational degree of freedom for acetylene is 7.

□ The number of vibrational degree of freedom for water is 3.

c. The potential energy at other geometry than the (minimum energy)

equilibrium geometry can be approximated using the displacements

(vib

,...,, 21 Nqqq ) from equilibrium position and a multidimensional

Taylor expansion:

vib vib

vib1 1

21 ...2

1)0,...,0,0(),...,,(

N

i

N

j

jiijN qqfVqqqVV

jiij

qq

Vf

2

are the second-order derivatives of V with respect to the coordinates.

d. A new set of coordinates {Qj} called normal modes or normal

coordinates can be found such that:

vib

1

2

2

1 N

jjjQFV

e. A normal mode is a synchronous motion of atoms or group of atoms

that does not change the center of mass of the molecule.

f. Using these normal mode coordinates, the vibrational Hamiltonian

operator becomes a sum of independent terms so the total wave

function (describing the nuclear motion) is a product of individual

wave functions and the energy is a sum of individual energies:

CHEM 3720

195

vibvib

1

22

22

1,vibvib

2

1

2ˆˆ

N

jjj

jj

N

jj QF

dQ

dHH

)()...()(),...,,(vibvibvib vib,2vib,21vib,121vib NNN QQQQQQ

vib

121

vib

N

jjj nhE where each nj = 0,1,2,…

g. The vibrational motion of a polyatomic molecule appears as Nvib

independent (harmonic or anharmonic) oscillators.

h. Being independent, one normal mode may be excited without leading

to the excitation of any other mode.

i. Without degeneracies, there will be Nvib characteristic fundamental

frequencies j.

j. A normal mode can be classified as:

□ stretching when it involves a change in a bond distance

□ bending when it involves a change in a bond angle

□ torsion when it involves a change in a bond dihedral angle

k. Stretching modes can be:

□ symmetric (when the symmetry of the molecule is conserved)

□ antisymmetric (when the symmetry is not conserved)

l. Examples: CO2 H2O

CHEM 3720

196

m. The case of CO2:

□ Asymmetric stretch (2349 cm–1):

○ The dipole moment oscillates parallel to molecular axis.

○ Such band is called a parallel band.

○ Selection rules: n = 1 and J = 1 leading to P and R

branches.

□ Symmetric stretch (1388 cm–1):

○ The dipole moment remains zero.

□ Two degenerate bending modes (667 cm–1):

○ The dipole moment oscillates perpendicular to molecular axis.

○ Such band is called a perpendicular band.

○ Selection rules: n = 1 and J = 0,1 leading to P, Q and R

branches.

n. Usually, the asymmetric stretch appears at higher frequency than the

symmetric stretch.

o. If the polyatomic molecule belongs to a point group, each normal mode

belongs to an irreducible representation of that point group.

□ This is a result of the fact that the vibrational properties of a

molecule change under any symmetry operation.

p. Example: H2O

□ Symmetric stretch (Qss) belongs to A1.

ssssˆ QQE ; ssss2

ˆ QQC

ssssˆ QQv ; ssssˆ QQv

□ Asymmetric stretch (Qas) belongs to B2.

asasˆ QQE ; asas2

ˆ QQC

asasˆ QQv ; asasˆ QQv

□ The bending mode belongs to A1.

O

HH

O

HH

CHEM 3720

197

q. When investigating the normal mode activity, the question is: Which

normal modes are active? or Which normal mode appears in the

vibrational absorption spectra?

r. The vibrations (normal modes) are infrared active only if the dipole

moment of the molecule changes when the atoms are displaced relative

to one another.

□ All the water normal modes and two of the CO2 are IR active.

□ One normal mode of CO2 (symmetric stretch) is IR inactive.

□ Homonuclear diatomics are IR inactive.

□ Heteronuclear diatomics are IR active.

3. Selection rules for transitions between two states

a. The probability of transition between 2 states (determined from

perturbation theory) is proportional to the transition dipole moment:

dzz 1*212

□ This is the z component of the transition dipole moment and there

are equivalent components for x and y.

b. If the transition dipole moment is zero the transition is not observed

(or not allowed or forbidden).

c. If the transition dipole moment is not zero the transition is observed

(or allowed).

d. If dipole moment is zero then transition dipole moment is zero.

e. Selection rules for rigid rotator can be deduced from this condition, and

they are:

)0or(0 and 1 KMJ

f. Selection rules for harmonic oscillator can be deduced from this

condition, and they are:

0 and 1

0

dq

dn

CHEM 3720

198

4. A general scheme for determining the IR/Raman activity of normal modes

a. Determine the reducible representation for the nuclear motion

□ Place the 3 coordinate axes on each atom.

□ Do each symmetry operation, write the result in a form of a matrix

of transformation, and determine its character.

○ Example of water:

O

HH

E 2C v v ˆ

N3 9 –1 1 3

□ An equivalent and probably simpler way is to count +1 for every

coordinate that does not change the place and the sign, –1 for every

coordinate that does not change the place but changes the sign, and

0 for every coordinate that changes the place.

□ In general, the entries for the various rotation axes can be deduced

from the rotation matrix.

○ For a rotation with degrees (C360/), the rotation matrix is

given by

z

y

x

z

y

x

100

0cossin

0sincos

.

○ For an improper rotation with degrees (S360/), the rotation

matrix is given by

z

y

x

z

y

x

100

0cossin

0sincos

.

□ One can also use the following table with contributions that each

unmoved atom makes to the character of the 3N-dimensional

CHEM 3720

199

representation obtained by operating on arbitrary (3-dimensional)

vectors attached to each of the N atoms in the molecule:

R E 2C i 233

ˆ,ˆ CC 344

ˆ,ˆ CC 566

ˆ,ˆ CC 2S 233

ˆ,ˆ SS 344

ˆ,ˆ SS 566 ,ˆ SS

Contribution

to )ˆ(R 3 1 –1 –3 0 1 2 –3 –2 –1 0

b. Reduce 3N to a sum of irreducible representations

□ Example of water:

v2C E 2C v v ˆ

1A 1 1 1 1 z 2x , 2y , 2z

2A 1 1 –1 –1 zR xy

1B 1 –1 1 –1 x, yR xz

2B 1 –1 –1 1 y, xR yz

N3 9 –1 1 3

313111)1(194

11

Aa

1)1(3)1(11)1(194

12

Aa

2)1(311)1()1(194

11

Ba

313)1(1)1()1(194

12

Ba

21213 323 BBAAN

c. From 3N, subtract the irreducible representations of the 3 translational

degrees of freedom denoted Tx, Ty, and Tz (same as x, y, and z) and the

irreducible representations of the 2 or 3 rotational degrees of freedom

denoted Rx, Ry, and Rz.

21vib 2 BA

□ This gives the types of irreducible representations that are

associated with vibration in the molecule or that are associated

with each normal mode.

CHEM 3720

200

d. Determining the IR activity

□ Look at the transition dipole moment for n = 0 n = 1 transition.

vibvibvib

....,...,,,...,, 2121121010 NN

z

y

x

N dQdQdQQQQQQQI

□ This integral is (should be) invariant to symmetry operations.

1010 )ˆ()ˆ()ˆ(ˆ,,1 IRRRIR

jzyx QA

○ 1)ˆ()ˆ()ˆ(,,1

RRRjzyx QA for the transition to be

allowed.

○ The product should transform as the totally symmetric

representation (A1, Ag, etc)

□ Example of water (C2v point group):

Initial state Dipole

moment

Final State

(normal mode)

Result

A1

x B1

y B2

z A1

A1

x B1

y B2

z A1

A1

x B1

y B2

z A1

B2

x A2

y A1

z B2

All normal modes of water are IR active.

e. Determining the Raman activity

□ Similar to IR activity but instead of the dipole moment in the

integral one uses polarizability that appear as a product of two

coordinates (x2, y2, etc)

□ Transition is Raman active if )ˆ()ˆ()ˆ(1

RRRjQA is equal to 1

for all R .

CHEM 3720

201

□ Example of water (C2v point group):

Initial state Polarizability Final State

(normal mode)

Result

A1

x2 A1

y2 A1

z2 A1

xy A2

xz B1

yz B2

A1

x2 A1

y2 A1

z2 A1

xy A2

xz B1

yz B2

A1

x2 A1

y2 A1

z2 A1

xy A2

xz B1

yz B2

B2

x2 B2

y2 B2

z2 B2

xy B1

xz A2

yz A1

All normal modes of water are Raman active.

f. Mutual Exclusion Rule: For molecules with a center of inversion, the

normal modes are active in either IR or Raman but not in both IR and

Raman.

□ Examples: CO2, trans-C2H2Cl2

g. IR/Raman activity of CCl4 normal modes

dT E 8 3C 3 2C 6 4S 6 d

1A 1 1 1 1 1 2x + 2y + 2z

2A 1 1 1 –1 –1

E 2 –1 2 0 0 ( 2222 yxz , 22 yx )

1T 3 0 –1 1 –1 ( xR , yR , zR )

2T 3 0 –1 –1 1 (x, y, z) ( xy , xz , yz )

N3 15 0 –1 –1 3

□ Determine the reducible

representation for nuclear

motion 3N

C

Cl Cl

Cl Cl

CHEM 3720

202

□ Reduce 3N to irreducible representations:

11361)1(61)1(310811524

11

Aa

0)1(36)1()1(61)1(310811524

12

Aa

10360)1(62)1(3)1(0821524

1Ea

1)1(361)1(6)1()1(300831524

11

Ta

3136)1()1(6)1()1(310831524

12

Ta

2113 3TTEAN

□ Determine the vibrational contribution to 3N

2trans T ; 1rot T 21vib 2TEA

○ There are 4 distinct vibrational frequencies where three of

them are degenerate: one doubly degenerate and two triply

degenerate.

C

Cl Cl

Cl Cl

C

Cl Cl

Cl Cl

C

Cl Cl

Cl Cl

C

Cl Cl

Cl Cl

□ Determine the IR activity:

○ The normal mode of A1 type is not active.

○ The normal mode of E type is not active.

○ The normal modes of T2 type are active.

□ Determine the Raman activity:

○ The normal mode of A1 type is active.

○ The normal mode of E type is active.

○ The normal modes of T2 type are active.

□ Only totally symmetric vibrations give rise to polarized lines (in

which the incident polarization is preserved).

CHEM 3720

203

5. Electronic spectroscopy

a. Once excited in an electronically excited state, a molecule does not

remain in that excited state but it decays back to the ground state.

b. There are few possible deexcitation processes.

c. A pictorial representation of the

pathways that electronically excited

(diatomic) molecules decay to their

ground state.

□ S0 is the singlet ground state, S1

is the first excited singlet state,

and T1 is the first excited triplet

state.

□ Assume that T1 state is lower in

energy that S1 state.

□ Assume also that

)()()( 1e1e0e TRSRSR .

d. There can be two types of transitions:

□ Radiative that are transitions between energy levels with

absorption or emission of radiation.

□ Nonradiative that are transitions between energy levels without

absorption or emission of radiation.

e. The activation/deactivation processes and their time scales:

□ Absorption from S0 to S1

□ Fluorescence = a radiative transition from S1 to S0

○ Timescale: 10–9 s

□ Internal Conversion = nonradiative transition from S1 to S0 due to

collisions with other molecules

○ Timescale: 10–7 – 10–12 s

CHEM 3720

204

□ Intersystem Crossing = nonradiative transition from S1 to T1

○ The spin state changes so this is typically a slower process.

○ Timescale: 10–6 – 10–12 s

□ Phosphorescence = a radiative transition from T1 to S0

○ Timescale: 10–5 – 10–7 s

□ Intersystem Crossing = nonradiative transition from T1 to S0

○ Timescale: 10–3 – 10–8 s

□ Vibrational Relaxation = nonradiative transition due to collisions

with other molecules

○ This is a very fast process.

○ Timescale: 10–14 s

f. Absorption and fluorescence spectroscopy

□ Consider a diatomic molecule with )()( 1e0e SRSR

□ Consider the vibrational levels of the ground electronic state

(denoted by n) and the excited electronic state (denoted by n)

□ Absorption spectrum

○ It is a series of lines reflecting transitions from n = 0 of the

ground electronic state to n = 0,1,2,… of the excited

electronic state.

○ The spacing between lines measures the energy gap between

vibrational states of the excited state.

□ Fluorescence spectrum

○ It is a series of lines reflecting transitions from n = 0 of the

excited electronic state to n = 0,1,2,… of the ground state

electronic state.

○ The spacing between lines measures the energy gap between

vibrational states of the ground state.

CHEM 3720

205

□ The transition from n = 0 to n = 0, called 0–0 transition, appear

in both absorption and fluorescence spectra.

□ The relative intensities of the absorption and emission lines are

determined by Frank-Condon principle.

□ Fluorescence spectrum appears at higher (or lower ) than the

absorption spectrum.

g. Phosphorescence typically occurs at lower energy (or frequency) than

fluorescence.

CHEM 3720

206

D. Lasers

1. Introduction

a. LASER stands for Light Amplification by Stimulated Emission of

Radiation.

b. Lasers are devices that exploit the amplification of light though

stimulated emission.

2. Principle of operation

a. Consider 2 energy levels and, for simplicity, assume no degeneracy.

□ Example: two electronic levels in an atom so there are no

vibrational and rotational contributions.

E1

E2

N1

N2

E1

E2

N1

N2

b. Consider that E1 and E2 are the ground and excited state energies.

c. Define N1 and N2 to be the number of atoms in each state so

Ntotal = N1 + N2.

d. Assume E2 – E1 >> kT (average energy at temperature T) so all atoms

are in state 1.

e. Define radiant energy density

□ radiant energy per unit volume: units of Jm–3

f. Define spectral radiant energy density

d

d

□ radiant energy density per unit frequency: units of Jsm–3

g. The rate of excitation from ground electronic state to excited electronic

state (if no emission):

dt

tdNtNB

dt

tdN )()()(

)( 211212

1

○ B12 is called Einstein coefficient.

CHEM 3720

207

h. Einstein proposed 2 pathways to treat relaxation back to ground state:

□ Spontaneous emission

)()(

2212 tNA

dt

tdN (if no other process)

○ A21 is an Einstein coefficient.

○ R21

1

A is the fluorescence lifetime (or radiative lifetime).

□ Stimulated emission (proportional to (12)

)()()(

212212 tNB

dt

tdN

○ This pathway amplifies the light intensity.

○ B21 is an Einstein coefficient.

i. If no other processes then:

)()()()()()()(

212212211121212 tNBtNAtNBdt

tdN

dt

tdN

j. At equilibrium:

0)()( 21

dt

tdN

dt

tdN

2112

2112

1

2)(

BB

A

N

Nv

□ But Tkh

eN

NB12 /

1

2 and

1

8)(

B12 /

312

312

Tkh

ec

h

(blackbody radiation formula)

2112 BB and 213

312

21

8B

c

hA

□ Einstein’s coefficients are related.

k. Consider the case of only 2 levels:

□ One can achieve stimulated emission if the stimulated emission is

greater than the rate of absorption.

1121221221 )()( NBNB 12 NN

CHEM 3720

208

○ The population in the excited state is

greater than in the lower state.

○ This is called population inversion,

and it is a non-equilibrium situation.

□ It can be shown that:

2

1

)(2

)(

12

12

21

2

total

)(2

BA

B

NN

N

N

N t

because 0A

□ The population inversion cannot occur in a two level system.

l. Consider the case of 3 levels:

□ A light beam h31 called the pump source is used to create excited-

state populations.

○ h31 is the pump light.

□ The rates of spontaneous emission from state 3 to state 2 and 1

(i.e., A31 and A21) can be different.

□ The total number of molecules )()()( 321total tNtNtNN .

□ At equilibrium: )(

)(

323232

323221

2

3

BA

BA

N

N

.

□ A population inversion 12

3 N

N can be obtained if 3221 AA .

□ Such a system is called a gain medium.

CHEM 3720

209

3. Laser components

a. Gain medium that amplifies light of desired wavelength

□ solid-state material

□ liquid solution

□ gas mixture

b. Pumping source that excites the gain medium

□ optical excitations (lasers convert a max of 50-70% of incident

light)

□ electrical excitations (more common for gas lasers)

c. Mirrors that direct the light beam back and forth through gain medium

□ One is 100 % reflective and the other is < 100 % reflective.

□ They create a laser cavity (resonator) without which the laser

would not work.

4. Laser properties and uses

a. Lasers can be continuous or pulsed.

b. Another important property of the laser light is that the light is

monochromatic (single color-wavelength) and that the light waves are

all in phase.

□ This property is called coherence.

□ This is due to the fact that stimulated

emission has the incident light wave

and stimulated light wave have the

same phase.

CHEM 3720

210

c. Lasers are used:

□ In everyday life, medicine, military (targeting), chemistry

□ In spectroscopy: the use of high-resolution lasers can resolve

absorption lines that cannot be distinguished by conventional

spectrometers

□ In photochemistry (the field of light-initiated reactions)

□ For studying chemical reactions with high spectral and time

resolution (order of 1510 seconds = femtoseconds)

5. Laser examples

a. Ruby laser

□ It was the first used laser.

□ Ruby is a crystal of Al2O3 with some Cr3+

impurities.

□ Electronic energy levels of Cr3+ in Al2O3 are

suitable for achieving a population inversion.

□ It is a pulsed laser that works at 694.3 nm.

b. Yttrium-aluminum garnet (YAG) laser

□ Y3Al5O15 (YAG) impurities with Nd3+

that is the active ion.

□ Can be both pulsed and continuous

lasers; work at 1064.1 nm.

□ Other lasers based on Nd can have

similar wavelength:

○ 1054.3 nm for Y3LixFy (YLF).

○ 1059 nm for Nd3+ in glass

CHEM 3720

211

c. N2–CO2 laser (gas lasers)

□ Population inversion can be achieved is some vibrational levels of

CO2 in ground electronic state (closer to each other than N2).

□ Continuous laser that can become tunable lasers through the

rotational levels.

d. He–Ne laser (gas-phase laser)

□ Contains a gas mixture in a glass cell.

□ Typical gas pressures 1.0 torr for He and 0.1 torr for Ne.

□ Continuous laser that can work at 4 different

wavelengths: 3391.3, 1152.3, 632.8, and

543.5 nm.

□ Excitation occurs through electrical

discharge that produces excited He atoms.

□ This is followed by a nonradiative energy

transfer through collisions from He to Ne.

□ The lifetimes of these excited states of Ne

are such that population inversion can be

achieved.

□ Several transitions are used to generate laser light. (There are few

Ne transitions that have been observed to lase.)

e. Chemical laser

□ It is based on the HHFHF *2 reaction.

□ F is formed by electric discharge in SF6 + H reaction.

□ The HF is formed in a vibrational excited state (so population

inversion is achieved).

□ Two transitions can lase.

CHEM 3720

212

E. Unit Review

1. Important Terminology

translational, rotational, vibrational degrees of freedom

rotational term

nonrigid rotator model

centrifugal distortion constant

vibrational term

overtones

anharmonic oscillator

anharmonicity constant

R , P, and Q branches

vibration-rotation interaction

dissociation energies: De and D0

n progression

Franck-Condon principle

CHEM 3720

213

principal axes

principal moments of inertia

linear, spherical, symmetric, or asymmetric rotor (or top)

oblate/prolate symmetric top

centrifugal distortion

normal modes or normal coordinates

stretching (symmetric/antisymmetric), bending, and torsion

parallel/perpendicular band

IR/Raman activity

transition dipole moment

deexcitation processes

radiative/nonradiative transitions

fluorescence

internal conversion

CHEM 3720

214

intersystem crossing

phosphorescence

vibrational relaxation

0–0 transition

LASER

stimulated emission

spontaneous emission

Einstein coefficient

fluorescence lifetime

population inversion

pump light

gain medium

pumping source

coherence

CHEM 3720

215

2. Important Formulas 22 )1(

~)1(

~)( JJDJJBJF

3)1(~

4)1(~

2)()1(~ JDJBJFJFv

...~~~)(2

21

ee21

e nxnnG

)1(~~~)0()(~eeeobs nnxnGnG

)1(~~)()(

~21

, JJBnJFnGE Jn

1~2~~~

)1(~,1,1obs JBEEJ JnJn

JBJ~

2~1~obs

)1(~~~

21

, JJBnE nJn

201011,01,1 )

~~()

~~3(

~2~~ JBBJBBBEE JJR

20101,01,1 )

~~()

~~(~~ JBBJBBEE JJP

21

ee~~~

nBBn

)1(~)1(~

)1(~

~~~~

21

e22

e

2

21

ee21

e,

JJnJJDJJB

nxnE Jn

32eee0 4~)~2

~2(~)(~ DmmmBm

222

21

ee21

eel

eltotal

1~

1~~~~~

)()(~

JJDJJBnxn

JFnGE

1~~~~~~~~~~~eeee4

1e2

1ee4

1e2

1eobs nnxnxxT e

ee41

e21

ee41

e21

0,0~~~~~~~~ xxTe

)1(~~~~~eee0,0obs nnxn

A28

~

cI

hA

,

B28

~

cI

hB

,

C28

~

cI

hC

CBA~~~

CHEM 3720

216

vib

1

2

2

1 N

jjjQFV

dzz 1*212

dt

tdNtNB

dt

tdN )()()(

)( 211212

1

)()(

2212 tNA

dt

tdN

)()()(

212212 tNB

dt

tdN

2112 BB

213

312

21

8B

c

hA

1121221221 )()( NBNB

)(2

)(

12

12

21

2

BA

B

NN

N

)(

)(

323232

323221

2

3

BA

BA

N

N

CHEM 3720

217

Unit XIII

Statistical Thermodynamics

A. Concepts of Statistical Thermodynamics

1. Introduction

a. Statistical thermodynamics is the link between molecular properties

(i.e., molecular energy levels) and the bulk thermodynamic properties

(i.e., properties of matter in bulk which deals with the average behavior

of a large number of molecules).

2. The Boltzmann distribution

a. Consider a macroscopic system that can be described by specifying:

□ the number of particles (N)

□ the volume of the system (V)

□ the forces between particles

b. In principle, the energy of a N-body system could be obtained by

solving the Schrödinger equation.

□ The energy will be written as Ej(N,V).

□ It depends on the number of particles and the volume.

□ j is a index of the various states of the system.

c. For the case of an ideal gas, the total energy is a sum of energies of

individual molecules:

Nj VNE ....),( 21

□ The energy of individual molecules is then a sum of electronic,

vibrational, rotational, and translational.

□ For a more general case when the molecules of the system interact

with each other, the energy Ej(N,V) cannot be written as a sum of

individual particles energy.

CHEM 3720

218

□ We will though still consider, for the purpose of discussion, the

energy of the system as a sum of individual molecular energies.

d. Deducing the Boltzmann distribution

□ Consider a collection of systems in thermal equilibrium with each

other.

○ This collection of macroscopic systems in thermal equilibrium

with a heat reservoir is called an ensemble.

○ More specifically, if N, V, and T are common, the ensemble is

called a canonical ensemble.

□ Determine the probability that a system will be in the state j having

energy Ej(N,V)

– The number of systems in state j is aj and the total number of systems is A.

– Determine the relative number of systems found in each state:

– Consider 2 states with energies 1E and 2E

)(),( 21211

2 EEfEEfa

a

where f(E1,E2) is a functional that should be determined and the difference (E1 –

E2) is used so that any arbitrary zero of energy will be canceled

– Consider a third state:

)()(

)(

)(

31122

3

311

3

122

1

EEfEEfa

a

EEfa

a

EEfa

a

)( 322

3 EEfa

a

– This is similar to Eyxyx eEfeee )(

)(

m

n nm EEe

a

a

jEj Cea

– Two constants (C and ) should be determined.

– Determine C by summing over all the number of systems:

j j

Ej AeCa j

j

E je

AC

CHEM 3720

219

j

E

E

jj

j

e

eAa

– Define A

a j as the fraction of systems in the ensemble that will be found in the state

j with energy Ej:

j

j

E

Ej

p

e

e

A

a

j

j

□ The aj/A fraction becomes, if the number of systems is very large,

the probability of a system to be in the state j with energy Ej(N,V),

denoted pj.

j

E

Ej

jj

j

e

e

A

ap

jEj ep

○ This is the Boltzmann distribution.

○ jEe

is called the Boltzmann factor.

□ The denominator is denoted by Q, and is a very important quantity

called the partition function:

j

VNE jeVNQ),(

),,(

□ It can be shown that TkB

1 where kB is the Boltzmann constant:

j

TkVNE jeTVNQ B/),(),,(

),,(

B/),(

TVNQ

ep

TkVNE

j

j

CHEM 3720

220

3. Correlations between the partition function and physical observables

a. Fundamental postulate in Physical Chemistry: Ensemble average of

any quantity, calculated using the probability distribution, is the same

as experimentally observed value.

b. The energy of a system

□ The observed energy of the system that is equal with the average

ensemble energy <E>:

j j

VNNj

jjVNQ

eVNEVNEVNpE

j

),,(

),(),(),,(

),(

j

VNEj

VN

VNE

VNQ

eVNEe

VNQ

VNQjj

),,(

),(

),,(

1),,(ln),(

,

),(

VN

QE

,

ln

– Considering that T

TkTk

2

BB

1

VNT

QTkE

,

2B

ln

VNVN T

QTk

QE

,

2B

,

lnln

□ A proton in a magnetic field Bz:

zBE 2

1 ;

TkBTkBz

zz eeBTQ BB 2/2/),(

TkBTkB

TkBTkBz

zz

zz

ee

eeBE

BB

BB

2/2/

2/2/

2

0as

2

as0

TB

T

E z

□ The monoatomic ideal gas:

!

),(),,(

N

VqVNQ

N and V

h

mVq

2/3

2

2),(

!lnln2

ln2

3ln

2

3!lnlnln

2NVN

h

mNNNqNQ

CHEM 3720

221

TNkQ

VNB

,2

3ln

nRTTNkE

2

3

2

3B ( ANnN )

RTUE2

3 (for 1 mole of gas)

○ Use U (i.e., internal energy) for the experimentally observed

energy of the system.

○ Use overbar notation to indicate a molar quantity.

□ Diatomic ideal gas (with rigid rotator-harmonic oscillator

approximation):

2/

2/

2

22/3

2 1

82),(

h

h

e

e

h

IV

h

mVq

on dependingnot terms)1ln(2

lnln2

3ln heN

hNN

NQ

h

h

e

eNhNhNNQ

122

3ln

h

h

e

eNhNhTNkEU

122

3B

h

hAA

e

ehNhNRTRTU

122

3

c. The heat capacity of a system

□ Determine the constant-volume heat capacity CV:

VNVNV

T

U

T

EC

,,

□ Example: The monoatomic gas:

RTUE2

3 RCV

2

3

□ Example: Diatomic gas:

2/

/2

B )1(2

5

12

5

B

B

Tkh

Tkh

h

h

AVe

e

Tk

hRR

e

e

ThNRC

□ Einstein model for atomic crystals (1905):

○ It treats the atoms in the crystal as 3-dimensional harmonic

oscillators vibrating with the same frequency.

CHEM 3720

222

N

h

hU

e

eeQ

32/

1

0

U0 is the sublimation energy at 0 K.

VN

QEU

,

ln

h

h

e

eNhNhUU

1

3

2

30

2/

/2

B )1(3

B

B

Tkh

Tkh

Ve

e

Tk

hRC

0as0

as3

T

TR

3R = 24.9 kJ/molK (giving the same result as Dulong-Petit rule)

This model was the first model to explained CV at low temperatures.

d. The pressure of a system

N

jj

V

EVNP

),(

j

jj VNPVNpP ),(),,(

PV

QTk

V

Q

VNQ

TkP

NN

,B

,

B ln

),,(

○ <P> is the ensemble average pressure.

○ P is the observed (experimental) pressure.

□ Example: The monoatomic ideal gas:

!

),(),,(

N

VqVNQ

N and V

h

mVq

2/3

2

2),(

!lnln2

ln2

3!lnlnln

2NVN

h

mNNqNQ

V

N

V

Q

ln

V

nRT

V

TNkP B (ideal gas law)

4. The relation between the partition function of a system and that of

individual molecules

a. Assume that the total energy of the system is the sum of individual

energies of particles in the system.

CHEM 3720

223

□ It is impossible to obtain the energies (or eigenvalues) by solving

the Schrödinger equation.

b. The partition function of a system of N independent and

distinguishable particles

□ An example is the particles in a crystal that can be distinguished by

their position.

□ Denote the individual particle energies as {ja} where j is the

energy state of the particle and a is the label of the particle.

□ The total energy of the system:

...)()()(),( VVVVNE ck

bj

ail

□ The partition function of the system:

)...,(),(),(

...

),,(,...,,

...)(

TVqTVqTVq

eee

eeTVNQ

cba

kji

kjil

E

ck

bj

ai

ck

bj

ail

i

Tk

ia

ii eeTVq B/),(

j

Tk

jb

jj eeTVq B/),(

and so on.

□ Usually {i} are a set of molecular energies so qa(V,T) is called a

molecular partition function.

□ The partition function of a system of independent, distinguishable

molecules is the product of molecular partition functions.

□ The molecular partition function qa(V,T) can be determined based

on the energies of only individual atoms or molecules.

□ If these energies are the same:

NTVqTVNQ ),(),,(

CHEM 3720

224

□ Interpretation: The molecular partition function gives an indication

of the average number of states that are thermally accessible at the

temperature of the system.

T

Tgq

as

0as0

○ g0 is the degeneracy of the ground state (usually is 1).

c. The partition function of a system of N independent and

indistinguishable particles

□ The total energy of the system:

...,..,, kjikjiE

□ The partition function of the system:

,...,,

...)(),,(

kji

kjieTVNQ

□ The summation cannot be made after i, j, k, … separately because

the particles are indistinguishable.

□ The partition function for fermions

○ Fermions are particles of half-integer spin (with antisymmetric

wavefunctions under the interchanged of two identical

particles).

○ No two identical fermions can occupy the same single-particle

energy state.

○ There should not be same terms in the sum over j = no two or

more same indices.

○ Example: 2 identical fermions with 4 possible energies

16 total possibilities but only 6 are allowed:

{1 + 2}; {1 + 3}; {1 + 4}; {2 + 3}; {2 + 4}; {3 + 4}

There are (42 – 4)/2! = 6 terms (almost 42/2! = 8).

CHEM 3720

225

□ The partition function for bosons

○ Bosons are particles of integer spin (with symmetric

wavefunctions under the interchanged of two identical

particles).

○ There are no restrictions so there can be the same indices in

the sum over j.

○ Example of 3 (or N) particles: {3 + 2 + 2}, {2 + 3 + 2},

and {2 + 2 + 3} (i.e, N terms) are possible but should not be

considered because the particles are indistinguishable.

○ For a case of different energies: NE ...21 there

will be N! arrangements representing same state.

□ For both fermions and bosons, with the condition that the number

of quantum states available to any particle is much greater than the

number of particles, the partition function can be approximated as:

!

),(),,(

N

TVqTVNQ

N

where

j

Tk

j

jj eeTVq B/),(

□ For both fermions and bosons, the partition function will be more

difficult to calculate in situations of particles with same energies

(i.e, same indices).

□ For the above equation to be true and therefore useful there should

be not only a large number of energy states in a molecule (which

they are) but there should be a large number of them with energies

roughly less than kBT which is usually the average energy of a

molecule at temperature T.

○ The translational states are usually sufficient to guarantee that

the number of states is greater than the number of particles.

CHEM 3720

226

○ One criteria that can be applied to check this is:

18

2/3

B

2

Tmk

h

V

N

○ The criteria is favored by:

– larger particle mass

– higher temperature

– lower density

○ Examples: )(KT 2/3

B

2

8

Tmk

h

V

N

27 2101 liquid Ne

27 5108.7 gaseous Ne

100 5103.3 gaseous He

20 29.0 liquid hydrogen

4 5.1 liquid He

300 1400 electrons in metals (Na)

(The last three entries are exceptions and special methods are needed.)

□ If the number of available molecular states is much greater than the

number of particles, the particles are said to obey Boltzmann

statistics.

○ More valid with increasing T.

d. Conclusion: With the condition that the number of quantum states

available to any particle is much greater than the number of particles,

the relation between the partition function of a system and that of

individual indistinguishable molecules can be approximated as:

!

),(),,(

N

TVqTVNQ

N

where

j

Tk

j

jj eeTVq B/),(

CHEM 3720

227

5. The molecular partition function

a. The relation with the average energy:

N

e

eN

T

qTNk

T

QTkE

jTk

Tk

jVVN j

j

B

B

/

/2

B,

2B

lnln

j

Tk

Tk

jj

j

e

e

B

B

/

/

is the average energy of any one

molecule.

b. The probability that a molecule is in its jth molecular state:

j

Tk

TkTk

jj

jj

e

e

TVq

e

B

BB

/

//

),(

c. Separation of molecular partition function on partition functions for

each degree of freedom:

□ Considering the molecular energy to be a sum of distinguishable

energies: elecvibrottranslkji

elecvibrottrans),( qqqqTVq

i

Tkieq Btrans /

trans

;

j

Tkjeq Brot /

rot

k

Tkkeq Bvib /

vib

;

l

Tkleq Belec /

elec

d. A molecular partition function can be decomposed into partition

functions for each degree of freedom.

e. Probability that a molecule is in ith translational state, jth rotational

state, kth vibrational state, and lth electronic state:

elecvibrottrans

//// Belec

Bvib

Brot

Btrans

qqqq

eeeeTkTkTkTk

ijkl

lkji

CHEM 3720

228

f. Probability that a molecule is in kth vibrational state:

vib

/

/

/

elecvibrottrans

////

vib

Bvib

Bvib

Bvib

Belec

Bvib

Brot

Btrans

q

e

e

e

qqqq

eeee

Tk

k

Tk

Tk

k

TkTk

j

Tk

i

Tk

k

k

k

k

lkji

g. Average translational, rotational, and vibrational energies of a

molecule:

T

qTk

q

q

e

i

Tk

i

i

trans2

Btrans

trans

/trans lnlnB

trans

T

qTk

q

q

e

j

Tk

j

j

rot2

Brot

rot

/rot lnlnB

rot

T

qTk

q

q

e

k

Tk

k

k

vib2

Bvib

vib

/vib lnlnB

vib

h. The molecular partition function written over the number of levels:

i

Tki

j

Tkij egeTVq BB //

),(

□ The first summation is for all states and the second is for all levels.

□ gi is the degeneracy of level j (i.e., the number of states that have

the same energy).

□ Example: rigid rotor

)1(2

2

JJI

J

and 12 JgJ (degeneracy of each level)

0

2/)1(rot

B2

)12()(J

TIkJJeJTq

CHEM 3720

229

B. Statistical Thermodynamics of Ideal Gases

1. Monoatomic ideal gas

a. Only translational and electronic energies are available:

electransatomic

)(),(),( electrans TqTVqTVq

b. Translational partition function:

– Translational energy in a cube of length a:

222

2

2

8zyx nnn

ma

h

3

12

22

trans8

exp),(

n ma

nhTVq

ah

mdne

ma

nh manh

n

2/1

20

8/

12

22 2

8exp

222

because

2/1

04

2

dne n

Vh

TmkTVq

2/3

2B

trans2

),(

□ Average translational energy:

TkT

qTk

VB

trans2Btrans

2

3ln

c. Electronic partition function:

□ Summation over the levels:

221elec )( eei eggegTq ee

iei

by choosing e1 as the zero of energy.

○ It depends on T but not on V.

□ Typical values: small very is so K, 100at 10

Kcm 695.0,cm 000,40

10elec

11B

1elec

e

k

CHEM 3720

230

□ Example of F atom:

– The 1s22s22p5 configuration gives 2P3/2 and 2P1/2 terms with 2P3/2 being the ground

state and 2P1/2 being 404 cm–1 above it.

– The 1s22s22p43s1 configuration gives 4P5/2, 4P3/2,

4P1/2, 2P3/2, and 2P1/2 terms but these

terms are much higher in energy (~100,000 cm–1) so their contribution to the

electronic partition function is insignificant.

22 2424)(elecee eeTq

– The fraction of F atoms in 2P1/2 state: 2

2

24

22

e

e

e

ef

f2 = 0.0672 at 300 K, 0.219 at 1000 K, and 0.272 at 2000 K.

□ Degeneracy of an electronic state: 12 Jge

□ In most cases: 1elec )( egTq or 221elec )( eeggTq ee

d. Properties of monoatomic gas:

!

),,( electrans

N

qqTVNQ

N

elec

22

,

22

2

3ln

q

egNTk

T

QTkU

eee

BVN

B

RdT

UdC

VN

V2

3

,

(If no contributions from exited electronic states.)

V

TNk

V

QTkP

TN

B

,B

ln

2. Diatomic ideal gas

a. Translational, rotational, vibrational, and electronic energies are

available:

elecvibrottrans

elecvibrottrans),( qqqqTVq

b. Translational partition function:

Vh

TkmmTVq

2/3

2B21

trans)(2

),(

○ qtrans (V,T) = (1.436 1032 m–3)V for N2 at 300 K.

CHEM 3720

231

c. Electronic partition function:

□ Choose the zero of electronic energy to be the separated atoms at

the rest in their ground electronic states:

De = -e1

e2

0

De = -e1

e2

0

Tk

eTkD

eee egegTq B2B /

2/

1elec )(

d. Vibrational partition function:

□ Use harmonic-oscillator approximation, and use the zero of the

vibrational energy to be the bottom of the internuclear potential

well of the lowest electronic state.

hnn

2

1; ( ,...2,1,0n )

T

T

h

h

n

nhh

n

e

e

e

e

eeeTq n

/

2/2/

0

2/vib

vib

vib

11

)(

□ Introduce the characteristic vibrational temperature: B

vibk

h

○ This is the temperature at which the characteristic frequency

equals kBT and is around 1.44 times bigger than e~ (in cm–1).

□ Average vibrational energy:

12 /

vibvibBvib

vib Te

NkE

CHEM 3720

232

□ Vibrational contribution to molar heat capacity:

2/

/2vib

vib,)1( vib

vib

T

T

Ve

e

TRC

○ High temperature limit: TRCV asvib,

□ Fraction of molecules in vibrational state n: TnT

n eef// vibvib )1(

□ Fraction of molecules in all vibrational excited states: hT

n eef

/0

vib

○ For N2 at 300 K, f0 1 and f1 10–5.

○ For Br2 at 300 K, f0 0.8 and f1 0.2.

□ At room temperature, most molecules are in their ground

vibrational state.

e. Rotational partition function (heteronuclear diatomics):

□ Use rigid-rotator approximation, and use the zero of the rotational

energy to be the J = 0 state:

I

JJJ

2

)1(2

; ( ,...2,1,0J )

Degeneracy: 12 JgJ

J

TJJ

J

IJJ eJeJTq/)1(2/)1(

rotrot

2)12()12()(

0

/)1(rot

rot)12()( dJeJTqTJJ

)(8

)( rot2B

2

rotrot T

h

TIkTTq

34.0at2

1

vibvib,

TCV

CHEM 3720

233

□ Introduce the rotational temperature: BB

2

rot2 k

hB

Ik

□ The sum-to-integral transformation is appropriate when rot << T.

○ This is not true for H2 for which rot = 85 K.

□ Average molar rotational energy: TNkE Brot

□ Rotational contribution to molar heat capacity: RC otV r,

□ Fraction of molecules in rotational level J: TJJ

J eTJf/)1(

rotrot)/)(12(

□ At room temperature, most molecules are in their excited rotational

levels.

□ This is possible because there are more states (than 1) in each level

and the number of states within a level (i.e., the degeneracy)

increases with J.

□ Most probable value of J: 2/1)2/( 2/1rot TJmp

f. Rotational partition function (homonuclear diatomics):

□ Use rigid-rotator approximation, and use the zero of the rotational

energy to be the J = 0 state but additional criteria should be

included due to the required symmetry of the wavefunction.

□ If the nuclei are bosons the wavefunction should be symmetric

while if the nuclei are fermions the wavefunction should be

antisymmetric.

2B

2

rotrot

4

2)(

h

TIkTTq

; ( Trot )

○ Only some (i.e., half) rotational levels are populated.

CHEM 3720

234

g. Rotational partition function:

□ Combine the homonuclear and heteronuclear formula above:

rotrot )(

TTq

□ is the symmetry factor of a molecule or symmetry number.

□ is the number of indistinguishable orientations of the molecule:

○ = 1 for heteronuclear diatomics.

○ = 2 for homonuclear diatomics.

h. Combining translational, rotational, vibrational, and electronic partition

functions:

TkDeT

Teeg

e

eTV

h

TMk

qqqqTVq

B

vib

vib/

1/

2/

rot

2/3

2B

elecvibrottrans

1

2

),(

□ Assumptions:

○ rot << T

○ Only ground electronic state is populated.

○ The zero of energy is considered the energy of separated atoms

at rest in their ground electronic state.

○ Zero of energy for vibrational energy is the bottom of the

internuclear potential well of the lowest electronic state.

i. Molar energy of diatomic ideal gas:

eT

T

VB

VNB

DNe

eRRRTRT

T

qTNk

T

QTkU

A/

/vibvib

2

,

2

vib

vib

122

3

lnln

○ 3RT/2 is the average translational energy.

○ RT is the average rotational energy.

– There is a RT/2 contribution from each degree of freedom.

CHEM 3720

235

○ Rvib/2 is the zero-point vibrational energy.

○ T

T

e

eR

/

/vib

vib

vib

1

is the vibrational energy in excess of ZPE.

○ NADe is the electronic energy relative to the zero of electronic

energy.

j. Molar heat capacity of diatomic ideal gas:

2/

/2vib

)1(2

5

vib

vib

T

TV

e

e

TR

C

3. Polyatomic molecules

a. Translational, rotational, vibrational, and electronic energies are

available:

elecvibrottrans

elecvibrottrans),( qqqqTVq

b. Translational and electronic partition functions are the same as for

diatomic molecules.

□ The zero of energy is the energy of each atom completely

separated, in their ground electronic state.

c. Vibrational partition function

□ The vibrational motion can be expressed as a set of independent

harmonic oscillators:

○ 3n – 5 for a linear molecule

○ 3n – 6 for a nonlinear molecule

○ n is the number of atoms in the molecule

□ The vibrational energy of a polyatomic molecule is:

1 2

1

jjjn hn ; ( ,...2,1,0jn )

CHEM 3720

236

□ Introduce the characteristic vibrational temperature for the normal

mode (or normal coordinate) j: B

vib,k

h jj

1/

2/

vib vib,

vib,

1jT

T

j

j

e

eq

□ Average molar vibrational energy of a polyatomic molecule:

1/

vib, vib,Bvib

12 vib,jT

jj

je

NkE

□ Vibrational contribution to molar heat capacity:

1 2/

/2 vib,

vib,)1( vib,

vib,

jT

Tj

Vj

j

e

e

TRC

□ Contribution of normal mode j to the vibrational heat capacity:

2/

/2 vib,

)1( vib,

vib,

T

Tj

j

j

e

e

TR

□ Example: CO2 vib (K) CV/R

bending (double degenerate) 954 0.635

asymmetric stretch 3360 0.016

symmetric stretch 1890 0.202

total 1.488

d. Rotational partition function for linear polyatomic molecules

□ Within rigid rotor approximation:

2B

2

rotrot

8)(

h

TIkTTq

□ The moment of inertia is:

n

jjjdmI

1

2 where dj is the distance

from the nucleus to the center of mass of the molecule.

□ is similar as for diatomics.

□ Examples: O=C=O ( = 2), H–CC–H ( = 2), S=C=O ( = 1)

CHEM 3720

237

e. Rotational partition function for nonlinear polyatomic molecules

□ The rotational properties of the molecule depend on the relative

magnitudes of their three principal moments of inertia.

□ Define three characteristic rotational temperatures in terms of the

three principal moments of inertia:

BB

2

rot,2 k

Bh

kI jj

, (j = A, B, C)

□ There are three cases:

○ spherical top (all three principal moments of inertia are equal)

Crot,Brot,Arot,

2/3

rot

2/1

rot )(

TTq

– Example of CH4: rot = 7.54; = 12.

– Example of CCl4: rot = 0.0823; = 12.

○ symmetric top (two of the three principal moments of inertia

are equal)

Crot,Brot,Arot,

2/1

Crot,Arot,

2/1

rot )(

TTTq

– Example of NH3: rot = 13.6, 13.6, and 8.92; = 3.

– Example of CH3Cl: rot = 0.637, 0.637, and 7.32; = 3.

○ asymmetric top (all three principal moments of inertia are

different)

Crot,Brot,Arot,

2/1

Crot,Brot,Arot,

32/1

rot )(

TTq

– Example of H2O: rot = 40.1, 20.9, and 13.4; = 2.

– Example of NO2: = 2.

CHEM 3720

238

□ Average molar rotational energy of a nonlinear polyatomic

molecule:

2

3ln)(ln 2/32rot2

Brot

RT

dT

TdRT

dT

TqdTNkU

– This is the rotational contribution to the heat capacity for nonlinear molecules.

f. Linear polyatomic molecules:

TkDe

n

jT

T

e

j

j

ege

eTV

h

TMkTVq B

vib,

vib,/

1

53

1/

2/

rot

2/3

2

B

1

2),(

Tk

D

e

T

TRT

U en

jT

jj

j B

53

1/

vib, vib,

1

/

22

2

2

3

vib,

53

12/

/2 vib,

)1(2

2

2

3

vib,

vib,n

jT

TjV

j

j

e

e

TR

C

g. Nonlinear polyatomic molecules:

TkDe

n

jT

T

e

j

j

ege

e

TV

h

TMkTVq

B

vib,

vib,/

1

63

1/

2/

2/1

Crot,Brot,Arot,

32/12/3

2

B

1

2),(

Tk

D

e

T

TRT

U en

jT

jj

j B

63

1/

vib, vib,

1

/

22

3

2

3

vib,

63

12/

/2 vib,

)1(2

3

2

3

vib,

vib,n

jT

TjV

j

j

e

e

TR

C

h. These formulas give relative good agreement with experiment but they

can be improve by including:

□ anharmonicity,

□ centrifugal distortions,

□ contributions from low lying electronic excited states.

CHEM 3720

239

C. Applications of Statistical Thermodynamics

1. Molecular interpretation of work and heat

a. Average energy of a macroscopic system:

j

j

N,VβE

jjj N,VE

N,V,Q

eN,VEN,V,βpU

j

)()(

)()(

)(

Differentiate: j j

jjjj dpEdEpdU

Consider: dVV

EdE

N

jj

j jjj

N

jj N,V,βdpN,VEdV

V

EN,V,βpdU )()()(

b. Comparing with revrev qwdU :

j N

jj dV

V

EN,V,βpδw )(rev

○ The work results from infinitesimal changes in the allowed

energies of the system without changing the probability

distribution.

j

jj N,V,βdpN,VEq )()(rev

○ The heat results from a change in probability distribution of

the states of a system without changing the allowed energies.

c. Compare with PdVw rev

Nj N

jV

E

V

EN,V,βpP

)(

○ This is the same formula as used before.

CHEM 3720

240

2. The relationship between entropy and partition functions

a. Without going through the demonstration:

WkS lnBensemble

j

jj ppkS lnBsystem (where pj = aj/A)

QkT

QTkS

N,V

lnln

BBsystem

□ Example: For a system of monoatomic gas particles:

Ne

NN

gVh

Tmk

NTVNQ

1

2/3

2B2

!

1),,(

A

e

N

gV

h

TmkRRS 1

2/3

2B2

ln2

5

○ The molar entropy of Ar at 298.2 K is molJ/K 154.8 S

(both experimental and calculated).

b. The determination of the entropy using partition functions:

N,VT

QTkQkS

lnln BB

j

TkVNE jTVNQ B/),(e),,(

j

TkE

j

TkEj

j

TkE

j

j

j

E

TkS

B

B

B

/

/

/B

e

e1

eln

□ The T 0 limit:

○ Assuming n states with same energy nEEE ...21

(ground state is n-fold degenerate) then m states with higher

energy nmn EE 1 :

CHEM 3720

241

...)e(e

...eee

B11B1

B1B1B

/)(/

///

TkEETkE

TkETkETkE

n

nj

mn

mn

0 as ee B1B k/k/

Tn

TE

j

TE j

nkT

E

T

Enk

n

nE

TnkS

TE

TETE

lnln

e

e1eln

B11

B

k/

k/1k/

BB1

B1B1

○ As T 0, S is proportional to the logarithm of the degeneracy

of ground state.

○ This means it is negligible even if n NA.

□ But

!

),(),,(

N

TVqTVNQ

N

for ideal gas (Assumption:

indistinguishable particles for which the available number of states

is much greater than the number of molecules)

VT

qTNkNkqNkS

ln!lnln BBB

(using Stirling formula: NNNN ln!ln )

VT

qTNk

N

TVqNkNkS

ln),(ln BBB

□ Example: diatomic ideal gas (example N2(g) at 298.15 K)

TkDeT

Teeg

e

eTV

h

TMkTVq B

vib

vib/

1/

2/

rot

2/3

2B

1

2),(

el/vib/

rotA

2/52/3

2B

ln/

ln

2ln

2ln

vib

vib ge

Ter

eT

N

V

h

TMk

R

S

T

T

e

CHEM 3720

242

K 88.2rot

K 3374vib

1el g

113

elecvibrottrans

molKJ 5.191)01015.113.414.150(

SSSSS

○ This is very close to 191.6 JK–1mol–1 based on calorimetric

data.

3. Evaluation of the chemical potential using partition functions

a. Recall that:

N,VT

QTkU

ln2B

QkT

QTkS

N,V

lnln

BB

b. The relation between the chemical potential and the partition function:

QTkTSUA lnB

VTVTVT N

QRT

n

QTk

n

A

,,B

,

lnln

c. The case of an ideal gas:

!

),(

N

TVqQ

N

NNNqNQ lnlnln

N

TVqNq

N

Q ),(ln11lnln

ln

N

TVqRT

),(ln

– Recall that VTfqqqqTVq )(),( elvibrottr

– Also P

Tk

N

V B for the ideal gas.

PRTTkV

qRT

N

V

V

TVqRT lnln

),(ln B

CHEM 3720

243

– Compare with

P

PRTTPT ln)(),(

P

PRTTGPTG ln)(),(

P

Tk

V

qRTPRTTk

V

qRTT B

B lnlnln)(

d. Example: Ar(g) at 298.15 K

3322/3

2B m10444.2

2),(

h

Tmk

V

TVq

326B m10116.4 P

Tk (P = 1bar)

kJ/mol39.97K)15.298(

○ This is the same as the experimental value.

e. Example of a diatomic molecule

□ For a diatomic molecule one should choose the ground-state

energy as the zero of energy for calculating chemical potential.

□ The form of the partition function changes when one changes the

zero of energy:

),(

...1

...),(

0

/)(

B/0

B01B/0

B/1B/0B/

TVqe

ee

eeeTVq

Tk

Tk

TkTkTkj

Tk

TkD

T

Teeg

e

eTV

h

TMkTVq B

vib

vib/

el/

2/

rot

2/3

2B

-1

2),(

el/rot

2/3

2B0

vib-1

12),( g

e

TV

h

TMkTVq

T

f. The chemical potential for a diatomic molecule:

PN

RT

V

qRTE

P

Tk

V

qRTET

A

0

0B

0 lnln)(

□ The use of q0 shows that the ground vibrational state of the

molecule is taken to be zero of energy.

CHEM 3720

244

□ Example: HI(g) at 298.15 K

326

A

3340

m10116.4;m1051.4),(

PN

RT

V

TVq

molkJ90.52)15.298( 0 E

○ Including corrections for anharmonicity and non-rigid rotator:

molkJ94.52)15.298( 0 E

4. Equilibrium constants in terms of partition functions

a. Deducing the expression:

□ Consider again a general reaction:

)(Z)g(Y)g(B)g(A ZYBA g

□ Equilibrium condition for a reaction at fixed volume and

temperature:

0

,

VTd

dA

0BBAAZZYY

□ Assume mixture of ideal gases (independent species):

!

),(

!

),(

!

),(

!

),(

),,,,,(

Z

Z

Y

Y

B

B

A

A

ZYBA

ZYBA

N

TVq

N

TVq

N

TVq

N

TVq

TVNNNNQ

NNNN

□ Introduce the chemical potential of each species:

x

x

,,xx

),(ln

ln

xyN

TVqRT

N

QRT

NVT

□ Substituting chemical potential in equation above:

BA

ZY

BA

ZY

BA

ZY

BA

ZY

qq

qq

NN

NN

CHEM 3720

245

□ Dividing each term by V:

BA

ZY

BA

ZY

)/()/(

)/()/()(

BA

ZY

BA

ZY

VqVq

VqVqTKc

)(/),(x TfVTVq

P

RTcTKTK cP )()(

○ Here, the unit for c is molecule/m3 or 1/m3.

b. Reaction involving diatomic molecules: H2(g) + I2(g) = 2HI(g)

RT

DDD

T

TT ooo

e

e

ee

mm

m

qq

q

VqVq

VqTK

2I2HHI

HIvib

2Ivib

2Hvib22

22

2222

2

2/

//

2HIrot

Irot

Hrot

2/3

HI

2HI

HI

2HI

HI

2HI

)1(

)1()1(

)(

4

)/)(/(

)/()(

□ KP can be calculated.

□ ln KP versus 1/T give Hr (–12.9

kJ/mol calculated compared to –13.4

kJ/mol experimental)

□ The discrepancy is due to the use of rigid rotator-harmonic

oscillator model.

c. The expressions for reactions involving polyatomic molecules get to be

more complicated.

5. Transition state theory

a. Transition state theory (TST) also called activated-complex theory is a

theory of the rate of elementary reactions.

b. The theory focuses on transient species (called activated complex or

transition state) located in the vicinity of the top of the barrier height

(activation energy) of a reaction.

1000 K/ T

ln KP

1000 K/ T

ln KP

CHEM 3720

246

c. Consider an elementary reaction: A + B Products

□ The rate law is given by: ]B][A[]P[

kdt

dv

□ According to the activated complex

theory, the model of the elementary

reaction is a two-step process:

PABBA ‡

○ AB‡ is the activated complex

□ The transition state (or activated

complex) quantities are denoted by

a double dagger sign, ‡ (not ).

□ Another important assumption is that the reactants and the

activated complex are in equilibrium with each other, and the

equilibrium constant for this equilibrium is:

]B][A[

]AB[

/]B[/]A[

/]AB[ ‡‡‡

c

cc

cKc

○ where c is the standard-state concentration (this is often 1.00

moldm–3)

□ The activated complexes are assumed to be stable within a small

region of width centered at the barrier top.

□ According to the transition-state theory, the rate of the reaction is

given by the rate of a unimolecular process, with a rate constant k‡,

that transform the activated complex into products:

][AB][ ‡‡k

dt

Pd

□ The rate constant k‡ is proportional to the frequency with which the

activated complexes cross over the barrier top (c), where the

Ener

gy

Reaction coordinate

Products

A+B

AB‡

Ener

gy

Reaction coordinate

Products

A+B

AB‡

CHEM 3720

247

proportionality constant () is called transmission coefficient (and

is assumed to be 1 if no other information is available):

cc‡ k

□ Because cKc /]B][A[]AB[‡‡

]B][A[/]B][A[][

‡c‡

c

c

KcK

dt

Pd cc

□ Compare to ]B][A[]P[

kdt

dv

c

Kk c

‡c

○ k has units of conc–1s–1.

□ The equilibrium constant written in terms of partition functions:

)/)(/(

)/(

]B][A[

]AB[

BA

‡‡‡

VqVq

cVqcKc

○ qA, qB, and q‡ are the partition functions of A, B and AB‡.

□ Assume that the motion of the reacting system over the barrier top

is a one-dimensional translational motion.

○ The translational partition function, qtrans, for this one-

dimensional motion (where m‡ is the mass of activated

complex containing all atoms of the system):

h

Tkmq

2/1B

trans)2(

○ Write the partition function of the activated complex as a

product of the partition function of the motion of passing over

the energy barrier and a partition function for the rest of

degrees of freedom of the activated complex: ‡inttrans

‡ qqq

CHEM 3720

248

)/)(/(

)/()2(

BA

‡int

2/1B

‡‡

VqVq

cVq

h

TkmKc

)/)(/(

)/()2(

BA

‡int

2/1B

cVqVq

cVq

hc

Tkmk

○ The quantities c and are hard to determine directly but the

product of them is a speed to which activated complexes

crosses the barrier for which we consider the average speed

<uac> = c.

○ Use the Maxwell-Boltzmann distribution to calculate the

average one-dimensional speed: 2/1

‡B

0

2/

B

0ac

22)( B

2‡

m

Tkduue

Tk

mduuufu

Tkum

○ The TST expression for the rate constant became:

‡B

BA

‡intB

)/)(/(

)/(K

hc

Tk

VqVq

cVq

hc

Tkk

○ K‡ is the equilibrium constant for the formation of the

transition state from the reactants but with one degree of

freedom (the motion along the reaction coordinate) excluded

from the activated complex partition function.

□ A better approximation is obtained when the motion of the reacting

system over the barrier top is considered to be a vibrational

motion.

○ The vibrational partition function, qvib, for this degree of

freedom:

c

Bvib

h

Tkq

CHEM 3720

249

○ Write the partition function of the activated complex as a

product of the partition function of the motion of passing over

the energy barrier and a partition function for the rest of

degrees of freedom of the activated complex:

‡intvib

‡ qqq ; )/)(/(

)/(

BA

‡int

c

B‡

VqVq

cVq

h

TkKc

‡B

BA

‡intB

‡c

)/)(/(

)/(K

hc

Tk

VqVq

cVq

hc

Tk

c

Kk c

○ K‡ is the equilibrium constant for the formation of the

transition state from the reactants but with one degree of

freedom (the vibrational mode describing the motion along the

reaction coordinate) excluded from the activated complex

partition function.

d. One can define a standard Gibbs energy of activation, ‡G, as the

change in Gibbs energy when going from the reactants at a

concentration c to transition state at concentration c.

‡‡ ln KRTG RTGe

hc

Tkk /B

○ But ‡G = ‡H – T‡S where ‡H is the standard enthalpy

of activation and ‡S is the standard entropy of activation:

RTHRS eehc

Tkk //B

‡‡

e. The relationship between the Arrhenius activation energy Ea and ‡H

and the relationship between the Arrhenius preexponential factor A and

‡S is obtained by comparing the equations above with RTE

Aek/a

:

dT

Kd

TdT

kdK

hc

Tkk

‡‡B ln1ln

CHEM 3720

250

○ For an ideal gas: 2

ln

RT

U

dT

Kd c

○ Similar to van’t Hoff equation: 2

ln

RT

H

dT

Kd P

2

‡1ln

RT

U

TdT

kd

RTUnRTUPVUH ‡‡‡‡‡‡

2a

2

‡ 2ln

RT

E

RT

RTH

dT

kd

RTEHRTHE 22 a‡‡

a

○ RTHE ‡a for reactions in solution.

RTERS eehc

Tkek

//B2

a‡

RSehc

TkeA /B

2 ‡

□ This is the thermodynamic interpretation of the Arrhenius A factor

based on the transition-state theory.

□ The value of ‡S gives information about the relative structure of

the activated complex and the reactants:

○ A positive value the structure of the activated complex is

less ordered than that of the reactants.

○ A negative value the structure of the activated complex is

more ordered than that of the reactants.

□ The value of ‡S depends on the choice of standard state, i.e., the

value of c.

f. Example: H(g) + Br2(g) HBr(g) + Br reaction 11311

a smoldm1009.1 kJ/mol;5.15 AE

K 1000at molKJ3.60 kJ/mol;13.1 11‡‡ SH

CHEM 3720

251

6. Potential energy surfaces

a. This is an important concept in understanding chemical reactivity and

transition state theory.

b. According to the Born-Oppenheimer approximation, the wavefunction

of a system composed of N nuclei and n electrons is written as a

product of nuclear wavefunction, depending on the positions of the

nuclei, and an electronic wavefunction, depending on the positions of

the electrons within a fixed nuclear configuration.

c. This allows solving the Schrödinger equation for the wavefunction of

the electrons alone at a specific nuclear configuration.

d. Modifying the nuclear configuration

and solving again the Schrödinger

equation one get a different energy.

e. The case of diatomic molecules:

representing the electronic energy

versus the interatomic distance one

obtains a potential energy curve.

f. The case of triatomic molecules there are three geometric parameters

that define the molecular geometry and representing all three of them

plus the energy requires a 4-dimensional representation.

g. Example: the geometry of water is defined by

bond lengths and one angle or 3 bond lengths or 1

bond length and 2 bond angles or 3 bond angles.

h. The minimum number of geometric parameters necessary to define the

geometry of a system of N atoms (N 3) is 3N – 6.

i. When the potential energy depends on more than one single geometric

parameter then use the phrase potential energy surface.

Energy

R

Energy

R

O

HA HB

O

HA HB

CHEM 3720

252

□ The entire potential energy function cannot be plotted because the

plotting is limited to 3 dimensions.

□ To overcome this, one usually represents the energy function of

only 2 geometric parameters keeping the other/others at a fixed

value.

□ Such a plot is a cross-sectional cut of the full potential energy

surface.

□ Example of water: a 3-dimensional plot )ct,,(BA HOHO rrV

versus AHOr and

BHOr gives information about how the

potential energy of water molecule changes when the bond lengths

are varied at a constant angle .

○ A series of cross-sectional plots at different values of give

information of how the potential energy depends on the .

□ The representation of the potential energy surface for the

molecules with more than 3 atoms is much more complicated.

j. The case of a triatomic reactive system:

□ Example of F(g) + DADB(g) DAF(g) + DB(g) reaction for which

one can look at three geometric parameters: the distance between F

and DA, rFD, the distance between DA and DB, rD2, and the angle

between these two internuclear distances (called collision angle).

DA DBF

F

DBDA

F

DBDA

□ Represent the potential energy surface for this reaction in a form of

a contour diagram for a collision angle of 180 degrees (which is

same as experimentally determined).

□ Each contour line corresponds to a constant value of the energy.

CHEM 3720

253

□ The zero of energy is arbitrarily

assigned for the reactants at

infinite separation.

□ When the reactants are at infinite

separation, the potential energy

surface is the same as for the D2

molecule.

□ The potential energy is the same as

for FD molecule when the

products are at infinite separation.

□ Look at the minimum energy path (reaction path or reaction

coordinate) from the reactants to products.

□ The maximum point along the minimum energy path is a saddle

point (the energy decreases along the reaction path in both

directions but increases on the direction perpendicular to the

reaction path).

k. The transition state is a hypersurface separating the reactant region

from the product region, and it contains the saddle point.

A–B

+ C

A +

B–C

A–B

+ C

A +

B–C

CHEM 3720

254

D. Unit Review

1. Important Terminology

Statistical thermodynamics

Ensemble

Canonical ensemble

Boltzmann distribution

Boltzmann factor

Partition function

Ensemble average of a quantity

Molecular partition function

Translational partition function

Rotational partition function

Vibrational partition function

Electronic partition function

Characteristic vibrational temperature

CHEM 3720

255

Rotational temperature

Symmetry factor (or number)

Translational contributions

Rotational contributions

Vibrational contributions

Electronic contributions

Chemical potential

Transition state theory

Transition state

Activated complex

Gibbs energy/enthalpy/entropy of activation

Potential energy surface

CHEM 3720

256

2. Important Formulas

j

E

Ej

jj

j

e

e

A

ap

; ),,(

B/),(

TVNQ

ep

TkVNE

j

j

j

VNE jeVNQ),(

),,(

j

TkVNE jeTVNQ B/),(),,(

VNVN T

QTk

QE

,

2B

,

lnln

VNVNV

T

U

T

EC

,,

PV

QTk

V

Q

VNQ

TkP

NN

,B

,

B ln

),,(

NTVqTVNQ ),(),,(

!

),(),,(

N

TVqTVNQ

N

j

Tk

j

jj eeTVq B/),(

j

Tk

TkTk

jj

jj

e

e

TVq

e

B

BB

/

//

),(

elecvibrottranslkji

elecvibrottrans),( qqqqTVq

i

Tkieq Btrans /

trans

;

j

Tkjeq Brot /

rot

k

Tkkeq Bvib /

vib

;

l

Tkleq Belec /

elec

CHEM 3720

257

elecvibrottrans

//// Belec

Bvib

Brot

Btrans

qqqq

eeeeTkTkTkTk

ijkl

lkji

i

Tki

j

Tkij egeTVq BB //

),(

TnTn eef

// vibvib )1(

hTn eef

/0

vib

TJJJ eTJf

/)1(rot

rot)/)(12(

TkDeT

Teeg

e

eTV

h

TMkTVq B

vib

vib/

1/

2/

rot

2/3

2B

1

2),(

TkDe

n

jT

T

e

j

j

eg

e

eTV

h

TMkTVq B

vib,

vib,/

1

53

1/

2/

rot

2/3

2B

1

2),(

TkDe

n

jT

T

e

j

j

eg

e

e

TV

h

TMkTVq

B

vib,

vib,/

1

63

1/

2/

2/1

Crot,Brot,Arot,

32/12/3

2B

1

2),(

Tk

D

e

T

TRT

U en

jT

jj

j B

63

1/

vib, vib,

1

/

22

3

2

3

vib,

63

1 2/

/2 vib,

)1(2

3

2

3

vib,

vib,n

jT

TjV

j

j

e

e

TR

C

CHEM 3720

258

j N

jj dV

V

EN,V,βpδw )(rev

j

jj N,V,βdpN,VEq )()(rev

WkS lnBensemble

QkT

QTkS

N,V

lnln

BBsystem

VT

qTNk

N

TVqNkNkS

ln),(ln BBB

VTVTVT N

QRT

n

QTk

n

A

,,B

,

lnln

P

Tk

V

qRTPRTTk

V

qRTT B

B lnlnln)(

BA

ZY

BA

ZY

)/()/(

)/()/()(

BA

ZY

BA

ZY

VqVq

VqVqTKc

][AB][ ‡‡k

dt

Pd

cc‡ k

c

Kk c

‡c

‡B Khc

Tkk

RTGehc

Tkk /B

RTHE 2‡a

RSehc

TkeA /B

2 ‡

CHEM 3720

259

Unit XIV

Gas-Phase Dynamics

A. The Kinetic Theory of Gases

1. Ideal gas equation of state

a. The kinetic theory of gases introduces a model for gases:

□ Molecules are in constant motion and collide with each other and

with the wall of the container.

□ Molecules behave as hard spheres with no interactions between

them except during collisions.

b. Ideal gas equation can be understood as a result of colliding molecules

with the walls of a container.

c. Pressure that a gas exerts on the walls is due to the collisions that the

gas particles of the gas make with the walls.

– Assume a rectangular parallelepiped of sides a, b, and c.

– The velocity of the molecule 1 has the components u1x, u1y, and u1z.

– The x-component of the momentum is mu1x before collision.

– The x-component of the momentum is –mu1x after collision (which

is assumed to be perfectly elastic).

– The rate of change of momentum due to collisions with the right-hand wall is a force (the

force that molecule 1 exerts on the right wall):

1

21

1

1)1

/2

2(F

a

mu

ua

mu

t

mux

x

xx

– The pressure that molecule 1 exerts on the right wall:

V

mu

abc

mu

bc

FP xx

21

211

1

– The total pressure (from all molecules) exerts on the right wall:

N

j

jx

N

j

j uV

mPP

1

2

1

– Define the average value of u1x2:

N

j

jxx uN

u

1

22 1

– Similar expression can be obtained for the y and z directions.

xu1

c

b

a

CHEM 3720

260

– The homogeneous gas is isotropic: < ux2 > = < uy

2 > =< uz2 >

– The speed u of any molecule: u2 = ux2 + uy

2 + uz2 and < u2 > = < ux

2 > + < uy2 > + < uz

2 >

– Average translational (kinetic) energy:

mole)(per molecule)(per 2

3

2

3

2

1

B2 RTTkum

TkNumNuNmuNmPV x B222

2

3

3

2

2

1

3

2

3

1

nRTTknNTNkPV BAB

2. The speed of a molecule in a gas

a. Average translational (kinetic) energy:

mole)(per molecule)(per 23

B232

21 RTTkum

M

RT

m

Tku

33 B2

b. The square root of <u2> has the units of speed (m/s), is called root-

mean-square speed, and denoted by urms: 2/1

2/12rms

3

M

RTuu

c. Example: Calculate urms for a nitrogen molecule at 25 C:

s

m515

kg

smkg1065.2

kg

J1065.2

g

kg10

mol

g2.28

K298Kmol

J314.83 2

122

52

1

5

2

1

3rms

u

d. The urms is not the average speed <u> (because <u2> <u>2) but is a

good estimate of it because they differ by less than 10 %.

e. Average molecular speeds are of the order of hundreds of m/s:

Gas <u> (m/s) at 25C urms (m/s) at 25C

H2 1770 1920

He 1260 1360

NH3 609 661

N2 475 515

O2 444 482

CO2 379 411

SF6 208 226

CHEM 3720

261

f. Assumptions made:

□ The collisions with the wall are perfectly elastic (true on average).

□ The molecules do not collide (results does not change).

g. More accurate treatments show same dependencies but usually differ

by constant factors.

h. Gas effusion and diffusion are dependent on (but not proportional to)

the molecular speed; they are faster as the molecular speed is greater.

3. The distribution of the components of molecular speeds

a. Let h(ux,uy,uz)duxduyduz be the fraction of molecules that have speeds

components between ux and ux + dux, uy and uy + duy, and uz and uz +

duz.

b. Probability distributions in each of the three directions are independent

of each other:

)()()(),,( zyxzyx ufufufuuuh

c. Probability distributions is the same in each of the three directions so

h(ux,uy,uz) must depend only upon the speed (or the magnitude of the

velocity u, where uu = u2 = ux2 + uy

2 + uz2.

d. It can be shown that 2)(ln)(ln

jj

j

duu

ufd

udu

uhd; (j = x, y, z)

2

)( juj Aeuf

e. Setting some conditions: 2/1)/(1)(

Aduuf xx

RT

M

M

RTduufuu xxxx

2)(22

f. The expression for the one-dimensional distribution:

TkmuRTMux

xx eTk

me

RT

Muf B

22 2/2/1

B

2/2/1

22)(

CHEM 3720

262

N2

0.0

0.5

1.0

1.5

-2000 -1000 0 1000 2000

u x (m/s)

f(u

x)*

1000

300 K

1000 K

g. Average value of ux:

0)( xxxx duufuu

h. Average value of ux2:

Tkumm

Tkduufuu xxxxx B2

1221B22 )(

i. Total kinetic energy: Tkum B232

21

□ The total energy of TkB23 is divided equally between the x, y, and

z components.

4. The Maxwell-Boltzmann distribution

a. It gives the distribution of molecular speeds.

b. The direction in which a molecule moves has no physical consequence,

only the magnitude of the speed is relevant.

c. Define a probability distributions of a molecule having a speed

between u and u + du:

zyxTkuuum

zzyyxx

dududueTk

m

duufduufduufduuF

zyx B222 2/)(

2/3

B2

)()()()(

d. A rectangular coordinate system with the axes ux, uy, and uz is a

velocity space.

e. Transform the infinitesimal volume element into spherical coordinates

rather than Cartesian coordinates: duxduyduz = 4u2du.

CHEM 3720

263

f. The Maxwell-Boltzmann distribution is:

dueuTk

mduuF

Tkmu B2 2/2

2/3

B24)(

N2

0.0

0.5

1.0

1.5

2.0

0 500 1000 1500 2000

u (m/s)

f(u

)*1

000

300 K

1000 K

□ The probability distribution is normalized:

0

1)( duuF

□ Average speed of a molecule:

0

2/18

)(M

RTduuuFu

□ Average of speed squared:

0

B22 33)(

M

RT

m

TkduuFuu

○ As before:

2/12/12

rms3

M

RTuu

○ Also: 92.03

82/1

rms

u

u

□ The most probable speed, obtained when 0)(

du

udF, is:

2/12/1B

mp22

M

RT

m

Tku

□ All the characteristic speeds (<u>, urms, ump) are of the form

ct(RT/M)1/2.

□ The Maxwell-Boltzmann distribution has been verified

experimentally.

CHEM 3720

264

g. The Maxwell-Boltzmann distribution in terms of kinetic energy

□ Use = mu2/2 u = (2/m)1/2 and du = d/(2m)1/2:

deTk

dFTkB/2/1

2/3B )(

2)(

□ The probability distribution is normalized:

0

1)( dF

□ Average kinetic energy of a molecule: TkdF B0

2

3)(

5. The frequency of collision with the wall

a. It is an important quantity for the theory of the rates of surface

reactions.

b. Deducing the expression:

– All the molecules that will strike an area A on the surface,

at an angle , with a speed u, in the time interval dt are the

molecules in the cylinder below.

– The volume of the cylinder: Aucosdt.

– The number of molecules in the cylinder: (Audt)cos.

The is the number density: = N/V (units of m–3).

– The fraction of molecules that have speeds between u and u + du is F(u)du.

– The fraction of molecules traveling within the solid angle between and + d and

between and + d is sindd/4.

– The number of molecules that strikes area A, from a specified direction in a interval dt:

4/sin)(coscoll ddduuFAudtdN

– The number of molecules (collisions) striking the wall per unit area per unit time whose

speeds are between u and u + du and directions lie within the solid angle sindd:

ddduuuF

dt

dN

Adz sincos)(

4

1 collcoll

Tkmueudz B

2 2/3coll

compared to

TkmueuduuF B

2 2/2)(

so the number of

collisions, zcoll, peaks at higher value of u.

c. The number of molecules (collisions) striking the wall per unit area per

unit time:

4sincos)(

4

1 2

0

2/

00

collcoll

uddduuuF

dt

dN

Az

udt uxdt

x

y

z

A

CHEM 3720

265

d. Example: N2 at 25C and 1 bar: 325

A m1043.2// RTPNVN ; 1sm475 u

2123coll cms1088.2 z

6. Molecular collisions

a. The number of collisions (or frequency of collision) between the

molecules of a gas

□ Consider the molecules as being hard spheres of diameter d (of

radius d/2).

□ The molecule of interest will collide with all the other molecules

whose center lies within the collision cylinder (a cylinder of

diameter d and length of ut or udt):

d

□ Each molecule presents a target of effective radius d, which is d2,

and is called collision cross section, denoted by .

○ The can be calculated or determined for more complicated

molecular shapes, and the values are tabulated.

□ The volume of the collision cylinder is: <u>dt.

□ The number of collisions in the interval dt is the product of the

volume of the collision cylinder and the density of molecules:

dtudN coll

□ The collision frequency zA considering that molecules are

stationary: 2/1

BcollA

m

8

Tku

dt

dNz

□ The collision frequency zA considering that molecules are not

stationary but are moving with respect to each other:

CHEM 3720

266

○ Replace the mass m by the reduced mass = m1m2/(m1 + m2) =

m/2 or replace <u> by <ur> = 21/2<u>: 2/1

B2/12/1rA

m

822

Tkuuz

□ Example: the collision frequency of a single molecule of N2 at

25C and 1 bar:

19

12203252/1A

s103.7

)sm475)(m100.45)(m1043.2(2

z

b. The average time between collisions is measured by the reciprocal of

the collision frequency zA.

□ Example: for a molecule of N2 at 25C and 1 bar:

s104.1/1 10A

z

c. The average distance a molecule travels between collisions is called

mean free path and is denoted by l:

2/12/1A 2

1

2

u

u

z

ul

□ The mean free path is few hundred times larger than the diameter

of a molecule.

□ For ideal gas: PN

RTlRTPN

A2/1A

2/

□ Example: for a molecule of N2 at 25C and 1 bar:

m105.6 8l

□ Example: The mean free path of a hydrogen atom in interstellar

space (considering that there is about one hydrogen atom per m3,

the diameter of a hydrogen atom is about 100 pm, and the average

temperature in interstellar space is 10 K):

m102)m10100(14.3)m1(2

1

2

1 19

21232/12/1

l

CHEM 3720

267

d. The probability of a collision

□ The number of molecules (targets) in a plane of unit area

perpendicular to the direction of movement of a molecule and of

thickness dx is dx.

□ The total target area presented by these molecules is a product of

the collision cross section of each target and the number of

molecules (targets) and is the probability of a collision dx.

□ Starting with n0 molecules, the number of molecules that undergo a

collision between x and dx is n(x)dx where n(x) is the number

of molecules that didn’t suffer collisions until x:

dxxndxxnxn )()()(

)()()(

xndx

dn

dx

dxxnxn

□ The solution of such equation: lxx enenxn /

00)(

□ The probability that one of the initial n0 molecules will collide in

the interval x and dx:

dxel

dxdx

xdn

nn

dxxnxndxxp lx /

00

1)(1)()()(

□ The probability that a molecule will collide before it travels the

distance x:

lxx

lx edxel

/

0

/ 11

e. The collision frequency per unit volume

□ The total collision frequency per unit volume among all molecules

in a gas: 2/1

22

rAAA22

1

2

1

uuzZ

□ Example: For N2 at 25C and 1 bar: 3134

AA ms109.8 Z

0.0

0.5

1.0

0 1 2 3 4

x/l

Pro

bab

ilit

y t

o c

oli

de

befo

re t

rav

els

x

CHEM 3720

268

f. The collision frequency per unit volume in a gas containing two types

of molecules, A&B:

BArABAB uZ

2BA

AB2

dd

2/1Br )/8( Tku

)/( BABA mmmm

g. The rate of collision between molecules whose relative kinetic energy

exceeds some critical value

□ The molecules traveling at higher speeds (identified by a factor of

dueuTkmu B

2 2/3 ) are more likely to collide with more molecules

(or to strike the wall).

□ The collision frequency per unit volume between molecules of

types A and B with relative speeds in the range ur and ur +dur:

r3r

2/2/12/3

BBAABAB

B2r

2duue

TkdZ

Tku

□ This distribution has a factor of ur3 that reflects the fact that the

molecules with higher relative speeds collide more frequently.

□ The collision frequency per unit volume between molecules of

types A and B with relative kinetic energies higher than some

critical value c:

r/

r

2/3

B

2/1

BAABABBr

18

deTk

dZTk

□ Integrate between c and :

Tke

Tk

TkZ Bc /

B

c2/1

BBAABcrAB 1

8)(

CHEM 3720

269

B. Gas-Phase Reaction Dynamics

1. Introduction

a. Introduce some of the current models that are used to describe the

molecular aspects of bimolecular gas-phase reactions.

b. Introduce new concepts such as reaction cross section and impact

parameter.

c. Give the expression for the rate constant of a bimolecular gas-phase

reaction.

□ Consider a general bimolecular elementary gas-phase reaction:

Products B(g)A(g)k

□ The rate of this reaction:

]B][A[]A[

kdt

dv

2. Hard-sphere collision theory

a. Estimating the rate of a reaction by the number of collisions:

BArABAB uZv

○ AB is the hard-sphere collision cross section of A and B

molecules.

○ <ur> is the average relative speed of the colliding pair A – B.

○ A and B are the number densities of A and B molecules.

○ ZAB has the units of collisionsm–3s–1 or m–3s–1.

b. Estimating the rate constant k using hard-sphere collision theory:

rAB uk

○ k has the units of ZAB/(AB) that is moleculesm3s–1.

○ To transform to more common units of moldm3s–1 multiply

by NA and 1000 dm3m–3:

rABA

33 )mdm1000( uNk

CHEM 3720

270

c. Example: Calculate k (in moldm3s–1) using hard-sphere collision

theory for H2(g) + C2H4(g) C2H6(g) reaction:

2192

2ABAB m1085.3

2

pm430pm270

d

13

2/1

27

1232/1B

r sm1083.1kg1012.3

K298KJ10381.188

Tku

)sm1083.1)(m1085.3)(mol10022.6)(mdm1000( 1321912333 k

11311 smoldm1024.4 k )smoldm1049.3 tal(experimen 11362 k

d. Hard-sphere collision theory predicts a temperature dependence of k as

T1/2.

e. Hard-sphere collision theory predicts that all the molecules collide

while having a relative speed of <ur>.

f. A better approximation is obtained assuming that the rate of reaction is

the rate of collisions between molecules whose relative kinetic energy

exceeds some critical value.

3. Activated hard-sphere collision theory

a. Improve the hard-sphere collision theory by assuming that reaction

occur only if the relative speed is sufficient to overcome the repulsive

forces.

b. Replace the collision cross section AB by a reaction cross section

r(uu) that depends on the relative speed of the reactants.

)()( rrr uuuk r

c. Calculate the observed rate constant by averaging over all possible

collision speeds:

rr0

rrrrr0

r )()()()( duufuuduufukk

d. The f(ur) is the distribution of relative speeds in the gas, and the ur f(ur)

dur is given by kinetic theory of gases:

CHEM 3720

271

r2/3

r

2/12/3

Brrr

B2r

2)( dueu

Tkduufu

Tku

e. Transform the dependent variable from relative speed ur to the relative

kinetic energy Er:

r

2/1

rr

2/1r

r2rr

2

12,

2

1dE

Edu

EuuE

r/

r

2/12/3

Brrr

Br12

)( dEeETk

duufuTkE

f. Substituting in the expression for k:

0rr

/r

2/12/3

B

)(12

Br dEEeETk

kTkE

g. A simple model for (Er) is to assume that only collisions with higher

relative kinetic energy than a threshold energy, E0, are reactive:

0r2AB

0rrr

0)(

EEd

EEE

h. The expression for k becomes:

Tk

Eeu

dEdeETk

k

TkE

E

TkE

B

0/ABr

r2AB

/r

2/12/3

B

1

12

B0

0

Br

i. Same result as obtained using collision of hard spheres whose kinetic

energy exceed a threshold energy but here this dependence is included

into the expression through the reaction cross section.

□ Explore different models of r(Er) that will give different values

for the rate constant.

CHEM 3720

272

j. Example: For the H2(g) + C2H4(g) C2H6(g) reaction, using the hard-

sphere collision theory k = 4.241011 moldm3s–1 while experimental

value is k = 3.4910–26 moldm3s–1. What is the value of 0E that give

the experimental value of k?

kJ/mol2231024.4

1049.31 011

26

B

0/ B0

ETk

Ee

TkE

○ The experimental activation energy is Ea = 180 kJ/mol.

4. The line-of-centers model and the impact parameter

a. Just a simple energy-dependent reaction cross section is unrealistic, the

two collision below occur at same relative collision energy but the one

depicted in the right side is less likely to generate products.

versus

b. A more realistic model for the reaction cross section is one that

considers the component of the relative kinetic energy along the line

connecting the centers of colliding molecules.

c. This model is called line-of-centers model of r(Er).

b

uA

uB

d. Denote the kinetic energy along the line of centers by Eloc, and assume

that the reaction occurs when Eloc > E0.

e. Look for the expression for r(Er) in this model:

□ Relative velocity: ur = uA – uB

□ Relative kinetic energy: Er = ur2/2

f. Define the impact parameter b as the perpendicular distance between

the lines of the trajectories before collision.

□ Collision occur if b < rA + rB = dAB.

□ Molecules will miss each other if b > dAB.

CHEM 3720

273

g. The kinetic energy along the line of centers depends on the impact

parameter:

□ When b = 0 Eloc = Er.

□ When b > dAB then r(Er) = 0.

h. Finally, after the derivation of the expression of the reaction cross

section:

0r

r

02AB

0r

rr 1

0

)(EE

E

Ed

EE

E

Collision energy

Cross

section

Collision energy

Cross

section

i. Substituting in the expression for k:

TkETkEeued

Tkk B0B0 /

ABr/2

AB

2/1B8

j. Correlation with Arrhenius parameters:

TkEdT

kdTkE B0

2Ba

2

1ln

2/1ABr euA

k. The values of A calculated using this model are bigger than the

experimental ones, the reaction cross sections determined

experimentally present a threshold energy but a different shape

compared to the calculated values base on the above model so the

simple hard-sphere picture is not accurate.

CHEM 3720

274

5. Other factors influencing the reactivity of molecular collisions

a. Depending on the shape of the reactants, the rate constant for a gas-

phase reaction depends on the orientations of the colliding molecules

□ Example: Rb(g) + CH3I(g) RbI + CH3(g)

reaction occurs only when the rubidium atom

collides with iodomethane in the vicinity of the

iodine atom, collisions with the methyl end are

nonreactive cone of nonreactivity.

b. The reaction cross section depends on the internal energy of the

reactants.

□ Example: H2+(g) + He(g) HeH+ + H(g)

□ The reaction cross section, at the same total energy, depends on the

vibrational state in which H2+ is.

□ For v 4 (v is the vibrational quantum number) there is no

threshold energy (Evib > E0 so no additional translational energy is

needed).

□ Chemical reactivity depends not only on the total energy but also

on its distribution among the internal energy levels.

I

HH

H

C

CHEM 3720

275

C. Unit Review

1. Important Terminology

Kinetic theory of gases

Root-mean-square speed

Average speed

Most probable speed

The Maxwell-Boltzmann distribution

The frequency of collision with the wall

The collision frequency

The average time between collisions

Mean free path

The probability of a collision

The collision frequency per unit volume

Hard-sphere collision theory

Reaction cross section

CHEM 3720

276

2. Important Formulas

TkmuRTMux

xx eTk

me

RT

Muf B

22 2/2/1

B

2/2/1

22)(

dueuTk

mduuF

Tkmu B2 2/2

2/3

B24)(

deTk

dFTkB/2/1

2/3B )(

2)(

2/12/12

rms3

M

RTuu ;

2/18

M

RTu

;

2/1

mp2

M

RTu

4coll

uz

2/1B2/12/1

rAm

822

Tkuuz

2/12/1A 2

1

2

u

u

z

ul ;

PN

RTl

PN

A2/1

A

2RT

dxel

dxxp lx /1)( ; lx

xlx edxe

l

/

0

/ 11

2/1

22

rAAA22

1

2

1

uuzZ

BArABAB uZ

2BA

AB2

dd ; 2/1

Br )/8( Tku ; BA

BA

mm

mm

Tke

Tk

TkZ Bc /

B

c2/1

BBAABcrAB 1

8)(

rAB uk

Tk

Eeuk

TkE

B

0/ABr 1B0


Top Related