PHYSICAL CHEMISTRY II CHEM 3720 Syllabus & Lecture Notes€¦ · PHYSICAL CHEMISTRY II CHEM 3720...
Transcript of PHYSICAL CHEMISTRY II CHEM 3720 Syllabus & Lecture Notes€¦ · PHYSICAL CHEMISTRY II CHEM 3720...
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
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
2ˆ
xmH x
, 2
22
2ˆ
ymH y
, 2
22
2ˆ
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
1ˆ
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
2ˆ
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(
4ˆ
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
1ˆ
2
1ˆ
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
1ˆ
2
1ˆ
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
1ˆ
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
1ˆ
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
1ˆ
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
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
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
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
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