Chemistry 481(01) Spring 2014
Instructor: Dr. Upali Siriwardane
e-mail: upali@latech.edu
Office: CTH 311 Phone 257-4941
Office Hours:
M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th, F 10:00 - 12:00 a.m.
April 10 , 2014: Test 1 (Chapters 1, 2, 3,)
May 1, 2014: Test 2 (Chapters 6 & 7)
May 20, 2014: Test 3 (Chapters. 19 & 20)
May 22, Make Up: Comprehensive covering all Chapters
Chemistry 481, Spring 2014, LA Tech
Chapter 6-1
Chapter 6. Molecular symmetry
An introduction to symmetry analysis
6.1 Symmetry operations, elements and point groups 179
6.2 Character tables 183
Applications of symmetry
6.3 Polar molecules 186
6.4 Chiral molecules 187
6.5 Molecular vibrations 188
The symmetries of molecular orbitals
6.6 Symmetry-adapted linear combinations 191
6.7 The construction of molecular orbitals 192
6.8 The vibrational analogy 194
Representations
6.9 The reduction of a representation 194
6.10 Projection operators 196
Chemistry 481, Spring 2014, LA Tech
Chapter 6-2
Symmetry
M.C. Escher
Chemistry 481, Spring 2014, LA Tech
Chapter 6-3
Symmetry Butterflies
Chemistry 481, Spring 2014, LA Tech
Chapter 6-4
Fish and Boats
Symmetry
Chemistry 481, Spring 2014, LA Tech
Chapter 6-5
Symmetry elements and operations
• A symmetry operation is the process of
doing something to a shape or an object so
that the result is indistinguishable from the
initial state
• Identity (E)
• Proper rotation axis of order n (Cn)
• Plane of symmetry (s)
• Improper axis (rotation + reflection) of order
n (Sn), an inversion center is S2
Chemistry 481, Spring 2014, LA Tech
Chapter 6-6
2) What is a symmetry operation?
Chemistry 481, Spring 2014, LA Tech
Chapter 6-7
E - the identity element
The symmetry operation corresponds to
doing nothing to the molecule. The E
element is possessed by all molecules,
regardless of their shape.
C1 is the most common element leading to E,
but other combination of symmetry
operation are also possible for E.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-8
Cn - a proper rotation axis
of order n
• The symmetry operation Cn corresponds to
rotation about an axis by (360/n)o.
• H2O possesses a C2 since rotation by 360/2o = 180o
about an axis bisecting the two bonds sends the
molecule into an
• indistinguishable form:
Chemistry 481, Spring 2014, LA Tech
Chapter 6-9
s - a plane of reflection
The symmetry operation corresponds to reflection in
a plane. H2O possesses two reflection planes.
Labels: sh, sd and sv.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-10
i - an inversion center
The symmetry operation corresponds to inversion
through the center. The coordinates (x,y,z) of every
atom are changed into (-x,-y,-z):
Chemistry 481, Spring 2014, LA Tech
Chapter 6-11
Sn - an improper axis of order n
The symmetry operation is rotation by (360/n)o and
then a reflection in a plane perpendicular to the
rotation axis.
operation is the
same as an
inversion is S2
=i
a reflection so
S1 = s.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-12
2) What are four basic symmetry elements and
operations?
Chemistry 481, Spring 2014, LA Tech
Chapter 6-13
3) Draw and identify the symmetry elements in:
a) NH3:
b) H2O:
c) CO2:
d) CH4:
e) BF3:
Chemistry 481, Spring 2014, LA Tech
Chapter 6-14
Point Group
Assignment
There is a systematic
way of naming most
point groups
C, S or D for
the principal
symmetry axis
A number for the
order of the
principal axis
(subscript) n.
A subscript h, d, or v
for symmetry planes
Chemistry 481, Spring 2014, LA Tech
Chapter 6-15
4) Draw, identify symmetry elements and assign the
point group of following molecules:
a) NH2Cl:
b) SF4:
c) PCl5:
d) SF6:
e) Chloroform
f) 1,3,5-trichlorobenzene
Chemistry 481, Spring 2014, LA Tech
Chapter 6-16
Special Point Groups
Linear molecules have a C∞ axis - there are an
infinite number of rotations that will leave a linear
molecule unchanged
If there is also a plane of symmetry perpendicular to
the C∞ axis, the point group is D∞h
If there is no plane of symmetry, the point group is
C∞v
Tetrahedral molecules have a point group Td
Octahedral molecules have a point group Oh
Icosahedral molecules have a point group Ih
Chemistry 481, Spring 2014, LA Tech
Chapter 6-17
Point groups
It is convenient to classify molecules
with the same set of symmetry elements
by a label. This label summarizes the
symmetry properties. Molecules with the
same label belong to the same point
group.
For example, all square molecules
belong to the D4h point group
irrespective of their chemical formula.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-18
5) Determine the point group to which each of
the following belongs:
a) CCl4
b) Benzene
c) Pyridine
d) Fe(CO)5
e) Staggered and eclipsed ferrocene, (η5-C5H5)2Fe
f) Octahedral W(CO)6
g) fac- and mer-Ru(H2O)3Cl3
Chemistry 481, Spring 2014, LA Tech
Chapter 6-19
Character tables
Summarize a considerable amount of information
and contain almost all the data that is needed to
begin chemical applications of molecule.
C2v
E
C2 sv sv'
A1
1
1
1
1
z
x2, y2, z2
A2
1
1
-1 -1
Rz
xy
B1
1
-1 1
-1
x, Ry xz
B2
1
-1 -1 -1
y, Rx yz
Chemistry 481, Spring 2014, LA Tech
Chapter 6-20
Character Table Td
E
8C3
3C2
6S4
6sd
1
1
1
1
1
1
1
1
-1
-1
2
-1
2
0
0
3
0
-1
1
-1
(Rx, Ry, Rz)
3
0
-1
-1
1
(x, y, z)
Chemistry 481, Spring 2014, LA Tech
x2+y2+z2
2z2-x2-y2, x2-y2
(xy, xz, yz)
Chapter 6-21
Information on Character Table
• The order of the group is the total number of
•
•
•
•
symmetry elements and is given the symbol h.
For C2v h = 4.
First Column has labels for the irreducible
representations. A1 A2 B1 B2
The rows of numbers are the characters (1,-1)of
the irreducible representations.
px, py and pz orbitals are given by the symbols x, y
and z respectively
dz2, dx2-y2, dxy, dxz and dyz orbitals are given by the
symbols z2, x2-y2, xy, xz and yz respectively.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-22
H2O molecule belongs to C2v point
group
Chemistry 481, Spring 2014, LA Tech
Chapter 6-23
Symmetry Classes
The symmetry classes for each point group and are
labeled in the character table
Label Symmetry Class
A
B
E
T
Singly-degenerate class, symmetric with
respect to the principal axis
Singly-degenerate class, asymmetric with
respect to the principal axis
Doubly-degenerate class
Triply-degenerate class
Chemistry 481, Spring 2014, LA Tech
Chapter 6-24
Molecular Polarity and Chirality
Polarity:
Only molecules belonging to the point groups Cn,
Cnv and Cs are polar. The dipole moment lies along
the symmetry axis formolecules belonging to the
point groups Cn and Cnv.
• Any of D groups, T, O and I groups will not be
polar
Chemistry 481, Spring 2014, LA Tech
Chapter 6-25
Chirality
Only molecules
lacking a Sn axis
can be chiral.
This includes
mirror planes
and a center of
inversion as
S2=s , S1=I and Dn
groups.
Not Chiral: Dnh,
Dnd,Td and Oh.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-26
Meso-Tartaric Acid
Chemistry 481, Spring 2014, LA Tech
Chapter 6-27
Optical Activity
Chemistry 481, Spring 2014, LA Tech
Chapter 6-28
Symmetry allowed combinations
• Find symmetry species spanned by a set of
orbitals
• Next find combinations of the atomic orbitals on
central atom which have these symmetries.
• Combining these are known as symmetry adapted
linear combinations (or SALCs).
• The characters show their behavior of the
combination under each of the symmetry
operations. several methods for finding the
combinations.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-29
Example: Valence MOs of Water
 H2O has C2v symmetry.
 The symmetry operators of the C2v group all
commute with each other (each is in its own
class).
 Molecualr orbitals should have symmetry
operators E, C2, sv1, and sv2.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-30
Building a MO diagram for H2O
z
y
x
Chemistry 481, Spring 2014, LA Tech
Chapter 6-31
a1 orbital of H2O
Chemistry 481, Spring 2014, LA Tech
Chapter 6-32
b1 orbital of H2O
Chemistry 481, Spring 2014, LA Tech
Chapter 6-33
b1 orbital of H2O, cont.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-34
b2 orbital of H2O
Chemistry 481, Spring 2014, LA Tech
Chapter 6-35
b2 orbital of H2O, cont.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-36
[Fe(CN)6]4-
Chemistry 481, Spring 2014, LA Tech
Chapter 6-37
Reducing the Representation
Use reduction formula
1
Chemistry 481, Spring 2014, LA Tech
Chapter 6-38
MO forML6 diagram Molecules
Chemistry 481, Spring 2014, LA Tech
Chapter 6-39
Group Theory and Vibrational
Spectroscopy
• When a molecule vibrates, the symmetry of the
molecule is either preserved (symmetric
vibrations) or broken (asymmetric vibrations).
• The manner in which the vibrations preserve or
break symmetry can be matched to one of the
symmetry classes of the point group of the
molecule.
• Linear molecules: 3N - 5 vibrations
• Non-linear molecules: 3N - 6 vibrations (N is the
number of atoms)
Chemistry 481, Spring 2014, LA Tech
Chapter 6-40
Chemistry 481, Spring 2014, LA Tech
Chapter 6-41
Hence we can deduce G3N for our triatomic molecule
in three lines:
E
3
3
unshifted atoms
c/unshifted atom
\
G3N
9
sxz
1
1
C2
1
-1
-1
1
syz
3
1
3
For more complicated molecules this shortened
method is essential!!
Having obtained G3N, we now must reduce it to find
which irreducible representations are present.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-42
Reducible Representations(3N) for
H2O: Normal Coordinate Method
• If we carry out the symmetry operations of C2v on this
set, we will obtain a transformation matrix for each
operation.
• E.g. C2 effects the following transformations:
• x1 -> -x2, y1 -> -y2, z1 -> z2 , x2 -> -x1, y2 -> -y1, z2 ->
z1, x3 -> -x3 , y3 -> -y3, z3 -> z3.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-43
Summary of Operations by a set of four
9 x 9 transformation matrices.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-44
Use Reduction Formula
Chemistry 481, Spring 2014, LA Tech
Chapter 6-45
Example H2O, C2v
Chemistry 481, Spring 2014, LA Tech
Chapter 6-46
Use Reduction Formula:
1
ap   c( R )c p ( R )
g R
to show that here we have:
G3N = 3A1 + A2 + 2B1 + 3B2
This was obtained using 3N cartesian coordinate vectors.
Using 3N (translation + rotation + vibration) vectors would
have given the same answer.
But we are only interested in the 3N-6 vibrations.
The irreducible representations for the rotation and
translation vectors are listed in the character tables,
e.g. for C2v,
Chemistry 481, Spring 2014, LA Tech
Chapter 6-47
GT = A1 + B1 + B2
GR = A2 + B1 + B2
i.e. GT+R = A1 + A2 + 2B1 + 2B2
But Gvib = G3N - GT+R
Therefore Gvib = 2A1 + B2
i.e. of the 3 (= 3N-6) vibrations for a molecule
like H2O, two have A1 and one has B2 symmetry
Chemistry 481, Spring 2014, LA Tech
Chapter 6-48
Further examples of the determination of Gvib, via G3N:
NH3
N
(C3v)
H
H
C3v
\ G3N
H
12
2C3
0
3 sv
2
12
0
2
E
Reduction formula G3N = 3A1 + A2 + 4E
GT+R (from character table) = A1 + A2 + 2E,
\ Gvib = 2A1 + 2E
(each E "mode" is in fact two vibrations (doubly degenerate)
Chemistry 481, Spring 2014, LA Tech
Chapter 6-49
H
CH4
(Td)
C
H
H
Td
E
15
\ G3N
15
H
8C3
0
3C2
-1
0
-1
6S4
-1
6 sd
3
-1
3
Reduction formula G3N = A1 + E + T1 + 3T2
GT+R (from character table) = T1 + T2,
\ Gvib = A1 + E + 2T2
(each E "mode" is in fact two vibrations (doubly degenerate),
and each T2 three vibrations (triply degenerate)
Chemistry 481, Spring 2014, LA Tech
Chapter 6-50
XeF4
F
(D4h)
F
Xe
F
D4h
E
15
\G3N
15
2C4
1
1
C2
-1
-1
F
2C2'
-3
-3
2C2" i 2S4
-1 -1 -1
-1
-1
-1
sh
5
2s v
3
5
3
2s d
1
1
Reduction formula 
G3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu
GT+R (from character table) = A2g + Eg + A2u + Eu,
\ Gvib = A1g + B1g + B2g + A2u + B2u + 2Eu
For any molecule, we can always deduce the overall symmetry
of all the vibrational modes, from the G3N representation.
To be more specific we need now to use the
INTERNAL COORDINATE method.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-51
INTERNAL COORDINATE METHOD
We used one example of this earlier - when we used
the "bond vectors" to obtain a representation
corresponding to bond stretches.
We will give more examples of these, and also the other
main type of vibration - bending modes.
For stretches we use as internal coordinates changes
in bond length, for bends we use changes in bond angle.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-52
Let us return to the C2v molecule:
r1
H
O
r2
H

C2v
Gstretch
Gbend
Use as bases for stretches:
Dr1, Dr2.
Use as basis for bend:
D
E
2
1
C2
0
1
sxz
syz
0
1
2
1
N.B. Transformation matrices for Gstretch
E, syz:
 1 0


 0 1
C2, sxz :
:
 0 1


 1 0
i.e. only count UNSHIFTED VECTORS (each of these +1 to c).
Chemistry 481, Spring 2014, LA Tech
Chapter 6-53
Gbend is clearly irreducible, i.e. A1.
Gstretch reduces to A1 + B2
We can therefore see that the three vibrational
modes of H2O divide into two stretches (A1 + B2)
and one bend (A1).
We will see later how this information helps
in the vibrational assignment.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-54
Other examples:
NH3
r1
r3
H
N
H
r2
H
1 oppos ite to r1
2 oppos ite to r2
3 oppos ite to r3
Bases for stretches: Dr1, Dr2, Dr3.
Bases for bends: D1, D2, D3.
C3v
Gstretch
Gbend
E
3
3
2C3
0
0
3s
1
1
Reduction formula Gstretch = A1 + E
Gbend = A1 + E
(Remember Gvib (above) = 2A1 + 2E)
Chemistry 481, Spring 2014, LA Tech
Chapter 6-55
H
r1
CH4
r4
C
r2
r3
H
6 angles 1,.....6, where 1
lies between r1 and r2 etc.
H
H
Bases for stretches: Dr1, Dr2, Dr3, Dr4.
Bases for bends: D1, D2, D3, D4, D5, D6.
Td
Gstretch
Gbend
E
4
6
8C3
1
0
3C2
0
2
6S4
0
0
6 sd
2
2
Reduction formula Gstretch = A1 + T2
Gbend = A1 + E + T2
But G3N (above) = A1 + E + 2T2
Chemistry 481, Spring 2014, LA Tech
Chapter 6-56
IR Absorptions
Infra-red absorption spectra arise when a molecular
vibration causes a change in the dipole moment of
the molecule. If the molecule has no permanent
dipole moment, the vibrational motion must create
one; if there is a permanent dipole moment, the
vibrational motion must change it.
Raman Absorptions
Deals with polarizability
Chemistry 481, Spring 2014, LA Tech
Chapter 6-57
Raman Spectroscopy
• Named after discoverer, Indian physicist C.V.Raman (1927).
It is a light scattering process.
• Irradiate sample with visible light - nearly all is transmitted;
of the rest, most scattered at unchanged energy
(frequency) (Rayleigh scattering), but a little is scattered at
changed frequency (Raman scattering). The light has
induced vibrational transitions in molecules (ground 
excited state) - hence some energy taken from light,
• scattered at lower energy, i.e. at lower wavenumber.
Raman scattering is weak - therefore need very powerful
light source - always use lasers (monochromatic, plane
polarised, very intense).
• Each Raman band has wavenumber:
where n = Raman scattered wavenumber
n0 = wavenumber of incident radiation
nvib = a vibrational wavenumber of the molecule
(in general several of these)
Chemistry 481, Spring 2014, LA Tech
Chapter 6-58
Molecular Vibrations
• At room temperature almost all molecules are in
their lowest vibrational energy levels with
quantum number n = 0. For each normal mode, the
most probable vibrational transition is from this
level to the next highest level (n = 0 -> 1). The
strong IR or Raman bands resulting from these
transitiions are called fundamental bands. Other
transitions to higher excited states (n = 0 -> 2, for
instance) result in overtone bands. Overtone
bands are much weaker than fundamental bands.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-59
If the symmetry label of a normal mode
corresponds to x, y, or z, then the
fundamental transition for this normal mode
will be IR active.
If the symmetry label of a normal mode
corresponds to products of x, y, or z (such as
x2 or yz) then the fundamental transition for
this normal mode will be Raman active.
Chemistry 481, Spring 2014, LA Tech
Chapter 6-60
Chemistry 481, Spring 2014, LA Tech
Chapter 6-61