2
2
3
1.
Symmetry Elements and Operations
3
2.
Groups
6
3.
Example of a triangular planar molecule (AB3) point group D3h.
11
4.
Matrix Representation for symmetry elements and corresponding operators.
13
5.
Example of a tetrahedral molecule.
17
6.
Construction of character table
19
7.
Decomposition of a reducible representation into irreducible representations
23
8.
Explanation of symbols used in a character table
27
9.
Method of classifying molecules into point group
28
10.
Properties of point groups
29
1. Symmetry Elements and Operations
What is a symmetry operation?
Symmetry operation is an operation (rotation, reflection, or a combination of both) that
upon its performance on any object (molecule, figure etc) brings a new configuration
which is indistinguishable (but not the same) from the original configuration. Identity
operation can be thought as an exception, because it is really a do-nothing operation.
Its presence has to be conceived in order to meet mathematical needs of a group.
This means that every molecule has at least on symmetry operation, that being
identity operation.
Kinds of symmetry operations
1. Rotation, Cn
3
4
2. Reflection, 
3. Rotation + Reflection (also known as Improper rotation), Sn
Rotation operation
Rotation is done along a line (known as axis and is represented as Cn, where n = 2 /
 ; is the angle of rotation. This is to say that a rotation by 1200 will be C3. (2 /1200 =
3). In order to identify a 2 /  rotation, its axis must be defined.
An example of a square planar molecule
It is easy to find that rotation by 900 along an axis passing
through the center of the square and perpendicular to its
plane (C4), is a symmetry operation. (The configuration
brought about by C4 cannot be distinguished from the
configuration before C4 was applied).
If the square is rotated, as above, by 900 two times i.e., net angle of rotation being
1800, then this double rotation through 900 can be represented by a single rotation
through 1800 (C2) C4.C4 = C42 = C2 .This is also a symmetry operation. Moreover, the
rules of group theory dictate that the combination of two symmetry operations must
also be a symmetry operation.
Similarly C4 . C4 . C4 = C43 is also a symmetry
operation.
Reflection
This operation is done along a plane whose both faces are mirrors. To find a
reflection symmetry in a molecule, the plane is passed through the body of the
molecule and the configuration of the image obtained is compared with the original
configuration. Let's consider the square planar molecule again. If you imagine that
the plane passes through the square such that it is perpendicular to the square and
bisects sides AB and DC, then the configuration after reflection will be
4
5
indistinguishable from the older configuration. Reflections are represented by .
Kinds of reflection Operations
1. v : In this kind of reflection, the plane contains the rotational axis of the highest
order. It is also called vertical reflection.
 h: Horizontal reflection. The plane in this case is perpendicular to the rotational
axis of the highest order.
 d: It is a diagonal reflection. For example, the plane passing through diagonal
points A & C is d.
[find out all the reflection operations in a square planar molecule]
Improper Rotations Sn
Sn is a combination of rotation by 2 / n followed by a reflection whose plane is
perpendicular to the axis of rotation OR vice versa.
Some features of Sn
1. The order is not important, i.e., it is immaterial whether you apply rotation first and
then reflection; or first reflection then rotation.
Cn . h = h . Cn
i.e., these two operations commute strictly only for improper rotations.
2.It is not necessary that either reflection or rotation are symmetry operations by
themselves. That means, the separate existence of symmetry rotation operation or
symmetry reflection operation is not a precondition for Sn.
5
6
3.S2 = C2. =  .C2 = i ; where is an inversion on a center.
An example
Square planar molecule is a very simple case to use. It already has symmetry
rotation operations (C4, C42, C43). It also contains symmetry reflection operation
perpendicular to the axis of rotation, h.
Therefore, combination of these results in:
C4 . h = S4 ; C42 . h = C2 . h = S2 ( inversion) ; C43 . n = S43
Now we know that symmetry operations are applied along a point, an axis or a plane.
Such a point, line or a plane is called symmetry elements of a group. For every
symmetry operation there is a symmetry element. Sometimes it is difficult to
distinguish between them. They are used synonymously. The following table shows
their difference :
Symmetry Element
Symmetry Operation
Identity
Plane
Center of Symmetry or inversion
center
Rotation through 3600 or doing nothing; E
Reflection in plane; 
n-fold axis
n-fold rotation reflection axis
Inversion of all atoms through a center; i
One or several rotations about on axis through an angle
2 /n; Cn
Rotation through an angle of 2/n followed by reflection
in a plane perpendicular to the rotation axis; Sn
2. Group
Group is a set of elements which satisfy the following properties:
1. There is an identity element, E, so that A . E = A for any A belonging to the group.
2. The product of two elements is also member of the group, i.e., if A and B belong to
6
7
the group then A.B will also be a member of the group. It means the group is closed
under the given operation.
3. Every element has its inverse as the member of the group. i.e., if A belongs to the
group, then A-1 also belongs to the group. If A.B = E it means a is the inverse of B
and vise versa.
4. Group members obey the associative law, that is A(B.C) = (A.B)C.
The order of a group is defined as the member of elements present in the group. As
for example, the order of the above group is 8 because it has 8 elements. The
element of a group can be several things as you define them. It can be integers,
vectors, matrices, symmetry operations (elements) etc. One has to define the
operation which goes on in the group which can be several things like addition,
multiplication, symmetry operations etc.
Examples of Groups
a) Collection of all positive and negative integers is a group under the operation of
addition. It is an infinite group.
b) Collection of all numbers forms a group under the operation of multiplication .
c) You will soon learn that collection of all the symmetry operations (elements)
belonging to any molecule from a group. (As a matter of fact this is the reason that
we are in the class on group theory and this property allows us to use mathematical
tools of groups for understanding the properties. of molecules).
Subgroups
They are smaller groups present in the group. They obey all the rules of a group.
Their order must divide the order of the group. That is, if you have a group of order
seven, then you cannot have any subgroup except a subgroup of order 1 that
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8
contains E only.
Similarly, if you have a group of order 12, you should not waste your time in looking
for a subgroup of order 5, 7, 8,9,10 and 11. Remember there is always a subgroup of
order 1.
Abelian Group: It is a group in which every element commutes with every other
element, i.e., A.B = B.A for every A and B.
Cyclic Group: It is a group which is generated by a single element called generator.
[A, A2. A3. ............. An = E] . The order of the group is n.
Cn is a generator of a group of order n and its members are Cn, Cn, Cn …… Cn = E
For Sn , things are a bit complicated.
If n is even, then Sn generates a group of order n and its members are :
Sn, Cn2, Sn3……… (Sn)n-1= Sn n-1, (Sn)n = E
S6 generates the group [S6 , C62 , S63 = S2 , C64 , S65, E ]
If n is odd, Sn generates a group of order 2n.
[Sn , Cn 2, Sn3…….(Sn)n = h , Cn, Sn2…… Sn n-1, (Sn)2n = E ]
For example, members of the group generated by S7 are :
S7, C72, S73, C74, S75, C76, h , C7, S72, C73, S74, C75, S76, E
Example :
Now, we consider the some of the symmetry elements belonging to a square planer
figure and also examine if together they from a group. The symmetry elements are as
follows:
Symmetry elements
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9
1. Identity element, E
2. Clockwise rotation thru 900, C4
3.Clockwise rotation thru 1800, C2
4. Clockwise rotation thru 2700, C3
(All rotations along the axis passing through the center)
5. Reflection about the horizontal plane thru mid points
of 1 & 4 and 2 & 3 , v(1)
6. Reflection about the horizontal plane thru mid points
of 1 & 2 and 3 & 4 , v(2)
7.Reflection about the diagonal plane thru points 1 and
3, d1
8. Reflection about the diagonal plane thru points 2 and
4, d2
Mathematical representations of these symmetry elements.
or,
(Numbers in the second row are the images of those in the first row)
;
;
1 2 3 4 
1 2 3 4 
 v (v2)  
;

;
4 3 2 1 
2 1 4 3
 v (1)  
;
Let us find out the products of elements:
1 2 3 4 1 2 3 4 
1 2 3 4 
1 2 3 4  1 2 3 4 
1 2 3 4 
 v (1).C2  


   v (2)
 4 3 2 1  3 4 1 2   2 1 4 3 
 v (1).C4  


  d2
 4 3 2 1   2 3 4 1  3 2 1 4 
Similarly, we can compute the product of all elements with each other and will find
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10
that the resulting elements are already members of the group. Inverse of C 4 (C4-1) is
C43 and vice-versa. Moreover, C2, v(1) ,v(2),d1, d2 and of course E are their own
inverses.
Multiplication Table
E
E
E
C2
C43
v(1)
v(2)
d1
d2
C4
C2
C43
v(1)
v(2)
d1
d2
E
d1
d2
v(2)
v(1)
C4
C4
C4
C2
C43
C2
C2
C43
E
C4
v(2)
v(1)
d2
d1
C43
C43
E
C4
C2
d2
d1
v(1)
v(2)
v(1)
v(1)
d2
v(2)
d1
E
C2
C43
C4
v(2)
v(2)
d1
v(1)
d2
C2
E
C4
C43
d1
d1
v(1)
d2
v(2)
C4
C43
E
C2
d2
d2
v(2)
d1
v(1)
C43
C4
C2
E
Classes
A group can be divided in several classes, also called conjugacy classes. The
importance of classes will be clear in our later studies. It is time consuming to find out
all classes. Choose any element, perform the so called similarity transformation, i.e.,
compute x-1ax, where x and a belong to the group. For each 'a' perform this
computation with x being all members of the group. Let us find out classes of the
above group.
E-1 C4 E = C4 ; C4-1 C4 C4 = C4 ; C2-1C4C2 = C4 ; (C43)-1 C4 C43 = C4 ;
v(1) -1 C4 h = v(1) C4 h = v(1)d1 = C43 ; v(2)-1 C4 v = v(2)d2 = C43 ; d1-1C4
d1 = C43 and d2-1 C4d2 = C43
This way we found out the whole class induced by C4. After the similar operations
with all elements we will find that there are altogether five classes and they are:
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11
E ; C4, C43 ; C2 ; v(1), v(2) ; d1, d2
How many classes an abelian group of order n will have?
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12
3. Example of a triangular planar molecule AB3 (Planar molecule);
(Point Group D3h)
Symmetry elements belonging to this group are:
1. Identity element, E
2. Rotation about the axis passing thru the mid point and perpendicular to the
molecular plane, by 1200, C32
3. Rotation about the above described axis by 2400, C32
4. Rotation thru 1800 about the axis joining the corners with the mid point, C2 and
there are three of them, called C2 (1), C2 (2), and C2 (3).
5. Reflection about the plane containing the line joining the corner with the mid point
and perpendicular to the molecular plane and again there are three of them, v(1),
v(2) and v(3).
6. Reflection about the plane, coplanar with the molecular plane, h.
7. Rotation thru 1200 and then reflection about the molecular plane or first reflection
and then rotation, S3
8. Rotation thru 2400 followed by reflection about the molecular plane or the other
way around, S32.
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13
GROUP MULTIPLICATION TABLE for D3h ( AB3 planar molecule)
E
C3
C32
C2(1)
C2(2)
C2(3)
v(1)
v(2)
v(3)
h
S3
S32
E
E
C3
C32
C2(1)
C2(2)
C2(3)
v(1)
v(2)
v(3)
h
S3
S32
C3
C3
C32
E
C2(2)
C2(3)
C2(1)
v(2)
v(3)
v(1)
S3
S32
h
C32
C32
E
C3
C2(3)
C2(1)
C2(2)
v(3)
v(1)
v(2)
S32
h
S3
C2(1)
C2(1)
C2(3)
C2(2)
E
C32
C3
h
S32
S3
v(1)
v(3)
v(2)
C2(2)
C2(2)
C2(1)
C2(3)
C3
E
C32
S3
h
S32
v(2)
v(1)
v(3)
C2(3)
C2(3)
C2(2)
C2(1)
C32
C3
E
S32
S3
h
v(3)
v(2)
v(1)
v(1)
v(1)
v(3)
v(2)
h
S32
S3
E
S32
C3
C2(1)
C2(3)
C2(2)
v(2)
v(2)
v(1)
v(3)
S3
h
S32
C3
E
C32
C2(2)
C2(1)
C2(3)
v(3)
v(3)
v(2)
v(1)
S32
S3
h
C32
C3
E
C2(3)
C2(2)
v(1)
h
h
S3
S32
v(1)
v(2)
v(3)
C2(1)
C2(2)
C2(3)
E
C3
C32
S3
S3
S32
h
v(2)
v(3)
v(1)
C2(2)
C2(3)
C2(1)
C3
C32
E
S32
S32
h
S3
v(3)
v(1)
v(2)
C2(3)
v(1)
C2(2)
C32
E
C3
INVERSES
C3-1 = C32 ; C2-1(1) = C2(1) ; C2-1(2) = C2(2); C2-1(3) = C2(3) ; v-1(1) = v(1) ; v-1(2) =
v(2); v-1(3) = v(3) ; h-1 = h ; S3-1 = S32 ; (S32) –1 = S3 ; S32 . S3 = E.
How to find all the classes ?
Let us pick any element, say C3, and find out all the products of type a-1C3 a where a
is any element of the group:
E-1C3E = C3 ; C3-1 C3C3 = C3 ; (C32)-1 C3C32 = C3
C2-1(1)C3.C2(1) = C2(1) .C3C2(1) = C2(1)C2(3) = C32 ; C2-1 (2)C3C2(2) =C2(2)C2(1) =
C32 ; C-1(3)CC2(3) = C(3)C2(2) = C32 ; v-1(1)C3v(1) =v(1) v(3) = C32 ;
v-1(2)C3v(2) = v(2) v(1) = C32 ; v-1 ((3)C3v(3) =v(3) v(2) = C32 ;
h-1C3h = hS3 = C3 ; S3-1C3 . S3 = S32C3 . S3 =S32.S32 = C3 ; (S32)-1C3S32 = S3h = C3
These results indicate that there is a class which contains only C3 and C32, because
similarity transformations on C3 yield either C3 or C32 only. Similarly it can be
demonstrated that similarity transformation on h yield only h; on C2(1), C2(2) or
C2(3) yield only C2(1), C2(2) or C2(3) ; on v yield v(1), v(2) or v(3) and on S3 or
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S32 yield only S3 or S32 .
Based on these observations the D3h group can be classified as follows:
{E}; {C3,C32}; {C2(1), C2(2), C2(3)}; {v(1), v(2), 2(3)} ; {h} and {S3, S32}.
Subgroups : Orders of the possible subgroups are 1, 2, 3, 4, and 6; all divisors of
the group of order 12. They are:
(E), (E, C3, C32); {E, C2(1)} {E, C2 (2)} etc.
There will be 6 subgroups of order 2.
4. Matrix Representation for symmetry elements and corresponding
operators
x1 = r cos (-) = r cos  cos  + r. sin .
sin  = x. cos  + y sin 
y1 = r sin (-) = r sin  cos  - r. cos  sin
 = y cos  - x sin 
cos 1200 = - ½ ; sin 1200 = 3/2
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15
EFFECTS OF THE SYMMETRY OPERATIONS ON ATOMS PLACED ON
THE CORNERS ( 1, 2 and 3) of AN EQUILATERAL TRIANGLE AND
ALSO ON A POINT ( x,y,z )IN THE SPACE .
E
C3
C32
C2(1)
C2(2)
C2(3)
v(1)
v(2)
v(3)
h
S3
1
1
2
3
1
3
2
1
3
2
1
2
2
2
3
1
3
2
1
3
2
1
2
3
3
3
1
2
2
1
3
2
1
3
3
1
x
x
½(-x+3y)
½(-x-3y)
x
½(-x-3y)
½(-x+3y)
x
½(-x-3y)
½(-x+3y)
x
½(-x+3y)
y
y
½(-3x – y)
½(3x – y)
-y
½( y -3x )
½( y +3x)
-y
½( y -3x )
½( y +3x)
y
½(-3x – y)
z
z
z
z
-z
-z
-z
z
z
z
-z
-z
S32
3
1
2
½(-x-3y)
½(3x – y)
-z
15
16
Matrix Representation of symmetry operations of D3h
16
17
17
18
5. Example of Tetrahedral Molecule, Td
Let ABCD be a tetrahedron
AB, BC, CD, DA, BD and AC are the edges of the tetrahedron. Z axis passes through the
midpoint of AB.
Y Axis passes through the midpoint of BD; X
The atom at the origin remains unshifted.
18
Axis passes through the midpoint of BD
19
EFFECT OF SYMMETRY OPERATIONS ( Tetrahedral molecule)
Td
1. E
2. C2 ( X )
3. C2 ( Y )
4. C2 ( Z )
5. C3 ( A )
6. C32 ( A )
7. C3 ( B )
8. C32 ( B )
9. C3 ( C )
10. C32 ( C )
11. C3 ( D )
12. C32 ( D )
13. S4 ( X )
14. S43 ( X )
15. S4 ( Y )
16. S43 ( Y )
17. S4 ( Z )
18. S43 ( Z )
19. AB = S43 ( Z ) c2 ( Z )
20. AC = S4 ( X ) c2 ( Y )
21. AD = S43 ( Y ) c2 ( X )
22. BC = S4 ( Y ) c2 ( X )
23. BD = S43 ( X ) c2 ( Y )
24. CD = S4 ( Z ) c2 ( X )
A
A
C
D
B
A
A
C
D
D
B
B
C
B
D
C
B
C
D
A
A
A
D
C
B
Question: Prepare the multiplication table for Td
19
B
B
D
C
A
D
C
B
B
A
D
C
A
C
A
A
D
D
C
B
D
C
B
B
A
C
C
A
B
D
B
D
D
A
C
C
A
B
D
B
D
A
B
A
D
C
B
C
A
C
D
D
B
A
C
C
B
A
C
B
A
D
D
A
C
B
C
A
B
C
B
D
A
D
D
X
X
X
-X
-X
-Z
Y
Z
Y
Z
-Y
-Z
-Y
-X
-X
Z
-Z
-Y
Y
Y
X
-Z
Z
X
-Y
Y
Y
-Y
Y
-Y
X
-Z
X
Z
-X
-Z
-X
Z
-Z
Z
-Y
-Y
X
-X
X
-Z
Y
Y
Z
-X
Z
Z
-Z
-Z
Z
-Y
-X
Y
X
-Y
X
Y
-X
Y
-Y
-X
X
-Z
-Z
Z
Y
-X
X
Y
Z
20
6. Construction of Character Table
Character Table for D3h point group
E
1
1
2
1
1
2
A1'
A2'
E'
A1"
A2"
E"
2C3
1
1
-1
1
1
-1
3C2
1
-1
0
1
-1
0
h
1
1
2
-1
-1
-2
3v
1
-1
0
-1
1
0
2S3
1
1
-1
-1
-1
1
x2+y2
Rz
(x,y)
Z
(Rx,Ry)
z2
(x2-y2, xy)
(xz,yz)
We have already described the matrix representations for the different symmetry
elements . Now, first we will find out the characters of those symmetry elements.
Character of a symmetry element will be defined as the sum of the diagonal
elements in the matrix representing the element. The characters, χ, of the symmetry
elements for the point group D3h whose matrix representations have been shown
before ( page 15 & 16) are given below:χ (E)=3,
χ (C3)=0,
χ { v(1)}=1,
χ(C32)=0,
χ {C2(1)} = -1,
χ { v(2)}=1, χ { v(3)}=1,
χ{c2(2)} = -1, χ {C2(3)} = -1
χ (h) = 1, χ (S3) = -2,
χ (S32) = -2
Mathematically it turns out that representations of a group can be expressed in
terms of these characters. As for example, the characters of D3h group, taken
together, will represent this group. And the representation will look like this:-
Γ
E
2C3
2C2
3v
h
2S3
3
0
-1
1
1
-2
But this representation is reducible and hence not suitable for group theoretical
treatment. A character table contains only irreducible representations. Therefore,
our task is to find out the characters of elements corresponding to the irreducible
20
21
representations. The following steps are taken in preparing the character table.
1.
Find the number of possible irreducible representations.
The number of irreducible representations belonging to a particular group is equal to
the number of classes present in that group. So, the group D3h will be represented
by 6 irreducible representations because this group has 6 classes.
2.
Find the dimensions of the irreducible representations
It turns out the sum of squares of the dimensions of irreducible representations is
equal to the number of elements present in the group (order of the group). For the
D3h group, I12 + I22 + I32 + I42 + I52 + I62 = 12; where I represents the dimension
of the irreducible representation. (Generally the dimension of a representation is
equal to the dimension of the matrices involved in the representation and it turns out
that the dimension of an irreducible representation is equal to the character of the
identity element belonging to that representation.) The only possible set of values
which will satisfy the above equation is I1=1, I2=1, I3=1, I4=1, I5=1 and I6=1. So the
D3h group has 6 representations, four with dimensions of one and two with
dimensions of two. (See the character table given above).
3.
Find the characters of elements for each irreducible representation
(a) Sum of squares of characters of elements in a particular irreducible
representation is equal to the number of elements present in the group.
For D3h χ (E2) + 2 χ (C3)2 + 3 χ (v)2 + 3 χ (C2)2 + 1 χ (h)2 + 2 χ (S3)2 = 12
There will be at least six sets of values for these characters which will satisfy the
above equation: the easiest one being where characters for all elements are unity.
The above mentioned condition is also useful in determining whether a certain
representation is reducible. If the sum of squares of characters of elements
21
22
corresponding to a representation is greater than the number of elements in the
group, the representation is reducible, that is, it can be further broken into smaller
representation. As for example, the representations for D3h derived above is a
reducible one because the sum of squares of characters is:
32 + 2x0 + 3 (-1)2 + 3 (1)2 + 1 (1)2 + 2 (-2)2 = 9+0+3+3+1+8 = 24
(b)
The sum of the products of characters of elements corresponding to two
different irreducible representations is zero, that is, two irreducible representations
are orthogonal. In essence what you do is find the product of characters of an
element in two representations and add the products obtained for all elements, and
the sum is zero if the representations are irreducible.
The above mentioned conditions will be used to prepare the character table.
If the matrices representing the similarity transformations outlined before are
separated in blocks as shown in page 15 & 16, then we will have two sets of
matrices representing the group, that is, the previous reducible representation is
divided into two representations as shown below. On considering the characters of
the matrices alone we can express the representations as shown below:Γ
2C3 C3
3v
h
2S3
1
1
-1
1
-1
-1
2
-1
0
0
2
-1
------------------------------------------------3
0
-1
1
1
-2
It can be easily tested that these two smaller representations are irreducible. It can
be also tested that these representations are orthogonal too.
1(2) (1) + 2(1) (-1) + 3(1) (0) + 3(-1) (0) + 1(-1) (2) + 2(-1) (-1) = 2 + (-2) + 0 + (-2) + 2 = 0
Therefore these two representations are in fact genuine representations of the
group, D3h. So far out of 6 irreducible representations we already know three of
22
23
them (marked first, second and fifth representation in the character table). To find
the rest of them we have to use hit and trial method, there is no easy way out. We
have to set up simultaneous equations and solve them. It needs more practice and
intuition than any thing else.
Let us find the remaining two one dimensional representations. To find them we
have to just change the signs in such a way that we generate a new representation
and check if it is orthogonal to any of the already known representations (irreducible
ones). Once you find two such representations, you have found all one dimensional
representations, see the table.
Now the remaining task is to find the irreducible representation of dimension two
because we already know one two-dimensional representation. To do this we solved
the following equations:1 (2) χ (E) + 2 (1) χ (C3) + 3 (0) χ (C2) + 3 (0) χ v) + 1 (2) χ (h + 2 (-1) χ (S3)
= 2.2 – 2 χ (C3) + 2 χ (h) –2 χ (S3) = 0 ....................................................................(1)
This equation derives from the othogonality relation between already known two
dimensional representation (fifth representation in the character table) and the
representation we are trying to find. The orthogonality relation between the fourth
representation and the representation we are trying to find gives:1 (1) χ (E) + 2 (1) χ (C3) + 3 (-1) χ (C2) + 3 (-1) χ( v) + 1 (1) χ (h )+ 2 (1) χ (S3) = 0
or 2 + 2 χ (C3) + 3 χ (C2) –3 χ (v) + χ (h) + 2 χ (S3) = 0 .......................................(2)
Similarly the orthogonality relation between the first representation and the
representation we are trying to find gives:2 + 2 χ (C3) + 3χ (C2) + 3 χ (v) + χ (h) + 2 χ (S3) = 0 .......................................................(3)
Add equations (2) & (3),
4 + 4 χ (C3) + 2 χ (h) + 4 χ (S3) = 0 .......................................................(4)
Multiply equation (1) by 2 and add it to equation (4), we get
23
24
8 + 4 + 6 χ (h) = 0, or χ (h) = -2.
Insertion of the value for the character of h in equation (1) indicates that χ (C3) = χ
(S3) in the last representation. With little manipulation by using the above conditions
one will arrive a set of values for characters which gives the last sixth representation.
THE CHARACTER TABLE of D3h Point Group
Γ1
Γ2
Γ3
Γ4
Γ5
Γ6
First representation
Second representation
Third representation
Fourth representation
Fifth representation
Sixth representation
E
1
1
1
1
2
2
2C3
1
-1
1
1
-1
1
3C2
1
-1
1
-1
0
0
3v
1
-1
-1
-1
0
0
h
1
-1
-1
1
2
-2
2S3
1
-1
-1
1
-1
1
7.DECOMPOSITION OF A REDUCIBLE REPRESENTATION INTO IRREDUCIBLE
REPRESENTATIONS
To apply the concepts of group theory to molecules it is very necessary to break a reducible
representation into irreducible ones. The importance of this step will be clear in next few
chapters. The nature and largeness of a reducible representation will depend upon the kind
of basis sets chosen to mathematically represent the effects of symmetry transformations.
The basis sets can be any of several parameters such as atoms themselves, the coordinate
of a point in the space, the atomic orbitals or their combinations, modes of vibrations, etc.
We have considered the first two in our previous treatment of point group D3h. As we have
seen earlier for D3h, the use of the coordinates of a point (x, y, z) for determining the effects
of symmetry transformations renders a reducible representation which looks as follows:-
Γ1
E
3
2C3
0
24
3C2
-1
3v
1
h
1
2S3
-2
25
Let us see what kind of representation we get if we use a pi orbital on each of
three corners of the triangle, points where atoms are placed. Each p z orbital will
have a positive and a negative sign and therefore there will be altogether 6 half
orbitals to be used for representation; they are P1+,P1-,P2+, P2-,P3+ and P3-.
Effect of symmetry operations on pi orbitals
P1+
E
P1+
C3
P2+
C2(1) P1v
P1+
h
P1S3.
P2-
P1P1P2P1+
P1P1+
P2+
P2+
P2+
P3+
P3P3+
P2P3-
P2P2P3P3+
P3P2+
P3+
P3-.
P3P1-.
P2+
P2P3+.
P1+
P3+
P3+
P1+
P2P2+
P3P1-
There is no need to find out the effect of C32, other C2 s, other v s and S32 because
their characters can be easily determined from the information given above. After
writing matrix representation of the symmetry elements as described earlier and by
adding the diagonal elements of the matrices we get the characters of the symmetry
operations as given below (a representation for the D3h point group).
E
6
Γ2
2C3
0
3C2
0
3v
2
h
0
2S3
0
And therefore we obtained a different representation by using a new basis set. This
illustrates the point we made earlier that a representation of a point group depends
upon the choice of the basis set.
Yet we can have a third representation if we place one pi orbital at each corner just
like before without further dividing them into half orbitals.
P1
E
C3
C2(1)
v
h
S3.
P1
P2
-P1
P1
-P1
-P2
P2
P3
P2
P3
P3
P3
-P2
-P3
P3
P1
P2
P2
-P3
-P3
Character (χ )
3
0
-1
-1
-3
0
25
26
Γ3
E
3
2C3
0
3C2
-10
3v
1
h
-3
2S3
0
A simple rule for finding the character of an symmetry operation
For each orbital or atom or any parameter unshifted by the symmetry operation, 1 is
added to its character; if the parameter is shifted to its negative value, -1 is added to
the character.
The character for C3 is zero because none of the orbitals remain unshifted due to this
operation and hence they contribute zero to the character. On the other hand, in
case of C2 one orbital is changed to its negative contributing –1 to the character
whereas the other two orbitals contribute nothing to the character because they are
changed to orbitals other than themselves.
These reducible representations are made of irreducible representations belonging
to the particular point group. One does not have to deduce irreducible
representations for a point group because they can be easily looked up into the
character tables supplied with group theory books. We have already gone thru the
procedures involved in finding out the irreducible representations belonging to a
group. Once you know the irreducible representations using a simple mathematical
rule you can determine how many times a particular irreducible representation
appears in a reducible representation derived on the basis of the method outlined
above. [Remember you will get different representations for different basis sets.]
26
27
How to break a reducible representation into irreducible ones
The simple mathematical formula is: Number of times an irreducible representation
appears, n = (Sum of products of characters of symmetry operation in reducible
representation and irreducible representation) / (number of symmetry elements
present in the group)
n = χ (R) χ i(R) / g; where g = Order of the group.
Let us decompose the first reducible representation.
Number of times the irreducible representation A1' is present in the reducible representation =
1/12(1x3x1 + 2x0x1 + 3x-1x1 + 3x1x1 + 1x1x1x + 2x-2x1) = 0, that is, A1 is not a
component of the reducible representation.
A2 ' : 1/12 (1x3x1 + 2x0x1 + 3x-1x-1 + 3x1x-1 + 1x1x1 + 2x-2x1) = 0; A2 ' is not present
in the reducible representation.
E : 1/12 (1x3x2 + 2x0x-1 + 3x-1x0 + 3x1x0 + 1x2x1 + 2x-2x1) = 1
Therefore E occurs once in the representation. This is a bi-dimensional representation.
Now we have to look for just one uni-dimensional representation.
A1" ; 1/12 (1x3x1 + 2x0x1 + 3x – 1x1 + 3x1x-1 + 1x1x-1 + 2x-2x-1) = 0
A2" ; 1/12 (1x3x1 + 2x0x1 + 3x – 1x1 + 3x1x-1 + 1x-1 + 2x-2x-1) = 1,
That is A2" is also present in the reducible representation.
Therefore, the reducible representation = A2' + E '
Similarly, it can be shown that the second representation
Γ2 = A1 + A2" + E' + E"
And the third representation Γ3 = A2" + E"
27
28
8. Explanation of Symbols used in the Character Table
Character Table for D3h point group
E
1
1
2
1
1
2
A1'
A2'
E'
A1"
A2"
E"
II
2C3
1
1
-1
1
1
-1
3C2
1
-1
0
1
-1
0
h
1
1
2
-1
-1
-2
3v
1
-1
0
-1
1
0
I
2S3
1
1
-1
-1
-1
1
x2+y2
Rz
(x,y)
z
(Rx,Ry)
III
z2
(x2-y2, xy)
(xz,yz)
IV
Area II
Symbol
A
Remarks
One – dimensional representation which are symmetric with respect to
the rotation about the principal axis of rotation, χ (Cn) = 1.
B
One – dimensional representation which are antisymmetric with respect
to the rotation about the principal axis of rotation, χ (Cn) = -1
E
Two – dimensional representations. Occur in molecules having an axis
higher than C2.
F
Three – dimensional representations. Occur in molecules having more
than two C3 axes.
Subscripts 1 and 2 to
A and B
Symmetric or antisymmetric with respect to a C2 axis (or a vertical
plane of symmetry) perpendicular to the principal axis.
Subscripts g and u to
A and B
Symmetric 'g' or antisymmetric 'u' with respect to a center of symmetry,
(inversion)
Primes and double
primes with A and B
Symmetric or antisymmetric with respect to h . χ (h) = + 1, χ (h) = 1
Area III
In this area we will always find six symbols: x, y, z, Ry, Ry, Rz. The first three present the
coordinates x, y, and z, while R's stand for rotation about the axes specified in the subscripts.
They form the basis for the irreducible representation specified. As for example, in Dh case,
28
29
the coordinate z forms a basis for the A2" representation. The another way of saying the
same thing is z transforms as (or according to) A2". Similarly, x and y together from the
basis for the E' representation, that is, x and y together transform as or according to E'
representation. In the same manner Rz will form the basis for the A2' representation. The
rotations around x and y axes together (Rx,Ry) from the basis for E" representation. i.e. they
transform as the E" representation.
Area IV
In this part of the table are listed all of the squares and binary products of coordinates
according to their transformation properties. In the present example, (x2 + y2) form a basis
for the A1' representation. The pair (xz, yz) transforms according to the representation E".
Similarly, (x2 – y2, z2 ) forms the basis for the representation of E'.
9. Method of Classifying Molecules into Point Groups
29
30
10. Properties of Point Groups
Point
group
C2v
C3v
C4v
C5v
C6v
C2h
C3h
C4h
C5h
C6h
D2h
D3h
D4h
D5h
D6h
D2d
D3d
D4d
D5d
D6d
Symmetry elements of the point group
No. of
elements
4
6
8
10
12
4
6
8
10
12
8
12
16
20
24
8
12
16
20
24
E; C2 ; v ; v
E; C3 ; C32 ; 3v
E; C4 ; C43 ; C42 = C2 ; 2 v ; 2 d
E; C5 ; C54 ; C52 , C53 ; 5 v
E; C6 , C65 ; C62 = C3 , C64 = C32 ; 5 v ; 5 d ; C2
E; C5 ; i ; h
E; 2C3 ; h; S3, S35 = (C32 . h)
E; 2C4; C2; h; i ; S4, S43
E; 2C5; 2C52 ; h; 2S5, 2S53
E; 2C6; 2C3 ; C2; h; i ; 2S6, 2S3
E; C2;C2 ; C2 ; h ; i ; v; 2v
E; C3 ; 3C2 ; h ; 2S3 ; 3v
E; 2C4; C2 ; 2C2; 2C2; h; i ; 2S4, 2v ; 2d
E; 2C5; 2C52 ; 5C2; 2S5, 2S53 ; 5v
E; 2C6; 2C3 ; 3C2; 3C2; 2S6, 2S3 ; h ; 3v ; 3d
E; C2; 2C2; 2S4; 2d
E; 2C3 ; 3C2; 2S6; i ; 3d
E; 2C4 ; C2; 4C2; 2S8; 4d
E; 2C5; 2C52; 5C2; 2S10; 2S103; i ; 6d
E; 2C6; 2C3 ; C2; 2C2; 2S12; 2S123; 2S125; 6d
Special Groups
Td
24
E; 8C3 ; 3C2; 6S4; 6d
30
5
No. of
classes
4
3
5
4
6
4
4
6
6
8
8
6
10
8
12
5
6
7
8
9
31
Some examples of Point Groups
Element
H2O
Benzene
ferrocene, Staggered
configuration
ethane, Staggered
planar boric acid
Methane
Dichloromethane
trans-CrBr2(H2O)4
1,3,5-tribromobenzene
cyclohexane (Chair form)
naphthalene
p-dichlorobenzene
phenanthrene
acetylene
anthracene
o-dichlorobenzene
p-dichlorobenzene
Point Group
(C2v)
(D6h)
(D5d)
(D3d)
(C3h)
(Td)
(C2V)
(D4h)
(D3h)
(D3d)
(D2h)
(D2h)
(C2V)
(Dαh)
(D2h)
(C2V)
D2h
Element
NH3
PF5 , trigonal bipyramidal
ferrocene, eclipsed
configuration
eclipsed ethane
ethylene
chloromethane
chloroform
gauche-CH2ClCH2Cl
IF5
B2H6
thiophene
CCl4
planar formic acid
H2
Chlorobenzene
CO
m-dichlorobenzene
31
Point Group
(C3v)
(D3h)
(D5h)
(D3h)
(D2h)
(C3V)
(C3V)
(C2)
(C4V)
(D2h)
(C2V)
(Td)
(Cs)
(Dαh)
(C2V)
(Cαv)
(C2v)
32
Analysis of Normal Modes of Molecular Vibration (ANMMV)
Normal vibrations belonging to a molecule form bases of Irreducible Representations of a
Character Table. A combination of group theory and experimental data of infra-red (IR) and
Raman Spectroscopy can help choose a structure of a molecule from a list of possible
structures that could be assigned to the molecule. For example, an AB3 type molecule can
have either planar triangular (D3h ) or triangular pyramidal ( C3v ) structures. Group theory
can help select one of these two possible structures, provided spectral data are available.
Moreover, characters for symmetry operations can give insight as to what irreducible
representations stand for symmetric and antisymmetric stretching and in-plane and out-ofplane bending vibrations. With the Character Table one can even predict that which
vibrations are IR active and which are Raman active.
Following steps are taken for ANVMM :
(1)
(2)
(3)
(4)
(5)
(6)
Determination of the Point Group for the molecule.
Regeneration of a Reducible Representation
Decomposition of the Reducible Representation into Irreducible Representations
Reduction of the Reducible Representation to the Vibrational Representation
Selections Rules to distinguish IR and Raman active vibrations
Determination of nature of Vibrational Modes
1. Determination of the Point Group for the molecule : Remember it will depend upon
possible structures that can be assigned to the molecule.
2. Regeneration of a Reducible Representation
As you know, a basis set will be required for this purpose. For application in the field of
ANMMV, a set of displacement coordinate vectors ( xi,yi,zi) is placed on each and every
atom. Matrices representing symmetry operations are constructed and their characters are
tabulated to create a reducible representation. In short, atoms that are moved by the symmetry
operation contribute zero to the character and one that remain unshifted contribute values to
the character as given in the following table :
Contribution made by an unshifted atom to the character of a symmetry operation
Symmetry Operation
E
σ
i
C2
C3
Character (χ)
3
1
-3
-1
0
Symmetry Operation
C4
S3
S4
S6
32
Character (χ)
1
-2
-1
0
33
Consider an example of an AB3 molecule having D3h point group
No. of unshifted atoms
Г
E
2C3
3C2
3σv
σh
2S3
4
1
2
2
4
1
12
0
-2
2
4
-2
3. Decomposition of the Reducible Representation into Irreducible Representations
A reducible representation can be broken down in Irreducible representations by following
method described in earlier chapters ( see page 23 ).
The above representation Г = A'1 + A'2 + 2A''2 + 3E' + E''
Note that sum of the dimensions ( A & B = 1, E = 2 ) is equal 12 given 3n rule.
4. Reduction of the Reducible Representation to the Vibrational Representation
Irreducible representations one each for translational modes ( Tx,Ty,Tz) and for rotational
modes ( Rx,Ry,Rz) are subtracted from the total reducible representation. The irreducible
representations having x, y & z in the corresponding third column of the Character Table are
marked for translational modes. Similarly, irreducible representations containing Rx, Ry & Rz
in their third column are marked for rotational modes. In case E has (x,y) or (Rx,Ry) in the
third column, removal of E will will take care of both Tx & Ty or Rx & Ry, respectively.
For AB3 molecule (D3h point group)
Translation : Irreducible representations (combined) for Tx, Ty is E' and for Tz is A''2
Rotation : Irreducible representation (combined ) for Rx, Ry is E'' and for Rz is A'2
Г = A'1 + A'2 + 2A''2 + 3E' + E''
Therefore, Гvib = A'1 + A''2 + 2 E' ; Note the sum of dimensions of Irreducible
representations is 6 = 3n – 6
5. Selections Rules to distinguish IR and Raman active vibrations
I. R. active : Remaining irreducible representations having x,y,z in their third column in the
Character Table represent normal modes of vibrations that are infra-red active.
33
34
Raman active : Remaining irreducible representations having sqare and binary products of
the coordinates and their combinations ( x2, y2, z2, xy, xz, yz, x2+ y2, x2- y2 etc.) in the fourth
column of the Character Table represent Raman active vibrations.
An irreducible representation can represent both I.R. and Raman active vibration.
For AB3 molecule (D3h point group)
A'1 – Raman active vibration; A''2 – I. R. active vibration
and E' – both Raman and I. R. active vibrations
6. Determination of nature of Vibrational Modes
The aim of this section is to get an idea about the nature of normal modes of vibration i.e.
whether they involve symmetric or assymetric stretching of bonds or in-plane or out-of-plane
bending of bond angles.
First it is determined whether one of the irreducible representations represent an out-of plane bending vibration. While these vibrations change the bond angles , they leave the
molecular plane. Out – of – plane bending ( deformation) vibration is represented by an
irreducible representation having negative value of χ (σh). Since out-of-plane bending
disturbs the symmetry of the molecule, it is not further considered for vibrational analysis
Next, bond distances and bond angles are chosen as internal coordinates for basis set. Effect
of symmetry operations on this basis set is determined ( bonds or angles that remain
unshifted contribute 1 and those that move contribute zero to the character) and then
compiled to generate a reducible representation which is decomposed into irreducible
representations by established method. The resulting irreducible representations represent
vibrations resulting from stretching or bending depending on whether bonds or angles were
selected as basis sets.
Let's consider again the example of AB3 (D3h)
Гvib = A'1 + A''2 + 2 E'
Identify the out-of-plane bending vibration and remove it.
From the examination of the Character Table we can easily determine that A'2 represents a
out – of – plane bending vibration
E
2C3
3C2
3σv
σh
2S3
Гvib
6
0
0
2
4
-2
A''2
1
1
-1
1
-1
-1
34
35
Гvib - A''2
5
-1
1
1
5
-1
Bond distance : Consider the three A-B bond distances as a basis set of the representation.
Compile the effects of symmetry operation to form a reducible representation ГAB.
ГAB
Гvib - A''2 - ГAB
3C2
3σv
σh
2S3
0
1
1
3
0
-1
0
0
2
-1
E
2C3
3
2
Decomposition will reveal that ГAB = A'1 + E' which would mean that A'1 and one of the
E' represent stretching vibrations; stretching vibration for A'1 is symmetric one.
Bond angles : We can consider only two bond angles for the representation because the third
would depend upon the other two ( not independent ). The logical conclusion is that the
remaining irreducible representation ( Гvib - A''2 - ГAB ) should represent the ГBAB. By
examination of the Character Table it is easily found that this representation is simply just
another E.' So , the following can be concluded regarding the nature of normal modes of
vibration :
A'1 --- Symmetric Stretching Vibration
E' ----
Assymetric Stretching Vibration
E' ----
In – Plane Bending Vibration
A''2 ---- Out – of – plane Bending Vibration
35
36
Problems
1. Determine point groups of (a) benzene, (b) planar boric acid,
(c) 1,3,5 tribromobenzene , (d) naphthalene ,
(f) p-dichlorobenzene, (g) PCl3 ( pyramidal) , (h) S2Cl2 (nonplanar),
(i) formaldehyde, (j) CH2FCl and (k) CO32-.
2. How many symmetry operations and how many classes of symmetry operations are
there in (a) C2v point group and (b) C3v point group ?
3. What is the relation between the number of symmetry species and the number of
classes of symmetry operations ?
.
4. For the point group C3v is it true, as it was for the C2v point group, that each
symmetry operation is its own inverse ? Give examples.
5. Generate a reducible representation for eclipsed and staggered ethanes.
6. Assume a molecule of PF5 can have (a) triangular bipyramidal or (b) square planar
pyramidal structure. Describe the application of group theory in the determination of
PF5 , provided data of vibrational spectroscopy are available.
7. Suppose H2O2 can have either one of the cis – or trans – planar structures. Describe
methods step be step for selecting one of the above structures if vibrational data are
available.
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8. None of the normal modes of vibration for a c3v molecule involves an out-of-plane
bending , verify.
9. Compare results obtained by analysis of normal modes of vibration of chloroform and
ammonia.
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