Transcript Ch4Lecture2

Ch 4 Lecture2 Applications of Symmetry
I.
Matrices
A.
Why Matrices? The matrix representations of the point group’s operations will
generate a character table. We can use this table to predict properties.
B.
Definitions and Rules
1) Matrix = ordered array of numbers
2)
 3 2
7 1 or 3 4 2 5


Multiplying Matrices
a) The number of columns of matrix #1 must = number of rows of matrix #2
b) Fill in answer matrix from left to right and top to bottom
c) The first answer number comes from the sum of [(row 1 elements of matrix
#1) X (column 1 elements of matrix #2)]
d) The answer matrix has same number of rows as matrix #1
The answer matrix has same number of columns as matrix #2
1 5 7 3  7  20 3  40 27 43
2 6  4 8  14  24 6  48  38 54

 
 
 

e)
Relevant example:
1 0 0
1 2 3 0  1 0  1  2 3
0 0 1
f)
II.
Exercise 4-4
Representations of Point Groups
A.
Matrix Representations of C2v
1) Choose set of x,y,z axes
a) z is usually the Cn axis
b) xz plane is usually the plane of the molecule
2) Examine what happens after the molecule undergoes each symmetry
operation in the point group (E, C2, 2s)
3)
Transformation Matrix = matrix expressing the effect of a symmetry
operation on the x,y,z axes to give x’,y’,z’
? ? ?  x   x' 
? ? ?   y    y'

    
? ? ?  z   z' 
transformation
matrix
4)
E Transformation Matrix
a. x,y,z  x,y,z
b. What matrix times x,y,z doesn’t change anything?
z
y
x
1 0 0  x   x 
0 1 0   y   y

    
0 0 1  z   z 
E Transformation Matrix
C2
5)
z
0

 0
y
x
6)
sv
y
x
7)
z
y
x
sv(xz) Transformation Matrix
a. x,y,z  x,-y,z
1
b. Correct matrix is:
0   x  - x 
 1 0   y   - y 
0 1  z   z 
0
0  x   x 
0  1 0   y  - y

    
0 0 1  z   z 
z
sd
C2 Transformation Matrix
a. x,y,z  -x, -y, z
1
b. Correct matrix is: 
sv(yz) Transformation Matrix
a. x,y,z  -x,y,z
b. Correct matrix is:  1
0

 0
0
0 0   x  - x 
1 0   y    y 
0 1  z   z 
8)
These 4 matrices are the “Matrix Representation” of the C2v point group
a. All point group properties transfer to the matrices as well
b. Example: Esv(xz) = sv(xz)
1 0 0 1 0 0 1 0 0
0 1 0   0  1 0    0  1 0 

 
 

0 0 1 0 0 1 0 0 1
B.
Reducible and Irreducible Representations
1) Character = sum of diagonal from upper left to lower right (only defined
for square matrices)
a)
The set of characters = a reducible representation (G) or shorthand
version of the matrix representation
a)
For C2v Point Group:
E
C2
sv(xz)
sv(yz)
3
-1
1
1
2)
Reducible and Irreducible Representations
a. Each matrix in the C2v matrix representation can be block diagonalized
b. To block diagonalize, make each nonzero element into a 1x1 matrix
1 0 0 
 0 1 0 


 0 0 1
E
c.
d.
Irreducible
Representations
Reducible Repr.
 1
 0

 0
0
 1
0
0
0 
1
1
0

 0
0
 1
0
0
0 
1
sv(xz)
C2
 1 0 0 
 0 1 0 


 0
0 1
sv(yz)
When you do this, the x,y, and z axes can be treated independently
i. Positions 1,1 always describe x-axis
ii. Positions 2,2 always describe y-axis
iii. Positions 3,3 always describe z-axis
Generate a partial character table from this treatment
Axis used
E
C2
sv(xz)
sv(yz)
x
1
-1
1
-1
y
1
-1
-1
1
z
1
1
1
1
G
3
-1
1
1
III. Character Tables
A.
The C2v Character Table
1) We have found three of the irreducible representations of the character
table through matrix math
2) One more (A2) irreducible representation is derived from the first three
due to the properties of character tables (below)
C2v
E
C2
sv(xz)
sv(yz)
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
3)
Rx, Ry, Rz stand for rotation about the x, y, z axes respectively
x’s are p and d orbitals
4)
Other symbols we need to know
a) R = any symmetry operation
b) c = character (#)
c) i,j = different representations (A1, B2, etc…)
d) h = order of the group (4 total operations in the C2v case)
B.
The C3v Character Table for NH3
1) The threefold symmetry of NH3 makes for complex transformation matrices
2)
C.
Though more complex, the C3v Character Table can be generated similarly to
that of the C2v group
Notes on Character Tables
1) Multiple operations in the same class are listed together
2) Different C2 axes are listed separately with primes (‘)
a) Those through outer atoms are ‘
b) Those not through outer atoms are ”
3) Symmetry of orbitals are listed except for s orbitals, which are always in the
first listed A irreducible representation
4) Irreducible Representation Labels
a) Degeneracy (dimension) is determined by the character of E operation
i. A if E = 1 and c of Cn = 1
ii. B if E = 1 and c of Cn = -1
iii. E if E = 2 (doubly degenerate)
iv. T if E = 3 (triply degenerate)
b)
a)
Subscripts
i. 1 if symmetric to perpendicular C2 axis (or sv)
ii. 2 if antisymmetric to perpendicular C2 axis (or sv)
iii. g if symmetric to i
iv. u if antisymmetric to i
Primes
i. ‘ if symmetric to sh
ii. “ if antisymmetric to sh
IV. Applications of Symmetry
A.
Chiral Molecules
1) Molecules not superimposable with their mirror images are called chiral
or dissymetric
2) They may still have some symmetry operations: E, Cn
3) Chiral molecules cannot have i, s, or Sn symmetry operations
B.
Molecular Vibrations
1. To use symmetry, we must assign axes
to each atom of the molecule
a. The z-axis is usually the Cn axis
b. The x-axis is in the molecular plane
c. The y-axis is perpendicular to the molecular plane
2.
Degrees of Freedom = possible atomic movements in the molecule
a. 3N degrees of freedom for a molecule of N atoms
b. Nonlinear molecules
i. 3 translations (along x, y, z)
ii. 3 rotations (around x, y, z)
iii. 3N – 6 vibrations
c. Linear molecules
i. Only 2 rotations change the molecule
ii. 3N – 5 vibrations
3.
We will use group theory to determine the symmetry of all nine motions
and then assign them to translation, rotation, and vibration
a)
Look at the C2v character table
C2v
E
C2
sv(xz)
sv(yz)
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
b)
Add up how many vectors stay the same after an operation
i. If the atom moves, none of its vectors stay the same = 0
ii. If the atom stays and the vector is unchanged = +1
iii. If the atom stays and the vector is reversed = -1
C2v
E
C2
sv(xz)
sv(yz)
G
9
-1
3
1
c)
Reduce the reducible representation to its irreducible components
The number of

 number of operations  character of the
 character of the

irreducibl e representa tions   1





 reducible representa tion  irreducibl e representa tion 

 order   in the class




of a given type

d)
i. nA1 = ¼[(1x9x1)+(1x-1x1)+(1x3x1)+(1x1x1)] = 3 A1
ii. nA2 = ¼[(1x9x1)+(1x-1x1)+(1x3x-1)+(1x1x-1)] = 1 A2
iii. nA1 = ¼[(1x9x1)+(1x-1x-1)+(1x3x1)+(1x-1x1)] = 3 B1
iv. nA1 = ¼[(1x9x1)+(1x-1x-1)+(1x3x-1)+(1x1x1)] = 2 B2
All motions of water match 3A1 + A2 + 3B1 + 2B2
e)
Use the character table to remove translations
x, y, z = A1 + B1 + B2
f)
Use the character table to remove rotations
Rx, Ry, Rz = A2 + B1 + B2
g)
The motions remaining are the vibrations = 2A1 + B1
i. A1 = totally symmetric
ii. B1 = antisymmetric to C2 and to reflection in yz plane
C.
4)
Symmetry and IR
a. IR only “sees” a vibration if the vibration changes the molecule’s dipole
b. Motion along the x, y, z axes creates a changed dipole
a. Infrared Active vibrations match up with x, y, z on character table
b. Infrared Inactive vibrations don’t
c. For water, all three vibrations are infrared active
5)
Examples and Exercises pages 113-116
Molecular Vibrations of ML2(CO)2 complexes
1) The symmetry of cis- ML2(CO)2 complexes is C2v
a) The C≡O stretch has only one possible direction of motion
b) Instead of using xyz vectors at each atom, we can use a single vector
c)
Reducible representation from the 2 vectors
C2v
E
C2
sv(xz)
sv(yz)
G
2
0
2
0
d)
2 possible vibrations from reduction formula: A1 + B1 (see both)
2)
The symmetry of trans- ML2(CO)2 complexes is D2h
a. Symmetry operations on the vectors generate a reducible
representation
b. Reduction formula give 2 irreducible representations
c.
Only the B3u representation is IR Active
d.
We can tell cis from trans by the number of C=O IR bands