topic 7 (group theory and spectroscopy)
DESCRIPTION
7th lecture of Monirul Islam sirTRANSCRIPT
Molecular symmetry and group theory
Dr. Md. Monirul Islam Department of Chemistry
University of Rajshahi
Molecular Vibrations
Determining the Reducible Representation
• to determine the symmetry of the molecular vibrations of a molecule, we must examine the irreducible representations spanned when every atom of a molecule is free to move in any direction
• this is achieved by building up a large matrix that has an x, y, z component for each atom
1 2
Z
X
Yv(xz)
v(yz)
Then the vector (xH1, yH1, zH1) specifies the displacement of one of the hydrogen atoms, and (xO, yO, zO) and (xH2, yH2, zH2) the displacements of the other atoms. We may combine these three 3-component vectors into one 9- component vector:
(xH1, yH1, zH1, xO, yO, zO, xH2, yH2, zH2 )
O
H2H1
xO
yO
zO
xH2
yH2
zH2
xH1
yH1
zH1
Consider the effect of the operations of C2v on the vector of displacements.The identity has no effect on any displacement, i.e.
O
H2H1
xO
yOzO
xH2
yH2zH2
xH1
yH1zH1 E O
H2H1
xO
yOzO
xH2
yH2zH2
xH1
yH1zH1
ExH1 = xH1
EyH1 = yH1
EzH1 = zH1
ExO = xO
EyO = yO
EzO = zO
ExH2 = xH2
EyH2 = yH2
EzH2 = zH2
These results may be summarized in matrix form
The character of this representation = 9
E
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2
=
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2
1 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 1
The rotation C2 leaves unchanged only the component zO. Its full effect is as follows:
O
H2H1
xO
yOzO
xH2
yH2zH2
xH1
yH1zH1 C2
O
H1H2
xO
yO
zO
xH1
yH1
zH1
xH2
yH2
zH2
C2xH1 = - xH2
C2yH1 = -yH2
C2zH1 = zH2
C2xO = -xO
C2yO = -yO
C2zO = zO
C2xH2 = -xH1
C2yH2 = -yH1
C2zH2 = zH1
These results may be summarized in matrix form
c2
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2
=
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2 The character of this representation = -1
0 0 0 0 0 0 -1 0 00 0 0 0 0 0 0 -1 00 0 0 0 0 0 0 0 10 0 0 -1 0 0 0 0 00 0 0 0 -1 0 0 0 00 0 0 0 0 1 0 0 0-1 0 0 0 0 0 0 0 00 -1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0
The effect of v(xz) is as follows:
O
H2H1
xO
yO
zO
xH2
yH2
zH2
xH1
yH1
zH1 v(xz)O
H2H1
xO
yO
zO
xH2
yH2
zH2
xH1
yH1
zH1
v(xz)xH1 = xH1
v(xz)yH1 = -yH1
v(xz)zH1 = zH1
v(xz)xO = xO
v(xz)yO = -yO
v(xz)zO = zO
v(xz)xH2 = xH2
v(xz)yH2 = -yH2
v(xz)zH2 = zH2
These results may be summarized in matrix form
v(xz)
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2
=
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2 The character of this representation = 3
1 0 0 0 0 0 0 0 00 -1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 1 0 0 0 0 00 0 0 0 -1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 -1 00 0 0 0 0 0 0 0 1
The effect of v(yz) is as follows:
O
H2H1
xO
yO
zO
xH2
yH2
zH2
xH1
yH1
zH1 v(yz) O
H1H2
xO
yO
zO
xH1
yH1
zH1
xH2
yH2
zH2
v(yz)xH1 = -xH2
v(yz)yH1 = yH2
v(yz)zH1 = zH2
v(yz)xO = -xO
v(yz)yO = yO
v(yz)zO = zO
v(yz)xH2 = -xH1
v(yz)yH2 = -yH1
v(yz)zH2 = zH1
These results may be summarized in matrix form
v(yz)
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2
=
xH1
yH1
zH1
xo
yo
zo
xH2
yH2
zH2 The character of this representation = 1
0 0 0 0 0 0 -1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 10 0 0 -1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 0-1 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0
The characters or trace of these matrices can be recorded in a representation table:
C2v E C2 v(xz) v(yz)
3N(H2O) 9 -1 3 1
However, the size of these matrices grows as 3N and thus they become very large, however as we are not interested in the particular matrix representation, but only in the trace, there is a short-cut to determining the reducible representation
Short-Cut Rule for Reducible Representation
• E always has a contribution of 3N
• When an atom moves it contributes nothing to the trace (look at H1 and H2 for the C2 and σ(yz) matrices of water)
• For each atom on an inversion center
• For each on a mirror plane (since the vector in the
plane remains unmoved and the other two are reflected
i
ii 3
1 0 00 1 00 0 1
-1 0 0 0 -1 0 0 0 -1
i
i
ii 1111
1 0 00 1 00 0 1
1 0 00 -1 00 0 1
v(yz)
Short-Cut Rule for Reducible Representation
For each atom on a rotation axis
i
ii 1cos21coscos
1 0 00 1 00 0 1
cos -sin 0 sin cos 0 0 0 1
Cn(z)
2cos + 1
C2 180 2(-1) +1 = -1
C31 , C3
2 120, 240 2(-1/2) + 1= 0
C41, C4
3 90, 270 2(0) + 1 = 1
For each atom on an improper rotation axis
i
ii 1cos21coscos
1 0 00 1 00 0 1
cos -sin 0 sin cos 0 0 0 -1
Sn(z) 2cos - 1
S31 , S3
2 120, 240 2(-1/2) - 1= -2
S41, S4
3 90, 270 2(0) - 1 = -1
Reducible Representation by Short-Cut Rule
summarizing: the contribution per un-shifted atom to the reducible representation
thus for H2O we can determine the reducible representation by considering the number of atoms unmoved by each symmetry class and then evaluating the contribution to the trace for that class
C2v E C2 v(xz) v(yz)
Un-shifted atom 3 1 3 1
(per atom) 3 -1 1 1
3N(H2O) 9 -1 3 1
N
H(1)(2)H
(3)H
C3
v(1)
v(2)
v(3)
Reducible Representation by Short-Cut Rule
Example 2
Example 3Trans-N2F2
Symmetry operationE, C2(z), i, h(xy)
C2h E C2 i h(xy)
Un-shifted atom 4 0 0 4
(per atom) 3 -1 -3 1
3N(Trans-N2F2) 12 0 0 4
C3v E 2C3 3v
Un-shifted atom 4 1 2
(per atom) 3 0 1
3N(NH3) 12 0 2
Reduction of the Reducible Representation
The reducible representation can then be broken up into
contributions from each of the irreducible representations in the way
that a vector
in Cartesian space can be defined by coefficients
times unit vectors (where the unit vectors completely define the
space).
A reducible representation can similarly be defined in terms of
coefficients times the "unit vectors" or irreducible representations of
the point group, which completely define the space.
Every reducible representation (R ) can be written as a sum of the
irreducible representations (IR) of a point group, where nIR=the
number of times a particular irreducible representation occurs:
kzjyixV ˆˆˆ
IR
IRIR
R n
Reduction Formula
The reduction formula is employed to determine the contribution of each irreducible representation to the reducible representation.
Q
RIRIR QQk
hn )()(
1
h = number of operation in the group
Q = a particular class of symmetry operation
k = the number of operations in that class
IR(Q) = the character of the irreducible representation under Q
R(Q) = the character of the reducible representation under Q
Use a reduction table and standard character table (what will be supplied in exam hall) for determining nIR
Reduction of the Reducible Representation: Example 1(H2O)
C2v E C2 v(xz) v(yz)
3N(H2O) 9 -1 3 1
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
34
121319
4
1)111()311()111()911(
4
11
An
14
41319
4
1)111()311()111()911(
4
12
An
34
121319
4
1)111()311()111()911(
4
11
Bn
24
81319
4
1)111()311()111()911(
4
12
Bn
21212 2313)( BBAAOHR
Reduction of the Reducible Representation: Example 2 (NH3)
C3v E 2C3 3v
3N(NH3) 12 0 2
A1 1 1 1 z x2+y2, z2
A2 1 1 -1 Rz
E 2 -1 0 (x,y), (Rx, Ry) (x2-y2, xy), (xz, yx)
36
186012
6
1)213()012()1211(
6
11
An
16
66012
6
1)213()012()1211(
6
12
An
46
240024
6
1)203()012()1221(
6
1En
EAANHR 413)( 213
Fundamental Vibration Modes of Molecules
There are 3 coordinates (x, y, z) contributing for each atom,
o 3N is the total modes of motion of molecules.
o There are 3N-6 vibrational motions for a molecule with N atoms
o There are 3N-5 vibrational motions for a linear molecules
The reduced representation still includes the collective motions that involve translational and rotational motion of the whole molecule. These are not vibrations, and must be subtracted from the full reducible representation before the IR spanned by the vibrational motions of the molecule are determined.
The irreducible representations of the translational and rotational motions for a molecule can easily be identified from standard character table.o Translation: irreducible representation with x (or Tx), y (or Ty) and z
(or Tz) basis vectors.o Rotation: irreducible representation with Rx, Ry and Rz basis vectors
Determination of Vibration Modes
Example 1 for water molecule
C2v E C2 v(xz) v(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
21212 2313)( BBAAOHRTotal
2111)nTranslatio( BBA
2121)Rotational( BBA
2121 2211)R&T( BBAA
11
21212121
2
)2211()2313(
)&(
BA
BBAABBAA
RTTotalvib
As part of a full assignment we need to identify which vibrational
modes are IR and Raman active
IR active modes
o to be infrared active a vibrational mode must produce a
dipole moment change
o the IR active modes are those that have the same
symmetry as the dipole moment (μ), which has the
same symmetry as the translational vectors {x or Tx, y or
Ty, and z or Tz}
o thus the IR active modes of water must be of B1, B2, or A1
symmetry
o hence all of the water modes are IR active, this is
indicated by adding (IR) in brackets after the listing the
symmetry of the mode
Full Assignment
Raman active modes
o to be Raman active a vibrational mode must produce a
change in the molecular polarizability of the molecule.
o the Raman modes are those that have the same symmetry as
the molecular polarizability (α) which has the same
symmetry as the binary functions {x2, y2, z2, xy, xz, yz}
o in addition Raman active modes will leave plane polarised
light polarised if they are totally symmetric, otherwise the
light will be depolarised.
thus the Raman active modes of water must be A1, A2, B1 or B2
symmetry, the Raman activity of A1 modes is identified by adding
(pol) in the brackets after the listed mode, and the activity of other
modes by adding (depol).
thus the full description for the vibrational modes of water is:
),(),(2)( 112 depolIRBpolIRAOHvib
The nature of the vibrational modes is shown in following Figure
Example 2 for ammonia molecule
C3v E 2C3 3v
A1 1 1 1 z x2+y2, z2
A2 1 1 -1 Rz
E 2 -1 0 (x,y), (Rx, Ry) (x2-y2, xy), (xz, yx)
EAANHR 413)( 213
EA 11)nTranslatio( 1
EA 11)Rotational( 2
EAA 211)R&T( 21
EA
EAAEAA
RTTotalvib
22
)211()413(
)&(
1
2121
The C3v character table shows that
),(2),(2)NH( Thus, 13 depolIREpolIRAvib