A lecture
on
“Group Theory and IR Spectra”
Department of Chemistry,
Deenbandhu Chhotu Ram University of Science and
Technology, Murthal,
Dr. Virender Bhatti
Deptt. of Chemistry, DCRUST
1
Group Theory and IR Spectra
✓ Functional group present in the organic molecule.
✓ At temp. above absolute zero, all atoms in molecules are in continuous vibrations
with respect to each other.
✓ When the frequency of specific vibration matches with the frequency of IR-
radiations, directed on the molecule, the molecule absorb IR-radiations.
Region Wavelength Wavenumbers Frequency (cps)
(M) (cm-1)
Near IR 0.78 to 2.5 12800-4000 3.8 x 1014 to 1.2 x 1014
Middle IR 2.5 to 50 4000-200 1.2 x 1014 to 6 x 1012
Far IR 50 to 1000 200-10 6 x 1012 to 3 x 1011
Most used IR 2.5 to 50 4000-600 1.2 x 1014 to 2 x 1013
• A molecule absorb IR radiations, it undergoes vibration due to -
✓ Bonds may rotate.
✓ Bonds may be flex or bend.
✓ Stretching of bonds.
✓ Group of atoms may wag or scissor.
✓ Molecule may rotate or vibrate in other way.
• Energy of molecule = Electronic + Vibrational + Rotational
➢ Requires periodic changes for transition .
➢ The absorption of infra-red radiations result in the excitation of molecules =Change
in the dipole moment of the molecule.
▪ Types of Molecular Vibration
1. Stretching vibrations
➢ Rhytmic compression and elongation of bonds
➢ Bond length increases /decreases
❑ Symmetric stretching
➢ Simultaneous vibration of two bonds, in which the bonds elongate together and
contract together
Elongation Compression
Symmetric stretching
❑ Asymmetric stretching
➢ An asymmetric stretch occurs when some atoms move in the same direction while
others move in the opposite direction.
Asymmetric stretching
2. Bending vibrations
➢ It may be defined as periodic changes in bond angles between bond formed by the
two atoms by central atoms.
❑ Scissoring
❑ Rocking
or
❑ Waggingg
Out of plane
❑ Twisting
Out of plane
Normal modes of vibrations
✓ The molecules have translational, rotational and vibrational motion.
✓ Total no. of degree of freedom can be determine for each motion depending upon the
type of motion
✓ Total degree of freedom = 3 N, where N is the no. of atoms presents in mlecule.
➢ Translational motion
✓ Molecule moves from one place to another but without cahnging its shape.
✓ The molecule move as a whole unit.
✓ Translational motion uses all the three coordinates, therefore, no. of translational
degree of freedom = 3
➢ Rotational motion
✓ Rotational degree of freedom, depends upon the shape of molecule
✓ The molecule move as a whole unit.
✓ Translational motion uses all the three coordinates, therefore, no. of translational
degree of freedom = 3
✓ In linear molecules
degree of freedom = 2
➢ Rotational motion
✓ In non-linear molecules
degree of freedom = 3
➢ Vibrational motion
✓ It is determined by the difference by the difference between the total degree of
freedom and the sum of translational and rotational degree of freedom.
[3N - (Trans. + Rot.)]
✓ Normal mode for a linear molecule = 3N - (3+2) = 3N - 5
✓ Normal mode for a non- linear molecule = 3N - (3+3) = 3N - 6
Degree of Monatomic Linear Non-linear
freedom molecules molecule
Translational 3 3 3
Rotational 0 2 3
Vibrational 0 3N - 5 3N - 6
Total 3 3N 3N
Group theoratical analysis to find vibrational spectra
▪ Geometry of molecule
✓ Point group of the molecule
✓ Order and no. of classes found in the molecule
▪ Total degree of freedom i.e; 3N
▪ ΓVib. = Γ3N - (ΓTrans.+ ΓRot.)
▪ IR and Raman spectral activity of all the vibrational modes so obtained is determined.
Determination of vibrational mode of H2O molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
✓ The H2O molecule belonging to C2v point group with symmetry operations E, C2, σv(xz) and σv(yz).
✓ The basis vectors of the atoms representing the degrees of freedom may be denoted as x1, y1, z1, x2,
y2, z2 and x3, y3, z3.
Directions of the nine degrees of freedom
Hence, total no. of modes for H2O
molecule = 3N = 3*3 = 9
No. of unshifted
atoms
1 Cartesian Coordinates
1
Character of reducible repersentation = No. of unshifted atoms x Character of operation
(i) Identity (E)
✓ No. of unshifted atoms = 3
✓ Character of E = 3
x x 1 0 0
E 0
y y
1 0 ɳE = +3
z z 0 0 1
✓ Matrix repersentation of operation =3
✓ So, 3x3 is the character of the reducible repersentation = 9
Character of reducible repersentation = No. of unshifted atoms x Character of operation
(ii) C2 axis-rotation about 1800 Z
1800
✓ No. of unshifted atoms = 1
✓ Character of C2 axis = -1 X
x -x -1 0 0 x
C2z
y -y 0 -1 0 y ɳ=C2z = -1 Y
Cartesian coordinates
z z 0 0 1 z
✓ So, character of the reducible repersentation = 1 x -1 = -1
Character of reducible repersentation = No. of unshifted atoms x Character of operation
(ii) XZ operation (molecular plane) Z
✓ No. of unshifted atoms = 3
✓ Character of XZ = 1
X
x x 1 0 0
y σxz − y 0 − 1 0 Y
ɳσxz = 1
z z 0 0 1 Cartesian coordinates
✓ So, character of the reducible repersentation = 3 x 1 = 3
Character of reducible repersentation = No. of unshifted atoms x Character of operation
(ii) YZ operation Z
✓ No. of unshifted atoms =1
✓ Character of YZ = 1
X
x - x − 1 0 0
y σyz y 0 1 0 Y
ɳσyz = 1
z z 0 0 1 Cartesian coordinates
✓ So, character of the reducible repersentation = 1 x 1 = 1
ELEMENT CHARACTER
E +3
σ (σv,σh,σd) +1
C2 -1
C3 0
C4 +1
S2 (i) -3
S3 -2
S4 -1
➢ Reducible repersentation for C2v point group, when 3N- Cartesian
coordinate system as basis set. Z
C2v E C2z σxz σyz
X
No. of Unshifted atoms 3 1 3 1
Y
Contribution per atom 3 -1 1 1
(Γoperations) O
Reducible representation 9 -1 3 1
H H
(Γ3N)
Character of each element
Character Table
Point group Symmetry operations, Classes x, y, z
Symmetry of translations (p orbitals)
Rx, Ry, Rz: rotations
Characters
+1 symmetric behavior
Mülliken symbols
-1 antisymmetric
Each row is an irreducible representation
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
C2v E C2z σxz σyz
C2v E C2z σxz σyz A1 1 1 1 1
(Γ3N) 9 -1 3 1 A2 1 1 -1 -1
B1 1 -1 1 -1
Reducible Repersentation
B2 1 -1 -1 1
Character Table
A1 = 1/4 (1x1x9) + (1x1x-1) + (1x1x3) + (1x1x1)
= 1/4 9 -1 + 3 + 1
1/h gpiR
= 1/4 x 12
standard reduction formula
= 3
i.e; 3A1
C2v E C2z σxz σyz
C2v E C2z σxz σyz A1 1 1 1 1
(Γ3N) 9 -1 3 1 A2 1 1 -1 -1
B1 1 -1 1 -1
Reducible Repersentation
B2 1 -1 -1 1
Character Table
A2 = 1/4 (1x1x9) + (1x1x-1) + (1x-1x3) + (1x-1x1) = 1
i.e; A2
B1 = 1/4 (1x1x9) + (1x-1x-1) + (1x1x3) + (1x-1x1) = 3
i.e; 3B1
B2 = 1/4 (1x1x9) + (1x-1x-1) + (1x-1x3) + (1x1x1) = 2
i.e; 2B2
Hence, reducible repersentation will be
= 3A1 + A2 + 3B1 + 2B2
=3+1+3 +2 (Dimension of A or B = 1)
= 9 (total no. of modes)
To check answer :
C2v E C2z σxz σyz
3A1 3 3 3 3
A2 1 1 -1 -1
3B1 3 -3 3 -3
2B2 2 -2 -2 2
(Γ3N) 9 -1 3 1
C2v E C2z σxz σyz
No. of Unshifted atoms 3 1 3 1
Contribution per atom (Γoperations) 3 -1 1 1
Reducible representation 9 -1 3 1
(Γ3N)
Total no. of mode = 3A1 + A2 + 3B1 + 2B2
Translational modes = - A1 - B1 - B2 (x, y, z)
Rotational modes = - A2 - B1 - B2 (Rx, Ry, Rz)
Vibrational modes = 2A1 + B1
C2v E C2z σxz σyz
To find which modes are IR active A1 1 1 1 1 z x2, y2, z2
and Raman active A2 1 1 -1 -1 Rz xy
➢ IR active = x, y, z B1 1 -1 1 -1 x, Ry xz
➢ Raman active = x2, y2, z2, xy, yz, xz
B2 1 -1 -1 1 y, Rx yz
Hence,
A1 = IR active ( z ) B1 = IR active ( x)
Raman active (x2, y2, z2 ) Raman active (xz )
Determination of vibrational mode of NH3 molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
✓ The NH3 molecule belonging to C3v point group with symmetry operations E, 2C3,
3σv
✓ Total no. of atoms in NH3 (N) = 4
✓ Total no. of modes in NH3 = 3N = 12
• 3N- Cartesian coordinate system as basis set
Z
C3v E 2C3 3σv
No. of Unshifted atoms 4 1 2
X
Contribution per atom 3 0 1
Reducible representation 12 0 2 Y
(Γ3N)
Character of each element
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
C3v E 2C3 3σv
C3v E 2C3 3σv
(Γ3N) 12 0 2 A1 1 1 1
A2 1 1 -1
Reducible Repersentation E 2 -1 0
Character Table
A1 = 1/6 (1x1x12) + (2x1x0) + (3x1x2)
= 1/6 12 +0 + 6
= 1/6 x 18
1/h gpiR
standard reduction formula
= 3
i.e; 3A1
C3v E 2C3 3σv
C3v E 2C3 3σv
A1 1 1 1
(Γ3N) 12 0 2
A2 1 1 -1
Reducible Repersentation E 2 -1 0
Character Table
A2 = 1/6 (1x1x12) + (2x1x0) + (3x-1x2) = 1
i.e; A2
E = 1/6 (1x2x12) + (2x-1x0) + (3x0x2) = 4
i.e; 4E
Hence, reducible repersentation will be
= 3A1 + A2 + 4E
=3+1+8 (Dimension of A or B = 1 and E = 2)
= 12 (total no. of modes)
To check answer :
C3v E 2C3 3σv
3A1 3 3 3 Reducible Repersentation
A2 1 1 -1
4E 8 -4 0 C3v E 2C3 3σv
(Γ3N) 12 0 2 (Γ3N) 12 0 2
Total no. of mode = 3A1 + A2 + 4E
Translational modes = - A1 - E (x, y, z)
Rotational modes = - A2 - E (Rx, Ry, Rz)
Vibrational modes = 2A1 + 2E
To find which modes are IR active
and Raman active
➢ IR active = x, y, z
➢ Raman active = x2, y2, z2, xy, yz, xz
Hence,
E = IR active ( x,y)
A1 = IR active ( z )
Raman active (x2-y2, xy ) (xz,yz)
Raman active (x2+y2, z2 )
Determination of vibrational mode of AB3 type molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
✓ The AB3 (ClF3) molecule belonging to C2v point group with symmetry operations E,
C2, σv(xz) and σv(yz)
✓ Total no. of atoms in AB3 (N) = 4
✓ Total no. of modes in AB3 = 3N = 12
➢ Reducible repersentation for C2v point group, when 3N- Cartesian
coordinate system as basis set. Z
C2v E C2z σxz σyz
X
No. of Unshifted atoms 4 2 2 4
Y
Contribution per atom 3 -1 1 1
(Γoperations)
Reducible representation 12 -2 2 4
(Γ3N)
Character of each element
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
C2v E C2z σxz σyz
C2v E C2z σxz σyz A1 1 1 1 1
(Γ3N) 12 -2 2 4 A2 1 1 -1 -1
B1 1 -1 1 -1
Reducible Repersentation
B2 1 -1 -1 1
Character Table
A1 = 1/4 (1x12x1) + (1x1x-2) + (1x1x2) + (1x1x4)
= 1/4 -2 + 2 + 4
1/h gpiR
= 1/4 x 16
standard reduction formula
= 4
i.e; 4A1
C2v E C2z σxz σyz
C2v E C2z σxz σyz A1 1 1 1 1
(Γ3N) 12 -2 2 4 A2 1 1 -1 -1
B1 1 -1 1 -1
Reducible Repersentation
B2 1 -1 -1 1
Character Table
A2 = 1/4 (1x1x12) + (1x1x-2) + (1x-1x2) + (1x-1x4) = 1
i.e; A2
B1 = 1/4 (1x1x12) + (1x-1x-2) + (1x1x2) + (1x-1x4) = 3
i.e; 3B1
B2 = 1/4 (1x1x12) + (1x-1x-2) + (1x-1x2) + (1x1x4) = 4
i.e; 4B2
Hence, reducible repersentation will be
= 4A1 + A2 + 3B1 + 4B2
=4+1+3 +4 (Dimension of A or B = 1)
= 12 (total no. of modes)
Total no. of mode = 4A1 + A2 + 3B1 + 4B2 C2v E C2z σxz σyz
Translational modes = - A1 - B1 - B2 (x, y, z) A1 1 1 1 1 z x2, y2, z2
Rotational modes = - A2 - B1 - B2 (Rx, Ry, Rz) A2 1 1 -1 -1 Rz xy
Vibrational modes = 3A1 + B1 + 2B2 B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
Determination of vibrational mode of AB3 type molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
✓ The AB3 (BF3) molecule belonging to D3h point group with symmetry operations E,
2C3, 3C2, σh, 2S3, 3σv
✓ Total no. of atoms in AB3 (N) = 4
✓ Total no. of modes in AB3 = 3N = 12
➢ Reducible repersentation for C2v point group, when 3N- Cartesian
coordinate system as basis set. Z
D3h E 2C3 3C2 σh 2S3 3σv
X
No. of Unshifted 4 1 2 4 1 2
atoms
Y
Contribution per 3 0 -1 1 -2 1
atom (Γoperations)
Reducible 12 0 -2 4 -2 2
representation
(Γ3N)
Character of each element
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
D3h E 2C3 3C2 σh 2S3 3σv
(Γ3N) 12 0 -2 4 -2 2
Reducible Repersentation
Character Table
A1’ = 1/12 (1x1x12) + (2x1x0) + (3x1x-2) + (1x1x4) + (2x1x-2) + (3x1x2)
= 1/12 + 0 - 6 + 4 - 4 + 6
= 1/12 x
standard reduction formula
1/h gpiR
= 1
i.e; A1
D3h E 2C3 3C2 σh 2S3 3σv
(Γ3N) 12 0 -2 4 -2 2
Reducible Repersentation
Character Table
A2’ = 1/12 (1x1x12) + (2x1x0) + (3x1x-2) + (1x1x4) + (2x1x-2) + (3x1x2) = 1
i.e; A2’
E ’ = 1/12 (1x2x12) + (2x-1x0) + (3x0x-2) + (1x2x4) + (2x-1x-2) + (3x0x2) = 3
i.e; 3E ’
A1’’ = 1/12 (1x1x12) + (2x1x0) + (3x1x-2) + (1x-1x4) + (2x-1x-2) + (3x-1x2) = 0
i.e; 0
A2’’ = 1/12 (1x1x12) + (2x1x0) + (3x-1x-2) + (1x-1x4) + (2x-1x-2) + (3x1x2) = 2
i.e; 2A2’’
E ’’ = 1/12 (1x2x12) + (2x-1x0) + (3x0x-2) + (1x-2x4) + (2x1x-2) + (3x0x2) = 1
i.e; E ’’
Hence, reducible repersentation will be
= A1’ + A2’ + 3E ’ + 2A2’’ + E ’’
= 1+1+6 +2+2 (Dimension of A or B = 1 and E = 2)
= 12 (total no. of modes)
Total no. of mode = A1’ + A2’ + 3E ’ + 2A2’’ + E ’’
Translational modes = - E1 - A2’’ (x, y, z)
Rotational modes = - A 2’ - E ’’ (Rx, Ry, Rz)
Vibrational modes = A1’ + 2E ’ + A2’’
To find which modes are IR active
and Raman active
➢ IR active = x, y, z
➢ Raman active = x2, y2, z2, xy, yz, xz
Hence,
E’ = IR active ( x,y)
A1’ = Raman active (x2+y2, z2 )
Raman active (x2-y2, xy )
A2’’ = IR active ( z)
Determination of vibrational mode of AB4 type molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
➢ AB4 molecules having different geometries/shapes
✓ SF4 molecule
✓ The SF4 molecule belonging to C2v point group with symmetry operations E, C2,
σv(xz) and σv(yz)
Z
✓ Total no. of atoms in SF4 (N) = 5
✓ Total no. of modes in SF4 = 3N = 15
X
➢ Reducible repersentation for C2v point group,
C2v E C2z σxz σyz Y
No. of Unshifted atoms 5 1 3 3
Contribution per atom 3 -1 1 1
(Γoperations)
Reducible representation 15 -1 3 3
(Γ3N)
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
C2v E C2z σxz σyz
C2v E C2z σxz σyz A1 1 1 1 1
(Γ3N) 15 -1 3 3 A2 1 1 -1 -1
B1 1 -1 1 -1
Reducible Repersentation
B2 1 -1 -1 1
Character Table
A1 = 1/4 (1x1x15) + (1x1x-1) + (1x1x3) + (1x1x3)
= 1/4 -1 + 3 + 3
1/h gpiR
= 1/4 x 20
standard reduction formula
= 5
i.e; 5A1
C2v E C2z σxz σyz
C2v E C2z σxz σyz A1 1 1 1 1
(Γ3N) 15 -1 3 3 A2 1 1 -1 -1
B1 1 -1 1 -1
Reducible Repersentation
B2 1 -1 -1 1
Character Table
A2 = 1/4 (1x1x15) + (1x1x-1) + (1x-1x3) + (1x-1x3) = 2
i.e; 2A2
B1 = 1/4 (1x1x15) + (1x-1x-1) + (1x1x3) + (1x-1x3) = 4
i.e; 4B1
B2 = 1/4 (1x1x15) + (1x-1x-1) + (1x-1x3) + (1x1x3) = 4
i.e; 4B2
Hence, reducible repersentation will be
= 5A1 + 2A2 + 4B1 + 4B2
=5+2+4 +4 (Dimension of A or B = 1)
= 15 (total no. of modes)
To check answer :
C2v E C2z σxz σyz
5A1 5 5 5 5
2A2 2 2 -2 -2
4B1 4 -4 4 -4 C2v E C2z σxz σyz
4B2 4 -4 -4 4 (Γ3N) 15 -1 3 3
(Γ3N) 15 -1 3 3
Total no. of mode = 5A1 + 2A2 + 4B1 + 4B2
Translational modes = - A1 - B1 - B2 (x, y, z)
Rotational modes = -A2 - B1 - B2 (Rx, Ry, Rz)
Vibrational modes = 4A1 + A2 + 2B1 + 2B2
To find which modes are IR active C2v E C2z σxz σyz
and Raman active A1 1 1 1 1 z x2, y2, z2
➢ IR active = x, y, z A2 1 1 -1 -1 Rz xy
➢ Raman active = x2, y2, z2, xy, yz, xz B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
Determination of vibrational mode of AB5 type molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
➢ AB5 molecules having different geometries/shapes
✓ PCl5 molecule
✓ The PCl5 molecule belonging to D3h point group with symmetry operations E, 2C3,
2C2, σh, 2S3 and σv
Z
✓ Total no. of atoms in PCl5 (N) = 6
✓ Total no. of modes in PCl5 = 3N = 18
X
➢ Reducible repersentation for D3h point group,
D3h E 2C3 3C2 σh 2S3 3σv
Y
No. of Unshifted 6 3 2 4 1 4
atoms
Contribution per 3 0 -1 1 -2 1
atom (Γoperations)
Reducible 18 0 -2 4 -2 4
representation
(Γ3N)
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
D3h E 2C3 3C2 σh 2S3 3σv
D3h E 2C3 3C2 σh 2S3 3σv
A1’ 1 1 1 1 1 1
(Γ3N) 18 0 -2 4 -2 4
A2’ 1 1 -1 1 1 -1
E’ 2 -1 0 2 -1 0
Reducible Repersentation A1” 1 1 1 -1 -1 -1
A2” 1 1 -1 -1 -1 1
E” 2 -1 0 -2 1 0
Character Table
A1’ = 1/12 (1x1x18) + (2x1x0) + (3x1x-2) + (1x1x4) + (2x1x-2) +
(3x1x4) 1/h gpiR
= 1/12 -0 - 6 + 4 - 4 + 12 standard reduction formula
= 1/12 x 24
= 2
i.e; 2A1’
D3h E 2C3 3C2 σh 2S3 3σv
D3h E 2C3 3C2 σh 2S3 3σv
A1’ 1 1 1 1 1 1
(Γ3N) 18 0 -2 4 -2 4
A2’ 1 1 -1 1 1 -1
E’ 2 -1 0 2 -1 0
Reducible Repersentation A1” 1 1 1 -1 -1 -1
A2” 1 1 -1 -1 -1 1
E” 2 -1 0 -2 1 0
Character Table
A2’ = 1/12 (1x1x18) + (2x1x0) + (3x-1x-2) + (1x1x4) + (2x1x-2) + (3x-1x4)
E’ = 1/12 (1x2x18) + (2x-1x0) + (3x0x-2) + (1x2x4) + (2x-1x-2) + (3x0x4) =
= A2 ’
4E’
1/h gpiR
A1” = 1/12 (1x1x18) + (2x1x0) + (3x1x-2) + (1x-1x4) + (2x-1x-2) + (3x-1x4) =0 standard reduction formula
A2” = 1/12 (1x1x18) + (2x1x0) + (3x-1x-2) + (1x-1x4) + (2x-1x-2) + (3x1x4)
=3A2”
E” = 1/12 (1x2x18) + (2x-1x0) + (3x0x0) + (1x-2x4) + (2x1x-2) + (3x0x4) = 2E’
Hence, reducible repersentation will be
= 2A1’ + A2’ + 4E’ + 3A2” + 2E”
=2+1+8 +3+4 (Dimension of A = 1, E = 2)
= 18 (total no. of modes)
To check answer :
D3h E 2C3 3C2 σh 2S3 3σv
2A1’ 2 2 2 2 2 2
A2’ 1 1 -1 1 1 -1
4E’ 8 -4 0 8 -4 0
3A2” 3 3 -3 -3 -3 3 D3h E 2C3 3C2 σh 2S3 3σv
2E” 4 -2 0 -4 2 0 (Γ3N) 18 0 -2 4 -2 4
(Γ3N) 18 0 -2 4 -2 4
Total no. of mode = 2A1’ + A2’ + 4E’ + 3A2” + 2E”
Translational modes = - E’ - A2” (x, y, z)
Rotational modes = -A2’ - E” (Rx, Ry, Rz)
Vibrational modes = 2A1’ + 3E’ + 2A2” + E”
To find which modes are IR active
and Raman active
➢ IR active = x, y, z
➢ Raman active = x2, y2, z2, xy, yz, xz
Determination of vibrational mode of AB6 type molecule
1. Writing the reducible repersentation
2. Reducing the reducible repersentation
3. Assigning of vibrational modes
➢ AB6 molecules having different geometries/shapes
✓ SF6 molecule
✓ The SF6 molecule belonging to Oh point group with symmetry operations E, 8C3,
6C2, 6C4, σh, 2S3 and σv
Z
✓ Total no. of atoms in SF6 (N) = 7
✓ Total no. of modes in SF6 = 3N = 21
X
➢ Reducible repersentation for D3h point group,
Oh E 8C3 6C2 6C4 3C2’ i 6S4 8S6 3σh 6σd Y
No. of Unshifted 7 1 1 3 3 1 1 1 5 3
atoms
Contribution per 3 0 -1 1 -1 -3 -1 0 1 1
atom (Γoperations)
Reducible 21 0 -1 3 -3 -3 -1 4 5 3
representation
(Γ3N)
No. of times an irreducible repersentation is repeated = 1/h gpiR
Here,
gp = No. of opreations in a class
i = Character of reducible repersentation
R = Character of irreducible repersentation
Oh E 8C3 6C2 6C4 3C2’ i 6S4 8S6 3σh 6σd
(Γ3N) 21 0 -1 3 -3 -3 -1 4 5 3
Reducible Repersentation
Character Table
A1g = 1/48 ( + 0 - 6 + 18 -9 -3 -6 +0 +15 +18)
= 1/48 x = 1 i.e; A1g 1/h gpiR
A2g = 1/48 ( + 0 + 6 - 18 -9 -3 +6 +0 +15 -18) standard reduction formula
= 1/48 x = 0
Eg = 1/48 (42+ 0 + 0 +0 -18 -6 +0 +0 +30 +0 )
= 1/48 x = 1 i.e; Eg
Oh E 8C3 6C2 6C4 3C2’ i 6S4 8S6 3σh 6σd
(Γ3N) 21 0 -1 3 -3 -3 -1 4 5 3
Reducible Repersentation
Character Table
T1g = 1/48 (63+0+6+18-9-9+6+0-15-18) = 1/48 x = 1 i.e; T1g
T2g = 1/48 (63+0-6-18+9-9+6+0-15+18) = 1/48 x = 1 i.e; T2g 1/h gpiR
A1u = 1/48 (27+0+6+8-4+3-6+0-15+18) = 1/48 x = 0 standard reduction formula
A2u = 1/48 (21+0-6+8-9+3-6+0+15+18) = 1/48 x = 0
Eu = 1/48 (42-0+0+0-18+6+0+0-30+0) = 1/48 x = 0
T1u = 1/48 (63+0+6-18+9-9+6+0+15+18) = 1/48 x = 3 i.e; 3T1u
T2u = 1/48 (63+0-6-18+9+9-6+0+15-18) = 1/48 x = 1 i.e; T2u
Total no. of mode = A1g + Eg + T1g + T2g + 3T1u + T2u
Translational modes = -T1u (x, y, z)
Rotational modes = - T1g
Vibrational modes = A1g + Eg + T2g + 2T1u + T2u
To find which modes are IR active
and Raman active
➢ IR active = x, y, z
➢ Raman active = x2, y2, z2, xy, yz, xz
