Chapter 4-Symmetry and Group Theory Applications

Symmetry Applications

Chem. 285 Lecture NotesAdvanced Inorganic Chemistry

Symmetry and Dipole Moments

Chirality

Optical Activity or Chirality

Vibrational Spectroscopy

Spectroscopy is “the study of the interaction of matter with energy (radiation in the electromagnetic spectrum).” A molecular vibration is a periodic distortion of a molecule from its equilibrium geometry. The energy required for a molecule to vibrate is quantized (not continuous) and is generally in the infrared region of the electromagnetic spectrum.

re = equilibrium distance between A and B

re

For a diatomic molecule (A-B), the bond between the two atoms can be approximated by a spring that restores the distance between A and B to its equilibrium value. The bond can be assigned a force constant, k (in Nm-1; the stronger the bond, the larger k) and the relationship between the frequency of the vibration, , is given by the relationship:

DAB

rAB0

DAB = energy required to dissociate into A and B atoms

k 2

c

k

or, more typically

Vibrational Spectroscopy

Molecule (cm-1) k (N/m) (amu)

HF 3962 878 19/20

HCl 2886 477 35/36 or 37/38

HBr 2558 390 79/80 or 81/82

HI 2230 290 127/128

Cl2 557 320 17.5

Br2 321 246 39.5

CO 2143 1855 6.9

NO 1876 1548 7.5

N22331 2240 7

2

ck

can be rearranged to solve for k (in N/m): k 5 89 10 5 2.

For a vibration to be active (observable) in an infrared (IR) spectrum, the vibration must change the dipole moment of the molecule. (the vibrations for Cl2, Br2, and N2 will not be observed in an IR experiment)

For a vibration to be active in a Raman spectrum, the vibration must change the polarizability of the molecule.

Vibrational Spectroscopy

For polyatomic molecules, the situation is more complicated because there are more possible types of motion. Each set of possible atomic motions is known as a mode. There are a total of 3N possible motions for a molecule containing N atoms because each atom can move in one of the three orthogonal directions (i.e. in the x, y, or z direction).

Translational modes

Rotational modes

A mode in which all the atoms are moving in the same direction is called a translational mode because it is equivalent to moving the molecule - there are three translational modes for any molecule.

A mode in which the atoms move to rotate (change the orientation) the molecule called a rotational mode - there are three rotational modes for any non-linear molecule and only two for linear molecules.

The other 3N-6 modes (or 3N-5 modes for a linear molecule) for a molecule correspond to vibrations that we might be able to observe experimentally. We must use symmetry to figure out what how many signals we expect to see and what atomic motions contribute to the particular vibrational modes.

Vibrational Spectroscopy and Symmetry

1. Determine the point group of the molecule.

2. Determine the Reducible Representation, tot, for all possible motions of the atoms in the molecule.

3. Identify the Irreducible Representation that provides the Reducible Representation.

4. Identify the representations corresponding to translation (3) and rotation (2 if linear, 3 otherwise) of the molecule. Those that are left correspond to the vibrational modes of the molecule.

5. Determine which of the vibrational modes will be visible in an IR or Raman experiment.

We must use character tables to determine how many signals we will see in a vibrational spectrum (IR or Raman) of a molecule.

Example, the vibrational modes in water.

The point group is C2v so we must use the appropriate character table for the reducible representation of all possible atomic motions, tot. To determine tot we have to determine how each symmetry operation affects the displacement of each atom the molecule – this is done by placing vectors parallel to the x, y and z axes on each atom and applying the symmetry operations. The sum for the vectors on all atoms is placed into the reducible representation.

Vibrational Spectroscopy and Symmetry

Make a drawing of the molecule and add in vectors on each of the atoms. Make the vectors point in the same direction as the x (shown in blue), the y (shown in black) and the z (shown in red) axes. We will treat all vectors at the same time when we are analyzing for molecular motions.

HO

H

H O Htop view

The E operation leaves everything where it is so all nine vectors stay in the same place and the character is 9.

The C2 operation moves both H atoms so we can ignore the vectors on those atoms, but we have to look at the vectors on the oxygen atom, because it is still in the same place. The vector in the z direction does not change (+1) but the vectors in the x, and y directions are reversed (-1 and -1) so the character for C2 is -1.

The v (xz) operation leaves each atom where it was so we have to look at the vectors on each atom. The vectors in the z and x directions do not move (+3 and +3) but the vectors in the y direction are reversed (-3) so the character is 3.

The ’v (yz) operation moves both H atoms so we can ignore the vectors on those atoms, but we have to look at the vectors on the oxygen atom, because it is still in the same place. The vectors in the z and y directions do not move (+1 and +1) but the vectors in the x direction is reversed (-1) so the character is 1. C2V E C2 v (xz) ’v (yz)

tot 9 -1 3 1

Example, the vibrational modes in water.

Vibrational Spectroscopy and Symmetry

H O H

C2

HO

H

H O H

HOH

H O H

v (xz)

H O H

HOH

’v (yz)

z y

x

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

From the tot and the character table, we can figure out the number and types of modes using the same equation that we used for bonding:

n 1

order# of operations in class (character of RR) character of XX

n 1

4A 1 1 9 1 1 1 1 1 3 1 1 1 1

This gives:

n 1

4B1 1 9 1 1 1 1 1 3 1 1 1 1

n 1

4B 2 1 9 1 1 1 1 1 3 1 1 1 1 n

1

4A 2 1 9 1 1 1 1 1 3 1 1 1 1

Which gives: 3 A1’s, 1 A2, 3 B1’s and 2 B2’s or a total of 9 modes, which is what we needed to find because water has three atoms so 3N = 3(3) =9.

Vibrational Spectroscopy and Symmetry

C2V E C2 v (xz) ’v (yz)

tot 9 -1 3 1

Vibrational Spectroscopy and SymmetryNow that we have found that the irreducible representation for tot is (3A1 + A2 + 3B1+ 2B2), the next step is to identify the translational and rotational modes - this can be done by reading them off the character table! The three translational modes have the symmetry of the functions x, y, and z (B1, B2, A1) and the three rotational modes have the symmetry of the functions Rx, Ry and Rz (B2, B1, A2).

Translational modes

Rotational modes

The other three modes (3(3)-6 = 3) that are left over for water (2A1 + B1) are the vibrational modes that we might be able to observe experimentally. Next we have to figure out if we should expect to see these modes in an IR or Raman vibrational spectrum.

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

Remember that for a vibration to be observable in an IR spectrum, the vibration must change the dipole moment of the molecule.

In the character table, representations that change the dipole of the molecule are those that have the same symmetry as translations. Since the irreducible representation of the vibrational modes is (2A1 + B1) all three vibrations for water will be IR active and we expect to see three signals in the spectrum.

For a vibration to be active in a Raman spectrum, the vibration must change the polarizability of the molecule. In the character table, representations that change the polarizability of the molecule are those that have the same symmetry as rotations and products of x,yand z ( x2,xy etc). All three vibrations for water will be Raman active and we expect to see three signals in the spectrum.

Vibrational Spectroscopy and Symmetry

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

The three vibrational modes for water. Each mode is listed with a (Greek letter ‘nu’) and a subscript and the energy of the vibration is given in parentheses. 1 is called the “symmetric stretch”, 3 is called the “anti-symmetric stretch” and 2 is called the “symmetric bend”.

How many vibrational modes belong to each of irreducible representation ?

Vibrational spectrum of H2O

Vibrational Spectrum of H2O

Sample Problem

MO of Polyatomic Molecules

MO of H2O

MO of H2O

MO of H2O

MO of H2O

MO of H2O

MO of H2O

MO of H2O

MO of H2O

MO of H2O

Now for the Million Question

Why is water Bent ????

MO Theory - Walsh Diagram

Linear XH2 molecules

MO of BeH2

Walsh Diagram of XH2

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of BH3

MO of NH3

MO of NH3

MO of NH3

MO of NH3

MO of NH3

MO of NH3

MO of NH3

MO of NH3

MO of CH4

MO of CH4

MO of CH4

MO of CH4

MO of CH4

MO of CH4

MO of CH4

MO of CH4

MO of CH4

MO of CH4

C 4 HCH4

2sA1

2p T2

1a1

2a1*

1t2

2t2*

C 2 H, 2 ClCH2Cl2

2s

A1,B1

2p

A1,B2

Molecular Orbital TheoryPolyatomic molecules

1a1

4a1*

1b1

1b2

3a1*

2b2*

2a1

2b1*

Consider what happens to the -bonding MO’s if the symmetry is reduced from Td to C2v: there will no longer be triply-degenerate MO’s. The four bonding orbitals are split into two sets - those with more H character (higher in energy) and those with more Cl character (lower in energy).

Since these diagrams are only concerned with the -bonding, the AO’s and MO’s from the 3p orbitals on Cl have been omitted. The correlation lines from the 2p orbitals are not drawn for clarity.

2 H

2 Cl

MO of AsF5

MO of ClF5 (C4V)

MO of SF6

MO of SF6

Matrix Representations

in Group Theory

Character Tables

Group theory makes use of the properties of matrices

Each symmetry operation may be expressed as a transformation matrix:[New coordinates] = [transformation matrix][old coordinates]

Example: in Cartesian coordinate system, reflection in x = 0 plane

• Changes the value of x to –x, multiplies it by -1

• Leaves y unchanged

• Leaves z unchanged

100

010

001

'

'

'

z

y

x

z

y

x

Original coordinates

Transformation matrix

Results of transformation.

=

Character Tables - 2

The matrix representation of the symmetry operations of a point group is the set of matrices corresponding to all the symmetry operations in that group. The matrices record how the x,y,z coordinates are modified as a result of an operation.

For example, the C2v point group consists of the following operations

E: do nothing. Unchanged.

C2: rotate 180 degrees about the z axis: x becomes –x; y becomes –y and z unchanged.

v (xz): y becomes –y

v’ (yz): x becomes -x

100

010

001

100

010

001

E

100

010

001

C2

100

010

001

v (xz): v’ (yz):

Matrix 3 - dimensional of an n – fold rotation

Matrix 3 - dimensional of a vertical mirror plane

The Group Multiplication Table of the C2v point group

E E C2 v (xz) ’v (yz)

E E C2 v (xz) v’ (xz)

C2 C2 E ’v (yz) v (yz)

v (xz v (xz) ’v (yz) E C2

v’ (xz) v

’ (xz) v (yz) C2 E

Exercise1.) Multiply the transformation matrix for C2 by

the transformation matrix for δv

C2δv = δv’

2.) Show that δv δv’ = C2

3.) Using matrix expression for C2 and E, show that C2 x C2 = E

Substituting n =2 and m =1 to C2

Representations A representation (V) for a point group any set of square matrices that multiply as the symmetry

operations of the group

A reducible representation contains matrices that can all be partitioned into the same

block – diagonal form consisting of a series of submatrices that lie along the main diagonal

The submatrices are the matrices for the irreducible representation

Block - diagonalization If a reducible representation (V) consists of

the matrices [A], [B] and [C], each of which can be partitioned into the block –diagonal form,

[X1] = [ X11 X12] , [X2] = X33 [X3] = X44

[ X21 X22]

V (direct sum) = V1 + V2 + V3

ExerciseFind the irreducible representation for this group ?

V(direct sum) = V3 + V4 + V1

C2V E C2 v (xz) ’v (yz)

A1 1 1 1 1 V1 z x2,y2,z2

A2 1 1 -1 -1 V2 Rz xy

B1 1 -1 1 -1 V3 x, Ry xz

B2 1 -1 -1 1 V4 y, Rx yz

Character Table for C3v

Theorems:

The number of irreducible representations is equal to the number of classes