Lecture 6

Character Tables

Group theory makes use of the properties of matrices

Idea: When an operation, O, (proper rotation, improper rotation, reflection, inversion)

is done on the function, f(x,y,z)O f(x,y,z)

the result is a value taken on by the original function at some other point (x’,y’,z’).

OrO f(x,y,z) = f (x’,y’,z’)

We know the function f(x,y,z) the problem is to determine the values (x’,y’,z’).

The nature of the operation tells us where to look.

Character Tables

Group theory makes use of the properties of matrices

Each 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 (multiplies it by 1)

• Leaves z unchanged (multiplies it by 1)

100

010

001

'

'

'

z

y

x

z

y

x

Original coordinates

Transformation matrix

Results of transformation.

=

To see the result of the operation at (x, y, z) look at the original object at (x’, y’, z’).

Recall Technique of Matrix multiplication

To get an element of the product vector a row in the operation square matrix is multiplied by the original vector matrix.

100

010

001

'

'

'

z

y

x

z

y

x=

V’ M V

jj

jii VMV ,'

For example

V’2 = y’ = M2,1 * V1 + M2,2 * V2 + M2,3 * V3

y’ = 0 * x + 1 * y + 0 * z = y

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):

Operations Applied to Functions - 1

Consider

f(x) = x2

v’ (f(x)) = v(x2) = (-x)2 = x2 = f(x)

v’ (f(x)) = 1 * f(x)

or

f(x) is an eigenfunction of this reflection operator with an eigenvalue of +1. This is called a symmetric eigenfunction.

Similarly

f(x) = x3

v’ (f(x)) = -1 * f(x)

f(x) is an eigenfunction of this reflection operator with an eigenvalue of -1. This is called a antisymmetric eigenfunction.

Transform the coordinates.

Plots of Functions, x2

0

5

10

15

20

25

30

-5

-4.6

-4.2

-3.8

-3.4 -3

-2.6

-2.2

-1.8

-1.4 -1

-0.6

-0.2 0.2

0.6 1

1.4

1.8

2.2

2.6 3

3.4

3.8

4.2

4.6 5

Series1

Here f(x) is x2. It can be seen to be a symmetric function for reflection at x = 0 because of mirror plane.

0

5

10

15

20

25

30

-5

-4.6

-4.2

-3.8

-3.4 -3

-2.6

-2.2

-1.8

-1.4 -1

-0.6

-0.2 0.2

0.6 1

1.4

1.8

2.2

2.6 3

3.4

3.8

4.2

4.6 5

Series1

Reflection yields.

The reflection carries out the mapping shown with the red arrows.

x2 is an eigenfunction of with eigenvalue 1

-150

-100

-50

0

50

100

150

-5

-4.6

-4.2

-3.8

-3.4 -3

-2.6

-2.2

-1.8

-1.4 -1

-0.6

-0.2 0.2

0.6 1

1.4

1.8

2.2

2.6 3

3.4

3.8

4.2

4.6 5

Series1

Plots of Functions, x3

Here f(x) is x3. It can be seen to be a antisymmetric function for reflection at x = 0.

The reflection carries out the mapping shown with the red arrows.

-150

-100

-50

0

50

100

150

-5

-4.6

-4.2

-3.8

-3.4 -3

-2.6

-2.2

-1.8

-1.4 -1

-0.6

-0.2 0.2

0.6 1

1.4

1.8

2.2

2.6 3

3.4

3.8

4.2

4.6 5

Series1

Reflection yields.

x3 is an eigenfunction of with eigenvalue -1

Operations Applied to Functions - 2

Now consider

f(x) = (x-2)2 = x2 – 4x + 4

v’ (f(x)) = v(x-2)2 = (-x-2)2 = x2 + 4x + 4

f(x) = (x-2)2 is not an eigenfunction of this reflection operator because it does not return a constant times f(x).

0

10

20

30

40

50

60

-5

-4.6

-4.2

-3.8

-3.4 -3

-2.6

-2.2

-1.8

-1.4 -1

-0.6

-0.2 0.2

0.6 1

1.4

1.8

2.2

2.6 3

3.4

3.8

4.2

4.6 5

Series1

0

10

20

30

40

50

60

-5

-4.6

-4.2

-3.8

-3.4 -3

-2.6

-2.2

-1.8

-1.4 -1

-0.6

-0.2 0.2

0.6 1

1.4

1.8

2.2

2.6 3

3.4

3.8

4.2

4.6 5

Series1

Reflection yields this function, not an eigenfunction.

Neither symmetric nor antisymmetric for reflection thru x = 0.

Let’s look at Atomic Orbitals

s orbital

Reflection

Get the same orbital back, multiplied by +1, an eigenfunction of the reflection, symmetric with respect to the reflection. The s orbital forms the basis of an irreducible representation of the operation

z

Atomic Orbitals

p orbital

Reflection

Get the same orbital back, multiplied by -1, an eigenfunction of the reflection, antisymmetric with respect to the reflection. The p orbital behaves differently from the s orbital and forms the basis of a different irreducible representation of the operation

z

Simplest ways that objects can behave for a group consisting of E and h , the reflection plane.

Cs E h

A’

A”

1 1

1 -1

x, y,Rz

z, Rx,Ry

x2,y2,z2,xy

yz, xz

s orbital is spherical behaves as x2 + y2 + z2. s orbital is A’. The s orbital is an eigenfunction of both E and h.

pz orbital has a multiplicative factor of z times a spherical factor. Behaves as A”. pz is an eigenfunction of both E and h.

Irreducible Representations. Basis of the Irreducible Reps.

sp Hybrids

hybrid

Reflection

Do not get the same hybrid back multiplied by +1 or -1 or some other constant. Not an eigenfunction.

Recall: the hybrid can be expressed as the sum of an s orbital and a p orbital.

=+

Reduction: expressing a reducible representation as a combination of irreducible representations.

zThe two hybrids form the basis of a reducible representation of the operation

Reducible RepresentationsUse the two sp hybrids as the basis of a representation

h1 h2

10

01

01

10

E operation. h operation.

2

1

h

h=

2

1

h

h

2

1

h

h=

1

2

h

h

The hybrids are unaffected by the E operation.

The reflection operation interchanges the two hybrids.

Proceed using the trace of the matrix representation.

1 + 1 = 2 0 + 0 = 0

h1 becomes h1; h2 becomes h2. h1 becomes h2; h2 becomes h1.

Our Irreducible Representations

Cs E h

A’

A”

1 1

1 -1

x, y,Rz

z, Rx,Ry

x2,y2,z2,xy

yz, xz

The reducible representation derived from the two hybrids can be attached to the table.

2 0 (h1, h2)

Note that = A’ + A”

h1h2

, ,h1 - h2 h1 + h2

Return to polynomials:

f(x) = (x-2)2 = x2 – 4x + 4

v (f(x)) = v(x-2)2 = (-x-2)2 = x2 + 4x + 4 =g(x)

Neither f nor g is an eigenfunction of but, taken together, they do form an reducible representation since they show what the operator does.

Approaching the problem in the same way as we did for hybrids we can carry out the reduction this wayu(x) = ½ (f(x) + g(x)) = ½ (f(x) + f(x)) = x2 + 4 symmetric, unchanged by the operator. Behaves as A’ v(x) = ½ (f(x) - g(x)) = ½ (f(x) - f(x)) = -4x, antisymmetric, multiplied by -1 by the operator. Behaves as A’’

Point group Symmetry operations, Classes

Characters+1 symmetric behavior

-1 antisymmetricMülliken symbols

Each row is an irreducible representation

Character Table

x, y, zSymmetry of translations (p orbitals)

Rx, Ry, Rz: rotations

Character Tables - 3

Irreducible representations are not linear combinations of other representation(Reducible representations are)

# of irreducible representations = # of classes of symmetry operations

Instead of the matrices, the characters are used (traces of matrices)

A character Table is the complete set of irreducible representations of a point group

Effect of the 4 operations in the point group C2v

on a translation in the x direction. The translation is simplymultiplied by 1 or -1. It forms a basis to show

what the operators do to an object.

Operation E C2 v v’

Transformation 1 -1 1 -1

Character Table

Verify this character. It represents how a function that behaves as x, Ry, or xz behaves for C2.

x, y, zSymmetry of translations (p orbitals)

Rx, Ry, Rz: rotations

Classes of operations

dxy, dxz, dyz, as xy, xz, yz

dx2- y

2 behaves as x2 – y2

dz2 behaves as 2z2 - (x2 +

y2)

px, py, pz behave as x, y, z

s behaves as x2 + y2 + z2

Another point group, C3v.

Symmetry of Atomic Orbitals

Naming of Irreducible representations

• One dimensional (non degenerate) representations are designated A or B.

• Two-dimensional (doubly degenerate) are designated E.

• Three-dimensional (triply degenerate) are designated T.

• Any 1-D representation symmetric with respect to Cn is designated A; antisymmétric ones are designated B

• Subscripts 1 or 2 (applied to A or B refer) to symmetric and antisymmetric representations with respect to C2 Cn or (if no C2) to v respectively

• Superscripts ‘ and ‘’ indicate symmetric and antisymmetric operations with respect to h, respectively

• In groups having a center of inversion, subscripts g (gerade) and u (ungerade) indicate symmetric and antisymmetric representations with respect to i

Character Tables

• Irreducible representations are the generalized analogues of or symmetry in diatomic molecules.

• Characters in rows designated A, B,..., and in columns other than E indicate the behavior of an orbital or group of orbitals under the corresponding operations (+1 = orbital does not change; -1 = orbital changes sign; anything else = more complex change)

• Characters in the column of operation E indicate the degeneracy of orbitals

• Symmetry classes are represented by CAPITAL LETTERS (A, B, E, T,...) whereas orbitals are represented in lowercase (a, b, e, t,...)

• The identity of orbitals which a row represents is found at the extreme right of the row

• Pairs in brackets refer to groups of degenerate orbitals and, in those cases, the characters refer to the properties of the set

Definition of a Group

• A group is a set, G, together with a binary operation : such that the product of any two members of the group is a member of the group, usually denoted by a*b, such that the following properties are satisfied :

– (Associativity) (a*b)*c = a*(b*c) for all a, b, c belonging to G. – (Identity) There exists e belonging to G, such that e*g = g = g*e

for all g belonging to G. – (Inverse) For each g belonging to G, there exists the inverse of

g, g-1, such that g-1*g = g*g-1 = e.

• If commutativity is satisfied, i.e. a*b = b*a for all a, b belonging to G, then G is called an abelian group.

Examples

• The set of integers Z, is an abelian group under addition.

What is the element e, identity, such that

a*e = a?

What is the inverse of the a element? 0

-a

As applied to our symmetry operators.

For the C3v point group

What is the inverse of each operator? A * A-1 = E

E C3(120) C3(240) v (1) v (2) v (3)

E C3(240) C3(120) v (1) v (2) v (3)

Examine the matrix represetation of the C2v point group

1 0 0

0 1 0 0 0 1

x y z

=

x y z

-1 0 0

0 -1 0 0 0 1

x y z

=

-x -y

z

C2

1 0 0

0 -1 0 0 0 1

x y z

=

x -y

z

v(xz)

E

- x y z

’v(yz)

Multiplying two matrices (a reminder)

a11 a12 a21 a22 a31 a32

b11 b12 b13 b21 b22 b23

=

c11 c12 c13 c21 c22 c23 c31 c32 c33

c11 = a11b11 + a12b21c12 = a11b12 + a12b22c13 = a11b13 + a12b23

c21 = a21b11 + a22b21c22 = a21b12 + a22b22c23 = a31b13 + a32b23

c31 = a31b11 + a32b21c32 = a31b12 + a32b22c33 = a31b13 + a32b23

±1 0 0

0 ±1 0 0 0 ±1

Most of the transformationmatrices we use have the form

-1 0 0

0 -1 0 0 0 1

x y z

=

1 0 0

0 1 0 0 0 1

x y z

=

1 0 0

0 -1 0 0 0 1

x y z

=

C2E v(xz) ’v(yz)

What is the inverse of C2? C2

What is the inverse of v? v

-1 0 0

0 -1 0 0 0 1

x y z

=

-1 0 0

0 -1 0 0 0 1

x y z

==

1 0 0

0 1 0 0 0 1

x y z

=

1 0 0

0 -1 0 0 0 1

x y z

=

1 0 0

0 -1 0 0 0 1

x y z

==

1 0 0

0 1 0 0 0 1

x y z

=

-1 0 0

0 -1 0 0 0 1

x y z

=

1 0 0

0 1 0 0 0 1

x y z

=

1 0 0

0 -1 0 0 0 1

x y z

=

C2E v(xz) ’v(yz)

What of the products of operations?

E * C2 = ? C2

1 0 0

0 1 0 0 0 1

x y z

=

-1 0 0

0 -1 0 0 0 1

x y z

==

-1 0 0

0 -1 0 0 0 1

x y z

=

v * C2 = ? ’v

1 0 0

0 -1 0 0 0 1

x y z

=

-1 0 0

0 -1 0 0 0 1

x y z

==

ClassesTwo members, c1 and c2, of a group belong to the same class if there is a member, g, of the group such that

g*c1*g-1 = c2

Properties of Characters of Irreducible Representations in Point Groups

• Total number of symmetry operations in the group is called the order of the group (h). For C3v, for example, it is 6.

1 + 2 + 3 = 6

• Symmetry operations are arranged in classes. Operations in a class are grouped together as they have identical characters. Elements in a class are related.

This column represents three symmetry operations having identical characters.

Properties of Characters of Irreducible Representations in Point Groups - 2

The number of irreducible reps equals the number of

classes. The character table is square.

3 by 3

The sum of the squares of the dimensions of the each irreducible rep equals the order of the group, h.

1 + 2 + 3 = 6

1

1

22

6

Properties of Characters of Irreducible Representations in Point Groups - 3

For any irreducible rep the squares of the characters summed over the symmetry operations equals the order of the group, h.

A1: 12 + (12 + 12 ) + = 6

A2: 12 + (12 + 12 ) + ((-1)2 + (-1)2 + (-1)2 ) = 6

E: 22 + (-1)2 + (-1)2 = 6

Properties of Characters of Irreducible Representations in Point Groups - 4

Irreducible reps are orthogonal. The sum of the products of the characters for each symmetry operation is zero.

For A1 and E:

1 * 2 + (1 *(-1) + 1 *(-1)) + (1 * 0 + 1 * 0 + 1 * 0) = 0

Properties of Characters of Irreducible Representations in Point Groups - 5

Each group has a totally symmetric irreducible rep having all characters equal to 1

Reduction of a Reducible Representation

Irreducible reps may be regarded as orthogonal vectors. The magnitude of the vector is h-1/2

Any representation may be regarded as a vector which is a linear combination of the irreducible representations.

Reducible Rep = (ai * IrreducibleRepi)

The Irreducible reps are orthogonal. Hence

(character of Reducible Rep)(character of Irreducible Repi) = ai * h

Or

ai = (character of Reducible Rep)(character of Irreducible Repi) / h

Sym ops

Sym ops

These are block-diagonalized matrices(x, y, z coordinates are independent of each other)

1 0 0

0 -1 0 0 0 1

1 0 0

0 1 0 0 0 1

-1 0 0

0 -1 0 0 0 1

C2 v(xz)E

-1 0 0

0 1 0 0 0 1

v'(yz)

C2 v(xz) v'(yz)

x

y

z

E

1

1

1

3

-1

-1

1

-1

1

-1

1

1

-1

1

1

1

Irreduciblerepresentations

Reducible Rep

Point group Symmetry operations

Characters+1 symmetric behavior

-1 antisymmetricMülliken symbols

Each row is an irreducible representation

C2v Character Table to be used for water

Let’s use character tables!Symmetry and molecular vibrations

# of atoms degrees of freedom

Translational modes

Rotational modes

Vibrational modes

N (linear) 3 x 2 3 2 3N-5 = 1

Example

3 (HCN)

9 3 2 4

N (non- linear)

3N 3 3 3N-6

Example

3 (H2O)

9 3 3 3

Symmetry and molecular vibrations

A molecular vibration is IR activeonly if it results in a change in the dipole moment of the molecule

A molecular vibration is Raman activeonly if it results in a change in the polarizability of the molecule

In group theory terms:

A vibrational mode is IR active if it corresponds to an irreducible representationwith the same symmetry of a x, y, z coordinate (or function)

and it is Raman active if the symmetry is the same asA quadratic function x2, y2, z2, xy, xz, yz, x2-y2

If the molecule has a center of inversion, no vibration can be both IR & Raman active

How many vibrational modes belong to each irreducible representation?

You need the molecular geometry (point group) and the character table

Use the translation vectors of the atoms as the basis of a reducible representation.

Since you only need the trace recognize that only the vectors that are either unchanged or have become the negatives of themselves by a symmetry operation contribute to the character.

Apply each symmetry operation in that point group to the moleculeand determine how many atoms are not moved by the symmetry operation.

Multiply that number by the character contribution of that operation:

E = 3 = 1C2 = -1i = -3C3 = 0

That will give you the reducible representation

A shorter method can be devised. Recognize that a vector is unchanged or becomes the negative of itself if the atom does not move. A reflection will leave two vectors unchanged and multiply the other by -1 contributing +1.For a rotation leaving the position of an atom unchanged will invert the direction of two vectors, leaving the third unchanged.Etc.

3x39

1x-1-1

3x13

1x11

Finding the reducible representation

(# atoms not moving x char. contrib.)

E = 3 = 1C2 = -1i = -3C3 = 0

Now separate the reducible representation into irreducible onesto see how many there are of each type

A1 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x1 + 1x1x1) = 3

A2 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x(-1) + 1x1x(-1)) = 1

9 -1 3 1

Symmetry of molecular movements of water

Vibrational modes

IR activeRaman active

Which of these vibrations having A1 and B1 symmetry are IR or Raman active?

ML C

CLO

OC2

E

1

2

ML C

CLO

O

1

2

ML C

CLO

O1

2

ML C

CLO

O

1

2

ML C

CLO

O1

2

C2 v(xz) v(yz)

Often you analyze selected vibrational modes

(CO)

Find: # vectors remaining unchanged after operation.

2 x 12

0 x 10

2 x 12

0 x 10

Example: C-O stretch in C2v complex.

A1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1

A2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x-1) = 0

2 0 2 0

B1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1

B2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x1) = 0

ML C

CLO

O Both A1 and B1 are

IR and Raman active

= A1 + B1

MC L

CLO

1

2O

What about the trans isomer?

C2(z) v(xz) v'(yz)

x

E

1

1

2

1

-1

0

1

-1

0

D2h

Ag

B3u

C2(y) C2(x)

1

1

2

i

1

-1

0

v(xy)

1

1

2

1

1

2

1

-1

0

Only one IR active band and no Raman active bands

Remember cis isomer had two IR active bands and one Raman active

Symmetry and NMR spectroscopy

The # of signals in the spectrumcorresponds to the # of types of nuclei not related by symmetry

The symmetry of a molecule may be determinedFrom the # of signals, or vice-versa

Molecular Orbitals

Atomic orbitals interact to form molecular orbitals

Electrons are placed in molecular orbitalsfollowing the same rules as for atomic orbitals

In terms of approximate solutions to the Scrödinger equation

Molecular Orbitals are linear combinations of atomic orbitals (LCAO)

caacbb (for diatomic molecules)

Interactions depend on the symmetry properties

and the relative energies of the atomic orbitals

As the distance between atoms decreases

Atomic orbitals overlap

Bonding takes place if:the orbital symmetry must be such that regions of the same sign overlapthe energy of the orbitals must be similarthe interatomic distance must be short enough but not too short

If the total energy of the electrons in the molecular orbitalsis less than in the atomic orbitals, the molecule is stable compared with the atoms

Combinations of two s orbitals (e.g. H2)

Antibonding

Bonding

More generally:ca(1sa)cb(1sb)]

n A.O.’s n M.O.’s

Electrons in bonding orbitals concentrate between the nuclei and hold the nuclei together(total energy is lowered)

Electrons in antibonding orbitals cause mutual repulsion between the atoms(total energy is raised)

Bothand notation means symmetric/antisymmetric with respect to rotation

zC2

zC2 zC2

zC2Not

Combinations of two p orbitals (e.g. H2)

and notation meanschange of sign upon C2 rotation

and notation means nochange of sign upon rotation

Combinations of two p orbitals

zC2

zC2

Combinations of two sets of p orbitals

Combinations of s and p orbitals

Combinations of d orbitals

No interaction – different symmetry means change of sign upon C4

NO NOYES

Is there a net interaction?

Relative energies of interacting orbitals must be similar

Strong interaction Weak interaction

Molecular orbitalsfor diatomic molecules

From H2 to Ne2

Electrons are placed

in molecular orbitals

following the same rules

as for atomic orbitals:

Fill from lowest to highest

Maximum spin multiplicityElectrons have

different quantum numbers including spin

(+ ½, - ½)

Bond order = # of electrons

in bonding MO's# of electrons in antibonding MO's

12

-

O2 (2 x 8e)

1/2 (10 - 6) = 2A double bond

Or counting onlyvalence electrons:1/2 (8 - 4) = 2

Note subscriptsg and u

symmetric/antisymmetricupon i

Place labels g or u in this diagram

g

g

u

u

g

u

g

u

u

g

g or u?

Orbital mixing

Same symmetry and similar energies !shouldn’t they interact?

orbital mixing

When two MO’s of the same symmetry mixthe one with higher energy moves higher and the one with lower energy moves lower

H2 g2 (single bond)

He2 g2 u2 (no bond)

Molecular orbitalsfor diatomic molecules

From H2 to Ne2

E (Z*)

E > E Paramagneticdue to mixing

C2 u2 u

2 (double bond)

C22- u

2 u2 g

2(triple bond)

O2 u2 u

2 g1 g1 (double bond)paramagneticO2

2- u2 u

2 g2 g2 (single bond)diamagnetic

Bond lengths in diatomic molecules

Filling bonding orbitals

Filling antibonding orbitals

Photoelectron Spectroscopy

h(UV o X rays) e-

Ionization energy

hphotons

kinetic energy of expelled electron= -

N2O2

*u (2s)

u (2p)

g (2p)

*u (2s)

g (2p)u (2p)

u (2p)

(Energy required to remove electron, lower energy for higher orbitals)

Very involved in bonding(vibrational fine structure)