The Irreducible reps are orthogonal. Hence for the reducible rep and a particular irreducible rep j

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

Or

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

Reduction of a Reducible Representation. Given a Reducible Rep how do we find what Irreducible reps it contains?

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, i = 1,2….

Reducible Rep = (ai * IrreducibleRepi)

Sym ops

Sym ops

i

Irred. Reps

Reducible Representations in Cs = E and h

Use 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.

Obtainthe trace of the matrix representation.

1 + 1 = 2 0 + 0 = 0

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

10

01

01

10

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 , h1; become themselves

h1 h2; do not become themselves, interchange

Let’s observe one helpful thing here.

Only the objects (hybrids) that remain themselves, appear on the diagonal of the transformation of the symmetry operation, contribute to the trace. They commonly contribute +1 or -1 to the trace depending whether or not they are multiplied by -1.

The Irreducible Representations for Cs.

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

The Irreducible Representations for Cs.

Cs E h

A’

A”

1 1

1 -1

x, y,Rz

z, Rx,Ry

x2,y2,z2,xy

yz, xz

Let’s verify some things.

Order of the group = # sym operations = 2

A’ and A’’ are orthogonal: 1*1 + 1+(-1) = 0

Sum of the squares over sym operations = order of group = h

The magnitude of the A’ and A’’ vectors are each (2) 1/2: magnitude2 = ( 12 + (+/- 1)2)

The Irreducible Representations for Cs.

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)

Now let’s do the reduction.

We assume that the reducible rep G can be expressed as a linear combination of A’ and A’’

= aA’ A’ + aA’’ A’’; our task is to find out the coefficients aA’ and aA’’

Cs E h

A’

A”

1 1

1 -1

x, y,Rz

z, Rx,Ry

x2,y2,z2,xy

yz, xz

2 0 (h1, h2)

aA’ = (1 * 2 + 1 *0)/2 = 1

aA’’ = (1 * 2 + 1 *0)/2 = 1

Or again = 1*A’ + 1*A’’. Note that this holds for any reducible rep as above and not limited to the hybrids in any way.

Another Example of Reduction…

Sulfur Difluoride belongs to the C2v point group.

What this means is that the molecular structure behaves at A1 for C2v This means that the molecular structure is transformed into itself by all the sym operations.

We can apply these symmetry operations to any kind of object and see how the object behaves.

Additional objects may be needed to show what the symmetry operations do.

F

SF x

z

We know that each irreducible rep is 1 dimensional but let’s assume that we do not know the dimensions of the irreducible reps.

We do know that the d orbitals on an isolated atom are equivalent. Let’s guess that we need all five for a representation. It will turn out to be reducible but again assume that we do not know that.

We have to find out how each d orbital transforms under the C2v operations. Formulate the matrices of the representation based on the d orbitals in this order dz2, dxz, dyz, dxy, dx2-y2.

Operation E C2 v(xz) v(yz)

Representation,

Trace 5 1 1 1

F

SF x

z

10000

01000

00100

00010

00001

Remember what these matrices mean!

Transformed d orbitals = matrix * original d orbitals

Now do Reduction.

We have four irreducible reps: A1, A2, B1, B2. The reducible rep, , should be expressible as a linear combination of the irreducible reps.

= aA1 A1 + aA2 A2 + aB1 B1 + aB2 B2

(5,1,-1,-1) = aA1 (1,1,1,1) + aA2 (1,1,-1,-1) + aB1 (1,-1, 1,-1) + aB2 (1,-1, -1,1)

Or using vector notation….

Find one of the a coefficents, say, aB1, by multiplying everything by the elements of the B1 rep and summing over sym elements.

(5,1,1,1) = aA1 (1,1,1,1) + aA2 (1,1,-1,-1) + aB1 (1,-1, 1,-1) + aB2 (1,-1, -1,1)

(1,-1, 1,-1) x

(1,-1, 1,-1) x

(1,-1, 1,-1) x

(1,-1, 1,-1) x

(1,-1, 1,-1) x

5 -1 +1 -1 = 4

0 0 aB1 4 0

a B1 = 1 and similarly: aA1 = 2; aA2 = 1; aB2 = 1

Or the d orbitals transform as 2A1 + A2 + B1 + B2

The d orbitals transform as 2 A1 + A2 + B1 + B2

Note thatdz2 and dx2-y2 transform as A1

dxy as A2

dxz as B1

dyz as B2

Point group Symmetry operations

Characters+1 symmetric behavior

-1 antisymmetricMülliken symbols

Each row is an irreducible representation

Vibrational Modes of water, C2v

# of Atoms Degrees of freedom

Translational modes

Rotational modes

Vibrational modes

Linear Molecules

N 3 x N 3 2 3N-5

Example

3 (HCN)

9 3 2 4

Non linear Molecules

N 3 x N 3 3 3N-6

Example

3 (H2O)

9 3 3 3

Let’s determine how many independent vibrations a molecule can have. It depends on how many atoms, N, and whether

the molecule is linear or non-linear.

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 increase the energy of the system.

Bothand notation means symmetric with respect to rotation about internuclear axis

zC2

zC2 zC2

zC2Not

p orbitals can combine in two ways, and .

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

Combination of s and p orbitals

Combination of two misaligned p orbitals in mode.

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 placedin molecular orbitalsfollowing the same

rulesas for atomic

orbitals:Fill from lowest to

highestMaximum 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

Notice the change in ordering here caused by the mixing.

H2 g2 (single bond)

He2 g2 u2 (no bond)

Molecular orbitalsfor diatomic molecules

From H2 to Ne2

In Li2 through N2 g above u.

O2 through Ne2 u above g.

-The orbitals are more

affected by effective nuclear charge than

are the -Less mixing for O, F and Ne due to smaller orbitals and

greater bond distance.

Paramagneticdue to mixing

Neutral: C2 u2 u

2 (double bond)

Dianion: C22- u

2 u2 g

2(triple bond)Neutral: O2 u

2 u2 g1 g1

(double bond, paramagnetic)Dianion: O2

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)