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

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 indicates 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 180 rotation

and notation means nochange of sign upon rotation

Combinations of two p orbitals

zC2

zC2

Both sigma and pi interactions of p orbitals

Combination of s and p orbitals

Combination of two misaligned p orbitals in mode.

No overlap, no bonding

d orbitals overlappingNo interaction – different symmetry type overlap: symmetric for any rotation

type overlap: antisymmetric for 180 rotation,

type overlap. Changes sign for rotation by 90

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 orbitals

following the same rulesas for atomic orbitals:

Fill from lowest to highestMaximum spin multiplicity

Electrons 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 mixingIn this diagram some of the MOs have the same symmetry.

This allows for interaction to take place. Mixing can occur.

Orbitals of g symmetry can mix together to yield a more stable and a less stable combination.

Likewise for u

sum

dif ference

For the orbitals of g symmetry.

Overall result: 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.

This revised ordering occurs Li N, but not O Ne

Notice the change in ordering here caused by the mixing.

O Ne Li N

The electronic configuratations

Li2 bond order 1

Be2 bond order ?

B2 bond order 1C2 bond order 2

N2 bond order 3

O2 bond order 2 F2 bond order 1

Expect the shortest bond length for N2

(But remember the atomic radii decrease with F smallest.)

Bond lengths in diatomic molecules

Filling bonding orbitals

Filling antibonding orbitals

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 u2 u

2 g1 g

1 (double bond, paramagnetic)Dianion: O2

2- u2 u

2 g2 g

2 (single bond, diamagnetic)

Photoelectron Spectroscopy

h(UV o X rays) e-

Ionization energy

hphotons

kinetic energy of expelled electron= -

Ionization is a transition between states

• Initial State: Neutral (or anion)

• Final State: Atom/Molecule/Anion after an electron is removed, plus the ejected electron

• M → M+ + e-

init = M ; final = M+ + e-

• For direct photoionization, transition probability is always > 0

h + Molecule

= Ionization Energy

e- + Molecule+

Eh - Ee- = EM+ - EM

IE = Difference in energy between states of M, M+

Ionization Energy (eV)

161718 15

↿⇂↿ ⇂H 1sH 1s

σ

σ*

What about molecules?

molecular rotationslower energy microwave radiation

electron transitionshigher energy visible and UV radiation

molecular vibrationsmedium energy IR radiation

Ground State

Excited StateDuring an electronic transition

the complex absorbs energy

electrons change orbital

the complex changes energy state

Timescale : ≈10-15 sec

Timescale of geometry changes (vibrations): ≈10-12 sec

As a result, observe vertical (Franck-Condon) transitions

In other words, we assume that we only have to consider the electronic portion of the ground- and excited-state wavefunctions to understand these transitions: Born-Oppenheimer approximation

Transitions between molecular potential energy surfaces

18

17

16

15

0

Ioni

zatio

n E

nerg

y (e

V)

H2+

r (Å)0 1 2

H2

Potential Energy Surface Description of the Ionization of Dihydrogen

Much more on this next time!!

Ground state (X) = 1g+

Ionization Energy (eV)151617181920

:N≡N:

2p

2s

2p

2s

1g+

2g+

1u+

2u

1u

1g

First ion state (X) = 2g+

Second ion state (A) = 2u

Third ion state (B) = 2u+

Consider Dinitrogen

N21g

2g+

2u+

2u+

Ground state (X) = 1g+

:N≡N:

2p

2s

2p

2s

1g+

2g+

1u+

2u

1u

1g

Potential Well Description

Models to describe molecular electronic structure

MO Theory compared to

Valence Bond Theory

Consider methane.

VSEPR gives 4 sp3 hybrid orbitals.

24 22 20 18 16 14 12

Ionization Energy (eV)

CH 4

24 22 20 18 16 14 12

Ionization Energy (eV)

CH 4

Photoelectron Spectroscopy

2s

2p

sp3

So why are there two valence ionizations separated by almost 10 eV?

Use of reducible representations in M.O. theory

Consider transformation properties ofvectors aligned with the 4 C-H bonds.

Td E 8C3 3C2 6S4 6σd

σ 4 1 0 0 2

1]120084[24

11

Aa

0]120084[24

12

Aa

0]00088[24

1Ea

0]1200012[24

11

Ta

1]1200012[24

12

Ta

C-H = A1 + T2

http://www.mpip-mainz.mpg.de/~gelessus/group.html

Apply Reduction Formula:

This is what the bonding MOs orbitals should be.

24 22 20 18 16 14 12

Ionization Energy (eV)

CH4

24 22 20 18 16 14 12

Ionization Energy (eV)

CH4

CH4

t2 (1, 2, 3)

a1 (1)

2p (t2)

2s (a1)

C H4

LCAO Description of Methane, Td

Obtained using the carbon AOs as basis objects. Obtained using the hydrogen

AOs as basis objects.Orbitals of same symmetry combine and bond.

N2 O2

*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)

Now back to our ordering of the MOs of diatomics.

Orbital EnergyOrbital Energy

Very Simple Molecular Orbital Theory

A molecular orbital, f, is expressed as a linear combination of atomic orbitals, holding two electrons.

The multi-electron wavefunction and the multi-electron Hamiltonian are

)...3()2(1...)3,2,1( 211 fffF

electrons

iihH ...)3,2,1(

Where hi is the energy operator for electron i and involves only electron i

AO

llluaf

Seek F such that...)3,2,1(...)3,2,1(...)3,2,1( EFFH

Divide by F(1,2,3…) recognizing that hi works only on electron i.

electrons

i j

ji Eif

ifh

)(

)(

jjj fefh Since each term in the summation depends on the coordinates of a different electron then each term must equal a constant.

)...3()2()1()...3()2()1()...3()2()1( 332132213211 fhffffhffffh

)...3()2()1( 321 ffEf

)...3()2()1( 321 fffhi

jjj fefh

Multiply equation at top by one of the atomic orbitals, uk k = 1,2,…#AOs, and integrate.

dvhuuh jkjk ,

dvfuedvhfu jkjjk

AO

lllj uaf

Recall the expansion of a molecular orbital in terms of the atomic orbitals.

Now use the expansion of the MO in the AOs.

dvuuS jkjk ,

0)( ,, AO

llklkl eSha

These integrals are fixed numerical values.

Now our attention is on the equation that only involves one electron.

AO

lllj uaf

Define

The al, l = 1, 2, 3…#AOs are unknowns.

These values are known numerical quantities.

AO

l

AO

lllkjllk dvuauedvuahu

AO

llkljl

AO

lkl dvuuaedvuhua

There are #AO such equations. Secular equations.

0)( ,, AO

llklkl eSha For k = 1 to AO

These are the secular equations. The number of such equations is equal to the number of atomic orbitals, AO.

For there to be a nontrivial solution (all al not equal to zero) to the set of secular equations then the determinant below must equal zero

The number of equations is equal to the number of AOs equations. The number of unknowns, the al, is also equal to the number of AOs.

0

)()(

)()(

,,1,1,

,1,11,11,1

AOAOAOAOAOAO

AOAO

eSheSh

eSheSh

0

)()(

)()(

,,1,1,

,1,11,11,1

AOAOAOAOAOAO

AOAO

eSheSh

eSheSh

Drastic assumptions can now be made. We will use the simple Huckel approximations.

hi,i = if orbital i is on a carbon atom.

Si,i = 1, normalized atomic orbitals,

hi,j = , if atom i bonded to atom j, zero otherwise

Sij = 0 if I is not equal to j

•Expand the secular determinant into a polynomial of degree AO in e, the energy.

•Obtain the allowed values of e by finding the roots of the polynomial.

•Choose one particular value of e, substitute into the secular equations.

•Obtain the coefficients of the atomic orbitals within the molecular orbital.

ExampleThe allyl pi system. 1

2 3

The secular equations:

(-e)a1 + a2 + 0 a3 = 0

a1 + (-e) a2 + a3 = 0

0 a1 + a2 + (-e) a3 = 0

Simplify by dividing every element by and setting (-e)/ = x

0)0(1)1(

10

11

012

xxx

x

x

x

2,0 x

For x = -sqrt(2)

e = sqrt(2)

0210

0121

0012

321

321

321

aaa

aaa

aaa

321 2 uuuf normalized

)2()121(

1321

5.0222uuuf

12 3

For x = 0; e =

Substitute into the secular equations

0010

0101

0010

321

321

321

aaa

aaa

aaa

31

2 0

aa

a

2/1

2/1

2/1

)2(

2/1

2/1

2/1

0

0

Verify that

h f = e f

Perturbation Theory

H0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e0, and the wavefunctions, f0.

H0f0 = e0f0

We now change the system slightly (change a C into a N, create a bond between two atoms). The Hamiltonian will be changed slightly.

For the changed system H = H0 + H1

H1 is the change to the Hamiltonian.

We want to find out what happens to the molecular orbital energies and to the MOs.

Changes (approximate)

Energy

Zero order (no correction): ei0

First Order correction: 1,

010iiii HdvfHf

Wave functions corrections to f0i

Zero order (no correction): f0i

First order correction: 000

1,

jij ji

ij fee

H

ExamplePi system only:

Perturbed system: allyl system

Unperturbed system: ethylene + methyl radical

12 3

12 3

2/)( 210

3 uuf 03e

30

2 uf 02e

2/)( 210

1 uuf 01e

00

0

00H

H

0

0

H

00

00

00001

HHH

0

0

21

21

00

00

000

02

1

2

111

e

03

02

03

2

03

03

01

11,3

02

02

01

11,2

11

02

1

))()/((

0

21

21

00

00

000

02

12

1

))/((

0

21

21

00

00

000

100

)/()/(

ff

f

f

feehfeehf

+

0

0

21

21

00

00

000

02

1

2

111

e

03

02

03

2

03

03

01

11,3

02

02

01

11,2

11

02

1

))()/((

0

21

21

00

00

000

02

12

1

))/((

0

21

21

00

00

000

100

)/()/(

ff

f

f

feehfeehf

+

Mixes in bonding

Mixes in bonding

Mixes in anti-bonding

Mixes in anti-bonding

Projection OperatorAlgorithm of creating an object forming a basis for an irreducible rep from an arbitrary function.

^^

RRh

lP

jj

jj

Where the projection operator results from using the symmetry operations multiplied by characters of the irreducible reps. j indicates the desired symmetry.

lj is the dimension of the irreducible rep.

1sA 1sB

z

y

Starting with the 1sA create a function of A1 sym

¼(E1sA + C21sA + v1sA + v’1sA) = ½(1sA + 1sB)