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 equationMolecular Orbitals are linear combinations of atomic orbitals (LCAO)

caacbb (for diatomic molecules)

Interactions depend on the symmetry propertiesand 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 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 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 u

2 (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 g

1 (double bond)paramagneticO2

2- u2 u

2 g2 g

2 (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)

Simple Molecular Orbital TheoryA 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

MO Theory - 2Seek F such that

...)3,2,1(...)3,2,1(...)3,2,1( EFFH

i

i ffEffffhFH )...3()2()1()...)3()2()1((...)3,2,1(...)3,2,1( 211211

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.

MO Theory - 3

jjj fefh

Multiply by uk and integrate.

dvhuuh jkjk ,

dvfuedvhfu jkjk

AO

llluaf

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

Define

dvuuS jkjk ,

Substituting the expansion for f

0)( ,, AO

llklkl eSha

These integrals are fixed numerical values.

MO Theory - 40)( ,,

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 (all al equal to zero) solution to the set of secular equations then the determinant below must equal zero

There are AO equations with AO unknowns, the al.

0

)()(

)()(

,,1,1,

,1,11,11,1

AOAOAOAOAOAO

AOAO

eSheSh

eSheSh

MO Theory 6

0

)()(

)()(

,,1,1,

,1,11,11,1

AOAOAOAOAOAO

AOAO

eSheSh

eSheSh

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

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

Si,i = 1, normalized atomic orbitals

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

Expand the secular determinant into a polynomial of degree AO in e. Obtain the allowed values of e by finding the roots of the polynomial. Choose one particular value of e, substitute into the secular equations and 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

2/1

2/1

2/1

)2(

2/1

2/1

2/1

0

0

Verify that

h f = e f

Perturbation Theory

The Hamiltonian is divided into two parts: H0 and H1

H0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e0, and the wavefunctions, f0. H0f0 = e0f0

H1 is a change to the system and the Hamiltonian which renders approximation desirable. The change to the energies and the wavefunctions are expressed as a summation.

Corrections

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)