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.

How are energies and wave functions affected by a change?

Energy to ei0:

Zero order (no change, 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

Example: Creating allyl system out of ethylene plus methyl radical

Pi system only:

“Unperturbed” system: ethylene + methyl radical1

2 31

2 3

2/)( 210

3 uuf 03e

30

2 uf 02e

2/)( 210

1 uuf 01e

0000

0H

0

0H

0000

00001

HHH

“Perturbed” system: allyl system

3

000

3

13,0

33j

jj

j fee

Hff

2

000

2

12,0

22j

jj

j fee

Hff

1

000

1

11,0

11j

jj

j fee

Hff

13,3

033 Hee

12,2

022 Hee

11,1

011 Hee

03

02

03

02

01

03

03

01

11,3

02

02

01

11,2

011

02

1

))()/((0

2121

0000

0000

21

21

))/((0

2121

0000

000100

)/()/(

ff

f

ff

feeHfeeHff

+

00

2121

0000

0000

21

21 0

1011

eee

Working out of the predicted values for the “perturbed” system. MO 1 only.

Energy e1

Orbital 1

Note the stabilizing interaction. Bonding!

Mixes in bonding

Mixes in bonding

Mixes in anti-bonding

Mixes in anti-bonding

Projection Operator

Algorithm to create an object forming a basis of an irreducible rep from an arbitrary function.

^^RR

hl

Pj

jj

j

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

lj is the dimension of the irreducible rep (number in the E column).

1sA 1sB

z

y

Starting with the 1sA create a function of A1 sym

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

Consider the bonding in NF3, C3v. We can see what Irreducible Reps can be obtained from the atomic orbitals. Then get the Symm Adapted Linear Combinations by using Projection Op.

A 3 0 -1

A B C D

B 3 0 1

C 3 0 1

D 3 0 1

A = A2 + E

B = C = D = A1 + E

1

23

}

N

F

F F

N

F

F F

N

F

F F

N

F

F FThese p orbitals are set-up as sigma and pi with respect to the internuclear axis.

Now we know which irreducible reps we can get out of each kind of atomic orbital. Have to get the correct linear combination of the AOs. The E irred. rep is the bad one.

Now construct SALCs first A2 then E.

^^RR

hl

Pj

jj

j

A = A2 + E

PA2(p1) = 1/6 (p1 + p2 + p3 + (-1)(-p)1 + (-1)(-1p3) + (-1)(-p2)

N

F

FF C3N

F

FF

N

F

FF C3'

N

F

FF

p1

p1

N

F

FF vN

F

FF

p1-p1

N

F

FFN

F

FFv'

p1

-p3

N

F

FFN

F

FFv''p1 -p2

No AO on N is A2

N

F

FF

Now construct an SALC of A2 sym

See what the different symmetry operations do to the one of the types of p orbitals.

Now get the SALC of E symmetry. Note that E is doubly degenerate.

PE(p1) = (2p1 - p2 - p3) = E1

Apply projection operator to p1

But since it is two dimensional, E, there should be another SALC. Apply PE to p2.

N

F

FF

PE(p2) = (2p2 - p3 - p1) = E’

But E1 and E’ should be orthogonal. We want sum of products of coefficients to be zero. Create a linear combination of

E2 = E’ + k E1.= (-1 +k*2) p1 + (2 + k(-1)) p2 + (-1 + k(-1))p3

Have to choose k such that they are orthogonal.

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

k = ½

E2 = (3/2 p2 - 3/2 p3) = p2 – p3

N

F

FF

N

F

FF

N

F

FF

N

F

FF

N

F

FF

For now we show the interaction of the N p orbitals (px and py) of E symmetry with the SALC of the F p orbitals of E symmetry.

We will return to C3v molecules later on….

Molecular orbitals of

heteronuclear diatomic molecules

The general principle of molecular orbital theoryInteractions of orbitals (or groups of orbitals) occur when

the interacting orbitals overlap. the energy of the orbitals must be similarthe interatomic distance must be short enough but not too short

A bonding interaction takes place when:regions of the same sign overlap

An antibonding interaction takes place when:regions of opposite sign overlap

Combinations of two s orbitals in a homonuclear molecule (e.g. H2)

Antibonding

Bonding

In this case, the energies of the A.O.’s are identical

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

n A.O.’s n M.O.’s

The same principle is applied to heteronuclear diatomic molecules

But the atomic energy levels are lower for the heavier atom

Orbital potential energies (see also Table 5-1 in p. 134 of textbook)

Average energies for all electrons in the same level, e.g., 3p(use to estimate which orbitals may interact)

The molecular orbitals of carbon monoxide

CO

cc(C)co(O)]

2s 2pC -19.43 -10.66O -32.38 -15.85

E(eV)

Each MO receives unequal contributions from C and O (cc ≠ co)

Group theory is used in building molecular orbitals

CO C°v (use C2v)

z

s pz

px py

A1

A1

A1

A1

B1 B2

B1 B2

“C-like MO”

“O-like MO”

Frontier orbitals

“C-like MO’s”

“O-like MO’s”

mixing

Larger homo lobe on C

Bond order 3

A related example: HF

H F C°v (use C2v)

s (A1) 2p(A1, B1, B2)

H -13.61 (1s)

F -40.17 (2s) -18.65

No s-s int.(E > 13 eV)

Non-bonding(no E match)

Non-bonding(no symmetry match)

Extreme cases: ionic compounds (LiF)

Li transfers e- to F

Forming Li+ and F-

A1

A1

Molecular orbitals for larger molecules

1. Determine point group of molecule (if linear, use D2h and C2v instead of D∞h or C∞v)

2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom)

3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for px, py, pz. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1)

4. Find the reducible representations (they correspond to the symmetry of group orbitals, also called Symmetry Adapted Linear Combinations SALC’s of the orbitals).

5. Find AO’s in central atom with the same symmetry

6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E

F-H-F-

D∞h, use D2h

1st consider combinations of2s and 2p orbitals from F atoms

8 GROUP ORBITALSDEFINED

Group orbitals can now be treated as atomic orbitalsand combined with appropriate AO’s from H

1s(H) is Ag so it matches two group orbitals 1 and 3

Both interactions are symmetry allowed, how about energies?

-13.6 eV

-40.2 eV

-13.6 eV

Good E matchStrong interaction

Poor E matchweak interaction

Bonding e

Non-bonding e

Lewis structureF-H-F-

implies 4 e around H !

MO analysisdefines 3c-2e bond

(2e delocalized over 3 atoms)

CO2

D∞h, use D2h

(O O) group orbitals the same as for F F

But C has more AO’s to be considered than H !

CO2

D∞h, use D2h

No match

Carbon orbitals

Ag-Ag interactions B1u-B1u interactions

All four are symmetry allowed

Primary Ag interaction

Primary B1u interaction

Bonding

Bonding

Non-bonding

Non-bonding

4 bondsAll occupied MO’s are 3c-2e

LUMO

HOMO

The frontier orbitals of CO2

Molecular orbitals for larger molecules: H2O

1. Determine point group of molecule: C2v

2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom - not necessary for H since s orbitals are non-directional)

3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for px, py, pz. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1)

4. Find the irreducible representations (they correspond to the symmetry of group orbitals,also called Symmetry Adapted Linear Combinations SALC’s of the orbitals).

5. Find AO’s in central atom with the same symmetry

6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E

For H H group orbitals

v’ two orbitals interchanged

E two orbitals unchanged

C2 two orbitals interchanged

2 20 0

v two orbitals unchanged

No match

pz

bonding

slightlybonding

antibonding

px

bonding

antibonding

py

non-bonding

3 10

Find reducible representation for 3H’s

Irreducible representations:

Molecular orbitals for NH3

pz

bonding

Slightlybonding

anti-bonding

bonding

anti-bonding

LUMO

HOMO

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

^^RR

hl

Pj

jj

j

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

lj is the dimension of the irreducible rep. h the order order of the group.

1sA 1sB

z

y

Starting with the 1sA create a function of A1 sym

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

Consider the bonding in NF3

N

F

F F

N

F

F F

N

F

F F

N

F

F F

A 3 0 -1

A B C D

B 3 0 1

C 3 0 1

D 3 0 1

A = A2 + E

B = C = D = A1 + E

1

23

}

Now construct SALC

^^RR

hl

Pj

jj

j

A = A2 + E

PA2(p1) = 1/6 (p1 + p2 + p3 + (-1)(-p)1 + (-1)(-1p3) + (-1)(-p2)

N

F

FF C3N

F

FF

N

F

FF C3'

N

F

FF

N

F

FF vN

F

FF

p1-p1

N

F

FFN

F

FFv'

p1

-p3

N

F

FFN

F

FFv''p1 -p2

No AO on N is A2

N

F

FF

Now construct an SALC of A2 sym

See what the different symmetry operations do to the one of the types of p orbitals.

E:

PA2(p1) = (2p1 - p2 - p3) = E1

Apply projection operator to p1

But since it is two dimensional, E, there should be another SALC

N

F

FF

PA2(p2) = (2p2 - p3 - p1) = E’

But E1 and E’ should be orthogonal. We want sum of products of coefficients to be zero.

Create a linear combination of

E2 = E’ + k E1.= (-1 +k*2) p1 + (2 + k(-1)) p2 + (-1 + k(-1))p3

Have to choose k such that they are orthogonal.

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

k = ½

E2 = (3/2 p2 - 3/2 p3) = p2 – p3

N

F

FF

N

F

FF

N

F

FF

N

F

FF

N

F

FF

The geometriesof electron domains

Molecular shapes:When we discussed

VSEPR theory

Can this be describedin terms of MO’s?

Hybrid orbitals

s + p = 2 sp hybrids (linear)

s + 2p = 3 sp2 hybridstrigonal planar

s + 3p = 4 sp3 hybridstetrahedral

s + 3p + d = 5 dsp3 hybridstrigonal bipyramidal

s + 3p + 2d = 6 d2sp3 hybridsoctahedral