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1 Molecular Orbital Theory I. Introduction. A. Ideas. 1. Start with nuclei at their equilibrium positions. 2. Construct a set of orbitals that cover the complete nuclear framework, called molecular orbitals (MO's) .3. Use the rules of quantum mechanics to arrange the molecular orbitals in order of increasing energy and add the electrons. 4. Fill the MO's with the molecule's electrons. a. Lowest energy MO's filled first. b. No more than two electrons in the same MO with their spins paired. c. Half fill a degenerate set with spins parallel before pairing up in the same MO. 5. Start with the atomic orbitals (AO’s). Can construct the MO's by taking Linear Combinations of the Atomic Orbitals using at least one AO from each atom (LCAO method). 6. Most of the time the only electrons that need be considered are the valence electrons and the AO's used in the linear combinations are the valence orbitals 7. Arrange the molecular orbitals in order of increasing energy and add the electrons. a. Lowest energy MO's filled first. b. No more than two electrons in the same MO with their spins paired. c. Half fill a degenerate set with spins parallel before pairing up in the same MO. B. Lowest energy states of homonuclear diatomic molecules. 1. Get the two lowest energy MO's by taking linear combinations of the lowest energy atomic orbitals, the 1s atomic orbitals. There are two ways to combine, can add them together or subtract them. The resulting MO's are sketched below.

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1

Molecular Orbital Theory

I. Introduction.

A. Ideas.

1. Start with nuclei at their equilibrium positions.

2. Construct a set of orbitals that cover the complete nuclear framework, called molecular

orbitals (MO's)

.3. Use the rules of quantum mechanics to arrange the molecular orbitals in order of increasing

energy and add the electrons.

4. Fill the MO's with the molecule's electrons.

a. Lowest energy MO's filled first.

b. No more than two electrons in the same MO with their spins paired.

c. Half fill a degenerate set with spins parallel before pairing up in the same MO.

5. Start with the atomic orbitals (AO’s). Can construct the MO's by taking

Linear Combinations of the Atomic Orbitals using at least one AO from

each atom (LCAO method).

6. Most of the time the only electrons that need be considered are the valence

electrons and the AO's used in the linear combinations are the valence orbitals

7. Arrange the molecular orbitals in order of increasing energy and add the electrons.

a. Lowest energy MO's filled first.

b. No more than two electrons in the same MO with their spins paired.

c. Half fill a degenerate set with spins parallel before pairing up in the same MO.

B. Lowest energy states of homonuclear diatomic molecules.

1. Get the two lowest energy MO's by taking linear combinations of the lowest energy atomic

orbitals, the 1s atomic orbitals. There are two ways to combine, can add them together or

subtract them. The resulting MO's are sketched below.

2

!1sA

"#1s

"1s

!1sB

!1s

A

! 1sA

!1s

B

!1sB

+

–Atom A Atom B

a. Ψ1s is a spherical function that is positive in all regions in space. When two 1s

functions are added the resulting wave function, labeled σ1s, is also positive in all

regions. (A sketch of σ1s is shown above.) On the other hand, when the two 1s atomic

orbital functions are subtracted, the resulting function, σ*1s , is positive around nucleus A,

negative around nucleus B, and is equal to zero halfway between the two nuclei

(see above). The wave function is said to have a nodal point between the two

nuclei.

b. The σ designation indicates that the wave function is symmetric with respect to

the inter-nuclear axis. That is, the wave function does not change sign as one

goes above or below the internuclear line. The subscript, 1s, tells the atomic

orbitals that are involved in the molecular orbital.

3. The relative energies of the orbitals.

a. The molecular orbital energy diagram.

3

!"1s

!1s

#1sA

#1sB

Atom A Atom BAB Molecule

$

$

Ener

gy

b. The σ1s orbital is lower in energy than the atomic orbitals by an amount β. Occupation

of this orbital will tend to stabilize the molecule and promote bonding between the two

atoms. The molecular orbital is said to be a bonding molecular orbital.

c. The σ*1s orbital is higher in energy (less stable) than the atomic orbitals by an

amount β. Occupation of this orbital will tend to destabilize the molecule and

detract from bonding. It is called an antibonding molecular orbital.

Antibonding molecular orbitals are identified by a right-hand superscript

asterisk (*).

d. In more complex molecules it is also possible to have nonbonding molecular

orbitals. These orbitals have the same energy at their input atomic orbitals.

Occupation of the orbitals should neither promote nor detract from bonding.

e. In general, the more electrons one has in bonding compared to antibonding

molecular orbitals, the more stable will be the molecule.

3. Filling of the two lowest energy orbitals.

a. Consider H+2 molecule-ion. This is a one electron system.

1) The electron will occupy the σ1s MO and H+2 will have a σ

!

1s

1

configuration and be

stabilized by an amount β compared to the separated atoms. Therefore, H+2 should be

stable.

4

2) This is a known substance whose bond dissociation energy is 255 kJ and

bond distance is 106 pm.

b. Consider the H2 molecule. This is a two electron system.

1) The two electrons will occupy the σ1s MO and will have their spins paired.

Since both electrons are stabilized by β, to give an over-all stability of 2β, the

molecule is stable.

2) H2 is the form of elemental hydrogen. It is diamagnetic, has a bond energy of

431 kJ, and a bond distance of 74 pm.

c. Consider the He+2 molecule-ion. This is a three electron system.

1) Two of the three electrons will occupy the σ1s MO while the third will be in

the σ*1s MO. Since two electrons are stabilized and only one is destabilized,

the ion should be stable.

2) He+2 is a known substance whose bond energy is 241 kJ.

d. Consider He2 molecule. This is a four electron system.

1) Two electrons will occupy the σ1s MO and two will be in the σ*1s MO. Since

two electrons are stabilized by β and two are destabilized by β, there should

be no net stability and the molecule should not be stable.

2) The He2 molecule has never been observed.

4. Molecular orbital electron configurations and bond order.

a. Molecular orbital electron configurations can be written in the same manner as atomic

electron configurations.

1) The molecular orbitals are written in order of increasing energy and the

population of each MO is given as a right-hand superscript.

2) Examples.

H+2 σ

1 1s H2 σ

2 1s He

+2 σ

2 1s σ

* 1s 1 He2 σ

2 1s σ

*1s 2

b. The stability will depend on the relative occupation of bonding MO's versus

antibondingMO's. This can be measured by the bond order (BO).

BO = 12 ( number of electrons in bonding MO's - number of electrons in antibonding MO's)

5

Examples:

BO in H+2 =

12 ( 1 - 0) =

12 BO in He

+2 =

12 ( 2 - 1) =

12

BO in H2 = 12 ( 2 - 0 ) = 1 BO in He2 =

12 ( 2 - 2 ) = 0

c. Note that the higher the BO,

1) the higher the bond energy.

2) the shorter the bond distance for the same atoms.

d. A substance with a BO = 0 should not be stable.

C. Higher energy MO's.

1. The next set of MO's can be obtained by taking linear combinations of the next higher

energy atomic orbitals. These are the 2s and 2p atomic orbitals. Recall that the 2s energy

is lower than that of the 2p atomic orbital energies.

2. The 2s AO's can be combined to give a bonding σ2s MO ( Ψ2sA+ Ψ2sB) and an

antibonding σ* 2s MO ( Ψ2sA- Ψ2sB). These will generally have the same shapes as do the

corresponding MO's obtained from the 1s AO's.

3. The 2p orbitals.

a. We must now worry about the orientations of the AO's. Assume that the bonding

axis is the x axis. The 2px will have σ symmetry and the 2py and 2pz will have π

symmetry with respect to this axis.

b. The 2px orbitals combine to form σ MO's.

!2pxA !

2pxB

add

subtract

!2pxA

!2pxA

!2pxB

!2pxB

+

"2p

"#2p

x

x

6

c. The 2py orbitals will combine to form π MO's as will the 2pz orbitals. The 2pz

orbitals could be combined similarly.

!2pyA

!2pyB

+

!2pyA

!2pyB

! 2pyA

!2pyB

!2pzB

!2pzA

!2pzB

!2pzA

!2pzA!2pzB

+

"#

2py

2py

"

"#2pz

2pz"

or

or

The π

!

2pyand the π

!

2pzwill have the same energy and the π

!

2py

* and the π

!

2pz

* will

also have the same energies.

4. Energies.

a. The energies change with atomic number as shown below.

7

a. At N2 and below, the relative energies of these MO's are

σ2s < σ*2s < π2p

y = π

2pz < σ

2px < π

*2py = π

* 2pz < σ

* 2px

b. Above N2 the relative energies are

σ2s < σ*2s < σ

2px < π2p

y = π

2pz < π

*2py = π

* 2pz < σ

* 2px

c. Note that the π 2py and π

2pz orbitals have the same energy, that is, are

degenerate, as are the π* 2py and π

* 2pz orbitals.

5. Orbital filling.

a. A total of 16 more electrons can be accommodated in these orbitals for a total

of 20 electrons.

b. Examples:

Substance N Electron Configuration B.E rx-x Magnetism BO

N2 14 σ2 1s σ

* 21s σ

2 2s σ

* 22s π

2 2py π

2 2pz σ

2 2px 941 110 diamagnetic 3

O2 16 σ2 1s σ

* 21s σ

2 2s σ

* 22s σ

2 2px π

2 2py π

2 2pz π

* 12py π

* 12pz 494 121 paramagnetic 2

F2 18 σ2 1s σ

* 21s σ

2 2s σ

2 2px σ

2 2px π

2 2py π

2 2pz π

* 22py π

* 22pz 151 142 diamagnetic 1

II. Other molecular systems.

A. Heteronuclear diatomic molecules.

1. The energies of the input atomic orbitals will not be the same. This will give rise to

unequal splitting as shown below.

Ψa* = C3ΦA– C4ΦB ΦB ENERGY ΦA Ψb = C1ΦA + C2ΦB Atom A Molecule AB Atom B

8

a. The bonding MO will have more of the character of ψA than ψB, that is, the

weighting factor C1 will be greater than C2. Conversely, the antibonding MO will

have more of ψB character than ψA character, that is, C4 > C3.

b. The greater the energy difference between the two atomic orbitals, the greater will be

the disparity in the weighting factors. If the energy difference is too much, then Ψb will

become identical with ψA and Ψa will become identical with ψB. This means that two

AO's do not interact. Therefore, only AO's of comparable energies can be combined

to form MO's.

2. This is the case in a molecule such as HF.

XX

XX

XXXX

XX

X

XXXXX

FHFH

ENERGY

2S

2p

1s

The 1s orbital of H is too high in energy interact with either the 1s or the 2s F orbitals. It

can interact with the F 2p orbital having σ symmetry, to form MO's as in the above

diagram. In this case the H 1s orbital would be ψB (the higher energy AO) and the F 2pσ

would be ψA.

a. The HF molecule has a total of 8 valence electrons, but only two are in delocalized MO's.

9

The 1s and 2s F AO's do not interact because of energy and only one of the

three F 2p AO has the correct symmetry.

b. There are only two electrons to be distributed in the MO's (one from H and one from F).

The electrons will occupy Ψb and the molecule should be stable. Note that the electron

density will be polarized towards the F (C1 > C2). The bond will be a polar bond.

3. In general, valence orbitals are of comparable energies and are the ones that can be

combined to form MO's. Therefore, for most molecules the inner core electrons can be

ignored and valence orbitals used to form the MO's that are filled by the valence electrons

of the constituent atoms.

2. Consider the SO3 molecule.

It is a trigonal planar molecule containing 24 valence electrons.

a. Assume that the atoms are sp2 hybridized and that the molecular plane is the xy plane.

Therefore, the pz orbital on each atom will have π symmetry. Of the 24

electrons, 6 will be involved in π bonding.

b. The qualitative π energy level diagrams and rough sketches of the MO's, in terms of

the input AO's, are shown on the following page.

c Note that two electrons occupy the bonding MO, Ψ1, while four are in the

nonbonding MO's, Ψ2 and Ψ3. Therefore, π interactions should stabilize the

molecule.

d. Also note that, considering the sum total of electron density from all the

occupied MO's, electron density is evenly distributed between the sulfur and

the three oxygens. Thus, the oxygens should be equivalent. In the valence

bond theory such equivalency could be obtained only by using resonance. In

molecular orbital theory the electrons are naturally delocalized and resonance

is not necessary. Molecular orbital theory is preferred in treating molecules

with delocalized π electrons

10

S

O

O

!1 !2

!3

!4

O

"MO = C1 !1 + C2 !2 + C3 !3 + C4 !4

ENERGY

SSO3 O's

!4

!1

!2

!3

!4

!3

!2

!1

"4

"1 "2 "3

QUALITATIVE ENERGY DIAGRAM AND ORBITAL SKETCHES

OF THE SO3 Π MOLECULAR ORBITALS.

11

e. Overall electron density in the molecule.

3. Consider benzene, C6H6.

It is a planar molecule consisting of six sp2 hybridized carbons that form a

prefect hexagonal ring. A hydrogen is bonded to each carbon, as shown below.

C

C

C

C

C

C

H

H

H

H

H

H

C

C

C

C

C

C

H

H

H

H

H

H

a. The molecule exhibits resonance. The two major resonance structures are

those shown above. The net effect of the resonance structures is that the π

electrons are evenly distributed about the ring.

b. To show this symmetric distribution of electron density, benzene is written as

shown below. The circle signifies that the π electrons are delocalized about

the planar ring. Benzene is a very stable molecule and forms the basis of

aromatic organic compounds. The bonding is best treated using MO

theory.

12

c. In the molecular orbital approach to describing the π electronic structure,

linear combinations of the following six pπ orbitals on benzene are taken to

construct six MO's.

!4

!6

!2 !

3

!5

!1

"MO

= C1!

1+ C

2!

2+ C

3!

3+ C

4!

4+ C

5!

5+ C

6!

6 d. There are six π electrons which can then be placed in the MO's. The

following diagram shows the relative energies, atomic orbital contributions,

and filling of these MO's.

Note that the net effect is that the π electron density is evenly distributed

around the six member ring.

A

E1

E2

B

!A

!E1(a)

!A

!A

!B

!E1(a) !E1(b)

!E2(a) !E2(b)

!E1(b)

!E2(a) !E2(b)

"2#

"#

#

2#

0

EN

ER

GY

Qualitative Energy Level Diagram and Orbital Sketches for the ! Orbitals of Benzene

13

e. π Molecular Orbitals.

f. Overall π electron density

14

15

16