+ -

+ +

Molecular Orbitals are simplyLinear Combinations of Atomic Orbitals

(“L.C.A.O.-M.O.s”)

σ bonding

σ anti-bonding (σ*)Example: H2

Next Question: Why does this work?

Bonding is driven by stabilization of electrons• Electrons are negatively charged

• Nuclei are positively charged

The bonding combination puts electron density between the two nuclei - stabilization

The antibonding combination moves electron density away from the nuclei - destabilization

= = nucleus+

σ* M.O. is raised in energy

σ M.O. is lowered in energy

H atom: (1s)1 electron configurationH2 molecule: (σ1s)2 electron configuration

Figure 13.1

Same as previous description of bonding (Ch 13):

σ

σ∗

Filling Molecular Orbitals with Electrons

1) Orbitals are filled in order of increasing Energy( Aufbau principle )

2) An orbital has a maximum capacity of two electrons with opposite spins ( Pauli exclusion principle )

3) Orbitals of equal energy are half filled, with spins parallel, before any is filled ( Hund’s rule )

Bond Order

Bond Order =

# bonding #antibondingelectrons electrons

2

The bond order is an indication of bond strength

Greater bond order Greater bond strength

Bond Order: Examples

H2

Bond order = (2-0)/2 = 1

Single bond

Stable molecule (450kJ/mol bond)

He2

Bond order = (2-2)/2 = 0

No bond!

Unstable molecule (0kJ/mol bond)

He2+

H2+

Bond order = (2-1)/2 = 1/2

Half of a single bond

Can be made, but its not very stable

(250kJ/mol bond)

Bond order = (1-0)/2 = 1/2

Half of a single bond

Can be made, but its not very stable

(255kJ/mol bond)

Covalent Bonding is Dominatedby the Valence Orbitals

Region of shared e-

density

Valence configurations of the 2nd row atoms:

Li Be B C N O F2s1 2s2 2s22p1 2s22p2 2s22p3 2s22p4 2s22p5

So far we have focused on bonding involving the sorbitals.

What happens when we have to consider the porbitals?

Figure 14.36

For diatomics of atoms with valence electrons in the p orbitals, we must consider three

possible bonding interactions:

σ−typeπ−type π−type

= nucleus

Example: the O2 Diatomic

Oxygen has the 2s22p4

valence configuration

Bond Order = (6-2)/2 = 2

O2 is stable(498kJ/mol bond strength)

(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2

A prediction from the M.O. diagram of O2

..

......O=O

The unpaired electrons predicted by the M.O. diagram should behave as small magnets-O2 should be magnetic!

σ(2p)

1s(O) –1s(O)1s(O) + 1s(O)

π*

σ(2s)

σ∗(2s)

π π

Oxygen Molecular OrbitalsE

nerg

y

π*

-50,658

-3,464

-2,233

-1,512-1,483

-845

Magnetism

Paramagnetic Sample-Unpaired ElectronsSlightly attracted to magnet

Diamagnetic Sample-All electrons pairedSlightly repelled from magnet

Faraday

A Complication…M.O. Diagram for O2

(similar for F2 and Xe2)M.O. Diagram for B2

(similar for C2 and N2)

A Complication…M.O. Diagram for O2

(similar for F2 and Xe2)M.O. Diagram for B2

(similar for C2 and N2)

s-p mixingNo s-p mixing

B, C, and N all have ≤ 1/2 filled 2p orbitals

When does s-p mixing occur?

O, F, and Xe all have > 1/2 filled 2p orbitals

• Having > 1/2 filled 2p orbitals raises the energies of these orbitals due to e- - e – repulsion: Electrons repel each other because they are negatively charged. • If 2 electrons are forced to be in the same orbital, their energies go up.(This is the basis of Hund’s Rule for orbital filling.)

s-p mixing only occurs when the s and p atomic orbitals are close in energy (≤ 1/2 filled 2p orbitals)

s-p mixing No s-p mixing

Relating the M.O. Diagrams toPhysical Properties

Sample Problem:Using MO Theory to Explain Bond Properties

Problem: Consider the following data for these homonuclear diatomic species: N2 N2

+ O2 O2+

Bond energy (kJ/mol) 945 841 498 623Bond length (pm) 110 112 121 112No. of valence electrons 10 9 12 11

Removing an electron from N2 decreases the bond energy of the resulting ion, whereas removing an electron from O2 increases thebond energy of the resulting ion. Explain these facts using M.O diagrams.

Plan: We first draw the MO energy levels for the four species, recalling that they differ for N2 and O2. Then we determine thebond orders and compare them with the data: bond order is related directly to bond energy and inversely to bond length.

Sample Problem - Continued

Solution: The MO energy levels are:N2 N2

+ O2 O2+

σ2p*σ2p

*

π2p*

σ2p

π2p

σ2s*

σ2sσ2s

σ2s*

π2p

π2p*

σ2p

BO: 1/2(8-2) = 3 1/2[7-2] = 2.5 1/2(8-4) = 2 1/2(8-3) = 2.5

Bonding in Heteronuclear vs. HeteronuclearDiatomic Molecules(Zumdahl Section 14.4)

Review from Chapter 13:

Figure 13.12Figure13.11

Covalent

Ionic

Ionic

Covalent

Homonuclear:H2

Heteronuclear:HF

Polar covalent bond(565kJ/mol bond)

Nonpolar covalent bond(450kJ/mol bond)

Electronegativity Electronegativity

Figure 14.45Figure 14.26

HF

Electronegativity

Because F (χ = 4.0) is more electronegative than H (χ = 2.2), the electrons move closer to F. (Table 13.2 for electronegativities)

This gives rise to a polar bond: H F

Figure 14.45

Electrons are not equally sharedin heteronuclear bonds

σ Antibonding (σ*)Mostly H(1s)

σ BondingMostly F(2p)

H F

H F

M.O.s of a Polar Covalent Bond: HF

When the electronegativities of the 2 atoms are more similar, the bonding becomes less polar.

Example: NO

Electronegativity

χ(N) = 3.0χ(O) = 3.4

.

......N=O

2s2s

2p2p

NON O

NO+

oxidation

NO

Sample Problem: Use the M.O. model to predict the changes in magnetism and bond order upon oxidation of NO to make NO+

Bond Order = (6-1)/2 = 2.5Paramagnetic

Bond Order = (6-0)/2 = 3Diamagnetic

Combining the Localized Electron and Molecular Orbital Models (into a convenient working model)

(Zumdahl Section 14.5)

Figure 14.47

Only the π bonding changes between these resonance structures- The M.O. model

describes this π bonding more effectively

Figure 14.51

Atomic Orbitals Molecular Orbitals

+ 3 others at higher energies

Another example: Benzene

π bonding:

σ bonding:

p atomic orbitals π molecular orbital