Molecular Orbitals are simply
Linear Combinations of Atomic Orbitals
(“L.C.A.O.-M.O.s”)
Example: H2
σ anti-bonding (σ*)
+
+
+
σ bonding
Next Question: Why does this work?
Bonding is driven by stabilization of electrons
• Electrons are negatively charged
• Nuclei are positively charged
= + = nucleus
The bonding combination puts electron
density between the two nuclei - stabilization
The antibonding combination moves electron
density away from the nuclei - destabilization
σ* M.O. is raised in energy
σ M.O. is lowered in energy
H atom: (1s)1 electron configuration
H2 molecule: (σ1s)2 electron configuration
Same as previous description of bonding (Ch 13):
σ∗
σ
Figure 13.1
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
electrons
#antibonding
electrons
2
The bond order is an indication of bond strength
Greater bond order
Greater bond strength
Bond Order: Examples
Bond order = (2-0)/2 = 1
H2
Single bond
Stable molecule
(450kJ/mol bond)
Bond order = (2-2)/2 = 0
He2
No bond!
Unstable molecule
(0kJ/mol bond)
He2+
Bond order = (2-1)/2 = 1/2
Half of a single bond
Can be made, but its not
very stable
(250kJ/mol bond)
H2
+
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 Dominated
by the Valence Orbitals
Region of
shared edensity
Valence configurations of the 2nd row atoms:
Li
2s1
Be B
C
N
O
2s2 2s22p1 2s22p2 2s22p3 2s22p4
F
2s22p5
So far we have focused on bonding involving the s
orbitals.
What happens when we have to consider the p
orbitals?
Figure 14.36
For diatomics of atoms with valence electrons
in the p orbitals, we must consider three
possible bonding interactions:
= nucleus
π−type
π−type
σ−type
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 magnetsO2 should be magnetic!
Oxygen Molecular Orbitals
-845
π*
Energy
π*
-1,483
-1,512
σ(2p)
π
π
-2,233
σ∗(2s)
-3,464
σ(2s)
-50,658
1s(O) + 1s(O)
1s(O) –1s(O)
Magnetism
Faraday
Paramagnetic SampleUnpaired Electrons
Slightly attracted to magnet
Diamagnetic SampleAll electrons paired
Slightly repelled from magnet
A Complication…
M.O. Diagram for B2
(similar for C2 and N2)
M.O. Diagram for O2
(similar for F2 and Xe2)
A Complication…
M.O. Diagram for B2
(similar for C2 and N2)
M.O. Diagram for O2
(similar for F2 and Xe2)
No s-p mixing
s-p mixing
When does s-p mixing occur?
B, C, and N all have ≤ 1/2 filled 2p orbitals
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 to
Physical 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
623
Bond length (pm)
110
112
121
112
No. 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 the
bond 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 the
bond 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*
π2p*
σ2p
π2p
π2p
σ2p
σ2s*
σ2s*
σ2s
σ2s
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. Heteronuclear
Diatomic Molecules
(Zumdahl Section 14.4)
Review from Chapter 13:
Covalent
Ionic
Covalent
Ionic
Figure13.11
Figure 13.12
Homonuclear:
H2
Electronegativity
Nonpolar covalent bond
(450kJ/mol bond)
Figure 14.26
Heteronuclear:
HF
Electronegativity
Polar covalent bond
(565kJ/mol bond)
Figure 14.45
Electrons are not equally shared
in heteronuclear bonds
HF
Electronegativity
Figure 14.45
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
M.O.s of a Polar Covalent Bond: HF
σ Antibonding (σ*)
Mostly H(1s)
H
F
H
F
σ Bonding
Mostly F(2p)
When the electronegativities of the 2 atoms are
more similar, the bonding becomes less polar.
Example: NO
2p
2s
N
2s
NO
O
Electronegativity
2p
χ(N) = 3.0
χ(O) = 3.4
. ..
..N=O..
Sample Problem: Use the M.O. model to predict the changes in
magnetism and bond order upon oxidation of NO to make NO+
NO+
NO
oxidation
Bond Order = (6-1)/2 = 2.5
Paramagnetic
Bond Order = (6-0)/2 = 3
Diamagnetic
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
Atomic Orbitals
Molecular Orbitals
+ 3 others at
higher energies
Figure 14.51
Another example:
Benzene
σ bonding:
π bonding:
p atomic orbitals
π molecular orbital