CHEMICAL BONDING II
MOLECULAR ORBITAL THEORY II
+
C2 MOLECULE
NO MOLECULE
BOND ORDER
Bond order = number of bonding e- minus number of antibonding e2
If the bond order is zero→ no bond!
divide by two because of pairs of electrons.
 Bond order is an indication of strength.
 Larger bond order means greater bond strength.
 H2 molecule: Bond order = (2-0)/2 = 1 and He2: Bond order = (2-2)/2 =
0 This implies it is not stable.
BOND ORDER
Calculate the Bond order for F2, C2+, and NO
BOND ORDER
F:
C:
NO:
(10-8)/2 = 1
(7-4)/2 = 1.5
(10-5)/2 = 2.5
HOMONUCLEAR DIATOMIC MOLECULES
those composed of two identical atoms.
Li2 Bond order = (2-0)/2 = 1.
Li2 is a stable molecule.
Be2 Bond order = (2-2)/2 = 0. This is not more stable than two Be atoms, so no molecule forms.
FILLING THE DIAGRAM
 It gets slightly more complicated when we leave Be and
move to 2p.
 The filling order for p’s is pi, pi, sigma all bonding,
followed by pi, pi, sigma all antibonding.
 Hund's Rule and the Pauli Exclusion Principle still apply.
General Energy Level Sequence for Filling
Orbitals Using the MO Theory
σ1s2 σ1s2* σ2s2 σ2s2*
Π2pxy4 σ2p2
π2pxy4* σ2p2*
MOLECULAR ORBITAL CONFIGURATIONS
For period 2 diatomic molecules up to and including N2:
σ1s2 σ1s2* σ2s2 σ2s2* π2pxy4 σ2p2 π2pxy4* σ2p2*
For period 2 diatomic molecules O2, F2, and Ne2 (hypothetical): π2p and σ2p
change order.
σ1s2 σ1s2* σ2s2 σ2s2* σ2p2 π2pxy4 π2pxy4* σ2p2*
MOLECULAR ORBITAL CONFIGURATIONS
Write the molecular orbital configurations for
F2, C2+, and NO.
MOLECULAR ORBITAL CONFIGURATIONS
F2: σ1s2 σ1s2* σ2s2 σ2s2* σ2p2 π2pxy4 π2pxy4*
C2+ : σ1s2 σ1s2* σ2s2 σ2s2* σ2p2 π2pxy3
NO: σ1s2 σ1s2* σ2s2 σ2s2* σ2p2 π2pxy4 π2pxy1*
PRACTICE SIX
Predict the bonding of B2.
PRACTICE SIX - ANSWER
B – 1s22s22p1
B
B
PARAMAGNETISM
 One of the most useful parts of this
model is its ability to accurately predict
paramagnetism and diamagnetism as
well as bond order.
 B2 is paramagnetic. That means that the
pi orbitals are of LOWER energy than
the sigmas and Hund’s rule demands
that the 2 electrons fill the 2 bonding pi
orbitals singly first before paring.
PRACTICE SEVEN
Write the appropriate energy diagram using the MO theory
for the nitrogen molecule. Find the bond order for the
molecule and indicate whether this substance is
paramagnetic or diamagnetic.
PRACTICE SEVEN
N2 electron configuration:
1s22s22p3
Bond order = (10-4)/2 = 6
diamagnetic
N
N
PRACTICE EIGHT
For the species O2, O2+, O2-, give the electron configuration and the
bond order for each. Which has the strongest bond?
PRACTICE EIGHT
O2
σ1s2 σ1s2* σ2s2 σ2s2* σ2p2
π2pxy4 π2pxy2*
Bond order = (8-4)/2 = 2
PRACTICE EIGHT
PRACTICE EIGHT
O2+
O2-
σ1s2 σ1s2* σ2s2 σ2s2* σ2p2
σ1s2 σ1s2* σ2s2 σ2s2* σ2p2
π2pxy4 π2pxy1*
π2pxy4 π2pxy3*
Bond order = (8-3)/2 = 2.5
Bond order = (8-5)/2 = 1.5
PRACTICE NINE
Use the molecular orbital model to predict the bond order and
magnetism of each of the following molecules.
Ne2
P2
PRACTICE NINE
Ne2
Bond order = (8 - 8)/2 = 0
Does not exist
PRACTICE NINE
P2
Bond order = (18 - 12)/2 = 3
diamagnetic
PRACTICE TEN
Use the MO Model to predict the magnetism and bond
order of the NO+ and CN- ions.
PRACTICE TEN
Each has the same number of valence electrons = 10 and
total electrons = 14
σ1s2 σ1s2* σ2s2 σ2s2* π2pxy4 σ2p2
Bond order = (10-4)/2 = 3
Both are diamagnetic.
The Molecular Orbtial
Energy-Level
Diagrams, Bond
Orders, Bond Energies,
and Bond Lengths for
the Diatomic
Molecules B2 Through
F2
COMBINING LE AND MO MODELS
LE model assumes that electrons are localized. This is not the
case for some molecules. This is apparent with molecules for
which we can draw several valid Lewis structures. So resonance
was constructed.
The best model is one with the simplicity of the LE model with
the delocalization characteristics of the MO model. So we
combine these two models to describe molecules that require
resonance. In O3 and NO3-1, the double bond changes position
in the resonance structure.
COMBINING LE AND MO MODELS
Since a double bond requires one σ and one
π bond, there is a σ bond between all bound
atoms in each resonance structure.
It is the π bond that has different locations.
Conclusion: σ bonds in a molecule can be
described as being localized with no
apparent problems, but the π bonding must
be treated as delocalized. So use LE to
describe the σ bonds in structures with
resonance, but use MO to describe π bonds
in these same structures.