The document discusses molecular orbital theory and hybridization. It begins by explaining how atomic orbitals combine to form molecular orbitals through constructive and destructive interference. This leads to bonding and antibonding molecular orbitals. Hybridization is then introduced to explain how atomic orbitals mix to form new hybrid orbitals used in bonding. Examples of sp, sp2, sp3, and sp3d hybridization are provided. The molecular orbital model is then described as an improvement over the localized electron model as it accounts for electron correlation. Bond order is defined in relation to molecular orbitals and shown to correlate with bond strength. Examples of homonuclear and heteronuclear diatomic molecules are used to demonstrate how molecular orbitals are formed and

1. Covalent Bonding Orbitals

2. Hybridization and the
Localized Electron Model

3. Localized Electron Model
The arrangement of valence electrons is represented by
the Lewis structure or structures, and the molecular
geometry can be predicted from theVSEPR model.
 Atomic orbitals are used to share electrons and form
bonds

4. Hybridization
In general we assume that bonding involves only the valence
orbitals.
 The mixing of atomic orbitals to form special bonding orbitals
is called hybridization.
 Carbon is said to undergo sp3
hybridization or is sp3
hybridized
because is uses one s orbital and three p orbitals to form four
identical bonding orbitals.
 The four sp3
orbitals are identical in shape each one having a
large lobe and a small lobe. The four orbitals are oriented in
space so that the large lobes form a tetrahedral
arrangement.

5. Hybridization

6. Hybridization

7. sp3
example: Methane

8. sp3
example: Ammonia

9. sp2
Hybridization

10. sp2
Hybridization

11. sp2
Hybridization
Ethylene (C2H4) is commonly used in plastics and has a
C=C double bond. Each carbon uses sp2
hybridization in
this molecule because a double bond acts as one
effective pair.
 In forming the sp2
orbitals, one 2p orbital on carbon has
not been used. This remaining p orbital is oriented
perpendicular to the plane of the sp2
orbitals.
 The double bond utilizes one sigma bond that is
hybridized and one pi bond with the unhybridized p
orbital.

12. sp2
Hybridization

13. sp2
Hybridization: Ethylene

14. Multiple Bonds
 Single bonds are sigma bonds (σ) and the electron pair is
shared in an area centered on a line running between
the atoms.These are hybridized bonding orbitals.
 With multiple bonds, a sigma bond is formed and then
one or two pi bond (π) form. These electrons occupy the
space above and below the sigma bond and use
unhybridized orbitals.

15. sp2
Hybridization: Ethylene

16. sp Hybridization
 sp hybridization involves one s orbital and one p orbital.
Two effective pairs will always require sp hybridization.
 CO2 is sp hybridized

17. sp Hybridization

18. sp Hybridization: CO2

19. sp Hybridization: CO2

20. sp Hybridization: CO2

21. sp Hybridization: N2

22. sp3
d Hybridization
 When a molecule exceeds the octet rule, hybridization
occurs using d orbitals. Also called dsp3
.
 PCl5 has sp3
d hybridization and is trigonal bipyramidal.

23. sp3
d Hybridization: PCl5
Each chlorine atom displays a tetrahedral arrangement
around the atom.

24. sp3
d2
Hybridization
 An octahedral arrangement requires six effective pairs
around the central atom.
 SF6 has sp3
d2
hybridization.

25. Example Problem
How is the xenon atom in XeF4 hybridized?
sp3
d2

26. Localized Electron Summary
 Draw the Lewis Structure
 Determine the arrangement of electron pairs using the
VSEPR model.
 Specify the hybrid orbitals needed to accommodate the
electron pairs.
 Do not overemphasize the characteristics of the
separate atoms. It is not where the valence electrons
originate that is important; it is where they are
needed in the molecule to achieve stability.

27. Effective Pairs andTheir
Spatial Arrangement

28. The Molecular Orbital
Model

29. Molecular Orbital Model
The localized electron model works very well with the
prediction of structure and bonding of molecules, but
the electron correlation problem still exists.
 Since we do not know the details of the electron
movements, we cannot deal with the electron-electron
interactions in a specific way
 The Molecular Orbital model helps us to deal with the
molecular problem.

30. Molecular Orbitals
Molecular orbitals (MOs) have many of the same
characteristics as atomic orbitals.Two of the most
important are:
 MOs can hold two electrons with opposite spins.
 The square of the MO’s wave function indicates electron
probability.

31. MOs
For simplicity we will first look at the H2 molecule.
 The combination of hydrogen 1s atomic orbitals results in 2
molecular orbitals.
 The wave phases of the atomic orbitals combine/overlap.
Since electrons move in wave functions, this causes
constructive and destructive interference in the wave pattern.
 When the orbitals are added, the matching phases produce
constructive interference and the opposite phases produce
destructive interference.

32. MOs
 A constructive combination gives a bonding MO. This
gives an enhanced electron probability between the
nuclei.
 The destructive combination gives an antibonding MO.
This interference produces a node between the nuclei.

33. MOs
Two MOs exist for H2:
 MO1= 1sH1 + 1sH2
 MO1 is constructive and therefore a bonding MO
 MO1 is lower energy
 MO2 = 1sH1 – 1sH2
 MO2 is destructive and therefore an antibonding MO
 MO2 is higher energy

34. MOs
The type of electron distribution described in these MOs is
called sigma as in the localized electron model. MO1 and
MO2 are sigma (σ) molecular orbitals.
 In this molecule only the molecular orbitals are available
for occupation by electrons. The 1s atomic orbitals of
the hydrogen atoms no longer exist, because the H2
molecule – a new entity – has its own set of new
orbitals.

35. MOs
The energy level of the bonding
MO is lower and more stable than
that of the antibonding MO. Since
molecule formation favors the
lowest energy state, this provides
the driving force for molecule
formation of H2. This is called
probonding.
If two electrons were forced to
occupy the higher-energy MO2 this
would be anti-bonding and the
lower energy of the separated
atoms would be favored.

36. Bonding and Antibonding

37. MOs
Labels are given to MOs indicate their symmetry (shape), the
parent atomic orbitals, and whether they are bonding or
antibonding.
 Antibonding character is indicated by an asterisk.
 Subscripts indicate parent orbitals
 σ and π indicate shape.
 H2 has the following MOs:
 MO1 = σ1s
 MO2 = σ1s*

38. MOs
Molecular electron configurations can be written in much
the same way as atomic (electron) configurations. Since
the H2 molecule has two electrons in the σ1s molecular
orbital, the electron configuration is: σ1s
2
 Each molecular orbital can hold two electrons, but the
spins must be opposite.
 Orbitals are conserved. The number of molecular
orbitals will always be the same as the number of
atomic orbitals used to construct them.

39. MOs
From this molecular electron configuration, we can determine a
molecules stability.
 Would H2
-
be stable?
 (σ1s)2
(σ1s*) 1
 The key idea is that H2
-
would exist if it were a lower energy than its
separated parts. Two electrons are in bonding and one is in
antibonding. Since more electrons favor bonding H2
-
is formed.
 This also is a good indicator of bond strength. H2 has a stronger
bond than H2
-
. The net lowering of the bonding electrons by one is a
direct relationship to bond strength. H2 is twice as strong.

40. Bond Order
To indicate bond strength, we use the concept of bond
order.
Example: H2 has a bond order of 1
H2
-
has a bond order of ½
Bond Order =
#οφβονδινγ ε−
− #οφαντιβονδινγ ε−
2

41. Bond Order
Bond order is an indication of bond strength because it
reflects the difference between the number of bonding
electrons and the number of antibonding electrons.
 Larger bond order means greater bond strength.
 Bond order of 0 gives us a molecule that doesn’t exist.

42. Bonding in Homonuclear
Diatomic Molecules

43. Homonuclear Bonding
When looking at bonding beyond energy level 1, we need to
consider what orbitals are overlapping and therefore
bonding.
 Li2 has electrons in the 1s and 2s orbitals; the 2s orbitals
are much larger and overlap, but the 1s orbitals are
smaller and do not overlap.
 To participate in molecular orbitals, atomic orbitals must
overlap in space. This means that only the valence
orbitals of the atoms contribute significantly to the
molecular orbitals of a particular molecule.

44. Li2 MO
What is the molecular electron configuration and bond
order of Li2?
 σ2s
2
with a bond order of 1
 Li2 is a stable molecule because the overall energy of the
molecule is lower than the separate atoms.

45. Be2 MO
What is the molecular electron configuration and bond
order of Be2?
 (σ2s)2
(σ2s*)2
with a bond order of 0
 Be2 has 2 bonding and 2 antibonding electrons and is not
more stable than the individual atoms. Be2 does not
form.

46. MOs from p orbitals
 p orbitals must overlap in
such a way that the wave
patterns produce
constructive interference.
As with the s orbitals, the
destructive interference
produces a node in the
wave pattern and
decreases the probability
of bonding.

47. MOs from p orbitals
 When the parallel p
orbitals are combined with
the positive and negative
phases matched,
constructive interference
occurs, giving a bonding π
orbital. When orbitals
have opposite phases,
destructive interference
results in an antibonding π
orbital.

48. MOs from p orbitals
 Since the electron probability
lies above and below the line
between the nuclei (with
parallel p orbitals), the
stability of a π molecular
bonding orbital is less than
that of a σ bonding orbital.
Also, the antibonding π MO is
not as unstable as the
antibonding σ MO. The
energies associated with the
orbitals reflect this stability.

49. B2 Example
 1s2
2s2
2p1
 1s2
does not bond
 2s2
and 2p1
bond
 (σ2s)2
(σ2s*)2
(σ2p)2
 Bond order: 1

50. Exceptions
 B2, C2, and N2 molecules use the
same set of molecular orbitals
that we expect but some
mixing of orbital energies
occurs. The s and p atomic
orbitals mix or hybridize in a
way that changes some MO
energy states. This affects
filling order and pairing of
electrons.

51. Paramagnetism
Most materials have no magnetism until they are placed in
a magnetic field. However, in the presence of such a
field, magnetism of two types can be induced:
 Paramagnetism – causes the substance to be attracted
into the magnetic field.
 Diamagnetism – causes the substance to be repelled
from the magnetic field.

52. Paramagnetism

53. Paramagnetism
 Paramagnetism is associated with unpaired electrons
and diamagnetism is associated with paired electrons.
 Any substance that has both paired and unpaired
electrons will exhibit a net paramagnetism since the
effect of paramagnetism is much stronger than that of
diamagnetism.
 ParamagnetismVideo

54. Summary
 There are definite correlations between bond order, bond
energy, and bond length. As bond order increases so does
bond energy and bond length decreases.
 Comparison of bond orders between different molecules
cannot predict bond energies of different molecules.
 B2 and F2 both have bond order of 1 but bond energies are
very different. B-B is a much stronger bond.
 N2 has a bond order of 3 and has a very large bond energy. N2 is
a very stable molecule and is used to drive powerful reactions.

56. Example Problem
For O2, O2
+
, and O2
-
, give the MO
electron configuration and the
bond order for each. Which has
the strongest bond?
O2 : (σ2s)2
(σ2s*)2
(σ2p)2
(π2p)4
(π2p*)2
BO=2
O2
+
: (σ2s)2
(σ2s*)2
(σ2p)2
(π2p)4
(π2p*)1
BO=2.5
O2
-
: (σ2s)2
(σ2s*)2
(σ2p)2
(π2p)4
(π2p*)3
BO=1.5
B2 C2 N2 O2 F2
σ2p* σ2p*
π2p* π2p*
σ2p π2p
π2p σ2p
σ2s* σ2s*
σ2s σ2s
σ1s* σ1s*
σ1s
σ1s

57. Example Problem
Use the molecular orbital model to predict the bond order
and magnetism of each of the following molecules: Ne2
and P2
Ne2 bond order is 0: does not exist
P2 bond order is 3 and diamagnetic
B2 C2 N2 O2 F2
σ2p* σ2p*
π2p* π2p*
σ2p π2p
π2p σ2p
σ2s* σ2s*
σ2s σ2s
σ1s* σ1s*
σ1s
σ1s

58. Bonding in Heteronuclear
Diatomic Molecules

59. Heteronuclear Molecules
 When dealing with different atoms within diatomic
molecules we can still use the MO model to determine
bond order and magnetism

60. NO example
 The valence electrons from
both atoms fill in the order
expected by the model.
 The bond order is 2.5 and is
paramagnetic.

61. Example Problem
Use the MO model to predict the
magnetism and bond order of
the NO+
and CN-
ions.
Both ions are diamagnetic and
have the same configuration.
Their bond order is 3

62. Heteronuclear Diatomics
What happens with the
diatomic molecules are
very different?
 A molecular orbital forms
between two different
atomic orbitals.
 HF example
 Note: energy level
difference vs.
electronegativity

63. Combining the Localized
Electron and MO models

64. Resonance
When a molecule has resonance. It is usually a double
bond that can have different positions around the
molecule.
 The single σ bonds remain localized and the π bonds are
said to be delocalized.

65. Resonance
 Benzene: All C-C bonds are known to be equivalent and
the molecule has resonance

66. Resonance
 Benzene: The σ bonds remain centered (on the plane)
between C atoms.

67. Resonance
 Benzene: The p orbitals are perpendicular to the plane and
form π bonds above and below the plane.The electrons in the
π bonds delocalize and give six equivalent C-C bonds that give
the structure true resonance.
 This is called delocalized π bonding.

68. NO3
-
 NO3
-
ion also displays
delocalized π bonding.

69. The End

70. MOWorksheet
B2 C2 N2 O2 F2
σ2p* σ2p*
π2p* π2p*
σ2p π2p
π2p σ2p
σ2s* σ2s*
σ2s σ2s
σ1s* σ1s*
σ1s
σ1s