1. Covalent Bonding: Orbitals
Chapter 09
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The four bonds around C are of equal length and Energy
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Can you explain this based on your knowledge
of electron energy levels? 6C
1s2 2s2 2p2
Bonding from s is
Different from bonding
From p. In addition the
Angles should be within
900!
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How to generate four equal orbitals?
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5. A key to wave mechanics is superposition
Hint:
A key to wave mechanics is
superposition
Which is creating new waves from interference of old ones
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Let’s do some mixing
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Cross section of an sp3 orbital.
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11. An energy-level diagram showing the formation of four sp3 orbitals.
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Hybridization
Ground state of C 1s2 2s2 2p2
Promote electron at n=2 2s12p3
Hybridize at n=2 sp3 sp3 sp3 sp3 s px py pz s px py pz
sp3 sp3 sp3 sp3
4sp3 orbitals of equal length, energy and in tetrahedral shape
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Hybridization
The mixing of atomic orbitals to form special orbitals for bonding.
The atoms are responding as needed to give the minimum energy for the molecule.
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14. Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1 2s
2px 2py 2pz
If use the
3 2p orbitals
predict 900
Actual H-N-H bond angle is
How do you explain this?
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Consider the n=2 for N
Original 2s
2px 2py 2pz
Mix 1s and 3p
And generate four
Equivalent sp3
Hybridized orbitals
sp3 sp3 sp3 sp3
1 sp3
lone
pair
3 bonding orbitals
Which can accommodate
The 1s1 electron from
hydrogen
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16. The nitrogen atom in ammonia is sp3 hybridized.
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An orbital energy-level diagram for sp2 hybridization. Note that one p orbital remains unchanged.
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When an s and two p orbitals are mixed to form a set of three sp2 orbitals, one p orbital remains unchanged and is perpendicular to the plane of the hybrid orbitals.
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Figure 9.13: (a) The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.
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Pi bond (p) – electron density above and below plane of nuclei of the bonding atoms
Sigma bond (s) – electron density between the 2 atoms
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The s bonds in ethylene. Note that for each bond the shared electron pair occupies the region directly between the atoms.
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A sigma () bond centers along the internuclear axis.
A pi () bond occupies the space above and below the internuclear axis.
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The orbitals used
to form the bonds
in ethylene.
(b) The Lewis structure for ethylene.
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The orbital energy-level diagram for the formation of sp hybrid orbitals on carbon.
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When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at 180 degrees results.
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32. The orbitals of an sp hybridized carbon atom.
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34. The orbital arrangement for an sp2 hybridized oxygen atom to form CO2.
8O 1s22s22p4
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35. The hybrid orbitals in the CO2 molecule.
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(a) The orbitals used to form the bonds in carbon dioxide. Note that the carbon-oxygen double bonds each consist of one s bond and one p bond. (b) The Lewis structure for carbon dioxide.
2sp orbitals from C to form two double bonds
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(a) An sp hybridized nitrogen atom. (b) The s bond in the N2 molecule. (c) The two p bonds in N2 are formed when electron pairs are shared between two sets of parallel p orbitals. (d) The total bonding picture for N2.
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38. Sigma (s) and Pi Bonds (p)
1 sigma bond
Single bond
Double bond
1 sigma bond and 1 pi bond
Triple bond
1 sigma bond and 2 pi bonds
How many s and p bonds are in the acetic acid
(vinegar) molecule CH3COOH? C H O
s bonds = 6
+ 1 = 7
p bonds = 1
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10.5

39. How to generate more than 4 bonds?
Remember the PCl5, SF6, etc…
The s and 3p can generate 4 orbitals
Include d orbitals to generate more!
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s p p p d d d d d
Hybridize 1s and 3p and 1d
Result in 5 dsp3 orbitals
dsp3 dsp3 dsp3 dsp3 dsp3
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A set of dsp3 hybrid orbitals on a phosphorus atom. Note that the set of five dsp3 orbitals has a trigonal bipyramidal arrangement. (Each dsp3 orbital also has a small lobe that is not shown in this diagram.)
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(a) The PCl5 molecule. (b) The orbitals used to form the bonds in PCl5. The phosphorus uses a set of five dsp3 orbitals to share electron pairs with sp3 orbitals on the five chlorine atoms. The other sp3 orbitals on each chlorine atom hold lone pairs.
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How to form 6 orbitals?
s p p p d d d d d
Hybridize 1s and 3p and 2d
Result in 6 d2sp3 orbitals
d2sp3 ds2p3 d2sp3 d2sp3 d2sp d2sp3
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The relationship of the number of effective pairs, their spatial arrangement, and the hybrid orbital set required.
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45. The Localized Electron Model
Draw the Lewis structure(s)
Determine the arrangement of electron pairs (VSEPR model).
Specify the necessary hybrid orbitals.
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46. Molecular Orbitals (MO)
Analagous to atomic orbitals for atoms, MOs are the quantum mechanical solutions to the organization of valence electrons in molecules. Remember bonds are waves and wave may be arranged in constructive or destructive interference.
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Types of MOs
bonding: lower in energy than the atomic orbitals from which it is composed.
antibonding: higher in energy (unstable) than the atomic orbitals from which it is composed.
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The combination of hydrogen 1s atomic orbitals to form molecular orbitals.
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51. Bonding and antibonding molecular orbitals (MOs).
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(a) The molecular orbital energy-level diagram for the H2 molecule. (b) The shapes of the molecular orbitals are obtained by squaring the wave functions for MO1 and MO2.
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A molecular orbital energy-level diagram for the H2 molecule.
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54. The molecular orbital energy-level diagram for the H2- ion.
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55. The molecular orbital energy-level diagram for the He2 molecule.
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56. The molecular orbital energy-level diagram for the Li2 molecule.
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The expected molecular orbital energy-level diagram resulting from the combination of the 2p orbitals on two boron atoms.
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The expected molecular orbital energy-level diagram for the B2 molecule.
However, B2 is found to be Paramagnetic!
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The correct molecular orbital energy-level diagram for the B2 molecule. When p-s mixing is allowed, the energies of the 2p and π2p orbitals are reversed. The two electrons from the B 2p orbitals now occupy separate, degenerate π2p molecular orbitals and thus have parallel spins. Therefore, this diagram explains the observed paramagnetism of B2.
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(a) The three mutually perpendicular 2p orbitals on two adjacent boron atoms. Two pairs of parallel p orbitals can overlap as shown in (b) and (c), and the third pair can overlap head-on as shown in (d).
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(a) The two p orbitals on the boron atom that overlap head-on produce two s molecular orbitals, one bonding and one antibonding. (b) Two p orbitals that lie parallel overlap to produce two p molecular orbitals, one bonding and one antibonding.
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The molecular orbital energy-level diagrams, bond orders, bond energies, and bond lengths for the diatomic molecules B2 through F2. Note that for O2 and F2 the  2p orbital is lower in energy than the π2p orbitals.
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Paramagnetic
Diamagnetic
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Note that O2 is paramagnetic
When liquid oxygen is poured into the space between the poles of a strong magnet, it remains there until it boils away. This attraction of liquid oxygen for the magnetic field demonstrates the paramagnetism of the O2 molecule.
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Paramagnetism
unpaired electrons
attracted to induced magnetic field
much stronger than diamagnetism
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Bond Order (BO)
Difference between the number of bonding electrons and number of antibonding electrons divided by two.
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bond order = 1 2
Number of electrons in bonding MOs
Number of electrons in antibonding MOs ( - )
bond order 1
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Molecular Orbital (MO) Configurations
The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.
The more stable the bonding MO, the less stable the corresponding antibonding MO.
The filling of MOs proceeds from low to high energies.
Each MO can accommodate up to two electrons.
Use Hund’s rule when adding electrons to MOs of the same energy.
The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms.
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Outcomes of MO Model
1. As bond order increases, bond energy increases and bond length decreases.
2. Bond order is not absolutely associated with a particular bond energy.
3. N2 has a triple bond, and a correspondingly high bond energy.
4. O2 is paramagnetic. This is predicted by the MO model, not by the LE model, which predicts diamagnetism.
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Heteronuclear Molecules and similar?
The molecular orbital energy-level diagram for the NO molecule. We assume that orbital order is the same as that for N2. The bond order is 2.5.
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Figure 9.42: The molecular orbital energy-level diagram for both the NO+ and CN- ions.
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Heteronuclear Molecules and similar?
A partial molecular orbital energy-level diagram for the HF molecule.
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The electron probability distribution in the bonding molecular orbital of the HF molecule. Note the greater electron density close to the fluorine atom.
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75. Combining LE and MO Models
 bonds can be described as being localized.
 bonding must be treated as being delocalized.
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Delocalized molecular orbitals are not confined between two adjacent bonding atoms, but actually extend over three or more atoms.
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(a) The benzene molecule consists of a ring of six carbon atoms with one hydrogen atom bound to each carbon. (b) Two of the resonance structures for the benzene molecule.
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78. The  bonding system in the benzene molecule.
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(a) The p molecular orbital system in benzene is formed by combining the six p orbitals from the six sp2 hybridized carbon atoms. (b) The electrons in the resulting p molecular orbitals are delocalized over the entire ring of carbon atoms, giving six equivalent bonds. A composite of these orbitals is represented here.
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Electron density above and below the plane of the benzene molecule.
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The resonance structures for O3 and NO3-. Note that it is the double bond that occupies various positions in the resonance structures.
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(a) The p orbitals used to form the π bonding system in the NO3- ion. (b) A representation of the delocalization of the electrons in the π molecular orbital system of the NO3- ion.
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