1. Overlap and Bonding
We think of covalent bonds forming through the sharing of electrons by adjacent atoms.
In such an approach this can only occur when orbitals on the two atoms overlap.
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2. Overlap and Bonding
Bond length (internuclear distance) corresponds to the minimum on the potential energy curve (most stable).
This represents the distance at which attractive forces between e- and p+ are balanced by repulsive forces between e- and e- and nucleus and nucleus (if atoms get too close, internuclear repulsion greatly raises the energy)
This value corresponds to bond strength as well as energy change for the formation of the H-H bond.
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3. Hybrid Orbitals
But it’s hard to imagine tetrahedral, trigonal bipyramidal, and other geometries arising from the atomic orbitals we recognize.
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4. Hybrid Orbitals Consider beryllium:
In its ground electronic state, it would not be able to form bonds because it has no singly-occupied orbitals.
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5. Hybrid Orbitals
But if it absorbs the small amount of energy needed to promote an electron from the 2s to the 2p orbital, it can form two bonds.
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6. Hybrid Orbitals
Mixing the s and p orbitals yields two degenerate orbitals that are hybrids of the two orbitals.
These sp hybrid orbitals have two lobes like a p orbital.
One of the lobes is larger and more rounded as is the s orbital.
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7. Hybrid Orbitals
These two degenerate orbitals would align themselves 180 from each other.
This is consistent with the observed geometry of beryllium compounds: linear.
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8. Hybrid Orbitals
With hybrid orbitals the orbital diagram for beryllium would look like this.
The sp orbitals are higher in energy than the 1s orbital but lower than the 2p.
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9. Hybrid Orbitals Using a similar model for boron leads to…
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10. Hybrid Orbitals …three degenerate sp2 orbitals.
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11. Hybrid Orbitals
With carbon we get…
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12. Hybrid Orbitals …four degenerate sp3 orbitals.
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13. Hybrid Orbitals
For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids.
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14. Hybrid Orbitals This leads to five degenerate sp3d orbitals…
…or six degenerate sp3d2 orbitals.
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15. Hybrid Orbitals
Once you know the electron-domain geometry, you know the hybridization state of the atom.
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16. Sample Exercise 9.5 Hybridization
Indicate the hybridization of orbitals employed by the central atom in (a) NH2–, (b) SF4 (see Sample Exercise 9.2).
Solution
Plan: To determine the hybrid orbitals used by an atom in bonding, we must know the electron-domain geometry around the atom. Thus, we first draw the Lewis structure to determine the number of electron domains around the central atom. The hybridization conforms to the number and geometry of electron domains around the central atom as predicted by the VSEPR model.
Solve:
The Lewis structure of NH2– is
Because there are four electron domains around N, the electron-domain geometry
is tetrahedral. The hybridization that gives a tetrahedral electron-domain geometry
is sp3 (Table 9.4). Two of the sp3 hybrid orbitals contain nonbonding pairs of
electrons, and the other two are used to make bonds with the hydrogen atoms.
(b) The Lewis structure and electron-domain geometry of SF4 are:
The S atom has five electron domains around it, giving rise to a trigonal-bipyramidal electron-domain geometry. With an expanded octet of ten electrons, a d orbital on the sulfur must be used. The trigonal-bipyramidal electron-domain geometry corresponds to sp3d hybridization (Table 9.4).

17. Sample Exercise 9.5 Hybridization
Predict the electron-domain geometry and the hybridization of the central atom in
(a) SO32– (b) SF6.
Practice Exercise

18. Answer: (a) tetrahedral, sp3 (b) octahedral, sp3d2

19. Valence Bond Theory
Hybridization is a major player in this approach to bonding.
There are two ways orbitals can overlap to form bonds between atoms.
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20. Sigma () Bonds Sigma bonds are characterized by Head-to-head overlap.
Cylindrical symmetry of electron density about the internuclear axis.
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21. Pi () Bonds Pi bonds are characterized by Side-to-side overlap.
Electron density above and below the internuclear axis.
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22. Single Bonds
Single bonds are always  bonds, because  overlap is greater, resulting in a stronger bond and more energy lowering.
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23. Multiple Bonds
In a multiple bond one of the bonds is a  bond and the rest are  bonds.
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24. Multiple Bonds
In a molecule like formaldehyde (shown at left) an sp2 orbital on carbon overlaps in  fashion with the corresponding orbital on the oxygen.
The unhybridized p orbitals overlap in  fashion.
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25. Multiple Bonds
In triple bonds, as in acetylene, two sp orbitals form a  bond between the carbons, and two pairs of p orbitals overlap in  fashion to form the two  bonds.
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26. Delocalized Electrons: Resonance
When writing Lewis structures for species like the nitrate ion, we draw resonance structures to more accurately reflect the structure of the molecule or ion.
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27. Delocalized Electrons: Resonance
In reality, each of the four atoms in the nitrate ion has a p orbital.
The p orbitals on all three oxygens overlap with the p orbital on the central nitrogen.
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28. Delocalized Electrons: Resonance
This means the  electrons are not localized between the nitrogen and one of the oxygens, but rather are delocalized throughout the ion.
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29. Resonance
The organic molecule benzene has six  bonds and a p orbital on each carbon atom.
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30. Resonance
In reality the  electrons in benzene are not localized, but delocalized.
The even distribution of the electrons in benzene makes the molecule unusually stable.
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