1. Hybridization Dr. Harris Lecture 11 (Ch 9.5-9.13) 9/25/12
HW: Ch 9: 21, 23, 39, 43, *75
*(show all carbon atoms, refer to pg 215)

2. Introduction
We now know that atoms can bond covalently through the sharing of electrons
VSEPR theory helps us predict molecular shapes. But, it does not explain what bonds are, how they form, or why they exist.
In ch 9, chemical bonding will be explained in terms of orbitals

3. Covalent Bonding Is Due to Orbital Overlap
In a covalent bond, electron density is concentrated between the nuclei.
Thus, we can imagine the valence orbitals of the atoms overlapping
The region of orbital overlap represents the covalent bond

4. Overlapping Valence Orbitals
Recall s and p orbitals (ch 5)
S orbitals are spherical. L = 0, mL = 0
Max of 2 electrons S
P orbitals consist of two lobes of electron density on either side of the nucleus.
L= 1, mL = -1, 0, 1 (3 suborbitals)
Max of 6 electrons px py pz

5. Forming Sigma (σ) Bonds
Energy H + H + H + H + σ
Covalent bond
stabilization (energy drop)
1s1
1s1 σ
Two overlapping atomic orbitals form a molecular bonding orbital.
A sigma (σ) bonding orbital forms when s-orbitals overlap.

6. X Hybridization Be 2 H 2p0 ENERGY 2s2 1s1 1s1
Imagine the molecule BeH2. We know that Be has a valence configuration of [He] 2s2 (Be-H bond is not ionic).
However, when we fill our orbitals in order as according to Hund’s rule, we notice that there are no unpaired electrons. Hence, we can not make any bonds. Stay mindful of the fact that a covalent bond involves electron sharing X Be
ENERGY
2p0
2 H
2s2
1s1
1s1

7. Hybridization
So how does BeH2 form? How can Beryllium make 2 bonds?
To bond with 2 hydrogen atoms, Beryllium mixes (hybridizes) two of its atomic orbitals. This creates two sp hybrid orbitals.
Energy
sp hybridization 2p
sp hybrid orbital
2s2
Each sp orbital is 50% s character and 50% p character

8. Hybridization s pz z + = z
The addition of an s-orbital to a pz orbital is shown above. The s orbital adds constructively to the (+) lobe of the pz orbital and adds destructively to the lobe that is in the opposite phase (-). The symbols indicate phase, not charge.
Remember, we are adding two atomic orbitals, so we will obtain two hybrid orbitals
2 sp hybrid orbitals

9. sp-Hybridized Bonding
Small negative lobes not shown. Recall that we have two unused p-orbitals along the x and y axes.
Now, 2 Hydrogen s-orbitals can overlap with the Be sp-hybrid orbitals to form BeH2. We would expect BeH2 to be linear, as is predicted by VSEPR

10. sp2 Hybridization 3 H B 2p1 2s2 1s1 1s1 1s1
Whenever we mix a certain number of s and p atomic orbitals, we get the same number of molecular orbitals. This is called the principle of conservation of orbitals.
The BH3 molecule gives us an example of sp2 hybrid orbitals.
Once again, we have a situation where we don’t have enough bonding sites to accommodate all of the hydrogens.
Energy
2p1
3 H
2s2 B
1s1
1s1
1s1

11. Hybridization B 2p1 ENERGY 2s2
So, to make 3 bonding sites, 3 hybrid molecular orbitals are formed by mixing the 2s-orbital with two 2p-suborbitals.
Each of these three hybrid orbitals are one-third s-character, and two-thirds p-character.
unused 2p suborbital
2p1
ENERGY B
sp2 hybrid orbitals
2s2

12. sp2 orbitals
The result of adding the s and p orbitals together is a trigonal planar arrangement of electron domains
This figure illustrates the 3 hybrid orbitals combined with the unused 2p orbital, which is perpendicular to the hybrid orbitals.

13. sp2 Geometry and Bonding
empty 2p orbital
σ bond H H + H + B H H H +

14. Four sp3 hybrid orbitals
sp3 Hybridization
Involves the mixing of an s-orbital and 3 p-orbitals. The resulting hybrid orbitals are one-fourth s-character and three-fourths p-character.
Ex. CH4. To accommodate 4 hydrogen atoms, 4 hybrid orbitals are created (C: [He] 2s22p2) C
ENERGY
2p2
Four sp3 hybrid orbitals
2s2

15. Formation of sp3 Hybrid Orbitals
The four hybrid orbitals arrange themselves tetrahedrally. Do you notice a trend yet?
4 σ-bonds

16. What about molecules with lone electron pairs?
Ex. What is the hybridization of H2O?
The valence electron configuration of O is [He]2s2 2p4
As you see, there are two unpaired O electrons. Does this mean that these two p-suborbitals can overlap with the two H 1s orbitals without hybridizing??
ENERGY
2p4
2s2
No. The reason is that we now have two sets of lone pairs of electrons that are substantially different in energy. The orbitals will hybridize to form degenerate (equal energy) sets of electrons.

17. Water has sp3 hybridization•• •• O H H
2p4
covalent bonding
lone pair
bonding electrons
2s2
2 H
ENERGY O
1s1
1s1
H2O
σ bonds

18. Water has sp3 hybridization
The angle between the sp3 hybrid orbitals in water is 104.5o, NOT 109.5o as expected in a normal tetrahedron
LONE PAIR REPEL THE ELECTRONS IN THE O-H BONDS
Strength of Repulsion
Lone pair – Lone pair > Lone pair – Bonding pair > Bonding pair- Bonding pair

19. sp3d and sp3d2 hybridization
Atoms like S, Se, I, Xe … etc. can exceed an octet because of sp3d and sp3d2 hybridization (combination of an ns, np, and nd orbitals where n>3).
This results in either trigonal bipyramidal or octahedral geometry
sp3d
sp3d2

20. Exceeding an Octet. Example: SF6
Energy
3d0
sp3d2 hybrid orbitals
3p4
Fluorine lone pair
3s2 S
sp3
6 F
SF6

21. Exceeding an Octet. Example: SF6
overlap
unpaired electron F
x 6 S
3 lone pair

22. Look Familiar ???

23. Examples: What is the hybridization of the central atom? NH3 NH4+ PCl5
CH3Cl
SeF6

24. Double and Triple Bonding
How can orbital overlap be used to explain double and triple bonds? What kind of interactions are these?
Lets look at ethene, C2H4 C H
The hybridization of each carbon is sp2 because each is surrounded by three electron domains. The geometry around each C is trigonal planar.
sp2
sp2

25. Forming Double Bonds C H Carbon atoms 2p2 2s2
unhybridized p-electron
2p2
sp2 hybrid orbitals
2s2
We can see that for each carbon atom, the three sp2 electrons will be used to make 3 bonds. But how is the double bond formed?

26. Double Bonds formed by simultaneous σ and π interaction• • H + H + H + H +
The remaining electrons form a π bond. This bond forms due to attraction between the parallel p-orbitals. The like-phase regions are drawn toward one another and overlap.
All double bonds consist of 1 σ-bond and 1 π-bond

27. Triple Bonds formed by 1 σ-bond and 2 π-bonds. Ex. HCNsp sp H C N •
Can you draw the orbital diagram for this molecule?

28. Examples
How many σ and π bonds are in each of the following molecules? Give the hybridization of each carbon.
CH3CH2CHCHCH3
CH3CCCHCH2