Lecture 11
Covalent Bonding Pt 3: Hybridization
(Ch. 9.5-9.13)
Suggested HW: Ch 9: 25, 29, 39, 43, 72
(For 25 and 43, you are illustrating the hybridization of the atomic
orbitals into hybrid orbitals and the overlapping of these hybrid orbitals
as described in the examples provided)
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
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
Overlapping Valence Orbitals
• Recall s and p orbitals (ch 5)
S
px
• S orbitals are spherical. L = 0, mL = 0
• Max of 2 electrons
py
pz
• P orbitals consist of two
lobes of electron density.
• L= 1, mL = -1, 0, 1
• (3 suborbitals)
• Max of 6 electrons
Forming Sigma (σ) Bonds
Energy
+
+
H
H
+
+
H σ H
Covalent bond
1s1
1s1
σ
• Two overlapping atomic orbitals form a molecular bonding orbital. Plus sign
indicates phase of electron wave, NOT CHARGE
• A sigma (σ) bonding orbital forms when s-orbitals overlap.
Introduction to Hybridization
 Imagine the molecule CH4. We know that carbon has 4 valence electrons (2s22p2).
 However, when we fill our orbitals in order as according to Hund’s rule, we notice
there are only enough unpaired electrons to make two bonds. Stay mindful of the
fact that a covalent bond involves the sharing of unpaired electrons
ENERGY
C
2p2
X
4H
2s2
1s1
1s1
1s1
1s1
sp3 Hybridization
 So how does CH4 form? How can carbon make 4 bonds?
 To make four bonds, carbon hybridizes four of its atomic orbitals. This creates
ENERGY
four equivalent sp3 hybrid orbitals, each containing one unpaired electron.
2p2
2s2
Four sp3 hybrid orbitals
• The name “sp3” originates from the fact that the hybrid orbitals form as a
result of the mixture of 1 s-orbital and 3 p-orbitals. Thus, each sp3 orbital is
25% s character and 75% p character
Formation of Sigma Bonding Orbitals
sp3 hybrid orbitals
C
ENERGY
1s1
1s1
1s1
atomic s-orbitals
4H
σ bonding orbitals
1s1
Illustration of Orbital Hybridization
pz
s
+
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.
• 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.
Illustration of sp3 Hybrid Orbitals and Orbital Overlap
4 σ-bonds
The four hybrid orbitals arrange
themselves tetrahedrally.
sp2 Hybridization
 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. (Remember, B is electron deficient!)
B
ENERGY
3H
2p1
2s2
1s1
1s1
1s1
sp2 Hybridization
 So, to make 3 bonding sites, 3 hybrid molecular orbitals are formed by
ENERGY
mixing the 2s-orbital with two 2p-suborbitals.
unused 2p suborbital
2p1
B
2s2
Three sp2 hybrid orbitals
 This forms an sp2 orbital. Each of these three hybrid orbitals are one-
third s-character, and two-thirds p-character.
sp2 orbitals
The result of adding
one s and two 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.
sp2 Geometry and Bonding
empty 2p orbital
σ bond
H
+
H
+
H
+
H
B
H
H
sp Hybridization
 Imagine BeH2 (the Be-H bond is covalent), with Be having the electron
configuration: [He]2s2
 Here, we have a situation where no bonding electrons are available. To
make two Be-H bonds, Be must create two hybrid orbitals by mixing
two atomic orbitals (the 2s orbital and one of the 2p orbitals). This
yields sp hybrid orbitals (50% s, 50% p)
ENERGY
Be
2H
2p0
2s2
1s1
1s1
sp Hybridization
ENERGY
Be
2p0
2s2
unused 2p suborbitals
Two sp hybrid orbitals
Hybridization of Lone Electron Pairs
 Ex. What is the hybridization of Oxygen in H2O?
 The valence electron configuration of O is [He]2s2 2p4
ENERGY
O
2p4
2s2
2H
1s1
1s1
As you see, there are two unpaired O electrons. Does this mean
that these two p-suborbitals can overlap with the two
Hydrogen 1s orbitals without hybridizing??
Hybridization of Lone Electron Pairs
O
ENERGY
2p electrons
2p4
2H
••
O
••
2s electrons
H
H
BAD!!
2s2
1s1
1s1
 No!! The reason is that we now have two sets of lone pairs of electrons
that are substantially different in energy (2s and 2p). The orbitals will
hybridize to form degenerate (equal energy) sets of electrons.
 Lone pair must always be equal in energy with each other, and with
bonding electrons.
Water has
sp3
sp3 electrons
hybridization
••
O
ENERGY
2p4
Lone pair
Bonding electrons
2H
1s1
H2O
H
H
Four sp3 hybrid orbitals
2s2
O
σ bonds
••
1s1
So What Do We Know So Far?
Total Electron Domains
Around Atom (Bond + LP)
Hybridization
2
3
4
sp
sp2
sp3
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
H
sp2
C
H
sp2
H
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.
Forming Double Bonds
H
C
H
C
C
H
H
unhybridized p-electron
2p2
sp2 hybrid orbitals
2s2
• We can see that for each carbon atom, we need three sp2 orbitals
and three unpaired electrons to make three sigma bonds. But how
is the double bond formed?
Double Bonds formed by simultaneous σ and π
interaction
+
H
+
H
•
•
+
H
+
H
The remaining p-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
Triple Bonds formed by 1 σ-bond and 2 π-bonds.
Ex. HCN
H
sp
sp
C
N ••
π
π
• Can you draw the orbital diagram for this molecule?
Examples
 How many σ and π bonds are in each of the following molecules? Give
the hybridization of each carbon.
 CH3CH2CHCHCH3
 CH3CCCHCH2
sp3d and sp3d2 hybridization
 Atoms like S, Se, I, Xe … etc. can exceed an octet because of sp3d and sp3d2
hybridization (combination of ns, np, and nd orbitals where n>3).
 This results in either trigonal bipyramidal or octahedral skeletal geometry
sp3d
sp3d2
Exceeding an Octet. Example: SF6
Energy
3d0
sp3d2 hybrid orbitals
3p4
Fluorine lone
pair
3s2
S
sp3
6F
SF6
Exceeding an Octet. Example: SF6
unpaired electron
S
F
3 lone pair
x6
Look Familiar ???
Examples:
 What is the hybridization of the central atom?
 CO2
 H2CO
 CH3CCH
 IF5
 PCl5
 SeOF4