Hybridization
and
its
Implications
Sp, sp2, sp3... If you’re reading this tutorial, you most likely know these terms refer to
hybridized orbitals. (If not, it’s okay. Keep reading.) But what exactly is going on behind
these combinations of letters and numbers? Hybridization is a complex topic that is
sometimes overly simplified, to the point where students may not fully understand its
concepts. This tutorial strives to present a concise explanation of hybridization theory and
its applications.
I.
II.
III.
IV.
V.
Basic Definitions
Why do Hybrid Orbitals Exist?
Determining Hybridization
Hybridization and Resonance
Practice Problems & Solutions
Basic Definitions
Orbital: A mathematical equation describing a region in space where there is a high
probability of finding an electron in an atom.1 There are many types of orbitals. In Chem
14C, we are mainly concerned about the s- and p-orbitals and their hybrids.
1
S
orbital
Px
orbital
Py
orbital
Pz
orbital
All definitions retrieved from Atkins and Jone’s Chemical Principles: The Quest for Insight, unless otherwise stated.
As shown above, the s-orbital is sphere-shaped, while the three p-orbitals each consist of
two lobes and can be shown as being aligned perpendicularly to each other as if on the x,
y, and z axes of a graph.
Bonds are formed when the orbitals of multiple atoms overlap. So what’s all the fuss over
hybridized orbitals? Why aren’t the normal s- and p-orbitals sufficient? As we are about
to see, without hybridized orbitals, many of the atomic bonds we chemistry students take
for granted would not exist.
Why do hybrid orbitals exist?
Atomic orbital: A mixed orbital formed by blending together atomic orbitals on the same
atom.
Chemist Linus Pauling developed hybridized orbital theory by studying the structure of
molecules such as methane (CH4). To answer this question, we will follow in his
footsteps.
Molecular orbital theory tells us that the C-H bonds in methane are formed by the overlap
of each hydrogen s-orbital with carbon’s 2s-orbital and px, py, and pz orbitals. But there’s
a problem with this assumption. We know (as Pauling did) that the structure of CH4 is
tetrahedral, and that its true bond angles are 109.5°. Carbon’s p-orbitals are lined up so
that they are perpendicular, at 90° angles from each other. So how are these tetrahedral
s/p-orbital overlaps possible?
Bonding with normal orbitals
would result in unstable 90° angles
between the C-H bonds.
Bonding with hybridized orbitals
results in the correct 109.5°
tetrahedral bond angles.
They aren’t. Atomic p-orbitals are fixed in their perpendicular position. Pauling
concluded that the only way for methane’s C-H bonds to have 109.5° an angle is if the
bonds were hybridized. By blending together its s- and p- orbitals, carbon can break the
restrictions of its perpendicular p-orbitals and expand the angle of its bonds to 109.5°. At
this angle, the atoms and bonds of methane are farthest apart, meaning that it is in the
lowest energy state possible. As with countless concepts in organic chemistry, the driving
force behind hybridization is stability (i.e. minimization of energy). Geometry is a key
factor in molecular stability, so it is important to remember that hybridization is
influenced by geometry, and not the other way around. Without hybridization, the
molecules that are the fundamental building blocks of our world would be much more
unstable.
The number of bonds an atom makes in a molecule determines the amount of hybridized
orbitals it needs. In the case of methane, carbon combines its one s-orbital and its three porbitals into four sp3 orbitals in order to bond to the four hydrogen atoms. The hybridized
orbitals encountered in Chem 14C also include sp and sp2. Their compositions are shown
in the table below.
Table I: Hybrid Orbital Composition
Hybrid Orbital
Orbitals Combined
Resulting Orbitals
sp
s-orbital + 1 p-orbital
2 sp orbitals + 2 p-orbitals
sp2
s-orbital + 2 p-orbitals
3 sp2 orbitals + 1 p-orbital
sp3
s-orbital + 3 p-orbitals
4 sp3 orbitals (no p-orbitals)
It is important to note that the sp3 hybridization leaves no p-orbitals left. This lack of free
p-orbitals has large implications, which will be discussed later.
Now that we have an understanding of how hybridized orbitals function and why they are
necessary, let’s proceed to their application.
Determining Hybridization
How can we tell what hybrid orbitals an atom needs to form the most stable bonds?
Generally, it is as simple as counting the regions of electron density around each atom in
order to determine how many orbitals we need in order to form the appropriate number of
bonds. In the case of methane, we knew that carbon needed four hybridized orbitals in
order to bond to accommodate all four hydrogen atoms; looking at the table above, we
see that only sp3 hybridization gives us the necessary amount of hybrid orbitals to form
four bonds. A second table condensing this information to a glance has been compiled
below.
Table II: General Hybridization Scheme
Number of Regions of
Electron Density
Hybridization
Shape
2
sp
Linear
3
sp2
Trigonal planar
4
sp3
Tetrahedral
5
dsp3 *
Trigonal pyramidal
6
d2sp3 *
Octahedral
*Atoms that are able to accommodate an expanded octet, such as silicon, phosphorus, and sulfur, hybridize
their d-orbitals along with their s- and - orbitals. You probably will not work with these hybridizations in
14C, but it is worth knowing.
Procedure for determining hybridization of an atom:
Step 1: Determine the number of regions of electron density around the atom.
Regions of electron density are either bonds or lone pairs. Double and triple
bonds each count as just one region.
Step 2: Match the number of regions with the corresponding hybridization.
Step 3: If hybridization is determined to be sp3, look for any possible resonance that
signifies that the atom is in fact sp2-hybridized (this step will be covered in depth in the
next section.)
Example #1:
Determine the hybridization of the indicated atoms.
This tutorial assumes that you are familiar with the use and interpretation of line
structures.
Solution:
Looking at the oxygen atom, we see that is has a double bond and two lone pairs.
Therefore we count three regions of electron density around the oxygen atom
(remember, a double bond counts as just one region), and, using the table, conclude
that the oxygen atom is sp2 hybridized. Looking at carbon now, there seem to be just
two attachments, but we should recognize that this is a line structure and there are two
hydrogen bonds not shown. Taking this into account, we count 4 regions of electron
density and conclude carbon is sp3 hybridized. The chlorine has 3 lone pairs and one
bond. 3 + 1 = 4, and we conclude that chlorine is also sp3 hybridized. The next step
would be to check for resonance, but we will discuss this in detail next. This molecule,
for the record, does not have resonance, so these hybridizations are correct.
A note on good practice: If you are given a question on the hybridization of an atom but
you are not given a line structure for it, draw a Lewis structure. You may be able to
contemplate the hybridization mentally, but if there are other things at work such as
resonance, your surest bet is to draw the structure.
We have the general method down now, but you should keep reading. In order to master
hybridization, we must know how to recognize when an “sp3” hybridized atom
participates in resonance.
Hybridization and Resonance
Resonance: a condition in which one Lewis structure is not sufficient to represent a
molecule; multiple Lewis structures are needed.2
Hybridization’s crucial exception is this: when an atom that appears to be sp3 hybridized
can participate in resonance, that atom is in fact sp2 hybridized. This is because resonance
requires an open p-orbital to occur. Sp3 atoms have 4 sp3-hybridized orbitals but no open
p-orbitals, and therefore cannot participate in resonance. Sp2-hybridized atoms can. Why
is it important for resonance to occur? Again, the answer lies in stability. Structures with
resonance are more stable than similar structures without resonance, so whenever an
atom has an opportunity to free up a p-orbital to accommodate resonance, it will do so.
How often will this exception come up? Pretty often. In fact, almost continuously.
Mastery of important Chem 14C topics such as conjugation and aromaticity depend on
your ability to discern an sp2 atom from an sp3 one. Let us work through an example.
2
This definition is derived from Dr. H’s Illustrated Glossary of Organic Chemistry.
Example 2:
Determine the hybridization of the indicated atoms.
Solution: The upper oxygen atom has three attached regions of electron density, and
we conclude that it is sp2 hybridized. Turning to the lower oxygen, we count four
regions of electron density, and conclude that it is sp3 hybridized. But before we leave
this as our final answer, we need to investigate the possibility that this oxygen can
participate in resonance.
A note on good practice: Drawing curve arrows is very helpful when drawing
resonance contributors.
Drawing this molecule’s only other significant resonance contributor shows that this
oxygen definitely participates in resonance. In order for this resonance to occur, this
oxygen atom must have an open p-orbital, and we therefore conclude that it is sp2
hybridized.
A note on good practice: When looking for other resonance structures, keep in mind
that definite sp3 orbitals (such as a carbon with four bonds) prevent resonance from
occurring through them due to their lack of a p-orbital. P-orbitals are like bridges that
electrons use as a path for resonance (delocalization). If there is no bridge, the path
cannot be traveled.
Practice Problems
Determine the hybridization of the indicated atoms. To be consistent with the problems
encountered in 14C, lone pairs and most hydrogen atoms are not shown.
1.
2.
3.
5.
7.
4.
6.
8.
9.
10.
11.
12.
13.
14.
Practice Problem Solutions
Circled in red are certain sp3 atoms that prevent resonance from occurring.
Many resonance contributors involve a negative formal charge on carbon.
1.
sp3
2. sp2 – resonanceabled
sp2
sp3
3.
4.
sp3
sp2 –
resonance
-abled
sp2
5. 6. sp3
sp
7.
sp2 – resonance-abled
9.
sp2 – resonance-abled
8.
sp2 – resonance-abled
sp3
10. sp3
sp3
2
sp – resonance-abled
Lone pairs are surrounded by
sp3 atoms, no available porbitals for resonance.
11. sp3
sp2 –
resonanceabled
sp3
13.
12. sp3
sp
All these atoms have lone pairs,
but nowhere to move them. Verify
this by trying to use curved arrows
to make something happen,
without breaking the octet rule.
Highly electronegative
atoms such as chlorine
are still resonance
participators.
sp2 – resonanceabled
3
14.
sp3
sp2 – resonanceabled
The carbon in red is still
The carbon with the + sign
sp3.
has three attachments,
only
and an open p-orbital.