Written by tutor Michael M.
The process by which we determine orbitals is a bit complex and relates to solutions of the Schrodinger equation. Luckily,
you won’t deal with this in an introductory chemistry class. Rather, we mostly focus on understanding the concepts of orbitals.
Essentially, these orbitals tell us the probability of finding an electron in a particular location around the nucleus.
Hybridization occurs when we combine two atomic orbitals to form hybrid orbitals. Hybridization helps us overcome the
problem of how molecules bond with one another. Bonding
solely with s and p orbitals would not necessarily yield the best spatial orientation for our compound. Atoms
like to be as far apart from each other to avoid electron repulsion.
We can relate hybridization to VSEPR
for simplicity; however they are not implicitly the same. To help with classifying though, you can approach molecules by looking at
the shape and think of hybridization from there. I will mention different VSEPR shapes that correspond to desired bond angles for hybridization.
In order to understand why an orbital will engage in hybridization, we need to first look at the electron configuration.
For a simple tetrahedral compound, such as CH4 we know that carbon’s ground state electron configuration is
1s22s22p2 or, written another way, [He]2s22p2.
We can write this out as:
The s orbital is spherical and can only hold two electrons. In contrast, we have three p orbitals which are lobed and are
aligned along each of the major three dimensional axes; x, y and z. Each p orbital can also hold a maximum of two electrons.
With the way the above set is written, we have a pairing of electrons in the s orbital and two lone electrons in two of the p
orbitals. The idea is that the p orbitals are of higher energy than the s orbitals, so we place the electrons in the s orbital before
moving to the p orbital. Since we can only engage in bonding with unpaired electrons, this would predict that carbon can only form two
bonds, one for each electron in two of the p orbitals. We know that this is not true, since carbon forms four bonds in CH4, not two.
We can also see that these p orbitals are not fully occupied, thus they can take up more electrons than the number currently assigned
to them. This means that we could combine (hybridize) the s and p orbitals to give us orbitals than can yield better bond angles, as
well as increase our number of unpaired electrons, allowing us to bind more atoms.
In the case of a tetrahedral compound, we want the bond distances to be equal at 109.5 degrees. To achieve this we would want
equivalent bonding orbitals. We can combine the s orbitals with all three p orbitals, since they are not filled and are not
involved in bonding. This would give us an sp3 orbital. The nomenclature is designed to give us a fraction designating
relative amounts of orbital character. In the case of sp3, our orbital is ¼ s character and ¾ p character. This hybrid
orbital has properties of both orbitals. Now we have four sp3 orbitals of equivalent energy levels, giving us the following energy diagram:
We have four unpaired electrons so now we can bind with four atoms. It is common for tetrahedral compounds to have sp3 hybrid orbitals.
For compounds that have double or triple bonds, we cannot get the same types of hybridization since now we have our p orbitals engaged in
pi bonding. Let’s consider O2.
Here we have a compound with a double bond. This means that we have one sigma bond and one pi bond. Since one of our 3 p orbitals is
involved in bonding, we only have 2 p orbitals left to engage in hybridization. This gives us sp2 hybrid orbitals.
Referring back to our nomenclature, this means that the hybrid orbital is 1/3 s character and 2/3 p character. There are also
cases where we can get sp2 hybrid orbitals in the absence of pi bonds. This normally occurs with compounds that have bond
angles that are 120 degrees apart. This means that trigonal planar compounds should have sp2 hybridization.
This concept can be applied to molecules with triple bonds as well, such as N2. Here we have two p orbitals involved in pi
bonding, leaving only one p orbital to hybridize with the s orbital. Therefore, the hybrid bond is sp. There are also cases where we
can get sp hybrid orbitals in the absence of pi bonds. This normally occurs with compounds that for bond angles that are 180 degrees
apart. This means that linear compounds should have sp hybridization.
We can get other types of hybridization as well. For some atoms, we can utilize the d orbitals to expand beyond our octet rule, and use
hybridization to yield hybrid orbitals such as dsp3 and d2sp3. The easiest way to determine when we
reach these types of hybridizations is to look at the molecular geometry of the compound. Trigonal bipyramidal compounds will exhibit
dsp3 hybridization, whereas octahedral compounds will exhibit d2sp3.
Note: There is currently debate about the contribution of the d orbitals in hybridization. Many traditional introductory chemistry
classes still teach it this way, however new scientific information may change the way this is taught in the future.
*Remember to take into account the spatial orientation of the lone electron pairs and not just the atoms when deciding the shapes of a
compound. This will help in determining the correct hybrid orbital.
*When determining the hybridization of an atom, look for molecular geometry and the presence of pi bonds.
Summary and VSEPR relation:
|Shape||Angle (Degrees)||Common Hybridization|
|Trigonal bipyramidal||90 and 120||dsp3|
What is the hybridization of beryllium in BeH2?
What is the hybridization of carbon in CO?
Carbon forms hybrid orbitals with fluorine in CF3. How much s character do these hybrid orbitals contain?
What is the hybridization of xenon in XeF2?
Zumdahl, Steven. Chemical Principles. Sixth. Boston, MA: Houghton Mifflin Company, 2009. Print.