Lewis structures and VSEPR are useful tools for predicting the shape of a molecule or ion, but
they really do not provide any information about the bonds that exist between the atoms; they
do not tell us why covalent bonds form nor do they describe what happens to the atomic
orbitals when the bond forms.
Valence bond theory, based on quantum mechanics, is used to describe the formation of
covalent bonds. Previously, we have discussed the formation of ionic bonds by showing the
transfer of electrons from one species (the metal) to another (the non-metal) – so see – there was
a point to that!! Now we will examine the sharing of electrons between two or more species.
The valence bond theory assumes that the each bond is localized, that is, each bond connects
two atoms together, just like we draw them in the Lewis structure. The valence bond theory
assumes that the atomic orbitals do not change during the bond formation – instead, they
simply overlap as the atoms approach one another. Electron density is higher in the region of
overlap, and the negative charge between the electrons in the overlapping orbitals and the
positive nuclei attracts and holds the two atoms together. We talked about covalent bond
formation and the distance between the two atoms in a bond must be one that balances out
repulsion of the nuclei, repulsion of the electron clouds, and the attraction of the nuclei with the
electrons. In general, the more extensive the overlap between the two orbitals, the stronger the
covalent bond.
As per the Pauli Exclusion Principle:
A covalent bond is formed by the pairing of a maximum of two electrons with opposite spins in
the overlap region of the atomic orbitals from the two atoms.
Some important points to remember about valence bond theory:
Most of the electrons in a molecule remain in the same orbital location that they occupied
in the separated atoms
Bonding electrons are localized in the region where the atomic orbitals overlap; the
greatest probability of finding the bonding electrons is in this region of overlap
For orbitals with directional lobes (the p and d and f orbitals), the maximum overlap
occurs when the atomic orbitals overlap end to end, along the axes of the orbitals. Recall
that the s orbital is a sphere and has no directional lobes.
The molecular geometry depends on the geometric relationships among the orbitals of
the central atom that participate in bonding. Thus, the valence bond theory combined
with VSEPR theory will account for the true geometric shape of the molecule.
The number of hybrid orbitals that results must equal the number of atomic orbitals that
are mixed together
The type of hybrid orbitals obtained varies depending on the types of atomic orbitals that
are mixed (do we get sp, sp2, sp3 etc. . . )
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What is hybridization?
In order to account for bonding, the valence bond theory is based on the hybridization of the
bonding orbitals. Hybridization is an imaginary mixing process in which the orbitals of an
atom rearrange to form new atomic orbitals called hybrid orbitals. In forming the hybrid
orbitals, if possible, shared pairs of electrons will be promoted so that all electrons are unpaired.
The promotion will occur between the s, p, d, and f orbitals in order to maximize the number of
unpaired electrons.
Electron promotion accounts for the bonding between certain atoms to be identical in energy.
Why, for instance, are all bonds in the CH4 molecule identical? The ground state electron
configuration for C is [He]2s22p2. It is easy to see where two bonds come from – by interaction
with the 2p electrons with the 1s1 electrons from H. But why are there 4 bonds and why are
they all equal? The 2s2 electrons are paired and at a different energy level than the 2p electrons.
Linus Pauling answered the question by stating that the overlapping atomic orbitals are actually
different in character than the atomic orbitals in the individual atoms. Quantum mechanical
calculations show that the mixing of atomic orbitals gives rise to new atomic orbitals. The
spatial organization of the overlapping orbitals leads to a more stable situation and leads to
consistency in assigning shape.
A single bond consists of two electrons of opposite spins, thus the two atomic orbitals that give
rise to a single bond can have no more than two electrons total in the overlap region. Many
times, each atom contributes an electron, but there are examples of single bonds bring formed
where one atom donates both electrons to the bond. When the atomic orbitals overlap to build
up the electron density between the two nuclei, the resulting localized bond is called a sigma
bond (, lower case Greek s).
Steps for forming hybrid orbitals:
1.)
2.)
3.)
4.)
Examine the central atom and its valence electrons.
Draw the orbital box diagrams for the valence electrons
Promote the electrons from the s to the p and from the p to the d if empty orbitals exist
This is the hybridization state that the central atom is in when participating in bonding
Write the electron configuration for C, and identify the valence electrons for C
C: 1s22s22p2
valence electrons are 2s22p2 electrons
Draw the orbital box diagram for the valence electrons
2s
2p
2
However, when hybridizing the orbitals, we will take the paired electrons, in this case an
electron from the 2s and promote it to the 2p. All electrons in the hybridized orbitals are
considered to be the same energy.
2s
2p
Thus, these three orbitals, the 2s and the 2p orbitals come together to make 4 orbitals, all of the
same energy, known as sp3 (notice there is an electron in the s and one electron in each of the 3 p
orbitals)
sp3 Hybridization
Tetrahedral geometry, indicating an AX4 geometry, (or an AX3E, or AX2E2). Notice there are 3 p
orbitals and 1 s orbital coming together. Notice 4 orbitals are available for bonding, and when 4
things – or 4 electron domains surround a central atom we know that we have a tetrahedral
VSEPR geometry.
sp3
Explain, using valence bond theory, why N wants to form 3 bonds and has a lone pair of
electrons.
N: 1s22s22p3
2s
2p
In this case, no electrons need to be promoted, all orbital boxes are filled with at least one
electron. Combining the orbital boxes as they are shows us the lone pair electrons and the 3
electrons that will be used for bonding
sp3
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Explain, using valence bond theory why O wants to form 2 bonds and indicate how many lone
pairs it has
O: 1s22s22p4
2s
2p
Again, in this case, no electrons need to be promoted, and the hybrid shows the lone pairs (2 of
them) and the two electrons available for bonding
sp3
Keep in mind that the hybridization explains the type of bonding that we see with these atoms
as the central atom. It can generally be used to explain the other atoms in covalent bonds, but it
is most often used to describe the central atom. It is also used to explain the type of bonding
that the central atom is participating in that particular molecule. Keep in mind that expanding
the valence shells will mean expanding the hybridization, thus, this method is used to describe
that after the fact, not really in predicting that expansion will occur.
sp2 Hybridization
Occurs when an s orbital and 2 p orbitals are of the same energy. The third p orbital is left
unhybridized (left alone). Examine Boron and its valence electrons.
B: 1s22s22p1
2s
2p
Hybridization occurs with the promotion of one of the paired electrons to an empty p orbital
sp2
2p unhybridized
3 bonding electrons means AX3 geometry which means trigonal planar VSEPR geometry.
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sp Hybridization
Two valence electrons, one s and one p come together having the same energy to participate in
bonding. In this case, examine Beryllium and its valence electrons.
Be: 1s22s2
2s
2p empty
In this case, we have a location to promote an electron to, as all the p orbital boxes are empty.
sp
2p unhybridized
2 bonding electrons means AX2 geometry which means linear VSEPR geometry
Any hybridization scheme that only allows for s and p orbitals means that we are confined by
the octet rule. However, we know that we do not always follow the octet rule when drawing
Lewis structures. Initially, you were just told that the d orbitals were available to participate in
bonding. Using valence bond theory you can determine exactly where the electrons are going
to go in the d subshell. It is also difficult to predict that this would happen unless you knew
that your molecule violated the octet rule. For example, consider Phosphorus.
Writing out the electron configuration for P: 1s22s22p63s23p3 we would predict phosphorus to
behave very similarly to Nitrogen – after all, they are in the same group, they each have 5
valence electrons, but P can have an expanded octet while N cannot. Thus, it is important to
know the molecule that you are dealing with – you cannot just know P and not what is attached,
as PH3 will have a different hybridization than PCl5.
Write out the hybridization for PCl5 knowing that P will have no lone pairs and have 5 electrons
available for bonding.
3s
3p
All the orbital boxes are filled. There are 3 electrons available for bonding, but we need 5
electrons available for bonding in order to be able to make PCl5. If we just promote a 3s electron
up to the 3p then we will just move our pair from the s to the p, we need to unpair the electrons
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as we did for CH4. Knowing that P will be surrounded by 5 bonds, which means 10 electrons,
which indicates we will be violating the octet rule. In order to do that, we will use the d
orbitals, and here is how:
3s
3p
3d
Creating a hybridization scheme like this:
sp3d
d unhybridized orbitals
This sp3d hybridization indicates one s, 3 p’s, and 1d orbital are used for bonding. That is a
total of 5 orbitals for bonding indicating the AX5 geometry or the trigonal bipyramidal VSEPR
geometry.
Finally, what about Sulfur. It is directly underneath oxygen, and will behave in the tetrahedral
manner (e.g. H2S) but what about when sulfur expands its octet making SF6?
Write out the hybridization for Sulfur knowing that it will form 6 bonds.
S: 1s22s22p63s23p4
3s
3p
In this case, S appears to only want to form 2 bonds (thus explaining H2S). But we need to move
the electrons around so that S will form 6 bonds.
3s
3p
3d
6
Hybridization indicates:
sp3d2
d unhybridized
An sp3d2 indicates 6 bonding orbitals for 6 bonding electrons which indicates AX 6 which
indicates octahedral VSEPR geometry.
A fast way to determine hybridization comes from looking at the molecule/Lewis structure and
counting the number of bonds and lone pairs. Once again, double and triple bonds are counted
as one electron domain. Then, when determining the hybridization the number of letters (s,p,
and d) must add up to the number of electron domains. There is a maximum of 1s, 3p’s and 5
d’s as we have seen when drawing orbital box diagrams.
Draw Lewis structures for CH4, H2SO4, BF3, SF6, PCl5, H2O, IF5 and determine hybridization and
shape (VSEPR and molecular) based on the central atom (C, S, B, S, P, O, and I).
When single bonds are formed, a sigma bond results and connects the atoms. When a double or
a triple bond is formed, another type of bond is formed, a pi bond ( bond) is formed. VSEPR
predicts that C2H4 will be trigonal planar in geometry. But it does not tell us how the CH 2
groups are orientated with respect to one another. Valence bond theory states that the atoms
will be orientated in space in order to maximize their orbital overlap.
Are the 2 CH2 groups in the same plane? Are they located perpendicular to one another? Or
are they located at some other angle?
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You can see that the sigma bond lies on the x axis between the two C atoms. This means that
the unhybridized p orbital is perpendicular (90o) to the bonding axis. This allows for pi bond
formation. Thus the 2 carbons are bonded together with 1 sigma bond and 1 pi bond.
Keep in mind that ultimately there are 3 p orbitals, a px, py, and pz and therefore you must be
aware which orbitals are involved in bonding.
For example, consider HCN. Draw a Lewis structure, predict VSEPR shape and hybridization
of the C atom. Then show the pi orbitals. In this case, there is 1 sigma bond and 2 pi bonds.


H 
C




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N
There is another way to classify orbitals in molecules, and it is called Molecular Orbital theory
(MO theory). In contrast to the localized orbitals of the valence bonding theory, MO theory is
based on the idea that pure atomic orbitals of a lone atom will mix and combine in a compound
or molecule to produce orbitals that are delocalized. These new orbitals are then called the
molecular orbitals.
The principle idea to remember from MO theory is that the total number of molecular orbitals is
the same as the total number of atomic orbitals that you add together. If you put two atomic
orbitals in, you will get two molecular orbitals out!!
When atomic orbitals combine, they can directly overlap, termed the bonding molecular orbital.
This is, in effect, addition of atomic orbitals. This is the same as the overlapping sigma bond in
the valence bond theory.
They can also repel one another, not overlap, and in effect, this is subtraction of atomic orbitals.
This is called the antibonding molecular orbital. Antibonding molecular orbitals have no
counterpart in the valence bond theory. So this is a place where the theories diverge.
The idea is that bonding molecular orbitals are lower in energy than either one of the atomic
orbitals that came together and they are also lower in energy than an antibonding orbital. This
makes some sense as we know that species that are bonded together do so because the bonding
situation is lower in energy (remember, nature is lazy!!)
We will talk more of MO theory as the ACS final approaches, for now, just know the above!
Pictures showing the additive and subtractive effects of the MO theory are provided below for
your visual enjoyment.
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