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To demonstrate why it is important to take the number of antibonding
electrons into account in our
bond order calculation, let us consider the possibility of making a
molecule of He2. An
orbital correlation diagram for He2 is provided in :
Figure %: An orbital correlation diagram for a hypothetical He-He molecule
From the orbital correlation diagram above you should
notice that the amount of
stabilization due to bonding is equal to the amount of destabilization due
to antibonding, because there
are two electrons in the bonding orbital and two electrons in the
antibonding orbital. Therefore, there
is no net stabilization due to bonding so the He2 molecule will
not exist. The bond
order calculation shows that there will be a bond order of zero for the
He2 molecule--exactly what we should predict given that helium is a
noble gas and does
not form covalent
compounds.
Both hydrogen and helium only have 1s atomic orbitals so they produce very
simple correlation
diagrams. However, we have already developed the techniques necessary to
draw a correlation
diagram for a more complex homonuclear diatomic like diboron,
B2. Before we can
draw a correlation diagram for B2, we must first find the
in-phase and out-of-phase
overlap combinations for boron's atomic orbitals. Then, we rank them
in order of increasing
energy. Each boron atom has one 2s and three 2p valence orbitals. Due to
the great difference in
energy between the 2s and 2p orbitals, we can ignore the overlap of these
orbitals with each other.
All orbitals composed primarily of the 2s orbitals will be
lower in energy than those
comprised of the 2p orbitals. shows the process of
creating the molecular
orbitals for diboron by combining orbitals of atomic boron. Note that the
orbitals of lowest energy have the most
constructive overlap (fewest
nodes) and the orbitals with the highest energy have the most destructive
overlap (most nodes).
Figure %: The molecular orbitals of diboron
Notice that there are two different kinds of overlap for p-orbitals--end-on
and side-on types of overlap.
For the p-orbitals, there is one end-on overlap possible which occurs
between the two
pz. Two side-on overlaps are possible--one between the two
px and one
between the two p y. P-orbitals overlapping end-on create s bonds. When p-orbitals bond in a side-on fashion,
they create p bonds. The difference between a p bond and a
s bond is the symmetry of the molecular orbital
produced. s bonds are cylindrically symmetric about the bonding
axis, the z-direction.
That means one can rotate the s bond about the
z-axis and the bond
remains the same. In contrast, p bonds lack that
cylindrical symmetry
and have a node passing through the bonding axis.
Now that we have determined the energy levels for B2, let's draw
the orbital correlation
diagram ():
Figure %: Orbital correlation diagram for diboron
The orbital correlation diagram for diboron, however, is not generally
applicable for all homonuclear
diatomic molecules. It turns out that only when the bond lengths are relatively
short (as in B2,
C2, and N2) can the two p-orbitals on the bonded
atoms efficiently
overlap to form a strong p bond. Some textbooks
explain this
observation in terms of a concept called s-p mixing. For any atom with an
atomic number greater
than seven, the p bond is less stable and higher in
energy than is the s bond formed by the two end-on overlapping p orbitals.
Therefore, the
following orbital correlation diagram for fluorine is representative of all
homonuclear
diatomic molecules with atomic numbers greater than seven.