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In a heteronuclear diatomic molecule, there is an electronegativity
difference between the atoms, which leads to an asymmetric distribution
of the electronic probability density, weighted more heavily toward
the element with the greater electronegativity.
Consider constructing a MO from two 2s orbitals for nuclei with
different values. Let atom A have an atomic number
and atom B have an atomic number . For a single
electron interacting with the two nuclei, an LCAO guess wave
function could be
Recall the energies and that were defined
for LCAO in the homonuclear case:
For the heteronuclear case,
. If atom B is
the more electronegative, then its will be larger (in the second
period), and the energy will be more negative due to the large
Coulomb attraction between the electrons and the nuclei.
Thus, in this case
(remember
and are both negative). Now, if we calculate
the guess of the groundstate energy
we find the more general result
and then perform the minimization of :
we obtain the conditions
If we make the simplifying approximation that the overlap , then
these conditions lead to the following relationship between and :
In order to make the analysis a bit easier, let us expand
the square root using the fact that
Then
Now, we know that
First, look at the
.
If , is more electronegative because its energy is
lower (think of the Bohr formula
, which
shows that as increases, the energy decreases). Generally,
a not so easy calculation of shows that .
Thus, if , then
and this means that or , which is what we would
expect if is more electronegative. Similarly, if ,
is more electronegative. We would then find
and , and , which is what we would expect if
is more electronegative.
Another consequence we can derive from the above result is that
if
or
,
then the coefficients will be very different, e.g.
or , in which case, the
resulting MO is not really an MO at all, but rather
more like the AO with the larger coefficient. Thus, it is
clear that and cannot be very different
if we are to have a reasonable amount of mixing of the
two AOs. This supports the idea that only orbitals with
similar energies can be combined in the LCAO scheme.
We can no longer use the ``g'' and ``u'' designators because the
orbitals have no particular symmetry when
.
That is
Thus, we denote
simply as and
simply as .
Let us now construct a correlation diagram for the heteronuclear
diatomic BO. Boron has an electronic configuration
while oxygen's is
Since we are interested in the chemical bond that forms between them,
we only consider the valence electrons explicitly, and these are the
electrons in the shell. In BO, oxygen is the more electronegative,
so its orbitals are lower in energy than those of boron. This must be
indicated on the correlation diagram. Thus, the correlation diagram
appears as in the figure below:
Figure:
Correlation diagram for the heteronuclear
diatomic molecule BO.

Note that the ordering of the MOs follows the pattern we would
expect for boron rather than oxygen. Only highlevel calculations
can predict this, but physically, the simple explanation is that
there is only one electron in the orbital, and
only one of the two atoms has a large nuclear charge (unlike in
O, where they both do). There is, therefore, insufficient
Coulomb attraction for this one electron to pull the energy
of the orbital below the energy of the
bonding orbitals.
The electronic configuration of BO is, therefore
and the bond order is (1/2)(72)=5/2. The fraction bond order
indicates that the molecule is paramagnetic, as with O.
As another example, consider the molecule NO. NO has 11 valence
electrons and has the electronic configuration:
The bond order is also 5/2, and the molecule is paramagnetic as well.
What about the molecule HF? Here, the 1s and 2s orbitals of F are so low
in energy compared to the 1s orbital in H that they cannot be combined
to form MOs. At the same time, the and orbitals
of F have an insignificant spatial overlap with the 1s orbital in H
(assuming that the two nuclei lie along the axis) that they
also do not form MOs. Only the orbital of F has significant
overlap with the 1s orbital in H and can mix with it energetically. Thus,
the LCAO guess wave function takes the form
where atom A is H and atom B is F. If H lies to the left of F,
then this orbital has antibonding character, while the orbital
has bonding character because there is significant amplitude
in the region between the nuclei. See the figure below:
Figure:
Illustration of the overlap between the
2s and 2p orbitals of F with the 1s of H in the molecule HF.

Figure:
MOs and correlation diagram for HF.

The orbital is denoted simply as while
the orbital is denoted for bonding
and antibonding, respectively. The 2s, and
orbitals of F do not mix with anything and are, therefore,
called nonbonding orbitals. They are denoted
and
and
,
respectively. The orbital ordering is
(since it is a lowenergy 2s orbital),
followed by , then the two nonbonding
orbitals and finally . Therefore, the correlation
diagram for HF is as shown in the figure below:
Figure:
Another look at the correlation diagram in HF.

The bond order of HF is easily seen to be 1.
Next: Frontier molecular orbitals
Up: lecture_15
Previous: lecture_15
Mark E. Tuckerman
20111121