Chemistry Reference
In-Depth Information
Term 3, 4
N
1
g
2
s
1
|
V
1
|
s
2
The electron density described by the overlap
integral
will have a Coulomb interaction with centre 1; this will be
attractive. The factor of 4 arises from an identical contribution
of the
s
1
|
s
2
density with centre 1 and both overlaps with cen-
tre 2. Figure A10.6e and f shows the integral and the overlap
density respectively for the bonding MO. The overlap builds
up charge between the centres, but this charge is spread wider
than the bond separation and so contributions to this term also
occur further out. The 'shielding' of the internuclear interac-
tion is really the sum of these terms and the nucleus-nucleus
potential given by Equation (A10.26), and so this may be less
effective than we might expect. In the energy integrals there
are, again, sharp turning points at the nuclear centre used in the
potential operator.
s
2
|
s
1
4
N
2u
2
−
s
1
|
V
1
|
s
2
In the antibonding orbital the overlap integrals correspond to
the loss of electron density from the internuclear region and a
minus sign appears for this contribution in Equation (A10.28).
The integral still evaluates to a negative value, and so this
becomes a positive term as plotted in Figure A10.8c. Once
again, the difference in the normalization constant means that
this term has a greater magnitude than that for the bonding
orbital.
An important check on the potential energy contributions suggested by Equa-
tions (A10.29) and (A10.30) is to consider what happens as
R
12
becomes large, i.e. we
break the molecular bond. In the limit of separated, noninteracting, nuclei the overlap
S
12
must become zero, and so term 3 vanishes, because it gives the interaction of the density
described by
S
12
with the nuclei. With no overlap we also obtain
N
1g
2
1
2
from
Equations (A10.18), and so term 1 becomes zero. Term 2 involves an interaction between
centre 1 and the remote charge density on atom 2, which will also be vanishingly small due
to the inverse power dependence of Coulomb's law. So, at the limit of large
R
12
there is no
difference between the potential energies of either MO and an isolated H atom infinitely
separated from H
+
, as required.
However, the H + H
+
system contains two different species and so cannot conform to
D
∞
h
symmetry. At some point, as the bond is stretched, the electron must decide 'which
way to jump', and this will destroy the molecular symmetry. By sticking to
D
∞
h
we are
forced to have a dissociated state of two H
0.5
+
ions. This has the same energy as the 'real'
reactants, but is not a correct physical picture.
The contributions to the total potential energy of the H
2
+
cation as a function of the
internuclear distance
R
12
with the electron in the ground-state MO,
N
2u
2
=
=
σ
g
+
,areshownin
Figure A10.9a. In the bonding orbital, the total potential energy is always positive and
so tends to favour dissociation of the molecule into its constituent atom and ion at all
internuclear separations.
In contrast, Figure A10.9b shows that if we place the electron in the first excited state,
|
1
|
2
σ
u
, then the total potential energy appears to be negative when this antibonding MO is
