Chemistry Reference
In-Depth Information
The symmetry of the molecule means that any integral involving
V
1
will have an equiv-
alent from the
V
2
potential; for example,
. This allows the set of
integrals obtained on multiplying out Equation (A10.24) to be reduced to just three terms:
s
1
|
V
1
|
s
1
=
s
2
|
V
2
|
s
2
σ
g
+
|
σ
g
+
=
2
N
1g
2
(
1
(
V
1
+
V
2
)
|
1
s
1
|
V
1
|
s
1
+
s
2
|
V
1
|
s
2
+
2
s
1
|
V
1
|
s
2
)
(A10.28)
The electronic potential energy of the single H atom on the reactant side of Equa-
tion (A10.1) is
, and so we can write down the potential energy change for the
electron moving from the atomic 1s to the 1
s
1
|
V
1
|
s
1
σ
g
orbital of the cationic molecule as
U
(H
2
+
)
1
σ
g
+
−
=
(2
N
1g
2
−
s
1
|
V
1
|
s
1
+
2
N
1g
2
(
s
2
|
V
1
|
s
2
U
(H)
1)
1
R
12
+
2
s
1
|
V
1
|
s
2
)
+
(A10.29)
This is the potential energy change most relevant to bond formation, as it compares the
molecular ion in its electronic ground state with the atomic reference, which is also an
electronic ground state.
A similar procedure for the antibonding MO gives the energy difference between the
molecular ion in its first excited state and the H atom reference as:
U
(H
2
+
)
(2
N
2u
2
2
σ
u
+
−
U
(H)
=
−
1)
s
1
|
V
1
|
s
1
1
R
12
2
N
2u
2
(
+
s
2
|
V
1
|
s
2
−
2
s
1
|
V
1
|
s
2
)
+
(A10.30)
Comparing results from Equations (A10.29) and (A10.30) would allow us assess the effect
of exciting the molecular ion, e.g. by irradiating with light of a specific frequency.
The first three terms in these expressions give the contributions to the change in the
potential energy of the electron as it moves from the atomic 1s orbital to either of the
molecular orbitals:
Term 1,
(2
N
1g
2
The origin of the integral
is illustrated in the plot of
Figure A10.6b. It involves the interaction between a nucleus
and the part of the electron density described by the associated
s-orbital, which is attractive and so gives a negative result. The
thin solid line in Figure A10.6a shows the integral over planes
plotted as a function of
z
. The corresponding integral for the
atomic
s
1
|
V
1
|
s
1
−
1)
s
1
|
V
1
|
s
1
Equation (A10.18)
says:
N
1g
=
1
√
2( 1
orbitals is shown as a dotted line on the same figure
and is split evenly between the two nuclei for comparison. In the
bonding MO the electron population is lower than in the atomic
state because the electron is now shared with the other nucleus.
This results in the positive difference shown as the thick black
line in Figure A10.6a.
|
s
+
S
12
)
