Chemistry Reference
In-Depth Information
A10.6 H
2
+
: the Electron Kinetic Energy
The kinetic energy for an electron in the
|
1
σ
g
+
bonding orbital will be given by
∞
∞
1
2
σ
g
+
|∇
σ
g
+
=
T
1
σ
g
+
=−
1
2
|
1
N
1g
2
(
s
1
+
s
2
)[
∇
2
(
s
1
+
s
2
)]
u
d
u
d
z
(A10.22)
π
−∞
0
And with the electron in the antibonding
|
2
σ
u
orbital we would have
∞
∞
1
2
T
2
σ
u
+
=−
2
σ
u
+
|∇
2
|
2
σ
u
+
=
N
2u
2
(
s
1
−
s
2
)[
∇
2
(
s
1
−
s
2
)]
u
d
u
d
z
(A10.23)
π
−∞
0
Now, we are in cylindrical polar coordinates and so must use the appropriate form of the
Laplacian:
u
∂
∂
1
u
∂
∂
u
2
∂
1
2
∂
2
2
∇
=
+
2
+
(A10.24)
u
u
∂φ
∂
z
2
We have chosen cylindrical polar coordinates to exploit the cylindrical symmetry of the
H
2
+
molecular ion. In the application of the Laplacian there will be no variation in the
wavefunctions with the angular coordinate, so the differential with respect to
will give
zero and so only the derivatives with respect to
u
and
z
need be considered. The results still
have to be integrated with respect to the
φ
, it is just that the same value of the integrand will
occur at any point on a circle centred on the molecular axis and in a plane perpendicular to
it. This has already been exploited in Equations (A10.22) and (A10.23) with the integration
φ
φ
as before. The full detail of applying the Laplacian to the molecular orbitals is
set out in the
Mathematica
sheet accompanying this appendix available from the Website.
The kinetic energy integrated over planes perpendicular to the molecular axis is plot-
ted in Figure A10.5. The bonding orbital (Figure A10.5a) shows a lower kinetic energy
giving 2
π
(a)
T
/Ha
(b)
T
/Ha
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
H
1
H
2
H
1
H
2
-3
-2
-1
1
2
3
-3
-2
-1
1
2
3
z
/
bohr
z
/
bohr
Figure A10.5
Plot of the kinetic energy integrated over planes perpendicular to the molecu-
lar axis against the z coordinate for (a)
. The dashed lines in each case
show the same data for an isolated H atom placed at the H
1
position.
|
1
σ
g
+
and (b)
|
2
σ
u
+
