838 20 The Electronic States of Diatomic Molecules
PROBLEMS
Section 20.2: LCAOMOs. Approximate Molecular
Orbitals That Are Linear Combinations of Atomic
Orbitals
20.7 Sketch the orbital region of theσg 3 sLCAOMO.
20.8 Sketch the orbital regions for the six LCAO molecular
orbitals that can be formed from the 3patomic orbitals.
20.9 Sketch the orbital region of the following LCAOMO,
with the bond axis on thezaxis:
ψcAψ 1 sA+cBψ 2 pyB
Explain why this orbital is unusable ifcAandcBare both
nonzero.
20.10Sketch the orbital region of the following LCAOMO:
ψσg 3 pc(ψ 3 pzA+ψ 3 pzB)
20.11By inspection of the orbital regions, predict which
united-atom orbital will result from each of the following
LCAOMOs:
a.πu 2 px
b.πu 2 py
20.12By inspection of the orbital regions, predict which
united-atom orbital will result from each of the following
LCAOMOs:
a.σg 2 pz
b.π∗g 2 py
20.13Show that the normalization constant for theσ∗u 1 s
LCAOMO is
cu
1
√
2 − 2 S
(20.2-16)
20.3 Homonuclear Diatomic Molecules
Homonuclear diatomic molecules have two nuclei of the same element. We base our
discussion of these molecules on the LCAO molecular orbitals of the H+ 2 molecule
ion in much the same way as we based our discussion of multielectron atoms on the
hydrogen-like atomic orbitals.
Nucleus A
rAB
x
y
z
r1A
r2A
r1B
r2B
r 12
Nucleus B
Electron 1
Electron 2
Figure 20.10 The Hydrogen Molecule
System.
The Hydrogen Molecule
Figure 20.10 depicts the hydrogen molecule, consisting of two hydrogen nuclei at
locations A and B and two electrons at locations 1 and 2. The distances between the
particles are labeled in the figure. The Born–Oppenheimer Hamiltonian is
ĤBO−h ̄
2
2 m
(
∇^21 +∇ 22
)
+
e^2
4 πε 0
(
1
rAB
−
1
r1A
−
1
r1B
−
1
r2A
−
1
r2B
+
1
r 12
)
ĤHMI(1)+ĤHMI(2)+
e^2
4 πε 0 r 12
+
e^2
4 πε 0 rAB
Ĥel+
e^2
4 πε 0 rAB
̂Hel+Vnn (20.3-1)
whereĤHMI(1) and̂HHMI(2) are the hydrogen-molecule-ion electronic Hamiltonian
operators denoted byĤelin Eq. (20.1-4). The nucleus–nucleus repulsion termVnnis a
constant in the Born–Oppenheimer approximation, sincerABis fixed. We omit it during
the solution of the electronic Schrödinger equation and add it to the resulting electronic
energy eigenvalue to obtain the Born–Oppenheimer energy, as in Eq. (20.1-6).