20.4 Heteronuclear Diatomic Molecules 859
Qualitative Description of Bonding in Molecules
The results of the Hartree–Fock–Roothaan treatment of lithium hydride conform to a
general pattern that enables us to make qualitative predictions about other molecules.
We will assume the following factors in making these predictions:1.Two atomic orbitals on different nuclei must have roughly equal orbital energies to
form a good bonding LCAOMO.If LCAOMOs are made from two atomic orbitals of
moderately different energies, a bonding LCAOMO and an antibonding LCAOMO
will result. The bonding LCAOMO will have a coefficient of larger magnitude for the
atomic orbital of lower energy, and the antibonding LCAOMO will have a coefficient
of larger magnitude for the higher-energy atomic orbital. If the atomic orbitals have
equal energies, the coefficients of the two atomic orbitals will have equal magnitudes
in both LCAOMOs. If the energies are greatly different, the lower-energy LCAOMO
will be almost the same as the lower-energy atomic orbital, making it nearly a
nonbonding orbital, like the 1slithium orbital in LiH.
2.Two atomic orbitals on different nuclei must have a fairly large overlap region to
form a good bonding LCAOMO.This rule is related to the fact that effective bonding
electrons must have a fairly large probability of being found between the nuclei in
order to attract both nuclei and stabilize the bond.
3.Two atomic orbitals on different nuclei must have the same symmetry around the
bond axis to form a good bonding LCAOMO.In a diatomic molecule, two orbitals
with different symmetry cannot form an eigenfunction of the appropriate symmetry
operators. We will assume the same behavior in a polyatomic molecule.We can relate the results of the lithium hydride calculation to this pattern. The
1 slithium orbital does not form a good bonding LCAOMO with the hydrogen 1sorbital
because the difference in energies is large and because the overlap region of these two
orbitals is small. The lithium 1sorbital therefore acts as a nonbonding orbital. The 2sp(1)
hybrid has a large overlap region with the hydrogen 1sorbital and forms a good bonding
orbital with it. The 2sp(2) orbital does not have a large overlap region with the hydrogen
1 sorbital and constitutes a nonbonding orbital that is vacant in the ground state.EXAMPLE20.12
Describe the bonding in HF. Include a statement about the polarity of the molecule.
Solution
The fluorine 1sorbital acts as a nonbonding orbital. The fluorine 2pxand 2pyatomic orbitals
do not have the same symmetry around thezaxis (the bond axis), as does the hydrogen 1s
orbital, and are not included in the bonding molecular orbital. They will act as nonbonding
“lone-pair” orbitals. We make two hybrid orbitals with the fluorine 2sand 2pzorbitals,
denoted by 2sp(1) and 2sp(2). The 2sp(1) orbital forms a bonding orbital called the 1σ
orbital with the hydrogen 1sorbital. There is also an antibonding orbital, which we denote
by 2σ* but which remains vacant in the ground state. The 2sp(2) orbital acts as a nonbonding
lone-pair orbital. The energy of the fluorine 2sp(1) hybrid is lower than that of the hydrogen 1s
orbital so the fluorine 2sp(1) atomic orbital will have a coefficient of larger magnitude in the
bonding molecular orbital than will the hydrogen 1sorbital. The corresponding antibonding
orbital (which remains vacant in the ground state) will have a coefficient of larger magnitude
for the hydrogen 2sorbital. The molecule will be polar with the fluorine end negative.