Chemistry Reference
In-Depth Information
6. The polarization of an MO will favour the energetically nearest AO reference orbital.
In bonding orbitals, this usually favours the more electronegative element.
7. When two reference orbitals match in symmetry but differ greatly in energy, the degree
of polarization means that the MO will be practically nonbonding. For example, the
core states of an atom may match in symmetry with the valence levels of a neighbour,
but the energies will be very different.
8. Reference orbitals that do not match any of the irreducible representations for the other
atom or set of atoms are nonbonding by symmetry.
Along the way we have also covered the idea of hybridization and taken the rather strict
approach that only orbitals of the same symmetry should take part in hybridization. This
makes it much easier to follow the symmetry arguments for building bonding/antibonding
orbitals. We have also seen how the idea of
-symmetry bonding derived from
diatomic molecules in the
D
∞
h
point group is generally extended to molecules in other
point groups, and even to metal complexes. In a similar way, the idea of hybridization to
discuss structure is often used outside of the restrictions imposed by symmetry.
σ
and
π
7.9 Self-Test Questions
1. In Table 7.3, both BO and NO have a longer bond length and lower bond dissoci-
ation energy than CO. Explain this observation based on the MO diagram for CO
(Figure 7.32).
2. The PES for CH
4
shown in Figure 7.21 shows two distinct peaks and a shoulder. To
explain this it is useful to consider the final state of the cation formed after ionization.
The electronic structure of the CH
4
+
cation in
T
d
geometry would contain a
t
2
level
with five electrons; show that distortion to
D
2d
symmetry will lower the system energy.
Explain how this structural distortion gives rise to the observed multiple peaks through
alternative arrangements of the electrons in the final state.
3. The ionization potential for the 3s state of Cl is 25.3 eV and that for the 3p is 13.7 eV.
Using this information, construct an MO diagram for Cl
2
. Table 7.3 gives the bond
length of Cl
2
as 1.988 Å and the bond dissociation energy as 239 kJ mol
−
1
. The molecu-
lar cation Cl
2
+
actually has a shorter bond (1.8917 Å) and a bond dissociation energy of
415 kJ mol
−
1
. Explain the differences between Cl
2
and Cl
2
+
based on your MO diagram.
4. The six p-orbitals that are perpendicular to the plane of benzene (
D
6h
) are the basis for
the
π
-symmetry orbitals of the molecule.
(a) Derive a reducible representation for this basis and apply the reduction formula to
obtain the irreducible representations for the
-symmetry six MOs.
(b) Apply the projection operator method to obtain the pattern of AOs in each of the
MOs from (a).
(c) Based on your sketches, estimate the relative energy of the MOs and, hence, obtain
the MO diagram for the
π
-symmetry orbitals of benzene. From your diagram you
should be able to explain why the HOMO state of benzene is doubly degenerate.
π
5. Derive the MO diagram for a tetrahedral metal complex with
-donor ligands. For the
ligand SALCs, reference to the case of methane (in Section 7.4) may be useful here;
σ
