Chemistry Reference
In-Depth Information
in a position to understand how these orbitals interact with the ligand orbitals to give MOs
with which to describe bonding in these complexes. We will concentrate on complexes in
which the metal-ligand interaction is that of a
-donor ligand, i.e. in the isolated ligand
the orbitals would be filled and the metal is sufficiently electronegative to receive charge.
σ
O
h
Symmetry Complexes
Octahedral complexes have six ligands equidistant from a metal centre at the corners
of an octahedron. In this class we include hexaaqua complexes such as [Ni(H
2
O)
6
]
2
+
shown in Figure 3.32a or [Mn(H
2
O)
6
]
2
+
in Figure 5.19a. The
-donor orbitals from the
H
2
O molecules can be thought of as the 1
b
1
levels in the MO diagram for isolated H
2
O
(Figure 7.26). These are the highest occupied states for H
2
O and so will most readily
donate electron density to vacant orbitals on the metal centre. Formally, the symmetry of
the complex is lower than
O
h
, since the H atoms of the H
2
O cannot be arranged to conform
to all of the point group operations. However, even at low temperature, the H
2
O molecule
orientation around metal-ligand bond will be randomized by thermal motion. For example,
the H
2
O molecules will be able to spin around the metal-ligand axis without disturbing the
σ
σ
-donor interaction with the metal.
A simplified basis of just the
-donor orbitals from H
2
O is shown in Figure 7.38, along
with some example symmetry axes from the
O
h
rotational classes. Appendix 12 gives the
rotational subgroup given for
O
h
as
O
.
σ
C
4
,
C
2
(
=
2
C
4
)
C
3
s
5
σ
h
s
2
s
3
C
2
s
1
s
4
C
4
,
C
2
C
2
C
4
,
C
2
s
6
Figure 7.38
The basis of
-donor orbitals for a six-coordinate complex in O
h
symmetry. The
ligand donor orbitals are represented by the s
1
σ
s
6
basis placed on the coordinate axis system.
Also shown are some examples of the symmetry axes present and one of the horizontial mirror
planes. The full set of operations is illustrated in Figure 3.28.
−
Problem 7.14:
The reducible representation for the six
-donor orbitals of Figure 7.38
in the
O
rotational subgroup is given in Table 7.13. Adding up the total number of
operations for
O
from the character table in Appendix 12 gives an order
h
σ
=
24. Using
the reduction formula (Equation (7.33)), show that
=
a
1
+
e
+
t
1
(7.68)
