Chemistry Reference
In-Depth Information
The sketches use the same atom layout as Figure 7.19, and so it should be clear that they
each complement a single C(2p) AO.
Problem 7.8:
Show that the three SALCs found in Equations (7.51) and (7.52) are
orthogonal to one another.
The simple projection of the
s
1
orbital in Table 7.8 does result in a function that could
be taken as one of an alternative set of three SALCs. However, we decided to abandon it
because it was not suitably aligned to bond to a single C(2p) orbital in the axis system of
Figure 7.19. Now we can see that it is related to the three
t
2
SALCs found by deliberately
constructing linear combinations aligned with the axis system at the bottom of Table 7.8.
Inspection of the formulae shows that
φ
1
(
t
)
+
φ
2
(
t
)
+
φ
3
(
t
)
=
3
s
1
−
s
2
−
s
3
−
s
4
(7.53)
i.e. the first function found is a linear combination of the final three SALCs and so is not
orthogonal to them. The three basic functions we have found are well suited to the MO
construction because they are aligned with the axes that are used to define the direction of
the C(2p) orbitals.
The MO diagram for CH
4
is shown in Figure 7.20; in this instance, C has a higher elec-
tronegativity than H (Table 7.4) and so its AOs are drawn lower than the H(1s) SALCs.
The lowest energy MO formed from the valence AOs is 2
a
1
, which is a linear combination
of C(2s) and the totally symmetric H(1s) SALC. This is lower in energy than the
t
2
MOs
2
t
2
3
a
1
t
2
a
1
2p
x
2p
y
2p
z
t
2
1
t
2
2s (
a
1
)
2
a
1
C
H, SALCs
CH
4
Figure 7.20
The MO diagram for CH
4
in the T
d
point group.
