Chemistry Reference
In-Depth Information
should be thought of as the orientation of the orbital and the length of the vector as its size.
The original p
x
and p
y
vectors are aligned with their respective axes. Figure 4.9a shows
that, for an arbitrary rotation of the p
x
orbital by an angle
, the new vector p
x
will not
usually be perfectly aligned with either the original p
x
or p
y
vectors. In fact, only
θ
θ
=
0
or 180
◦
will align p
x
with p
x
to give characters of
+
1 and
−
1, and only
θ
=
90
◦
or 270
◦
will align p
x
with p
y
, either giving a character of 0. For any other values of
we can use a
linear combination of the original basis functions to produce the new p-orbital orientation.
In the BF
3
example,
θ
θ
=
120
◦
, but for now we will continue with the general case and
insert this angle for
at the end.
The general formula for the linear combination can be written:
θ
p
x
=
c
x
1
p
x
+
c
x
2
p
y
(4.16)
where
c
x
1
and
c
x
2
control the amount of the original p
x
and p
y
orbitals used in the trans-
formed function and are coefficients we have to find. In this case
c
x
1
is the character for
the p
x
orbital, and the diagram shows that it represents the component of p
x
along the p
x
direction. This limits the range of characters that are possible for the p
x
orbital to between
−
1, because the component of a vector can never be greater than its magnitude.
The length of the vectors representing p
x
and p
y
are the same and are not affected by the
rotation. With this in mind we can use the trigonometry of the diagram in Figure 4.9a to
find the coeffiecients. Figure 4.9c is a reminder of the general definitions of the sine and
cosine functions; applying these to the right-angle triangle with
1 and
+
θ
marked in Figure 4.9a
gives
c
x
2
p
y
|
c
x
1
|
p
x
|
cos(
θ
)
=
=
c
x
1
and
sin(
θ
)
=
=
c
x
2
(4.17)
|
p
x
|
p
x
|
Here, we have used the convention that vertical lines mean the length or magnitude of a
vector. Equation (4.16) now becomes
p
x
=
p
x
cos(
θ
)
−
p
y
sin(
θ
)
(4.18)
Here the minus sign indicates that the opposite edge of the triangle in Figure 4.9a is in
the opposite direction to
p
y
,the
c
x
2
coefficient in the linear combination must take this into
account. Similar considerations using Figure 4.9b show that the p
y
vector can be written as
p
y
=
p
x
sin(
θ
)
+
p
y
cos(
θ
)
(4.19)
Equations (4.18) and (4.19) can be combined into a single form using matrix algebra:
p
x
p
y
cos(
p
x
p
y
θ
−
θ
)
sin(
)
=
(4.20)
θ
θ
sin(
)
cos(
)
For the particular example of the p-orbitals on B in BF
3
,
θ
is 120
◦
;so:
p
x
p
y
cos( 120)
p
x
p
y
−
−
√
3
p
x
p
y
−
sin( 120)
1
/
2
/
2
√
3
=
=
(4.21)
sin( 120)
cos( 120)
/
2
−
1
/
2
