250
H Solutions to Problems
4.4 The tangential π -orbitals transform as Γ π in the C 5v site group of I h . According to Sect. C.2, one has:
Γ π C 5v ↑ I h = T 1g + T 1u + G g + G u + H g + H u
4.5 When the projector that generated the component is characterized as ˆ
P
Γ i
kl ,the
other components may be found by varying the k index.
4.6 Act with an operator ˆ
S on the projector and carry out the substitution ˆ
R =
ˆ
S −1 ˆ
T :
ˆ
S ˆ
P
Γ 0
11 = ˆ
S
1
|G|
R
ˆ
R
=
1
|G|
R
ˆ
S ˆ
R =
1
|G|
T
ˆ
T = ˆ
P
Γ 0
11
4.7 Applying the inverse transformation to the SALCs of the hydrogens in ammonia yields
|sp 2
A | sp 2
B | sp 2
C
=
|2s| 2p x | 2p y
⎛
⎜
⎜
⎝
1
√
3
1
√
3
1
√
3
2
√
6
−
1
√
6
−
1
√
6
0
1
√
2
−
1
√
2
⎞
⎟
⎟
⎠
4.8 This mode transforms as E y . It can be written as a linear combination of a
radial and a tangent mode:
Q =
−1
√
2
Q
rad
y +
1
√
2
Q
tan
y
with
Q
rad
y =
1
√
2
((R B − R C )
Q
tan
y =
1
√
6
R(2φ A − φ B − φ C )
This mode preserves the center of mass and is a genuine normal mode.