6.8 Application: Linear and Circular Dichroism
141
The three vectors of the ligand positions can be expressed in a row notation for the
primed x ′ ,y ′ ,z ′ coordinate system as:
R A = ρ(1, 0, 0)
R B = ρ
−
1
2
,
√
3
2
, 0
R C = ρ
−
1
2
, −
√
3
2
, 0
(6.93)
The transfer term then becomes:
µ
e ǫ (t 2g ) → ψ
A
=−eκR A = κµ A
(6.94)
where the parameter κ is an overlap factor which indicates what fraction of the
charge is actually transferred:
κ =−
e ǫ (t 2g )|H|ψ A
E ψ − E t 2g
=−
√
2H π
E ψ − E t 2g
(6.95)
Note that the transfer term is always polarized in the direction of the transferred
charge.
This parametrization can now be used to calculate the transfer term for the relevant trigonal transitions. The Hamiltonian operator is of course totally symmetric, so
allowed interactions can take place only between orbitals with the same symmetry,
and are independent of the component; hence:
e ǫ (t 2g )|H|e ǫ (ψ)
=
√
3H π
e θ (t 2g )|H|e θ (ψ)
=
√
3H π
(6.96)
Symmetry prevents interaction between the a 1 (t 2g ) and a 2 (ψ) orbitals. The
metal-ligand π acceptor interaction will thus stabilize the e-component of the t 2g
shell, while leaving the a 1 -orbital in place, as shown in the simple orbital-energy
diagram in the left panel of Fig. 6.6. We can now calculate the transfer term for the
e → e and e → a 2 orbital transitions. In each case only one component needs to be
calculated. The interaction element in this case is obtained from Eq. (6.96) and the
transfer fraction reads:
−
√
3H π
E ψ − E t 2g
=
3
2
κ
(6.97)
The transfer-dipole element is given by:
1
6
2ψ
A − ψ
B − ψ
C |µ|2ψ
A − ψ
B − ψ
C
=
1
6
[4µ A + µ B + µ C ]=
1
2
µ A (6.98)
141
The three vectors of the ligand positions can be expressed in a row notation for the
primed x ′ ,y ′ ,z ′ coordinate system as:
R A = ρ(1, 0, 0)
R B = ρ
−
1
2
,
√
3
2
, 0
R C = ρ
−
1
2
, −
√
3
2
, 0
(6.93)
The transfer term then becomes:
µ
e ǫ (t 2g ) → ψ
A
=−eκR A = κµ A
(6.94)
where the parameter κ is an overlap factor which indicates what fraction of the
charge is actually transferred:
κ =−
e ǫ (t 2g )|H|ψ A
E ψ − E t 2g
=−
√
2H π
E ψ − E t 2g
(6.95)
Note that the transfer term is always polarized in the direction of the transferred
charge.
This parametrization can now be used to calculate the transfer term for the relevant trigonal transitions. The Hamiltonian operator is of course totally symmetric, so
allowed interactions can take place only between orbitals with the same symmetry,
and are independent of the component; hence:
e ǫ (t 2g )|H|e ǫ (ψ)
=
√
3H π
e θ (t 2g )|H|e θ (ψ)
=
√
3H π
(6.96)
Symmetry prevents interaction between the a 1 (t 2g ) and a 2 (ψ) orbitals. The
metal-ligand π acceptor interaction will thus stabilize the e-component of the t 2g
shell, while leaving the a 1 -orbital in place, as shown in the simple orbital-energy
diagram in the left panel of Fig. 6.6. We can now calculate the transfer term for the
e → e and e → a 2 orbital transitions. In each case only one component needs to be
calculated. The interaction element in this case is obtained from Eq. (6.96) and the
transfer fraction reads:
−
√
3H π
E ψ − E t 2g
=
3
2
κ
(6.97)
The transfer-dipole element is given by:
1
6
2ψ
A − ψ
B − ψ
C |µ|2ψ
A − ψ
B − ψ
C
=
1
6
[4µ A + µ B + µ C ]=
1
2
µ A (6.98)