5.04, Principles of Inorganic Chemistry II

Lecture 17: AOM for MLn Ligand Fields

Consider a ligand positioned arbritrarily about the metal

z

M

�

�

We can imagine placing the ligand on the metal z axis

(with x and y axes of M and L also aligned) and then L

rotate it on the surface of a sphere (thus maintaining (xL, yL, zL) M-L distance) to its final coordinate position.

y

x

Within the reference frame of the ligand:

z1

zL

xL

zL

yL yLyL z2 xL

R(�) R(�)

rock rotate y2about Z about Z

x y1

x1 x2

related by a coordinate

transformation

SML in complex SML (� and �) = 1

z

y

�

�

zL

xL

2 z

yz

xz

xy

2 2 x -y

F (�, �)

2 2 2 z2 y2z2 x2z2 x2y2 x2 -y2

1

4 1 + 3cos2�( ) 0 �

3

2 sin2� 0 3

4 1 � cos2�( )

3 sin� sin2� cos�cos� sin�cos2� -cos�sin� �

1 sin� sin2�

2 2

3 cos� sin2� -sin�cos� cos�cos2� sin�sin� �

1 cos� sin2�

2 2

3

4 sin2� 1 � cos2�( ) cos2�sin�

1

2 sin2� sin2� cos2�cos�

1

4 sin2� 3 + cos2�( )

3

4 cos2� 1 � cos2�( ) -sin2�sin�

1

2 cos2� sin2� -sin2�cos�

1

4 cos2� 3 + cos2�( )

5.04, Principles of Inorganic Chemistry II Lecture 17 Page 1 of 5

MIT Department of Chemistry

For ligands in an octahedral complex

Ligand 1 2 3 4 5 6

y � �

0

0

90

0

90

90

90

180

90

270

180° 0

1

4

3

6 2

5

z

x

For dz2 for L

y2z2 �

2

dz

2

= 1

(1+ 3cos2�)dz2

2 + 0d 3

sin2�dx2z2 � d 2

� �� 1 0 0 0 3� � dz

2 �

4 2 � z � � 2 2 � � 2 � � dyz � � 0 0 0 �1 0 � � dy2z2

�

+ 0dx2y2 +

4

3 (1� cos2�) +

2

3 d

x22 � y2

2 ��

dxz ��

= �� 0 0 �1 0 0

�� �� dx2z2

��

� dxy � � 0 1 0 0 0 � � dx y�

2 2

2 2 2= � 1

d 2 + 0dy2z2 + 0dx2z2

+ 0dx2y2 +

3d 2 2 �

�dx � y �� ��

23 0 0 0

21 �� ��� d

x2 � y2 ���

2 z2 2 x2 � y2 2

continuing other elements of the transformation

matrix may be obtained

2 1 3 1 3 Ed 2 = �SML = d 2 + d 2 2 = �� + ��z

4 z2 4 x2 � y2 4 4

Following the same procedure…. Other transformation matrices of Li:

for other d-orbitals, E(di) is…

E dyz �

-1 3�( ) = e

� � � 0 0 0 1 0 0 0 0 � 2 2 �� � E dxz ) = e

� �0 1 0 0 0 � � 0 0 -1 0 0 �( � �

E dxy) = e �

L1: �� 0 0 1 0 0 �

� L

3: � 0 0 0 1 0

�( �0 0 0 1 0 � � 0 -1 0 0 0 �

( 2 2 ) = 3

�

1 �

� � �� 2

3 1

2

��E d e + e �0 0 0 0 1 � � 0 0 0 -

x � y 4 4

� -1 0 0 0 3� �

-1 0 0 0 3� �1 0 0 0 0 � � 2 2 � � 2 2 � � �

� 0 0 0 1 0 � � 0 0 1 0 0 � �0 -1 0 0 0 � L

4: � 0 0 1 0 0

� L5: � 0 0 0 -1 0

� L6: �0 0 1 0 0 � � � � � � �

� 0 1 0 0 0 � � 0 -1 0 0 0 � �0 0 0 -1 0 � � 3 � � 3 1 � � � � 2

0 0 0 12 � �� 2

0 0 0 -2 � �0 0 0 0 1 �

5.04, Principles of Inorganic Chemistry II Lecture 17 Page 2 of 5

Squaring the coefficients for each of the ligands:

L1 L2 L3 L4 L5 L6 ETOTAL

1 3 1 3 1 3 1 3 E(d 2) e� e

� + e � e

� + e � e

� + e � e

� + e � e� = 3e� + 3e�

z 4 4 4 4 4 4 4 4

E(dyz) e� e� e� e� e� e� = 4e� + 2e� E(dxz) e� e� e� e� e� e� = 4e� + 2e� E(dxy) e� e� e� e� e� e� = 4e� + 2e�

3 1 3 1 3 1 3 1E(dx2 � y2) e� e

� + e � e

� + e � e

� + e � e

� + e � e� = 3e� + 3e�

4 4 4 4 4 4 4 4

As mentioned above… e�<<e� or e�… thus e� may be ignored. The Oh energy level

diagram is:

�-donor �-donor �-acceptor

M-L�*

4e ' �

9 9 other L�

M-L�* M-L�* M-L�*

2 2 2 2 2 2 2 2 2eg (dx -y , dz ) eg (dx -y , dz )eg (dx -y , dz )

3e3e �� M-L�* 3e� t2g (dxy, dxz, dyz)

n.b. 4e� ndnd t2g (dxy, dxz, dyz) nd

4e�

t2g (dxy, dxz, dyz)

M-L�

6L�6L� 6L�

3e� ' 3e ' 3e ' � �

M-L� egM-L� egM-L� eg

Note the d-orbital splitting is the same result obtained from the CFT model taught

in Freshman chemistry. In fact the energy parameterization scales directly between

CFT and AOM

10 Dq = �0 = 3e� - 4e�

5.04, Principles of Inorganic Chemistry II Lecture 17 Page 3 of 5

A table of angular scaling factors for e and e has been removed due to � �

copyright considerations.

A list of common structures for two through six

ligands has been removed due to copyright

considerations.

5.04, Principles of Inorganic Chemistry II Lecture 17 Page 4 of 5

Let’s use e� and e� parameters to determine the d-energy level splitting diagram for

Td complexes

M-L�* M-L�* M-L�* M-L�

t2 (dxy, dxz, dyz) t2 (dxy, dxz, dyz)

M-L�* 4 8 4 8

t2 (dxy, dxz, dyz) e + e e + e� �3 9 3 � 9 �

M-L�* M-L�* 4

2 2 2 2 2 2e (dx -y , dz )� e (dx -y , dz )3 e

8 8en.b. e� �3 3nd nde (dx2-y

2, dz

2) nd

8L��

M-L�� t2

t1

8 9

e�

8

3 e�

e

8L�8 e�9

8 e� t23 M-L�

e

nbnb nb4L� 4L� 4L�

44 e ' e� ' 3

� 3

M-L� t2 M-L� t2 M-L� t2

Note that � �

�Td

= 4

e � + 8

e � � 8

e � = 4

e � � 16

e � = 4

(3e � � 4�) = 4 �

0

� 3 9 � 3 3 9 9 9�

This is the result from CFT

(inversion of e/t2 leads to

the (-) sign)

�

4 �T

d = � �

0

9

5.04, Principles of Inorganic Chemistry II Lecture 17 Page 5 of 5