Download Lecture Notes 9 - La Salle University

Crystal Field Theory,
Electronic Spectra and MO of
Coordination Complexes
Or why I decided to become an
inorganic chemist
or
Ohhh!!! The Colors!!!
Gemstone owe their color from
trace transition-metal ions
• Corundum mineral, Al2O3: Colorless
•
Cr  Al : Ruby
•
Mn  Al: Amethyst
•
Fe  Al: Topaz
•
Ti &Co  Al: Sapphire
• Beryl mineral, Be3 Al 2Si6O18: Colorless
•
Cr  Al : Emerald
•
Fe  Al : Aquamarine
d x -y
2 2
d z2
Do
or
10 Dq
dxy
dyz
dxz
d x -y
2 2
d z2
+ 0.6 Do
or
+ 6 Dq
- 0.4 Do
or
- 4 Dq
dxy
dyz
dxz
Let’s Look at 4 Co 3+ complexes:
Config.
[Co(NH3)6]3+
[Co(NH3)5(OH2
Color of Complex
Absorbs
d6
)]3+
[Co(NH3)5Br]2+
d6
[Co(NH3)5Cl]2+
d6
350-400
400-500
Values are in nm
Greater
d6
520-570
Splitting
600-700
600-650
570-600
OTHER QUESTIONS
So there are two ways to put the electrons
Low Spin
High Spin
Which form for our 4 cobalt(III) complexes?
And why the difference between Cl- and Br-?
R. Tsuchida (1938) noticed a trend in while
looking at a series of Cobalt(III) Complexes.
With the general formula : [Co(NH3)5X]
look at that! The same ones we just looked at….
He arrived a series which illustrates the effect of ligands on Do (10Dq)
He called it:
The Spectrochemical Series
Tsuchida, R. Bull. Chem. Soc. Jpn. 1938, 13, 388
The Spectrochemical Series
Ligand effect on Do :
Small Do
I- < Br- < S2- < Cl- < NO3- < F- < OH- < H2O < CH3CN < NH3 < en <
bpy < phen < NO2- < PPh3 < CN- < CO
Large Do
Or more simply :
X<O<N<C
Metals also effect Do :
Mn2+ < Ni2+ < Co2+ < Fe2+ < V2+ < Fe3+ < Co3+ < Mn4+ < Mo3+ <
Rh3+ < Ru3+ < Pd2+ < Ir3+ < Pt2+
Fe3+ << Ru3+
Ni2+ << Pd2+
Important consequences result!!!
Spectrochemical Series
I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO
Weak field ligands
Small D
Strong field ligands
Large D
[Fe(H2O)6]3+
[Ni(H2O)6]2+
[Co(H2O)6]2+
[Zn(H2O)6]2+
[Cu(H2O)6]2+
S=5/2
S=5/2
S=1/2
S=1
S=2
Spectrochemical Series
Another important question arises:
How does filling electrons into orbitals effect the stability
(energy) of the d-orbitals relative to a spherical environment
where they are degenerate?
We use something called Crystal Field Stabilization Energy
(CFSE) to answer these questions
For a t2gx egy configuration : CFSE = (-0.4 · x + 0.6 · y)Do
So Lets take walk along the d-block…….and calculate the CFSE
d1 config. [t2g1]:
S=1/2
CFSE = –0.4 Do
d2 config. [t2g2]:
S=1
CFSE = –0.8 Do
d3
config. [t2g3]:
S=3/2
CFSE = -1.2 Do
BUT WHEN YOU GET TO:
d4
THERE ARE TWO OPTIONS!!!!!
Low Spin
High Spin
CFSE = -1.6 Do + P
CFSE = -0.6 Do
When is one preferred over the other ?????
It depends. (P  14,900 cm-1 / e- pair)
P = Do
both are equally stabilized
P > Do
high spin (weak field) stabilized
NOTE: the text uses the symbol P, for spin pairing energy
P < Do
low spin (weak field) stabilized
P , Spin Pairing Energy is composed of two terms
(a) The coulombic repulsion –
This repulsion must be overcome when forcing electrons to occupy
the same orbital. As 5-d orbitals are more diffuse than 4-d
orbitals which are more diffuse than 3-d orbitals, the pairing
energy becomes smaller as you go down a period. As a rule 4d
and 5d transition metal complexes are generally low spin!
(b) The loss of exchange energy –
The exchange energy (Hünd’s Rule) is proportional to the number of
electrons having parallel spins. The greater this number, the more
difficult it becomes to pair electrons. Therefore, d5 (Fe3+ , Mn2+)
configurations are most likely to form high spin complexes.
Pairing energy for gaseous 3d metal ions
M2+
P (cm-1)
M3+
P (cm-1)
d4
Cr2+
23,500
Mn3+
28,000
d5
Mn2+
25,500
Fe3+
30,000
d6
Fe2+
17,600
Co3+
21,000
d7
Co2+
22,500
Ni3+
27,000
Pairing energies in complexes are likely to be 15-30% lower, due to covalency in
the metal-ligand bond.
These values are on average 22% too high.
C. K. Jørgensen’s f and g factors
Do = f (ligand) · g (metal)
Do in 1000 cm-1 (Kkiesers)
g factors
f factors
3d5 Mn(II)
8.0
Br -
0.72
3d8 Ni (II)
8.7
SCN -
0.73
3d7 Co(II)
9.0
Cl -
0.78
12.0
N3 -
0.83
F-
0.90
3d3 V(II)
3d5 Fe(III) 14.0
3d3 Cr(III) 17.4
oxalate2-
0.99
3d6 Co(III) 18.2
H2O
1.00
3d9 Cu(II)
9.5
NCS -
1.02
3d4 Cr(II)
9.5
CH3CN
1.22
4d6 Ru(II)
20.0
pyridine
1.23
NH3
1.25
3d3 Mn(IV) 23.0
3d3 Mo(III) 24.6
en (ethylenediamine)
1.28
4d6 Rh(III) 27.0
bipy (2,2’-bipyridine)
1.33
4d3 Tc(IV) 30.0
5d6 Ir(III)
32.0
5d6 Pt(IV)
36.0
Phen (1:10-phenanthroline)
CN -
1.34
1.70
Note: Rh3+ and Ir3+ are a lot different than Co3+
g3d < g4d ≤ g5d
EXAMPLE:
Calculate the Do (10Dq) for [Rh(OH2)6]3+ in cm-1 and nm.
for [Rh(pyr)3Cl3]
Tetrahedral Coordination
Dt = 4/9Do
All tetrahedral compounds are
High Spin
Why do d8 metal compounds often form square planar compounds
z
Thought experiment: Make a square planar
compound by removing two ligands from an
octahedral compound
L
L
L
M
L
y
L
L
dx2-y2
dz2
x
L
L
M
L
L
dx2-y2
dxy
dxy
dxz,dyz
dz2
dxz,dyz
Octahedral
Square Planar
dx2-y2
dx2-y2
dz2
dxy
dxz,dyz
dxy
dxy
dxz,dyz
dx2-y2
dz 2
H2O
H2O
Ni
dxz,dyz
Tetrahedral
Octahedral
OH2
dz 2
2
OH2
Cl
OH2
H2O
Octahedral
Coordination number =6
Ni(II) d8 S = 1
Cl
Square Planar
2-
N
N
Cl
C
C
Ni
2-
Ni
Cl
Tetrahedral (CN=4)
C
N
C
N
Square Planar (CN=4)
Ni(II) d8 S =1
Ni(II) d8 S = 0
The Energy Levels of d-orbitals in Crystal Fields of Different Symmetries