Bonding in Transition Metal Complexes
Crystal Field Theory
1) Crystal Field Theory (ligand field theory)
Treat Ligands as negative charges
(they repel the e- in the d orbitals
deals only with d orbitals
2) Molecular orbital theory
Mn+
The six negative charges are equally distributed
in a sphere around the metal
Bonding in Coordination Compounds
Isolated
transition metal
atom
Bonded
transition metal
atom
Crystal field splitting ( Δ) is the energy difference between
two sets of d orbitals in a metal atom when ligands are present
S=5/2
S=5/2
S=1/2
S=2
S=1
S = 1/2
S = 5/2
S=1
S=2
S=0
S=2
S = 3/2
S = 1/2
Bonding in Coordination Compounds
Magnetic Susceptibility
Δo
Unpaired electrons:
Paramagnetic compounds
attracted to magnetic field
“weigh more”
Paired electrons:
Diamagnetic compounds
repelled by magnetic field
“weigh less”
Δο = hν
Bonding in Coordination Compounds
Spectrochemical Series
Spectrochemical Series
I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO
Weak field ligands
Small Δ
Strong field ligands
Large Δ
Sc
Y
La
3
Ti V
Zr Nb
Hf Ta
4 5
Cr Mn
Mo Tc
W Re
6 7
Fe Co Ni Cu Zn
Ru Rh Pd Ag Cd
Os Ir Pt Au Hg
8 9 10 11 12
Sc
Y
La
3
Valence electron count for neutral metal
Quiz:
A. 1
Sc
Y
La
3
B. 2
C. 3
Ti V
Zr Nb
Hf Ta
4 5
D. 4
Cr Mn
Mo Tc
W Re
6 7
Quiz:
F. 6
G. 7
H. 8
Fe Co Ni Cu Zn
Ru Rh Pd Ag Cd
Os Ir Pt Au Hg
8 9 10 11 12
Valence electron count for neutral metal
Quiz:
Fe Co Ni Cu Zn
Ru Rh Pd Ag Cd
Os Ir Pt Au Hg
8 9 10 11 12
How many d electrons in?
Ti(H2O)6+4
d electrons = ?
E. 5
Cr Mn
Mo Tc
W Re
6 7
Valence electron count for neutral metal
How many d electrons in?
Fe(H2O)6+2
Ti V
Zr Nb
Hf Ta
4 5
A. 1
Sc
Y
La
3
B. 2
C. 3
Ti V
Zr Nb
Hf Ta
4 5
d electrons = ?
D. 4
E. 5
Cr Mn
Mo Tc
W Re
6 7
F. 6
G. 7
H. 8
Fe Co Ni Cu Zn
Ru Rh Pd Ag Cd
Os Ir Pt Au Hg
8 9 10 11 12
Quiz: What is spin state? S = ?
How many d electrons in?
Co(NH3)4Cl2+1
d electrons = ?
Fe(H2O)6+2
I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO
A. 1
Sc
Y
La
3
B. 2
C. 3
Ti V
Zr Nb
Hf Ta
4 5
D. 4
E. 5
Cr Mn
Mo Tc
W Re
6 7
F. 6
G. 7
H. 8
Fe Co Ni Cu Zn
Ru Rh Pd Ag Cd
Os Ir Pt Au Hg
8 9 10 11 12
Quiz: What is spin state? S = ?
A. 1/2
B. 1
Sc
Y
La
3
Ti V
Zr Nb
Hf Ta
4 5
C. 3/2
D. 2
Cr Mn
Mo Tc
W Re
6 7
E. 5/2
F. 3
Fe Co Ni Cu Zn
Ru Rh Pd Ag Cd
Os Ir Pt Au Hg
8 9 10 11 12
Quiz: What is spin state? S = ?
Fe(CN)6-3
Co(NH3)6+3
low spin.
I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO
I- < Br- < Cl- < OH- < F- < H2O < NH3 < en < CN- < CO
A. 1/2
A. 1/2
B. 1
C. 3/2
D. 2
E. 5/2
F. 3
B. 1
C. 3/2
D. 2
E. 5/2
F. 3
Tetrahedral Coordination
Δt = 4/9Δo
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
x
L
L
M
Square Planar
Complexes
L
L
trans Pt(NH3)2Cl2
dx2-y2
dx2-y2
dz 2
dxy
dxy
dxz,dyz
dz2
[Pt(NH3)4]2+
dxz,dyz
Octahedral
Square Planar
H2O
H2O
OH2
Ni
2
OH2
OH2
H2O
Octahedral
Coordination number =6
Ni(II) d8 S = 1
Cl
Cl
2-
Ni
Co(I) Ni(II)
Pd(II)
Rh(I)
Pt(II)
Ir(I)
cis Pt(NH3)2Cl2
Pt2+ d8
Cl
N
N
C
C
Ni
C
N
C
N
Square Planar (CN=4)
Ni(II) d8 S = 0
Cl
Tetrahedral (CN=4)
Ni(II) d8 S =1
2-
dxy
dxz,dyz
dx2-y2
dz2
OH2
Ni
2
OH2
Ni(II) d8 S = 1
n+
L
L
4s
dxz,dyz
Square Planar
3d
Cl
OH2
H2O
Octahedral
Coordination number =6
L
L
M
L
L
dz 2
Tetrahedral
Octahedral
H2O
4p
dxy
dxy
dxz,dyz
H2O
Molecular Orbital Theory for ML6
dx2-y2
dx2-y2
dz 2
Cl
2-
N
N
Cl
C
C
Ni
2-
Ni
Cl
Tetrahedral (CN=4)
C
N
N
Square Planar (CN=4)
Ni(II) d8 S =1
Ni(II) d8 S = 0
σ antibonding
4p
salcs
C
Metal
ligands
The 4s an 4p orbitals is much more diffuse
in space than are the 3d orbitals. They have
a higher quantum number n and the have
fewer nodes (0 or 1) versus 2 for the d orbitals.
4s
This means that when a transition metal
atom bonds to other atoms, the largest
interactions are with the s and
p orbitals, not the d orbitals.
3d
salcs
σ bonding
dxy, dxz, dxy can’t over lap with ligand orbitals
σ Antibonding
eg*
y
y
nonbonding
3d
x
x
t2g
dxy
dxy
dxy, dxz, dxy become nonbonding
nonbonding
σ bonding
salcs
eg
σ bonding between the d orbitals and the ligand orbitals
is less than that between the ligand orbitals and the s and p
orbitals on the metals
[Co( NH3)6]
4p
4s
4p
4s
eg
3+
Co3+ d6
2 x 6L = 12eTotal 18e-
eg*
Δo
3d
t2g
3d
salcs
t2g
salcs
6 M-L σ bonding
MO’s (12 e-)
pi bonding
+
4p
dxy
4s
3d
eg
pi*
-
pi antibonding
t2g
pi
salcs
sigma
Other combinations
are non bonding.
Nodes don’t match
4p
4s
4p
4s
eg
3d
eg
3d
t2g
t2g
pi
pi
salcs
salcs
sigma
sigma
4p
4s
4p
eg
3d
no pi
eg
3d
t2g
pi*
t2g
salcs
salcs
sigma
sigma
Quiz. For each of the following show the d orbital electron
occupancies. Count d electrons? What is the spin, S?
Spectrochemical Series
pi
4s
pi*
pi antibonding
Fe(Cl)6-3
A. 0 B. 1/2
Mn(CN)6-4
C. 1
Cu(Cl)4-2
D. 3/2
E. 2
Pd(CN)4-2
F. 5/2
G. 3
The following complexes are not colored?
Why not?
Ti(H2O)6+4
Zn(H2O)6+2
Mn(H2O)6+2