Download Crystal Field Theory

Crystal Field Theory
Method used to explain some physical properties that occur in transition metal complexes. This involves a simple
electrostatic argument which can yield reasonable results and predictions about the d orbital interactions in metal
complexes. Consider the different orbitals:
Note the orientation of the orbitals
relative to the axes.
m+
Consider metal ion, M , lying at the centre of an octahedral set of point charges.
3+
4+
Suppose the metal atom has a single d electron outside of the closed shells (Ti or V ). In the free ion, the
electron can be in any one of the 5 orbitals, since all are equivalent. Wrong! To explain this, we need to look at the
2
2 2
shape of the d orbitals again. The dxy, dxz and dyz lie between the axes and are equivalent whereas dz and dx - y lies
on the plane of the axes and are equivalent to each other. As a result, 2 set of orbitals: dxy , dyz , dzx belong to t2g
2
2 2
and dz , dx - y to eg.
eg
barycentre
0.4 Δo
Δo
0.6 Δo
t 2g
Δo is the difference in energy between eg and t2g. The net energy of a t2gx egy configuration relative to the
barycentre is called the ligand field stabilization energy (LFSE):
LFSE = (0.4x – 0.6y)Δo
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Let us see what happens when we withdraw the 2 trans ligands in an Oh complex (z ligands).When this happens,
m+
we have a tetragonally distorted octahedral complex. As soon as the distance from M to these 2 ligands
2
becomes greater than the other 4 ligands, new energy differences are established. z orbital becomes more stable
2 2
than x -y orbital. yz and xz are equivalent more stable than xy
dx2-y2
eg
dxy
Δo
E
dz2
t2g
dzy , dzx
II
II
II
Square complexes of Co , Ni and Cu lead to energy level diagrams shown as follows:
dx2-y2
eg
Δo exactly
Δo
2/5 Δo
t2g
dz2
1/12 Δo
octahedral
MX6
dyz , dzx
square
MX4
High- Spin vs Low- Spin in Oh complexes
d 1, d 2, d 3 – simple
d4
High-spin
High- spin d 4: t2g3 eg1: x = 3 , y = 1: E = (0.4x – 0.6y)Δo = 0.6 Δo
4
4
0
Low- spin d : t2g eg : x = 4 , y = 0: E = (0.4x – 0.6y)Δo = 1.6 Δo + P
2
Low-spin
3
Example 1: What is the LFSE for octahedral ions of the following configurations: (a) d (b) high-spin d
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Solution:
Exercise for the Idle Mind
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What is LFSE for both high- and low-spin d configuration?
The spectrochemical series
The splitting of d orbitals in the CF model not only depends on the geometry of the complex, it also depends on
the nature of the metal ion, the charge on this ion and the ligands that surround this ion. When the geometry and
the ligands are held constant, this splitting decreases in the following order:
4+
3+
3+
3+
3+
3+
2+
2+
2+
2+
Pt > Ir > Rh > Co > Cr > Fe > Fe > Co > Ni > Mn
When the geometry and the metal are held constant, the splitting of the d- orbitals increases in the following
order:
-
-
-
-
-
-
-
I < Br < [NCS] < Cl < F < OH < H2O < NH3 < en < CN < CO
The ligand- field splitting parameter, Δo varies with the identity of the ligand. In the series of complexes
n+
[CoX(NH3)5] with X = I , Br , Cl H20 and NH3, the colours range from purple (for X = I ) through pink (X = Cl ) to
yellow (with NH3). This observation indicates that energy of the lowest electronic transition increases as the
ligands are varied along the series. Ligands that give rise to high energy transition (such as CO) is referred to as a
strong-field ligand. Ligands that give rise to low energy transitions (such as Br-) referred to as weak-field ligand.
Magnetic measurements
Used to determine the number of unpaired spins in a complex, hence identify its ground-state configuration.
Compounds are classified as diamagnetic if they are repelled by a magnetic field and paramagnetic if they are
accepted by a magnetic field. The spin-only magnetic moment, μ, of a complex with total spin quantum number is
given by:
μ = 2 {S (S + 1)}½ μB
μB = Bohr magneton.
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Calculated spin-only magnetic moments
Ion
N
S
μ / μB
Calc.
Expt
3+
1
½
1.73
1.7-1.8
3+
2
1
2.83
2.7-2.9
3
1½
3.87
3.8
4
2
4.90
4.8-4.9
5
2½
5.92
5.9
Ti
V
Cr
3+
3+
Mn
3+
Fe
Example 2: The magnetic moment of a certain Co(II) complex is 4.0 μB . What is its d- electron configuration?
Solution
Exercise for the Idle Mind
4-
The magnetic moment of the complex [Mn(NCS)6] is 6.06 μB. What is its electron configuration?
4