Chapter 5 Crystal field theory

CHAPTER 5: CRYSTAL FIELD THEORY
RECALL
The elements in the periodic table are often divided into four categories:
(1) main group elements, (2) transition metals, (3) lanthanides, and
(4) actinides.
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TRANSITION METALS vs. MAIN-GROUP ELEMENTS
There is some controversy about the classification of the elements on the
boundary between the main group and transition-metal elements on the
right side of the table.
The elements in question are zinc (Zn), cadmium (Cd), and mercury (Hg).
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THE ELECTRON CONFIGURATION OF TRANSITION-METAL IONS
The relationship between the electron configurations of transition-metal
elements and their ions is complex.
EXAMPLE
Consider the chemistry of cobalt which forms complexes that contain
either Co2+ or Co3+ ions.
Co: [Ar] 4s2 3d7
Co2+: [Ar] 3d7
Co3+: [Ar] 3d6
In general, electrons are removed from the valence shell s orbitals before
they are removed from valence d orbitals when transition metals are ionized.
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THE ORIGIN OF COLOUR - ABSORPTION
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The colour can change depending on a number of factors e.g.
• Metal charge
• Ligand
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Are there any simple theories to explain the colours in transition metal
complexes?
There is a simple electrostatic model used by chemists to rationalize the
observed results
THIS THEORY IS CALLED CRYSTAL FIELD THEORY
It is NOT A RIGOROUS BONDING THEORY but merely a simplistic
approach to understanding the possible origins of photo- and
electrochemical properties of the transition metal complexes.
Method of explaining some physical properties that occur in transition metal
complexes.
Involves a simple electrostatic argument which can yield reasonable results
and predictions about the d orbital interactions in metal complexes.
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THE OCTAHEDRAL CRYSTAL FIELD
Consider metal ion, Mm+, lying at the centre of an octahedral set of point
charges.
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Suppose the metal atom has a single d electron outside of the closed shells
(Ti3+ or V4+)
In the free ion, the electron can be in any one of the 5 orbitals, since all are
equivalent (degenerate).
Recall the shapes of the d orbitals
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2 groups of orbitals
dxy , dyz , dzx
t2g
dz2 , dx2- y2
eg
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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
HIGH- SPIN VS LOW- SPIN IN Oh COMPLEXES
d4
d 1, d 2, d 3 - simple
high- spin
low- spin
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High-spin d 4
Low-spin d 4
t2g3 eg1
t2g4 eg0
x=3,y=1
x=4,y=0
E = (0.4x – 0.6y)Δo = 0.6 Δo
E = (0.4x – 0.6y)Δo = 1.6 Δo + P
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EXAMPLE
What is the LFSE for octahedral ions of the following configurations:
(a) d 3
(b) high-spin d 5
SOLUTION
(a) electronic configuration : t2g3eg0, x = 3, y = 0
Therefore, LFSE = (0.4x – 0.6y)Δo = [(0.4)(3) – (0.6)(0)]Δo = 1.2 Δo
(b) electronic configuration : t2g3eg2, x = 3, y = 2
Therefore, LFSE = (0.4x – 0.6y)Δo = [(0.4)(3) – (0.6)(2)]Δo = 0
EXERCISE FOR THE IDLE MIND
What is LFSE for both high- and low-spin d 6 configuration?
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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:
Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+
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 [CoX(NH3)5]n+ with X = I-, Br-, Cl- H20 and NH3,
the colours range from purple (for X = I-) through pink (X = Cl-) to yellow
(with NH3).
Ligand that give rise to high energy transition (such as CO) is referred to as a
strong-field ligand; low energy transitions (such as Br-) referred to as weakfield ligand.
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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 =
𝑆 𝑆 + 1 𝜇𝐵
μB = Bohr magneton
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CALCULATED SPIN-ONLY MAGNETIC MOMENTS
ION
N
μ/μB
S
CALC.
EXPT.
Ti3+
1
½
1.73
1.7-1.8
V3+
2
1
2.83
2.7-2.9
Cr3+
3
1½
3.87
3.8
Mn3+
4
2
4.90
4.8-4.9
Fe3+
5
2½
5.92
5.9
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EXAMPLE
The magnetic moment of a certain Co(II) complex is 4.0 μB . What is its delectron configuration?
SOLUTION
A Co(II) complex is d 7.
Two possible configurations: t2g5eg2 (high-spin, S = 1½) with 3 unpaired
electrons or t2g6eg1 (Low-spin, S = ½) with 1 unpaired electron.
The spin-only magnetic moments are 3.87 μB and 1.73 μB.
Therefore, the only consistent assignment is the high-spin configuration
t2g5eg2.
EXERCISE FOR THE IDLE MIND
The magnetic moment of the complex [Mn(NCS)6]4- is 6.06 μB. What is its
electron configuration?
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