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 Co 2+ or Co 3+ ions.
Co: [Ar] 4 s 2 3 d 7
Co 2+ : [Ar] 3 d 7
Co 3+ : [Ar] 3 d 6
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 photoand 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, M m+ , 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
(Ti 3+ or V 4+ )
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 d xy
, d yz
, d zx d z
2 , d x
2
- y
2 e t
2g g
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Δ o is the difference in energy between e g and t
2g
.
The net energy of a t
2g x e g y configuration relative to the barycentre is called the ligand field stabilization energy (LFSE).
LFSE = (0.4x – 0.6y)Δ o
HIGH- SPIN VS LOW- SPIN IN Oh COMPLEXES d 1 , d 2 , d 3 - simple d 4 high- spin low- spin
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High-spin d 4 Low-spin d 4 t
2g
3 e g
1 t
2g
4 e g
0 x = 3 , y = 1 x = 4 , y = 0
E = (0.4
x – 0.6
y )Δ o
= 0.6 Δ o
E = (0.4
x – 0.6
y )Δ o
= 1.6 Δ o
+ 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 : t
2g
3 e g
0 , x = 3, y = 0
Therefore, LFSE = (0.4
x – 0.6
y )Δ o
= [(0.4)(3) – (0.6)(0)]Δ o
= 1.2 Δ o
(b) electronic configuration : t
2g
3 e g
2 , x = 3, y = 2
Therefore, LFSE = (0.4
x
– 0.6
y )Δ o
= [(0.4)(3) – (0.6)(2)]Δ o
= 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:
Pt 4+ > Ir 3+ > Rh 3+ > Co 3+ > Cr 3+ > Fe 3+ > Fe 2+ > Co 2+ > Ni 2+ > Mn 2+
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 < H
2
O < NH
3
< en < CN < CO
The ligand- field splitting parameter, Δ o varies with the identity of the ligand.
In the series of complexes [CoX(NH
3
)
5
] n+ with X = I , Br , Cl H
2
0 and NH
3
, the colours range from purple (for X = I ) through pink (X = Cl ) to yellow
(with NH
3
).
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:
𝐵
μ
B
= Bohr magneton
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CALCULATED SPIN-ONLY MAGNETIC MOMENTS
ION
Ti 3+
V 3+
Cr 3+
Mn 3+
Fe 3+
N
3
4
5
1
2
S
½
1
1½
2
2½
μ/μ
B
CALC. EXPT.
1.73 1.7-1.8
2.83 2.7-2.9
3.87 3.8
4.90 4.8-4.9
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 d electron configuration?
SOLUTION
A Co(II) complex is d 7 .
Two possible configurations: t
2g
5 e g
2 electrons or t
2g
6 e g
1
(high-spin, S = 1½) with 3 unpaired
(Low-spin, S = ½) with 1 unpaired electron.
The spin-only magnetic moments are 3.87 μ
B and 1.73 μ
B
.
Therefore, the only consistent assignment is the high-spin configuration t
2g
5 e g
2 .
EXERCISE FOR THE IDLE MIND
The magnetic moment of the complex [Mn(NCS)
6
] 4is 6.06
μ B. What is its electron configuration?
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