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chapter
Crystal Field Splitting
>> Parts of this equation/concept include:
In an atom, all d orbitals have the same energy because
their only difference is the orientation of the orbitals. However,
in a molecule, the outermost electrons may interact with other electrons
if they are oriented correctly. The orientation of d orbitals
is
dxy = between the x and y
axes
dxz = between the x and z
axes
dyz = between the y and z
axes
 = on the z-axis with a doughnut
in the xy plane
 = on the
x and y axes
In an "octahedral field," six atoms interact on either side of
all three axes. Therefore, the electrons of these other atoms will
repel the d orbitals that lie on the axes,
and .
The law of conservation of energy says that if these orbitals are
destabilized (higher in energy), the other orbitals, dxy,
dxz, dyz, must
be stabilized (lower in energy) so that the net effect is no change
in energy. The energy difference between dxy,
dxz, and dyz,
and
and
is the crystal field splitting energy,  o
(o stands for octahedral), Figure 10.24.
In a "tetrahedral field," four interacting atoms are located between
the axes. Therefore the
and
are stabilized and the dxy, dxz,
and dyz are destabilized. The splitting
energy is t
(opposite of the octahedral arrangement, Figure 10.26).
A "square planar field" is like an octahedral field, but without
one axis. Consequently, each level of the octahedral arrangement
is resplit (Figure 10.27). The splitting energy is sp.
The magnitude of the split depends on the type of splitting (o
> t), the type of atom being
split and the type of atom doing the splitting. Because t
is always small, it is more energetically favorable for an electron
to go to a higher-energy d orbital than to pair up (despite
Hund's rule). With o, it
depends on the magnitude of o
whether the electron would prefer to pair (o
is larger) or go to the higher orbital (o
is small). Compounds where the electrons prefer to pair are called
low spin. Compounds where electrons prefer a higher-energy
orbital are called high spin. For atoms with less than three
or more than seven d electrons, it will not make a difference.
>> Example 1
Draw the crystal field splitting of the following ions
- Iron(II), low spin in an octahedral field
- Iron(III) in a tetrahedral field
- High-spin manganese(II) in an octahedral field
- Platinum(II) in a square planar field
Solution:
-
First you must determine how many d electrons are
present in Fe2+. (Review Chapter
3.)
The electron configuration of Fe is [Ar]4s23d6.
Since higher n values are always lost first, the electron
configuration of Fe2+ = [Ar]3d6. Consequently,
iron(II) has six d electrons.
Since iron(II) is in an octahedral field, the d orbitals
split as
Since it is a low-spin compound, electrons will fill the
bottom row before the top one. Therefore the crystal field
diagram is
-
For Fe3+ the electron configuration is [Ar]3d5.
Since this is a tetrahedral field, the splitting is
Since it is a tetrahedral field, it is automatically "high
spin." Therefore the five d electrons arrange as
-
Mn2+ has an electron configuration of [Ar]3d5.
It is high spin and in an octahedral field. Therefore the
diagram looks like
-
Pt2+ has an electron configuration of [Xe]4f145d8.
Therefore it has eight d electrons. The square planar
splitting diagram is
The eight d electrons arrange as if they were "low
spin."
The number of unpaired electrons determines the magnetic properties
of a molecule. A substance without unpaired electrons is "diamagnetic."
A substance with unpaired electrons is "paramagnetic." Obviously,
crystal field splitting will affect the number of unpaired electrons.
>> Example 2
How many unpaired electrons are in each substance listed in the
previous example? Which substances are diamagnetic?
Solution:
Refer to the preceding splitting diagrams.
- Low-spin Fe2+ (octahedral field) is diamagnetic
with no unpaired electrons.
- Fe3+ in a tetrahedral field has five unpaired electrons (paramagnetic).
- High-spin Mn2+ (octahedral field) has five unpaired electrons (paramagetic).
- Pt2+ in a square planar field is diamagnetic with no unpaired electrons.
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An electron can move from one orbital to a higher-energy orbital
by absorbing light that has exactly the same energy as the energy
difference between the orbitals. For transition metals this energy
difference is often the crystal field splitting energy. To relate
this energy to the wavelength absorbed, Equation 10.2 is used.
= hc/
Remember that if this wavelength of light is absorbed, you can't
see it! What you see instead is a mixture of all the wavelengths
that are not absorbed.
>> Example 3
What wavelength is absorbed if o
= 3.60 x 10–19 J?
Solution:
h = 6.626 x 10–34 J•s
c = 2.998 x 108 m/s
From Equation 10.2,
|
= |
| hc |
|
|
|
so
|
= |
| hc |
|
|
|
| |
|
|
|
= |
| (6.626 x 10–34 J•s)(2.998 x 108 m/s) |
|
| 3.60 x 10–19 J |
|
| |
|
|
|
= |
5.52 x 10–7 m |
>> Example 4
What is t if the compound absorbs at 690 nm?
Solution:
|
= |
690 nm |
 |
|
= |
6.90 x 10–7 m |
Using Equation 10.2,
|
= |
| hc |
|
|
|
= |
| (6.626 x 10–34 J•s)(2.998 x 108 m/s) |
|
| 6.90 x 10–7 m |
|
= |
2.88 x 10–19 J |
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