Download Chapter 24 Transition Metals and Coordination Compounds part 2

Chapter 24 Transition Metals and Coordination Compounds part 2
Bonding in Coordination Compounds
Valence Bond Theory
Coordinate covalent bond is between:
completely filled atomic orbital
from
and
an empty atomic orbital.
in complex.
The metal ion orbital becomes
Coordination #
Hybridization
Geometry
Example
2
sp
linear
[Ag(NH3)2]+
4
sp3
tetrahedral
[Zn(CN)4]2-
4
dsp2 of sp2d
square planar
[Ni(CN)4]2-
5
dsp3
trigonal bipyrimidal
[CuCl5]3-
5
d2sp2
square pyramidal
[Ni(CN)5]3-
6
d2sp3 or sp3d2
octahedral
[Fe(CN)6]4-
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Crystal Field Theory – color
Focuses on interactions between ligands and unhybridized d orbitals as the
ligands approach during bond formation.
 Originally there are __ degenerate (same energy level) d orbitals.
 3 orbitals have configurations
between the axes:
2 orbitals have configurations
that put them along the axes;)
If ligands approach the metal cation on a path that makes them overlap with the
electrons in the d orbitals, this will cause ________________
This will cause those orbitals to be ________________
This will result in them having ______________
A. For an octahedral complex, there are 6 ligand attachment points
(along the x, y, & z axes)
 Three of the orbitals are farther
away from the ligands because they
lie between the axes, not on them.
These feel _____ destabilizing
repulsive forces from the ligand
electrons.
 Two of the orbitals are closer to the ligands
because they lie on the axes.
These feel _____ destabilizing
repulsive forces from the ligand electrons.
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For an octahedral complex, degenerate d orbitals _____ into __ levels.
- All split levels are __________ in energy than the original orbitals
due to ______________ from ligand electrons,
some more than others.
- The orbitals along the axes, with _____ destabilization due to greater repulsive
forces end up ___________ in energy than those that are between the axes.
- The difference between the two energy levels is referred to a ____, known as
the ____________________________ .
Crystal Field Theory – magnetic properties
A substance that contains unpaired electrons and
is attracted by a magnetic field is _____________.
A substance that has electrons that are all paired and
is weakly repelled by a magnetic field is _____________.
Valence bond theory cannot always account for the observed magnetic
behavior of complexes.
Hund’s rule – electrons occupy degenerate orbitals singly as long as an empty
orbital is available.
When dealing with original metal atomic d orbitals this gives:
____ ____ ____ ____ ____
When dealing with non-degenerate d orbitals, the decision is more complex
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____ ____
____ ____ ____
OR
___ ___
____ ____ ____
B. Tetrahedral and Square Planar complexes
1. Tetrahedral Complexes:
 There are only ____ ligands instead of six.
 The ligands approach the metal cation ______ axis.
 energy splitting is just the opposite of that in octahedral complexes.
 (tetrahedral complexes)  ½(octahedral complexes).
  < Pairing energy (P)
These terms will be explained in a
few minutes.
 all complexes are high-spin
2. Square Planar Complexes:
 two trans ligands along the z axis are missing (compared to octahedral).
 Large energy gap between the x2-y2 orbital and the four lower-energy
orbitals.
 Most common for metal ions with ____ electronic configurations.
 Favors __________ complexes in which all four lower-energy orbitals
are filled and the higher energy x2-y2 orbital is vacant
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Problem: Consider [Ni(NH3)6]2+
a) What is the shape and configuration of the complex?
b) What metal ion orbitals are used?
a) To find the shape, start by looking at the coordination number
b) To find the orbital used, start by finding the electron configuration.
Ni atom (28 valence e-) = ________________
Ni2+
(26 valence e-) = ________________
Octahedral splits
Filling in the 8 e- in the d orbitals:
Fill lower level first? We’ll come back to this!
All of the 3d orbitals are occupied.
Is the complex diamagnetic or paramagnetic?
Ligands must put their pair of electrons into _________ orbitals!
Is the complex d2sp3 or sp3d2?
If there are empty orbitals left in the current
d level, it will be ______ .
If there are NO empty orbitals left in the
current d level, it will be ______.
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The complex would look like:
Color In Transition Metal Complexes
A. The color of transition-metal complexes depends on the identity of
_____________________________.
B. Metal complex can absorb light by undergoing an electronic transition from its
lowest energy state (E1) to a higher energy state (E2).
 Wavelength of absorbed light depends on the energy separation between
the two states ______________.
 Visible light contains a continuum of wavelengths, one of which is the
correct wavelength to accomplish the promotion of an electron from a
lower energy d orbital to a higher energy d orbital. :
  =E= hc/ where is the wavelength of light absorbed.
Complexes always absorb a range of wavelengths, not just a single one.
The color we see approximates the compliment of
the peak of the absorbance spectrum, known as
______.
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Example:
What wavelength (in nm) is absorbed if o = 2.75 x 10–19 J?
What color will the solution be?
h = 6.626 x 10–34 J•s
c = 2.998 x 108 m/s
According to the color wheel, it should appear to be ________.
Note:  is the energy to move 1 electron. If the question asks about the energy per mole,
you need to multiply by _________________!
High Spin & Low Spin Complexes
In octahedral geometry complexes, with ions containing ____ d electrons:
We have 2 possible choices for electron configurations.
A. High Spin Complexes:
1. Makes early use of ________ energy orbitals.
2. Puts d electrons in as many different d orbitals as possible.
3. Creates as many unpaired e- as possible.
B. Low Spin Complexes:
1. Completely fills __________ energy orbitals first.
2. Puts as many d electrons in the lowest energy d orbitals as possible.
3. Reduces the number of unpaired e-.
a) Substances with unpaired electrons are ___________.
b) Those without unpaired electrons are ___________
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C. How do molecules choose high or low spin? ________________________
1. There is an energetic penalty for putting 2 negative particles (e-) in the
same space.
A repulsive force must be overcome.
2. There is an energetic penalty for placing an e- in a higher energy level.
3. Decision Process
The complex that forms will be the one that takes _______ energy.
If we have 6 e- to place, the first 3 will always go in the
bottom level.
If P <  Then we have ______ spin:
because of bigger .
If P >  Then we have ______ spin:
because of smaller .
P (pairing energy) does not change. It is the size of  that changes.
D. Spectrochemical Series of Ligands
1. Size of  depends on the nature of the ligands.
weak-field ligands - produce a relatively ________ value of 
resulting in __________________ complexes.
strong-field ligands - produce a relatively ________ value of 
resulting in __________________ complexes.
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Spectrochemical Series of Ligands
I- < Br- <Cl- <:SCN <F- OH- < C2O42- < H2O <:NCS < NH3 < en < phen < CN-  CO
The place of a ligand in the spectrochemical series is determined
largely by its donor atoms. Thus, all N-donor ligands are close to
ammonia in the spectrochemical series, while all O-donor ligands
are close to water. The spectrochemical series follows the positions
of the donor atoms in the periodic table as:
C
N
O
F
P
S
Cl
?
very little
data on
P-donors –
may be higher
than N-donors
S-donors ≈
between Br
and Cl
Br
spectrochemical
series follows
arrows around
starting at I and
ending at C
I
More complete list that is on the formula sheet
Weak field ligands
strong field ligands
I-< Br- < S2- < :SCN- < Cl- < F- < OH– < C2O42– < :ONO– < H2O < :NCS– < EDTA4- < NH3 < en < phen < :NO2– < CN– ≈ CO
en = ethylenediamine
phen = phenanthrolene
C2O42- = oxalate ion
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Small 
a. caused by _________ field ligands(i.e., halides , oxalate ion)
b. lower energy wavelengths absorbed
(longer wavelengths absorbed)
(closer to the red end of the spectrum absorbed)
(colors appear closer to the violet end of the spectrum)
Large 
a. caused by ____________ field ligands (i.e., CN-, CO)
b. higher energy wavelengths absorbed
(shorter wavelengths absorbed)
(closer to the violet end of the spectrum absorbed)
(colors appear closer to the red end of the spectrum)
Example: Compare
[CoF6]3-
vs
[Co(CN)6]3-
These are Co3+ complexes.
Co3+ = [Ar] 3d6
6 ligands = octahedral orientation and orbital splitting
 Weak field ligands will result in ________ spin complexes
 Strong filed ligands will result in ________ low spin complexes
Starting orientation for original 6 electrons is:
[CoF6]3-
[Co(CN)6]3-
Complex is
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Practice___________________________________________________________
How many unpaired electrons are there in the tetrahedral complex ion [CoCl4]2? Is it diamagnetic or paramagnetic?
Practice___________________________________________________________
How many unpaired electrons are there in the complex ion [Cr(CN)6]4- ? Is it
diamagnetic or paramagnetic?
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