Download The Synthesis and Color of Cr Complexes

.CHEM 122L
General Chemistry Laboratory
Revision 4.0
The Synthesis and Color of Cr3+ Complexes
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To learn about Coordination Compounds and Complex Ions.
To learn about the Color of Transition Metal Complexes.
In this laboratory exercise, we will synthesize a Coordination Compound of Chromium,
tris (2,4-pentanedionato) chromium (III), Cr(acac)3, a brightly colored solid, from CrCl3•6H2O,
another brightly colored solid. We will then examine the absorption spectrum of the aqueous
form of this compound to determine the magnitude of the splitting of the Cr3+'s d-orbitals, such
splitting being the source of its bright color..
CrCl3•6H2O(aq) + 3 acac-(aq)
Cr(acac)3(s) + 3 Cl-(aq) + 6 H2O
(Eq. 1)
A couple of things are to be noted about this reaction. First, acac- is short hand for the 2,4pentanedionate ion, which has the formula:
This anion is generated by treating 2,4-pentanedione, otherwise known as Acetyl Acetone
(acacH), with Ammonia:
acacH(aq) + NH3(aq)
acac-(aq) + NH4+(aq)
(Eq. 2)
Thus, in order to carry-out the reaction of (Eq. 1), we first treat acacH with Ammonia and the
resulting acac- ion attacks the Cr(H2O)63+ complex, which is itself the dissociation product of
CrCl3•6H2O in Water:
CrCl3•6H2O(aq)
Cr(H2O)63+(aq) + 3 Cl-(aq)
(Eq. 3)
In order that the reaction should proceed at a manageable rate, instead of adding the Ammonia
directly to the solution, we will instead generate the Ammonia in situ via the decomposition of
Urea:
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NH2COCH2(aq) + H2O
2 NH3(aq) + CO2(aq)
(Eq. 4)
Second, Cr(acac)3 is a Coordination Compound. Coordination Compounds are substances that
contain at least one Complex, a species containing a central metal cation that is bonded to
molecules or anions, called Ligands. In the present case, the coordination compound Cr(acac)3
is formed from Cr3+ and three acac- ligands. In many cases, the complex formed is an ion, so it
would be referred to as a Complex Ion. For instance, Ammonia (NH3) complexes with Co3+ to
form the Co(NH3)63+ complex cation. When precipitated from an aqueous solution, this 3+
cation combines with 3 Cl- anions to form the [Co(NH3)6]Cl3 coordination compound; much like
the Na+ cation combines with the Cl- anion to form the compound NaCl. (Note the use of [ ]’s
around the Complex Ion Co(NH3)63+ in the chemical formula.) In our case the Cr(acac)3
complex is neutral and so precipitates out of an aqueous solution directly.
The covalent bonding between the metal ion and its ligands, in our case, between the Cr3+ and
the acac-’s, is via Coordinate Covalent Bonds; the ligands act as Lewis Bases and donate pairs
of electrons to the metal, which acts as a Lewis Acid. A general example of this type of bonding
involves the coordination of Co3+ and NH3:
Since this example complex contains six NH3 ligands, it has six coordinate covalent bonds
arranged octahedrally about the metal center:
The number of coordinate covalent bonds formed is referred to as the Coordination Number of
the complex. In the above example, the coordination number is six. The coordination number
determines the geometry of the complex. Typical coordination geometries are:
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Coordination Number
2
Geometry
Linear
4
Square Planar
or
Tetrahedral
6
Octahedral
For our complex, Cr(acac)3, the coordination number is also six. 2 electron pairs are donated to
the Cr3+ ion by each 2,4-pentanedionate ion, acac-, ligand. The 2,4-pentanedionate ligand is
referred to as bidentate because it forms two bonds to the metal.
Three acac- ligands will thus bind to the Cr3+ metal cation with 6 coordinate covalent bonds in an
octahedral arrangement:
(Note the cartoon used to represent the acac- ligand. Only the O-Cr bonds are important for our
purposes.)
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Finally, you should note during the laboratory that the starting material CrCl3•6H2O is a brightly
colored green substance, whereas the product is a deep purple color. This bright coloring is
fairly common for transition metal complexes. Electrons in partially filled d-orbitals of a
transition metal center, a distinctive feature of the transition metals, can absorb light of visible
wavelengths. The color absorbed is then the complement of the color we see.
The absorption of relatively low energy visible light occurs because of splitting of the d-orbitals’
energy levels of the Cr3+ ion. Qualitatively, this is due to an interaction between the donated
electron pair of the ligands with the lobes of the d-orbitals on the metal center; the splitting
pattern being determined by the geometry of the ligands about the d-orbitals. Crystal Field
Theory provides for a crude model of how this splitting occurs.
Recall, the d-orbitals have the following shapes:
According to Crystal Field Theory, if the negatively charged bonding electron pairs of the
ligands interact with high electron density regions of say the dx2-y2 orbital along the x and y axes:
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the interactions are unfavorable and the energy of the system rises. This unfavorable interaction
is true for all the d-orbitals. However, if the electron pairs of the ligands approach a dxy orbital
along these same axes, the approach is along a nodal plane and the interaction is less
unfavorable, causing the energy of the system to not rise as much.
This leads to a splitting in the energy of the d-orbitals. If we consider how each of the d-orbitals
interacts with the ligands, we find that for an octahedral complex, the d-orbital energies will split
according to the following pattern:
As is reasonable, the geometry of the ligands will determine the nature of the their interaction
with the d-orbitals and hence the nature of the splitting pattern. The splitting pattern for
tetrahedral, square planar and octahedral geometries is given below.
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(It should be noted this diagram only shows the splitting patterns of the various geometries, and not
the relative energies of the orbitals. In fact, if the ligands are the same, then Tetrahdral < Octahedral.)
The relative strength of the splitting energy depends on the nature of the ligands surrounding
metal cation and the metal cation itself. For a given metal, the Spectrochemical Series represents
an ordering of the ability of a ligand to influence . Ligands on the low end of the series, so
called "weak field ligands", split the d-orbitals minimally, whereas those at the high end, "strongfield ligands", split the orbitals significantly.  increases by roughly a factor of 2 from one end
of the series to the other.
Spectrochemical Series
I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN-< CO
Now consider what happens when photons interact with a metal complex. As an example, we
will examine, the Ti(H2O)63+ complex, which contains a d1 metal ion:
Ti = [Ar] 4s2 3d2
Ti3+ = [Ar] 3d1
(Recall, for Transition Metal Cations, s-orbital electrons are lost before d-orbital electrons. The
rule of thumb is: "first in-first out".) So, it is an octahedral complex with one d electron. If
Ephoton = , then the complex will absorb the photon.
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(Recall also, the relationship between the energy of a photon Ephoton and its wavelength  is
Ephoton = hc/, where h is Planck's Constant and c is the Speed of Light.) It is found  ~ 500 nm
for this complex, meaning it absorbs blue-green photons. We, of course, "see" the color of this
complex as the complement of the absorbed photons; red-violet.
Color Wheel for Determining Complementary Colors
For a slightly more complex case, let's look at the color of CoF63- vs. Co(NH3)63+. Both involve a
Co3+, or d6 metal ion and both are octahedral.
Co = [Ar] 4s2 3d7
Co3+ = [Ar] 3d6
Because F- is a weak field ligand, we expect a small splitting energy . On the other hand NH3
is a much stronger field ligand, so will be much larger.
Note that the arrangement of the electrons in the d-orbitals changes from one complex to the
other. In CoF63- the energy to pair the electrons is greater than  and so we have a "high spin"
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complex; the electrons still obey Hund's Rule. In Co(NH3)63+ the pairing energy is small
compared to  and so we have a "low spin" complex. (Paired electrons have a lower net spin
than do unpaired electrons.)
Because  is small for CoF63-, it absorbs long wavelength red photons ( ~ 700 nm) and appears
green. For Co(NH3)63+, because  is much larger, it absorbs shorter wavelength blue photons
( ~ 475 nm) and appears yellow-orange.
We will be determining the wavelength at which our complex absorbs light spectroscopically.
Recall, a Visible Spectrometer consists of a White Light source, dispersive optics to separate the
wavelengths of the light, a sample compartment and a detector.
Light from the source passes through an Entrance Slit and is focused on the dispersive element,
such as a diffraction grating. This separates out the various wavelengths comprising the White
Light into a rainbow of colors. The desired wavelength is selected by rotating the dispersive
element such that the desired color passes through an Exit Slit, through the Sample and then onto
a detector.
The Transmittance of the light is then defined as:
T = P / Po
(Eq. 5)
This is simply a measure of how many photons pass through the sample without being absorbed.
This is then related to the Absorbance by:
A = - log T
(Eq. 6)
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A major complication occurs if an absorbing or emitting species is in a condensed phase; a liquid
or a liquid solution. If the absorbing molecule/atom is in a solution, it is surrounded by
constantly jostling solvent molecules. Thus, each molecule/atom finds itself in a slightly
different environment than its brothers. This causes the energy gap between the quantum states
responsible for the absorbance of photons to be slightly different for each molecule/atom. This
means we will have a series of very, very closely spaced absorbance lines. Practically, this
means the Absorbance Spectrum will be a broad band, rather than a sharp line. This is as
diagramed below:
In this case, we usually identify the absorbance band by the wavelength of maximal absorbance,
max.
The d-orbial splitting energy  can then be determine from max via:
 = Ephoton = hc / max
(Eq. 7)
Now, for the complex we will be synthesizing, Cr(acac)3, one final and rather nasty complication
occurs. The spectrum will contain not one absorbance peak but instead two.
An octahedral complex like Ti(H2O)63+ contains a Ti3+ metal center with a [Ar] 3d1 electron
configuration. The absorbed photon will kick the single d electron from any of the lower three dorbitals to any of the upper two d-orbitals. This is as was depicted above:
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Our complex Cr(acac)3 contains a Cr3+ metal center with a [Ar] 3d3 electron configuration, In its
ground state the electrons will line up according Hund's Rule in the lower three d orbitals. (I've
included the specific orbital designations because we will need them momentarily.)
When a photon is absorbed and an electron is promoted, it will result in any of six possible
electron-orbital configurations.
In the first three configurations above the excited electron lies in an orbital that is in a different
plane than that of the unexcited electrons. In the first configuration, the excited electron is in a
3dz2 orbital, which lies along the z-axis. The other two electrons are in 3dyz and 3dxz orbitals,
which lie in the x-y plane. Hence the e--e- repulsions will be fairly minimal. This is true of the
next two configurations as well. However, in the next three configurations, the excited electron
is in an orbital that lies in the same plane as the unexcited electrons, resulting in fairly significant
e--e- repulsions. So, the first set of excited state electrons will have a lower total energy than will
the second set. Hence, the ground state configuration can be excited into a lower or higher
energy configuration. And thus we will observe two absorbance lines in the absorbance
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spectrum for the complex. The splitting energy  is associated with the longest wavelength
(lower energy) absorbance peak in the spectrum.
Whew! A lot to digest.
To recap, we will be synthesizing the coordination compound Cr(acac)3, measuring its
absorbance spectrum and then determining max for the compound. This measurement will be
used to determine the d-orbital splitting energy. This will then be compared with the splitting
energy of another Cr3+complex to confirm the spectrochemical series is in fact correct.
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Pre-Lab Safety Questions
1.
What is "Hexavalent Chromium"? What are some of the health effects of ingesting even
small amounts of Hexavalent Chromium? Do we need to worry about Hexavalent
Chromium during this laboratory exercise?
2.
What are the health hazards associated with working with Acetylacetone? How are
mitigating these hazards in this laboratory exercise?
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Procedure
[This procedure was adapted by TA extraordinaire Allen Erickson, from a procedure adapted from Z. Szafran; R.M.
Pike; M.M. Singh; Microscale Inorganic Chemistry, Wiley: New Your, 1991. This, in turn, was adapted from T.G.
Dunne, "Spectra of Cr(III) Complexes, J. Chem. Ed. 44 (1967) 101.)
This synthesis requires that you heat your reaction mixture to boiling (to “reflux”) on top
of a magnetic stirring plate. Reflux is a common technique in chemistry used to supply
heat to get the solution to its boiling point in order to accelerate the reaction. In this
technique, a liquid reaction mixture is placed in a vessel that is open only at the top. This
vessel is connected to a condenser so that any vapors given off from the vessel are
condensed and fall back into the vessel. A picture of a reflux apparatus is shown below.
Note the water flows into the condenser near the bottom so as to maximally cool the hot
vapors from the reaction vessel and to force the water up the condenser and out neat the
top.
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1.
Obtain a small magnetic stir bar from your Teaching Assistant and add to a 25 mL
round-bottomed flask. The flask can be found in the organic microlaboratory kit at
your station. Aside from the reflux procedure, please keep the flask in the cork
support provided for you to avoid knocking it over.
2.
Obtain 260 mg of CrCl3·6H2O and add to your flask. Dissolve it in 4.0 mL of
distilled water.
3.
Add 1 g of urea to the flask.
4.
In the fume hood, measure out 0.8 mL of acetylacetone using the 1 mL graduated
pipette provided. Please note that the pipette is not calibrated to-drain.
5.
Assemble the reflux apparatus as shown below. The condenser can also be found in
the organic kit. Be sure that the direction of water flow is the same as is shown in
the diagram above. Have your Teaching Assistant check your apparatus before you
start refluxing.
6.
Turn on the magnetic stirring slowly to a moderate setting. If the solution sputters
onto the sides of the flask, it may result in lower yields.
7.
Apply heat to begin refluxing. The solution should be kept at a mild-to-moderate
boil. If the solution is boiled too quickly, some vapors may escape the condenser.
8.
Reflux your reaction for one hour (from the point it begins to boil). During this
time, you may work on the two addendum exercises, but keep an eye on the vessel
to keep the boil under control.
9.
After one hour, cool the flask thoroughly in an ice bath for 8 minutes. Crystals
should form at the bottom of the flask. If they do not appear, continue to cool the
flask.
10. Collect the crystals by suction filtration. Wash the crystals with small amounts of
distilled water, and allow to dry at the pump.
11. Measure the yield of your synthesis.
12. Cover the weigh boat containing the product with another weigh boat. A laboratory
assistant will bring you to the instrument room to take the UV-Vis Spectrum of
your product. Record the wavelength of maximum absorbance.
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Data Analysis
1.
Calculate the Theoretical Yield of your Product.
2.
Calculate the Percentage Yield of your Product.
3.
Sketch a Crystal Field Theory energy level diagram for the product Cr(acac)3 complex.
What color light does this a solution of this complex absorb?
4.
Determine max for your product.
5.
Calculate  for your product.
6.
Below is the absorbance spectrum for Tris(ethylenediamine) chromium (III) chloride
dihydrate, [Cr(en)3]Cl3•2H2O.
(Source: T.G. Dunne, "Spectra of Cr(III) Complexes, J. Chem. Ed. 44 (1967) 101.)
Note that "en" is short-hand for Ethylenediamine:
NH2-CH2CH2-NH2
a bidentate ion with coordinate covalent bonds involving the Nitrogen:
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i) What is the complex ion in this compound?
ii) Estimate max from the spectrum of this compound and then determine .
iii) Compare the  for your compound with that for this one. Do they conform to the
spectrochemical series? Note that your compound's ligands bind to the metal center via
a O-Metal bond, whereas the "en" ligand binds via N-Metal bonds.
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Post Lab Questions
1.
Determine the Oxidation State and the Coordination Number of the Transition Metal Ion in
each of the following Complex Ions:
a)
b)
c)
d)
2.
Ni(H2O)62+
Mn(CN)64CuCl42Cr(NH3)63+
Write a chemical equation representing what happens when each of the following water
soluble Coordination Compounds dissolves in water.
a) [Cr(NH3)4Cl2]Cl
b) K2[PtCl4]
c) [Co(en)3](NO3)3
3.
Cr(NH3)63+ is yellow-orange whereas Cr(NH3)5Cl2+ is purple. Explain this using Crystal
Field Theory energy level diagrams.
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Addendum
The Color of Some Complex Ions
In this addendum we return to some reactions we examined earlier. We repeat these
experiments now with an eye toward explaining the observed colors of the solutions
involved. We will re-examine the formation of the Copper Ammine complex and the
equilibrium between two Cobalt complexes. In each case, carefully note the colors of the
solutions.
Procedure
Perform all Addendum reactions in a Fume Hood!!!
Copper Ammine Complex
Cu2+(aq) + 2 OH-(aq)
Cu(OH)2(s) + 4 NH3(aq)
Cu(OH)2(s)
Cu(NH3)42+(aq) + 2 OH-(aq)
Put 10 drops 0.1M Cupric Sulfate (CuSO4) into a test tube. Note the color of the
solution. Add 3-4 drops of 6M Ammonia (NH3). Note what happens. You should recall
that Ammonia (NH3) acts as a weak Base in Water:
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
So, an Aqueous Ammonia solution contains both NH3, which can complex with Cu2+ to
form the ammine complex (Cu(NH3)42+), and OH-, which will form the Cupric Hydroxide
(Cu(OH)2) precipitate.
Data Analysis
1.
Use Crystal Field Theory energy level diagrams to explain the color changes in
each case. One cautionary note: Cu2+ is short-hand for Cu(H2O)62+ and Cu(NH3)42+
is short-hand for Cu(NH3)4(H2O)22+.