Download REEL IA - OSU Chemistry

REEL IA – ELECTRONIC STRUCTURE, COLOR & PIGMENTS
REEL – RESEARCH EXPERIENCES TO ENHANCE LEARNING
PURPOSE
To prepare solutions containing Cr3+ and Cr(VI) and to probe the electronic excitations of
these species using UV-Visible absorption spectroscopy, and to synthesize and characterize
pigments of general formula Pb1-xSrxCrxMo1-xO4 and analyze them using X-Ray powder
diffraction and UV-Visible diffuse reflectance spectroscopy.
DISCUSSION
In this lab you’ll explore the relationship between the electronic structure, bonding and color. The lab
focuses largely on the chemistry of chromium ions bonded to oxygen. We will see how changes in the
oxidation state and coordination environment of chromium impact the optical absorption spectrum, which
in turn impacts the color. In addition to examining the colors and absorption spectra of species in solution
you will also synthesize a pigment in powdered form. This is a precursor to the next lab where you will
analyze your pigment sample using X-ray powder diffraction and UV-Visible diffuse reflectance
spectroscopy.
REACTIONS OF CHROMIUM IN AQUEOUS SOLUTIONS
The chemistry of chromium is dominated by two oxidation states, Cr3+ and Cr6+. Because of the very
high oxidation state of Cr6+ this ion forms highly covalent bonds with anions such as oxygen. Therefore,
the value of 6+ is meant to convey the oxidation state, rather than an actual charge. To acknowledge this
it we will denote this ion as Cr(VI) rather than Cr6+. *
When Cr(NO3)3·9H2O dissolves in water the Cr3+ and NO3– ions are separated. The Cr3+ ion is
surrounded by solvent (water) molecules. Because water is a polar molecule the oxygen end of the
molecule has a partial negative charge. Consequently, the oxygen end of the water molecule is attracted to
the positively charged Cr3+ ion. We know from the Lewis structure that there are two nonbonding electron
pairs on the water molecule. The lone pairs on the H2O molecule can and do interact with empty orbitals
on the Cr3+ ion to form a bond. These bonds are what hold the coordination complex together. The
preferred geometry is for six water molecules to coordinate to the chromium ion in such a way that the
oxygen atoms form an octahedron around the chromium. This leads to the formation of the complex ion
hexaaquachromium(III) or [Cr(H2O)6]3+, as shown in Figure 1. †
*
The reactions you will carry out in this lab are adapted from “The chemistry involved in the preparation of a paint
pigment. An experiment for the freshman laboratory”, T. L. Daines, J. Chem. Educ. 1976, 53, 117.
†
Often times we neglect the water molecules that coordinate to cations in aqueous solution. Thus [Cr(H2O)6]3+ is
often simply written as Cr3+(aq).
1
Figure 1: The [Cr(H2O)6]3+ complex. The figure on the left shows the chromium ion (in green) surrounded by six
water molecules (oxygen in red, hydrogen in gray). In the figure on the right the hydrogen atoms have been omitted
to stress the interactions between chromium and oxygen.
Upon addition of OH– ions to a solution containing Cr3+ initially the precipitate Cr(OH)3 forms.
However, because chromium(III) hydroxide is amphoteric further addition of OH– leads to the formation
of the complex ion, tetrahydroxochromium(III) or [Cr(OH)4]– according to the following reaction
sequence:
Cr3+(aq) + 3OH–(aq)
blue/violet
solution
→ Cr(OH)3 (s) + OH–(aq)
gray-green
ppt
→ Cr(OH)4–(aq)
green
solution
In the next step we add the oxidizing agent hydrogen peroxide, H2O2, to oxidize the Cr3+ to Cr(VI). Upon
adding H2O2 the solution should turn a brownish color, due to the presence of both Cr3+ and Cr(VI) in the
solution. Upon gentle heating in a water bath the conversion of Cr(OH)4– to CrO42– occurs according to
the following reaction:
Δ
2 OH– + 2 Cr(OH)4–(aq) + 3 H2O2 ⎯
8H2O + 2 CrO42–(aq)
⎯→
In this reaction the peroxide ion (O22–) is reduced while chromium is oxidized. The change in oxidation
state of the chromium ion removes all of the electrons from the Cr 3d orbitals, which we will see in the
next section has a profound impact on the optical absorption spectrum.
Finally the addition of Pb(NO3)2 triggers the formation of insoluble PbCrO4 through the following
precipitation reaction:
Pb2+(aq) + CrO42–(aq)
→
PbCrO4 (s)
The product should be a bright yellow to yellowish-orange solid. Over the years PbCrO4 has been
extensively used as a pigment for its brilliant color, but concerns about the toxicity both of Pb2+ and
CrO42– have curtailed its use in more recent times. You should make a note of the color and save your
sample for comparison with the PbCrO4 sample(s) made by a solid state reaction route in a later part of
this lab.
2
ELECTRONIC STRUCTURE AND OPTICAL ABSORPTION SPECTROSCOPY: INTRA-ATOMIC (D TO D)
TRANSITIONS
The color we see in the chromium complexes originates from selective absorption of light. Because the
complexes absorb some but not all wavelengths of visible light the light that passes through the solution
and reaches our eyes appears colored. Absorption of visible light occurs because the energy associated
with a photon of light is transferred to an electron, exciting it to a higher energy molecular orbital (MO).
Such a process is called an electronic excitation. Light is absorbed selectively because only those photons
whose energies are equal to the energy difference between an occupied MO and an empty MO are
absorbed. Therefore, if we want to understand the origin of color we need to know something about the
electronic structure of the complexes in question.
To understand the electronic structure we need to consider two factors (a) the shape or geometry of the
molecule or complex ion, and (b) the electron configuration of the atoms involved. Let’s begin with the
complex ion that is most relevant to this lab: [Cr(H2O)6]3+. The Cr3+ ion has an electron configuration of
[Ar]3d3 with three valence electrons residing in the 3d orbitals. For a free Cr3+ ion with no neighboring
atoms the five d-orbitals would all be at the same energy, as shown on the left hand side of Figure 2. Of
course we are not dealing with a free metal cation in this experiment. The metal is surrounded by solvent
molecules, and the nonbonding electron pairs on the oxygen atoms of the water molecules interact with
the metal to form bonds. These bonds alter the energy of the d-orbitals on chromium, which in turn
impacts the color.
Energy
dz2
Δ
dxy
dy z
dx z
dz2
dx2-y2
Cr3+ ion with a
spherical crystal field
dxy
dx2-y2
dyz
dxz
dz2
dx2-y2
dxy
dyz
dxz
Cr3+ ion in an
octahedral crystal field
Free Cr3+ ion
Figure 2: The energy levels and filling of the 3d-orbitals of a Cr3+ ion. The diagram on the left hand side refers to a
free Cr3+ ion (in a vacuum). The diagram in the middle shows how the d-orbitals are destabilized by interactions
with nonbonding electron pairs on the ligands, assuming a spherical distribution of the surrounding ligands. The
diagram on the right shows the energy levels of the d-orbitals in an octahedral environment (i.e. [Cr(H2O)6]3+). The
splitting of the lower three d-orbitals from the upper two d-orbitals is labeled Δ, the crystal-field splitting energy.
The electrons in the d-orbitals are repelled by the nonbonding electron pairs on oxygen, which in turn
raises the energies of the d-orbitals. If the Cr3+ ion was surrounded by water molecules equally from all
directions (a spherical crystal field) the 3d orbitals would all be destabilized by the same amount, as
shown in the middle of Figure 2. However, in an octahedral environment the energies of the d-orbitals
3
are not affected equally by the ligands (water molecules in this case). The dz2 and dx2-y2 orbitals point
directly at the ligands, whereas the dxy, dxz and dyz orbitals point in-between the ligands (see Figure 3).
Consequently the ligands destabilize the dz2 and dx2-y2 orbitals to a greater extent than the other d-orbitals,
and these two orbitals have a higher energy than the remaining d-orbitals (dxy, dxz and dyz). ‡ The energy
difference between these two sets of d-orbitals referred to as the crystal field splitting energy, Δ,. As you
might imagine the octahedral geometry is quite stable for a metal cation with 3 d-electrons, such as Cr3+,
because we can put one electron into each of the three lower energy d-orbitals.
dxy
dz2
z2
dx2-y2
dyz
yz
dxz
xz
Figure 3: The orientations of the d-orbitals with respect to an octahedral crystal field. The small blue spheres
represent the negative charge associated with the lone pair of electrons on the ligands. Notice that the dz2 and dx2-y2
orbitals point directly at the ligands, while the dxy, dxz and dyz orbitals point between the ligands. (taken from
Chemistry: The Central Science 10th Edition by Brown, Lemay & Bursten, Prentice Hall, 2006)
Now let’s take a closer look at the relationship between metal-ligand bonding, electronic absorptions and
color. To illustrate the important concepts consider the Cu2+ ion which we have seen produces striking
blue and green colors in solution. Cu2+ has an electron configuration of [Ar]3d9, which is only one
electron shy of completely filling the 3d subshell. This missing electron (or hole) in the 3d-subshell is
responsible for colors of Cu2+ compounds and complexes. If the d-orbitals were filled, as they are for
‡
A more accurate description would be to say that the dz2 and dx2-y2 orbitals form sigma antibonding (σ*) interactions
with the ligands, while the dxy, dxz and dyz orbitals form pi antibonding (π*) interactions with the ligands.
4
Zn2+, there would be no electronic excitations that could take place (with visible light anyway). This the
reason why solutions containing Zn2+ ions are colorless.
Like Cr3+ the preferred geometry of a Cu2+ ion can be roughly described as an octahedron, but unlike Cr3+
two of the bonds (the ones opposite of each other) are elongated so that they are much longer than the
other four bonds. These bonds are quite weak and it is a reasonable approximation to ignore these bonds
altogether, in which case we can approximate the coordination geometry of the Cu2+ ion as square planar,
as shown in Figure 4c.
Long
Bond
O
Long
Bond
≅
O
Cr3+
(a)
Cu2+
(b)
(c)
Figure 4: (a) The octahedral coordination geometry of the Cr3+ ion in [Cr(H2O)6]3+, where only the oxygen atoms
(in red) of the water molecules are shown. All six Cr–O bonds in this molecule are of equal length (b) The distorted
octahedron surrounding the Cu2+ ion in [Cu(H2O)6]2+, the bonds above and below the copper ion are much longer
than the other four bonds. (c) If we neglect the long Cu-O bonds, the complex ion is written as [Cu(H2O)4]2+ and the
coordination geometry about Cu2+ can be described as square planar.
Let’s consider how the distortion of the geometry from octahedral to square planar changes the energy
levels of the d-orbitals. We take the d-orbital energy levels in an octahedron as the starting point. The
most obvious effect is that the major lobes of the dz2 orbital no longer point directly at ligands. This
lowers the energy of this orbital. The absence of ligands along the z-axis also lowers the energy of the dxz
and dyz orbitals. At the same time if we move the remaining four ligands closer to copper (to compensate
for the loss of bonding that comes along with losing two ligands) it raises the energy of the dxy and dx2-y2
orbitals. The net effect of this distortion is shown in Figure 5.
Figure 5 also shows what happens when a [Cu(H2O)4]2+ ion absorbs a photon of light. Remember from
the Pauli Exclusion Principle that each orbital can only hold 2 electrons and those two electrons must be
of opposite spin. Therefore, the only excitations that can occur are those that cause a spin down electron
from one of the d-orbitals with two electrons to move up into the dx2-y2 orbital. For this to occur, the
energy of light must be equal to the energy difference between two orbitals. The lowest energy photon
that can trigger such a transition is a photon with energy equal to the energy difference between the dxy
and dx2-y2 orbitals.
To determine the energy difference between the dxy and dx2-y2 orbitals we use a technique called absorption
spectroscopy. The concept is quite simple. We pass monochromatic light (light of a single wavelength)
through the sample and measure how much light is absorbed by the sample (see Figure 6). When the
energy of the light is different than the energy difference between orbitals, the light is not absorbed and
that color (wavelength) of light passes through the sample. On the other hand, when energy of light
corresponds to the energy difference between dxy and dx2-y2 orbitals the light is absorbed and its energy is
transferred to an electron in the dxy orbital exciting it to the dx2-y2 orbital. By measuring the wavelength
5
where the maximum absorption occurs we can determine the energy difference between the dxy and dx2-y2
orbitals.
Light
dx2-y2
dz2
dx2-y2
Energy
dxy
dxy
dz2
dxy
dx2-y2
dyz
dz2
dxz
dyz
Cu2+ ion in an
octahedral crystal field
dxz
Cu2+ ion in a square
planar crystal field
dyz
dxz
Cu2+ ion after
excitation by light
Figure 5: The energy levels and filling of the 3d-orbitals of a Cu2+ ion. The diagram on the left hand side
corresponds to a Cu2+ ion in an octahedral environment (i.e. [Cu(H2O)6]2+). The diagram in middle shows how the
energies of the d-orbitals change when the coordination environment is distorted from octahedral to square planar
(i.e. [Cu(H2O)4]2+). The diagram on the right hand side shows what happens when a photon of light is absorbed and
its energy is transferred to an electron (shown in red), exciting it from the dxy orbital to the dx2-y2 orbital.
The absorption spectrum for [Cu(H2O)4]2+ is shown in Figure 7. Notice the presence of a peak centered at
approximately 795 nm. This peak in the absorption spectrum occurs because light with that wavelength
has energy equal to the energy separation of the dxy and dx2-y2 orbitals. We can calculate the energy
separation of the d-orbitals by converting the wavelength to energy:
E = hc/λ = {(6.626 × 10–34 J-s)(2.98 × 108 m/s)}/(795 × 10–9 m) = 2.48 × 10–19 J
Figure 6: An absorption spectroscopy experiment (taken from Chemistry: The Central Science 10th Edition by
Brown, Lemay & Bursten, Prentice Hall, 2006)
6
A more convenient unit of energy for discussing electronic excitations is the electron volt (eV). A
electron volt is equal to 1.602 × 10–19 J. Converting from Joules to electron volts:
E = 2.48 × 10–19 J × (1 eV/1.602 × 10–19 J) = 1.55 eV
The visible spectrum runs from approximately 400 – 700 nm, which corresponds to an energy range of
3.10 eV to 1.77 eV. Thus we see that the absorption peak for [Cu(H2O)4]2+ is located in the infrared
region of the spectrum. Why then are aqueous solutions of Cu2+ colored? The reason is because the
absorption peak is broad enough that the absorption spills into the visible region of the spectrum,
absorbing some red and orange light. If we use the color wheel and the concept of complimentary colors
we conclude that a compound which absorbs red light should be green, while one that absorbs orange
light should be blue. This explains the light bluish-green color associated with aqueous solutions of Cu2+.
2.0
1.8
1.6
absorbance
1.4
[Cu(H2O)4]
2+
1.2
2+
[Cu(NH3)4]
1.0
0.8
0.6
0.4
0.2
0.0
200
300
400
500
600
700
800
900
wavelength (nm)
Figure 7: The absorption spectra for [Cu(H2O)4]2+ and [Cu(NH3)4]2+. The photos show the colors of the respective
solutions.
We have seen in earlier experiments that Cu2+ solutions turn a deep blue color when NH3 is added to the
solution. We know the change in color can be attributed to the formation of the complex ion
[Cu(NH3)4]2+. However, there are two questions surrounding this color change that we have not
considered. How does the absorption spectrum change and why does [Cu(NH3)4]2+ absorb different
wavelengths of light than does [Cu(H2O)4]2+?
The absorption spectrum of [Cu(NH3)4]2+ is also shown in Figure 7. Notice that the absorption maximum
has shifted from ~795 nm to ~625 nm, which falls squarely in the red-orange region of the visible
spectrum. Because we are once again dealing with a fairly broad peak there is also significant absorption
of yellow light and even some absorption of green light. Consequently, only violet and blue (and some of
the green) light can pass through without being absorbed. When these colors of light (not pigments) are
blended the result is the deep blue color of the [Cu(NH3)4]2+ (see Figure 7).
7
Based on the absorption maximum of 625 nm, we can calculate the energy difference between dxy and
dx2-y2 orbitals to be 3.04 × 10–19 J or 1.99 eV. This value is larger than the 1.55 eV separation in
[Cu(H2O)4]2+. Why does the energy difference between these two orbitals increase when we replace the
water molecules with ammonia molecules? The answer to this question arises from differences in the
way oxygen and nitrogen bond to copper. The nitrogen atom of the ammonia molecule interacts more
strongly with copper than does the oxygen atom of the water molecule. As a general rule of thumb
whenever we increase the bonding between the metal and the ligands the energy separation of the dorbitals increases. The bonding interaction can increase in a variety of ways: (a) due to an increase in the
oxidation state of the cation, (b) due to a change to a more electronegative cation, or (c) due to a change
in the ligand. Common ligands can be arranged in order of increasing ability to increase the energy
separations or splitting of the d-orbitals. This arrangement is called the spectrochemical series:
(smallest splitting) Cl– < F– < H2O < NH3 < en < NO2– < CN– (largest splitting)
Generally the increase in splitting is due to an increase in the bonding interactions between the ligand and
the metal. For example, you can see that going from fluorine (F–) to oxygen (H2O) to nitrogen (NH3, en,
NO2–) the electronegativity of the anion bonding to the cation decreases. Because anions are more
electronegative than cations this causes the difference in electronegativity to become smaller, thereby
increasing covalency of the bond. For a complete analysis there are other factors to should be taken into
account, however, but the cation-anion electronegativity difference is one important factor. For our
purposes it is sufficient to know that NH3 is higher than H2O in the spectrochemical series. Therefore, we
would expect the energy separation between dxy and dx2-y2 orbitals to increase and the absorption
maximum to shift to higher energy (lower wavelengths), as observed.
In this lab you will collect absorption spectra for the coordination compounds [Cr(H2O)6]3+ and
Cr(OH)4(H2O)2]–. Note that although we write tetrahydroxychromium(III) ion as [Cr(OH)4]– there are
two water molecules bonded to chromium that are often left off. Therefore, a more accurate formula
would be [Cr(OH)4(H2O)2]–. Because both complexes contain a Cr3+ ion in an octahedral crystal field the
energy level splitting shown on the right hand side of Figure 2 is applicable. You will calculate the
octahedral crystal field splitting energy, Δ, for both species from these spectra. Note that the absorption
spectra for cations with only 1 electron (like Ti3+) or those with only one electron missing from a filled dsubshell (like Cu2+) are simpler than spectra associated with cations that have an intermediate number of
electrons. For ions, like Cr3+, with 3 d-electrons three peaks are expected in the spectrum. In order to
estimate the crystal field splitting energy, Δ, you will need to determine the wavelength and energy
that correspond to the lowest energy (highest wavelength) peak in the absorption spectrum.
ELECTRONIC STRUCTURE AND OPTICAL ABSORPTION SPECTROSCOPY: INTERATOMIC
(CHARGE TRANSFER) TRANSITIONS
The electronic transitions that we have been discussing so far are intra-atomic transitions. That is to say
they are transitions between different orbitals on the same atom. It is also possible for absorption of light
to cause an electron to be transferred from one atom to a different atom. These types of transitions are
called charge transfer transitions. Charge transfer excitations from a ligand to a metal only occur in the
visible for cases where the metal-ligand bond is highly covalent. This situation tends to be met only when
the cation is in a high oxidation state, as it is in the polyatomic oxyanions CrO42– and MnO4–, which
contain Cr(VI) and Mn(VII), respectively. We will discuss the absorption spectrum and electronic
structure of the permanganate ion here, and you will collect and analyze data for CrO42– during the course
of this experiment.
The oxidation state of manganese in MnO4– is Mn(VII), which means that formally manganese doesn’t
have any electrons in its d-orbitals. Unfortunately this means that it is not possible to understand the
8
absorption spectrum and color simply by investigating the splitting of the d-orbitals. Instead we must
consider a more complete molecular orbital (MO) diagram. The MO diagram for the permanganate ion is
shown below in Figure 8.
t2* orbitals
(antibonding)
Δtetr
e* orbitals
(antibonding)
[MnO4]–
CT2
Energy
Mn 3d-orbitals
CT1
Nonbonding
Oxygen 2p MO’s
e orbitals
(bonding)
12 Oxygen 2p orbitals
(4 oxygens x 3 p
orbitals)
t2 orbitals
(bonding)
Figure 8: The molecular orbital diagram for a permanganate ion.
In all likelihood this is the most complicated MO diagram than you have seen (even though Figure 8 is a
somewhat simplified version of the full MO diagram for MnO42–). § So it would be understandable if this
diagram seemed a little intimidating on first encounter. For the time being suppress your urge to run away
from chemistry forever. This MO diagram follows the same basic rules as do the MO diagrams you have
already encountered for simple molecules like H2 and O2. To analyze the diagram we will break the
molecule into fragments, one fragment being the manganese atom and one fragment being the four
oxygen ligands. With this approach we can think of this molecule as a pseudo-diatomic molecule. We
only need to consider the interactions between the manganese atom and the oxygen ligands. Let’s
consider the basic principles molecular orbital theory and how they apply to the MO diagram for MnO42–.
1. The number of molecular orbitals in a molecule must be equal to the number of atomic orbitals
associated with the atoms that form the molecule.
In this case the central manganese atom has five 3d-orbitals, which are shown on the left. Each of the
four oxygen atoms in the molecule has three 2p-orbitals, which leads to 4 (atoms) × 3 (orbitals/atom) = 12
§
We have neglected the 2s orbitals on oxygen and the 4s and 4p orbitals on manganese. For the purposes
of understanding the absorption spectrum of a MnO42– molecule this is not a problem, because all of the
electronic transitions of interest are described by the molecular orbitals shown in Figure 8.
9
ligand based orbitals. These are shown on the right (remember we are neglecting the 4s and 4p orbitals of
manganese and the 2s orbitals of oxygen).
2. If both parts of the molecule have orbitals of the same symmetry they can overlap to form a bonding
molecular orbital and an antibonding molecular orbital. The bonding MO has a lower energy than
either of the atomic orbitals that mix to form it, and the antibonding MO has a higher energy than
either of the atomic orbitals that mix to form it.
In the permanganate ion each of the five 3d-orbitals on Mn mix with an oxygen based molecular orbital to
form five bonding MO’s (labeled e and t2 at the bottom of the MO diagram) and five antibonding MO’s
(labeled e* and t2* at the top of the MO diagram).
3. The energy separation between a bonding MO and its corresponding antibonding MO increases as
the overlap of the orbitals increases.
You may be wondering why there are two different energies for the bonding MO’s and two different
energies for the antibonding MO’s. The reason for this is because the Mn dxy, dyz and dxz orbitals (labeled
as the t2 orbitals in the diagram) overlap more effectively with the oxygen orbitals than do the Mn dz2 and
dx2-y2 orbitals (labeled the e orbitals in the diagram). Because of this the bonding t2 orbitals are lowest in
energy and the antibonding t2 orbitals are highest in energy.
Keep in mind that this description is only accurate when the geometry is a tetrahedron. You will probably
notice that the d-orbitals that overlap the most effectively for a tetrahedron (dxy, dyz and dxz) are the
opposite of those that overlap most effectively for an octahedron (dz2 and dx2-y2). We will explore this
further in the report questions using an online resource. Although it was discussed in a slightly different
light the d-orbitals shown in Figures 2 and 5 correspond to antibonding MO’s arising from interactions
between the metal d-orbitals and the ligand orbitals.
4. If an orbital on one side of the molecule cannot find a partner of the same symmetry on other side of
the molecule, then that orbital becomes a non-bonding MO. Its energy is unchanged by formation of
the molecule.
Because there are only five orbitals on the metal (Mn) available for bonding and twelve orbitals for the
ligands (O) there are 12 – 5 = 7 ligand orbitals that do not have a Mn orbital with which to bond.
Therefore, we have seven nonbonding oxygen 2p orbitals in the MO diagram for MnO42–.
Now we are ready to consider the absorption spectrum for a permanganate ion. The lowest energy
electronic excitation is labeled CT1 and marked with a blue arrow in Figure 8. If a photon of light with
an energy equal to CT1 strikes the molecule it will excite an electron from a nonbonding oxygen 2p
orbital to an antibonding Mn 3d – O 2p orbital. Because the antibonding orbital has significant
manganese character, we could loosely describe this transition as an excitation of an electron from an
oxygen orbital to a manganese orbital. That is why this transition is called a charge transfer excitation.
The second lowest energy electronic excitation is labeled CT2 and marked with a red arrow in Figure 8.
The absorption spectrum of the permanganate ion in aqueous solution is shown in Figure 9. The CT1
peak (centered at roughly 530 nm) absorbs strongly in the green and yellow regions of the spectrum. The
complimentary colors are red and violet, which explains the deep purple, almost maroon color of a
permanganate solution. The CT2 peak falls in the UV, and therefore, it does not have a significant effect
on the color.
10
0.5
CT1
Oxygen nonbonding to e*
0.4
CT2
absorbance
Oxygen nonbonding to t2*
0.3
0.2
0.1
0.0
250
350
450
550
650
750
wavelength (nm)
Figure 9: The absorption spectra for a KMnO4 solution. The two broad peaks correspond to the electronic
transitions labeled CT1 and CT2 in figure 8.
From the MnO42– absorption spectrum we can determine the energy separation between the oxygen 2p
nonbonding orbitals and the antibonding dx2-y2 and dz2 orbitals (labeled e*) as well as the energy separation
between oxygen 2p nonbonding and the dxy, dyz, dxz orbitals (labeled t2*).
Oxygen nonbonding → e*:
E = hc/λ = {(6.626 × 10–34 J-s)(2.98 × 108 m/s)}/(530 × 10–9 m)
E = 3.7 × 10–19 J
E = 3.7 × 10–19 J × (1 eV/1.602 × 10–19 J) = 2.3 eV
Oxygen nonbonding → t2*:
E = hc/λ = {(6.626 × 10–34 J-s)(2.98 × 108 m/s)}/(330 × 10–9 m)
E = 6.0 × 10–19 J
E = 6.0 × 10–19 J × (1 eV/1.602 × 10–19 J) = 3.7 eV
Knowing the energies of both sets of Mn 3d orbitals allows us to indirectly calculate the tetrahedral
crystal field splitting energy, Δtetr:
Δtetr = E(t2*) – E(e*) = 3.7 eV – 2.3 eV = 1.4 eV
One last note about interatomic (charge transfer) vs. intra-atomic (d to d) transitions. The charge transfer
excitations absorb light much more effectively. For example, a MnO4– solution absorbs light (at its
maximum absorbance) more than 200 times more effectively than does a Mn3+ ion in solution (at its
maximum absorbance). Therefore, compounds where the absorption of visible light is due to a charge
11
transfer excitation make much more effective pigments than do compounds where the electronic
absorptions are due to d-to-d transitions.
SOLID STATE SYNTHESIS OF A PIGMENT
The second part of this lab is to prepare a powdered sample of a pigment. Each person in your lab section
will prepare a different composition. In the next period, you will characterize the composition, absorption
spectrum and color of your pigment. By combining your results with those of your classmates we will be
able to map out the behavior of an entire 4-component system. The discussion and procedure for the next
lab will describe the characterization of your pigment in detail. For now we only need to be concerned
about the preparation or synthesis of your sample.
Most of the compounds that you have synthesized in lab this year have been prepared by precipitation
reactions. While precipitation routes are quick, and can be very effective, they can also be difficult to
control. Some of the ions in solution may precipitate while others do not. This makes it difficult to
control the stoichiometry. Another issue is the fact that we will be using X-ray diffraction to characterize
the composition of your samples, yet precipitation routes do not always lead to crystalline products. For
these reasons and others you will prepare your sample by direct reaction of solid phase reactants. This
will require you to measure out the exact amounts of your reagents, grind those reactants together with a
mortar and pestle, and heat the resulting powder to high temperature (500 °C – 900 °C) for several hours.
The peer mentors and TA’s will assist with the heating of your sample, but it will be up to you to measure
out the reagents in the correct quantities and grind them into a fine mixture.
You have been assigned one member of the SrxPb1-x(CrO4)y(MoO4)1-y solid solutions series to synthesize.
The starting materials are the solids PbCO3, SrCO3, Cr2O3 and MoO3. Your assignment is to determine
how much of each of these reagents is needed to prepare 1.500 g of the pigment you have been assigned.
To do so you need to go through the following steps before you come to lab.
1. Write out the balanced equation that represents the reaction that will be used to prepare your
pigment.
2. Determine the formula weight of your product.
3. Given the target yield of 1.500 grams and the formula weight of your compound, determine the
number of moles of product that you want to prepare.
4. Assume a stoichiometric reaction (where each reactant is equally a limiting reactant) and
calculate how many grams of each reactant are needed for the reaction.
To show how this is done let’s go through these steps one by one for an example composition,
Pb(CrO4)0.5(WO4)0.5.
1. Using PbCO3, Cr2O3, and WO3 write out the balanced chemical equation to represent the reaction
that will lead to the formation of your pigment.
There are several points to keep in mind regarding solid state syntheses from oxides and carbonates.
-
First of all, at the high temperatures where the reactions occur (in this lab we will be heating to
~700 °C), the carbonates such as PbCO3 and SrCO3 decompose by releasing gaseous CO2 into the
atmosphere. This leaves a metal oxide, either PbO or SrO, behind, which is incorporated into the
product. Nonetheless, CO2 must be included as a product in order to balance the equation.
-
Secondly, the chromium containing reactant, Cr2O3, contains Cr3+ ions. Yet the product contains
Cr(VI) ions. An oxidizing agent is needed to oxidize the chromium. In these reactions the
12
oxidizing agent is O2 from the atmosphere. The O2 is reduced to oxide ions that are incorporated
into the product.
Taking these two factors into account give the following unbalanced chemical equation:
PbCO3(s) + Cr2O3(s) + WO3(s) + O2(g) →
Pb(CrO4)0.5(WO4)0.5(s) + CO2(g)
Balancing the equation from here is straightforward, with one twist. Normally we do not encounter
fractional subscripts in the formula for a compound. Such a formula would never be used for a molecular
substance, but as an empirical formula to represent the overall composition of a crystalline solid solution
this is acceptable. The process is simpler if we use a coefficient in front of the pigment product that leads
to an integer number of atoms on the product side. With that in mind lets start by assuming a coefficient
of 10 in front of the Pb(CrO4)0.5(MoO4)0.5.
PbCO3(s) + Cr2O3(s) + WO3(s) + O2(g) →
10 Pb(CrO4)0.5(WO4)0.5(s) + CO2(g)
Next we adjust the coefficients to balance for Pb, Cr and W. If necessary we can use fractional
coefficients. At the end we will multiply by a common factor to obtain integer coefficients:
10 PbCO3(s) + 2.5 Cr2O3(s) + 5 WO3(s) + O2(g) →
10 Pb(CrO4)0.5(WO4)0.5(s) + CO2(g)
Next, we can choose coefficients for the gases that balance the number of oxygen and carbon atoms:
10 PbCO3(s) + 2.5 Cr2O3(s) + 5 WO3(s) + 3.75 O2(g) → 10 Pb(CrO4)0.5(WO4)0.5(s) + 10 CO2(g)
Finally, we multiply by a factor that gives integer coefficients. In this case we need to multiply all of the
coefficients by a factor of 4:
40 PbCO3(s) + 10 Cr2O3(s) + 20 WO3(s) + 15 O2(g) → 40 Pb(CrO4)0.5(WO4)0.5(s) + 40 CO2(g)
2. Determine the formula weight of your product.
This step is straightforward based on the atomic weights of Pb (207.2 g/mol), Cr (52.00 g/mol), W
(183.84 g/mol) and O (16.00 g/mol). It is probably a bit simpler to write the empirical formula as
PbCr0.5W0.5O4.
FW = 1 × (207.2 g/mol) + 0.5 × (52.00 g/mol) + 0.5 (183.84 g/mol) + 4 (16.00 g/mol)
FW = 389.1 g/mol
3. Determine the number of moles of product that you want to prepare.
This is a straightforward calculation given the target yield of 1.50 g and the formula weight we calculated
in step 2:
# of Moles = 1.500 g × (1 mol/389.1 g) = 3.855 × 10–3 mol
4. Calculate the number of grams of each reactant that are needed to prepare the product in a
stoichiometric reaction.
We can use the number of moles of product, the coefficients from the balanced equation and the formula
weights of the reactants, PbCO3 (267.2 g/mol), Cr2O3 (152.0 g/mol) and WO3 (231.8 g/mol). We do not
13
need to concern ourselves with the amount of oxygen, because the oxygen will come from the
atmosphere:
Grams PbCO3
(3.855 × 10–3 mol PbCr0.5W0.5O4) × (40 mol PbCO3/40 mol PbCr0.5W0.5O4) × (267.2 g/mol)
= 1.030 g PbCO3
Grams Cr2O3
(3.855 × 10–3 mol PbCr0.5W0.5O4) × (10 mol Cr2O3/40 mol PbCr0.5W0.5O4) × (52.0 g/mol)
= 0.0501 g Cr2O3
Grams WO3
(3.855 × 10–3 mol PbCr0.5W0.5O4) × (20 mol PbCO3/40 mol PbCr0.5W0.5O4) × (231.8 g/mol)
= 0.4468 g PbCO3
Once all of the reagents are weighed out you will need to mix them well by grinding in a mortar and
pestle. Be sure to grind for the recommended amount of time to insure the products are sufficiently
mixed to react completely when heated.
MATERIALS REQUIRED
Equipment
Chemicals
four 4-mL test tubes
test tube rack
Bunsen burner, wire mesh
ring stand, ring
thin-walled rubber tubing
small stirring rod
100-mL beaker (water bath)
0.0200M chromium (III) nitrate, Cr(NO3)3 (aq)
3M sodium hydroxide, NaOH (aq)
30% hydrogen peroxide, H2O2 (aq)
0.1M lead (II) nitrate, Pb(NO3)2 (aq)
PbCO3 (s) & SrCO3 (s)
Cr2O3 (s) & MoO3 (s)
CAUTIONS
Be careful to avoid burns from the open flame and the ring. Salts containing Cr(VI) as well as those
containing Pb2+ are toxic; be sure to wash your hands thoroughly.
14
PROCEDURE
A. Preparation of a Pigment Solid Solution (solid state reaction route)
1. Obtain a mortar and pestle from your lab instructor.
2. Using the analytical balance weigh out the proper quantities of the starting materials (SrCO3,
PbCO3, MoO3 and/or Cr2O3) that were calculated in question 2 of the prelaboratory exercise.
Place them in your mortar and pestle. At the same time weigh an empty crucible and record
the mass in your notebook.
3. Grind the powders together for at least 15 minutes. Note the color in your notebook.
4. After you finish grinding, place the powders in a ceramic crucible, weight the crucible and
record the mass in your notebook.
5. Label the crucible with a sticker, then give the labeled crucible to the assigned peer mentor or
teaching assistant.
B. Preparation of [Cr(H2O)6]3+, [Cr(OH)4]–, CrO42–, and PbCrO4 (wet chemical route)
6. Obtain four clean, dry 4-mL test tubes. Label them Tube 1, Tube 2, Tube 3 & Tube 4.
7. Prepare a water bath in a 100-mL beaker. The beaker should be only half full of water so the
test tubes will not float. Use a gentle flame for the bath.
8. Add 10 drops of 0.0200M Cr(NO3)3 solution to each of the four test tubes. Note the color of
the solution in your notebook.
9. Add 1 drop of 3M NaOH to Tubes 2, 3, and 4 only. Observe the hazy precipitate that forms.
10. Add 2 more drops of 3M NaOH to Tubes 2, 3, and 4 to dissolve the precipitate and form
complex ions. You may need to stir the solution with a clean, small stirring rod. Note the
color of the solution in your notebook.
11. Add 6 drops of 30% H2O2 to Tubes 3 and 4. Stir the solution with a clean, small stirring rod.
The solution should turn a brownish color.
12. Place Tubes 3 and 4 in the water bath. You will see the solution slowly turn a bright yellow
while heating. Heat gently in the water bath for 20 minutes.
**NOTE: It’s important to heat the solution for 20 minutes to destroy any excess H2O2. Any
remaining H2O2 will reduce CrO42– to Cr3+ if the solution becomes acidic.
13. After heating, carefully remove Tubes 3 and 4 from the water bath and allow them to cool.
Turn off the Bunsen Burner.
14. Add 1 mL of 0.1M Pb(NO3)2 to Tube 4. You may wish to centrifuge Tube 4 to view the
precipitate. Note the color of this precipitate in your notebook.
15
C. Collection of Absorption Spectra for [Cr(H2O)6]3+, [Cr(OH)4]– and CrO42–
USB Cable
Blue Fiber Assembly
USB 4000
ISS
Cuvette Holder
15. Take Tubes 1-4 in a test tube rack to one of the Spectrometer Stations in your room.
16. Match your Tubes 1-3 with Cuvettes 1-3 at the Spectrometer Stations and observe the color.
Make a note of the similarities and/or differences in your notebook.
17. Standardize the spectrometer: (to save lab time, your TA may have already done this step)
a. Open the SpectraSuite software on the laptop (making sure the above assembly is
correct).
b. Check the Data Sources window (top left of screen) to make sure software recognizes
spectrometer.
c. Turn on the light source (switch all the way to the left).
d. Insert a cuvette with distilled water into the cuvette holder (marked “W”).
e. Click “File” → “New” → “Absorbance Measurement”.
f.
Follow the Absorbance Measurement Wizard:
i. Step 1: Use default and Click “Next”
ii. Step 2: Increase/decrease the integration time until the “Last Peak Value” is in
black lettering and closely matches the “Recommended Peak Value”. Use the up
& down arrows to adjust the time. Click “Next” when complete.
iii. Step 3: Make sure the cuvette with the distilled water is inserted, and click on the
light bulb picture to take a light reference. Click “Next”.
iv. Step 4: Remove the cuvette. Block the light by placing an obstruction in the
cuvette holder, and click the black light bulb to take a dark reference. Click
“Finish”.
18. Take an absorbance spectra for Cuvette 1 ([Cr(H2O)6]3+):
a. Place the cuvette with sample in the cuvette holder. You should see the spectrum on the
screen.
16
b. Click anywhere on the spectra to see the absorbance at the chosen wavelength. You will
see a green line where you click. (The values will be at the bottom of the screen wavelength in the box, absorbance value in color.)
c. Click the
button to automatically adjust the axes to fit the screen.
d. Click the
button to manually adjust the scale on the x and y axes. (You may
want to start the x-axis at 300 nm.)
e. To freeze the pattern, first click anywhere on the spectra. (Again, you will see a green
line.) Click the Pause button
f.
to freeze the pattern.
To save data, first freeze the pattern. Click the “Copy Spectral Data to Clipboard” button
on the right side of the toolbar above the spectra. This button is similar to the “copy”
button in MS Word.
“Copy Spectral Data to
Clipboard” button
g. Click anywhere on the spectra to see the absorbance at the chosen wavelength. You will
see a green line where you click. (The values will be at the bottom of the screen wavelength in the box, absorbance value in color.)
h. Open MS Excel on the computer.
i.
Click on an empty cell in a new spreadsheet. Click the “Paste” button in Excel. You
should see your data from the spectra. Save your data where your instructor designates.
j.
You can save and graph your data using Excel.
k. Use the cursor to determine the approximate wavelength of any peaks in the spectrum.
Record those values on the printed version of your spectrum.
l.
Go back to the SpectraSuite window on the laptop.
m. Click the Play button
to un-freeze the pattern.
19. Repeat Step 13 for Cuvettes 2 ([Cr(OH)4]–) and 3 (CrO42–).
17
WASTE DISPOSAL
Pure acids and bases should be washed down the sink drain with a large amount of water. All other
solutions and solids must be collected in a beaker at your desk and added to the inorganic waste
beaker in the hood.
18
REEL I-A REPORT FORM
ELECTRONIC STRUCTURE, COLOR AND PIGMENTS
Last Name
First Name
Day Lab Meets / Hours Lab Meets
Lab Room Number
Quarter/Year
Date Submitted
RESULTS
1. Describe the color of
a. [Cr(H2O)6]3+
______________________________
b. [Cr(OH)4(H2O)2]– ______________________________
c. CrO42–
______________________________
d. PbCrO4
______________________________
2. List the wavelengths (in nm) corresponding to absorbance peaks in spectra of the following solutions:
a. [Cr(H2O)6]3+
____________
____________
____________
b. [Cr(OH)4(H2O)2]– ____________
____________
____________
c. CrO42–
____________
____________
3. Calculate energies (in eV) corresponding to absorbance peaks in spectra of the following solutions:
a. [Cr(H2O)6]3+
____________
____________
____________
b. [Cr(OH)4(H2O)2]– ____________
____________
____________
c. CrO42–
____________
____________
4. Calculate the crystal field splitting energy (in eV) for each of the following complexes:
a. [Cr(H2O)6]3+
Δoct =
____________
b. [Cr(OH)4(H2O)2]–
Δoct =
____________
c. CrO42–
Δtetr =
____________
19
REPORT QUESTIONS
1. Based on your results would you expect OH– to be located higher (larger Δ) or lower (smaller Δ) in
than H2O in the spectrochemical series? Explain.
2. What color would you expect a solution containing [CrCl6]3– to be? Explain.
3. The red color of a ruby comes from the presence of Cr3+ impurities in Al2O3. In this structure the Cr3+
impurities are surrounded by an octahedron of oxygen atoms, much like Cr3+ ions in aqueous
solution. The absorption peaks in ruby are measured to occur at 555 nm, 389 nm and 256 nm. (a)
Make a semiquantitative sketch of the absorption spectrum for ruby. Is this spectrum consistent with
the red color of a ruby? (b) What is the energy shift (in eV) for each peak with respect to the values
you measured [Cr(H2O)6]3+. (c) Would you expect the Cr–O bond distance in ruby to be shorter or
longer than in [Cr(H2O)6]3+? Explain.
4. Go to the following website http://undergrad-ed.chemistry.ohio-state.edu/pigments/ and click on the
CrO42– link to see calculated images of molecular orbitals in a chromate ion.
a. Describe each of the molecular orbitals HOMO1, HOMO2, HOMO2, LUMO1 and LUMO2
in terms of the atomic orbitals that are involved in the MO (i.e. O 2p, Cr 3dxy, Cr 3dxz, Cr
3dyz, Cr 3dz2 or Cr 3dx2-y2) and the bonding interaction (bonding, nonbonding or antibonding).
[Note: If you select axes on it will help you determine which d-orbital you are looking at.]
b. The energy of the molecular orbital is shown below the image (the energies are given in
atomic units, au, where 1 au = 27.21 eV). Given your assignments in part (a) what energy
would you assign for the lowest energy charge transfer transition (CT1 in Figure 8). At what
wavelength would you expect to find the peak in the absorption spectrum corresponding to
CT1?
c. Carry out a similar analysis for the MoO42– ion, which can also be found at this site. What is
the energy of the CT1 transition for a MoO42– ion? At what wavelength would you expect to
find the peak in the absorption spectrum corresponding to CT1? What color would you expect
for a solution containing MoO42– ions?
20