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).

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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) Cr(OH)3 (s) + OH–

(aq) → Cr(OH)4–(aq) →

blue/violet gray-green green solution ppt 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:

2 OH– + 2 Cr(OH)4–(aq) + 3 H2O2 8H2O + 2 CrO4

2–(aq) ⎯→⎯Δ

In this reaction the peroxide ion (O2

2–) 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) + CrO4

2–(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 CrO4

2– 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.

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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.

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

ddxxyy ddyyzz ddxxzz ddzz22 ddxx22--yy22

ddxxyy ddyyzz ddxxzz ddzz22 ddxx22--yy22

ddyyzz ddxxzz ddxxyy

ddzz22 ddxx22--yy22

Ene

rgy

Δ

Cr3+ ion with a spherical crystal field

Cr3+ ion in an octahedral crystal field

Free Cr3+ ion

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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.

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.

ddz ddxx22--yy22z22

ddxxyy ddy ddxyzz xzz

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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

Long Bond

O

Cu2+O

Cr3+

(a) (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

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where the maximum absorption occurs we can determine the energy difference between the dxy and dx2-y2 orbitals.

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)

ddyyzz ddxxzz ddxxyy

ddzz22 ddxx22--yy22

Cu2+ ion after excitation by light

Cu2+ ion in an octahedral crystal field

Ene

rgy

ddyyzz ddxxzz

ddxxyy

ddzz22

ddxx22--yy22

Cu2+ ion in a square planar crystal field

ddyyzz ddxxzz

ddxxyy

ddzz22

ddxx22--yy22 Light

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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+.

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0.4

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1.4

1.6

1.8

2.0

200 300 400 500 600 700 800 900

wavelength (nm)

abso

rban

ce

[Cu(H2O)4]2+

[Cu(NH3)4]2+

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).

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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 d-orbitals 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 d-subshell (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 CrO4

2– 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 CrO4

2– 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

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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.

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 MnO4

2–).§ 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 MnO4

2–. 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 MnO4

2– molecule this is not a problem, because all of the electronic transitions of interest are described by the molecular orbitals shown in Figure 8.

EE nn ee

rr ggyy

Mn 3d-orbitals

12 Oxygen 2p orbitals (4 oxygens x 3 p

orbitals)

Nonbonding Oxygen 2p MO’s

CT1

t2* orbitals

(antibonding)

e* orbitals

(antibonding)

[[MMnnOO44]]––

t2 orbitals (bonding)

e orbitals (bonding)

Δtetr

CT2

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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 MnO4

2–.

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.

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wavelength (nm)

abso

rban

ceCT1

Oxygen nonbonding to e*

CT2 Oxygen nonbonding to t2*

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 MnO4

2– 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

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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

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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

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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 0.0200M chromium (III) nitrate, Cr(NO3)3 (aq) test tube rack 3M sodium hydroxide, NaOH (aq) Bunsen burner, wire mesh 30% hydrogen peroxide, H2O2 (aq) ring stand, ring 0.1M lead (II) nitrate, Pb(NO3)2 (aq) thin-walled rubber tubing PbCO3 (s) & SrCO3 (s) small stirring rod Cr2O3 (s) & MoO3 (s) 100-mL beaker (water bath)

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.

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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]–, CrO4

2–, 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 CrO4

2– 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.

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C. Collection of Absorption Spectra for [Cr(H2O)6]3+, [Cr(OH)4]– and CrO42–

USB 4000ISS

Cuvette Holder

USB Cable

Blue Fiber Assembly

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.

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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 to freeze the pattern.

f. 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–).

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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.

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REEL I-A REPORT FORM ELECTRONIC STRUCTURE, COLOR AND PIGMENTS Last Name First Name Quarter/Year Day Lab Meets / Hours Lab Meets Lab Room Number Date Submitted RESULTS 1. Describe the color of

a. [Cr(H2O)6]3+ ______________________________

b. [Cr(OH)4(H2O)2]– ______________________________

c. CrO4

2– ______________________________

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. CrO4

2– ____________ ____________

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. CrO4

2– ____________ ____________

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. CrO4

2– Δtetr = ____________

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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 MoO4

2– ion, which can also be found at this site. What is the energy of the CT1 transition for a MoO4

2– 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 MoO4

2– ions?

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