The Synthesis and Color of trans-dichlorobis(ethylenediamine)cobalt(iii) Chloride

Transcription

1 CHEM 122L General Chemistry Laboratory Revision 1.0 The Synthesis and Color of trans-dichlorobis(ethylenediamine)cobalt(iii) Chloride 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 Cobalt, trans-dichlorobis(ethylenediamine)cobalt(iii) Chloride, [Co(en) 2 Cl 2 ]Cl, from Cobalt(II) Chloride Hexahydrate, CoCl 2 6H 2 O. 4 CoCl 2 6H 2 O + O HCl + 8 en 4 [Co(en) 2 Cl 2 ]Cl + 26 H 2 O (Eq. 1) "en" is short-hand for Ethylenediamine: NH 2 -CH 2 CH 2 -NH 2 This synthesis involves oxidation of the Cobalt from Co +2 to Co +3 by Oxygen dissolved in the reaction solution: 4 Co 2+ + O H + 4 Co H 2 O (Eq. 2) In an aqueous environment, Co 3+ is unstable and is readily reduced back to Co 2+. However, by adding Ethylenediamine ligands to the Co 3+ results in a complex that stabilizes the Co 3+ oxidation state. Hence, if the oxidation is carried out in the presence of Ethylenediamine, the Co 3+ formed is stabilized and retains the higher oxidation state. One of the key characteristics of the resulting compound is its dark green color. As part of this exercise, we will examine why it is so distinctly colored. [Co(en) 2 Cl 2 ]Cl is a Coordination Compound. Coordination Compounds are substances that contain at least one Complex Ion, a species containing a central metal cation that is bonded to molecules or anions, called Ligands. In the present case, the coordination compound [Co(en) 2 Cl 2 ]Cl is formed from the Co(en) 2 Cl 2 + complex cation and the Cl - anion. The Complex itself contains four Ligands; two Cl - s and two Ethlyenediamines (en). (Note the [ ] s around the Complex Ion Co(en) 2 Cl 2 + in the chemical formula.) Like other Water soluble compounds, this ionic substance can dissolve in Water according to:

2 P a g e 2 [Co(en) 2 Cl 2 ]Cl(aq) Co(en) 2 Cl 2 + (aq) + Cl - (aq) (Eq. 3) The covalent bonding between the metal ion and its ligands, in our case, between the Co 3+ and the Cl - s and en s, is via Coordinate Covalent Bonds; the ligands act as Lewis Bases and donate pairs of electrons to the metal, acting as a Lewis Acid. As a general example of this type of bonding, consider the formation of the Co(NH 3 ) 6 3+ complex ion. The NH 3 ligand has one lone pair of electrons which can coordinate covalent bond to the Co 3+ metal center: Since this example complex contains six NH 3 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: Coordination Number Geometry 2 Linear 4 Square Planar or Tetrahedral 6 Octahedral

3 P a g e 3 For our complex ion, Co(en) 2 Cl 2 +, the coordination number is also six. 2 electron pairs are donated to the Co 3+ ion by each chloride ion, Cl -, ligand. The remaining 4 electron pairs are donated by the two Ethylenediamine ligands, the Ethylenediamine ligand being bidentate; forming two bonds to the metal per en. Because each en ligand in our Complex is bidentate, two different Isomers of our Complex are possible; a cis form and a trans form. These are pictured below: We will be synthesizing the trans isomer (Green) of this complex, although some of the cis isomer (Purple) will be formed as an impurity. Isomers are substances that have the same chemical formula, but are different compounds; i.e., each has its own set of physical and chemical properties. Each of the above pictured isomers has the chemical formula CoCl 2 C 4 H 16 N 4 + and yet the trans form is distinctly different than the cis form, as can be seen from the fact that the former is a Green color and the later is a Purple color. As you will note during the laboratory, the starting material, Cobalt (II) chloride hexahydrate, is a brightly colored pink-purple substance, whereas our product is a deep forest green. 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. This is due to an interaction between the donated electron pair of the ligand with the lobes of the d-orbitals on the metal center; with the splitting pattern determined by the geometry of the ligands about the d-orbitals.

4 P a g e 4 Recall, the d-orbitals have the following shapes: So, if the negatively charged bonding electron pairs of the ligands interact with high electron density regions of say the d x2-y2 orbital along the x and y axes: 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 d xy orbital along these same axes, the approach is along a nodal plane and the interaction is less unfavorable, causing the energy of the system not to rise as much.

5 P a g e 5 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 can be seen, 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. (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.)

6 P a g e 6 These qualitative results are the product of Crystal Field Theory, which postulates how a metal cation interacts with its ligands. The details of this theory can be found in any general chemistry textbook. Finally, 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, "strong-field 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 - < H 2 O < SCN - < NH 3 < en < NO 2 - < CN - < CO Consider what happens when light interacts with a metal complex. The complex Ti(H 2 O) 6 3+ contains a d 1 metal ion: Ti = [Ar] 4s 2 3d 2 Ti 3+ = [Ar] 3d 1 (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 E photon =, then the complex will absorb the photon. (Recall also, the relationship between the energy of a photon E photon and its wavelength is E photon = 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.

7 P a g e 7 Color Wheel for Determining Complementary Colors Consider a slightly more complex case; CoF 6 3- vs. Co(NH 3 ) Both involve a Co 3+, or d 6 metal ion and both are octahedral. Co = [Ar] 4s 2 3d 7 Co 3+ = [Ar] 3d 6 Because F - is a weak field ligand, we expect a small splitting energy. Whereas NH 3 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 CoF 6 3- the energy to pair the electrons is greater than and so we have a "high spin" complex; the electrons still obey Hund's Rule. In Co(NH 3 ) 6 3+ 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 CoF 6 3-, it absorbs long wavelength red photons ( ~ 700 nm) and appears green. For Co(NH 3 ) 6 3+, because is much larger it absorbs shorter wavelength blue photons ( ~ 475 nm) and appears yellow-orange.

8 P a g e 8 Thus, we will be synthesizing a particular isomer (trans) of the coordination compound [Co(en) 2 Cl 2 ]Cl where the complex ion Co(en) 2 Cl 2 + adopts an octahedral geometry due to the six coordinate covalent bonds formed with the ligands. Further, this complex is Green because of the particular splitting of the d-orbitals of the Co 3+ center.

9 P a g e 9 Pre-Lab Questions 1. The Ti(H 2 O) 6 3+ complex absorbs light at 490nm. What is the energy of separation of the d- orbitals? 2. Sketch a Crystal Field Theory energy level diagram for the complex ion that forms when we dissolve CoCl 2 6H 2 O(s) in water; Co(H 2 O) 6 2+ (aq). (When the solid is dissolved in Water, the Chloride Ions are released and the Co 2+ ion is hydrated by 6 Water molecules.)

10 P a g e 10 Procedure At some point during the lab, when you are waiting for the reaction to proceed, build models of both the cis and the trans Dichlorobis(ethylenediamine)cobalt(III) complex. Perform the following steps in a fume hood. 1. Heat about 300mL of water in a 600mL beaker and maintain at a moderate boil. 2. In an evaporating dish, combine 4.0g of Cobalt (II) Chloride Hexahydrate with 10mL of distilled water. To this, add 15mL of 10% ethylenediamine (en). Place the evaporating dish on top of the beaker of boiling water. Stir the mixture over this steam bath for 40 minutes; maintain the volume of the solution at about 20mL by occasionally adding small portions of water. During this process, the Co 2+ is oxidized to Co 3+ by the Oxygen in the Air. Good agitation is necessary to promote solvation of the oxygen. 3. Add 12mL of Concentrated 12M HCl. Continue heating and stirring, without addition of water, until a thin slurry of crystals has formed. (Note: Stopping the evaporation at the right time is critical to the success of the synthesis. Stopping too soon will give a poor yield of the product; stopping too late will give an impure product.) Cool the slurry to Room Temperature by setting the evaporating dish on the lab bench. Stir occasionally for 15 minutes. 4. Filter the mixture using a Buchner funnel with Side-Arm Flask attached to an aspirator. When the draining from the Buchner funnel has essentially stopped, add 5mL of 6M HCl to the funnel and gently stir up the mixture with a spatula, taking care not to tear the filter paper. (Do not use Pure Water to wash your product. If you do, the complex will decompose.) If the product is still brown or blue, repeat the washing process with 6M HCl. If the product is pure green, skip the washing to avoid loss of product. 5. Transfer the moist crystals of the product to a clean watch glass. Dry the material on the steam bath. During the last stages of drying, powder the material on the watch glass by scrapping and pressing with a spatula. 6. Transfer the dried product to a previously weighed 3-dram vial. Label this vial with your Name(s), Product Name, Product Formula, Mass Product Obtained and the Date.

11 P a g e 11 Data Analysis 1. Calculate the Theoretical Yield of your Product. 2. Calculate the Percentage Yield of your Product. 3. Identify the points in the Procedure at which loss of Product is likely. 4. Sketch a Crystal Field Theory energy level diagram for the product Co(en) 2 Cl 2 + complex ion. What color light does this a solution of this complex absorb?

12 P a g e 12 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) 2+ Ni(H 2 O) 6 b) 4- Mn(CN) 6 c) 2- CuCl 4 d) 3+ Cr(NH 3 ) 6 2. Write a chemical equation representing what happens when each of the following water soluble Coordination Compounds dissolves in water. a) [Cr(NH 3 ) 4 Cl 2 ]Cl b) K 2 [PtCl 4 ] c) [Co(en) 3 ](NO 3 ) 3 3. Cr(NH 3 ) 6 3+ is yellow-orange wehreas Cr(NH 3 ) 5 Cl 2+ is purple. Explain this using Crystal Field Theory energy level diagrams.

13 P a g e 13 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!!! 1. Copper Ammine Complex Cu 2+ (aq) + 2 OH - (aq) Cu(OH) 2 (s) + 4 NH 3 (aq) Cu(OH) 2 (s) Cu(NH 3 ) 4 2+ (aq) + 2 OH - (aq) Put 10 drops 0.1M Cupric Sulfate (CuSO 4 ) into a test tube. Note the color of the solution. Add 3-4 drops of 6M Ammonia (NH 3 ). Note what happens. You should recall that Ammonia (NH 3 ) acts as a weak Base in Water: NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) So, an Aqueous Ammonia solution contains both NH 3, which can complex with Cu 2+ to form the ammine complex (Cu(NH 3 ) 4 2+ ), and OH -, which will form the Cupric Hydroxide (Cu(OH) 2 ) precipitate. 2. Cobalt Equlibrium CoCl H 2 O Co(H 2 O) Cl - Put 5 drops (no more) 1M Cobalt Chloride (CoCl 2 ) into a test tube. Add 8 small drops conc. Hydrochloric Acid (HCl). (Do not add too much HCl.) This forces the Co 2+ to form the Chloride comples. Note the color of the solution. Add no more than 8 drops Distilled Water, dropwise, until you just observe a color change. This now forces the Co 2+ to form the Aquo complex. Note the color of the solution.

14 P a g e 14 Data Analysis 1. Use Crystal Field Theory energy level diagrams to explain the color changes in each case. One cautionary note: Cu 2+ is short-hand for Cu(H 2 O) 6 2+ and Cu(NH 3 ) 4 2+ is short-hand for Cu(NH 3 ) 4 (H 2 O) 2 2+.