Download The Synthesis and Color of trans-Dichlorobis

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)2Cl2]Cl, from
Cobalt(II) Chloride Hexahydrate, CoCl2•6H2O.
4 CoCl2•6H2O + O2 + 4 HCl + 8 en
4 [Co(en)2Cl2]Cl + 26 H2O
(Eq. 1)
"en" is short-hand for Ethylenediamine:
NH2-CH2CH2-NH2
This synthesis involves oxidation of the Cobalt from Co+2 to Co+3 by Oxygen dissolved in the
reaction solution:
4 Co2+ + O2 + 4 H+
4 Co3+ + 2 H2O
(Eq. 2)
In an aqueous environment, Co3+ is unstable and is readily reduced back to Co2+. However, by
adding Ethylenediamine ligands to the Co3+ results in a complex that stabilizes the Co3+
oxidation state. Hence, if the oxidation is carried out in the presence of Ethylenediamine, the
Co3+ 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)2Cl2]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)2Cl2]Cl is formed from the Co(en)2Cl2+ 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)2Cl2+ in the chemical formula.) Like other Water soluble compounds, this
ionic substance can dissolve in Water according to:
Page |2
[Co(en)2Cl2]Cl(aq)
Co(en)2Cl2+(aq) + Cl-(aq)
(Eq. 3)
The covalent bonding between the metal ion and its ligands, in our case, between the Co3+ 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(NH3)63+ complex
ion. The NH3 ligand has one lone pair of electrons which can coordinate covalent bond to the
Co3+ metal center:
Since this example complex contains six NH3 ligands, it has six coordinate covalent bonds
arranged octahedrally about the metal center:
The number of coordinate covalent bonds formed is referred to as the Coordination Number of
the complex. In the above example, the coordination number is six. The coordination number
determines the geometry of the complex. Typical coordination geometries are:
Coordination Number
2
Geometry
Linear
4
Square Planar
or
Tetrahedral
6
Octahedral
Page |3
For our complex ion, Co(en)2Cl2+, the coordination number is also six. 2 electron pairs are
donated to the Co3+ 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 CoCl2C4H16N4+ 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.
Page |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 dx2-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 dxy orbital
along these same axes, the approach is along a nodal plane and the interaction is less
unfavorable, causing the energy of the system not to rise as much.
Page |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.)
Page |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- < H2O < SCN- < NH3 < en < NO2- < CN-< CO
Consider what happens when light interacts with a metal complex. The complex Ti(H2O)63+
contains a d1 metal ion:
Ti = [Ar] 4s2 3d2
Ti3+ = [Ar] 3d1
(Recall, for Transition Metal Cations, s-orbital electrons are lost before d-orbital electrons. The
rule of thumb is: "first in-first out".) So, it is an octahedral complex with one d electron. If
Ephoton = , then the complex will absorb the photon.
(Recall also, the relationship between the energy of a photon Ephoton and its wavelength  is
Ephoton = hc/, where h is Planck's Constant and c is the Speed of Light.) It is found  ~ 500 nm
for this complex, meaning it absorbs blue-green photons. We, of course, "see" the color of this
complex as the complement of the absorbed photons; red-violet.
Page |7
Color Wheel for Determining Complementary Colors
Consider a slightly more complex case; CoF63- vs. Co(NH3)63+. Both involve a Co3+, or d6 metal
ion and both are octahedral.
Co = [Ar] 4s2 3d7
Co3+ = [Ar] 3d6
Because F- is a weak field ligand, we expect a small splitting energy . Whereas NH3 is a much
stronger field ligand, so will be much larger.
Note that the arrangement of the electrons in the d-orbitals changes from one complex to the
other. In CoF63- the energy to pair the electrons is greater than  and so we have a "high spin"
complex; the electrons still obey Hund's Rule. In Co(NH3)63+ the pairing energy is small
compared to  and so we have a "low spin" complex. (Paired electrons have a lower net spin
than do unpaired electrons.)
Because  is small for CoF63-, it absorbs long wavelength red photons ( ~ 700 nm) and appears
green. For Co(NH3)63+, because  is much larger it absorbs shorter wavelength blue photons
( ~ 475 nm) and appears yellow-orange.
Page |8
Thus, we will be synthesizing a particular isomer (trans) of the coordination compound
[Co(en)2Cl2]Cl where the complex ion Co(en)2Cl2+ 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 Co3+ center.
Page |9
Pre-Lab Questions
1.
The Ti(H2O)63+ complex absorbs light at 490nm. What is the energy of separation of the dorbitals?
2.
Sketch a Crystal Field Theory energy level diagram for the complex ion that forms when
we dissolve CoCl26H2O(s) in water; Co(H2O)62+(aq). (When the solid is dissolved in Water,
the Chloride Ions are released and the Co2+ ion is hydrated by 6 Water molecules.)
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 Co2+ is oxidized to
Co3+ 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.
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)2Cl2+ complex
ion. What color light does this a solution of this complex absorb?
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)
b)
c)
d)
2.
Ni(H2O)62+
Mn(CN)64CuCl42Cr(NH3)63+
Write a chemical equation representing what happens when each of the following water
soluble Coordination Compounds dissolves in water.
a) [Cr(NH3)4Cl2]Cl
b) K2[PtCl4]
c) [Co(en)3](NO3)3
3.
Cr(NH3)63+ is yellow-orange wehreas Cr(NH3)5Cl2+ is purple. Explain this using Crystal
Field Theory energy level diagrams.
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
Cu2+(aq) + 2 OH-(aq)
Cu(OH)2(s) + 4 NH3(aq)
Cu(OH)2(s)
Cu(NH3)42+(aq) + 2 OH-(aq)
Put 10 drops 0.1M Cupric Sulfate (CuSO4) into a test tube. Note the color of the
solution. Add 3-4 drops of 6M Ammonia (NH3). Note what happens. You should
recall that Ammonia (NH3) acts as a weak Base in Water:
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
So, an Aqueous Ammonia solution contains both NH3, which can complex with
Cu2+ to form the ammine complex (Cu(NH3)42+), and OH-, which will form the
Cupric Hydroxide (Cu(OH)2) precipitate.
2.
Cobalt Equlibrium
CoCl42- + 6 H2O
Co(H2O)62+ + 4 Cl-
Put 5 drops (no more) 1M Cobalt Chloride (CoCl2) into a test tube. Add 8 small
drops conc. Hydrochloric Acid (HCl). (Do not add too much HCl.) This forces
the Co2+ 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 Co2+ to form the Aquo complex. Note the color of the
solution.
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: Cu2+ is short-hand for Cu(H2O)62+ and Cu(NH3)42+
is short-hand for Cu(NH3)4(H2O)22+.