Inorganic Chemistry 2 Lab (2_2566)

Direction
242332

Inorganic Chemistry II

Name.............................................................................................................
Code......................................................................................................

Division of Chemistry, School of Science
University of Phayao

Laboratory Glassware and Equipments

Beaker Cylinder Erlenmeyer Flask Dropper

Burette Burette clamp Condenser clamp Clamp holder
= Boss head

Test tube Test tube rack Stand Stirring rod

Funnel Plastic Wash bottle Watch glass Brush test tube

Buchner Funnel Filter Flask Measuring pipette Transfer pipette
= Suction Flask = Graduate pipette = Volumetric

Pipette

Syringe ball Filter paper Forceps Thermometer

Spatula Evaporating dish Crucible tongs Crucible and cover

Clay triangle Wire gauze Tripod Volumetric Flask

Mortar and Pestle Ring support Safety glasses Para film
= O - ring = Goggle

Volt meter Calorimeter Glove Pasture pipette

Stopwatch Stopper Test tube screw Separating funnel
cap

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Contents

Infrared Spectroscopy – Characteristic Frequencies
a) Terms
b) Frequencies of various vibration modes of some simple
Inorganic ions
c) Other useful frequencies
d) Listing the bands of an infrared spectrum

The decision tree for identifying a molecular point group

Experiment 1 10

Crystal structure

Experiment 2 17

Molecular Symmetry and Point Groups

Experiment 3 21

Synthesis of Coordination Compound

Experiment 4 24

The Reactions of Copper and Its Compounds

Experiment 5 27

Determination of the Empirical Formula of a metal Iodide

Experiment 6 31
Determination of the Composition of Iron – Phenanthroline Complex
by Job’s Method

Experiment 7 36

Characterization of a Copper Oxalate Complex , KaCub(C2O4)cdH2O

Experiment 8 40
Electronic Absorption Spectrum of Cu(II) complexs: Crystal Field Theory

Experiment 9 45

Preparation and Identification of Geometrical Isomers of Cobalt(III)-Iminodiacetate

References 50

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Infrared Spectroscopy – Characteristic Frequencies
(a) Terms

The following term are used in the tables:
V stretching i.e.v(s-o) is a sulfur – oxygen stretching vibration
deformation

w wagging

r rocking

t twisting

out of plane bending

(b) Frequencies of various vibration modes of some simple inorganic ions

CO32- v(m-o) v(o-m-o) v(mo3)
NO3- 1460 700 860
1400 690 830
v(s-o) v(o-s-o)

SO32- 1000 630

SO42- 1100 600
PO43- v(p-o) v(o-p-o)
ClO4- 1080 500
v(cl-o) v(o-cl-o)
1100 630

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(d) Listing the bands of infrared spectrum

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1. Both Nujol and Fluorolube have absorbances in the range of the infrared spectrometer.
The bands due to these mulling agents should not be included when listing the bands in the
spectrum.
2. Intensities of infrared bands are assigned somewhat arbitrarily, as the height of a
particular band is dependent on the amount of compound in the mull and the
thickness of the film of mull between the plates. The relative intensities of the various bands
due to a specific compound remain the same in spite of variations in concentrations and
thickness of the mull.

Diagram point group

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Experimental 1
Crystal Structures
(Packing of Spheres)

Objective
1. To study the packing of sphere and understanding of the concepts of the packing of
spheres by building and examining models of some of these structures.

Theory
A crystalline solid is a substance which is composed of a basic unit which is repeated

in three dimensions (3D) in an ordered fashion. The basic 3D repeating pattern is called the
unit cell and can be composed of atoms, ions, or molecules. The collection of particles which
make up the unit cell must meet three requirements:
1. they must indicate the coordination number
2. they must be consistent with the empirical formula
3. they must generate the crystal structure when repeated in three dimensions

These particles can also be considered as uniform, hard spheres which may form
different structures depending on how they are packed. One of the controlling aspects of the
packing of spheres is how they are arranged within each layer. In some layers all of the
spheres are touching and each sphere is surrounded by six others. This is the most efficient
way to pack spheres and is called closest packing (see figure A). Layers can also be
fashioned from spheres which do not make such efficient use of the available space; in these
layers one sphere will be in contact with only four other spheres (see figure B). The least
efficient packing of sphere layers occurs when no contact at all exists between the spheres
(see figure C). The arrangement of the spheres in two dimensional sheets represents only one
aspect of the overall crystal structure; how the two dimensional sheets are stacked must also
be considered.

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Two different types of closest packed structures can be generated depending on how
close-packed layers of spheres are stacked. The hexagonal close-packed structure is built by
placing one layer of spheres in the depressions formed by a first layer and then adding a third
directly above the first. A hexagonal prism in generated by the repeating pattern defined by
the hexagonal unit cell (see figure D).

The cubic closed-packed structure is similar to the hexagonal close-packed structure
except that the third layer sits in different depressions than the first. This structure is
represented by the face-centered cubic unit cell (see figure E).

In the close-packed structures holes are formed as the result of the spheres being
placed on top of each other; they are referred to as voids or interstices. There are two types of
voids depending on the number of spheres surrounding the void:
• Tetrahedral voids are formed when one sphere in one layer is fit over three spheres in
another generating a void which is surrounded by four spheres.
• Octahedral voids are formed when three spheres in one layer are fit over three spheres in
another generating a void which is surrounded by six spheres.

The body-centered cubic cell (see figure F) is generated from packing of layers
similar to that for the hexagonal close-packed structure except that the layers are composed of
spheres that do not touch (see figure C). Here the body diagonal (that which connects the top
left back corner of the cube to the bottom right front corner) is the only place in the unit cell
where spheres are touching.

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Placement of layers of spheres which have four nearest neighbors (see figure B)
directly on top of each other generates the simplest of three structures and is composed of a
simple cubic unit cell (see figure G).

Procedure
Working in pairs, carefully insert wires into the centers of Styrofoam balls. Note

which size balls you are instructed to use for each structure. Answer each question in the
space provided. It may be helpful to remember the following geometric relationships. In
reality there are seven different types of unit cells, but we will only be dealing with the cubic
unit cell. So, remember that it has to be a cube. SHOW ALL WORK!

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Simple Cubic
1. Using your largest spheres, construct a single simple cubic unit cell lattice. Using dots to
represent the center of each atom, draw a diagram to represent your model.

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2. How many atoms are contained in the simple cubic unit cell?
3. Express the length of the side of the cube in terms of sphere radii.
4. Express the body diagonal (BD) of your unit cell in terms of sphere radii.
5. Calculate the volume of the simple cube unit cell in terms of sphere radii.
6. Calculate the volume actually occupied by spheres in the simple cubic unit cell in terms
of sphere radii.
7. The spheres occupy what % of the total available volume in the simple cubic cell

Body Centered Cubic
1. Construct a body centered cubic cell. (start with the center sphere and work outwards).
Using dots to represent the center of each atom, draw a diagram to represent your model.

2. How many atoms are contained in the body centered cubic unit cell?

3. Are the spheres in contact along the edge, face diagonal, or body diagonal?

4. Express the body diagonal (BD) of your unit cell in terms of sphere radii.

5. Express the length of the side of the cube in terms of sphere radii.

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6. Calculate the volume of the body centered cubic unit cell in terms of sphere radii.
7. Calculate the volume actually occupied by spheres in the body centered cubic unit cell in
terms of sphere radii.
8. The spheres occupy what % of the total available volume in the body centered cubic cell.

Face Centered Cubic
1. Using your spheres, construct a face centered cubic unit cell lattice. Using dots to
represent the center of each atom, draw a diagram to represent your model.

2. How many atoms are contained in the face cubic unit cell?
3. Express the face diagonal (FD) of your unit cell in terms of face radii.

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4. Express the length of the side of the cube in terms of sphere radii.

5. Calculate the volume of the face cube unit cell in terms of sphere radii.

6. Calculate the volume actually occupied by spheres in the face cubic unit cell in terms of
sphere radii.

7. The spheres occupy what % of the total available volume in the face cubic cell.

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Experiment 2

Molecular Symmetry and Point Groups

Introduction

Symmetry greatly enhances the beauty of nature, as is very evident in flowers, leaves,
and crystals. Molecular structure reveals remarkable ramifications of symmetry, which are
developed further in the orderly packing of molecules in crystals. In a chemical compound
orbital shapes are dictated by symmetry, and even energy states can be described in terms of
symmetry. When we say that a molecule has symmetry, we mean that certain parts of it can
be interchanged with others without altering either the identity or the orientation of the
molecule. The interchangeable parts are said to be equivalent to one another by symmetry.

Symmetry operations and symmetry elements

Symmetry operations are geometrically defined ways of exchanging equivalent parts
of a molecule. There are five kinds which are used conventionally and these are sufficient for
all our purposes.

1. Axis of Rotation (Cn)

An objective has axial symmetry when it is invariant to rotation by some fraction of
3600( or 2 radian) Simple rotation about an axis passing through the molecule by an angle
3600/n (or 2 /n). This operation is called a proper rotation (n-fold axis of symmetry) and is
symbolized Cn. if it is repeated n times , of course the molecule comes all the way back to the
original orientation.

2. Mirror Plane( )

Reflection of all atoms through a plane that passed through the molecule. This
operation is called reflection and is symbolized ( or by the letter m in crystallography) .
Reflection through the yz plane takes an arbitrary point in Cartesian three dimensional space
from the position ( x, y, z) to a new position ( -x, y, z) only the out –of-plane coordinate
changes sign.

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3. Center of symmetry (i)

Reflection of all atoms through a point in the molecule. This operation is called
inversion and is symbolized i. the inversion operation takes a point on a line through the
origin (the inversion center) to an equal distance on the other side , thus transforming a point
with the coordinates (x, y, z) to one with coordinates (-x, -y, -z).

4. Improper Rotation (Sn)

The combination, in either order , of rotation the molecule about an axis passing
through it by 2 /n and reflecting all atoms through a plane that is perpendicular to the axis of
rotation is called improper rotation (rotation - reflection). Both the symmetry element and the
operation are given the symbol Sn for a rotation – reflection of 3600/n ( or 2 /n radians) to
obtain the product of two symmetry operations [ xy(Cn)z].

5. Identity

These operations are symmetry operations if, and only if, the appearance of the
molecule is exactly the same after one of them is carried out as it was before.

A symmetry element is an axis ( line) , plane, or point about which symmetry
operations are performed. The existence of a certain symmetry operations implies the
existence of a corresponding symmetry , the presence of a symmetry element means that a
certain symmetry operation or set of operations is possible.

Assigning Molecule to Symmetry Groups

The rotation group include Cn, Dn and the Platonic solid rotation group (T , O ,or I).To
assign a molecule to a point group, we first assign the rotation group by identifying the
highest-order proper rotation axis and then looking for other proper rotating ; then we look
for other symmetry elements to identify the full symmetry group following Fig. 1.1.( The
highest-order Cn axis is conventionally taken as the z axis. )

If a molecule has only one Cnaxis , the rotation group is Cn(left of Fig. 1.1).If there is
a C2 axis perpendicular to Cn , the rotation group is Dn and there must be n C2 axes. If a

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molecule has more than one Cn axis with n > 2,it belongs to one of the platonic solid rotation
group: I if n=5, O if n = 4, and T if n = 3. In each there are several C5, C4, or C3 axes

After identifying the rotation group , we look for other symmetry properties. If the
rotation group is Cn, we look for a mirror plane perpendicular to C n ( h)to identify the full
group as Cnh. If a molecule belonging to a Cn rotation grouphas a vertical plane ( slicing
through Cn) , the group is and there must be n v.

If the molecular belonging to a Cn rotation group has no mirror planes, we look for Sn
axes, but there are few cases to be examined. For a molecule belonging to the C2 rotation
group , if there is an S4 axis, coincident with the C2 axis, the full group is S4. For a molecule
belonging to the C3 rotation group, if there is an S6. Alternatively, identification of the highest
Cn as a C3 axis and a center of symmetry is adequate to characterize the S6 full group. For the
case of a C4 rotation group, the presence of an S8 axis coincident with the C4 axis identifies
the S8 full group. The S4 group is rare, and no examples of the S8 group (or there with higher
n) are known. Sn(n odd) group are called Cnh(C3h and C5h ). If there are no ,I, or Sn elements,
the full group is Cn.

If a molecule with the rotation group Dn has a mirror plane ( h) perpendicular to the
major axis (Cn), the group is Dnh. A mirror plane bisecting the angle formed by two C2axes ,
called a dihedral plane ( d), give the Dnhgroup. There must be n d planes present.

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Flow chart for assignment of molecules to point group
Procedure

1. Draw the symmetry elements found in molecule and write the flow chart for
assignment point group

2. What is the conformation of 1,2-dibromoethene if the point group of this
molecule is C2h (Draw the symmetry elements)

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Experiment 3

Synthesis of a Coordination Compound of Copper

Objectives
1. To synthesis of a coordination compound of copper
2. To study the synthesis techniques

Theory
The general formula of the unknown compound is Cux(NH3)y(SO4)z • a H2O Four

principal species are present initially in the reaction mixture: copper (II) ions [actually
Cu(H2O)62+ ions], ammonia molecules (NH3), sulfate ions (SO42-), and water. The product of
the synthesis is therefore presumed to be formed by the reaction of two or more of these
species. Ethanol is also present, but it is an indirect participant in the reaction. In aqueous
solutions ethanol, which is miscible with water but of lower dielectric constant (less polar),
decreases the solubility of ionic compounds. The marked color change that occurs in the
reaction is an important clue to the nature of the product. The product is analyzed for copper
(II) ions, sulfate ions, and ammonia molecules. Water is determined as the mass of a sample
of the compound that is not accounted for as one of these three species. The analyses to be
performed in this experiment are quantitative and are of three types: gravimetric, volumetric,
and spectrophotometric. The gravimetric analysis is for sulfate ions; the volumetric analysis
is for ammonia molecules; and the spectrophotometric analysis is for copper ions. With
careful attention to detail and to technique one can obtain excellent results for each part of the
analysis.

**You performed volumetric, and spectrophotometric

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Figure 3.1: Complex ions containing bound water and ammonia.
The structure of Cu2+ ion in water is on the left. Note that the water molecules are
bound, through oxygen atoms, to the copper ion. In this experiment you will make a
compound that has NH3 molecules bound to the copper(II) ion through N atoms, similar to
the complex ion [Co(NH3)6]3+.

Procedure
Part 1: Synthesis of the Compound
1. Weigh out 10 grams of copper sulfate pentahydrate, CuSO4 • 5 H2O, to 0.1 mg on an
analytical balance and place the crystals in a 250 mL beaker.
Obtain and wear a pair of gloves! Do the next 3 steps in a FUME HOOD.
2. Add 10-15 mL of water to the solid and then add 20 mL of 15 M NH3 (concentrated
ammonia). Stir to dissolve the crystals. You can determine if the solid is dissolved by holding
the beaker up to the light.
3. Once the solid has been dissolved, over a period of 1 minute, slowly add 20 mL of 95%
ethanol to the solution, stir, and cool to room temperature.
4. Meanwhile, prepare 30 mL of a solution using 15 mL each of concentrated ammonia and
95% ethanol.
5. Cover all your solutions containing ammonia with watch glasses (to prevent the fumes of
NH3 from saturating the lab atmosphere), and return to your desk. These solutions will smell
and may cause nausea/headaches, make sure to keep the solution covered at all times when
not in use!
6. Set up an apparatus for vacuum filtration. Moisten the filter paper with a small amount of
your ethanol/ammonia solution and turn on the aspirator. Carefully filter the slurry of crystals
that has formed in the copper-containing solution and suck off all the solution.

In the event that a significant amount of crystalline product remains in the beaker, you
should use the filtrate from the filter flask to wash the crystals onto the filter paper. To
prevent backup of tap water into the filter flask, which would render the filtrate useless as a
wash, pull the hose off the aspirator while the water is still running. Remove the filter
funnel and pour the filtrate from the filter flask back into the beaker that contains the crystals.
Reassemble the filter apparatus and collect the remaining product on the filter paper.

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7. Disconnect the hose, and then turn off the water aspirator and carefully pour 10 mL of the
ammonia-ethanol solution onto the crystals. Gently swirl* the solid with the wash to permit
the liquid to penetrate the mass completely and then turn on the aspirator to suck off the
liquid. Be careful NOT to tear your filter paper. Repeat the washing procedure twice with
additional 10 Ml portions of ammonia-ethanol.

*NOTE: Poor swirling technique will result in giant clumps of solid crystals which coagulate
on the filter paper and do not allow for a “seal” to form which makes it difficult for the
apparatus to suck the moisture from the solid. If you clump your solid, you will/may need to
leave the sample in your lab drawer for a longer period of time to dry (it will not dry during
filtering so do not waste the acetone trying!)

8. Next, wash twice with 10-mL portions of 95% ethanol and finally with two 10-mL
portions of acetone, gently swirling the crystals with a spatula in each step before turning on
the aspirator. (At this point, your crystals should appear nearly “dry”, that is not moistened
with liquid to any great extent. If you feel they could be dried more, add additional amounts
of acetone as described above.) To remove the last traces or moisture and other solvents from
your solid, draw air through the crystals for at least 5 minutes.

9. Put the crystals on a large watch glass, and leave them in your drawer to dry thoroughly
until the next laboratory period. Your crystals must dry for at least 24 hours!
10. When you come to the laboratory at least ONE day following the one in which the
crystals were prepared, inspect them to insure they are dry and then weigh the entire sample
to the nearest 0.1 mg. Record the mass. Once the crystals are thoroughly dry, transfer your
crystals from the filter paper to a weighing bottle in your lab drawer.

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Experiment 4

The Reactions of Copper and Its Compounds

Objective
1. To prepare a copper compound
2. To study the reactions of Copper

Theory
The law of conservation of mass states that matter is neither created nor destroyed

during a chemical reaction. This was first observed by Antoine Lavoisier in the late
eighteenth century, in which he observed that the total mass of all substances present after a
chemical reaction is the same as the total mass before the reaction. Any changes which occur
during a reaction merely involve the rearrangement of atoms. In theory, this means that an
element may go through several reactions until it is transformed back into its original state
without the loss of any mass. In practice, however, this may not always appear to be the case.
In this experiment, Lavoisier’s theory is tested by reacting solid copper with certain reagents
and synthesizing various compounds in order to arrive at the original mass of copper at the
conclusion of the experiment.

Procedure
1. Weigh approximately 0.35 g of copper into a 250 mL beaker. Describe the color and form
of the sample. Record the mass of the sample.
2. Part A: WORK IN THE HOOD! Add 3-5 mL of concentrated HNO3 slowly to dissolve the
copper; warm gently on a hot plate if all of the copper does not dissolve. After the copper has
dissolved, remove the solution from the hot plate and add 15 mL of deionized H2O slowly.
Note the color of the solution as well as the reaction that is occurring.
3. IN THE FUME HOOD, prepare a boiling water bath for Part C: Fill a 400 mL beaker with
about 200 mL of water and add 2–3 boiling stones. Heat this solution at about 50 % power on
a hot plate. While waiting for the water to boil, continue to Part B.
4. Part B: Use NO HEAT during this step! Add 6 M NaOH(aq) very slowly while stirring
until the solution turns red litmus paper blue (use a stirring rod to test a drop of solution). Do

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not confuse the color of the precipitate with the color of the litmus paper. Note any color
changes; a precipitate will form. Describe the color and the supernatant. Note the reaction
that is occurring.
5. Part C: Dilute the solution to 100 mL total with DI water. Place your 250 mL beaker in the
boiling water bath that you prepared in the fume hood. If the water bath is not boiling yet,
you may increase the power on the hot plate to bring the water to a boil. Allow your solution
to heat up and note any color changes. When all of the solid material appears to have changed
color, allow the solution to heat for an additional 3 minutes, then turn off the hot plate,
remove the 250 mL beaker from the water bath and allow the solution to cool. While the
solution is cooling, prepare a filter by folding a piece of filter paper in half twice to fit your
funnel. Filter the cooled mixture and discard the liquid. Rinse the filter with deionized water.
You will return the solid to the original beaker, so it is not necessary to completely transfer
the solid. Record the color of the residue and the filtrate. Note the reaction that is occurring.
6. Part D: Transfer the filter paper and solid back to the original beaker. Dissolve the solid
copper oxide by adding approximately 10 mL of 3 M H2SO4(aq) to the filter paper containing
the residue from the previous step. Remove the filter paper and rinse it with 10 – 20 mL of
deionized water once the solid has dissolved, add the washings to the acid solution, and save
the solution for the next step. Note any color change. Also note the reaction that is occurring.
7. Part E: WORK IN THE HOOD! Add about 0.40 g of zinc metal to the acidic copper
solution. If any blue color remains after the zinc has dissolved, a bit more zinc may need to
be added (record in your report!). Note the reaction that is occurring. Dissolve any excess
zinc with a small amount (approximately 5 mL) of 3 M H2SO4(aq).
8. Filter the supernatant liquid from the solid and wash the solid 3 times with 20 mL portions
of deionized water.
9. Transfer the solid copper onto a large watch glass and place it in the laboratory oven for 15
to 20 minutes at around 100 °C or until dry. Weigh to determine the mass of recovered
copper.

Data Analysis and Calculations
Initial color and form of the copper __________________________
Mass of Cu(s) at the beginning of the experiment __________________________
Mass Recovered __________________________

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% Recovery __________________________

Comment and discussion:
How does your percent recovery deviate from the expected 100%? Briefly
explain.

Post-lab Questions
1. Write the symbol or formula for the form of copper that is present in the
following parts of the experiment:
A. after adding nitric acid _____________________________
B. after adding NaOH, litmus paper
turns blue _____________________________
C. after boiling _____________________________
D. after adding sulfuric acid _____________________________
E. after adding zinc _____________________________
2. Now give the color of the copper substances in each of the steps above:
A. _________ D. __________
B. _________ E. __________
C. _________
3. What should the student do if the solution in step E is still blue?

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Experiment 5

Determination of the Empirical Formula of a Metal Iodide

Objectives

1. To prepare a metal iodide compound
2. To use the method of the gravimetric analysis in the determination the percent
composition and empirical formula of a compound

Theory

Atoms of different elements chemically combine to form different compounds
following two fundamental laws. The Law of constant composition states that a sample of a
pure compound is always composed of the same elements combined in the same proportions
by mass. If we analyze samples of water, we will find that the ratio of the masses of hydrogen
and oxygen in each sample is the same, regardless of the source of the water. The law of
multiple proportions states that if two elements combine to form a series of compounds, the
mass ratios of the elements are small whole numbers. For example, hydrogen forms two
compounds with oxygen: water (H2O) and hydrogen peroxide (H2O2). In the first compound,
16.0 grams of oxygen (one mole of O) combine with 2.016 grams of hydrogen (two moles of
H ) ; and in the second compound, 32. 0 grams of oxygen ( two moles of O) combine with
2.016 grams of hydrogen (two moles of H). The ratio of masses of oxygen that combine with
the hydrogen in these compounds is 16.0 g/32.0 g or 1:2, which supports the law of multiple
proportions.

When a new compound is prepared, its formula must be determined. This task is most
commonly achieved from the percent composition of the compound-the percent by mass of
each element relative to the total mass of the compound. It is obtained by dividing the mass
of each element in the compound by the molar mass of the compound, and multiplying by
100. In one mole of water (MW = 18.02 g/mole) are two moles of H (2.016 g) and one mole
of O ( 16. 0 g) . Therefore, % H = ( 2. 016/ 18. 016) 100 = 11. 19 percent and % O =
(16.0/18.016)100 = 88.81 percent.

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The simplest whole-number ratio of the number of atoms of each element in a
compound is called the empirical formula of the compound. The molecular formula shows
the actual number of each type of atom present in a compound. For example, suppose that the

empirical formula CH is obtained experimentally from the percent composition of a
compound. This formula may represent a molecule having the molecular formula from the
empirical formula one must know the molar mass of the compound.

The empirical formula of a compound can be determined either by chemical analysis
or by synthesis. In chemical analysis, a known mass of the compound is decomposed to
obtain the masses of the elements themselves or some of their known derivatives (for carbon,
the derivative is CO2 and for hydrogen, it is water) . From these masses, the percent
composition of the compound is determined. In synthesis, known masses of the elements are
allowed to from a compound and the mass of the compound is determined. From these data, it
is possible to derive the empirical formula of the compound.

Example 1. 1 In an experiment, a mixture of copper and sulfur was heated to produce a
sample of copper sulfide. The following data were obtained:

Weight of the empty crucible 2.077 g
Weight of the crucible + copper 2.289 g
Weight of the crucible + copper sulfide 2.396 g

Calculate the percentages of copper and sulfur in copper sulfide and derive its empirical
formula.

Answer:

Copper mass = 2.289 – 2.077 = 0.212 g
Copper sulfide mass = 2.396 − 2.077 = 0.319 g
Sulfur mass = 2.396 – 2.289 = 0.107 g
% Copper in copper sulfide = (0.212/0.319) x 100 = 66.5%
% Sulfur in copper sulfide = (0.107/0.319) x 100 = 33.5%
Mole of Cu = 66.5 g (1 mol/63.55g) = 1.05 mol Cu
Mole of S = 33.5 g (1mol/32.0 g) = 1.05 mole S

Empirical formula 29
= CuS

Procedure

Clean the 25 ml beaker and the stirring bar with detergent, then with deionized water,
and finally rinse with acetone three times. Dry them in an oven at 110C. Cool the beaker and
the stirring bar to room temperature ( preferably in a desiccator) for 10 min. Accurately
weight the beaker containing the micro stir bar. Place 200 mg of zinc powder in the beaker,
reweight it, and record the mass. Transfer about 120 to 130 mg of I2 crystals to the beaker and
reweight the beaker accurately. Calculate the exact amount of iodine taken. In place of a stir
bar, one can use a glass rod to stir the mixture.

Add 2 - 3 ml of water to the mixture. Place the beaker on a magnetic stirring hotplate
and cover it with a watch glass. With continuous stirring, warm the contents of the beaker (do
not boil) . Make sure that iodine do not sublime ( vaporize) . Within a minute or two, the
aqueous layer will become yellow as a result of the reaction between zinc and iodine. The
reaction is exothermic and may become brisk. If that happens, remove the beaker from the
heat and cool it in an ice-water bath. Continue to stir the mixture (heat if necessary) until the
brown color of the aqueous layer almost disappears. Add water to make up the loss due to
evaporation, washing down the sides with a few drops of water.

When the aqueous layer becomes colorless, decant the solution to a small flask
(making sure that no zinc powder is transferred). Rinse the unreacted zinc powder five or six
times with 2 ml of deionized water, decanting the aqueous layer to a waste container. While
transferring the wash liquid, you must not lose any zinc. Finally, wash the metal with acetone
four or five times, using 1 ml of acetone each times, decant the solvent, and keep all the zinc
in the beaker. Place the beaker with zinc over a warm sand bath. Heat slowly to dry the
powder ( to prevent spattering). When it is dry, cool the beaker in a desiccator. Weight the
beaker with the dried zinc in it. Empty the beaker by transferring the unreacted zinc into a
designated contaminated container and rinse the beaker.

Do not throw away the leftover zinc; it must be recovered, recycled, and reused.

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Calculate the amount of zinc that has reacted with the given amount of iodine.
Determined the percent zinc and percent iodine from the data and obtain the empirical
formula of zinc iodine.

Prelaboratory Problems

1. Balance the following reactions

Fe(s) + Br2(g) Fe2Br3(s)
C6H6(l) + O2(g) CO2(g) + H2O(g)

2. A 1.275 g sample of Al was heated strongly in pure oxygen and the combustion product
weighed 2.409 g. Calculate the percent O in the product and find the empirical formula of the
compound formed.
3. Cobalt forms two different chlorides. One contained 54.7 percent Cl and the other 64. 4
percent Cl. Determine the empirical formulas of the two compounds.

Postlaboratory Problems
1. A compound contains 90.6 percent lead and 9.40 percent oxygen. Determine the empirical
formula of the compound.
2. Calculate the percent H and percent O in hydrogen peroxide (H2O2)

31

Experiment 6

Determination of the Composition of Iron-Phenanthroline
Complex by Job’s Method

Objective

To establish the formula of a colored complex by the spectrophotometric method

Theory

The formation of complex in solution is often accompanied by the appearance of
color. Measurement of the absorbance of such a solution will afford a measure of the amount
of complex ion in solution. For a reaction of the type.

Mn+ + yL [MLy]n+

the amount of complex ionic solution can be determined colorimetrically for various ratios of
[Mn+] to [L] ; the total concentration of metal ion and ligand is kept constant. Measurements

of absorbance at a suitable wavelength will show a maximum when the ratio of ligand to
metal is equal to that in the complex. The method is known as ‘ Job’ s Method’ , after the
originator, and is often referred to as ‘ The Method of Continuous Variation’ . Measurement
may be made at any wavelength where the complex shows appreciable absorption. A portion
of maximum absorption is preferred.

If the measurements are made at only one wavelength then the system must such that
only one complex is formed. This may be verified by making measurements at a number of
wavelengths within the absorption spectrum of the complex( es) . If measurements at all
wavelengths give the same result, it may be concluded that only a single compound is
formed. Note that concurrent formation of a colorless compound may be overlooked by this
method.

In general, transition metal ions form a large number of complex compounds. For
example, the almost colorless iron( II) cation reacts with 1,10-phenanthroline ( o-phen,
colorless) to form a red complex cation:

32

xFe2+(aq) + y(o-phen) [Fex(o-phen)y]2+

Job’ s method of continuous variation will be used in conjunction with visible
spectroscopy to establish the formula of the complex formed by Fe( II) and 1,10-
phenanthroline. Job’s method gives accurate results only under these circumstances.

1. The reaction is quantitative and complete.

2. A single complex species is formed, and the species is stable under the condition of
the reaction.

3. The λmax for the complex species is known and is at a different wavelength from either
the ligand or the metal ion.

4. The pH and ionic strength of the solution remain constant.

Using this method, one makes a series of absorbance measurements in which the
concentration of Fe( II) and o-phen are varied, while the total number of moles remains
constant. The requirement is most easily achieved by preparing solutions of Fe(II) and o-phen
at identical concentrations and then mixing them in various volume ratios, keeping the total
volume constant.

For example, the mole fraction of Fe( II) in a series if solutions may be varied in
increasing order (0, 0.10, 0.25, 0.40, 0.60, 0.75, 0.90, and 1.00) while the mole fraction of o-
phen is varied in decreasing order (1.00, 0.90, 0.75, 0.60, 0.40, 0.25, 0.10 and 0). Note that
the sum of the mole fractions of the metal ion and the ligand in each pair is constant.

The mole fraction of the metal and the ligand are easily calculated. By definition,

so that XM = nM/nT and XL = nL/nT
XM + XL = 1 or XM = 1 - XL

Where nM and nL = number of moles of the metal cation and the ligand, respectively;
nT = total number of moles (= nM + nL) and XM and XL = moles fraction of the metal and the
ligand, respectively.

Under ideal reaction conditions, the maximum amount of the complex
[Fex( o-phen) y] 2+ will form when the mole fractions of Fe(II) and o-phen are in the correct

33

stoichiometric ratio. All other combinations result in the formation of lesser amounts of the
complex. Since the product complex is colored, the absorbance of the solution mixture
indicates the amount of the complex that has formed. A plot of absorbance versus the mole
fraction of the metal ion ( XM) and the ligand ( XL) generates a graph ( Figure 2. 1) whose
maximum indicates the stoichiometric composition. In Figure 3.1, the absorbance maximum
occurs at about XL = 0.75 and XM = 0.25. The ratio between them XL/XM = 0.75/0.25 = 3,
indicating that the ligand-to-metal mole ratio is 3:1.

Figure 3.1 Absorption vs. mole fraction

Procedures
1. Preparation of Stock Solutions

The following solutions will be prepared and labeled as A, B, C, and D,
corresponding as described:

Solution A : 50 ml of 1.0 x 10-3 M Fe2+ in 0.1 M HCl
Solution B : 60 ml of 1.0 x 10-3 M o-phen
Solution C : 10 ml of freshly prepared aqueous 1% hydroquinone solution
Solution D : 10 ml of 1 M HCl solution

2. Preparation of Complex Solutions

34

Using 10 ml graduate pipettes ( use a separate pipette for each solution) , add the
amounts of solutions A through D shown for solution #1 (in the following table) to a 25 ml
volumetric flask. Add 2.5 percent sodium citrate solution to the volumetric flask to adjust the
pH to  3.5 (test this by applying a drop of solution to a piece of universal pH paper). Finally,
add deionized water to the mark. Shake the solution well, and allow it to stand for 10
minutes. Transfer the solution to the 50 ml beaker labeled “ solution #1” . Cover the beaker
with a watch glass. Rinse the volumetric flask, and repeat the procedure for Solution #2
through Solution #7.

Beaker # Solution (ml)

AB C D

Solution 1 0 5.0 0.5 0.5
Solution 2 0.5 4.5 0.5 0.5
Solution 3 1.0 4.0 0.5 0.5
Solution 4 2.0 3.0 0.5 0.5
Solution 5 3.0 2.0 0.5 0.5
Solution 6 4.0 1.0 0.5 0.5
Solution 7 5.0 0.5 0.5
0

Now, prepare seven blank solutions identical to the proceeding, except for leaving
out solution B, in 25 ml volumetric flask. Transfer the solutions to labeled Erlenmeyer flasks.

3. Absorbance Measurement and Calculation

Set the wavelength of the spectrophotometer to 508 nm. Fill one with the “Blank #1”
solution and the other with “ Solution #1” . Insert the “ Blank #1” cell in the cavity and the
absorbance. Remove the blank cell and insert the cell with “ Solution #1” . Record the
absorbance, A. Repeat the procedure with all over sets of solutions.

Determine the number of millimoles of Fe2+ (nM) and o-phen (nL) in each solution
and obtain the total millimoles (nT). Calculate the mole fractions of Fe2+ (XM) and the ligand
(XL) for each solution. Make a plot of absorbance versus the mole fractions (XM and XL) as
shown in Figure 3.1 From the maximum of the curve, determine the mole ratio between Fe2+
and o-phen and establish the formula of the complex [Fex(o-phen)y]2+.

35

Prelaboratory Problems

1. A student records a transmittance of 40 percent for a particular solution. What would

the transmittance be if the solution concentration were doubled?

2. A metal cation M n+ reacts with a ligand L to from the complex

aMn+ + bL [MaLb]n+

Using a 0.005 M solution of Mn+ ( A) and 0. 005 M solution of L ( B) , a series of
solution was prepared as shown below. Each solution was diluted to 50 ml and the
absorbance was measured.

Sample # Sample A Sample B Absorbance
(ml) (ml)
1 0.100
2 0 10 0.238
3 1 9 0.380
4 3 7
5 0.510
6 5 5 0.376
7 7 3 0.240
9 1 0.098
10 0

Calculate the mole fractions of the metal cation and the ligand for each solution.
Show your calculations. Make a plot of absorbance (y axis) versus mole fraction of the metal
cation and the mole faction of the ligand (x axis). Determine the composition of the complex.

Postlaboratory Problems

1. What effect will a dirty cuvette have on the absorbance reading?
2. If the meter on a spectrophotometer read 30 percent transmittance, what is the absorbance?
3. A sample of 10 ml of 1.0 x 10-3 M Cu2+ solution is mixed with 40 ml of 0.001 M NH3
solution. Calculate the mole fractions of Cu2+ and NH3

36

Experiment 7

Characterization of a Copper Oxalate Complex; KaCub(C2O4)cdH2O

Objectives

1. To synthesize a copper (II) oxalate complex.
2. To study how to characterize the complex using the redox titrimetric analysis.

Theory

Copper, a transition metal, has an electronic configuration of [ Ar] 3d10 4s1. Copper
exhibits two common oxidation states, Cu( II) and Cu( I). Cu(II) and Cu(I) have electronic
configuration of [Ar] 3d9 4s0 4p0 4d0 and [Ar] 3d10 4s0 4p0 4d0, respectively. The Cu(II) uses
its two vacant 3d, one 4s, and three 4p atomic orbitals to create a total of six vacant metal
orbitals (hybridized d2sp3), which can receive as many as 12 electrons (6 electron pairs) from
ligands.

The solution chemistry of these ions is extensive. When a solution of potassium
oxalate ( K2C2O4) is added to Cu( II) in aqueous solution, a copper oxalate complex forms.
Here, copper is bonded to several oxalate ligands. When CaCl2 is added to an aqueous
solution of a simple oxalate salt such as K2C2O4, solid calcium oxalate (CaC2O4) immediately
precipitates. However, when CaCl2 solution is added to an aqueous solution of an oxalate
complex, no immediate precipitation of calcium oxalate occurs. This observation
demonstrates that the oxalate ions are differently bonded to copper( II) than they are in a
simple salt. In the oxalate complex, each oxalate ligand bonds to the copper through its two
oxygen atoms, each of which donates a pair of electrons to the copper. Oxalate is therefore a
bidentate ligand. In this experiment, a sample of the hydrate potassium salt of oxalato-
copper(II) complex will be prepared.

Copper(II) is nearly always determined by reduction to the monovalent state by iodide
ion and followed by titration the liberate iodine with sodium thiosulphate. Oxalate, however,
would interfere and must be determined first. The half reactions involved in this experiment
are

37

2Cu2+ + 2I− + 2e− 2CuI
[S4O6]2− + 2e− 2[S2O3]2−
2CO2 + 2e− [C2O4]2−
I2 + 2e−
[MnO4] − + 8H+ + 5e− 2I−
Mn2+ + 4H2O

The copper(II) ion is mildly oxidizing and can be used as the starting material for the
synthesis of complexes of copper(I) as well as copper(II). Since the chemistry of the copper
ion is modified by the presence of the surrounding ligands, in this experiment, we will
prepare and characterize the complex and investigate the oxalate ligand on the chemistry of
the copper ion.

Procedures
1. Preparation of Copper oxalate complex

Caution: oxalic acid and oxalate are toxic and must not be ingest.

Dissolve 1.2 g of copper(II)sulphate pentahydrate in 3 ml of hot water and add to this
a solution of 3.5 g of potassium oxalate hydrate 20 ml in hot water. Allow to cool to room
temperature and finally to about 10C in ice. Filter the product by suction and dry thoroughly
in air. Record the weight obtained.

2. Determination of the Oxalate Content

The complex was prepared in Procedure 1. Weigh accurately about 125 mg of the
complex into a conical flask and add 25 ml of dilute sulphuric acid to dissolve. Heat to 60-
70C and titrate with approximately 0.02 M potassium permanganate solution provided. Note
that during the initial stages of the titration the solution is cloudy due to undissolved complex.
This clears during the titration. Ensure that the solution is still hot as the endpoint is
approached. Heat to boil if the reaction appears to be slow. Take care not overshoot the
endpoint as the presence of a large excess of permanganate makes the copper determination
difficult. Preserve the resulting liquid for the determination of copper.

38

The permanganate solution should be standardized against sodium oxalate in the
following manner. Weight accurately two portions of about 200 mg of ‘ AnalarR’ sodium
oxalate into conical flasks. Dissolve each in water, acidify the solution with sulphuric acid
and titrate with permanganate as above, at 60.

Caution: Potassium permanganate solution is a powerful oxidant and the dilute solution is
a disinfectant. Avoid skin contact, but if you do spill it on yourself, simply wash with cold
water. Any remaining brown stain will soon disappear.

3. Determination of the solution obtained above

Ensure that the solution obtained above ( 2 ) contains no excess permanganate by
boiling until the purple color is discharged. To the cooled solution, add 1 g of potassium
iodide and titrate the liberated iodine with 0.05 M sodium thiosulphate. When the color of the
resulting suspension is pale yellow, add a few drops of starch and continue titrating until the
blue color begins to fade. Add 250 mg of potassium thiocyanate and swirl the contents. The
blue color should intensify as iodine, adhering to the solid CuI, is liberated. Continue the
titration until a white suspension is obtained. Note that the addition of potassium thiocyanate
should be carried out very near to the end point. Calculate the copper to oxalate ratio in the
complex. From these results deduce the likely values of a, b and c.

4. Determination of the water of hydration in the complex

Using a steel spatula, crush the dried sample of the complex to a powder on watch
glass. Weigh accurately about 200 mg of the complex taken. Place the watch glass and its
contents in an oven at 110 to 120C for one hour. Cool the sample in a desiccator and weigh
it again (if necessary, repeat the procedure of weighing, heating, cooling, and drying until a
constant mass is obtained) . Record the mass of the complex. Calculate the mass of the
anhydrous compound and the mass of water lost. Calculate the percent water of hydration in
the complex. Finally, calculate the moles of water per mole of the complex, and deduce the
value of d.

5 IR study of copper oxalate

39

IR spectra of free oxalate and copper oxalate are recorded and assigned in mid and
far IR.

Discuss and conclude the structure of complex using chemical analysis and infrared
data
Prelaboratory Problems

1. Complete overall reactions, which occur in the determination of, oxalate content and
copper content steps. (Hint: using the half reaction mentioned in the theory part)
2. If 2.00 ml of a 0.0200 M solution of sodium oxalate requires 1.58 ml of KMnO4 solution
for the complete oxidation of oxalate, calculate the molarity of KMnO4 solution.
3. Can sodium thiosulphate be used as a primary standard for determination of copper
content?
4. A 1.201 g sample of hydrated lead acetate was heated to drive off the water of hydration.
The cooled mass of the anhydrous lead acetate, Pb(C2H3O2)2,is 1.030 g. calculate the value of
x in the formula Pb(C2H3O2)2.xH2O.

Postlaboratory Problems

1. Suggest the structure of copper oxalate complex, corresponding with your obtained
formula.
2. Suggest the reason( s) why the presence of a large excess of permanganate makes the
copper determination difficult.
3. Instead of using dilute sulphuric acid, can we use dilute hydrochloric acid for the
determination of oxalate content.
4. Instead of oxalate anion, if we use ethylenediamine ( en = H2NCH2CH2NH2) as the
bidentate ligand, what would be the product? What are the donor atoms in en?

40

Experiment 8

Electronic absorption Spectrum of Cu(II) complex: Crystal Field Theory

Objective

To study the effect of ligands on crystal field splitting energy (E)

Theory

The color of coordination compounds of the transition elements is one of their
characteristic properties. These colors are due to the absorption and subsequent emission of
light in the visible part of the spectrum. Light in this region of the spectrum caused promotion
of d-electrons from a lower to a higher energy level. The spectra which result are generally
referred to as Electronic Spectra. Note that electronic transition may also be effected by
ultraviolet light. Because of the size of the quanta involved, electronic transitions in
molecules are always accompanied by vibrational and rotational changes, and hence a band
spectrum is observed. In general, the bands which arise are much broader than bands in an
infrared spectrum and are little used for identification purposes.

The crystal field theory of bonding in transition metal complexes has help
appreciably to rationalize many of the physical properties of such complexes. Much of the
data required for crystal field theory calculation is obtained from a study of the absorption
spectra of transition metal complexes.

Regularly six-coordination is most readily pictured by placing the ligands at the plus
and minus ends of the three coordinate axes. An electron in the dx2-y2 and dx2 orbitals is
therefore most effected by the field of the ligands and is raised in energy relative to an
electron in the dxy, dyz and dzx orbitals. The combined energy level diagram is therefore
composed of two upper orbitals, of equal energy, and three lower orbitals, which are also
degenerate ( Figure 5.1). the energy zero is conveniently taken as the weighted mean of the
energies of these two sets of orbitals; the lower trio are thus stabilized by-2/ 5E while the
upper pair are destabilized by 3/5E, where E is the total energy separation. It is termed the
crystal field splitting energy. The eg and t2g symbols are symmetry labels arising from Group
theory and are now the most commonly used symbols.

41

The size of E is most readily measured spectrochemically by observing the energy
of the electronic transitions between the t2g and eg orbitals. The energy usually lies in the

42

visible or near ultra-violet region of the spectrum and it is such d-d transitions, which are
responsible for the colors of most transitions metal compounds. The magnitude of E
depends on the ligand and on the nature of the transition metal ion.

Consider the simplest possible case, namely a complex in which there is one electron
in a d level, as in the ion [Ti( H2O) 6]3+ . The transition from the t2g to eg level of the single
electron, give rise to single absorption band in the visible region (figure 5.2). The spectrum
appears as a single absorption band at 20,000 cm-1. The water ligands have split the
degeneracy of the free gaseous ion in two: the t2g and eg level. Irradiation of the complex
with light of an appropriate frequency results in excitation of the electron in the lower t2g
level to the higher eg level. Knowing that light of frequency 20,000 cm-1 is required the
crystal field splitting energy for a d1 system in an octahedral field of water ligand can be
calculated. However, in this experiment, we will observe this phenomenon by using
copper(II) complex, due to their vivid colors.

Procedures
1. Preparation of Cis - diaquabis (glycinato) copper(II)monohydrate

Dissolve about 1.5 g of copper sulphate, CuSO4.5H2O, in 8.5 ml of 1 M HCl and add
700 mg of glycine. Cautiously warm the mixture in the water bath about 30 minutes. Slowly
add solid NaHCO3 to the solution until a crystalline precipitate is completely formed (Do not
overheat). Filter by suction. Allow the crystal to dry in the oven at 100 C.

2. Preparation of Bis(acetylacetonato)copper(II)

Dissolve 1.25 g of acetylacetone 50 ml of 0.25 M NaOH. Add a solution of 1.55 g of
CuSO4. 5H2O in 50 ml of H2O. The sparingly insoluble Bis( acetylacetonato) copper( II) is
immediately formed as a crystalline precipitate. Filter by suction. Allow the crystal to dry at
room temperature.

3. Absorbance Measurement of Complex Solutions
The following stock solutions will be prepared;

1. 0.10 M Cu(NO3)2 in 2 M NH4NO3 (Use for preparation of solution # 1-5)
2. 0.10 M Cu(NO3)2 in 1 M KNO3 (Use for preparation of solution # 6-7)
3. 0.10 M NH3

43

4. 0.10 M ethylenediamine (en)
Then, the Copper( II) complexes solution will be prepared in the different ratios as shown
below.

Solution # Complex The mole ratio

M-L M:L

1 Cu(II)-H2O 1:1

2 Cu(II)-NH3 1:1

3 Cu(II)-NH3 1:2

4 Cu(II)-NH3 1:3

5 Cu(II)-NH3 1:4

6 Cu(II)-en 1:1

7 Cu(II)-en 1:2

8 Cu(II)-glycine (0.01 M in H2O)

9 Cu(II)-acetylacetone (0.01 M in CHCl3)

Measure the absorbance of these solutions in the range of 500-850 nm. Compare

the λmax values obtained when different ligands are used and suggest the spectrochemical

series.

4. IR and NMR spectra of copper complexes

IR spectra of free glycine, acetylacetone and copper complexes are recorded and
assigned in mid and far IR.

1H and 13C NMR spectra of free glycine, acetylacetone and copper complexes are
examined and assigned. For the 13C spectra, several different modes of the spectrum were
obtained eg. 13C-{1H}, DEPT-135.

Discuss and conclude the structure of complexes using spectroscopic techniques.

Prelaboratory Problems

1. Determine the position of maximum absorption and hence calculate the crystal field
splitting energy for a d9 system in an octahedral environment of water ligands.

2. What effect does this have on the absorption band?

44

3. Predict the position of acetylacetonato ligand in the spectrochemical series, as compared to
water ligand. Suggest your reasons.

Postlaboratory Problems
1. If [Ti(H2O) 4]3+ can exist, which frequency do you expect in the absorption band of this
tetrahedral complex? Moreover, how about the spectrochemical series, the same or different
as compared to octahedral environment?
2. Write the structures of Cis-diaquabis( glycinato) copper( II) monohydrate and
Bis(acetylacetonato)copper(II) complex.
3. Does the mole ratio M:L effect the frequency of the band? Why?

45

Experiment 9

Preparation and Identification of Geometrical Isomers of
Cobalt(III)-Iminodiacetate

Theory

Compounds with the same formula, but different structures are called isomers, of
which two types can be classified. These include structural isomer and stereoisomer. One
type of structural isomerism that may arise when a ligand has more than one chemically
distinct donor atom is linkage isomerism. Co(NH3)5(NO2)2+ and Co(NH3)5(ONO)2+ can
exemplify the case. It may be noteworthy here that the linkage isomers frequently adopt
different names for different coordination modes.

pentaamminenitro cobalt (III) pentaamminenitrito cobalt (III)

The other structural isomerism finding importance in coordination chemistry is
coordination sphere isomerism, e.g.

[Cr(H2O)6]Cl3 violet
[Cr(H2O)5Cl]Cl2.H2O green
[Cr(H2O)4Cl2]Cl.2H2O green

Stereoisomerism is the most common and most important class of isomerism,
including geometrical isomerism, used to distinguish the phenomena that the same ligand

46

binds to different sites on the metal. There are several types of geometrical isomer, namely
cis and trans, fac- and mer-. Facial (fac) isomerism describes three of the same ligands lying
on one face of an octahedron while meridional ( mer) isomerism refers to three of the same
ligands lying on a plane passing through the center of the complex (a meridian).

trans-diamminedichloro platinum (II) cis-diamminedichloro platinum (II)

trans-tetraamminedichloro cobalt (III) cis-tetraamminedichloro cobalt (III)

fac-triamminetrichloro cobalt(III) mer-triamminetrichloro cobalt (III)

In this experiment, the geometrical isomerism will be studied. The ligand of interest is

47

iminodiacetic acid (H2IDA)

Name: Iminodiacetic acid
Synonyms: N-(Carboxymethyl)-glycine
Molecular Formula: (NH(CH2COOH)2), C4H7NO4
Molecular Weight: 133.10

When the ligand is coordinated to the metal ion, it becomes a tridentate ligand and its
stereochemistry is of five membered amino acid ring. Coordinated to the CoIII ion with give
the coordination compound with the formula [Co(IDA)2]-, with three possible isomers.

trans (fac) cis trans (mer)

The cobalt ( III) system is chosen due to the highly colored isomers, i.e. purple and
brown, both of which can be prepared at different temperatures. According to the possible
straining of the C-N-C bond angle in the IDA2- ligand, the trans(mer) isomer can be expected
to be rather unstable and a reasonable assignment for the trans compound will be of

48

trans(fac) geometrical configuration. It is also apparent that solubility of the trans isomer is
considerably less than that of the corresponding cis isomer. This may be due to a possible
intermolecular hydrogen bonds in the trans crystal, formed between the amine protons on one
molecule and the carboxyl group on a neighboring molecule.

To identify the isomers, we would normally use molecular models, proton magnetic
resonance ( PMR) and visible absorption spectra. However in this experiment we will only
use visible absorption spectra. On the basis of PMR experiment, the brown isomer can be
assigned as the trans( fac) and therefore the purple isomer as the cis isomer. From the
respective visible absorption spectra, a discussion of the expected splitting patterns of the
1A1g → 1T1g ( Oh) and 1A1g → 1T2g ( Oh) d-d transitions of the low-spin Co( III) ( d6)
complexes under the lower C2v or D4h symmetries of the isomers can be used to interpret the
band profiles in the visible absorption spectra.

Materials

1. Solid potassium hydroxides

2. Iminodiacetic acid

3. Cobaltous chloride hexahydrate

4. Hydrogen peroxide (15 % and 30 %)

Preparation of purple isomer of cis-K[Co(IDA)2].2.5H2O
Dissolve 1.6 g of KOH using 6.5 ml of water in a 100 ml beaker. After completely

dissolve KOH, dissolve further 2. 0 g of iminodiacetic acid ( H2IDA) , and then 1. 7 g of
CoCl2. 6H2O in to a solution. Cool the solution to approximately 12C using an ice bath.
Then, add 5.0 ml of 15% H2O2 into a cool solution over a period of 2-3 min with stirring.
While maintaining a temperature at about 12C for 2-3 hour, the mixture should be stirred
periodically. The purple crystalline product is finally collected using a Buchner funnel,
washed first with one 4-5 ml portion of ice cold water, and then with a 10 ml portion of

49

ethanol. Allow the solid product to dry at 80C in an air-oven for one day. (1.5-2 g of solid
should be obtained)

Preparation of brown isomer of trans-K[Co(IDA)2].2H2O

Dissolve KOH 1.6 g using 28 ml of water in a 100 ml beaker. Then dissolve 2.0 g of
iminodiacetic acid and 1.7 g of CoCl2.6H2O, respectively, into the solution. After completely
dissolve the reactants, heat the solution using a water bath to 80C, before slowly adding 1 ml
of 30% H2O2 (CAUTION! The reaction is vigorous.) After the reaction has ceased, cover the
beaker with a watch glass and continue heating the solution for approximately 45 minutes.
Place the beaker and cover in dark locker until the next laboratory period (for about a week).
After a specified duration, collect the brown crystals by filtration using a Buchner funnel,
wash first with a 5 ml portion of ice cold water, and then with a 10 ml portion of ethanol.
Allow the crystals to dry at 80C in an air-oven for one day. (Approximately 1.0 g of solid
should be obtained)

Collect visible absorption spectra

Dissolve each isomer in water to the fix concentration of 0.01M for the purple isomer,
[ cis-K[ Co( IDA) 2] .2.5H2O, and 0.025M for the brown isomer, [ trans-K[ Co( IDA) 2] .2H2O,
using 25 ml volumetric flasks. Collect the visible absorption spectra of each isomer in a 800-
320 nm range using 1 cm glass cells. After this, submit the remained crystals to the instructor.

Answer the following questions:
1. Based on the experimental results, calculate the mole ratio between the metal and ligand.
2. Draw all possible isomers for Ma2bcd assuming a square pyramidal coordination.
References
1. G.A. Lawrance and C.J. Rix, J. Chem. Educ., 1979, 56, 211-212.
2. J.A. Weyh, J. Chem. Edu., 1970, 47, 715-716.
3. J. Hidaka, Y. Shimura, and R. Tsuchida, Bull Chem. Soc. Japan, 1966, 35, 567.
4. D.W. Cooke, Inorg. Chem., 1966, 5, 1411.
5. J.I. Legg and D.W. Cooke, Inorg. Chem., 1966, 5, 594.

50

References

1. Sing, M.M., Pike, R.M., and Szafran, Z., Microscale& Selected Mcroscale Experiments for
General & Advanced Chemistry; An Innovetive Approach. 1st ed., john wiley& Sons, USA,
1995.
2. Pass, G. and Sutcliffe, H. Practical Inorganic chemistry: Preparations, Reactions and
Instrumental Methods, Chapman and hall, London, 1974.
3. Sing, M.M., Pike, R.M., and Szafran, Z., Microscale & Selected Macroscale Experiments
for General &Advance General Chemistry : An Innovative Approach. 1sted., John Wiley &
Sons, USA, 1995.
4. Sing, MM.,Pike, R.M., and Szafran, Z., microscale& Selected MacroscaleExperimenth for
General & Advanced General Chemistry; An Innovative Approach.1st ed., John wiley& Sons,
USA, 1995.
5. Woollins,J.D., Inorganic Experimenth. VCH, Tokyo, 1994.

6.Potts, R.A.1974.Journal of chemical Education,51,539

7.Mackay, K.M., Mackay, R.A. and Henderson, W., Modern Inorganic Chemistry, 5th ed.,
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