Chemistry assignment essay help theory writing: Solutions of transition metal salts

Chemistry assignment essay help theory writing: Solutions of transition metal salts

The immediately obvious property of solutions of transition metal salts is their colour.  All ranges of colour are observed at intensities varying from the palest pink of manganese (III) salts to the intense blues of some copper (II) complexes.

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In the last thirty years the concepts of crystal field theory and molecular orbital theory, and the empirical methods of ligand field theory, have allowed the electronic spectral transitions that give rise to the colours to be identified and characterised.

 

While the two theories: crystal field theory and molecular orbital theory are based on totally different concepts of the nature of chemical bonding, both theories assume the existence of molecular structure and that the symmetry properties of this molecular structure play a significant role in determining the nature of spectroscopically observable parameters such as the absorption of electromagnetic radiation.

To provide  some  insight  into  the  theoretical  basis  of  the  electronic  spectra  of  transition  metal complexes, the  basic theory of nomenclature of the crystal field will be developed and the results applied to an interpretation of the  spectra of a range of copper(II) complexes.   At points where the crystal field theory is inadequate to ‘explain’ the observed phenomena, concepts of molecular orbital theory will be introduced in a qualitative fashion.  It is this combination of crystal field and molecular orbital concepts that have assumed the title ligand field theory.

Buy Sample AssignmentCrystal Field Theory

In crystal field theory, a transition metal complex, eg [Co(NH3)6]3+; is treated as though the only interaction  between  the central metal atom (Co3+) and the set of nearest molecules or ions (NH3) is electrostatic.

 

The electronic spectral transitions observed then are those attributed to the central metal atom, with the frequencies  observed for the free metal ion perturbed due to the electrostatic potential field of the surrounding ligands.

To appreciate the effect of this potential generated by the ligands, imagine that a group of ligands with spherical  symmetry  is brought up to a charged ion from a far distance.                      Firstly, the electrostatic repulsions between the ligand electrons and those in the d-orbitals of the metal increases the energy of all 5 d orbitals equally.

Then, as the ligands approach bonding distance from the metal, the electron- electron repulsions assume a directional character.  Each d orbital, having different axes of quantization will be affected by the electrons in a different fashion.

 In the specific case of metal ion  Mn+  at the centre of an octahedral set of anions, X-, the  d orbitals that point directly along the x, y, and z axes towards the ligands.  (dx2-y2 and dz2) experience a greater repulsive interaction than do the  d orbitals that point between the ligands: dxy , dxz , dyz .

In terms of orbital energy levels then, the effect of an octahedral crystal field is to split the originally degenerate set of five  d orbitals into two sets.  The higher energy orbitals  (dx2-y2 , dz2  ) are labelled eg  ; the lower energy orbitals (dxy , dxz , dyz ) are labelled  t2g.

eg (dx2

2 ,dz2)

 

 

0.6 D0

 

 

 

0.4 D0

 

         t2g (dxy, dxz, dyz, )


Crystal Field Splitting

 

Degenerate d orbitals

(symmetric field)


Splitting of d orbitals in an octahedral field of ligands

 

 

The crystal field splitting is the energy difference between the t2g  and eg   orbital levels and it is frequently measured in terms of the parameter Do.  The magnitude of the splitting depends on the base strength of the ligands and is of the order of energy corresponding  to the energy of the visible region of electromagnetic radiation (hence the colours). Absorption of light  by the complex corresponds to an excitation of electrons from the lower to the higher energy levels.   Thus, the spectrum of d1  ion in a regular octahedral crystal field (one electron in the d orbitals) such as Ti3+  should  exhibit only one absorption band corresponding to excitation of the electron from the t2g    to the eg                                                                                  orbitals.    The

spectrum of [Ti(H2O)6]3+ (d1) does indeed have a single broad band with a maximum at 20,000 cm-1.

The crystal field splitting, Do, is equal to 20,000 cm-1.

 

The inadequacy of the quantitative application of crystal field theory is obvious even here however; using the purely  electrostatic model with totally separable metal and ligand wave functions, Do   is calculated to be only  ≈ 2000 cm-1 or only 10% of the observed value.

 

It is clear that the metal and ligand orbitals must interact as postulated in molecular orbital theory – nuclear magnetic resonance and electron spin resonance experiments show this – and so the crystal field theory must be modified to take  into  account this interaction.                               The modified theory which has a semi-empirical basis (i.e. the spectral parameters are derived from experiment) is known as the ligand field theory.

Spectrochemical Series

Buy Assignment AustraliaCorrelations of the electronic spectral parameters of a large number of complexes containing various metal ions and ligands have shown that ligands can be arranged in a series according to their capacity to affect the magnitude of d-orbital splitting.  Such a series, called the spectrochemical series, for some common ligands is:

Cl- <  oxygen donors  <  nitrogen donors  <  PPh3   < CN-  <  CO

Molar Absorptivities (M-1 cm-1)                  Examples

 10-2 – 1            d-d bands in many octahedral and tetrahedral complexes of d5  ions.

1 – 102                  d-d bands of 6-coordinate complexes in general and of some square planar complexes.

102  – 103           d-d bands in many tetrahedral complexes, in square planar complexes, particularly with organic ligands, and in some 6-coordinate complexes of very low symmetry.

Above 103        charge-transfer bands, bands resulting from allowed transitions.

EXTRACTION OF CHLOROPHYLL FROM LEAVES: REPLACEMENT OF BOUND MAGNESIUM BY COPPER

 

1.

§Aims

To extract and purify chlorophyll from leaves §

§To remove magnesium from chlorophyll and replace with copper

To record and interpret the UV visible spectrum of the apo, Mg and Cu chlorophyll compounds.

2.

References

E. Dujardin, P. Lazlo and D. Sacks, J. Chem. Ed., (1975)  52, 742.

3.

Introduction

The  tetrapyrrolic  macrocyclic  ligand,  porphyrin,  forms  complexes  with  metal  ions  known  as porphyrins.

Buy Sample AssignmentThe extensive  conjugation  in  these  macrocyclic  systems  is  responsible  for  low  energy  electronic transitions.  Porphyrins are accordingly intensely coloured.  Substituted porphyrins are very important biological compounds, being found in a wide variety of proteins and enzymes.  e.g. the bright red heme iron complexes in the transport of oxygen while vitamin B12, also bright red, contains cobalt ion at the centre  of  the  porphyrin.   Chlorophylls  are  magnesium  substituted  porphyrins  that  are  utilized  in photosynthesis to transform the photo energy of the sun into chemical energy through photosynthesis.

 Chlorophylls are the pigments responsible for the green colour of plants and are a mixture of two compounds, chlorophyll-a, pictured on the right and chlorophyll-b in which the methyl group marked with an asterisk is replaced by an aldehyde.

 The two  components  can  be  separated  by  shaking  a  light petroleum  solution  of  chlorophyll  with  aqueous  methanol. Chlorophyll-a  remains                          in              the   light  petroleum                while chlorophyll-b transfers to the aqueous methanol.  Chlorophyll- a is a  bluish  black solid and chlorophyll-b is a dark green solid.  Both give green solutions in organic solvents.

Treatment of chlorophyll-a,b with acid readily removes the magnesium ion and replaces it with two hydrogen atoms, giving an olive-brown solid, phaeophytin-a,b.  Hydrolysis of this splits of the phytol group to yield phaeophorbid-a,b.

 

N                N                                                    NH              N


+                         NH              N

H+                                                                                               H2O / H

Mg

N                  N                                                   N               HN


N               HN

 

 

H

 

H

 

H

 

H

 

H                 O CO2H

O           OC28H33


O           OC28H33


O            OH

Chlorophyll- a                                                        Phaeophytin- a                                                Phaeophorbid- a

 

 

 

In this experiment, a mixture of chlorophyll-a and chlorophyll-b will be extracted from leaves with ether, and purified by thin layer chromatography (TLC).  Visible spectra will be recorded.  Magnesium will be removed by addition of acid and replaced by copper.

Experimental Procedure

 Obtain at least two green leaves.  Spinach leaves are very suitable.

Break the leaves into small pieces, and place in a mortar. Add some sand and diethyl ether and crush (this should take about 5 to 10 minutes).  Use a total of approximately 50 ml of diethyl ether.  Filter the resultant green solution into a 100  ml conical flask and add some sodium sulphate drying agent (approximately 1 g) to remove any water from the solution.  Swirl the flask and allow to stand for 10 to

Re-filter the resultant green solution into a 100ml quick fit flask and reduce the volume on a rotary evaporator  to less than 5 ml.

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Take one of the 10 cm X 10 cm pre-prepared TLC and lightly rule a pencil line 2.5 cm from one edge. Dip a fine glass capillary into the crude chlorophyll solution.  Lightly touch with the capillary a point on the pencil line on one of your plates about 1.5 cm from the edge.  Some solution will drain from the capillary onto the plate to form a spot.  Form a second spot adjacent to and overlapping the first, and so on, to about 1.5 cm from the other edge.  This effectively introduces a band of material onto the plate. Superimpose several layers (depending on concentration of the solution) over the first to ensure that sufficient material has been transferred to the plate. A similar procedure is applied to the second plate.

 

Place the plate in the tank, and place in the dark (e.g. in a locker) to develop (20 – 30 min).  Remove the plates when the solvent front is near the top of the plates and immediately mark the position of the solvent front.  Observe the plates under visible and UV radiation, noting the colour and intensity of the bands.  Calculate the Rf  value for each band, using the formula:

Rf     =              distance moved by band

distance moved from solvent front

Both distances are measured from the pencil line where substance was applied.  The Rf  is characteristic of the substance for the particular type of plate and solvent.

Buy Assignments OnlineThe plates may be stored, if necessary, in the dark.  With a metal spatula, scrape off the green layer (or layers separately if the chlorophyll-a and chlorophyll-b are resolved) and place in a centrifuge tube.

 

Add 5 ml acetone, stir, then centrifuge (balancing with a tube of equal weight).   Transfer the green solution into a test tube.

Run the visible/near UV spectrum from 700 to 350 nm (using a water reference).  Record the maxima in nm.              Since the  concentration is unknown, extinction coefficients cannot be calculated, but you should give some indication of comparative intensities.

Removal of Magnesium

To 3 ml of the above solution add 500mL of 0.1 M HCl.  warm briefly on the steam bath.  Note any visible colour change.  Run the visible/near UV spectrum of the pheophytin solution.

Introduction of Copper

Transfer 2 ml of the pheophytin solution into a small quick fit flask.   Add 1 ml saturated aqueous copper(II) nitrate solution.  Place a condenser on the flask and reflux on the steam bath for about 5 minutes.  Allow to cool.  Shake.  Collect the (upper) hexane layer in a test tube.  Note the colour.  Run the visible / near UV spectrum.  Compare your results with the spectra obtained from the chlorophyll and pheophytin samples.

In your discussion, suggest reasons why peaks due to the copper ions are much less intense than those due to the chlorophyll and pheophytin samples.

Direct replacement of Mg by Copper or Zinc

Buy Assignments OnlineHeat leaf strips in diluted acetic acid (10%).  Separately heat other leaf strips in diluted acetic acid in the presence of a copper or zinc salt.  Observe the colour changes in the leaf strips.

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