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The aim of this experiment was to synthesize HgCo(SCN)4 , [Ni(en)3]S2O3 , NiCl2(PPh3)2 , K2[Ni(CN)4] , K3[Fe(C2O4)3] and Cu(acac)2 but only K2[Ni(CN)4] , K3[Fe(C2O4)3] and Cu(acac)2 were synthesized and explained in this report. Characterization was then done on each of the 3 synthesized complexes by obtaining their UV-Vis spectrum. The UV-Vis spectrum of each complex would identify different properties and would help to explain the observations made when they were synthesized (for example their colour and how they reacted).
Table 1: Synthesis of K2[Ni(CN)4]
Amount of Reagents used Method Observations
NiSO4 • 7H2O : 6.00 g
KCN : 3.01g In 20ml water NiSO4 • 7H2O was dissolved and the solution containing KCN (dissolved in 7ml water) was added to the Nickel solution. This solution was filtered and the precipitate was washed with small quantities of water. This precipitate was added to the solution of KCN (dissolved in 3ml of water). This solution was heated and swirled so that the crystals that formed dissolved. It was heated again and left so that crystals started forming again. This solution was then placed in an ice bath to enhance crystallization and the crystals were removed by filtration. The water was removed by drying the crystals in the oven at 100°C. A small amount of the sample was dissolved in water and the UV-Vis spectrum was obtained. Nickel (II) sulphate hexahydrate was used in place of heptahydrate.
When the precipitate was added to the KCN solution, it turned dark red. As it was heated it darkened even more but as the crystals started forming the colour changed to a light red.
As the crystals dried more its colour changed from light red-orange to a light yellow colour.
These crystals were dissolved in water for the UV-vis spectrum. The solution was a yellow colour.
Table 2: Synthesis of K3[Fe(C2O4)3]
Amount of Reagents used Method Observations
(NH4)2Fe(SO4)2 • 6H2O: 5.03g To a solution of (NH4)2Fe(SO4)2 • 6H2O (dissolved in 20ml water) 1 ml of sulphuric acid was added and stirred. The oxalic acid (dissolved in 25ml water) was also added and this combined solution was slowly heated to boiling, resulting in the formation of yellow iron(II) oxalate precipitate. The liquid was decanted and 15ml of hot water was added again to the precipitate, this was stirred and filtered. The precipitate (iron(II) oxalate) was transferred to another beaker and a potassium oxalate solution (dissolved in 10ml hot water) was added.
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"Synthesis and Characterization of K2[Ni(CN)4] , K3[Fe(C2O4)3] and Cu(acac)2 Complexes." 123HelpMe.com. 09 Dec 2018
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After the hot 15ml water was added to the precipitate and this solution was filtered, the filtration took very long, this is because the precipitate was very fine powder.
When the second oxalic acid solution was added (20ml and 10ml later) the solution turned to a blue-green colour. When the 30 ml ethanol was then added to the filtered solution it turned dark green. Giving rise to green crystals.
The crystals were a light green colour, they were dissolved in water for the UV-vis spectrum. The solution was a light-green to yellow colour.
Sulphuric acid (H2SO4): 1 ml
Oxalic Acid (C2O4H2) : 2.54g
Potassium oxalate: 3.5g
Hydrogen Peroxide H2O2 20%: 8 ml
Ethanol : 30 ml
Table 3: Synthesis of Cu(acac)2
Amount of Reagents used Method Observations
Acetyl acetone : 2.5 ml Acetyl acetone was neutralised with the diluted ammonia and to this solution the Cu solution (dissolved in 50ml water) was added. This was then filtered and the precipitate was washed with water, acetone and a small amount of ether. Recrystallization was performed using ethanol: the precipitate was 1.2g thus 120ml of ethanol was added to this precipitate (100ml ethanol per 1 g complex). This solution was then cooled on ice and the crystals were dried by using filtering. A small amount of the sample was dissolved in water and the UV-Vis spectrum was obtained When the acetyl acetone was neutralised with 0.5 ml diluted ammonia, precipitate immediately formed when the ammonia was added to the acetyl acetone. When the acetyl acetone solution was added to the Cu solution a purple pastel colour was obtained when filtered. As this precipitate was recrystallized with ethanol it turned to a darker purple.
As this solution was filtered and dried, dark purple-blue crystals were obtained.
These crystals were dissolved in DMSO for the UV-vis spectrum because it did not dissolve in water. A blue solution was obtained.
Diluted ammonia NH3 : 0.5 ml
CuSO4 • 5H2O: 3.20 g
Table 4: Results obtained from each synthesis
Synthesis of: Structure Grams obtained % Yield
(Cfm Oskar Tropitzsch GmbH, 2014) 3.26 g 63.22 %
(Saritha, et al, 2012) 3.17 g 56.66 %
(Börnstein and Heidelberg, 2009) 1.02 g 31.83 %
Calculations for % yield:
n(moles) = m d(density) = m
M (g/mol) V (ml)
a. Synthesis of K2[Ni(CN)4]
NiSO4 • 7H2O + 4KCN K2[Ni(CN)4] + 7H2O + K2SO4
NiSO4 • 7H2O : 6g / 280.75 g/mol = 0.0214 moles
KCN : 6.01g / 65.12 g/mol = 0.0923 moles
Limiting reagent: NiSO4 • 7H2O
NiSO4 • 7H2O : K2[Ni(CN)4]
1 : 1
0.0214 moles : 0.0214 moles
Expected grams of K2[Ni(CN)4] (240.97 g/mol):
0.0214 moles x 240.97 g/moles
= 5.157 g
3.26g / 5.157 g x 100
b. Synthesis of K3[Fe(C2O4)3]
ox = C2O42-
2(NH4)2Fe(SO4)2•6H2O + 3K2C2O4 + 2H2C2O4 2K3[Fe(C2O4)3] + 2(NH4)2SO4 + 2H2SO4 + 6H2O
(NH4)2Fe(SO4)2•6H2O : 5 g / 392.14 g/mol = 0.0128 moles
K2C2O4 : 3.5g / 166.22 g/mol = 0.02106 moles
H2C2O4 : 2.5 g / 90.02 g/mol = 0.0278 moles
Limiting reagent: (NH4)2Fe(SO4)2•6H2O
(NH4)2Fe(SO4)2•6H2O : K3[Fe(C2O4)3]
2 : 2
0.0128 moles: 0.0128 moles
Expected grams of K3[Fe(C2O4)3] (437.12 g/mol) :
0.0128 moles x 437.12 g/mol
= 5.595 g
3.17 g / 5.595 g x 100
c. Synthesis of Cu(acac)2
acac = CH3C(O)CHC(OH)CH3 (M= 100.13 g/mol)
2acac + CuSO4 • 5H2O Cu(acac)2 + 5H2O + SO42-
acac = 2.5 ml x 0.9732 g/ml = 2.433 g
2.433 g / 100.13 g/mol = 0.0243 moles
CuSO4 • 5H2O : 3.2 g / 249.69 g/mol = 0.0128 moles
Limiting reagent: acac
2acac : Cu(acac)2
2 : 1
0.0243 moles : 0.01215 moles
Expected grams of Cu(acac)2 (263.81 g/mol):
0.01215 moles x 263.81 g/mol
= 3.205 g
1.02 g / 3.205 g x 100
= 31.83 %
1. Answer to question 1:
a. Synthesis of [Ni(CN)4]2-:
Ni(II) : high spin metal
CN-: strong field ligand
Thus Δ0 is large based on the ligand (CN) and electrons would pair
Low spin = Square planar
Hybridization (orbitals involved in bonding): dsp2
Ligand: CN- : X- type ligand
2 unpaired electrons (paramagnetic)
b. Synthesis of [Fe(C2O4)3]3-
Fe3+: high spin metal
Ox: weak field ligand
Δ0 is small based on the high spin and weak field ligand thus electrons would jump
Hybridization(orbitals involved in bonding): sp3d2
Ligand: ox = X-type (X-)
5 unpaired electrons
c. Synthesis of Cu(acac)2
Cu(II) : Low spin metal
acac: strong field ligand
Thus Δ0 is large based on the metal and ligand and electrons would pair
Low spin = Square planar
Hybridization(orbitals involved in bonding): sp3
Ligand: acac = X-type (X-)
1 unpaired electron
2. Answer to question 2:
3. Answer to question 4:’
Examples of Mn(II) complexes:
MnCl2 , MnF2 (Atkins & Shriver, 2010: 455).
Mn(OH2) (Atkins & Shriver, 2010: 505).
Extra examples from 2013 Notes CMY 285 (Bezuidenhout):
Strong field ligands are ligands that produce a large Δ in the crystal field diagram meaning that the Δ(gap) between the 2 different energy levels are too large thus electrons would first pair with each other rather than jumping to the next energy level because they would need a lot of energy to jump the large gap. This would result in more paired electrons.
Strong field ligand – CN-
Weak field ligands are ligands that produce a small Δ in the crystal field diagram meaning that the Δ(gap) between the 2 different energy levels is small enough to let the electrons jump to the next energy level rather than pairing with other electrons. This would result in more unpaired electrons.
Weak field ligand – Br-
4. Answer to question 5:
Ligand CN- is a strong field ligand causing the Δsp to be large thus the electrons would pair. Because the ligand is a strong field ligand the metal is then low spin. This gives rise to a square planar structure. This complex is also diamagnetic. Ni(II) = d8. dsp2
Ni(II) is a high spin metal. PPh3 is a strong field ligand. Cl- is a weak field ligand. Because Ni and Cl is both on the left side of the spectrochemical series it can be concluded that the complex would have a small ΔT thus the electrons would jump. From this it can be said that this complex would be tetrahedral. e4 t24.
The ligand en is a strong field ligand causing that the Δ0 is large thus the electrons would rather pair. The structure would be octahedral and paramagnetic. t2g6 eg2. It is bound to 6 atoms(3 en, en is a bidentate ligand) = octahedral.
5. Answer to question 6:
[Fe(CN)6]3- - low spin complex
Orbital occupancy = t2g5 eg0
[Fe(ox)3]3- - high spin complex
Orbital occupancy = t2g3 eg2
Fe(III) = d5
[Ar] 4s2 3d3 (for both cases)
[Fe(ox)3]3- : This complex is optical active and this is because this complex has a mirror image.
[Fe(CN)6]3- : This complex is not optical active because it has no mirror image.
A solution ferrioxalate complex (when crystals form) is decomposed when left in direct sunlight. This causes the crystals to change colour from green to yellow.
[Fe(C2O4)3]3- + hν → [Fe(C2O4)2]2- + 2 CO2
The light allows the Iron-III to oxidize one of the oxalate ligands (reduce Iron-III complex) to carbon dioxide and give rise to the yellow ferro-oxalate complex ion which coordinate around an Iron-II centre, but when placed in the dark the Iron-II is re-oxidized to Iron-III by oxygen in the atmosphere and the green ferri-oxalate complex ion re-forms. The yellow Iron-II complex starts to appear after around ten minutes exposure to light. The re-oxidation in the dark is very slow. (Arellanes, 2009: 125).
6. Answer to question 7:
NCS- : SCN- :
NCS- ligand binds to the metal through the N where SCN- binds to the metal through the S and this means that this is an ambidentate ligand.
NCS- and SCN- can be seen as resonance structures thus they don’t have different steric features. The steric features is the same for both of them because the atoms sit at the position in both NCS- and SCN- thus in both cases each atom takes up the same space.
Ligand cone angle is the measurement of the size of the ligand that is bound to the metal. It is the direct angle formed with the metal and the hydrogen atoms at the perimeter of the cone.
Table 5: Cone angles
Ligands Cone angles (°)
P tBu3 182
(Phosphine Donors, 2014)
PH3 had the lowest cone angle as expected because only Hydrogens is present in the ligand and they take only a small space because they are the smallest atom.
As expected in table 5, the cone angles increases down the table because the ligands also increases in size down the table. P tBu3 is the largest ligand between all 4 and as expected it had the largest cone angle.
7. Answer to question 8:
a. For K2[Ni(CN)4] :
There were no isomers because this complex is achiral. Ni was bound to 4 of the same ligands thus if a mirror image of this complex is obtained, then they are super imposable on each other thus it is achiral.
b. For K3[Fe(C2O4)3]:
For this complex two isomers were obtained.
c. For Cu(acac)2:
There were no isomers because this complex is achiral. Cu was bound to 2 of the same ligands thus if a mirror image of this complex is obtained, then they are super imposable on each other thus it is achiral.
8. Answer to question 9:
Magnetic moments: µ = √N(N+2) N = unpaired electrons
a. K2[Ni(CN)4] :
No unpaired electrons. (see question 1 crystal field structure)
µ = √N(N+2)
Ni(II) : [Ar] 4s2 3d6
yz2, zx2 , z2, 2 , xy2
There were no unpaired electrons (also see question 1 with the electron configuration) thus as expected the magnetic moment would be zero. Because there is only paired electrons this means that the electrons spin in opposite direction thus the magnetic field of the electrons cancel out, giving rise to no net magnetic field as seen in the calculation.
5 unpaired electrons. (see question 1 crystal field structure)
µ = √N(N+2)
Fe(III) : [Ar] 4s2 3d3
There were 5 unpaired electrons (also see question 1 with the electron configuration) thus as expected the magnetic moment would be high because it is dependent on the unpaired electrons. There were 5 unpaired electrons this means that the electrons attract a magnetic field thus the magnetic field of the electrons would cause a dipole moment, giving rise to a large net magnetic field as seen in the calculation.
Because there were a lot of unpaired electrons it was expected that the magnetic moment would be high because a magnetic moment comes from unpaired electrons.
This is compared with the electron configuration and with the electron configuration it can be seen that there is unpaired electrons thus it was expected that the magnetic moment would be high.
1 unpaired electron. (see question 1 crystal field structure)
µ = √N(N+2)
Cu(II) : [Ar] 4s2 3d7
yz2, zx2 , z2, 2 , xy2 , (x2 – y2)1
There was 1 unpaired electron (also see in question 1 and with the electron configuration) thus as expected the magnetic moment would be present but not as high. There was 1 unpaired electron this means that the single electron attracted a magnetic field thus the magnetic field of the electron would cause only a small dipole moment, giving rise to a small net magnetic moment as seen in the calculation.
Because there was only 1 unpaired electron it was expected that the magnetic moment would be small because a magnetic moment comes from unpaired electrons.
This is compared with the electron configuration and with the electron configuration it can be seen that there was an unpaired electron thus it was expected that the magnetic moment would be small.
In this experiment 3 complexes were synthesized. Different colours were observed for each complex. The % yield for each complex was calculated. The last complex had the lowest % yield and this could be because too much ether was used when the crystals were washed or because the crystals were not cooled long enough so all the crystals did not form completely thus when they were dried, a lot of the crystals were lost.
When K3[Fe(C2O4)3] was synthesized it should have been kept in the dark to prevent photo reduction and this was done correctly because when the crystals formed they were a shocking green colour and no yellow crystals formed.
By using the crystal field structure each structure of the complexes was determined. From this it can be concluded that the experiment was done correctly and results were obtained that corresponded with the theories used.
Arellanes, C. 2009. Measurements of reactive Oxygen Species in the Particle Phase. ProQuest LLC: United States. [Online]. Available:
Bezuidenhout, D.I. 2013. Coordination Compounds. University of Pretoria: Pretoria.
Börnstein, L. & Heidelberg, S.V.B. 2009. Bis(acetylacetonato)copper(II) (C10H14CuO4). Landolt-Börnstein Substances/Property Index. [Online]. Available: < http://lb.chemie. uni- hamburg.de/static/RN/1_43143-72-2%20...%2046376-57-2.php?content=151/ EntvWz01> [Accessed 10 April 2014].
Cfm Oskar Tropitzsch GmbH. 2014. Potassium tetracyanonickelate-II-hydrate. [Online]. Available:< http://www.cfmot.de/en/kalium-tetracyano-nickelate-ii-hydrat.html> [Accessed 10 April 2014].
Hund, I. 2014. Chapter 8: Nucleophilic Substitution. Department of Chemistry. [Online]. Available: < http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch08/ch8-8.html> [Accessed 16 April 2014]. University of Calgary: Canada.
Saritha, A., Raju, B., Ramachary, M., Raghavaiah, P. & Hussain, K.A. 2012. Physica B. Synthesis, crystal structure and characterization of chiral, three-dimensional anhydrous potassium tris(oxalato)ferrate(III), 407(21): 4208-4213. [Online]. Available:
Shriver, D.F. & Atkins, P.W., 2010. Inorganic Chemistry. Oxford University Press: New York.
Stanley. 2014. Phosphine Donors. Chempub. [Online]. Available: < http://chem-faculty.lsu.edu/stanley/webpub/4571-chap4-phosphines.pdf> [Accessed 18 April 2014].
I enjoyed the experiment a lot. It could have been interesting to also synthesize the other 3 complexes in order to cover the more different structures and reactions.