### Lab Report on Preparation and Spectrophotometric analysis of a Ni(II) Complex

The lab report below was submitted as part of the coursework for CM2111 Inorganic Chemistry. Please do not plagiarise from it as plagiarism might land you into trouble with your university. Do note that my report is well-circulated online and many of my juniors have received soft copies of it. Hence, please exercise prudence while referring to it and, if necessary, cite this webpage.

Aim

To prepare a nickel(II) amine  complex and identify it thereafter with volumetric titration and UV/VIS spectroscopy.

Results and calculations
 Mass of NiCl2.6H2O used 3.500 g Mass of Nickel (II) complex obtained 2.458 g

Observation
Green NiCl2.6H2O solid was dissolved in water. Upon adding concentrated aqueous ammonia, the solution turns purple. Violet crystals formed in the pale navy blue solution, which was cooled in an ice bath. Subsequently, these violet crystals were retrieved upon suction filtration.

Analysis of ammonia
Standardisation of NaOH:
Concentration of HCl = 0.2820M
 1 2 Final volume of NaOH /cm3 27.1 27.1 Initial Volume of NaOH /cm3 0 0 Volume of NaOH used /cm3 27.1 27.1

No. of moles of HCl = 10.0/1000 x 0.2820 = 2.820 x 10-3 mol
NaOH + HCl à NaCl + H2O
Since 1 mole of NaOH reacts with 1 mole of HCl, no. of moles of NaOH reacted = 2.855 x 10-3 mol
[NaOH]= no. of moles / volume = 2.820 x 10-3 / (27.10 / 1000) = 0.1041 mol /L
Titration of sample with NaOH:
 Trial 1 2 3 Mass of sample used /g 0.2025 0.2028 0.2026 Final volume of NaOH /cm3 24.95 24.9 24.95 Initial Volume of NaOH /cm3 0 0 0 Volume of NaOH used /cm3 24.95 24.9 24.95

From the 1st titration, No. of moles of NaOH = (24.95/ 1000) × 0.1041 = 2.597 × 10-3 mol
No. of moles of HCl reacted with NaOH = No. of moles of NaOH = 2.597 × 10-3 mol
Initial no. of moles of H+ in HCl = (25.00 / 1000) × 0.2820 = 7.050 × 10-3 mol

[Ni(NH3)x]2+ + xH+ + 6H2O → [Ni(H2O)6]2+ + xNH4+

No. of moles of H+ reacted with [Ni(NH3)x]2+ = 7.050 × 10-3  - 2.597 × 10-3 = 4.453 × 10-3 mol
No. of moles of NH4+ formed = No. of moles of H+ reacted with [Ni(NH3)x]2+ = No. of moles of NH3 present in complex = 4.453 x 10-3 mol
Mr of NH3 = 14.01 + 1.01 x 3 = 17.04
Mass of ammonia in [Ni(NH3)x]2+ = 4.453 × 10-3  × 17.04 = 0.07588 g
Weight % of ammonia   = 0.07588 / 0.2025 x 100 % = 37.4 %

Similarly, for 2nd and 3rd titration, the percentages by weight of ammonia are 37.5% and 37.4%.
Hence, the average percentage by weight of ammonia
= (37.4 + 37.5 + 37.4)/ 3 37.43% 37.4%

Analysis of Nickel
Mass of sample used = 0.2999 g
Wavelength of Maximum absorbance = 394.00 nm

Concentration of NiCl2.6H2O = 240.0410 g/L
Concentration of NiCl2.6H2O in sample 1 = 10.0 /100.0 x 240.0410 = 24.00410 g/L
Concentration of NiCl2.6H2O in sample 2 = 20.0 / 25.0 x 24.00410 = 19.20328 g/L
Concentration of NiCl2.6H2O in sample 3 = 10.0 / 25.0 x 24.00410 = 9.60164 g/L

From the spectra, the concentration of NiCl2.6H2O found in sample 4 = 12.375 mg/mL
Mass of NiCl2.6H2O in 25.0 mL of sample 4 = 12.375 x 25.00 = 309.375 mg = 0.3093 g
Mr. of NiCl2.6H2O = 58.69 + 2 × 35.45 + 6 × [2 × (1.01) +16.00]= 237.71
No. of moles of NiCl2.6H2O = 0.3093 / 237.71 = 1.301 x 10-3 mol
No. of moles of [Ni(NH3)x]2+ = 1.301 x 10-3 mol
Mass of Ni in [Ni(NH3)x]2+ = 1.301 x 10-3 x 58.69 = 0.07638 g

Hence, average percentage by weight of Ni = 0.07636 / 0.2999 x 100 % 25.46% 25.5 %
Ammonia : Nickel molar ratio and determination of complex
 Ni NH3 Weight percentage (%) 25.46 37.43 Molar mass (gmol-1) 58.69 17.04 Mole 25.5 / 58.69 = 0.434 37.4 / 17.04 = 2.197 Molar ratio 1 5
From the above table, possible identities of the complex formed are [Ni(NH3)5Cl]Cl and [Ni(NH3)5H2O]Cl2.
If the complex is [Ni(NH3)5Cl]Cl,
Mr of [Ni(NH3)5Cl]Cl  =58.69 + 17.04 x 5 + 35.45 x 2 = 214.79
weight % of ammonia = 5 x 17.04 / 214.79 x 100 % = 39.7 %
weight % of nickel = 58.69 / 214.79 x 100 % = 27.3 %

If the complex is [Ni(NH3)5H2O]Cl2,
Mr of [Ni(NH3)5H2O]Cl2 = 58.69 + 17.04 x 5 + 1.01 x 2 + 16.00 + 35.45 x 2 = 232.81
weight % of ammonia = 5 x 17.04 / 232.81 x 100 % = 36.6 %
weight % of nickel = 58.69 / 232.81 x 100 % = 25.2 %

The calculations are summarised in the table below to facilitate comparison of the weight percentages of the possible complexes with experimental values:
 Experimental value [Ni(NH3)5Cl]Cl [Ni(NH3)5H2O]Cl2 % NH3 37.4 39.7 36.6 % Ni 25.5 27.3 25.2
By comparing the percentage by weight of Ni and NH3, the product formed is deduced to be [Ni(NH3)5H2O]Cl2.
Percentage yield
NiCl2.6H2O + 5NH3 à [Ni(NH3)5H2O]Cl2 + 5H2O
No. of moles of NiCl2.6H2O =3.50 / 237.71 = 0.01472 mol
Since 1 mole of NiCl2.6H2O produces 1 mole of [Ni(NH3)5H2O]Cl2, no. of mole of [Ni(NH3)5H2O]Cl2 formed = 0.01472 mol
Theoretical mass of [Ni(NH3)5H2O]Cl2 formed = 0.01472 x 232.81 = 3.428 g
Percentage yield of [Ni(NH3)5H2O]Cl2 = 2.458 / 3.428 x 100 % 71.7 %

Molar extinction coefficient for Ni(II) Complex
Beer Lambert Law: A = e l c, where A is absorbance, e is the molar extinction coefficient, l is the path length and c is the concentration.
ε = A / (l × c) = 0.267 / ( 12.375 / 232.81 x 1 ) = 5.02 L mol-1 cm-1

Discussion
Colour of compounds
Nickel complexes are coloured because they absorb light in the visible region. From empirical observation, NiCl2.6H2O is green while [Ni(NH3)5H2O]Cl2 is violet. Since NH3 is higher in the spectrochemical series than H2O, the crystal field splitting, Δo of the ammine complex would be larger than that for the aqua complex. With a greater Δo, the wavelength of absorption is shorter, accounting for the purple of [Ni(NH3)5H2O]Cl2.
Assignment of absorption bands
In transition metal complexes, absorption of energy from the UV-visible light is sufficient to cause electronic transitions. 3 types of electronic transitions are possible:  d-d, charge transfer and ligand-ligand transitions.
Ni(II) has a d8 electronic configuration at ground state with 3F term. The associated energy levels due to crystal field splitting are A2g, T2g, T1g(F) and T1g (P), arranged in increasing energy.  There are only 3 possible transitions, arranged from small energy gap to big energy gap: 3T2g ß3A2g,   3T1g (F) ß 3A2g and 3T1g (P)ß 3A2g. The energy of the radiation is inversely proportional to the wavelength at which the radiation is absorbed. Hence, 3T1g (P)ß 3A2g occurs at a smallest wavelength followed by 3T1g (F) ß 3A2g then 3T2g ß3A2g. The spectrum of Ni (II) complex contains 3 peaks at 719.00nm, 656.50nm and 394.00nm. . The peak at 394.00nm correspond to the transition 3T1g (P)ß 3A2g with largest energy gap and, hence the smallest wavelength. The peak at 656.50nm corresponds to the second transition 3T1g (F) ß 3A2g. The peak at 719.00nm corresponds to the lowest energy transition of 3T2g ß3A2g.
The three transitions are all spin allowed transitions as the spin multiplicity remains as 3 in all the above transitions. However, by the Laporte selection rule, these transitions are forbidden as there is no change in parity. In addition, from orbital selection rule, d-d transitions are forbidden. This contributes to the weak absorption bands and the small molar extinction coefficient around 5 L mol-1 cm-1 which correlates to d-d transitions in centrosymmetric molecules. Even though the selection rules had been contradicted, the d-d transition still occurred. This is because the selecton rules are relaxed by molecular vibrations which cause temporary removal of the center of symmetry and spin-orbit coupling. d-d transitions could be induced by ligands which leads to distortion of octahedral symmetry or due to the mixing of the p-d orbitals as the 2 orbitals are in close proximity, which causes interactions between them.
Volumetric analysis of NH3
Volumetric analysis measures liquid volumes to determine the quantitative amount of a dissolved substance in solution. In this experiment, the sample was titrated with NaOH – which was previously standardized with known concentration of HCl. This allowed the number of ammonia ligands coordinated to the nickel metal centre to be determined. Standardisation of sodium hydroxide solution is required as NaOH reacts with atmospheric carbon dioxide, thereby affecting its concentration.
Spectrophotometric analysis of nickel
One of the most common applications of spectrophotometry is to determine the concentration of an analyte in a solution. This experimental approach requires Beer's Law, which predicts a linear relationship between the absorbance of the solution and the concentration of the analyte (assuming all other experimental parameters do not vary).
A solution of the synthesized compound was scanned from 330 to 800nm and its maximum absorbance wave is 394.00nm.
Then, 3 standard solutions with accurately known concentration were prepared. The absorbances of the standard solutions were measured to prepare a calibration curve that indicates how absorbance varies with concentration. By carrying out the calibration at maximum absorbance wavelength, this allows the absorbance to be amplified and result in a more accurate calibration graph. The concentration of nickel (II) ions was then determined by reading from the calibration graph representing the 3 standard solutions.
Formation of complexes
In this experiment, the complex formation involves the displacement of ligands. With NH3 being a stronger ligand than water, NH3 will displace the water ligand and form bond with the central metal ion. The Cl- will remain in the outer coordination sphere though there is a chance for Cl- to enter the inner coordination sphere. However, since H2O and NH3 are stronger ligands than Cl-, Cl- will be displaced by either of the ligands even if it formed bond with the central metal ion. As the number of H2O molecules available for substituting by NH3 decreases, it is more and more difficult for NH3 ligand to come in to replace H2O, thus the probability of the formation of [Ni(NH3)6] Cl2 is low as shown in the increase in pK of around 0.5 with each successive reaction.
[Ni(OH2)6]Cl2 + NH3 ↔ [Ni(OH2)5(NH3)]Cl2                                        pK1 = -2.72
[Ni(OH2)5(NH3)] Cl2 + NH3 ↔ [Ni(OH2)4(NH3)2] Cl2                                 pK2 = -2.17
[Ni(OH2)4(NH3)2] Cl2 + NH3 ↔ [Ni(OH2)3(NH3)3] Cl2                          pK3 = -1.66
[Ni(OH2)3(NH3)3] Cl2 + NH3 ↔ [Ni(OH2)2(NH3)4] Cl2                          pK4 = -1.12
[Ni(OH2)2(NH3)4] Cl2 + NH3 ↔ [Ni(OH2)(NH3)5] Cl2                           pK5 = -0.67
[Ni(OH2)(NH3)5] Cl2 + NH3 ↔ [Ni(NH3)6] Cl2                                      pK6 = -0.03

Identifying the Nickel (II) complex
From the volumetric and spectroscopic data recorded, the mole ratio between nickel and ammonia turned out to be 1:5. This narrowed down the identity of the compound to either [Ni(NH3)5Cl]Cl or [Ni(NH3)5(H2O)]Cl2.
By comparing the weight percentage of NH3 and Ni2+ with the respective theoretical complexes, it was concluded that [Ni(NH3)5(H2O)]Cl2 is the most probable product.

There are some possible reasons that may cause the deduction of different structures, especially since the experimental results do not exactly match the expected results. Foremost, the displacement of H2O ligands with NH3 ligands was not complete as the complex formed in the reaction is liabile to ligand substitution. Also, when dissolved in an aqueous medium especially in titration when water was added, water may displace ammonia from the complex, thereby reducing product yield.

Handling of experiment
The crystals were thoroughly dried before spectroscopy was performed as water impurities may interfere with the spectrometric results. The produced crystals underwent suction filtration, were rinsed with ethanol three times and thereafter rinsed with ether. Ethanol and ether were used because they are good drying reagents.
Deionised water was not used in the washings of the crystals as the ammonia ligands present in the product were labile and may be displaced by water molecules. In addition, drying of the crystals was not done under IR lamp to prevent the crystals from decomposing.
Percentage yield
The percentage yield of [Ni(NH3)5H2O]Cl2 was a reasonable 71.7 %. The yield was not 100% because ligand substitution reactions are reversible. The yields are also dependent on the concentration of the ligands. With a higher concentration of ligand, NH3, there is more final complex being formed due to the Le Chatelier’s Principle. Some of the crystals may have also been lost during the transferring processes. To improve the yield, the crystals should be allowed to stand for a longer period of time as well as using a higher concentration of the ligands. Minimal volume of solvent should be used for washing as the crystals may dissolve and hence, reduce the yield.
Improvement/Modification to experiment
The identity of the nickel complex can be further confirmed by making use of the fact that the pair [Ni(NH3)5Cl]Cl and [Ni(NH3)5(H2O)]Cl2 has different number of ionisable Cl- with [Ni(NH3)5Cl]Cl having 1 and [Ni(NH3)5(H2O)]Cl2 having 2 such as measuring the conductance of the compound  or precipitation with lead nitrate.
Conclusion
A prepared nickel (II) complex was analysed via volumetric and spectrophotometric methods. The ammonia: nickel molar ratio in the synthesized compound is 5: 1. The product formed is deduced to be [Ni(NH3)5H2O]Cl2 with weight percentage of nickel 25.5% and ammonia 37.4%.  The percentage yield of product is 71.7%. The molar extinction coefficient of Ni(II) complex is  5.02 L mol-1 cm-1.
Reference
1)   Crystal Field Theory. University of West Indies, article retrieved on 25 Jan 2012: http://wwwchem.uwimona.edu.jm/courses/CFT.html
2)   Colour, solubility, and complex ion equilibria of nickel (II) species in aqueous solution. Bassam Z. Shakhashiri , Glen E. Dirreen and Fred Juergens . J. Chem. Educ., 1980, 57 (12), p 900.
3)   Inorganic Chemistry. Shriver and Atkins, 3rd Edition, Oxford University Press, 1999.
4)  Quantitative Chemical Analysis (6th edition). Daniel C. Harris. Pg 549 – 552.
5)   The Interactive Chemistry Laboratory Manual (ICLM), reference retrieved on 26 Jan 2012: http://courseware.nus.edu.sg/iclm/default.asp

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