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	27.10	27.10
Initial Volume of NaOH /cm3	0.00	0.00
Volume of NaOH used /cm3	27.10	27.10

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	24.90	24.95
Initial Volume of NaOH /cm3	0.00	0.00	0.00
Volume of NaOH used /cm3	24.95	24.90	24.95

From the 1^st 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, Δ_o of the ammine complex would be larger than that for the aqua complex. With a greater Δ_o, 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 A_2g, T_2g, T_1g(F) and T_1g (P), arranged in increasing energy.  There are only 3 possible transitions, arranged from small energy gap to big energy gap: ³T_2g ß³A_2g,   ³T_1g (F) ß ³A_2g and ³T_1g (P)ß ³A_2g. The energy of the radiation is inversely proportional to the wavelength at which the radiation is absorbed. Hence, ³T_1g (P)ß ³A_2g occurs at a smallest wavelength followed by ³T_1g (F) ß ³A_2g then ³T_2g ß³A_2g. 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 ³T_1g (P)ß ³A_2g with largest energy gap and, hence the smallest wavelength. The peak at 656.50nm corresponds to the second transition ³T_1g (F) ß ³A_2g. The peak at 719.00nm corresponds to the lowest energy transition of ³T_2g ß³A_2g.

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 (6^th 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