The ligand field spectra of octahedral nickel(II) complexes were investigated an

Question

The ligand field spectra of octahedral nickel(II) complexes were investigated and the UV-visible absorption spectra of all the products was to be recorded including the hexaaquanickel(II) chloride from which some of the complexes were made. Determine the amount/mass or how much of the products/complexes (show calculations) is needed in order to prepare the following molar aqueous solutions in measuring their absorption spectra in the 300 - 1000 nm range:

1. ~ 0.1 M of aqueous solutions of NiCl2.6H2O, [Ni(en)3]Cl2.2H2O and [Ni(pn)3]Cl2.2H2O

2. ~ 0.1 M [Ni(NH3)6]Cl2 aqueous solution using a 2 M ammonia solution as the solvent (and also therefore for the blank)

3. ~ 0.05 M [Ni(bipy)3]2+ aqueous solution

NOTE: The procedure for producing the complexes are shown below:

[Ni(NH3)]Cl2 Prepare about 3 g of [Ni(NH3)]Cl2 by the dropwise addition of concentrated aqueous ammonia to a concentrated solution of aqueous nickel chloride. Take note any colour changes. The reaction is complete when an excess of NH3 has been added to the reaction mixture. (How can you tell when an excess of NH3 is present?)

Explanation / Answer

1. Molecular weight of NiCl2.6H2O = 58 + (2*35.5) + (6*18) = 237u

Let x1 be its amount required to prepare the solution.

Given its moalrity = 0.1 M which means 0.1 mole per litre.

Therefore, 0.1 = x1 / 237

x1 = 23.7 g

   Molecular weight of [Ni(en)3]Cl2.2H2O = 58 + (3*60) + (2*35.5) + (2*18) = 345u

Let x2be its amount required to prepare the solution.

Given its moalrity = 0.1 M which means 0.1 mole per litre.

Therefore, 0.1 = x2 / 345

x2 = 34.5 g

Molecular weight of [Ni(pn)3]Cl2.2H2O = 58 + (3*74) + (2*35.5) + (2*18) = 387u

Let x3 be its amount required to prepare the solution.

Given its moalrity = 0.1 M which means 0.1 mole per litre.

Therefore, 0.1 = x3 / 387

x1 = 38.7 g

2. Molecular weight of [Ni(NH3)6]Cl2 = 58 + (6*17) + (2*35.5) = 231u

Let x4 be its amount required to prepare the solution.

Given its moalrity = 0.1 M which means 0.1 mole per litre

And NH3 is 2M solution which means 2 moles per litre and we require 0.6 moles of ammonia for 0.1 mole of the product..

Therefore, 0.1 = x4 / 231

x4 = 23.1 g

3. Molecular weight of [Ni(bipy)3]2+ = 58 + (3*156) = 526u

Let x5 be its amount required to prepare the solution.

Given its moalrity = 0.05 M which means 0.05 mole per litre.

Therefore, 0.05 = x5 / 526

x5 = 26.3 g

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