Question

Ni-en complex
If I desire to make these products, and I start with 5 mL of [Ni(H2O)6] 2+ in each case, what is the amount of (en) ligand that should be added to create each of these complexes?

[Ni(H₂O)₄(en)]²⁺

[Ni(H₂O)₂(en)₂]²⁺ :

[Ni(en)₃]²⁺

Answer 1

Nickel-ethylenediamine complexes will be synthesize with the general formula, [Ni(en)_x(H2O)_6-2x]SO₄•YH2O.

Nickel(II) is a good Lewis acid and prefers to bind six Lewis base donor atoms (atoms with lone pair of electrons). In pure water nickel(II) sulfate dissolves in water to form [Ni(OH2)6]²⁺, also known as Ni²⁺_(aq). When ethylenediamine is added, (abbreviated en; which contains 2 donor atoms per molecule) some or all of the water may be replaced in [Ni(OH2)6]²⁺

[Ni(H₂O)₆]²⁺ + H₂NCH₂CH₂NH_{2(aq) -->}[Ni(H₂O)₄(en)]²⁺

[Ni(H₂O)₄(en)]²⁺ + H₂NCH₂CH₂NH_{2(aq) -->}[Ni(H₂O)₂(en)₂]²⁺

[Ni(H₂O)₂(en)₂]²⁺ + H₂NCH₂CH₂NH_{2(aq) -->} [Ni(en)₃]²⁺

Ni2+ in aqueous solution exists as the octahedral complex ion hexaaquanickel (II), [Ni(H2O)6]²⁺. When the bidentate ethylenediamine ligand (en) is added to the solution in a 1:1 concentration ratio, it replaces two water ligands of the [Ni(H2O)6]²⁺ to form [Ni(H2O)4(en)]²⁺:

If 5ml of 1M [Ni(H2O)6]²⁺ is used we have to use equal concentration of en ligand to from above given species.

So we have to add 5 ml of 5M En ligand solution at first step to produce [Ni(H₂O)₄(en)]²⁺,

And then at each step 5 ml of 1M solution of en ligand should be added till water ligands are replace by en [Ni(en)₃]²⁺ therefore total 15 ml of ligand solution should be added to fully replace all water ligand