Question

Looking at the nickel complexes as an example, it is found that those formed with ammoniagive the complex [Ni(NH3)6]2 and when formed with ethylenediamine the complex [Ni(en)3]2 is the result. Even though both are octahedral complexes, why do you think that nickel iscoordinated with six ammonia ligands in one case and only three ethylenediamine ligands inthe other

Answer 1

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Step-by-step explanation:

There is a concept in chemistry called the denticity of a ligand. The denticity of a ligand refers to the number of donor groups in a given ligand that can bond to a central metal atom or ion. In other words, it is the number of bonding positions available in a ligand.

Ammonia has only one bonding position(monodentate ligand); that is, the lone pair on its nitrogen atom. Hence, six ammonia ligands are required to form an octahedral complex with the nickel center.

However, ethylenediamine has two positions available for bonding to a central metal atom or ion. These are the two nitrogen atoms present in the molecule. Hence, each molecule of ethylenediamine is bidentate, bonding to the nickel using two "teeth". Hence, three ethylenediamine molecules form six bonds to the nickel center as required in an octahedral complex.