Question

Why is $\left[Ni\right(en{)}_3{]}^{2+}$, nearly ${10}^{10}$ times more stable then $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$?

• A. $N{H}_3$ evaporates easily and causes instability to $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$ complex
• B. Six $N{H}_3$ ligands cause steric hindrance around the $N{i}^{2+}$ centre
• C. 'en' is a chelating ligand and forms thermodynamically more stable complexes
• D. $N{H}_3$ is the weakest ligand known

Answer 1

The correct option is C 'en' is a chelating ligand and forms thermodynamically more stable complexes

Option (C) is correct. $\left[Ni\right(en{)}_3{]}^{2+}$, is nearly ${10}^{10}$ times more stable then $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$ because '$en$' is a chelating ligand and forms thermodynamically more stable complexes while ammonia is a weak ligand.

In general cases, a cyclic or chelate complex is more stable than open complex due to chelating effect or macrocyclic effect.