Question

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

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

Answer 1

Answer: (B)

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