Question

Explain $${\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_6\right]}^{+3}$$ is an inner orbital complex whereas $${\left[\text{Ni}{\left(\text{N}{\text{H}}_3\right)}_6\right]}^{+2}$$ is an outer orbital complex.

Answer 1

Hint:Ammonia is a ligand with medium field strength. In certain cases it acts as a strong field ligand like with cobalt but in other cases like with iron, manganese and nickel, it acts as a weak field ligand.

Complete answer:
According to valence bond theory, the central metal atom will provide a vacant orbital according to coordination number. This vacant orbital undergoes hybridization and forms a coordinate bond with donor atom or ligand. The pairing of $$\left(\text{n}-1\right)\text{d}$$ electron of central metal atom is possible before hybridization in presence of strong field ligand and complex formation involves that newly available vacant orbital, such complexes are known as inner orbital complex. On the other hand, pairing will not occur in the presence of weak field ligands and complex formation does not involve such an orbital with unpaired electrons in it, such complexes are known as outer orbital complexes.
Generally, ligands with carbon or nitrogen donors are strong field ligands. The ligands with halogen, sulphur or oxygen donors are weak field ligands. Monodentate ligands are those species which form only one coordinate bond with a central metal atom like ammonia and water. Bidentate ligands are those species which form two coordinate bonds with metal. For example: ethylene diamine.
In $${\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_6\right]}^{+3}$$ complex, oxidation state of cobalt is $$+3$$ and coordination number is 6 as ammonia is monodentate ligand. Electronic configuration of $$\text{C}{\text{o}}^{+3}$$ is $$3{\text{d}}^{6}4{\text{s}}^0$$ . Since ammonia is strong field ligand, it will pair up the unpaired electron of $$\left(\text{n}-1\right)\text{d}$$ electron and results into availability of two vacant orbital in 3d. Thus, the complex will have $${\text{d}}^2\text{s}{\text{p}}^3$$ hybridization, octahedral geometry and diamagnetic. The complex will be an inner orbital complex and low spin complex.
In $${\left[\text{Ni}{\left(\text{N}{\text{H}}_3\right)}_6\right]}^{+2}$$ complex, oxidation state of nickel is $$+2$$ and coordination number is 6 as ammonia is a monodentate ligand. Electronic configuration of $$\text{N}{\text{i}}^{+2}$$ is $$3{\text{d}}^{8}4{\text{s}}^0$$ . Being strong field ligand, ammonia tries to pair up the unpaired electron of $$\left(\text{n}-1\right)\text{d}$$ but fails to do so due to avoid unnecessary pairing. Thus, the complex will have $$\text{s}{\text{p}}^3{\text{d}}^2$$ hybridization, octahedral geometry and paramagnetic. The complex will be an outer orbital complex and high spin complex.
Thus, $${\left[\text{Co}{\left(\text{N}{\text{H}}_3\right)}_6\right]}^{+3}$$ is an inner orbital complex and $${\left[\text{Ni}{\left(\text{N}{\text{H}}_3\right)}_6\right]}^{+2}$$ is an outer orbital complex.

Note:
Paramagnetic compounds or complexes are also known as high spin complexes as they have unpaired electrons. On the other hand, diamagnetic compounds which do not possess unpaired electrons are known as low spin complexes.