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Question: Which is more stable and why? \[{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}\]and \[{...

Which is more stable and why?
[Co(NH3)6]3+{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}and [Co(CN)6]3−  {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}\;

Explanation

Solution

We have to know that in a coordination complex, there are metal, coordination complexes and ligands attached to it. The metal, ligands and counter ions together form a coordination complex. Bidentate ligands are the one in which two atoms coordinate directly to the central atom forming a coordination complex.

Complete answer:
The given complexes are coordination complexes. The given complexes have one thing in common that they both are cobalt metal complexes. One is having different ligands i.e. ammonia while other is having cyanide. Both complexes have the same coordination number i.e. six as the number of ligands is same in both. Out of ammonia and cyanide, Ammonia is a weak field ligand whereas cyanide is a strong field ligand. When we look at cobalt it has atomic number 2727 having electronic configuration[Ar]3d74s2\left[ {Ar} \right]3{d^7}4{s^2}. The hybridization of [Co(NH3)6]3+{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}is sp3d2s{p^3}{d^2} that means outer d orbitals are used in this case. In case of [Co(NH3)6]3+{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} this complex four unpaired electrons are left since ammonia is a weak field ligand and this complex is high spin complex. The hybridization of [Co(CN)6]3−{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }} is d2sp3{d^2}s{p^3}. [Co(CN)6]3−{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}, is diamagnetic and has no unpaired electrons. Thus we can infer that since [Co(CN)6]3−{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}is a low spin complex having no unpaired electron thus is considered to be more stable as compared to [Co(NH3)6]3+{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}this ligand which is a high spin complex having four unpaired electrons.

Note:
We must have to know that when an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule defines oxidation state. Coordination number is derived for the central atom present in any complex, that central atom is bonded to many ligands around it.