Question

Which of the following complexes is expected to be inert to the ligand substitution?
A.${\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)}_{\text{6}}}} \right]^{{\text{3 - }}}}$
B.${\left[ {{\text{Ni}}{{\left( {{\text{en}}} \right)}_{\text{3}}}} \right]^{{\text{2 + }}}}$
C.${\left[ {{\text{Mg}}\left( {{\text{EDTA}}} \right)} \right]^{{\text{2 - }}}}$
D.${\left[ {{\text{Sc}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}$

Answer 1

(1) The complexes which allow very quick exchange of ligands from their coordination sphere by other ligands are called labile complexes and the complexes which do not allow exchange of ligands or very slow exchange of ligands are called as inert complexes.
(2) The crystal field stabilization energy or CFSE plays a key role in determining the lability and inertness of octahedral complexes.
(3) According to the CFSE values, amongst the weak field octahedral complexes, the complexes having metal ions with ${{\text{d}}^{\text{0}}}{\text{,}}{{\text{d}}^{\text{1}}}{\text{,}}{{\text{d}}^{\text{2}}}{\text{,}}{{\text{d}}^{\text{4}}}{\text{,}}{{\text{d}}^{\text{5}}}{\text{,}}{{\text{d}}^{\text{6}}}{\text{,}}{{\text{d}}^{\text{7}}}{\text{,}}{{\text{d}}^{\text{9}}}$ and ${{\text{d}}^{{\text{10}}}}$ configurations would be labile and those having ${{\text{d}}^{\text{3}}}$ and ${{\text{d}}^{\text{8}}}$ configurations would be inert. Amongst strong field octahedral complexes, the complexes having metal ions with ${{\text{d}}^{\text{0}}}{\text{,}}{{\text{d}}^{\text{1}}}{\text{,}}{{\text{d}}^{\text{2}}}{\text{,}}{{\text{d}}^{\text{7}}}{\text{,}}{{\text{d}}^{\text{9}}}$ and ${{\text{d}}^{{\text{10}}}}$ configurations would be labile and those having ${{\text{d}}^{\text{3}}}{\text{,}}{{\text{d}}^{\text{4}}}{\text{,}}{{\text{d}}^{\text{5}}}{\text{,}}{{\text{d}}^{\text{6}}}$and ${{\text{d}}^{\text{8}}}$ configurations would be inert.

Complete step by step answer:
According to the spectrochemical series, ${\text{N}}{{\text{O}}_{\text{2}}}^{\text{ - }}{\text{,en,EDTA}}$ are strong field ligands and water is a weak field ligand.
In ${\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)}_{\text{6}}}} \right]^{{\text{3 - }}}}$complex, cobalt is in +3 oxidation state as the six nitrite ions have charge of -1 each and the charge on the overall complex is -3. Cobalt in +3 oxidation state has the ${{\text{d}}^{\text{6}}}$ configuration and the nitrite ligand is a strong field ligand. So, the ${\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)}_{\text{6}}}} \right]^{{\text{3 - }}}}$complex is a strong field complex having ${{\text{d}}^{\text{6}}}$ configuration. So, from CFSE values, it will be inert. So, option A is correct.
In ${\left[ {{\text{Ni}}{{\left( {{\text{en}}} \right)}_{\text{3}}}} \right]^{{\text{2 + }}}}$ complex, nickel is in +2 oxidation state as the three ethylene diamine ligands have charge zero and the overall charge is +2. Nickel in +2 state has ${{\text{d}}^8}$configuration and the en ligand is a strong field ligand. So, according to CFSE, the ${\left[ {{\text{Ni}}{{\left( {{\text{en}}} \right)}_{\text{3}}}} \right]^{{\text{2 + }}}}$ complex is an inert complex whatever maybe the ligand strength. So, option B is also correct.
${\left[ {{\text{Mg}}\left( {{\text{EDTA}}} \right)} \right]^{{\text{2 - }}}}$ is a labile complex as EDTA is a moderately strong field ligand.
In ${\left[ {{\text{Sc}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}$, scandium is in +3 state which means it has ${{\text{d}}^0}$ configuration and water is a weak field ligand. So, it is a labile complex. So, option D is incorrect.

Note:
For non-transition metal complexes, the factors that control lability are charge on the central metal atom, size of the central metal atom, charge to size ionic ratio and geometry of the complex. Greater positive charge, smaller metal ion size and greater charge to size ratio will lead to less labile complexes.