Chemistry - Coordination Chemistry Question with Solution | TestHub

I.${{\text{[Co(NH}}_{\text{3}}{\text{)}}_{\text{6}}\text{]}}^{\text{3+}}$
Here$\text{Co}$is present as${\text{Co}}^{\text{3+}}$ion.${\text{NH}}_{\text{3}}$being strong field ligand pair up the unpaired electrons of$\text{Co}$.
${}_{\text{27}}\text{Co}\text{=}\left[\text{Ar}\right]{\text{3d}}^{\text{7}}{\text{4s}}^{\text{2}}$
$\therefore {\text{Co}}^{\text{3+}}\text{=}\left[\text{Ar}\right]{\text{3d}}^{\text{6}}{\text{4s}}^{\text{0}}$



II.$\left[\text{Ni}{\left(\text{CO}\right)}_{\text{4}}\right]$
Here$\text{Ni}$is present in its ground state.$\text{CO}$being a strong field ligand, pair up the unpaired electrons of$\text{Ni}$
${}_{\text{28}}\text{Ni}\text{=}\left[\text{Ar}\right]{\text{3d}}^{\text{8}}{\text{4s}}^{\text{2}}$



III.$\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{\text{Cl}}_{\text{2}}\right]$
Here$\text{Pt}$is present as${\text{Pt}}^{\text{2+}}$ion.
${}_{\text{78}}\text{Pt}\text{=}\left[\text{Xe}\right]{\text{ 5d}}^{\text{9}}{\text{6s}}^{\text{1}}$
${\text{Pt}}^{\text{2+}}\text{=}\left[\text{Xe}\right]{\text{ 5d}}^{\text{8}}{\text{6s}}^{\text{0}}$



IV.${\left[{\text{CoF}}_{\text{6}}\right]}^{\text{3}-}$
Here$\text{Co}$is present as${\text{Co}}^{\text{3+}}$ion,$\text{F}$being weak field ligand is unable to pair up its unpaired electrons.
${\text{Co}}^{\text{3+}}\text{=}\text{[Ar]}{\text{3d}}^{\text{6}}{\text{4s}}^{\text{0}}$