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.
${\text{Co}}^{\text{3+}}\text{=}\text{[Ar]}{\text{3d}}^{\text{6}}{\text{4s}}^{\text{0}}$