$\mathrm{PtCl}_4 \cdot 6 \mathrm{NH}_3$ [hexammine platinum (IV) chloride]. It gives $4 \mathrm{Cl}^{-}$ions and $6 \mathrm{NH}_3$, not involved in its ion because $\mathrm{NH}_3$ are present out of the complex of $\mathrm{PtCl}_4$.
$\begin{aligned} & \text { Hence, } \mathrm{PtCl}_4 \cdot 6 \mathrm{NH}_3 \longrightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_4 \longrightarrow \\ & \underbrace{\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}+4 \mathrm{Cl}^{-}}_{\text {Total 5 ions }} \\ & \end{aligned}$
In $\mathrm{Ni}(\mathrm{CO})_4, \mathrm{CO}$ is very strong binding ligand, it is difficult to remove all $4 \mathrm{CO}$ from nickel.
On other hand, $\mathrm{CoCl}_3 \cdot 5 \mathrm{H}_2 \mathrm{O}$ dissociates into $\mathrm{Co}^{3+}$, $3 \mathrm{Cl}^{-}$ions and $5 \mathrm{H}_2 \mathrm{O}$ molecules.
Hence, only $\mathrm{PtCl}_4 \cdot 6 \mathrm{NH}_3$ gives maximum number of ions in aqueous solution.