The number of chloride ions which would be precipitated when the complex${\mathrm{PtCl}}_{4 }. 4{\mathrm{NH}}_3$ is treated with silver nitrate is : (here coordination number of platinum is 6).

There are 2 chloride ions which would be precipitated when the complex  $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]{\mathrm{Cl}}_2$

$\begin{array}{l}\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]{\mathrm{Cl}}_2\stackrel{aq}{\iff }{\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]}^{2+}\left(aq\right)+2{\mathrm{Cl}}^{-}\left(aq\right)\\ 2{\mathrm{Ag}}^{+}+2{\mathrm{Cl}}^{-}\to 2\mathrm{AgCl}\downarrow \text{ (white) }\end{array}$

Hence correct option is D.