How many total possible isomers are present for

${\left[\mathrm{Pt}(\mathrm{CN}{)}_2{\left({\mathrm{NO}}_2\right)}_2\right]}^{2-}$?

$\begin{array}{r}\left[\mathrm{Pt}{\left({\mathrm{CN}}_2{\left({\mathrm{NO}}_2\right)}_2\right]}^{2-}\right.\\ \leftarrow \mathrm{C}\equiv {\mathrm{N}}^{\mathrm{\prime }}\to \\ \mathrm{O}=\mathrm{N}-{\mathrm{O}}^{\ominus }\to \\ \perp \end{array}$

Here

$(\mathrm{C},\mathrm{C})\left[\right(\mathrm{N},\mathrm{N})$ or $(\mathrm{N},\mathrm{O})$ or $(\mathrm{O},\mathrm{N})$ or $(\mathrm{O},0)]=4$

$\begin{array}{l}\left(\mathrm{C},{\mathrm{N}}^{\mathrm{\prime }}\right)\Rightarrow 4\\ \left({\mathrm{N}}^{\mathrm{\prime }},{\mathrm{N}}^{\mathrm{\prime }}\right)\Rightarrow 4\\ (\mathrm{C},\mathrm{N})\Rightarrow \left[(\mathrm{C},\mathrm{N}),\left({\mathrm{N}}^{\mathrm{\prime }},\mathrm{N}\right),(\mathrm{C},0),\left({\mathrm{N}}^{\mathrm{\prime }},0\right)\right]=4\\ (\mathrm{C},\mathrm{O})\Rightarrow \left[\left({\mathrm{N}}^{\mathrm{\prime }},\mathrm{N}\right),\left({\mathrm{N}}^{\mathrm{\prime }},\mathrm{O}\right)\right]=2\\ \left({\mathrm{N}}^{\mathrm{\prime }},\mathrm{N}\right)\Rightarrow \left[{\mathrm{N}}^{\mathrm{\prime }},\mathrm{O}\right]=1\end{array}$

Total = 19

Thus, the total number of total possible isomers present in given complex is 19.

Hence correct option is D