Precipitation with NaI

I dissolved $\ce{[Pt(NH3)4](NO3)2}$ in water then I added small amount of $\ce{NaI}$ ($1~\mathrm{M}$) to it. Precipitation occurred. I want to know what happened?

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• You precipitated platinum(II) iodide. – MaxW Dec 1 '15 at 18:32
• Just out of curiosity, did you see any gas evolution? – Joel J. Dec 1 '15 at 20:15
• I thought it is PtI2. but when I took xrd for the precipitation, I saw too many peaks which I got confused on assigning the peaks to PtI2. – Sonia Dec 4 '15 at 20:25
• I didn't see any gas evolution. – Sonia Dec 4 '15 at 20:26

The original solution contains the square planar $\ce{Pt(II)}$ complex ion $\ce{[Pt(NH3)4]^2+}$. When sodium iodide is added to the solution, one or more $\ce{NH3}$ molecules are replaced by $\ce{I-}$ ions in a ligand exchange reaction. In particular, replacing two $\ce{NH3}$ molecules with $\ce{I-}$ ions yields the neutral complex $\ce{[PtI2(NH3)2]}$, which is poorly soluble in water, and therefore a precipitate is formed.
Remarkably, two isomers of square planar $\ce{[PtI2(NH3)2]}$ exist: $\ce{$cis${-}[PtI2(NH3)2]}$ and $\ce{$trans${-}[PtI2(NH3)2]}$.
When starting from $\ce{[Pt(NH3)4]^2+}$, the first $\ce{I-}$ can replace any of the four equivalent $\ce{NH3}$ ligands. Because of the larger trans effect of $\ce{I-}$ compared to $\ce{NH3}$, the second $\ce{NH3}$ is preferably replaced trans to the first $\ce{I-}$ ligand and therefore the trans product $\ce{$trans${-}[PtI2(NH3)2]}$ is obtained.
(Note that, if one starts from $\ce{[PtI4]^2- }$ and replaces two $\ce{I-}$ ligands with $\ce{NH3}$, the cis product $\ce{$cis${-}[PtI2(NH3)2]}$ is obtained instead.)