$\left[\mathrm{PtBrCl}\left({\mathrm{NO}}_2\right)\left({\mathrm{NH}}_3\right)\right]\mathrm{I}$ on ionization gives the ion

• In a coordination compound, the atoms or molecules present inside the square bracket are not ionizable and are called secondary valency. The atoms or molecules present outside the square bracket are primary valency and are ionizable. 

Ionization of $\left[\mathrm{PtBrCl}\left({\mathrm{NO}}_2\right)\left({\mathrm{NH}}_3\right)\right]\mathrm{I}$ results in on ${\mathrm{I}}^{-}$since I is the only ionizable valency.