How is the oxidation state in this[PtCl2(NH3)2] is II and in this [PtCl(NH3)4]Br2 is IV??
In [PtCl2(NH3)2], overall charge on the complex = 0, charge on Cl = -1 x 2 , charge on NH3 = 0
Therefore, -2 charge on Cl is balanced by +2 charge on Pt.Thus, the oxidation state of Pt is +2
In [PtCl2(NH3)4]Br2, charge on [PtCl(NH3)4] = +2 due to -2 charge of Br
Now, charge on Cl = -2, NH3 = 0, therefore, -2 charge on Cl and +2 charge on the coordination sphere is balanced by +4 charge on Pt.Thus, the oxidation state of Pt is +4.