How to find oxidation number and IUPAC name in compounds like [Pt(NH3)4 Cl2] [PtCl4] (both are components of a same compound)
here we use the concept of coordination no. for Pt having 4 as coordination no. the oxidation state is +2 and for having 6 as coordination no. the oxidation state is +4
[Pt(NH3)4 Cl2]+y [PtCl4]-y for coordination no. 6 Pt has oxidation state +4
4-2=y
y=2 similarly lets check for coordination no. 4 the oxidatn state of Pt =+2
2-4=-y
y=2
therefore [Pt(NH3)4Cl2]+2 [PtCl4]-2 this check the oxidation state of Pt as +4 and +2 then the naming would be same as before cations first and anions later just like before but individually then the name would be Tetra ammine dichlorido platinum(IV) tetra chorido platinate(II)
[Pt(NH3)4 Cl2]+y [PtCl4]-y for coordination no. 6 Pt has oxidation state +4
4-2=y
y=2 similarly lets check for coordination no. 4 the oxidatn state of Pt =+2
2-4=-y
y=2
therefore [Pt(NH3)4Cl2]+2 [PtCl4]-2 this check the oxidation state of Pt as +4 and +2 then the naming would be same as before cations first and anions later just like before but individually then the name would be Tetra ammine dichlorido platinum(IV) tetra chorido platinate(II)