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)

 
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there should be no spacing while writing the name but its wasn't allowing to submit therefore ...
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Thanks
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tetra ammine dichlorido platinum(lV) tetra chlorido platinate(ll)
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The iupac name of the compound is tetra ammino platnium (II) tetra chlorido platinate (II)    

see for these types of compound take the first complex [Pt(NH3)4Cl2] the oxdiation no of Pt is  

4(0)ammonia + xplatnium +2(-1)chlorine = 0neutral    ===> x= 2 

the second [PtCl]

xplatnium + 4(-1)chlorine  = 0neutral  =>  x=4

the x=4 is shared bw the both equally so each gets 2 (dont worry abt 1st complex)
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