[emissivity]

   Hi! I'm stuck in these problems so I've just come to ask you some favor..
well.. let's get started..

I've just tried to name these compounds. I'll show you the procedure of mine..
1. [Ag(NH₃)₂][Ag(CN)₂]
First, left side is cation and the other side is anion. Next, each of them have +1, -1 charge(I think this procedure might have had my answer wrong). So, ammine has no charge at all,
diamminesilver(1) 
cyano has -1 charge so x-2=-1, x=1
dicyanoargentate(1)

thus its name is diamminesilver(1) dicyanoargentate(1)


2. [Pt(NH₃)₄)][Pt(Cl)₄]
-> tetraammineplatinum(1) tetrachloroplatinate(3)


3. [Fe(NH₃)₆][Cr(CN)₆]
-> hexaammineiron(1) hexacyanochromate(5)

Well.. However the answers were diamminesilver(I) dicyanoargentate(I), tetraammineplatinum(II) tetrachloroplatinate(II), hexaammineiron(III) hexacyanochromate (III) in order. I think it was the main problem that I assigned each anion and cation +1, -1. How can I overcome this problem? Please help me!



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The first one was right. But the others were wrong though I have solved them in same way.

[Schrödinger]

The problem is that you're starting off by assigning +1 oxidation state to the leftmost metal. This is not how it works, since it's not an algorithm that men have devised. These are common oxidation states found in nature.

You need to be able to read off data from the periodic table given the compounds. For instance, Fe(I) is not a common oxidation state. The most common ones for Fe are +2 and +3. And for Cr coordinated in an octahedral fashion, the most common oxidation state is +3, which gives rise to the d³ configuration.

Take the case of Pt. The most common oxidation states are +2, +4. +1 and +3 are, as far as I know, non-existent (although if they do exist, they must be very rare)

I would suggest familiarizing yourself with common oxidation states. You should be able to find this in any inorganic text.

[emissivity]

Thank you, Schrödinger.
As you have already said that, I might have to get used to common oxidation states.

Thank you again!