Which coordination compound will most likely form a precipitate when treated with silver(II) nitrate (aqueous)?
[Cr(NH3)3Cl3]
[Cr(NH3)6Cl3]
Na3[Cr(CN)6]
Na3[CrCl6]
I tried the solubility rules, but that didn't work out. Any help would be great.
I made a mistake about, it's suppose to be silver(I) nitrate not (II)
For AgNO3 to react with chloride in a molecule, the Cl need to be attached outside the coordination sphere. That will be the one in which Cl is more ionic in character and will react with the AgNO3 to produce AgCl. The Cl atoms inside the coordination sphere usually are covalently bound and don't react with AgNO3, or at least if they do, they do so very slowly. Which Cl do you think is outside the coordination sphere?
The solubility rules, by the way, tell you that AgCl is insoluble but they don't help with this kind of problem.
I just started this section, so I am still learning about it and am still a bit confused...
Would it be
[Cr(NH3)]6Cl3
Exactly.
I think the original post has some brackets out of order; howewver, in this case, you see the Cr is coordinated with 6 NH3 molecules and when it is placed in water it ionizes
[Cr(NH3)6]Cl3 ==> [Cr(NH3)6]^+3 + 3Cl^- and the Cl^- are readily available to react with AgNO3 to form AgCl.
For one of the others, say, [Cr(NH3)3Cl3], you see that 3 NH3 are coordinated with Cr as are the 3 Cl atoms (coordination number for Cr is 6), so the Cl is covalently bound and is not available in water solution to react with AgNO3.
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