Which of the following coordination compounds will most likely form a precipitate when treated with an aqueous solution of silver(I) nitrate?
[Cr(NH3)3Cl3]
[Cr(NH3)6]Cl3
[Cr(NH3)Cl]SO4
Na3[Cr(CN)6]
Na3[CrCl6]
The idea here is that AgCl is a white insoluble ppt in most solutions. We know that. However, in coordinations complexes, the Cl that is free and clear (from ionic bonding) will be free to form the AgCl. That Cl that is WITHIN the coordination sphere is covalently bound and will not be free to form AgCl. How do you know it is is ionic or covalent? Cl within the brackets are within the coordination sphere. Those Cl outside the brackets probably are ionic and will be free to form AgCl. For example, the first one the Cl is within the brackets so no ppt. Same thing for the last one.
To determine which of the given coordination compounds is most likely to form a precipitate when treated with silver(I) nitrate, we need to consider the solubility of the resulting compounds.
Silver chloride (AgCl) is insoluble in water, so any compound that contains chloride ions (Cl-) may form a precipitate when combined with silver(I) ions (Ag+). We need to identify the compounds that contain chloride ions (Cl-) and then check their solubility.
Let's go through each compound one by one:
1. [Cr(NH3)3Cl3]: This compound contains chloride ions (Cl-), so it has the potential to form a precipitate. However, we need to check its solubility. Complexes of a transition metal like chromium (Cr) tend to be soluble in water. Therefore, it is not likely to form a precipitate.
2. [Cr(NH3)6]Cl3: This compound also contains chloride ions (Cl-). Similar to the previous compound, the complex of chromium (Cr) tends to be soluble, so it is not likely to form a precipitate.
3. [Cr(NH3)Cl]SO4: This compound does not contain chloride ions (Cl-), so it will not form a precipitate with silver(I) nitrate. It contains sulfate ions (SO4^2-), which are typically soluble in water.
4. Na3[Cr(CN)6]: This compound does not contain chloride ions (Cl-) and cyanide (CN-) is generally a stable and soluble complex ion. Therefore, it is not likely to form a precipitate.
5. Na3[CrCl6]: This compound contains chloride ions (Cl-), which can potentially react with silver(I) nitrate to form a precipitate of silver chloride (AgCl). It is likely to form a precipitate.
Therefore, the compound that is most likely to form a precipitate when treated with an aqueous solution of silver(I) nitrate is Na3[CrCl6].