Consider aqueous solutions of the following coordination compounds: \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}, \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}, \mathrm{Na}_{2} \mathrm{Pt} \mathrm{I}_{6}\), and \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3} .\) If aqueous \(\mathrm{AgNO}_{3}\) is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.

According to the solubility rules, AgI is an insoluble compound in aqueous solutions. Therefore, in any of these coordination compounds, if an iodide (I-) ion is present as a counterion, it will react with the Ag+ ion from AgNO3 to form a precipitate of AgI. We can see that all four coordination compounds contain iodide ions: 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): Co(III) with 3 I- ions as counterions. 2. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): Pt(II) with 4 I- ions as counterions. 3. \(\mathrm{Na}_{2} \mathrm{Pt} \mathrm{I}_{6}\): Pt(IV) with 6 I- ions as counterions. 4. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): Cr(III) with 3 I- ions as counterions. So, all of these coordination compounds will react with AgNO3 to form a precipitate of AgI.

To find the number of moles of AgI that will form per mole of transition metal present in the solution, we need to determine the ratio of moles of I- ions per mole of the transition metal in each of the coordination compounds: 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): For every mole of Co(III), there are 3 moles of I- ions. So, there will be 3 moles of AgI precipitated per mole of Co(III) present. 2. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): For every mole of Pt(II), there are 4 moles of I- ions. So, there will be 4 moles of AgI precipitated per mole of Pt(II) present. 3. \(\mathrm{Na}_{2} \mathrm{Pt} \mathrm{I}_{6}\): For every mole of Pt(IV), there are 6 moles of I- ions. So, there will be 6 moles of AgI precipitated per mole of Pt(IV) present. 4. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): For every mole of Cr(III), there are 3 moles of I- ions. So, there will be 3 moles of AgI precipitated per mole of Cr(III) present. So, the number of moles of AgI that will precipitate per mole of transition metal present when aqueous \(\mathrm{AgNO}_{3}\) is added to separate beakers containing solutions of each coordination compound are as follows: - \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}\): 3 moles of AgI per mole of Co(III) present - \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}\): 4 moles of AgI per mole of Pt(II) present - \(\mathrm{Na}_{2} \mathrm{Pt} \mathrm{I}_{6}\): 6 moles of AgI per mole of Pt(IV) present - \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\): 3 moles of AgI per mole of Cr(III) present