Question

Consider aqueous solutions of the following coordination compounds: \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}, \operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}, \mathrm{Na}_{2} \mathrm{Pt}_{6},\) and \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3}\) . If aqueous 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.

Short answer

Upon adding AgNO₃ to the separate beakers containing the coordination compounds, the moles of AgI precipitated per mole of transition metal are: For \(\mathrm{Co}(\mathrm{NH}_{3})_{6}\mathrm{I}_{3}\): 3 moles of AgI, For \(\mathrm{Pt}(\mathrm{NH}_{3})_{4}\mathrm{I}_{4}\): 4 moles of AgI, For \(\mathrm{Na}_{2}\mathrm{Pt}_{6}\): 0 moles of AgI, and For \(\mathrm{Cr}(\mathrm{NH}_{3})_{4}\mathrm{I}_{3}\): 3 moles of AgI.

Step-by-step solution

Identify and count the I⁻ ions in each coordination compound.

: The four coordination compounds are: 1. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\mathrm{I}_{3}\): The coordination sphere is \([Co(\mathrm{NH}_{3})_{6}]\), and there are 3 I⁻ ions present. 2. \(\mathrm{Pt}(\mathrm{NH}_{3})_{4} \mathrm{I}_{4}\): The coordination sphere is \([Pt(\mathrm{NH}_{3})_{4}]\), and there are 4 I⁻ ions present. 3. \(\mathrm{Na}_{2} \mathrm{Pt}_{6}\): The coordination sphere is \([Pt(\mathrm{NH}_{3})_{6}]\), and there are 0 I⁻ ions present. 4. \(\mathrm{Cr}(\mathrm{NH}_{3})_{4}\mathrm{I}_{3}\): The coordination sphere is \([Cr(\mathrm{NH}_{3})_{4}]\), and there are 3 I⁻ ions present. Now, we will determine the molar ratio of AgI formed per mole of the transition metal in each compound.

Determine the molar ratio of AgI formed per mole of transition metal.

: 1. For \(\mathrm{Co}(\mathrm{NH}_{3})_{6}\mathrm{I}_{3}\), the 3 I⁻ ions will react with 3 moles of AgNO₃ to produce 3 moles of AgI. Therefore, the molar ratio of AgI formed per mole of Co is 3:1. 2. For \(\mathrm{Pt}(\mathrm{NH}_{3})_{4} \mathrm{I}_{4}\), the 4 I⁻ ions will react with 4 moles of AgNO₃ to produce 4 moles of AgI. Therefore, the molar ratio of AgI formed per mole of Pt is 4:1. 3. For \(\mathrm{Na}_{2}\mathrm{Pt}_{6}\), there are no I⁻ ions, and therefore no AgI will precipitate. The molar ratio of AgI formed per mole of Pt is 0:1. 4. For \(\mathrm{Cr}(\mathrm{NH}_{3})_{4}\mathrm{I}_{3}\), the 3 I⁻ ions will react with 3 moles of AgNO₃ to produce 3 moles of AgI. Therefore, the molar ratio of AgI formed per mole of Cr is 3:1. So, the moles of AgI precipitated per mole of transition metal for each coordination compound are as follows: \(\mathrm{Co}(\mathrm{NH}_{3})_{6}\mathrm{I}_{3}\): 3 moles of AgI \(\mathrm{Pt}(\mathrm{NH}_{3})_{4}\mathrm{I}_{4}\): 4 moles of AgI \(\mathrm{Na}_{2}\mathrm{Pt}_{6}\): 0 moles of AgI \(\mathrm{Cr}(\mathrm{NH}_{3})_{4}\mathrm{I}_{3}\): 3 moles of AgI

Key concepts

Aqueous Solutions

When discussing the chemistry of coordination compounds, it's important to note the role that aqueous solutions play. An aqueous solution is a liquid mixture where water is the solvent. In coordination chemistry, many reactions and syntheses occur in these solutions.
Water often interacts with metal ions and ligands, helping dissolve or react substances due to its polar nature. This makes it essential for studying coordination compounds.
A coordination compound such as \( \mathrm{Co} \left( \mathrm{NH}_{3} \right)_{6} \mathrm{I}_{3} \) dissociates in water, releasing ions. The complex ion stays intact while the iodide ions become free in solution. For reaction predictions, understanding the composition of these solutions aids in determining which ions will interact when other reactants are introduced.

Precipitation Reactions

Precipitation reactions are a type of reaction where two soluble compounds in a solution react to form an insoluble solid, known as a precipitate. In our context, when \( \mathrm{AgNO}_3 \) is added to aqueous solutions of coordination compounds, a precipitation reaction occurs if iodide ions are available.
In these exercises,

• Adding silver nitrate \( (\mathrm{AgNO}_3) \) to the solutions results in a reaction with free iodide \( (\mathrm{I}^- ) \) ions.
• The reaction forms silver iodide \( (\mathrm{AgI}) \), a characteristic yellow solid that precipitates out of the solution.

The presence of a precipitate indicates the formation of a product that's less soluble in water. Monitoring these reactions helps to identify the number of iodide ions present in the original coordination compound, making precipitation a useful tool in coordination chemistry.

Molar Ratios

The concept of molar ratios is fundamental in chemistry, especially in balancing chemical reactions and determining quantities of reactants or products. A molar ratio shows the proportion of reactants and products in a reaction, based on the coefficients from a balanced chemical equation.
For coordination complexes in aqueous solutions, understanding molar ratios helps to predict the quantity of a precipitate, like \( \mathrm{AgI} \), formed during a reaction.Take for example:

• \( \mathrm{Co} \left( \mathrm{NH}_{3} \right)_{6} \mathrm{I}_{3} \) possesses 3 iodide ions. These react with 3 moles of \( \mathrm{AgNO}_3 \) to form 3 moles of \( \mathrm{AgI} \), leading to a molar ratio of 3:1 for silver iodide to cobalt.
• Similarly, \( \mathrm{Pt} \left( \mathrm{NH}_{3} \right)_{4} \mathrm{I}_{4} \) reacts with silver nitrate to form 4 moles of \( \mathrm{AgI} \), resulting in a molar ratio of 4:1 for silver iodide to platinum.

These ratios are crucial for predicting outcomes in chemical processes, and understanding them can help ensure that chemical reactions are both efficient and complete, with no excess reactants remaining.