One molecule of \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{4}\) gives \(\mathrm{X}\) ions in solution and requires \(\mathrm{Y}\) of \(\mathrm{AgNO}_{3}\) for complete precipitation of chloride ions. Here \(X\) and \(Y\) are respectively. (a) 4,4 (b) 5,5 (c) 5,4 (d) 4,5

When dealing with coordination compounds in chemistry, it's important to understand how they behave in solution. A complex ion comprises a central metal ion surrounded by molecules or anions called ligands. In the example of \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{4}\), platinum (Pt) is the central metal ion with ammonia (\(\mathrm{NH}_3\)) acting as ligands. Complex ions are notable for their stability: they tend not to disassociate in solution.

In solution, the complex ion and the counter ions behave independently. For our complex \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]\mathrm{Cl}_4\), this means that upon dissolving, the \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}\) complex ion will remain intact. It's crucial to recognize that, despite yielding multiple particles, this entire complex behaves as one ion in solution.

Understanding this concept is pivotal in predicting the properties of the compound in different reactions and environments.

One way to identify the presence of certain ions in a solution is by precipitation reactions. During such reactions, ions react to form an insoluble compound, known as the precipitate. In the given exercise, we focus on the precipitation of chloride ions when they react with silver ions coming from \(\mathrm{AgNO}_3\).

Here's the reaction to consider:
\(\mathrm{Cl}^- + \mathrm{Ag}^+ \rightarrow \mathrm{AgCl}\)(s)

Each chloride ion pairs with a silver ion to form a solid silver chloride, which precipitates out of the solution. The entire reaction is reliant on the stoichiometry of the ions involved; a 1:1 ratio of chloride to silver ions is necessary. It's important for students to recognize the signs of a precipitation reaction, as this will often be needed to identify reaction products or to purify compounds in a mixture.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the context of the exercise, stoichiometry informs us about the proportions of reactants required to proceed with the reaction. From the provided compound \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]\mathrm{Cl}_4\), we deduce that four moles of \(\mathrm{AgNO}_3\) are needed to completely react with the four moles of chloride ions. Each mole of \(\mathrm{AgNO}_3\) provides one mole of silver ions, which in turn reacts with one mole of chloride ions to form a precipitate.

The stoichiometrically-informed problem-solving steps involve: understanding the compound's makeup, determining the number of resultant ions in a solution, and then calculating the exact amount of a reactant required. Through this, we find that the correct answer to the exercise is the one where \(X\) (the number of ions from the complex in the solution) is 5 and \(Y\) (the moles of \(\mathrm{AgNO}_3\) required) is 4, clearly pushing students to marry the practicalities of stoichiometry with theoretical knowledge.