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Chapter 20: Problem 88

The number of ions produced from one molecule of \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{3}\) in the aqueous solution will be (a) 4 (b) 5 (c) 6 (d) 7

Short Answer

Expert verified
The correct answer is (a) 4 ions.

Step by step solution

01

Understanding the Compound

The compound given is \([ ext{Pt}( ext{NH}_3)_5 ext{Br}] ext{Br}_3\). This is a coordination complex with one platinum ion surrounded by five ammonia molecules and one bromine inside the coordination sphere, and three additional bromine ions outside the bracket (counter ions).
02

Identifying the Coordination Sphere

In the compound \([ ext{Pt}( ext{NH}_3)_5 ext{Br}] \), the components inside the bracket (\( ext{Pt}( ext{NH}_3)_5 ext{Br}\)) form the coordination sphere. The bromine inside this coordination sphere is not an ionizable bromine.
03

Counting the Ionizable Parts

The bromine atoms outside the coordination sphere, \( ext{Br}_3\), are ionizable and exist as separate ions in the solution.
04

Dissociation in an Aqueous Solution

When the compound is dissolved in water, it dissociates into the complex ion \([ ext{Pt}( ext{NH}_3)_5 ext{Br}]^{+3}\) and its counter ions \(3 ext{Br}^-\) which are three separate bromide ions.
05

Summing Up the Ions

Adding the ions from dissociation: one \([ ext{Pt}( ext{NH}_3)_5 ext{Br}]^{+3}\) ion and three \( ext{Br}^-\) ions give a total of four ions in solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization
Ionization is a process where an atom or a molecule acquires a positive or negative charge by gaining or losing electrons. In the context of coordination complexes, ionization involves the release of ions into a solution when the complex is dissolved in water.

Consider a coordination complex given by the compound \([\mathrm{Pt}(\mathrm{NH}_3)_5 \text{Br}] \text{Br}_3\). When this compound is introduced into an aqueous solution, the solid lattice breaks apart, causing a separation between the coordination complex ion and any counter ions. In this case, the counter ions are three bromide ions, denoted as \(3\text{Br}^-\).

It is vital to understand that the ionization you're seeing involves only the breaking of soluble bonds, leading to a distribution of the ions into the solution. This process highlights the essential role of solvent interaction in the complete separation and stabilization of ions.
Dissociation
Dissociation is the process where a compound breaks down into its constituent ions. This is a crucial concept for understanding the behavior of coordination complexes in water or other solvents.

In the case of \([\mathrm{Pt}(\mathrm{NH}_3)_5 \text{Br}]\text{Br}_3\), when the compound dissolves in water, it undergoes dissociation. This results in the separation into the primary complex ion \([\mathrm{Pt}(\mathrm{NH}_3)_5 \text{Br}]^{+3}\) and the counter ions \(3\text{Br}^-\), which are the ionizable parts of the compound.

Dissociation does not break the bonds within the coordination sphere, meaning the connections between the central metal atom, ammonia, and the bromine within the brackets remain intact. Thus, the compound dissociates into four individual ions:
  • One complex ion: \([\mathrm{Pt}(\mathrm{NH}_3)_5 \text{Br}]^{+3}\)
  • Three bromide ions: \(\text{Br}^-\)
Understanding this process is pivotal for calculating the number of ions a compound releases into an aqueous solution.
Aqueous Solution
An aqueous solution is a solution in which water is the solvent. It serves as the medium wherein compounds are dissolved or dispersed. For coordination complexes, an aqueous solution facilitates the dissociation and ionization of complex molecules.

When \([\mathrm{Pt}(\mathrm{NH}_3)_5 \text{Br}]\text{Br}_3\) is introduced into an aqueous solution, water molecules surround the compound. The water interacts with the ions, helping to stabilize them as they dissociate from the solid form.

The resulting process involves breaking the ionic bonds between the coordination sphere and the counter ions, releasing them into the water as freely moving entities. This environment allows complex compounds to interact with other chemicals or ions, leading to potential chemical reactions or changes in physical properties.
  • Aqueous solutions enable the dispersion of ions.
  • They are essential for many chemical reactions.
  • They provide a medium for conductivity due to free ions.
Through this, the true behavior of coordination complexes is realized, helping to predict their roles in various chemical applications.

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Most popular questions from this chapter

Tetrahedral complexes of the types of \(\left[\mathrm{Ma}_{4}\right]\) and \(\left[\mathrm{Ma}_{3} \mathrm{~b}\right]\) (here \(\mathrm{M}=\) Metal, a, b = Achiral ligands) are not able to show optical isomerism because (a) these molecules/ions have non super imposable mirror images (b) these molecules possess a centre of symmetry (c) these molecules/ions possess a plane of symmetry and hence are achiral (d) these molecules/ions possess \(\mathrm{C}\) axis of symmetry

Which one of the following has a square planar geometry? [2007] (a) \(\left[\mathrm{FeCl}_{4}\right]^{2}\) (b) \(\left[\mathrm{NiCl}_{4}\right]^{2}\) (c) \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\)

The possible number of co-ordination isomers of \(\mathrm{Pt}(\mathrm{Py})_{4} \mathrm{CuCl}_{4}\) are

While \(\mathrm{Ti}^{3+}, \mathrm{V}^{3+}, \mathrm{Fe}^{3+}\) and \(\mathrm{Co}^{2+}\) can afford a large number of tetrahedral complexes, \(\mathrm{Cr}^{3+}\) never does this, the reason being (a) crystal field stabilisation energy in octahedral vis-à-vis tetrahedral \(\mathrm{Cr}^{3+}\) system plays the deciding role (b) \(\mathrm{Cr}^{3^{3}}\) forces high crystal field splitting with a varieties of ligands (c) electronegativity of \(\mathrm{Cr}^{3+}\) is the largest among these trivalent 3 d-metals and so chromium prefers to be associated with as many ligands as its radius permits (d) both (b) and (c)

The complex with spin-only magnetic moment of \(\sim 4.9\) B.M. is (a) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3+}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\) (d) \([\mathrm{Fe}(\mathrm{H}, \mathrm{O})]^{2^{+}}\)
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