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How many isomers are there of the complex ion \(\left[\mathrm{CoCl}_{2}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} ?\) Sketch their structures.

Short Answer

Expert verified
There is only one isomer of the complex ion \([\mathrm{CoCl}_{2}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\]^{+}\), which results from ionization isomerism, and is \([\mathrm{Co}\mathrm{Cl}\left(\mathrm{en}\right)\left(\mathrm{NH}_{3}\right)_{3}\]^{+}\).

Step by step solution

01

Analyze the Complex Ion

First, analyze the complex ion. It consists of cobalt (Co) as the central atom, two chloride ions (Cl-), one Ethylenediamine (en) which is a bidentate ligand, and two Ammonia (NH3). Due to the presence of the bidentate ligand, geometrical isomerism is not possible here.
02

Identify Structural Isomerism

Determine if there is structural Isomerism, which involves rearrangement of the ligands. It can be of three types - Ionization Isomerism, Hydrate Isomerism, or Linkage Isomerism. In this case, only Ionization Isomerism is possible where a ligand can be replaced by an ion from the solution. Here, we can replace one Cl- with NH3.
03

Draw the Isomer Structures

Having identified the possible isomers, they are sketched. The original complex ion is \([\mathrm{CoCl}_{2}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\]^{+}\). The ionization isomer would be \([\mathrm{Co}\mathrm{Cl}\left(\mathrm{en}\right)\left(\mathrm{NH}_{3}\right)_{3}\]^{+}\), where one Cl- is replaced by NH3 and Cl- becomes a counter ion.

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Key Concepts

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

Structural Isomerism in Complex Ions
Structural isomerism in coordination compounds, such as complex ions, occurs when the arrangement of the ligands is different, but the overall composition of the complex remains the same. This can result in significantly different physical and chemical properties despite sharing the same formula.

For instance, in the context of the given exercise, structural isomerism involves rearrangement of the chloride ions, ammonia molecules, and the ethylenediamine (en) around the central cobalt ion. One particular form of structural isomerism that appears in this exercise is 'ionization isomerism,' where one of the chloride ligands is replaced by an ammonia molecule from the surrounding environment. This creates a counter ion in solution, which considerably alters the properties of the complex such as solubility, reactivity, and color.
Coordination Chemistry
Coordination chemistry is the branch of chemistry that focuses on the study of coordination compounds featuring central atoms or ions (typically metals) and surrounding ligands, which can be ions or molecules. These ligands are bound to the central atom by coordinate bonds, which involve the donation of a pair of electrons from the ligand to the metal.

As a case in point, the complex ion from the exercise involves cobalt as a central atom surrounded by multiple ligands: chloride ions, ammonia molecules, and ethylenediamine. The coordination number, which is the number of ligand attachment points to the central atom, is critical in determining the geometry and possible isomers of a complex ion.
Ligand Replacement
Ligand replacement plays a crucial role in structural isomerism. It is the process by which one ligand in a complex ion is swapped for another. This change often leads to the formation of an isomer.

Take for example the exercise at hand – we can replace one chloride ion (Cl-) within the coordination sphere with an ammonia (NH3) molecule, which results in isomerization. Such substitutions can markedly influence the properties of the complex and can even lead to new compounds with distinct reactivities and functions, pivotal in areas ranging from industrial catalysis to medicinal chemistry.
Ionization Isomerism
Ionization isomerism arises when the ligands and counter ions in a complex ion swap roles but the overall composition of the complex doesn't change. In our exercise, by replacing a bound chloride ion with an ammonia molecule that was initially not part of the coordination sphere, an ionization isomer is produced.

The resulting ionization isomer, \[\mathrm{Co}\mathrm{Cl}(\mathrm{en})(\mathrm{NH}_3)_3\]^+\, sees a chloride ion become a counter ion outside the coordination sphere. This type of isomerism affects the ionic behavior of the compound, which can lead to differences in things like solubility, electrical conductivity, and the way the compound responds to other reactants in solution.

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Most popular questions from this chapter

Explain the important distinction between each of the following pairs: (a) coordination number and oxidation number; (b) monodentate and polydentate ligands; (c) cis and trans isomers; (d) dextrorotatory and levorotatory compounds; (e) low-spin and highspin complexes.

In both \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) ions, the iron is present as \(\mathrm{Fe}(\mathrm{II}) ;\) however, \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is paramagnetic, whereas \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) is diamagnetic. Explain this difference.

Draw plausible structures corresponding to each of the following names. (a) pentamminenitrito- \(N\) -cobalt(III) ion (b) ethylenediaminedithiocyanato-S-copper(II) (c) hexaaquanickel(II) ion

If \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) are four different ligands, (a) how many geometric isomers will be found for square-planar \([\mathrm{PtABCD}]^{2+} ?\) (b) Will tetrahedral \([\mathrm{ZnABCD}]^{2+}\) display optical isomerism?

The compound \(\mathrm{CoCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \cdot 4 \mathrm{NH}_{3}\) may be one of the hydrate isomers \(\left[\mathrm{CoCl}\left(\mathrm{H}_{2} \mathrm{O}\right)\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl} \cdot \mathrm{H}_{2} \mathrm{O}\) or \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2} .\) A \(0.10 \mathrm{M}\) aqueous solution of the compound is found to have a freezing point of \(-0.56^{\circ} \mathrm{C} .\) Determine the correct formula of the compound. The freezing-point depression constant for water is \(1.86 \mathrm{mol}\) \(\mathrm{kg}^{-1}\) \(^{\circ} \mathrm{C}\), and for aqueous solutions, molarity and molality can be taken as approximately equal.

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