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Consider the following three complexes: \(\left(\right.\) Complex 1) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Cl}\) 2) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{ONO})_{2}\right]\) (Complex (Complex 3) \(\left[\mathrm{V}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+},\) Which of the three complexes can have (a) geometric isomers, (b) linkage isomers, (c) optical isomers, (d) coordinationsphere isomers?

Short Answer

Expert verified
Complex 1 can have geometric isomers (cis and trans) and optical isomers (cis-form). Complex 2 can have geometric isomers (cis and trans) and linkage isomers. Complex 3 can have geometric isomers (cis and trans) and optical isomers (cis-form).

Step by step solution

01

Complex 1: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Cl}\)

Complex 1 is a six-coordinate complex with tetraammine and two bromine ligands. To check for the presence of different types of isomers, we can consider the following: (a) Geometric isomers: Since it's an octahedral complex, it can exhibit geometrical isomerism in the form of cis and trans configurations. In this case, there are cis and trans isomers possible due to the different arrangement of bromine ligands around the cobalt center. (b) Linkage isomers: As there are no ligands with ambidentate bonding in this complex, it cannot form any linkage isomers. (c) Optical isomers: Due to the presence of two bromine ligands, the cis-form of this complex will exhibit optical isomerism (it has a non-superimposable mirror image). (d) Coordination-sphere isomers: This complex cannot have coordination-sphere isomers as all the ligands are within the same coordination sphere.
02

Complex 2: \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{ONO})_{2}\right]\)

In Complex 2, there are four ligands around the center metal: two ammonia and two nitro ligands. (a) Geometric isomers: Given the square-planar geometry of this complex, it can have cis and trans configurations due to the difference in the arrangement of ammonia and nitro ligands around the palladium center. (b) Linkage isomers: The nitro ligands can form linkage isomers when bonded through nitrogen instead of oxygen, forming the nitrito ligand. (c) Optical isomers: Since it is a square-planar complex, it cannot have optical isomers. (d) Coordination-sphere isomers: This complex cannot have coordination-sphere isomers as all the ligands are within the same coordination sphere.
03

Complex 3: \(\left[\mathrm{V}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\)

Complex 3 is an octahedral complex with two ethylenediamine (en) ligands and two chloride ligands. (a) Geometric isomers: This complex shows geometric isomerism, with cis and trans isomers differing in the arrangement of chloride ligands around the vanadium center. (b) Linkage isomers: As there are no ligands with ambidentate bonding in this complex, it cannot form any linkage isomers. (c) Optical isomers: The cis-form of this complex is optically active and will have non-superimposable mirror images, making it an optically active complex. (d) Coordination-sphere isomers: This complex cannot have coordination-sphere isomers as all the ligands are within the same coordination sphere. In summary, the three complexes exhibit the following isomers: Complex 1: (a) Geometric isomers, (c) Optical isomers Complex 2: (a) Geometric isomers, (b) Linkage isomers Complex 3: (a) Geometric isomers, (c) Optical isomers

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Key Concepts

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

Geometric Isomerism
Geometric isomerism is a type of stereoisomerism where compounds with the same formula have different spatial arrangements. This isomerism is common in coordination complexes. Here, the arrangement of ligands around the central atom gives rise to different isomers. For instance, in an octahedral complex, ligands arranged next to each other can form a 'cis' isomer, while those directly opposite each other form a 'trans' isomer. In our exercise:
  • Complex 1, \([\mathrm{Co}(\mathrm{NH}_{3})_{4} \mathrm{Br}_{2}] \mathrm{Cl}\), has a cobalt center with bromine ligands that can be arranged in cis and trans positions, creating geometric isomers.
  • Complex 2, \([\mathrm{Pd}(\mathrm{NH}_{3})_{2}(\mathrm{ONO})_{2}]\), is square-planar, allowing it to exhibit geometric isomerism between cis and trans configurations with ammonia and nitro ligands.
  • Complex 3, \([\mathrm{V}(\mathrm{en})_{2} \mathrm{Cl}_{2}]^{+}\), with two chloride ligands in an octahedral complex, also shows geometric isomerism.
The spatial distribution of ligands in these structures leads to different isomers with distinct properties.
Optical Isomerism
Optical isomerism arises when a complex has a non-superimposable mirror image, very much like the difference between left and right hands. This type of isomerism mainly occurs when there is an asymmetric arrangement in a molecule, often referred to as chirality. In our discussed complexes:
  • Complex 1, with its cis form, presents optical isomerism due to the distinct arrangement of bromine ligands around the cobalt center, allowing it to have non-superimposable mirror images.
  • Complex 3 also shows optical isomerism, particularly in its cis configuration with two ethylenediamine ligands surrounding vanadium, making it chiral.
Optical isomers, also known as enantiomers, can rotate plane-polarized light in opposite directions, a property greatly used in different fields of chemistry.
Linkage Isomerism
Linkage isomerism occurs when a ligand can bind to the central metal atom or ion through different atoms. This type of isomerism is specifically relevant to ligands that have more than one potential bonding atom, termed "ambidentate" ligands. Considering the complexes in focus:
  • Complex 2, \([\mathrm{Pd}(\mathrm{NH}_{3})_{2}(\mathrm{ONO})_{2}]\), can exhibit linkage isomerism through its nitro ligands. These ligands can attach to the palladium center either through the nitrogen atom (nitro) or through the oxygen atom (nitrito), resulting in linkage isomers.
This creates an alternative structure with potentially different chemical properties while retaining the same chemical formula.
Coordination-Sphere Isomerism
Coordination-sphere isomerism involves isomers that differ due to an exchange of ligands between the coordination sphere and the surrounding environment in a compound. This type of isomerism substantially changes the position of ligands relative to the central atom or ion. In the given exercise:
  • None of the named complexes exhibit coordination-sphere isomerism. This happens because all the ligands are contained within the same coordination sphere for each complex.
Coordination-sphere isomerism typically involves more complex structures where ligands might switch places with counter ions, offering different configurations and often different chemical properties as a result.

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Most popular questions from this chapter

The complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains five unpaired electrons. Sketch the energy-level diagram for the \(d\) orbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

Which species are more likely to act as ligands? (a) Positively charged ions or negatively charged ions? (b) Neutral molecules that are polar or those that are nonpolar?

Identify each of the following coordination complexes as either diamagnetic or paramagnetic: (a) \(\left[\operatorname{CoBr}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}\) (b) \(\left[\mathrm{W}(\mathrm{CN})_{6}\right]^{3-}\) (c) \(\left[\mathrm{VF}_{6}\right]^{3-}\) (d) \(\left[\mathrm{Rh}(\mathrm{o}-\mathrm{phen})_{3}\right]^{3+}\)

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin Fe(II) complex; deoxyhemoglobin, without the \(\mathrm{O}_{2}\) molecule, is a high- spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of \(\mathrm{O}_{2}\) in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15-minute exposure to air containing 400 ppm of CO causes about \(10 \%\) of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and \(\mathrm{O}_{2}\) to hemoglobin? (e) CO is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{NO}_{3}\) (b) \(\mathrm{Hg}\left[\mathrm{Co}(\mathrm{SCN})_{4}\right]\) (c) \(\left[\mathrm{Ru}\left(\mathrm{PPh}_{3}\right)_{3} \mathrm{Cl}_{3}\right]\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{CO}_{3}\right]_{2} \mathrm{SO}_{4}\)

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