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Determine if each of the following complexes exhibits geometric isomerism. If geometric isomers exist, determine how many there are. (a) tetrahedral \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right],(\mathbf{b})\) square-pla- \(\operatorname{nar}\left[\operatorname{IrCl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-},(\mathbf{c})\) octahedral \(\left[\mathrm{Fe}(o \text { -phen })_{2} \mathrm{Cl}_{2}\right]^{+} .\)

Short Answer

Expert verified
For the complexes given: (a) \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right]\) has a tetrahedral geometry and does not exhibit geometric isomerism. (b) \(\left[\operatorname{IrCl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-}\) has a square planar geometry and exhibits geometric isomerism with 2 possible isomers: cis and trans. (c) \(\left[\mathrm{Fe}(o\text{-phen})_{2} \mathrm{Cl}_{2}\right]^{+}\) has an octahedral geometry and exhibits geometric isomerism with 2 possible isomers.

Step by step solution

01

(a) Tetrahedral complex: \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right]\)

For this complex, the geometry around the central metal ion (Cd) is tetrahedral. In a tetrahedral complex, we do not have the possibility for geometric isomerism, as there is no unique axis for swapping ligands. Thus, this complex does not exhibit geometric isomerism.
02

(b) Square planar complex: \(\left[\operatorname{IrCl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-}\)

In this complex, the geometry around the central metal ion (Ir) is square planar. For square planar complexes, we need to check if it is possible to arrange ligands in a different way. In this case, we have two Cl ligands and two PH3 ligands. If we place the two Cl ligands adjacent to each other (cis) and the two PH3 ligands adjacent to each other (cis), we obtain one arrangement. Alternatively, we can place the Cl ligands opposite of each other (trans) with the PH3 ligands also opposite of each other (trans), obtaining another arrangement. Therefore, this complex exhibits geometric isomerism, and there are 2 possible isomers (cis and trans).
03

(c) Octahedral complex: \(\left[\mathrm{Fe}(o\text{-phen})_{2} \mathrm{Cl}_{2}\right]^{+}\)

In this complex, the geometry around the central metal ion (Fe) is octahedral. We have two Cl ligands and two o-phen ligands. To determine if geometric isomers exist, we can examine the axial and equatorial positions. One arrangement could be with both Cl ligands in the axial positions and both o-phen ligands in the equatorial plane. Another arrangement could be with one Cl ligand and one o-phen ligand in the axial positions and the other Cl ligand and o-phen ligand in the equatorial plane. Since we have found two distinct arrangements, this complex exhibits geometric isomerism, and there are 2 possible isomers.

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Key Concepts

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

Tetrahedral Complexes
Tetrahedral complexes are coordination compounds where the central metal ion is surrounded by four ligands at the corners of a tetrahedron. This shape is quite common in coordination chemistry. A crucial aspect to understand is that due to the symmetry of the tetrahedral shape, there is no unique axis that can cause the arrangement of ligands to be different. As a result, tetrahedral complexes generally do not exhibit geometric isomerism. This is because all the positions around the central metal ion are equivalent, meaning any swap of ligands doesn't lead to a distinct isomer.
When dealing with tetrahedral complexes, it's important to remember that no matter how you position different ligands, resulting arrangements will be equivalent. An example from textbook exercises includes the complex ext{f{Cd(H}_2 ext{O})_2 ext{Cl}_2 ext{]}} . This complex does not show geometric isomerism because the coordination environment is non-discriminatory towards different ligand positioning.
Square Planar Complexes
Square planar complexes are a type of coordination compounds where four ligands are arranged around a central metal ion in the form of a square plane. These coordination complexes are notable because they can exhibit geometric isomerism. In square planar geometry, the positions of the ligands can be changed, leading to different spatial arrangements known as isomers.
A typical form of isomerism in square planar complexes is the cis-trans isomerism.
  • **Cis Isomers**: If identical ligands are next to each other, they form a cis isomer. This placement leads to one isomer where similar ligands are on the same side of the plane.
  • **Trans Isomers**: If identical ligands are placed opposite each other, they form a trans isomer. Here, ligands of the same type lie across from each other in the square plane.
For example, the complex ext{f{IrCl}_2 ext{(PH}_3 ext{)}_2 ext{]}} from the exercises has two possible isomers: one where both Cl ligands are adjacent (cis) and one where they are opposite (trans). This diversity in ligand arrangement is what makes square planar complexes capable of geometric isomerism.
Octahedral Complexes
Octahedral complexes are characterized by having six ligands symmetrically arranged around a central metal ion at the vertices of an octahedron. This geometry is prevalent in coordination compounds with transition metals. Unlike tetrahedral complexes, octahedral complexes do have the potential to exhibit geometric isomerism.
In octahedral complexes, ligands can be distributed in the axial (longitudinal) and equatorial (planar) positions, leading to varying configurations. Two primary geometric isomers are possible depending on how ligands are distributed:
  • **Cis Isomers**: Identical ligands are adjacent to each other in the octahedral structure.
  • **Trans Isomers**: Identical ligands are placed directly opposite each other.
An example here is the complex ext{f{Fe(o-phen})_2 ext{Cl}_2 ext{]}} from the exercises. This compound allows for different isomeric possibilities: for instance, the chlorine atoms could either both be axial or one axial and one equatorial, leading to two distinct geometric isomers. This ability to have variable arrangements makes octahedral complexes quite interesting in terms of geometric isomerism.

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Most popular questions from this chapter

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) \(\left[\mathrm{Rh}(\mathrm{o}-\mathrm{phen})_{3}\right]^{3+}\)

Indicate the likely coordination number of the metal in each of the following complexes: (a) \(\left[\mathrm{Ru}(\text { bipy })_{3}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\operatorname{Re}(\text { o-phen })_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\) (c) \(\mathrm{Pd}(\mathrm{PPh} 3)_{3} \mathrm{Cl}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mn}(\mathrm{EDTA})\)

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit d-d transitions? Explain. (c) As mentioned in "A Closer Look" on charge-transfer color, the violet color of \(\mathrm{MnO}_{4}\) is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\) occurs at a wavelength of \(565 \mathrm{nm}\). The \(\mathrm{CrO}_{4}^{2-}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for \(\mathrm{MnO}_{4}^{-}\) ? Explain. (e) The \(\mathrm{VO}_{4}^{3-}\) ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

A four-coordinate complex \(\mathrm{MA}_{2} \mathrm{~B}_{2}\) is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?

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