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Chapter 23: Problem 42

Consider an octahedral complex \(\mathrm{MA}_{3} \mathrm{B}_{3} .\) How many geo- metric isomers are expected for this compound? Will any of the isomers be optically active? If so, which ones?

Short Answer

Expert verified
There are two possible geometric isomers for the given \(\mathrm{MA}_{3}\mathrm{B}_{3}\) octahedral complex. Geometric isomer 1 is not optically active, while geometric isomer 2 is optically active, as it has an enantiomeric/non-superimposable mirror image.

Step by step solution

01

Determine the possible arrangements of ligands around the central atom

First, let's draw an octahedron with positions for ligands numbered 1 to 6. We have three A ligands and three B ligands to place around the central metal atom. We will start with A ligands and consider symmetry to avoid duplicating isomers. One possible arrangement is placing three A ligands in a plane, with the B ligands filling the other three positions. For example, we can place A ligands in positions 1, 3, and 4, and B ligands in positions 2, 5, and 6. We'll call this geometric isomer 1. The second possible arrangement is placing the A ligands at every other position, such as 1, 3, and 5. With B ligands in the other three positions, at 2, 4, and 6. We'll call this geometric isomer 2. There are no other unique arrangements possible for the \(\mathrm{MA}_{3}\mathrm{B}_{3}\) complex, so there are two possible geometric isomers.
02

Determine if any of the isomers are optically active

To investigate optical activity, we'll look for enantiomers (non-superimposable mirror images) for each geometric isomer. Geometric isomer 1 (three A ligands in a plane and three B ligands in a plane) has a mirror plane of symmetry containing M and dividing the A and B ligands. This means that any mirror image of the complex would still be superimposable on the original complex. Thus, geometric isomer 1 is not optically active. Geometric isomer 2 (three pairs of A and B ligands in alternate positions) does not have any symmetry planes. If we draw a mirror image of this isomer, we get an enantiomeric/non-superimposable complex. Thus, geometric isomer 2 has an enantiomer and is optically active. #Summary# There are two possible geometric isomers for the given \(\mathrm{MA}_{3}\mathrm{B}_{3}\) octahedral complex. Geometric isomer 1 is not optically active, while geometric isomer 2 is optically active, as it has an enantiomeric/non-superimposable mirror image.

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Key Concepts

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

Octahedral Complex
In the context of coordination chemistry, an octahedral complex is a compound where a central metal atom is surrounded by six ligands in a geometric arrangement that resembles an octahedron. Each ligand is connected to the metal by a coordinate bond, and they are spaced around the metal atom at 90-degree angles to one another along the Cartesian axes.

Understanding the structure of an octahedral complex is crucial, as the spatial arrangement of ligands can affect the complex's chemical properties, including reactivity and optical activity. In our exercise example, the octahedral complex is described by the formula \(\mathrm{MA}_{3} \mathrm{B}_{3}\), indicating that the central metal atom, M, is surrounded by three A ligands and three B ligands. The question is how these ligands can be arranged and what properties, such as geometric or optical isomerism, may result from different arrangements.
Geometric Isomers
Geometric isomers, also known as cis-trans isomers, are compounds with the same formula but different spatial arrangements of atoms. This type of isomerism is particularly pertinent to coordination complexes like the octahedral complex, where the arrangement of different ligands around the central metal can lead to isomers with distinctive properties.

For the \(\mathrm{MA}_{3} \mathrm{B}_{3}\) complex, we consider two different arrangements of the ligands A and B. If the three A ligands are adjacent to one another, forming a plane, and the B ligands occupy the opposite positions, we have one geometric isomer. Alternatively, if A and B alternate around the central atom, we have another distinct geometric isomer. Geometric isomers can result in vastly different chemical and physical behaviors, including their responses to polarized light, which is pivotal in the study of optical activity.
Enantiomers
Enantiomers are a type of stereoisomers that are mirror images of each other but cannot be superimposed. This non-superimposability makes enantiomers chiral, meaning they have handedness, much like left and right hands. Enantiomers often arise in coordination chemistry, where the spatial arrangement of ligands around the central metal can generate mirror-image forms.

In the \(\mathrm{MA}_{3} \mathrm{B}_{3}\) example, the second geometric isomer, with alternating A and B ligands, does not possess a plane of symmetry and therefore can exist as two enantiomers, which are optically active. Optical activity is detected when such compounds rotate the plane of polarized light, a characteristic exploited for applications ranging from pharmaceuticals to the food industry. In contrast, a compound like the first isomer, with a plane of symmetry, will not exhibit optical activity as its mirror image is superimposable on itself.

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Most popular questions from this chapter

Determine if each of the following complexes exhibits geometric isomerism. If geometric isomers exist, determine how many there are. (a) [ Rh(bipy) \((o-\) phen \()_{2} ]^{3+},\) \((\mathbf{b})\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\mathrm{bipy}) \mathrm{Br}\right]^{2+},(\mathbf{c})\) square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right].\)

Determine if each of the following metal complexes is chiral and therefore has an optical isomer: (a) square planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right],(\mathbf{b})\) octahedral \(\left[\mathrm{Ni}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},(\mathbf{c})\) octahe- dral \(\operatorname{cis}-\left[\mathrm{V}(\mathrm{en})_{2} \mathrm{ClBr}\right]\)

The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a 0.100-L sample of the water with a solution of EDTA \(^{4-} .\) The EDTA \(^{4-}\) chelatesthe two cations: $$\begin{array}{c}{\mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-}} \\\ {\mathrm{Ca}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-}}\end{array}$$ It requires 31.5 \(\mathrm{mL}\) of 0.0104 \(\mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second 0.100-L sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with 18.7 \(\mathrm{mL}\) of 0.0104 \(M[\mathrm{EDTA}]^{4-} .\) Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in \(\mathrm{mg} / \mathrm{L} .\)

The \(E^{\circ}\) values for two low-spin iron complexes in acidic solution are as follows: \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \Longrightarrow\) \(\quad\quad\quad\quad\quad\quad\quad\) \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+}(a q) \quad E^{\circ}=1.12 \mathrm{V}\) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) \quad E^{\circ}=0.36 \mathrm{V}\) (a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}\) or \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\) is more difficult to reduce? (c) Suggest an explanation for your answer to (b).

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 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 MnO \(_{4}^{-}?\) Explain. (e) The 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.
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