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A How many geometric isomers of the complex ion \(\left[\mathrm{Cr}(\mathrm{dmen})_{3}\right]^{3+}\) can exist? (dmen is the bidentate ligand 1,1 -dimethylethylenediamine.)

Short Answer

Expert verified
There are 2 geometric isomers (\(\Delta\) and \(\Lambda\)).

Step by step solution

01

Understand the Composition

The complex ion \([\mathrm{Cr}(\mathrm{dmen})_{3}]^{3+}\) contains chromium (Cr) as the central metal ion and three bidentate ligands: dmen (1,1-dimethylethylenediamine). Each dmen ligand coordinates to the chromium through two donor atoms.
02

Recognize the Coordination Geometry

The coordination number for the chromium ion is 6, as each dmen ligand donates two pairs of electrons. This suggests an octahedral geometry for the complex ion.
03

Determine Possible Geometric Isomers

In an octahedral complex with three bidentate ligands, the ligands can arrange such that they form geometric isomers. For this specific structure, we can have \(\Delta\) (right-handed) and \(\Lambda\) (left-handed) optical isomers, leading to two geometric isomers.
04

Double-Check for Other Symmetries

Given the symmetry and identical nature of the three ligands, no additional geometric isomers exist beyond the enantiomers (\(\Delta\) and \(\Lambda\)). Thus, only the two optical isomers are present.

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Key Concepts

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

Octahedral Geometry
In coordination chemistry, octahedral geometry is a common and important shape. It is characterized by six atoms or groups symmetrically positioned around a central atom, resembling an octahedron. This geometry is prevalent among transition metal complexes. The \([\mathrm{Cr}(\mathrm{dmen})_3]^{3+}\) complex showcases this arrangement, as chromium is surrounded by three bidentate ligands, totaling six coordination sites. Each ligand attaches through two donor sites, contributing to the full coordination sphere of six. Octahedral geometry allows for various isomer possibilities, like geometric and optical isomers. Understanding this foundational structure helps in predicting the properties and behavior of metal complexes.
Bidentate Ligands
Bidentate ligands are an interesting component in coordination chemistry. A bidentate ligand has two donor atoms that can simultaneously bind to a central metal ion. In \([\mathrm{Cr}(\mathrm{dmen})_3]^{3+}\), 'dmen' (1,1-dimethylethylenediamine) serves as the bidentate ligand.
  • Each dmen provides two nitrogen atoms to form coordinate bonds with the chromium center.
  • This two-point attachment increases the stability of the complex.
Using bidentate ligands, such complexes display distinct geometrical arrangements due to the fixed angle between the donor atoms. This rigidity plays a crucial role in the formation of geometric and optical isomers. Therefore, understanding bidentate ligands is key for predicting the structure and reactivity of such complexes.
Chromium Complex
The chromium complex \([\mathrm{Cr}(\mathrm{dmen})_3]^{3+}\) is a notable example of an octahedral coordination compound.
  • Chromium acts as the central metal ion.
  • This compound is a trivalent complex with a 3+ charge.
These characteristics affect both the physical and chemical properties of the complex. Crucially, the choice of chromium affects the electronic configuration, impacting things like color and magnetic properties. In an octahedral environment, the specific arrangement of ligands around chromium allows us to explore the concept of isomerism. Therefore, chromium complexes like this serve as critical study subjects in coordination chemistry due to their varied and intricate behavior.
Optical Isomers
Optical isomers, also known as enantiomers, are a type of stereoisomer that are non-superimposable mirror images of each other. In the octahedral \([\mathrm{Cr}(\mathrm{dmen})_3]^{3+}\) complex, optical isomerism occurs.
  • The arrangement of the bidentate dmen ligands leads to two distinct configurations: \(\Delta\) (right-handed) and \(Λ\) (left-handed) forms.
  • These forms rotate plane-polarized light in different directions.
The presence of optical isomers enriches the stereochemistry of a complex. Although both optical isomers have the same connectivity of atoms, their spatial arrangement differs, leading to differences in how they interact with other chiral molecules or systems. This aspect is extremely valuable in fields like pharmaceuticals, where isomer specificity can determine the effectiveness of a drug.

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Most popular questions from this chapter

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: $$\begin{aligned} &\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\\\ &K_{f}=10^{8} \end{aligned}$$ $$\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq}) \quad \quad K_{\mathrm{f}}=10^{18}$$ The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{r} H^{\circ}\) for the NH seaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn},\) and \(\Delta_{r} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}\) -rxn. Is the difference in \(\Delta_{r} H^{\circ}\) sufficient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

Give the oxidation number of the metal ion in each of the following compounds. (a) \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{SO}_{4}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (d) \(\operatorname{Cr}(\text { en })_{2} \mathrm{Cl}_{2}\)

As mentioned on page \(1047,\) transition metal organometallic compounds have found use as catalysts. One example is Wilkinson's catalyst, a rhodium compound \(\left[\mathrm{RhCl}\left(\mathrm{PR}_{3}\right)_{3}\right]\) used in the hydrogenation of alkenes. The steps involved in the catalytic process are outlined below. Indicate whether the rhodium compounds in each step have 18 - or 16 -valence electrons. (See Study Question \(34 .)\) Step \(1 .\) Addition of \(\mathrm{H}_{2}\) to the rhodium center of Wilkinson's catalyst. (For electron-counting purposes \(\mathrm{H}\) is considered a hydride ion, \(\left.\mathrm{H}^{-}, \text {a two-electron donor. }\right)\) Step \(2 .\) Loss of a PR \(_{3}\) ligand (a two-electron donor) to open a coordination site. (PR \(_{3}\) is a phosphine such as \(\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) triphenylphosphine.) Step \(3 .\) Addition of the alkene to the open site. Step 4. Rearrangement to add H to the double bond. (Here the \(-\mathrm{CH}_{2} \mathrm{CH}_{3}\) group is a two-electron donor and can be thought of as a \(\left[\mathrm{CH}_{2} \mathrm{CH}_{3}\right]^{-}\) anion for electron counting purposes.) Step \(5 .\) Loss of the alkane. Step \(6 .\) Regeneration of the catalyst. $$\text { Net reaction: } \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}$$

Give the name or formula for each ion or compound, as appropriate. (a) pentaaquahydroxoiron(III) ion (b) \(\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]\) (c) \(\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (d) ammonium tetrachloroplatinate( (11)

The following equations represent various ways of obtaining transition metals from their compounds. Balance each equation. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s})\) (b) \(\operatorname{Ti} \mathrm{Cl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{MgCl}_{2}(\mathrm{s})\) (c) \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow\) \(\mathrm{Ag}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})\) (d) \(\mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})\)

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