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Chapter 24: Problem 5

Which of the following complexes are chiral? Explain. [Section 24.4]

Short Answer

Expert verified
The chiral complexes among the given options are \(\text{[Co(en)}_{3}]\), \(\text{[CoCl(NH}_{3})_{5}]\), and \(\text{[CoBr(NH}_{3})_{5}]\). This is because they have unique 3D arrangements that create non-superimposable mirror images. The complex \(\text{[Co(NH}_{3})_{6}]\) is not chiral, as all its ligands are identical and cannot form any non-superimposable mirror images.

Step by step solution

01

Identify the complexes

The given complexes are: 1. \(\text{[Co(en)}_{3}]\) 2. \(\text{[Co(NH}_{3})_{6}]\) 3. \(\text{[CoCl(NH}_{3})_{5}]\) 4. \(\text{[CoBr(NH}_{3})_{5}]\)
02

Analyze each complex and determine chirality

1. \(\text{[Co(en)}_{3}]\): The complex has three bidentate en (ethylenediamine) ligands. Since there are three bidentate ligands, they can create different 3D arrangements around the central Co atom. The complex will have a non-superimposable mirror image, so it is chiral. 2. \(\text{[Co(NH}_{3})_{6}]\): This complex has six monodentate NH3 ligands. The ligands are all identical and arranged octahedrally around the central Co atom. As all the ligands are the same, there can't be any non-superimposable mirror images. Therefore, this complex is not chiral. 3. \(\text{[CoCl(NH}_{3})_{5}]\): This complex has one Cl and five NH3 ligands. The five NH3 ligands are indistinguishable, but the Cl ligand is different. In the octahedral arrangement of this complex, there will be a non-superimposable mirror image. Thus, this complex is chiral. 4. \(\text{[CoBr(NH}_{3})_{5}]\): Similar to the previous complex, this complex has one Br and five NH3 ligands. As with the [CoCl(NH3)5] complex, this complex will have a non-superimposable mirror image, so it is chiral.
03

Summarize the results

Analyzing the given complexes, we find that the following complexes are chiral: 1. \(\text{[Co(en)}_{3}]\) 2. \(\text{[CoCl(NH}_{3})_{5}]\) 3. \(\text{[CoBr(NH}_{3})_{5}]\) The complex \(\text{[Co(NH}_{3})_{6}]\) is not chiral.

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Key Concepts

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

Chirality in Coordination Compounds
Chirality is a fascinating concept commonly encountered in coordination chemistry. A chiral molecule is one that is not superimposable on its mirror image. Imagine your left and right hands; they are mirror images but cannot perfectly align when superimposed.

In coordination compounds, chirality arises when the spatial arrangement of the ligands around the central metal atom leads to non-superimposable mirror images. This often occurs in complexes with an arrangement like octahedral, where different orientations of ligands can create "left-handed" or "right-handed" forms, known as enantiomers.

To determine if a coordination compound is chiral, consider the following:
  • Check if the central metal is symmetrically surrounded by identical ligands. If they are all the same, as in \[\text{[Co(NH}_{3})_{6}]\], it's symmetric, hence achiral.
  • See whether the ligands can create non-superimposable arrangements due to different positions or identities, like in \[\text{[CoCl(NH}_{3})_{5}]\].
  • Keep in mind the presence of bidentate ligands, which can significantly influence chirality due to their ability to "clasp" the central atom forming specific 3D orientations.
Bidentate Ligands
Bidentate ligands are a specific type of complexing agent in coordination chemistry that can clasp onto a central metal ion with two donor atoms. This kind of attachment creates more stable and often more interesting coordination compounds.

Consider ethylenediamine (en), a classic example of a bidentate ligand used in the complex \[\text{[Co(en)}_{3}]\]. With three en ligands coordinating to a cobalt center, it showcases an elaborate structure where each en ligand contributes to bonding through its two nitrogen atoms.

The involvement of bidentate ligands can enhance the potential for chirality because they create specific orientations around the metal center. Since these ligands occupy more "space" compared to monodentate ligands, they can lead to unique geometric configurations that result in non-superimposable mirror images.

When studying coordination compounds, remember:
  • Bidentate ligands bind more strongly due to the "chelate effect," stabilizing the compound.
  • They influence the spatial arrangement of the ligand, potentially causing chirality if three-dimensional asymmetry is introduced.
  • These ligands are excellent at promoting ligand exchange reactions, further diversifying compound properties.
Octahedral Complexes
In coordination chemistry, octahedral complexes represent one of the most common structural types. These complexes have six ligands symmetrically arranged around a central metal ion, forming an octahedral geometry.

This arrangement is highly versatile and can encompass both simple symmetrical structures and more complex asymmetrical configurations.

Octahedral complexes, like \[\text{[Co(NH}_{3})_{6}]\], show their beauty and simplicity, being symmetric and hence achiral. However, introducing different ligands or combinations of bidentate ligands, such as in \[\text{[Co(en)}_{3}]\], can create fascinating asymmetrical structures that are chiral.

Key points about octahedral complexes include:
  • They can accommodate a variety of ligands, enabling a range of chemical behaviors.
  • Their geometry can lead to different types of isomerism, including geometric (cis/trans) and optical isomerism, which is directly related to chirality.
  • Understanding these complexes is crucial for exploring coordination chemistry's depth, as they are found throughout transition metals.

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Most popular questions from this chapter

(a) Sketch a diagram that shows the definition of the crystal-field splitting energy \((\Delta)\) for an octahedral crystal field. (b) What is the relationship between the magnitude of \(\Delta\) and the energy of the \(d-d\) transition for a \(d^{1}\) complex? (c) Calculate \(\Delta\) in \(\mathrm{kJ} / \mathrm{mol}\) if a \(d^{1}\) complex has an absorption maximum at \(590 \mathrm{~nm}\).

For each of the following metals, write the electronic configuration of the atom and its \(3+\) ion: (a) Ru, (b) Mo, (c) Co. Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(3+\) ion, assuming a weak-field complex. How many unpaired electrons are there in each case?

Sketch all the possible stereoisomers of (a) tetrahedral \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right]\), (b) square-planar [IrCl \(\left._{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-}\), (c) octa- hedral \(\left[\mathrm{Fe}(o \text { -phen })_{2} \mathrm{Cl}_{2}\right]^{+}\)

Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex \(\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}\), on treatment with water over a long time, converts to a brown-orange complex. Reaction of \(\mathrm{AgNO}_{3}\) with a solution of the product precipitates \(3 \mathrm{~mol}\) of \(\mathrm{AgCl}\) per mole of Cr present. (Write two chemical equations.) (c) When an \(\mathrm{NaOH}\) solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{a}\) precipitate forms. Addition of excess \(\mathrm{NaOH}\) solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) turns deep blue on addition of concentrated hydrochloric acid.

The complex \(\left[\mathrm{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\) undergoes substitution reactions with several ligands, replacing the water molecule with the ligand. \(\left[\mathrm{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}+\mathrm{L} \longrightarrow[\mathrm{Ru}(\mathrm{EDTA}) \mathrm{L}]^{-}+\mathrm{H}_{2} \mathrm{O}\) The rate constants for several ligands are as follows: $$ \begin{array}{ll} \hline \text { Ligand, } \mathrm{L} & k\left(M^{-1} s^{-1}\right) \\ \hline \text { Pyridine } & 6.3 \times 10^{3} \\ \text { SCN }^{-} & 2.7 \times 10^{2} \\ \mathrm{CH}_{3} \mathrm{CN} & 3.0 \times 10 \\ \hline \end{array} $$ (a) One possible mechanism for this substitution reaction is that the water molecule dissociates from the complex in the rate-determining step, and then the ligand \(\mathrm{L}\) fills the void in a rapid second step. A second possible mechanism is that \(L\) approaches the complex, begins to form a new bond to the metal, and displaces the water molecule, all in a single concerted step. Which of these two mechanisms is more consistent with the data? Explain. (b) What do the results suggest about the relative basicities of the three ligands toward Ru(III)? (c) Assuming that the complexes are all low spin, how many unpaired electrons are in each?
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