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Chapter 24: Problem 67

When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was because of the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we won't worry about that for now.)

Short Answer

Expert verified
A chiral cobalt(III) complex that contains no carbon atoms can be constructed using the following ligand set: chlorine (Cl-), ammonia (NH3), water (H2O), and a chiral anionic bidentate ligand derived from NO-OH, called nitrito-N,ONO (charge of -1). The complex can be represented as: \[ \text{[Co(Cl)(NH}_3\text{)(H}_2\text{O)(NO-OH)]^2- \] This complex exhibits chirality due to the tetrahedral arrangement of the four different ligands surrounding the cobalt(III) center and does not contain any carbon atoms.

Step by step solution

01

(Step 1: Determine the coordination number of cobalt(III))

Cobalt(III) typically forms octahedral complexes, which means it has a coordination number of 6. This means that cobalt(III) is surrounded by six ligand groups.
02

(Step 2: Design a suitable ligand set)

To ensure the complex is chiral and does not contain any carbon atoms, we will need to select ligands that satisfy these conditions. Let's consider chlorine (Cl-), ammonia (NH3), and water (H2O) as potential ligands for our complex, as they do not contain carbon atoms.
03

(Step 3: Apply the rules for chirality)

For our cobalt(III) complex to be chiral, it must have non-superimposable mirror images, or enantiomers, which can be achieved by having the central cobalt atom surrounded by four different ligands in an octahedral geometry. In this case, we can adopt the following arrangement for our ligand set: three different ligands (Cl-, NH3, and H2O) and an additional ligand that is a chiral ligand. For the chiral ligand, we can use an anionic bidentate ligand derived from NO-OH, called nitrito-N,ONO (charge of -1).
04

(Step 4: Construct the chiral cobalt(III) complex)

Adopting the ligand set from step 3, we can now construct the chiral cobalt (III) complex: \[ \text{[Co(Cl)(NH}_3\text{)(H}_2\text{O)(NO-OH)]^2- \] This cobalt(III) complex will exist in two enantiomeric forms due to the tetrahedral arrangement of the four different ligands surrounding the cobalt(III) center, thus making it chiral. Furthermore, the complex does not contain any carbon atoms as required in the problem statement.

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Most popular questions from this chapter

Indicate the coordination number of the metal and the oxidation number of the metal in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CdBr}_{4}\right]\) (c) \(\left[\mathrm{Pt}(\mathrm{en})_{3}\right]\left(\mathrm{ClO}_{4}\right)_{4}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right]^{+}\) (e) \(\mathrm{NH}_{4}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{NCS})_{4}\right]\) (f) \(\left[\mathrm{Cu}(\mathrm{bipy})_{2} \mathrm{I}\right] \mathrm{I}\)

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?

(a) Draw the structure for \(\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\). (b) What is the coordination number for platinum in this complex, and what is the coordination geometry? (c) What is the oxidation state of the platinum? [Section 24.1]

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\), abbreviated py, is the following molecule: (a) Why is pyridine referred to as a monodentate ligand? (b) Consider the following equilibrium reaction: \(\left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\mathrm{bipy}\) What would you predict for the magnitude of the equilibrium constant for this equilibrium? Explain the basis for your answer.

A manganese complex formed from a solution containing potassium bromide and oxalate ion is purified and analyzed. It contains \(10.0 \% \mathrm{Mn}, 28.6 \%\) potassium, \(8.8 \%\) carbon, and \(29.2 \%\) bromine by mass. The remainder of the compound is oxygen. An aqueous solution of the complex has about the same electrical conductivity as an equimolar solution of \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\). Write the formula of the compound, using brackets to denote the manganese and its coordination sphere.
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