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\(\begin{array}{lllll}\text { Sketch all the possible } & \text { stereoisomers } & \text { of }\end{array}\) (a) \(\left[\mathrm{Rh}(\text { bipy })(o \text { -phen })_{2}\right]^{3+}\), (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\text { bipy }) \mathrm{Br}\right]^{2+},\) (c) square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right]\).

Short Answer

Expert verified
The possible stereoisomers for the given complexes are: (a) \(\left[\mathrm{Rh}(\text { bipy })(o \text { -phen })_{2}\right]^{3+}\): 1. cis - both o-phen ligands are adjacent to each other, and bipy is opposite to one of them. 2. trans - both o-phen ligands are opposite to each other. (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\text { bipy })\mathrm{Br}\right]^{2+}\): 1. fac - the three ammine ligands are in a facial arrangement. 2. mer - the three ammine ligands are in a meridional arrangement. (c) square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right]\): 1. cis - both cyanide ligands are adjacent to each other. 2. trans - both cyanide ligands are opposite to each other.

Step by step solution

01

(a) Identify the ligands and geometry for \(\left[\mathrm{Rh}(\text { bipy })(o \text { -phen })_{2}\right]^{3+}\)

To start, we must identify the ligands and geometry. Bipyridine (bipy) is a bidentate ligand, and ortho-phenanthroline (o-phen) is also a bidentate ligand. Thus, the complex has two bidentate ligands: bipy and two o-phen molecules. The Rhodium center is octahedral with a coordination number of 6.
02

(a) Determine the possible isomeric forms for \(\left[\mathrm{Rh}(\text { bipy })(o \text { -phen })_{2}\right]^{3+}\)

In this complex, all bidentate ligands form a ring. Rings are able to exhibit atropisomerism, a type of stereoisomerism resulting from restricted rotation around a single bond. However, in this case, the presence of the two different ligands on Rhodium distinguishes the stereoisomers. We can recognize that the isomers formed will be cis- and trans-isomers. The complex will have two stereoisomers: 1. cis - both o-phen ligands are adjacent to each other, and bipy is opposite to one of them. 2. trans - both o-phen ligands are opposite to each other.
03

(b) Identify the ligands and geometry for \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\text { bipy })\mathrm{Br}\right]^{2+}\)

The complex contains three ammine (NH3), one bipyridine (bipy), and one bromide (Br) ligands. The cobalt center is octahedral with a coordination number of 6.
04

(b) Determine the possible isomeric forms for \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\text { bipy })\mathrm{Br}\right]^{2+}\)

In this case, due to the presence of the different ligands in the complex, we can identify facial (fac) and meridional (mer) isomers: 1. fac - the three ammine ligands are in a facial arrangement. 2. mer - the three ammine ligands are in a meridional arrangement.
05

(c) Identify the ligands and geometry for square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right]\)

The complex contains one ethylenediamine (en) and two cyanide (CN) ligands. The geometry is given as square-planar, meaning the coordination number of the palladium center is 4.
06

(c) Determine the possible isomeric forms for square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right]\)

In this case, we can identify cis- and trans-isomers due to the presence of the two different ligands: 1. cis - both cyanide ligands are adjacent to each other. 2. trans - both cyanide ligands are opposite to each other.

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Key Concepts

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

Isomerism
Isomerism is a fascinating concept in chemistry where compounds have the same chemical formula but different structures or arrangements of atoms, leading to distinct properties. In coordination chemistry, isomers can be broadly classified into structural and stereoisomers. Structural isomers differ in the bonds or connectivity of atoms, while stereoisomers have the same bonds but differ in the spatial arrangement of atoms.
  • **Geometric Isomers:** These are a type of stereoisomerism, where ligands are arranged differently around the central atom. A classic example is the cis-trans isomerism found in octahedral and square-planar complexes.
  • **Optical Isomers:** These isomers are non-superimposable mirror images of each other, much like left and right hands. They are significant in compounds with chiral centers.
Understanding these types of isomerism helps in predicting the reactivity and biological activity of compounds. Some isomers might have the same chemical reactivity yet vastly different biological impacts.
Coordination Compounds
Coordination compounds are complex structures where a central metal atom or ion is bonded to a set of molecules or ions, called ligands. These ligands donate a pair of electrons to the metal, forming a coordination bond. The chemistry of these compounds primarily involves transition metals and various ligands, which can be simple ions or complex organic molecules.

Key Features of Coordination Compounds:

  • **Coordination Number:** This indicates the number of ligand atoms directly bonded to the central atom. Common coordination numbers range from 2 to 8.
  • **Geometry:** Depending on the coordination number and the ligands, the geometry of these compounds can vary. Common shapes include octahedral, tetrahedral, and square-planar geometries.
  • **Ligands:** These can be neutral or charged species and can act as either monodentate (binding through a single site) or polydentate (binding through multiple sites) entities.
Coordination chemistry is vital in biochemical processes, industrial catalysis, and material science due to the versatile nature of these compounds.
Stereoisomers
Stereoisomers are a subgroup of isomers that have the same chemical and structural composition but differ in the three-dimensional orientation of their atoms. This subtle difference can significantly influence the chemical and physical properties of a compound.

Types of Stereoisomers in Coordination Compounds:

  • **Cis-Trans Isomerism:** Found in coordination compounds, especially with octahedral and square-planar geometries. For instance, in the square-planar ext{[Pd(en)(CN)}_{2} ext{]}, the difference between cis and trans arrangements of CN ligands results in distinct properties.
  • **Facial-Meridional Isomerism:** In octahedral complexes such as ext{[Co(NH}_3)_3( ext{bipy})Br]^2+, the ligands can occupy adjacent or non-adjacent positions, leading to fac or mer isomers respectively.
Recognizing stereoisomerism in coordination chemistry is crucial for understanding how ligands influence reaction mechanisms and the resulting properties. This knowledge is applicable in designing new drugs and catalytic processes.

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Most popular questions from this chapter

Sketch a plot of atomic radius versus number of valence \(d\) electrons for the period 5 transition metals, and explain the trend.

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 AgCl per mole of Cr present. (Write two chemical equations.) (c) When an NaOH solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},\) 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.

Sketch the structure of the complex in each of the following compounds and give the full compound name: (a) cis- \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{Ru}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{5}\right]\) (c) trans- \(\mathrm{NH}_{4}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (d) cis- \(\left[\mathrm{Ru}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\)

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{array}{ll} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

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

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