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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: \(\left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\) bipy What would you predict for themagnitude of the equilibrium constant for this equilibrium? Explain the basis for your answer.

Short Answer

Expert verified
Pyridine is referred to as a monodentate ligand because it can form only one coordination bond with a metal ion, using its lone pair of electrons on the nitrogen atom. For the given equilibrium reaction, the magnitude of the equilibrium constant, \(K\), is expected to be less than 1, as the chelate complex with the bidentate bipy ligand is more stable and the equilibrium will tend to favor the reactant side.

Step by step solution

01

(a) Monodentate ligand

A ligand is a molecule or ion that donates one or more pairs of electrons to a central metal ion, helping it achieve a stable electron configuration. When a ligand is able to bond to the central metal ion through only one atom, it is called a monodentate (single-tooth) ligand. For a ligand to be considered monodentate, it should have only one atom that can participate in the formation of a coordination bond. In the given molecule, pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) consists of a nitrogen atom bonded to a five-carbon aromatic ring. The nitrogen atom has a lone pair of electrons, making it available for forming a coordination bond with a metal ion. Other atoms in the pyridine molecule do not have any lone pairs available for coordination bonds. Therefore, pyridine is referred to as a monodentate ligand because it can form only one coordination bond with a metal ion, using its lone pair of electrons on the nitrogen atom.
02

(b) Equilibrium constant prediction

The equilibrium reaction can be written as: \(\left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+} + \mathrm{bipy}\) We need to predict the magnitude of the equilibrium constant, \(K\), for this reaction. The equilibrium constant is a measure of the extent of the reaction and is dependent on the relative stability of products and reactants. In this case, the reaction involves the substitution of a bidentate ligand (bipy) with two monodentate ligands (py). In general, chelate complexes (formed by bidentate or polydentate ligands) are more stable than those formed by monodentate ligands due to the chelate effect. This means that the initial complex with bipy is likely more stable than the final complex with six pyridines. Based on this understanding, we can predict that the equilibrium will tend to favor the reactant side, as the complex with the bidentate bipy ligand is more stable. Therefore, the magnitude of the equilibrium constant, \(K\), is expected to be less than 1.

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Key Concepts

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

Monodentate Ligand
A monodentate ligand is a simple yet foundational concept in coordination chemistry. It refers to a ligand that can form only one bond with a central metal ion. This happens because the ligand has only one donor atom with a lone pair of electrons that can be donated to a metal ion to form a coordination bond.
Pyridine \((\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N})\) is an example of a monodentate ligand. The nitrogen atom in pyridine possesses a lone pair of electrons that it can use to form a bond with a metal ion, stabilizing the metal complex.
In pyridine, even though it is made up of multiple atoms, only the nitrogen atom acts as the binding site. This reflects the key characteristic of monodentate ligands:
  • Only one pair of electrons available for bonding.
  • Forms a single coordination bond with a metal ion.
Understanding monodentate ligands is essential when examining the properties and behaviors of metal complexes. They often participate in substitution reactions where one ligand type replaces another.
Equilibrium Constant
The equilibrium constant, denoted as \(K\), is a crucial part of understanding chemical reactions. It represents the ratio of product concentrations to reactant concentrations at equilibrium, each raised to their respective stoichiometric coefficients. The value of \(K\) gives insight into the position of equilibrium and the relative stability of reactants and products.
In the context of the equilibrium reaction involving pyridine and the ruthenium complex:
\[\left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+ \mathrm{bipy}\] it involves the substitution of a bidentate ligand (bipy) by monodentate ligands (py).
When predicting \(K\) for such reactions, we consider the relative stabilities:
  • Bidentate (chelating) ligands often form more stable complexes.
  • The chelate effect suggests increased stability attributable to bidentate bonds, meaning they are less likely to be replaced by monodentate ligands.
Thus, due to the stability provided by bidentate ligands, we expect equilibrium to favor the reactant side \([\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})]^{2+}\), indicating that \(K < 1\).
Chelate Effect
The chelate effect is a fascinating principle that explains why complexes formed with chelating ligands (like bidentate ligands) are typically more stable than those with monodentate ligands.
This effect occurs because chelating ligands bind to a metal ion at multiple points, creating a ring-like structure. This gives rise to two main advantages:
  • Increased entropy: Forming a chelating ring generally results in fewer particles on the product side, increasing disorder and thus favorability.
  • Enthalpic stabilization: Multiple bonds with the same ligand enhance the overall stability of the complex.
In the equilibrium reaction with \([\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})]^{2+}\), we observe that replacing a bidentate bipy with two monodentate py ligands decreases stability due to lesser entropic and enthalpic benefits.
Moreover, the chelate effect is often noticeable in the magnitude of the equilibrium constant, where \(K < 1\) suggests strong binding of the chelated complex and reluctance in breaking such a bond in favor of monodentate alternatives. Understanding this principle provides deep insights into reaction mechanisms and the design of stable coordination compounds.

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Most popular questions from this chapter

Give the number of \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\) (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4}\), (e) \([\mathrm{Sr}(\mathrm{EDTA})]^{2-}\)

(a) What is meant by the term chelate effect? (b) What thermodynamic factor is generally responsible for the chelate effect? (c) Why are polydentate ligands often called sequestering agents?

Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the \(3+\) rather than in the \(2+\) oxidation state. Furthermore, for a given ligand the complexes of the bivalent metal ions of the first transition series tend to increase in stability in the order \(\mathrm{Mn}(\mathrm{II})<\mathrm{Fe}(\mathrm{II})<\mathrm{Co}(\mathrm{II})<\) \(\mathrm{Ni}(\mathrm{II})<\mathrm{Cu}(\mathrm{II})\). Explain how these two observations are consistent with one another and also consistent with a crystal-field picture of coordination compounds.

Two different compounds have the formulation \(\mathrm{CoBr}\left(\mathrm{SO}_{4}\right) \cdot 5 \mathrm{NH}_{3} .\) Compound \(\mathrm{A}\) is dark violet, and compound \(\mathrm{B}\) is red-violet. When compound \(\mathrm{A}\) is treated with \(\mathrm{AgNO}_{3}(a q)\), no reaction occurs, whereas compound \(\mathrm{B}\) reacts with \(\mathrm{AgNO}_{3}(a q)\) to form a white precipitate. When compound \(\mathrm{A}\) is treated with \(\mathrm{BaCl}_{2}(a q)\), a white precipitate is formed, whereas compound \(B\) has no reaction with \(\mathrm{BaCl}_{2}(a q)\). (a) Is Co in the same oxidation state in these complexes? (b) Explain the reactivity of compounds \(\mathrm{A}\) and \(\mathrm{B}\) with \(\mathrm{AgNO}_{3}(a q)\) and \(\mathrm{BaCl}_{2}(a q)\). (c) Are compounds \(A\) and \(B\) isomers of one another? If so, which category from Figure \(24.17\) best describes the isomerism observed for these complexes? (d) Would compounds \(A\) and \(B\) be expected to be strong electrolytes, weak electrolytes, or nonelectrolytes?

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

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