Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The complex ion \(\mathrm{Ru}(\text { phen })_{3}{ }^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) ? b. \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+}\) ). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.

Short Answer

Expert verified
a. The type of isomerism found in \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) is geometrical isomerism, specifically facial (fac) and meridional (mer) isomerism. b. The crystal field diagram for the d orbitals in the \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) complex ion is as follows: ─── eg ─── ↑↑ dx2-y2 ↑↑ d(z2) _________ ─── t2g ─── ↑↑ d_xy ↑↑ d_xz ↑↑ d_yz

Step by step solution

01

Understand Isomerism Types

Isomerism refers to the existence of two or more chemical species with the same molecular formula but different arrangements of atoms or groups within their respective structures. Common types of isomerism in coordination chemistry are: 1. Structural isomerism: When isomers have different connections or bonds between atoms (e.g., linkage isomerism, coordination isomerism, and ionization isomerism). 2. Stereoisomerism: When isomers have different spatial arrangements of atoms or groups (e.g., geometrical isomerism and optical isomerism).
02

Analyze the Complex Ion

The given complex ion has a central metal atom - Ruthenium (Ru) and three bidentate ligands - phenanthroline (phen). The metal ion \(\mathrm{Ru^{2+}}\) is surrounded by three bidentate ligands, each contributing two electron pairs, making the coordination number 6.
03

Determine the Type of Isomerism

Since each phenanthroline ligand has two bonding sites, the complex cannot change its connections or bonds without rearranging the geometry. Thus, no structural isomerism is observed. However, it can form stereoisomers, specifically geometrical isomers. Geometrical isomers in a complex ion with a coordination number of 6 can have two forms or geometries - octahedral and square planar. The \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) complex has an octahedral geometry due to its 3 bidentate ligands. In the case of octahedral complexes, typically those formed by tris-chelated complexes, the isomerism is called facial (fac) and meridional (mer) isomerism. a. The type of isomerism found in \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) is geometrical isomerism, specifically facial (fac) and meridional (mer) isomerism.
04

Draw Crystal Field Diagram for d Orbitals in the Complex

In octahedral complexes, the crystal field splits the d orbitals into two sets with different energies: three with lower energy (t2g orbitals: dxy, dxz, dyz) and two with higher energy (eg orbitals: dx2-y2, d(z2)). To draw the crystal field diagram, we should remember that \(\mathrm{Ru}^{2+}\) has a [Kr]4d6 electron configuration. b. Since the complex is diamagnetic, it means all d orbitals are fully filled, or in other words, all electrons are paired. Below is the crystal field diagram for the d orbitals in this complex ion: ─── eg ─── ↑↑ dx2-y2 ↑↑ d(z2) _________ ─── t2g ─── ↑↑ d_xy ↑↑ d_xz ↑↑ d_yz In conclusion, the given complex ion \(\mathrm{Ru}(\mathrm{phen})_{3}{ }^{2+}\) exhibits geometrical isomerism, specifically fac and mer isomers due to its octahedral geometry. The crystal field diagram shows that it is diamagnetic, with all d orbital electrons being paired.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Octahedral Complexes
In coordination chemistry, octahedral complexes are one of the most common geometric arrangements. This spatial configuration occurs when a metal ion is surrounded by six ligands, forming an octahedral shape. The complex \(\mathrm{Ru} (\text{phen})_3^{2+}\) is a perfect example of an octahedral complex. Here, each phenanthroline ligand, which is bidentate, coordinates with the ruthenium metal ion, creating this six-coordinate structure.
Octahedral complexes can exhibit isomerism, especially geometrical isomerism. Geometrically, these complexes may present two types of isomers: **facial (fac)** and **meridional (mer)**. In **facial isomerism**, three identical ligands form a triangular face. In **meridional isomerism**, three ligands form a plane that includes the metal center. Because of these geometrical arrangements, octahedral complexes like \(\mathrm{Ru}(\mathrm{phen})_3^{2+}\) can have distinct isomeric forms.
Octahedral complexes play an important role in diverse fields, such as bioinorganic chemistry and materials science, making their study essential for understanding complex chemical behaviors.
Crystal Field Theory
Crystal Field Theory (CFT) is an essential concept in understanding the electronic structure of transition metal complexes. It provides a model to explain how ligands affect the energies of the d orbitals of the central metal ion. When ligands approach a metal ion, they interact with its d orbitals, splitting them into different energy levels.
In **octahedral complexes**, like \(\mathrm{Ru}(\mathrm{phen})_3^{2+}\), the incoming ligands create a field that splits the five d orbitals into two groups. The **lower energy group** consists of the \(\text{t}_{2g}\) orbitals (d_xy, d_xz, d_yz), while the **higher energy group** comprises the \(\text{e}_g\) orbitals (d_{x^2-y^2}, d_{z^2}).
A crucial point in CFT is the **electron configuration** in these orbitals. In \(\mathrm{Ru}(\mathrm{phen})_3^{2+}\), which is known to be diamagnetic, all electrons in d orbitals are paired, resulting in no unpaired electrons. This specific pairing is responsible for its magnetic properties. Understanding CFT helps in predicting and explaining the color, magnetism, and reactivity of metal complexes.
Bidentate Ligands
Bidentate ligands are ligands that can form two bonds with a central metal ion. Each bidentate ligand possesses two donor atoms capable of donating a pair of electrons each to bind to the metal ion. In the case of \(\mathrm{Ru}(\text{phen})_3^{2+}\), phenanthroline is the bidentate ligand.
These ligands are important because they create more stable complexes by forming ring structures, known as "chelate rings," with the metal ion. This formation is often referred to as the "chelate effect," which increases the stability of a complex compared to those formed with monodentate ligands.
In applications such as catalysis or therapeutic treatments, bidentate ligands are favored due to the stability they impart to metal complexes. Their ability to stably occupy coordination sites affects the physicochemical properties of the complexes they form, making them especially significant in designing compounds with specific functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation-reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: \(\begin{aligned} \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q) &+\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\ & \longrightarrow \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned}\) Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} .\) Is this consistent with the reaction proceeding through formation of \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\) as an intermediate? Explain.

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosulfate. The equation for the reaction is: \(\begin{aligned} \mathrm{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q) & \longrightarrow \\\& \quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q) \end{aligned}\) c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}(a q)$$

What is the electron configuration for the transition metal ion in each of the following compounds? a. \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) b. \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}\) c. \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}\) d. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\)

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When \(0.105 \mathrm{~g}\) of compound \(\mathrm{A}\) was strongly heated in \(\mathrm{ex}\) cess \(\mathrm{O}_{2}, 0.0203 \mathrm{~g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took \(32.93 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{~g}\) compound \(\mathrm{A}\). iii. Compound A was found to contain \(73.53 \%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when \(0.601 \mathrm{~g}\) compound \(\mathrm{A}\) was dissolved in \(10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(\left.1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be sixcoordinate, with \(\mathrm{NH}_{3}\) and possibly \(\mathrm{I}^{-}\) as ligands. The \(\mathrm{I}^{-}\) ions will be the counterions if needed.)

When \(6 M\) ammonia is added gradually to aqueous copper(II) nitrate, a white precipitate forms. The precipitate dissolves as more \(6 M\) ammonia is added. Write balanced equations to explain these observations. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\left.\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+} .\right]\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free