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Chapter 20: Problem 80

The complex ion \(\mathrm{Ru}\) (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 \(\mathrm{Ru}(\text { phen })_{3}^{2+} ?\) b. \(\mathrm{Ru}(\text { phen })_{3}^{2+}\) 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 Ru(phen)₃²⁺ is linkage isomerism. b. The crystal field diagram for the d orbitals in Ru(phen)₃²⁺: t₂g: ↑↓ , ↑↓ , ↑↓ e_g: , This low-spin configuration corresponds to the diamagnetic nature of the complex ion, with all d electrons of Ru²⁺ being paired and the e_g orbitals remaining empty.

Step by step solution

01

Identify the type of isomerism in Ru(phen)3²⁺

The complex ion Ru(phen)3²⁺ has ruthenium (Ru) as the central metal ion, and phenanthroline (phen) as a bidentate ligand. Bidentate ligands can form coordination bonds through two sites. When multiple bidentate ligands are attached to a central metal ion, they give rise to isomerism called linkage isomerism. In this case, Ru(phen)3²⁺ has three phen ligands, thus resulting in linkage isomerism.
02

Determine the oxidation state and electron configuration of the central metal ion

In Ru(phen)3²⁺, the overall oxidation state of the complex is 2+ while the oxidation state of the phen ligands is neutral. The central metal ion Ru must have an oxidation state of 2+, i.e., Ru²⁺ to maintain the overall charge balance. The ground-state electron configuration of Ru is [Kr] 4d⁷ 5s¹. When it loses two electrons, the electron configuration of Ru²⁺ will be [Kr] 4d⁵.
03

Assign the complex ion as high-spin or low-spin

Knowing that the complex ion is diamagnetic allows us to conclude that all the d electrons of Ru²⁺ (which has 5 d electrons) must be paired. This implies that the complex has a low-spin configuration.
04

Determine the shape of the complex and the crystal field splitting diagram

Since Ru(phen)3²⁺ has three bidentate ligands, it forms a total of six coordinate bonds with the Ru²⁺ ion and has an octahedral shape. In an octahedral crystal field, the d orbitals split into two sets: the lower-energy d(z²) and d(x²-y²) orbitals, referred to as e_g orbitals, and the higher-energy d(xy), d(xz), and d(yz) orbitals, referred to as t₂g orbitals.
05

Draw the crystal field diagram for d orbitals in Ru(phen)3²⁺

In low-spin octahedral complexes, the electrons fill the lower-energy orbitals before filling the higher-energy orbitals. As the complex has five d electrons (from Ru²⁺) to be placed in the orbitals, filling starts from the lowest energy t₂g orbitals. All the electrons will pair up in the t₂g orbitals, leaving the e_g orbitals empty. The crystal field diagram for the d orbitals in Ru(phen)3²⁺ is as follows: t₂g: ↑↓ , ↑↓ , ↑↓ e_g: , Here, the t₂g orbitals are fully occupied, with each containing a pair of electrons, while the e_g orbitals are empty. This corresponds to the diamagnetic nature of the complex ion (no unpaired electrons) as required in the problem.

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Most popular questions from this chapter

a. In the absorption spectrum of the complex ion \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{cm}^{-1}\). Given 1 \(\mathrm{cm}^{-1}=1.986 \times 10^{-23} \mathrm{J},\) what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) is predicted to be \(180^{\circ} .\) What is the hybridization of the \(\mathrm{N}\) atom in the NCS - ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ}\) \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) undergoes substitution by ethylenediamine (en) according to the equation \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}+2 \mathrm{en} \longrightarrow \mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}+4 \mathrm{NCS}^{-}\) Does \(\operatorname{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit geometric isomerism? Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit optical isomerism?

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How many bonds could each of the following chelating ligands form with a metal ion? a. acetylacetone (acacH), a common ligand in organometal:atalysts: b. diethylenetriamine, used in a variety of industrial processes: $$ \mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} $$ c. salen, a common ligand for chiral organometallic catalysts: d. porphine, often used in supermolecular chemistry as well as catalysis; biologically, porphine is the basis for many different types of porphyrin- containing proteins, including heme proteins:

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. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\)

A compound related to acetylacetone is 1,1,1 -trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise \(47 .\) Both \(\mathrm{Be}^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa \(^{-}\) having the formula \(\mathrm{M}\) (tfa) \(_{2}\). Two isomers are formed for each metal complex. a. The \(\mathrm{Be}^{2+}\) complexes are tetrahedral. Draw the two isomers of Be(tfa)_2. What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by $\mathrm{Cu}(\mathrm{tfa})_{2} ?
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