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Give the number of \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{TiCl}_{6}\right]\), (b) \(\mathrm{Na}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right]\), (c) \(\left[\operatorname{Ru}(\mathrm{en})_{3}\right] \mathrm{Br}_{3}\), (d) \([\mathrm{Mo}(\mathrm{EDTA})] \mathrm{ClO}_{4},(\mathrm{e}) \mathrm{K}_{3}\left[\mathrm{ReCl}_{6}\right]\).

Short Answer

Expert verified
The number of \(d\) electrons associated with the central metal ions in the given complexes are as follows: (a) K3[TiCl6]: 1 \(d\) electron, (b) Na3[Co(NO2)6]: 6 \(d\) electrons, (c) [Ru(en)3]Br3: 5 \(d\) electrons, (d) [Mo(EDTA)]ClO4: 1 \(d\) electron, and (e) K3[ReCl6]: 2 \(d\) electrons.

Step by step solution

01

(a) K3[TiCl6]

Step 1: Determine the oxidation state of the central metal ion Since K has a +1 oxidation state and there are 3 K ions, the [TiCl6] complex has a -3 overall charge. Since Cl has a -1 oxidation state and there are 6 Cl ions, the oxidation state of Ti is: \[ 6(-1) + \text{oxidation state of Ti} = -3 \] \[ \text{oxidation state of Ti} = +3 \] Step 2: Calculate the number of \(d\) electrons Ti has an atomic number of 22, which means its electron configuration is [Ar]4s2 3d2. As Ti3+, the electron configuration is [Ar] 3d1, so there is only 1 \(d\) electron associated with the central metal ion in K3[TiCl6].
02

(b) Na3[Co(NO2)6]

Step 1: Determine the oxidation state of the central metal ion Since Na has a +1 oxidation state and there are 3 Na ions, the [Co(NO2)6] complex has a -3 overall charge. Since NO2 has a -1 oxidation state and there are 6 NO2 ions, the oxidation state of Co is: \[ 6(-1) + \text{oxidation state of Co} = -3 \] \[ \text{oxidation state of Co} = +3 \] Step 2: Calculate the number of \(d\) electrons Co has an atomic number of 27, which means its electron configuration is [Ar]4s2 3d7. As Co3+, the electron configuration is [Ar] 3d6, so there are 6 \(d\) electrons associated with the central metal ion in Na3[Co(NO2)6].
03

(c) [Ru(en)3]Br3

Step 1: Determine the oxidation state of the central metal ion Since Br has a -1 oxidation state and there are 3 Br ions, the [Ru(en)3] complex has a +3 overall charge. Therefore, the oxidation state of Ru is +3. Step 2: Calculate the number of \(d\) electrons Ru has an atomic number of 44, which means its electron configuration is [Kr]5s1 4d7. As Ru3+, the electron configuration is [Kr] 4d5, so there are 5 \(d\) electrons associated with the central metal ion in [Ru(en)3]Br3.
04

(d) [Mo(EDTA)]ClO4

Step 1: Determine the oxidation state of the central metal ion Since ClO4 has a -1 oxidation state and there is only 1 ClO4 ion, the [Mo(EDTA)] complex has a +1 overall charge. Since EDTA has a -4 oxidation state, the oxidation state of Mo is: \[ -4 + \text{oxidation state of Mo} = +1 \] \[ \text{oxidation state of Mo} = +5 \] Step 2: Calculate the number of \(d\) electrons Mo has an atomic number of 42, which means its electron configuration is [Kr]5s1 4d5. As Mo5+, the electron configuration is [Kr] 4d1, so there is only 1 \(d\) electron associated with the central metal ion in [Mo(EDTA)]ClO4.
05

(e) K3[ReCl6]

Step 1: Determine the oxidation state of the central metal ion Since K has a +1 oxidation state and there are 3 K ions, the [ReCl6] complex has a -3 overall charge. Since Cl has a -1 oxidation state and there are 6 Cl ions, the oxidation state of Re is: \[ 6(-1) + \text{oxidation state of Re} = -3 \] \[ \text{oxidation state of Re} = +3 \] Step 2: Calculate the number of \(d\) electrons Re has an atomic number of 75, which means its electron configuration is [Xe]6s2 5d5. As Re3+, the electron configuration is [Xe] 5d2, so there are 2 \(d\) electrons associated with the central metal ion in K3[ReCl6].

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Key Concepts

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

Understanding Oxidation States in Transition Metal Complexes
In chemistry, the oxidation state of a metal in a compound reflects the degree of its electron loss or gain relative to its elemental form. Determining the oxidation state is crucial for analyzing transition metal complexes, as it directly affects the metal's d-electron count and, consequently, the complex's properties.

The oxidation state can be deduced by considering the charges of the ligands and the overall charge of the complex. For example, in a complex ion where the central metal is accompanied by negatively charged ligands and the total charge is known, we can calculate the metal's oxidation state by balancing the charges. As seen in the exercise solutions, this is achieved by adding the charges of the ligands and the metal to match the overall charge of the complex.

The importance of determining the correct oxidation state cannot be overstated—it dictates not only the number of d electrons but also the metal's chemistry, including its reactivity, magnetism, and even color.
D-block Elements and Their Role in Complexes
The d-block elements, often referred to as transition metals, reside at the center of the periodic table. These elements possess a partially filled d-subshell, which allows the formation of a diverse range of complexes with various colors, geometries, and properties.

In the context of the given exercise, each complex contains a central d-block element such as titanium (Ti), cobalt (Co), ruthenium (Ru), molybdenum (Mo), or rhenium (Re). Their unique electron configurations allow these metals to form stable compounds despite undergoing multiple oxidation states. Moreover, d-block elements often exhibit variable oxidation states, further influencing the properties of the resulting complexes.

It is fascinating to note that the coordination compounds formed by these elements often have technological and biological significance, ranging from catalysis to the active sites in enzymes. The versatility of d-block elements stems from their ability to adopt numerous coordination geometries with different ligands.
Crystal Field Theory: An Insight into Complex Structures
Diving into Crystal Field Theory (CFT) gives us a macroscopic perspective on how ligands interact with the central metal ion's d-orbitals, splitting them into different energy levels. This theory is pivotal for explaining the color and magnetism of transition metal complexes.

In simplistic terms, when ligands approach the central ion, they repel the electrons in its d orbitals due to electrostatic forces. The degree of this splitting, known as crystal field splitting, depends on the geometry of the complex and the nature of the ligands. For instance, in an octahedral complex, the d-orbitals split into two groups with different energy levels, often denoted as t2g and eg.

For learners, a practical understanding of CFT involves grasping how the number of d electrons and the splitting of these orbitals influence properties like absorption of light, paramagnetism, and diamagnetism. Overall, CFT is an essential tool for predicting and rationalizing the electronic structure of transition metal complexes.
Ligand Field Theory and Its Impact on Complex Stability
Building on CFT, Ligand Field Theory (LFT) incorporates the more modern concepts of molecular orbitals and covalency. It not only considers the repulsion between ligands and metal d electrons but also takes into account the direct overlap of their orbitals. This gives a more accurate depiction of the bonding in transition metal complexes.

LFT explains why some complexes are more stable than others by looking at the stabilization provided by the ligands’ interaction with the metal’s d orbitals. It clarifies how the strength of the metal-ligand bond can affect the overall stereochemistry and electronic transitions, influencing the compound's reaction mechanisms. In the problem-solving exercise, understanding LFT can help students not only count d electrons but also predict how these electrons will be distributed among the d orbitals and how the ligands can stabilize certain oxidation states.

This robust theory is crucial for advanced students who are delving into the nuances of inorganic chemistry, as it provides a deeper insight into the electronic behavior of coordination compounds, which is vital for applications in catalysis, material sciences, and medicine.

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Most popular questions from this chapter

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit \(d-d\) transitions? Explain. (c) As mentioned in "A Closer Look" on charge-transfer color, the violet color of \(\mathrm{MnO}_{4}^{-}\), is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\) occurs at a wavelength of \(565 \mathrm{~nm}\). The \(\mathrm{Cr} \mathrm{O}_{4}^{2-}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for \(\mathrm{MnO}_{4}^{-} ?\) Explain. (e) The \(\mathrm{VO}_{4}{ }^{3-}\) ion is colorless. Is this observation consistent with the wavelengths of the LMCT transitions in \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{CrO}_{4}^{2-}\) ?

Sketch all the possible stereoisomers of (a) tetrahedral \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right]\), (b) square-planar \(\left[\mathrm{IrCl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-}\), (c) octa- hedral \(\left[\mathrm{Fe}(0 \text { -phen })_{2} \mathrm{Cl}_{2}\right]^{+}\)

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.

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin \(\mathrm{Fe}(\mathrm{II})\) complex; deoxyhemoglobin, without the \(\mathrm{O}_{2}\) molecule, is a high-spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of \(\mathrm{O}_{2}\) in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15-minute exposure to air containing 400 ppm of CO causes about \(10 \%\) of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and \(\mathrm{O}_{2}\) to hemoglobin?

The \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) ion has an absorption maximum at about \(725 \mathrm{~nm}\), whereas the \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) ion absorbs at about \(570 \mathrm{~nm} .\) Predict the color of each ion. (b) The \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\) ion absorption maximum occurs at about \(545 \mathrm{~nm}\), and that of the [Ni(bipy) \(\left._{3}\right]^{2+}\) ion occurs at about \(520 \mathrm{~nm}\). From these data, indicate the relative strengths of the ligand fields created by the four ligands involved.

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