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The ion \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) has one unpaired electron, whereas \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) has five unpaired electrons. From these results, what can you conclude about whether each complex is high spin or low spin? What can you say about the placement of NC

Short Answer

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The given complexes \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) have one and five unpaired electrons, respectively. The \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is low spin since it has minimal unpaired electrons, whereas the \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) complex is high spin since it has the maximum possible unpaired electrons for iron(III). From these results, we can conclude that the \(\mathrm{NCS}^{-}\) ligand is on the left side of the spectrochemical series as it forms a high spin complex, meaning it is a weak-field ligand.

Step by step solution

01

Understanding high spin and low spin complexes

High spin and low spin complexes refer to the arrangement of electrons in the d-orbitals of a metal ion in a coordination complex. High spin complexes have the maximum number of unpaired electrons possible, whereas low spin complexes have the minimum number of unpaired electrons possible. This difference arises due to the varying strengths of the ligands to split the d-orbitals.
02

Determine if the given complexes are high spin or low spin

We are given that the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex has one unpaired electron, and the \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) complex has five unpaired electrons. Iron(III) has a d5 high spin electron configuration which means that all the d-orbitals have single electrons and no orbital pairing occurs. Since the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex has only one unpaired electron, we can conclude that the ligand \(\mathrm{CN}^{-}\) has resulted in a low spin complex. In contrast, the \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) complex has five unpaired electrons, which is the maximum possible for an iron(III) complex, giving us a high spin complex.
03

Determine the placement of \(\mathrm{NCS}^{-}\) in the spectrochemical series

The spectrochemical series is a list of ligands ordered according to their ability to split the d-orbitals of the central metal ion. Ligands on the left side of the series are weak-field ligands and result in high spin complexes, while ligands on the right side are strong-field ligands and result in low spin complexes. Since the \(\mathrm{NCS}^{-}\) ligand forms a high spin complex with iron(III) as shown in the \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) complex, we can conclude that the \(\mathrm{NCS}^{-}\) ligand is on the left side of the spectrochemical series (i.e., it is a weak-field ligand).

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Key Concepts

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

Coordination Complexes
Coordination complexes, sometimes simply called complexes, are compounds that consist of a central metal atom or ion bonded to multiple molecules or ions, known as ligands. The ligands have lone pairs of electrons that they donate to the metal atom, creating a coordination bond. These complexes can have a variety of shapes, colors, and chemical reactivity, depending on the metal and ligands.

Most commonly studied within inorganic chemistry, coordination complexes play a critical role in many biological processes and industrial applications. Examples include hemoglobin, which contains iron coordination complexes, and catalysts used in processes to manufacture important chemicals.
Ligand Field Theory
Ligand Field Theory (LFT) is a cornerstone of inorganic chemistry that describes how ligands affect the distribution of electrons in the d-orbitals of transition metal ions within coordination complexes. According to LFT, when ligands approach the central metal ion, they interact with its d-orbitals, which creates an energy difference called crystal field splitting. This concept is crucial to understand the magnetic properties, color, and reactivity of complexes.

Depending on the relative energies of the split d-orbitals, electrons may be arranged in different ways, resulting in high spin or low spin configurations. Large field splits will typically lead to low spin configurations, where electrons pair up in the lower energy orbitals, while small field splits favor high spin configurations with more unpaired electrons.
Spectrochemical Series
The spectrochemical series is a ranking of ligands based on their ability to split the d-orbitals of a central metal ion. Ligands that cause a large splitting are called strong-field ligands and are typically found on the right side of the series. These strong-field ligands often lead to low spin configurations. On the other hand, weak-field ligands, located on the left side of the series, cause less splitting and typically result in high spin complexes with more unpaired electrons.

Understanding where a ligand falls within the spectrochemical series can help predict whether a coordination complex will be high spin or low spin. This classification is fundamental when assessing the magnetic properties or certain aspects of the reactivity of the complex.
Unpaired Electrons
In coordination chemistry, the number of unpaired electrons in a complex is a key factor determining its magnetic properties. Unpaired electrons contribute to magnetic moments, leading to paramagnetism, where the substance is attracted to magnetic fields. Conversely, substances with all paired electrons exhibit diamagnetism and are slightly repelled by magnetic fields.

To determine if a metal complex has unpaired electrons, one needs to consider the electronic structure of the metal ion as well as the field strength of the ligands, as explained by ligand field theory. This concept ties directly into the high spin and low spin classifications, with high spin complexes having more unpaired electrons and thus stronger magnetic properties.

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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 \(\mathrm{CrO}_{4}^{2}\) 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. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

Crystals of hydrated chromium(III) chloride are green, have an empirical formula of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\), and are highly soluble, (a) Write the complex ion that exists in this compound. (b) If the complex is treated with excess \(\mathrm{AgNO}_{3}(a q)\), how many moles of \(\mathrm{AgCl}\) will precipitate per mole of \(\mathrm{CrCl}_{3} * 6 \mathrm{H}_{2} \mathrm{O}\) dissolved in solution? (c) Crystals of anhydreus chromium(III) chloride are violet and insoluble in aqueous solution. The coordination geometry of chromium in these crystals is octahedral as is almost always the case for \(\mathrm{Cr}^{3+}\). How can this be the case if the ratio of \(\mathrm{Cr}\) to \(\mathrm{Cl}\) is not 1:6?

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) \(\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{2}\left[\mathrm{TiCl}_{6}\right]\) (c) \(\mathrm{MoOCl}_{4}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{Br}_{2}\)

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 (for metals that form stable \(+3\) ions in the first place). Suggest an explanation, keeping in mind the Lewis acid-base nature of the metalligand bond.

As shown in Figure 23.26, the \(d-d\) transition of \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) produces an absorption maximum at a wavelength of about \(500 \mathrm{~nm}\). (a) What is the magnitude of \(\Delta\) for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) in \(\mathrm{kJ} / \mathrm{mol}\) ? (b) How would the magnitude of \(\Delta\) change if the \(\mathrm{H}_{2} \mathrm{O}\) ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) were replaced with \(\mathrm{NH}_{2}\) ligands?

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