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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 \(\mathrm{NCS}^{-}\) in the spectrochemical series?

Short Answer

Expert verified
In conclusion, \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is a low spin complex due to its strong-field ligand \(\mathrm{CN}^{-}\) and one unpaired electron, while \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) is a high spin complex due to its weak-field ligand \(\mathrm{NCS}^{-}\) and five unpaired electrons. In the spectrochemical series, \(\mathrm{NCS}^{-}\) should be placed earlier than \(\mathrm{CN}^{-}\).

Step by step solution

01

Recall the principles of Crystal Field Theory (CFT)

Crystal field theory explains the interaction between the central metal ion and the ligands in a coordination complex. High spin and low spin complexes arise due to the difference in electron configurations that can be favored when the ligands approach the central metal ion. In a high spin complex, the ligands have a weak splitting effect, allowing the electrons to occupy higher energy orbitals. In a low spin complex, the ligands have a strong splitting effect, causing the electrons to pair up in the lower energy orbitals.
02

Identify high spin or low spin based on unpaired electrons

The given complexes are \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\), which has one unpaired electron, and \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\), which has five unpaired electrons. A complex with fewer unpaired electrons, such as 1, is a low spin complex because it has a strong splitting effect, causing electrons to pair up in the lower energy orbitals. Therefore, \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is a low spin complex. A complex with a higher number of unpaired electrons, such as 5, is a high spin complex because it has a weak splitting effect, allowing electrons to occupy the higher energy orbitals. Therefore, \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) is a high spin complex.
03

Compare the ligands and determine the placement of NCS- in the spectrochemical series

The spectrochemical series is a list of ligands ordered by their ability to split the d-orbitals of the central metal ion. Strong-field ligands, which cause a large splitting of the d-orbitals and result in low spin complexes, are found towards the end of the spectrochemical series. Weak-field ligands, which cause a small splitting of the d-orbitals and result in high spin complexes, are found at the beginning of the series. Since \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is a low spin complex, the \(\mathrm{CN}^{-}\) ligand is a strong-field ligand, and it is found towards the end of the spectrochemical series. On the other hand, \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) is a high spin complex, so the \(\mathrm{NCS}^{-}\) ligand is a weak-field ligand. Therefore, \(\mathrm{NCS}^{-}\) must be placed earlier than \(\mathrm{CN}^{-}\) in the spectrochemical series. In conclusion, \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is a low spin complex, and \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) is a high spin complex. \(\mathrm{NCS}^{-}\) should be placed earlier than \(\mathrm{CN}^{-}\) in the spectrochemical series according to the information provided.

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Key Concepts

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

High Spin Complexes
High spin complexes are characterized by having a larger number of unpaired electrons. This happens when the ligands surrounding the central metal ion cause a smaller splitting of the d-orbitals. In simpler terms, the energy gap between the lower (t2g) and higher (eg) energy levels is quite small. This small energy gap allows electrons to populate higher energy orbitals instead of pairing up in lower ones.
For example, in the complex \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\), the presence of five unpaired electrons indicates it is a high spin complex. The \(\mathrm{NCS}^{-}\) ligand causes a weaker splitting, leading to unpaired electrons in several d-orbitals.
When analyzing any complex:
  • Check the number of unpaired electrons
  • More unpaired electrons typically indicate a high spin configuration
Low Spin Complexes
Low spin complexes are distinguished by having fewer unpaired electrons. The ligands in this case exert a strong field, causing a significant splitting of the d-orbitals. This large energy gap encourages electrons to pair up in the lower energy levels (t2g) before filling the higher ones (eg).
In the given example of \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\), the presence of only one unpaired electron indicates a low spin complex. The \(\mathrm{CN}^{-}\) ligand is responsible for a stronger field and larger splitting of the d-orbitals, facilitating electron pairing.
To determine if a complex is low spin, consider:
  • Fewer unpaired electrons often mean a low spin configuration
  • Look for ligands known to cause significant splitting
Spectrochemical Series
The spectrochemical series is an incredibly insightful lineup of ligands arranged according to the field strength they generate. This list helps predict whether a given ligand will form high spin or low spin complexes.
Key points about the spectrochemical series include:
  • Strong-field ligands: Located towards the end, cause significant d-orbital splitting, typically forming low spin complexes.
  • Weak-field ligands: Found earlier in the series, they cause minimal splitting and often lead to high spin complexes.
When evaluating \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\), we observe that \(\mathrm{CN}^{-}\) produces a stronger field leading to low spin, while \(\mathrm{NCS}^{-}\) forms a weaker field indicating a high spin. Hence, \(\mathrm{NCS}^{-}\) appears earlier in the spectrochemical series compared to \(\mathrm{CN}^{-}\). Recognizing where ligands stand in this series provides a predictive edge in understanding complex behaviors.

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Most popular questions from this chapter

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 AgCl will precipitate per mole of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) dissolved in solution? (c) Crystals of anhydrous 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 Cl is not \(1: 6 ?\)

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (both octahedral), and \(\left[\mathrm{CoCl}_{4}\right]^{2-}(\) tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in octahedral ones, assign a color to each complex.

The value of \(\Delta\) for the \(\left[\mathrm{MoI}_{6}\right]^{3-}\) complex is \(198.58 \mathrm{~kJ} / \mathrm{mol}\). Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lower energy to the higher-energy \(d\) -orbital set in this complex. Should the complex absorb in the visible range?

Which periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 and \(8 ?(\mathbf{a})\) The number of valence electrons reaches a maximum at group 8. (b) The effective nuclear charge increases on moving left across each period. (c) The radii of the transition-metal elements reach a minimum for group \(8,\) and as the size of the atoms decreases it becomes easier to remove electrons.

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\), abbreviated py, is the molecule (a) Would you expect pyridine to act as a monodentate or bidentate ligand? (b) For the equilibrium reaction $$ \left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\mathrm{bipy} $$ would you predict the equilibrium constant to be larger or smaller than one?

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