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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Most popular questions from this chapter

For each of the following metals, write the electronic configuration of the atom and its \(2+\) ion: $(\mathbf{a}) \mathrm{Mn},(\mathbf{b}) \mathrm{Ru},(\mathbf{c}) \mathrm{Rh} .$ Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(2+\) ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

Write out the ground-state electron configurations of (a) \(\mathrm{Sc}^{2+}\) (b) \(\mathrm{Mo}^{2+}\) (c) \(\mathrm{Rh}^{3+}\), (d) \(\mathrm{Fe}^{3+}\).

Carbon monoxide, CO, is an important ligand in coordination chemistry. When CO is reacted with nickel metal, the product is \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\) which is a toxic, pale yellow liquid. (a) What is the oxidation number for nickel in this compound? (b) Given that \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is a diamagnetic molecule with a tetrahedral geometry, what is the electron configuration of nickel in this compound? (c) Write the name for \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) using the nomenclature rules for coordination compounds.

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How many electrons are in the valence \(d\) orbitals in these transition-metal ions? $(\mathbf{a}) \mathrm{Ru}^{3+},(\mathbf{b}) \mathrm{Pd}^{2+},(\mathbf{c}) \mathrm{Ti}^{2+},\( (d) \)\mathrm{W}^{6+}$.

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