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
The \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is a low spin complex with one unpaired electron, while the \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) complex is a high spin complex with five unpaired electrons. This indicates that the NCS- ligand has a weaker field strength than the CN- ligand, and is placed lower in the spectrochemical series.

Step by step solution

01

Understanding high spin and low spin complexes

In transition metal complexes, electrons are distributed between bonding and antibonding molecular orbitals. The energy difference between these orbitals, known as the crystal field splitting energy (\( \Delta\)), determines whether a complex is high spin or low spin. Weak field ligands result in a small \( \Delta\) and high spin complexes, while strong field ligands result in a large \( \Delta\) and low spin complexes. High spin complexes have more unpaired electrons, as electrons prefer to occupy degenerate orbitals before pairing up in lower-energy orbitals. Low spin complexes have fewer unpaired electrons, as they occupy lower-energy orbitals before moving to degenerate orbitals.
02

Determine high spin and low spin for given complexes

The complex \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) has one unpaired electron, while the complex \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) has five unpaired electrons. The larger number of unpaired electrons in the second complex indicates that it has a smaller \( \Delta\), making it a high spin complex. Conversely, the smaller number of unpaired electrons in the first complex indicates that it has a larger \( \Delta\), making it a low spin complex.
03

Placement of NCS- in the spectrochemical series

The spectrochemical series is the order of ligands based on their ligand field strength. The stronger the ligand, the larger the \( \Delta\), and the fewer the unpaired electrons (low spin complex). In this case, the CN- ligand forms a low spin complex, while the NCS- ligand forms a high spin complex. This indicates that the NCS- ligand has a weaker field strength than the CN- ligand. Thus, we can conclude that the NCS- ligand is placed lower in the spectrochemical series relative to the CN- ligand. In summary, the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is a low spin complex, while the \(\left[\mathrm{Fe}(\mathrm{NCS})_{6}\right]^{3-}\) complex is a high spin complex. The NCS- ligand is placed lower in the spectrochemical series as compared to the CN- ligand, as it results in a high spin complex with a smaller \( \Delta\).

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Most popular questions from this chapter

Given the colors observed for \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion (see Exercise \(23.84 ),\) what can you say about how the energy separation between the ligand orbitals and the empty \(d\) orbitals changes as a function of the oxidation state of the transition metal at the center of the tetrahedral anion?

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CdBr}_{4}\right]\) (c) \(\left[\mathrm{Pt}(\mathrm{en})_{3}\right]\left(\mathrm{ClO}_{4}\right)_{4}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right]^{+}\) (e) \(\mathrm{NH}_{4}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{NCS})_{4}\right]\) (f) \(\left[\mathrm{Cu}(\mathrm{bipy})_{2} \mathrm{I}\right] \mathrm{I}\)

Write names for the following coordination compounds: (a) \(\left[\mathrm{Cd}(\mathrm{en}) \mathrm{Cl}_{2}\right]\) (b) \(\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]\) (c) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{CO}_{3}\right)\right] \mathrm{Cl}\) (d) \(\left[\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{3}\)

For each of the following pairs, identify the molecule or ion that is more likely to act as a ligand in a metal complex: (a) acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) or ammonium \(\left(\mathrm{NH}_{4}^{+}\right)\) (b) hydride \(\left(\mathrm{H}^{-}\right)\) or hydronium \(\left(\mathrm{H}_{3} \mathrm{O}^{+}\right),(\mathbf{c})\) carbon monoxide \((\mathrm{CO})\) or methane \(\left(\mathrm{CH}_{4}\right)\) .

For each of the following metals, write the electronic configuration of the atom and its \(3+\) ion: (a) Fe, (b) Mo, (c) Co. 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 \(3+\) ion, assuming a weak-field complex. How many unpaired electrons are there in each case?

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