Question

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

In conclusion, the complex \([\mathrm{Fe}(\mathrm{CN})_{6}]\mathrm{^{3-}}\) is a low spin complex due to its single unpaired electron, while the complex \([\mathrm{Fe}(\mathrm{NCS})_{6}]\mathrm{^{3-}}\) is a high spin complex with five unpaired electrons. This indicates that the thiocyanate ligand (NCS-) has a lower field strength than the cyanide ligand (CN-) in the spectrochemical series.

Step-by-step solution

Determine the oxidation state of the central metal ion

In both complexes, the central metal ion is Fe. In the first complex, the overall charge is -3, and the cyanide ligand's charge is -1, which means that Fe must have an oxidation state of +3 (since there are 6 cyanide ligands). Similarly, in the second complex, the overall charge is -3, and the thiocyanate ligand's charge is -1, which means that Fe must also have an oxidation state of +3.

Write the electron configurations for the Fe(III) ion

The electron configuration of Fe is [Ar]3d^6 4s^2. When it loses 3 electrons to become Fe(III), the electron configuration becomes [Ar]3d^5 with 5 empty 4s and 4p orbitals.

Determine the electronic configuration of the complexes

For \([\mathrm{Fe}(\mathrm{CN})_{6}]\mathrm{^{3-}}\), there is one unpaired electron. This means that out of the five d-orbitals in Fe(III), four are completely filled, and one is half-filled. This configuration is consistent with a low spin complex, as all the lower energy orbitals are filled before any higher energy orbitals are occupied. For \([\mathrm{Fe}(\mathrm{NCS})_{6}]\mathrm{^{3-}}\), there are five unpaired electrons. This means that each of the five d-orbitals in Fe(III) has one electron. This configuration aligns with a high spin complex, as the electrons occupy higher energy orbitals before completely filling the lower energy orbitals.

Determine the position of NCS- in the spectrochemical series

Since the \([\mathrm{Fe}(\mathrm{CN})_{6}]\mathrm{^{3-}}\) complex is low spin, we can deduce that the CN- ligand is a strong field ligand. On the other hand, the \([\mathrm{Fe}(\mathrm{NCS})_{6}]\mathrm{^{3-}}\) complex is high spin, meaning that the NCS- ligand is a weak field ligand. Therefore, NCS- must be placed on the weaker field side of the spectrochemical series compared to CN-. In conclusion, the complex \([\mathrm{Fe}(\mathrm{CN})_{6}]\mathrm{^{3-}}\) is a low spin complex, while the complex \([\mathrm{Fe}(\mathrm{NCS})_{6}]\mathrm{^{3-}}\) is a high spin complex. The thiocyanate ligand (NCS-) has a lower field strength than the cyanide ligand (CN-) in the spectrochemical series.