Question

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

Short answer

Both statements are incorrect: a. \(\mathrm{CoCl}_{4}^{2-}\) is actually a high-spin complex (since \(\mathrm{Cl}^{-}\) is a weak-field ligand) and has three unpaired electrons, not two. b. \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) is a low-spin complex (because \(\mathrm{CN}^{-}\) is a strong-field ligand), but has three unpaired electrons, not four.

Step-by-step solution

Understand crystal field theory

Crystal field theory is used to describe the electronic structures of complex metal ions. It is based on the idea that the interaction between the metal ion and the ligands is purely electrostatic, and it helps us understand the energy levels of the d orbitals. The d orbitals in transition metal ions are not all of the same energy when in the presence of ligands; their energies are split due to the distribution of the ligands around the metal.

Recall the tetrahedral and octahedral splitting diagrams

In a tetrahedral complex, the splitting of the d orbitals results in three energy levels (d\(_{z^2}\), d\(_{x^2-y^2}\)) lower in energy than the other two (d\(_{xy}\), d\(_{xz}\), d\(_{yz}\)). In an octahedral complex, the opposite occurs, where the d\(_{xy}\), d\(_{xz}\), and d\(_{yz}\) orbitals are lower in energy, while the d\(_{z^2}\) and d\(_{x^2-y^2}\) orbitals are higher in energy.

Identify strong-field and weak-field ligands

Strong-field ligands, or those that produce a large splitting of the d orbitals, generally lead to low-spin complexes, where electrons are more likely to pair up in the lower-energy orbitals before occupying the higher-energy orbitals. Weak-field ligands, on the other hand, lead to high-spin complexes, where there is a smaller energy difference between the split orbitals, and electrons prefer to occupy all of them singly before pairing up.

Analyze statement a

Statement a claims that \(\mathrm{CoCl}_{4}^{2-}\) is a strong-field case with two unpaired electrons. However, in reality, \(\mathrm{Cl}^{-}\) is considered to be a weak field ligand, which means that \(\mathrm{CoCl}_{4}^{2-}\) should be a high-spin complex. Tetrahedral complexes generally have high-spin configurations, and in this case, the Co(III) ion has three d electrons. Since \(\mathrm{Cl}^{-}\) is a weak-field ligand, there would be one unpaired electron in each of the three lower-energy orbitals, which results in three unpaired electrons, not two as the statement suggests.

Analyze statement b

Statement b claims that \(\mathrm{CN}^{-}\) is a weak-field ligand, resulting in four unpaired electrons in the \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) complex and considering it as a low-spin case. However, \(\mathrm{CN}^{-}\) is, in reality, a strong-field ligand, which generally leads to low-spin complexes. In this case, the Co(III) ion has three d electrons. Because \(\mathrm{CN}^{-}\) is a strong-field ligand, these electrons would prefer to occupy all three of the lower-energy orbitals in an octahedral complex (d\(_{xy}\), d\(_{xz}\), d\(_{yz}\)) without pairing up, resulting in one unpaired electron in each lower-energy orbital, for a total of three unpaired electrons, not four as mentioned in the statement.