Problem 36 - Chemistry: An Atoms-Focused Approach

Question

On average, the \(d\) orbitals of a transition metal ion in an octahedral field are higher in energy than they are when the ion is in the gas phase. Why?

Step-by-Step Solution

Answer

Answer: The energy levels of \(d\) orbitals in a transition metal ion are higher in an octahedral field compared to the gas phase due to the interaction between ligands and the ion's \(d\) orbitals, which causes electron repulsion and increases energy levels. Additionally, the \(d\) orbitals split into two groups with different energy levels, further increasing the overall energy.

Step 1: Understand the energy levels of \(d\) orbitals in the gas phase

In the gas phase, the five \(d\) orbitals of a transition metal ion are degenerate, meaning they have the same energy. When the ion is not interacting with ligands, these energy levels are relatively low.

Step 2: Identify the change in energy levels when the ion is in an octahedral field

When a transition metal ion is in an octahedral field, the energy levels of its \(d\) orbitals change. The ligands in the octahedral field approach the ion along its \(x, y, z\) axis, which results in the \(d\) orbitals splitting into two energy levels: three lower-energy \(t_{2g}\) orbitals (\(d_{xy}, d_{yz}, d_{zx}\)) and two higher-energy \(e_g\) orbitals (\(d_{x^2-y^2}, d_{z^2}\)).

Step 3: Understand the effect of ligand interaction on the \(d\) orbitals

As ligands approach the ion along the \(x, y, z\) axis, electron repulsion between the ligands and the ion's \(d\) orbitals increases. The ligands push the \(d\) electrons away, causing an increase in energy levels.

Step 4: Explain the overall energy increase

Although the \(d\) orbitals split into two groups, with the \(t_{2g}\) orbitals having a lower energy than the \(e_g\) orbitals, the overall energy is increased for all five \(d\) orbitals in an octahedral field compared to the gas phase. This energy increase is mainly due to the electron repulsion between ligands and the transition metal ion's \(d\) orbitals. In conclusion, the \(d\) orbitals of a transition metal ion in an octahedral field are higher in energy compared to the gas phase due to the interaction between ligands and the ion's \(d\) orbitals, as well as the splitting of the orbitals into two different energy levels.