Problem 70 - Chemistry Central Science

Question

Suppose that a transition-metal ion was in a lattice in which it was in contact with just two nearby anions, located on opposite sides of the metal. Diagram the splitting of the metal \(d\) orbitals that would result from such a crystal field. Assuming a strong field, how many unpaired electrons would you expect for a metal ion with six \(d\) electrons? (Hint: Consider the linear axis to be the z-axis).

Step-by-Step Solution

Answer

In a linear crystal field with two nearby anions located on opposite sides of the metal along the z-axis, the d orbitals split into two energy levels: higher energy level with \(d_{z^2}\) and lower energy level with \(d_{xy}\), \(d_{xz}\), \(d_{yz}\), and \(d_{x^2-y^2}\). For a metal ion with six d electrons in a strong field, the electron distribution would first occupy the lower energy level orbitals, and the remaining electron goes to the higher energy level orbital. This results in two unpaired electrons, one in the \(d_{yz}\) orbital and another in the \(d_{x^2-y^2}\) orbital.

Step 1: Understand the splitting of d orbitals in a crystal field

Typically, in a crystal field, the electrostatic interactions between the metal ions and the nearby anions result in splitting the metal's d orbitals into two energy levels. This splitting is determined by the geometry of the surrounding anions. In our case, since there are only two anions, and they are located on opposite sides of the metal, we can treat it as a linear crystal field, with the anions located along the z-axis.

Step 2: Diagram the splitting of d orbitals in a linear crystal field

In a linear crystal field, the d orbitals will split into two energy levels, with the \(d_{z^2}\) orbital at a higher energy level, while the \(d_{xy}\), \(d_{xz}\), \(d_{yz}\), and \(d_{x^2-y^2}\) orbitals remain at a lower energy level. The splitting can be represented as: - Higher energy level: \(d_{z^2}\) - Lower energy level: \(d_{xy}\), \(d_{xz}\), \(d_{yz}\), and \(d_{x^2-y^2}\)

Step 3: Determine the number of unpaired electrons for a strong field

A strong field indicates that the electrostatic interactions between the metal ions and the anions are strong enough to cause a significant energy difference between the two split energy levels. In a strong field, the electrons will occupy the lower energy level orbitals before occupying the higher energy level orbitals. Since we have six d electrons, in a strong field, they will first occupy all the lower energy level orbitals: - \(d_{xy}\): 2 electrons (paired) - \(d_{xz}\): 2 electrons (paired) - \(d_{yz}\): 1 electron (unpaired) - \(d_{x^2-y^2}\): 1 electron (unpaired) The remaining electron will occupy the higher energy level orbital: - \(d_{z^2}\): 0 electrons

Step 4: Count the number of unpaired electrons

Based on the strong field distribution, we have the following unpaired electrons: - \(d_{yz}\): 1 unpaired electron - \(d_{x^2-y^2}\): 1 unpaired electron In a strong field linear crystal with six d electrons, we would expect two unpaired electrons.