Problem 85 - Chemistry: An Atoms-Focused Approach

Question

What is the formula of the oxide that crystallizes with \(\mathrm{Fe}^{3+}\) ions in one-fourth of the octahedral holes, \(\mathrm{Fe}^{3+}\) ions in one- eighth of the tetrahedral holes, and \(\mathrm{Mg}^{2+}\) in one-fourth of the octahedral holes of a cubic closest-packed arrangement of oxide ions \(\left(\mathrm{O}^{2-}\right) ?\)

Step-by-Step Solution

Answer

Answer: Fe₂MgO₄

Step 1: Identify ion types and holes

In this exercise, there are three types of ions: Fe^3+ ions in one-fourth of the octahedral holes, Fe^3+ ions in one-eighth of the tetrahedral holes, and Mg^2+ ions in one-fourth of the octahedral holes.

Step 2: Count the number of ions in each hole type

In a cubic closest-packed structure, there are an equal number of octahedral holes and O^2- ions, and twice the number of tetrahedral holes as O^2- ions. If we assume that there is 1 O^2- ion in the structure, we can determine the numbers of ions in each hole type: - Fe^3+ ions in one-fourth of the octahedral holes: \((1/4)(1\,\mathrm{O}^{2-}) = 1/4\,\mathrm{Fe}^{3+}\) - Fe^3+ ions in one-eighth of the tetrahedral holes: \((1/8)(2(1 \,\mathrm{O}^{2-})) = 1/4\,\mathrm{Fe}^{3+}\) - Mg^2+ ions in one-fourth of the octahedral holes: \((1/4)(1\,\mathrm{O}^{2-}) = 1/4\,\mathrm{Mg}^{2+}\)

Step 3: Determine the ions ratio and adjust with respect to the oxide ions

Now we'll find the ratio of the ions: - \(\mathrm{Fe}^{3+}\) ions from octahedral holes: \(1/4:1\Rightarrow 1:\mathrm{Fe}^{3+}\) - \(\mathrm{Fe}^{3+}\) ions from tetrahedral holes: \(1/4:1\Rightarrow 1:\mathrm{Fe}^{3+}\) - \(\mathrm{Mg}^{2+}\) ions from octahedral holes: \(1/4:1\Rightarrow 1:\mathrm{Mg}^{2+}\) Therefore, in one unit cell, we have 1 oxide ion \(\left(\mathrm{O}^{2-}\right)\), 2 \(\mathrm{Fe}^{3+}\) ions (1 from each type of hole), and 1 \(\mathrm{Mg}^{2+}\) ion.

Step 4: Determine the formula of the oxide

Since all ion ratios can be expressed as whole numbers, we can now write the formula of the oxide: \(\textbf{Fe}_2\textbf{Mg}\textbf{O}_4\) So, the formula of the oxide that crystallizes with the given ions distribution is \(\textbf{Fe}_2\textbf{Mg}\textbf{O}_4\).

Key Concepts

Cubic Closest-Packed Structure

The cubic closest-packed (CCP) structure, also known as the face-centered cubic (FCC) structure, is one of the most efficient ways to pack equal-sized spheres in three dimensions. This arrangement is found in many metallic and ionic compounds, where it provides a stable configuration.
CCP structures are characterized by layers of atoms or ions stacked in an ABC pattern. This means the third layer of spheres is offset from the first, creating a dense and strong packing order. Every layer of spheres sits in the grooves formed by the layers beneath them, representing the optimal use of space.

• The packing efficiency of a CCP structure is about 74%, indicating the percentage of volume occupied by the spheres.
• CCP structures in crystals often have octahedral and tetrahedral holes, which accommodate other ions in ionic compounds.

Understanding the CCP helps in determining how smaller ions fit into the larger structure, influencing the compound's chemical formula and properties.

Octahedral Holes

In the CCP structure, octahedral holes are a key feature, crucial to the understanding of ionic compounds. These are spaces created between six ions arranged at the corners of an octahedron. They are larger than tetrahedral holes, making them suitable for larger ions.
Each sphere in a CCP structure is surrounded by 12 neighbors, and in this highly packed arrangement, there are as many octahedral holes as there are spheres.

• An octahedral hole sits at the center of a cluster formed by one layer of three spheres sitting directly over three spheres in the layer below.
• In oxide structures like the one presented, cations like \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mg}^{2+}\) can occupy octahedral holes.

These holes are strategically filled based on the ion's size and charge, affecting the composition and resultant properties of the crystal.

Oxide Ions

Oxide ions \(\mathrm{O}^{2-}\) are a common component in many oxidic compounds and play a pivotal role in determining the structure based on ionic radii. They often form the primary lattice in many crystal structures, providing the basis upon which metal ions are arranged.
The size and charge of oxide ions influence how metal cations are accommodated in crystal structures. In the cubic closest-packed structure, oxide ions are typically in a framework arrangement, offering octahedral and tetrahedral sites for other ions to occupy.

• The presence of oxide ions determines the potential size of the available holes, dictating which cations can fit into tetrahedral or octahedral holes.
• Due to their charge, oxide ions establish strong ionic bonds within the structure, helping to stabilize the compound.

Recognizing the role of oxide ions helps understand the intricacies of ionic compounds and their crystalline forms, as seen in the hypothetical formula \(\text{Fe}_2\text{Mg}\text{O}_4\).