Problem 61 - Chemistry The Science in Context

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: The formula of the oxide is Fe₂MgO₄.

Step 1: Determine the number of O²⁻ ions in the unit cell

In a cubic closest-packed arrangement, each unit cell contains 4 O²⁻ ions.

Step 2: Calculate the number of Fe³⁺ ions occupying octahedral holes

Since Fe³⁺ ions occupy one-fourth of the octahedral holes, there are 1/4 × 4 = 1 Fe³⁺ ions occupying octahedral holes in the unit cell.

Step 3: Calculate the number of Mg²⁺ ions in the unit cell

As Mg²⁺ ions also occupy one-fourth of the octahedral holes, there are 1/4 × 4 = 1 Mg²⁺ ions in the unit cell.

Step 4: Calculate the number of Fe³⁺ ions occupying tetrahedral holes

There are 8 tetrahedral holes in a cubic closest-packed arrangement, and since Fe³⁺ ions occupy one-eighth of the tetrahedral holes, there are 1/8 × 8 = 1 Fe³⁺ ion occupying tetrahedral holes in the unit cell.

Step 5: Calculate the total number of cations in the unit cell

We have 1 Fe³⁺ ion in an octahedral hole, 1 Mg²⁺ ion in an octahedral hole, and 1 Fe³⁺ ion in a tetrahedral hole. So, there are a total of 3 cations in the unit cell.

Step 6: Calculate the charge balance of the oxide

The charge of the ions in the unit cell must be balanced. We have: - 1 Fe³⁺ ion from an octahedral hole: +3 charge - 1 Fe³⁺ ion from a tetrahedral hole: +3 charge - 1 Mg²⁺ ion from an octahedral hole: +2 charge - 4 O²⁻ ions: -2 × 4 = -8 charge Adding these charges together, we get +3 + 3 + 2 - 8 = 0, which means the charge is balanced.

Step 7: Find the formula of the oxide

Since there are 1 Fe³⁺ from an octahedral hole, 1 Fe³⁺ from a tetrahedral hole, 1 Mg²⁺, and 4 O²⁻ ions, the formula of the oxide is \(\mathrm{Fe_2MgO_4}\).

Key Concepts

Cubic Closest-Packed Structure

The cubic closest-packed (ccp) structure is a type of atomic arrangement that provides high packing efficiency in crystallography. Imagine stacking layers of spheres, where each sphere represents an atom or ion. In the ccp structure, the layers are stacked in a repeating sequence of three, which can be described as ABCABC.

This arrangement is also known as face-centered cubic (fcc) because, if you visualize a cube, each face of the cube has a sphere at its center. Due to this dense packing, it achieves a packing efficiency of about 74%. This means the structure uses most of its available space, which is why it's common in many metal crystals like copper, silver, and gold. Such high density is efficient and allows room for additional ions or atoms to fit into the spaces or 'holes' left in the ccp arrangement.

Octahedral and Tetrahedral Holes

In the ccp structure, atoms or ions do not completely fill the space, leaving voids or 'holes' that can be occupied by other smaller atoms or ions. These holes are termed based on their geometric shape and coordination number.

Octahedral Holes: These are voids formed between six atoms—four in one plane and two more atoms above and below it (like an octahedron). Each ccp unit cell has four octahedral holes. In the featured problem, ions such as Mg\(^{2+}\) and Fe\(^{3+}\) occupy these holes.

Tetrahedral Holes: These are smaller voids created by four atoms arranged in a tetrahedron shape. Each unit cell of ccp contains eight such holes. They are typically occupied by smaller ions, such as Fe\(^{3+}\), in our example, which fits into one of these spaces. These holes are strategically used to create various ionic compounds, influencing their stability and properties.

Charge Balance in Compounds

For any compound, maintaining a charge balance is crucial. This means that the sum of the positive and negative charges must equal zero, ensuring the compound is electrically neutral.

To solve this specific problem, we have:

• Fe\(^{3+}\) Ions: Two Fe\(^{3+}\) ions contribute a charge of +6 (3 + 3) from one octahedral and one tetrahedral hole.
• Mg\(^{2+}\) Ion: One Mg\(^{2+}\) ion adds a +2 charge.
• O\(^{2-}\) Ions: Four oxide ions, each with a -2 charge, provide a total of -8.

Adding these charges gives us +3 + 3 + 2 from the cations and -8 from the anions; thus, we achieve a net charge of zero.

Through this approach, we understand the role of each ion in balancing the structure. This method ensures we derive the correct chemical formula, \(\mathrm{Fe_2MgO_4}\), capturing both the elemental composition and structural arrangement within the compound.