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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 \(\left(\mathrm{O}^{2}\right)\) ?

Short Answer

Expert verified
Answer: The formula of the oxide is Fe鈧侻gO鈧.

Step by step solution

01

Determine the number of O虏鈦 ions in the unit cell

In a cubic closest-packed arrangement, each unit cell contains 4 O虏鈦 ions.
02

Calculate the number of Fe鲁鈦 ions occupying octahedral holes

Since Fe鲁鈦 ions occupy one-fourth of the octahedral holes, there are 1/4 脳 4 = 1 Fe鲁鈦 ions occupying octahedral holes in the unit cell.
03

Calculate the number of Mg虏鈦 ions in the unit cell

As Mg虏鈦 ions also occupy one-fourth of the octahedral holes, there are 1/4 脳 4 = 1 Mg虏鈦 ions in the unit cell.
04

Calculate the number of Fe鲁鈦 ions occupying tetrahedral holes

There are 8 tetrahedral holes in a cubic closest-packed arrangement, and since Fe鲁鈦 ions occupy one-eighth of the tetrahedral holes, there are 1/8 脳 8 = 1 Fe鲁鈦 ion occupying tetrahedral holes in the unit cell.
05

Calculate the total number of cations in the unit cell

We have 1 Fe鲁鈦 ion in an octahedral hole, 1 Mg虏鈦 ion in an octahedral hole, and 1 Fe鲁鈦 ion in a tetrahedral hole. So, there are a total of 3 cations in the unit cell.
06

Calculate the charge balance of the oxide

The charge of the ions in the unit cell must be balanced. We have: - 1 Fe鲁鈦 ion from an octahedral hole: +3 charge - 1 Fe鲁鈦 ion from a tetrahedral hole: +3 charge - 1 Mg虏鈦 ion from an octahedral hole: +2 charge - 4 O虏鈦 ions: -2 脳 4 = -8 charge Adding these charges together, we get +3 + 3 + 2 - 8 = 0, which means the charge is balanced.
07

Find the formula of the oxide

Since there are 1 Fe鲁鈦 from an octahedral hole, 1 Fe鲁鈦 from a tetrahedral hole, 1 Mg虏鈦, and 4 O虏鈦 ions, the formula of the oxide is \(\mathrm{Fe_2MgO_4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攆our 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.

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