Problem 47 - Chemistry: An Atoms-Focused Approach

Question

An interstitial alloy is prepared from metals \(\mathrm{A}\) and \(\mathrm{B}\), where \(\mathrm{B}\) has the smaller atomic radius. The unit cell of metal \(\mathrm{A}\) is fec. What is the formula of the alloy if \(\mathrm{B}\) occupies (a) all of the octahedral holes; (b) half of the octahedral holes; (c) half of the tetrahedral holes?

Step-by-Step Solution

Answer

Answer: The formula of the alloy is: (a) AB when B occupies all the octahedral holes (b) A2B when B occupies half of the octahedral holes (c) AB when B occupies half of the tetrahedral holes

Step 1: Understanding fcc unit cell and counting atoms in it

A face-centered cubic unit cell has atoms located at the corners and faces of the cube. There are 8 corner atoms and 6 face atoms in the unit cell, but as each corner atom is shared by 8 unit cells, and each face atom by 2 cells, there are effectively 8 * 1/8 + 6 * 1/2 =1+3= 4 atoms of A in the unit cell.

Step 2: Determine number of octahedral and tetrahedral holes per unit cell

In an fcc unit cell, each atom of A is associated with four octahedral holes and eight tetrahedral holes. In total there are: 4 atoms of A * 1 octahedral hole/atom = 4 octahedral holes 4 atoms of A * 2 tetrahedral holes/atom = 8 tetrahedral holes So, in the unit cell, there are 4 octahedral and 8 tetrahedral holes.

Step 3: Calculate the formula for Case (a) - All octahedral holes occupied by B

Since B occupies all 4 octahedral holes in the unit cell, there are 4 atoms of B present. Therefore, the formula for the alloy in this case is A4B4, which can be simplified to AB.

Step 4: Calculate the formula for Case (b) - Half of octahedral holes occupied by B

Since half of the 4 octahedral holes are occupied by B, there are 2 atoms of B present in the unit cell. Thus, the formula for the alloy in this case is A4B2, which can be simplified to A2B.

Step 5: Calculate the formula for Case (c) - Half of tetrahedral holes occupied by B

Since half of the 8 tetrahedral holes are occupied by B, there are 4 atoms of B present in the unit cell. So, the formula for the alloy in this case is A4B4, which can also be simplified to AB. In conclusion, the formula of the alloy is: (a) AB when B occupies all the octahedral holes (b) A2B when B occupies half of the octahedral holes (c) AB when B occupies half of the tetrahedral holes

Key Concepts

Octahedral Holes

In crystallography, octahedral holes are voids or empty spaces found in crystal lattices, particularly within the face-centered cubic (fcc) structures. Imagine a cube. When positioned at the center of each edge and in the interior center of the cube, atoms form these poles, which create octahedral holes. These are essential in forming alloys, especially with interstitial elements.
To visualize, if you have an atom at each corner of a cube and another at the center of each face, the octahedral holes are found in the gaps in between. They can accommodate small atoms or ions, adding to the structural integrity or creating new materials.
In the given exercise, if an element B occupies all octahedral holes, each atom of metal A surrounded by these holes gives a coordination number of 6. This results in 4 atoms of B filling all octahedral positions in the unit cell, making the alloy formula A4B4, which simplifies to AB.

Tetrahedral Holes

Tetrahedral holes are smaller gaps within a crystal lattice, such as a face-centered cubic (fcc) structure. These locations are situated where an atom of the main lattice is surrounded by four other atoms, forming a tetrahedron.
Picture four atoms lying in the corners of a triangular pyramid; the small space in the center of this pyramid is the tetrahedral hole. These spaces can accommodate smaller atoms and occasionally accommodate other species, which influences the creation of interstitial alloys.
In the context of the exercise, if half of these tetrahedral holes are filled by a smaller atom B, it means 4 out of 8 of these spaces are occupied. Thus, in the unit cell, chemical B occupies half of the 4 possible tetrahedral holes, resulting in the same alloy formula as when all octahedral holes are occupied, simplifying to AB.

Face-Centered Cubic

The face-centered cubic (fcc) structure is a type of crystal arrangement where atoms are located at each of the cube's corners and the centers of all cube faces. This setup results in a highly efficient packing arrangement with the highest possible packing density of 0.74.
In an fcc lattice, each unit cell contains effectively 4 atoms, calculated by adding up the fractional counts of shared atoms across cell boundaries. This structure is not only important in understanding how metals pack in solid form, but it also lays the foundation for understanding vacant spaces within the cells, namely the octahedral and tetrahedral holes mentioned above.
The efficient arrangement of atoms in the fcc structure allows both octahedral and tetrahedral holes to be naturally present, influencing the properties of compounds such as interstitial alloys.