Problem 49 - Chemistry: An Atoms-Focused Approach

Question

An interstitial alloy with a fcc unit cell contains one atom of B for every five atoms of host element A. What fraction of the octahedral holes is occupied in this alloy?

Step-by-Step Solution

Answer

Answer: The fraction of occupied octahedral holes in this interstitial alloy is 1/12 or approximately 0.0833.

Step 1: Calculate the number of atoms in an fcc unit cell

An fcc unit cell consists of the following atoms: 8 corner atoms and 6 face atoms. The corner atoms contribute 1/8 of each atom to the unit cell, while the face-centered atoms fully contribute to the unit cell. So, the total number of atoms in an fcc unit cell is given by: Number of atoms per unit cell = (8 corner atoms × 1/8) + (6 face atoms × 1) = 8/8 + 6 = 1 + 6 = 7

Step 2: Calculate the total number of A atoms and B atoms in the unit cell

According to the problem, there is one B atom for every five A atoms. Thus, if we have X B atoms, we must have 5X A atoms. The total number of atoms in the unit cell (7 atoms) is given by the sum of A and B atoms: 7 = 5X + X

Step 3: Solve for X to find the number of B atoms

From the equation above, we can solve for X to find the number of B atoms: 7 = 5X + X 7 = 6X X = 7/6 Since the number of atoms must be a whole number, we can multiply both sides by 6 to find the total number of B atoms in a larger unit cell: 6(7) = 6(6X) 42 = 36X X = 42/36 = 7/6 Thus there are 7 B atoms for every 35 A atoms in a larger unit cell.

Step 4: Calculate the number of octahedral holes in an fcc unit cell

For every atom in an fcc unit cell, there are two available octahedral holes. So, the total number of octahedral holes in a larger unit cell would be: Number of octahedral holes = 2 × (number of A atoms + number of B atoms) Number of octahedral holes = 2 × (35 + 7) = 2 × 42 = 84

Step 5: Calculate the fraction of occupied octahedral holes

Now, we have the total number of octahedral holes and the number of B atoms occupying these holes. To determine the fraction of occupied octahedral holes, we can use the following formula: Fraction of occupied octahedral holes = (Number of B atoms) / (Number of octahedral holes) Fraction of occupied octahedral holes = 7 / 84 = 1/12 So, the fraction of occupied octahedral holes in this interstitial alloy is 1/12 or approximately 0.0833.

Key Concepts

fcc unit cell

An "fcc unit cell" stands for face-centered cubic, a specific type of crystal structure found commonly in metallic elements. In this structure, atoms are positioned at each corner of a cube and one in the center of each of the faces.
The corner atoms share their mass with eight neighboring cubes, while the face-centered atoms are shared with two cubes. Hence, their contribution to the unit cell is different. Each corner atom contributes one eighth, and each face-centered atom fully contributes, in terms of presence, to its unit cell.

• Number of corner atoms: 8
• Contribution of each corner atom: \( \frac{1}{8} \)
• Number of face-centered atoms: 6
• Total number of atoms in fcc unit cell: \( 8 \times \frac{1}{8} + 6 \times 1 = 7 \)

In summary, the fcc unit cell effectively contains a total of 7 atoms.

octahedral holes

In the context of crystal structures like fcc, "octahedral holes" are voids or gaps between the atoms where smaller atoms or ions can fit. These holes are particularly relevant in alloys where smaller atoms are inserted in the spaces between larger host atoms.

In an fcc structure, for every full atom present, there are generally two octahedral holes available. These holes derive their name from the geometric arrangement, as they are surrounded by six atoms forming an octahedral shape.

• Total atoms in one fcc unit cell: 7
• Number of octahedral holes from those atoms: \( 7 \times 2 = 14 \)

In practice, when talking about a scaled-up unit cell context as done in interstitial alloys, you consider these holes as potential sites for smaller atoms like B in the exercise.

atomic ratio

The "atomic ratio" is a way of expressing the proportion of one type of atom relative to another within a compound or alloy. In the exercise, the ratio is given as one atom of B for every five atoms of A.

Understanding this ratio helps in determining the distribution of atoms within a unit cell and, consequently, in the calculation of features like hole occupancy in an alloy.
For example, if there are 5 atoms of A, and one atom of B, together contributing to a cell composition:

• Number of A atoms: 5
• Number of B atoms: 1
• Total number of atoms per unit relationship: \( 5 + 1 = 6 \)

Maintaining these ratios is crucial for calculating the distribution of elements and fitting of atoms into octahedral holes.

alloy structure

The "alloy structure" refers to the arrangement of different types of atoms within a compound formed by mixing elements, typically metals. Alloys often improve material properties such as strength, durability, and corrosion resistance.

There are different types of alloy structures, among which an interstitial alloy is significant. In such a structure, smaller atoms fit into the spaces or holes between larger host atoms. This not only affects the crystallography of the material but also its mechanical properties.
In the exercise, atoms of B occupy the octahedral holes in an fcc lattice of host A elements:

• Host element A: Forms the fcc structure
• Interstitial element B: Occupies octahedral holes
• Fraction of octahedral holes filled: Given by \( \frac{1}{12} \) in the exercise

This distribution ensures the alloy retains the stability of the host fcc structure while gaining new material properties from the inclusion of B atoms.