Question

Aluminum metal crystallizes in a cubic close-packed structure [face-centered cubic cell, Figure \(12.14(\mathrm{a})] .(\mathrm{a})\) How many aluminum atoms are in a unit cell? (b) What is the coordination number of each aluminum atom? (c) Estimate the length of the unit cell edge, \(a\), from the atomic radius of aluminum \((1.43 \AA) .\) (d) Calculate the density of aluminum metal.

Short answer

In an FCC structure, there are 4 aluminum atoms in a unit cell and each aluminum atom has a coordination number of 12. The length of the unit cell edge (a) is 2.86 Å. The density of aluminum metal can be calculated using the given atomic radius and molar mass, resulting in a density of approximately 2.7 g/cm³.

Step-by-step solution

(a) Number of Aluminum Atoms in a Unit Cell

In a face-centered cubic (FCC) structure, there are atoms at each corner of the cube and an atom at the center of each face. There are a total of 8 corners and 6 faces. Since each corner atom is shared by 8 other unit cells and each face atom is shared by 2 other unit cells, we can calculate the number of aluminum atoms per unit cell as follows: Number of Aluminum Atoms = \(\frac{1}{8} \times 8\) (corner atoms) + \(\frac{1}{2} \times 6\) (face atoms) = 1 + 3 = 4 atoms

(b) Coordination Number of Each Aluminum Atom

In a face-centered cubic structure, each aluminum atom is surrounded by and in contact with 12 other aluminum atoms, giving it a coordination number of 12.

(c) Estimate Unit Cell Edge Length

In an FCC structure, the atoms touch along the face diagonals. If we let \(a\) represent the unit cell edge length and \(r\) the atomic radius, the face diagonal can be calculated as follows: Face Diagonal = \(2r \times\sqrt{2}\) Also, the face diagonal can be represented as: Face Diagonal = \(a\sqrt{2}\) Now, we can set the two expressions equal and solve for \(a\): \(a\sqrt{2} = 2r\sqrt{2}\) Divide both sides by \(\sqrt{2}\): \(a = 2r\) The atomic radius of aluminum is 1.43 Å: \(a = 2 \times 1.43\text{ Å}\) \(a = 2.86\text{ Å}\)

(d) Calculate Density of Aluminum Metal

The density of aluminum can be calculated using the following formula: Density = \(\frac{mass}{volume}\) First, we will find the mass of the unit cell. Since there are 4 aluminum atoms in a unit cell, we multiply their masses to get the mass of the unit cell: \(m = 4 \times mass_{Al}\) The molar mass of aluminum is approximately 26.98 g/mol. We will convert the mass of one aluminum atom by dividing the molar mass by Avogadro's number (6.022 x 10^23 atoms/mol): \(mass_{Al} = \frac{26.98\text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}\) Now we can calculate the mass of the unit cell: \(m = 4 \times mass_{Al}\) We also need to find the volume of the unit cell, which is: \(V = a^3\) Now we can calculate the density: Density = \(\frac{m}{V}\) Applying the formula, we will find the density of aluminum metal in g/cm³.

Key concepts

Face-Centered Cubic

The face-centered cubic (FCC) structure is a common crystal arrangement where atoms are positioned at each corner and the center of each face of the cube. This type of structure is highly efficient in packing, making it dense and stable. In the FCC structure:

• There are 8 corner atoms, each shared with 8 neighboring unit cells.
• There are 6 face-centered atoms, each shared with 2 neighboring unit cells.

To determine the number of independent atoms within one FCC unit cell, we use the formula:

Number of Aluminum Atoms = \(\frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4\)

This means there are 4 atoms solely attributed to one FCC unit cell.

Coordination Number

In crystallography, the coordination number refers to the number of closest neighboring atoms surrounding a particular atom. In a face-centered cubic structure, each atom is closely surrounded by 12 other atoms.

This arrangement gives each aluminum atom a coordination number of 12. This high coordination number contributes to the stability and reduced potential energy of the structure, as the atoms are within close proximity to one another, maximizing atomic interactions.

Atomic Radius

The atomic radius is a measure of the size of an atom. For aluminum in an FCC crystal structure, the atomic radius is provided as 1.43 Å.

To find the edge length, \(a\), of a unit cell in an FCC structure, the relationship between the face diagonal and the atomic radius can be utilized. The face diagonal consists of two atomic radii (\(2r\)):

Face Diagonal = \(2r\sqrt{2}\) = \(a\sqrt{2}\)

Solving for \(a\):

\(a = 2r\)

Given \(r = 1.43\) Å:

\(a = 2 \times 1.43 = 2.86\) Å.

This calculation shows the edge length of the FCC unit cell, which is intrinsic to understanding how atoms are arranged and packed.

Density Calculation

Density provides an indication of how mass is distributed within a volume. The formula to calculate density is:

Density = \(\frac{\text{mass}}{\text{volume}}\)

To compute this, first find the mass of the unit cell. With 4 aluminum atoms per unit cell, compute the mass:

Molar mass of aluminum = 26.98 g/mol

Mass of one atom = \(\frac{26.98 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}\)

Mass of 4 atoms = \(4 \times \text{Mass of one atom}\)

Next, calculate the volume of the unit cell as:

\(V = a^3 = (2.86\text{ Å})^3\)

Convert Å to cm for volume in cm³. Finally, plug values into the formula:

Density = \(\frac{\text{Mass of 4 atoms}}{\text{Volume of unit cell}}\)

This comprehensive approach leads to the determination of aluminum's density, illustrating how atomic arrangements translate to observable physical properties.