Question

Aluminum metal crystallizes in a face-centered cubic unit cell. (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 ?). (d) Calculate the density of aluminum metal.

Short answer

(a) In a face-centered cubic (fcc) unit cell, there are 7 aluminum atoms. (b) The coordination number of aluminum in this structure is 6. (c) The unit cell edge length, 'a', is estimated to be 4.05 Å using the atomic radius of aluminum (1.43 Å). (d) The density of aluminum metal is calculated to be 2.70 g/cm³.

Step-by-step solution

(a) Number of Aluminum Atoms in a Unit Cell)

In a face-centered cubic (fcc) unit cell, there are atoms located at the 8 corners of the cube and one atom at the center of each of the 6 faces. So, to find the total number of aluminum atoms in a unit cell, we need to account for the atoms at the corners and the atoms at the center of each face. There are 8 corner atoms in a unit cell, and each contributes 1/8 to the total unit cell. For the face-centered atoms, there are 6 faces and each has an atom in the center, which fully contributes to the unit cell. So, the number of aluminum atoms in a unit cell is: \(8 \times \frac{1}{8} + 6 \times 1 = 1 + 6 = 7\) atoms.

(b) Coordination Number of Aluminum)

The coordination number of an atom is the number of nearest neighbors it has. In a face-centered cubic unit cell, an atom at the center of a face is in contact with 4 corner atoms in the same plane (in the four directions) and one atom above and one below the face. Thus, the coordination number of aluminum in this structure is 6.

(c) Estimating Unit Cell Edge Length)

From the atomic radius of aluminum (1.43 Å), we can estimate the length of the unit cell edge, 'a'. In an fcc structure, the unit cell edge length is related to the atomic radius (r) as follows: \(a = 2\sqrt{2}r\) Plugging the value of r (1.43 Å) into the equation, we get: \(a = 2\sqrt{2} \times 1.43 = 4.05\) Å

(d) Calculating Aluminum Density)

To calculate the density of aluminum, we can use the formula: \(\rho = \frac{mass}{volume}\) The mass of one aluminum atom is given by its atomic weight divided by Avogadro's number: \(mass = \frac{26.98\, g/mol}{6.022 \times 10^{23}\, atoms/mol}\) We know that there are 7 aluminum atoms in a unit cell and the volume of a unit cell is given by \(V = a^3\), where 'a' is the unit cell edge length. So, the mass of 7 aluminum atoms in the unit cell is: \(mass_{7-atoms} = 7 \times mass\) Now we can find the density of aluminum metal: \(\rho = \frac{mass_{7-atoms}}{V} = \frac{7 \times mass}{a^3}\) Substituting the values of 'mass' and 'a' from previous steps, we get the density of aluminum: \(\rho = \frac{7 \times (26.98\, g/mol) / (6.022 \times 10^{23}\, atoms/mol)}{(4.05\, Å)^3 \times (10^{-8}\, cm/Å)^3}\) \(\rho = 2.70\, g/cm^3\) So, the density of aluminum metal is 2.70 g/cm³.

Key concepts

Coordination Number

Understanding the coordination number of an atom is crucial when studying crystal structures such as the face-centered cubic (fcc) unit cell typical of aluminum. In essence, the coordination number is the total count of nearest neighbor atoms surrounding a selected atom. In the fcc lattice, each aluminum atom connects with 12 immediate neighboring atoms: 4 in the same layer, 4 in the layer above, and 4 in the layer below.

This spatial configuration offers a stable structure due to the uniform distribution of atomic contacts. For students grappling with visualizing complex 3D structures, imagine the central atom as the center of a clock face, with neighbors located at 3, 6, 9, and 12 o'clock positions on both the upper and lower planes, effectively surrounding the central atom.

Atomic Radius

The atomic radius is a measure from the center of an atom to the outer boundary of its electrons. This distance is pivotal when making calculations related to the unit cell dimensions, especially in metals like aluminum with an atomic radius of 1.43 Ångströms (Å).

In a face-centered cubic unit cell, the relationship between the atomic radius and the unit cell edge (a) can be visualized by drawing a diagonal across a face of the cube that connects the centers of atoms, forming a right triangle. The hypotenuse is the face diagonal, which equals to the sum of the radii of 4 atoms. By applying Pythagoras' theorem, this forms the basis for calculating the edge length of the unit cell.

Unit Cell Calculations

Unit cell calculations provide a means to delve deeper into the crystal structure of materials. When we determine the edge length of a unit cell from the atomic radius in a face-centered cubic structure, the formula is as intuitive as it is essential: \(a = 2\sqrt{2}r\).

Once the unit cell edge length is known, various other properties, such as the cell's volume and the spacing between atoms, become calculable, making it possible to predict how atoms will pack together in a solid. These calculations are fundamental to materials science as they allow predictions of material properties based on their microstructures.

Density of Aluminum

Density, often denoted by the Greek letter rho (\(\rho\)), is a property of materials that describes mass per unit volume. In the case of aluminum, calculated using the unit cell approach, density is a function of the mass of the 7 aluminum atoms present in the fcc unit cell and the volume of the cell itself.

Calculating the density of aluminum involves determining the mass of a single atom, multiplying by the number of atoms in the unit cell, and then dividing by the volume of the unit cell. Given that density directly correlates with how tightly atoms are packed in a material, it plays a key role in understanding and predicting a material's mechanical and physical properties. For instance, a higher density often implies a material with higher strength and weight, critical factors in materials engineering and selection for specific applications.