Question

Crystalline Structures and Unit Cells Platinum crystallizes with the face-centered cubic unit cell. The radius of a platinum atom is 139 pm. Calculate the edge length of the unit cell and the density of platinum in \(\mathrm{g} / \mathrm{cm}^{3}\) .

Short answer

The edge length of the unit cell is approximately 389.71 pm, and the density of platinum is approximately 21.45 g/cm^3.

Step-by-step solution

Calculate the Edge Length of the Unit Cell

For a face-centered cubic (fcc) unit cell, the atoms on the face centers and corners touch along the face diagonal. The face diagonal can be represented in terms of the atom's radius (r) with the formula: face diagonal = 4r. Since the face diagonal of the cube also corresponds to the square root of 2 times the edge length (a), we can write \( \sqrt{2} \cdot a = 4r \). Solve for a to find the edge length: \( a = \frac{4r}{\sqrt{2}} \). Now, substitute the value of the radius of platinum (139 pm): \( a = \frac{4 \cdot 139 \mathrm{pm}}{\sqrt{2}} \) and calculate the edge length.

Calculate the Volume of the Unit Cell

The volume (V) of the unit cell is the cube of the edge length (a), \( V = a^3 \). Use the edge length calculated from the previous step to find the volume of the unit cell.

Determine the Number of Platinum Atoms per Unit Cell

In a face-centered cubic unit cell, there are 8 corner atoms each contributing 1/8 to the unit cell, and 6 face atoms each contributing 1/2 to the unit cell. Therefore, there are a total of \( 8 \cdot \frac{1}{8} + 6 \cdot \frac{1}{2} = 4 \) platinum atoms per unit cell.

Calculate the Mass of Platinum Atoms in the Unit Cell

Knowing the number of platinum atoms per unit cell and that one mole of platinum atoms has a mass of approximately 195.08 grams, determine the mass of platinum atoms in the unit cell by using Avogadro's number (6.022 \( \times \) 10^23 atoms/mole): \( \text{Mass of 4 atoms} = \frac{4 \times 195.08 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \).

Calculate the Density of Platinum

Density (D) is mass (m) divided by volume (V). Use the mass of platinum atoms in the unit cell obtained from the previous step and the volume of the unit cell calculated in step 2 to find the density of platinum in \( \mathrm{g} / \mathrm{cm}^{3} \): \( D = \frac{m}{V} \). Convert the unit cell volume from \( \mathrm{pm}^{3} \) to \( \mathrm{cm}^{3} \) by remembering that 1 pm = 10^{-12} cm before doing the division.

Key concepts

Crystalline Structure

When we talk about the crystalline structure of a material, we refer to the highly ordered arrangement of atoms, molecules, or ions that extend in all three spatial dimensions. In the case of platinum, we are dealing with a face-centered cubic (fcc) unit cell, a common type of crystalline structure for metals. In such a structure, atoms are located at each of the eight corners and the centers of all the faces of the cube, resulting in a highly symmetrical layout.

This arrangement greatly affects the material's properties, such as density and strength. The fcc structure is particularly dense, as it allows more atoms to be packed in a given volume compared to other structures, and this contributes to the overall high density of platinum.

Atomic Radius

The atomic radius is a measure of the size of an atom's ion core—the distance from the center of the nucleus to the boundary of the surrounding cloud of electrons. Knowing the atomic radius of an element, such as platinum, allows us to deduce several important physical properties.

For the face-centered cubic unit cell, the atom’s diameter can be directly related to the edge length of the cell. This is due to the geometric considerations of how atoms in an fcc structure pack together: specifically, that atoms along the face diagonal touch each other. As the radius of the platinum atom is given as 139 picometers (pm), we can calculate the cell's edge length using the geometric relationship that exists in an fcc crystal.

Unit Cell Volume

Understanding the volume of a unit cell is crucial when characterizing the crystalline structure of a substance. The unit cell volume is simply the cube of the edge length (in the case of cubic unit cells).

Since we can express the edge length 'a' of the cube in terms of the atomic radius for a face-centered cubic lattice, we're also able to calculate the volume 'V' by cubing this edge length: \( V = a^3 \). Knowing the volume is essential for further calculations, such as finding the density of the substance.

Density Calculation

Density is one of the most fundamental physical properties of a material, defined as mass per unit volume. In the context of a crystalline solid like platinum, calculating density requires knowing both the mass of atoms within a unit cell and the volume of the unit cell.

We can determine the mass of the unit cell by considering the number of atoms it contains and the mass of each atom. For platinum, there are 4 atoms per unit cell in its face-centered cubic lattice. Knowing the atomic mass of platinum and utilizing Avogadro’s number allows us to calculate the total mass of these atoms. Afterwards, the density is found by dividing this mass by the volume of the unit cell (converted into cubic centimeters), yielding the substance’s density in \(\text{g}/\text{cm}^3\), a critical value for various engineering and scientific applications.