Problem 41 - Chemistry and Chemical Reactivity

Question

The very dense metal iridium has a face-centered cubic unit cell and a density of \(22.56 \mathrm{g} / \mathrm{cm}^{3} .\) Use this information to calculate the radius of an atom of the element.

Step-by-Step Solution

Answer

The atomic radius of an iridium atom is approximately \(1.356 \times 10^{-8}\) cm.

Step 1: Understanding the Unit Cell

In a face-centered cubic (FCC) unit cell, there are atoms at each corner of the cube and one atom on each face of the cube. This configuration effectively results in 4 atoms per unit cell because each corner atom is shared by 8 unit cells and each face atom is shared by 2 unit cells.

Step 2: Calculate the Volume of the Unit Cell

Given that the density \(\rho\) of iridium is \(22.56 \mathrm{g/cm}^{3}\), we can use the relationship \(\rho = \frac{M}{V}\), where \(M\) is the mass of the unit cell and \(V\) is its volume. The mass of one mole of iridium atoms (its molar mass) is \(192.22 \mathrm{g/mol}\). We first need to find the mass of the unit cell: \(M = \frac{192.22 \mathrm{g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \times 4 \text{ atoms} = 1.282 \times 10^{-21} \text{ g}\). Now solve for \(V\): \(V = \frac{M}{\rho} = \frac{1.282 \times 10^{-21} \text{ g}}{22.56 \text{ g/cm}^{3}} = 5.683 \times 10^{-23} \text{ cm}^{3},\) which is the volume of the unit cell.

Step 3: Determine the Edge Length of the Unit Cell

The volume \(V\) of a cube is given by \(a^3\), where \(a\) is the edge length of the cube. Therefore, \(a = \sqrt[3]{V} = \sqrt[3]{5.683 \times 10^{-23} \text{ cm}^3} = 3.833 \times 10^{-8} \text{ cm}.\)

Step 4: Calculate the Atomic Radius

In a face-centered cubic lattice, the relationship between the atomic radius \(r\) and the edge length \(a\) is \(a = \sqrt{8}r\). Rearrange this to get \(r = \frac{a}{\sqrt{8}}\). Substituting in \(a = 3.833 \times 10^{-8} \text{ cm}\), we get \(r = \frac{3.833 \times 10^{-8} \text{ cm}}{\sqrt{8}} = 1.356 \times 10^{-8} \text{ cm}.\)

Key Concepts

Face-Centered Cubic Unit Cell

The face-centered cubic (FCC) unit cell is a highly efficient arrangement of atoms in a crystal structure. In this configuration, atoms are positioned at each of the cube's eight corners and in the center of each of the cube's six faces. It may seem like there are more atoms than that, but let's break down how many atoms actually belong to one FCC unit cell.

• Each corner atom is shared among 8 adjacent unit cells. Hence, each corner contributes only 1/8th of an atom to the cell.
• The face-centered positions are shared between two unit cells, accounting for half an atom per face.
• As a result, an FCC unit cell effectively contains 4 complete atoms: 8 corners × 1/8 + 6 faces × 1/2 = 4 atoms.

Understanding this arrangement is crucial for calculating various properties, such as density and atomic radius, as it defines how much space the atoms take up.

Atomic Radius Calculation

In the context of the face-centered cubic structure, calculating the atomic radius involves understanding its relation to the edge length of the unit cell. The distance between the centers of two adjacent atoms on the face diagonals gives insight into this measurement. This geometric relationship is based on the Pythagorean theorem in three dimensions.

• The diagonal across a face of the cube consists of a path through two atomic radii, lying through the center atom and half of the other two corner atoms.
• The equation in a face-centered cubic unit cell is expressed as: \( a = \sqrt{8}r \), where \(a\) is the edge length and \(r\) is the atomic radius.

To find the atomic radius \(r\), simply rearrange this equation: \[ r = \frac{a}{\sqrt{8}} \] Plugging in the known edge length \(a\) into this formula will yield the radius, allowing further calculations and insights into the material's properties.

Volume of a Unit Cell

When calculating the density or volume of a unit cell, the notion of volume for the cube arrangement is crucial. The volume \( V \) of a cube can be determined easily once the edge length \( a \) is known.

• The relationship is direct: \( V = a^3 \), where \(a\) is the cube’s edge length.
• To find \(a\), you can derive it from the unit cell's mass \(M\), as \( M = \rho \times V \), where \( \rho \) is the density.

Using given values, you can rearrange this expression to find the volume: \[ V = \frac{M}{\rho} \] Finally, knowing the volume of the unit cell is vital for deducing the spatial arrangement and properties of substances, particularly in crystalline materials.