Problem 55 - 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 radius of the iridium atom is approximately \(135.4 \, pm\).

Step 1: Understand the Face-Centered Cubic (FCC) Unit Cell

In a face-centered cubic (FCC) unit cell, there are atoms located at each corner of the cube and one atom centered on each face. Each corner atom is shared among eight unit cells, and each face-centered atom is shared with just one adjacent unit cell. Therefore, the total number of atoms in one FCC unit cell is 4.

Step 2: Relationship Between Edge Length and Atomic Radius

In a face-centered cubic unit cell, the relationship between the edge length of the cube ()L) and the radius of the atom ()r) is given by the formula \(L = 2\sqrt{2}r\).

Step 3: Calculate the Volume of the Unit Cell

The volume of the unit cell (V) can be expressed in terms of the edge length, \( V = L^3 \). To find \( L \), we will use the formula \(L = 2\sqrt{2}r\). However, we need the atomic radius, which we will find using density.

Step 4: Use Density Formula to Find Edge Length

The density of the substance ()\(\rho\)) is given by \(\rho = \frac{mass}{volume}\). Therefore, for a unit cell, \(\rho = \frac{Z \times M}{N_A \times V}\), where:- \(Z\) is the number of atoms per unit cell (4 for FCC structure) - \(M\) is the molar mass of iridium (approx. 192.22 g/mol)- \(N_A\) is Avogadro's number (6.022 x 10^23 atoms/mol)- \(V\) is the volume of the unit cellRearranging gives \(V = \frac{Z \times M}{N_A \times \rho}\). Substitute the known values to find \(V\).

Step 5: Solve for Edge Length and Then for Atomic Radius

Once you have found \(V\), solve for \(L\) using \(L = V^{1/3}\). Then substitute \(L\) back into the relationship \(L = 2\sqrt{2}r\) to solve for \(r\). This will give you the atomic radius of an iridium atom.

Key Concepts

Understanding Atomic Radius

The atomic radius is a crucial concept in the study of materials, particularly in determining the dimensions of atoms within a crystalline structure. To envision the atomic radius, imagine an atom as a tiny sphere. The radius is essentially the distance from its nucleus to the outer boundary of its electron cloud. This measurement helps in understanding how atoms stack together in a solid form.
The atomic radius is pivotal in the context of crystal structures like the face-centered cubic (FCC) unit cell. In an FCC cell, the atomic radius helps define the dimensions and geometry of the cube. To calculate it from the cube's edge length, use the formula: \( L = 2\sqrt{2}r \).

• The edge length \( L \) is the distance across the cube's side.
• The atomic radius \( r \) is then derived from the known measurements of the cube itself.

In case of iridium, calculating its atomic radius gives insights into its density and how tightly packed its atoms are.

Mastering Density Calculations

Density calculation is an important exercise in understanding the compactness of matter, especially in metals like iridium. Density (\(\rho\)) is defined as mass per unit volume and provides insights into how closely packed the atoms in a substance are. For iridium, with its highly dense nature, this calculation can indicate how many atoms fit into a specific volume.
To determine the density of a unit cell, use the formula:

• \( \rho = \frac{Z \times M}{N_A \times V} \), where:
• \( Z \) is the number of atoms per unit cell, which is 4 for FCC structures.
• \( M \) is the molar mass of iridium, approximately 192.22 g/mol.
• \( N_A \) is Avogadro's number, approximately 6.022 x 10²³ atoms/mol.
• \( V \) is the unit cell volume.

The value of these components can be integrated into the formula to find the density, reinforcing the understanding of how tightly atoms are packed in the crystalline structure.

Iridium's Distinct Characteristics

Iridium is a rare, dense metal known for its impressive physical and chemical properties. It sits in the platinum group and displays high corrosion resistance and melting point, making it valuable in industrial applications and scientific research.
Being in the platinum group, iridium has a significant molar mass of approximately 192.22 g/mol, contributing to its high density. This high atomic weight, combined with its face-centered cubic (FCC) unit cell structure, results in an exceptionally dense metal.
When approaching exercises involving iridium, its density plays a crucial part. It signifies how atoms are compactly packed within its structure, resulting in remarkable material characteristics. The analysis of iridium, such as calculating its atomic radius, benefits from assuming its highly organized and dense atomic packing.

Calculating Unit Cell Volume

The unit cell volume is a significant aspect of crystallography, crucial in understanding the spatial arrangement of atoms in a solid. For a face-centered cubic (FCC) unit cell, the volume (\(V\)) is determined by the cube's edge length (\(L\)) using the formula: \( V = L^3 \).
To calculate the unit cell volume:

• First, determine the edge length with two key components: the atomic radius \(r\) and the formula \(L = 2\sqrt{2}r\).
• With this edge length, compute the volume by cubing it \(L^3\).

This calculated volume is then used in density formulas to reveal how atoms fit together in a solid. Understanding unit cell volume helps to comprehend material physical properties, pointing to how substances like iridium maintain their distinct densities and structures.