Problem 86 - Chemistry and Chemical Reactivity

Question

A The density of lead is \(11.350 \mathrm{g} / \mathrm{cm}^{3},\) and the metal crystallizes in a face-centered cubic unit cell. Estimate the radius of the lead atom.

Step-by-Step Solution

Answer

The radius of a lead atom is approximately 0.175 nm.

Step 1: Understanding the Relationship Between the Density and Atomic Radius

To find the radius of a lead atom in a face-centered cubic (FCC) unit cell, we first must understand the relationship between the density of a substance and the dimensions of its unit cell. The density is related to the mass and volume of the unit cell, and through the geometry of the FCC lattice, we can connect this to atomic radius.

Step 2: Calculating the Mass of the Unit Cell

In an FCC lattice, there are 4 atoms per unit cell. The atomic mass of lead is approximately 207.2 g/mol. To find the mass in grams of a single lead atom, use Avogadro's number: \[\text{Mass of one Pb atom} = \frac{207.2 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 3.44 \times 10^{-22} \text{ g}.\]The total mass of the 4 atoms in one unit cell is:\[4 \times 3.44 \times 10^{-22} \text{ g} = 1.376 \times 10^{-21} \text{ g}.\]

Step 3: Relating Density to Volume of the Unit Cell

The density formula gives us:\[\text{Density} = \frac{\text{mass}}{\text{volume}}.\]Reorganize to obtain the volume using the density of lead:\[\text{Volume} = \frac{1.376 \times 10^{-21} \text{ g}}{11.350 \text{ g/cm}^3} \approx 1.213 \times 10^{-22} \text{ cm}^3.\]

Step 4: Calculating the Unit Cell Edge Length

The volume of a cube with edge length \(a\) is \(a^3\). So, solve for \(a\):\[a^3 = 1.213 \times 10^{-22} \text{ cm}^3 \implies a = \sqrt[3]{1.213 \times 10^{-22}} \approx 3.50 \times 10^{-8} \text{ cm}.\]

Step 5: Finding the Atomic Radius

In an FCC structure, the relationship between the atomic radius \(r\) and the edge length \(a\) is given by \[a = 2\sqrt{2}r.\]Rearrange this to solve for \(r\):\[r = \frac{a}{2\sqrt{2}} = \frac{3.50 \times 10^{-8}\text{ cm}}{2\sqrt{2}} \approx 1.75 \times 10^{-8} \text{ cm}.\]

Step 6: Final Result: Connecting Calculation to Physical Meaning

The mathematical procedures above yield an atomic radius for lead of approximately \(1.75 \times 10^{-8} \text{ cm}\), which can be converted to nanometers for easier comprehension: 1 nm = \(10^{-7}\) cm, so \(1.75 \times 10^{-8} \text{ cm} = 0.175 \text{ nm}\).

Key Concepts

Face-Centered Cubic Unit Cell

The face-centered cubic (FCC) unit cell is one of the most common crystal structures for metals. In an FCC structure, each cube unit cell has one atom at each of its corners and one atom at the center of each face.
This arrangement results in a total of 4 atoms per unit cell. The FCC structure allows atoms to pack tightly, which is efficient and maximizes density.

• The edge length of the FCC unit cell is related to the atomic radius by the formula: \[a = 2\sqrt{2}r\]
• The structure is highly symmetrical and close-packed, meaning it has one of the highest packing efficiencies.

Understanding this structure is crucial for determining properties such as density and atomic radius.

Density

Density is defined as mass per unit volume and is a fundamental property that helps to identify substances and their physical characteristics.
In the context of a face-centered cubic unit cell, density connects the macroscopic world (how heavy a substance feels) to the atomic world. For a metal like lead, density can be calculated using the formula: \[\text{Density} = \frac{\text{mass of unit cell}}{\text{volume of unit cell}}\]
Given the density and mass of individual atoms, one can calculate the unit cell's volume, an essential step in finding atomic dimensions. This relationship helps us to understand why specific metals are heavier or lighter based on their atomic arrangement.

Atomic Mass

Atomic mass is the mass of a single atom, usually expressed in atomic mass units (amu) or grams per mole (g/mol). It's crucial for calculating various physical properties.
For lead, with an atomic mass of approximately 207.2 g/mol, we use Avogadro's number to find the mass of a single atom in grams. This allows us to compute how many lead atoms contribute to the mass of the unit cell.

• One mole of lead consists of \(6.022 \times 10^{23}\) atoms, providing a conversion factor for calculating atomic mass in practical terms.
• The calculated mass of one lead atom helps determine the mass within an entire unit cell, affecting the overall density of the material.

Understanding atomic mass is essential for bridging individual atomic properties and their effects on macroscopic scales.

Unit Cell Volume

The volume of the unit cell is a critical concept for understanding material properties at the atomic level.
For an FCC unit cell, the volume is calculated by determining the edge length \(a\) of the cube and using the equation: \[a^3 = \text{Volume of the unit cell}\]
The balance between mass, volume, and density in calculating the unit cell volume helps us deduce how tightly atoms are packed in a crystal lattice.

• By rearranging the density equation, we can solve for the volume, connecting it directly to measurable properties like density and computed mass.
• Knowing the volume enables us to determine other geometric details, such as the atomic radius, through the lattice structure's relationships.

Comprehending unit cell volume is vital for predicting and explaining the physical behavior of materials.