Problem 36 - Chemistry The Central Science In Si Units

Question

Barium crystallizes in a body-centered cubic structure. (a) How many Ba atoms are contained in each unit cell? (b) How many nearest neighbors does each Ba atom possess? (c) Estimate the length of the unit cell edge, \(a\), from the atomic radius of barium \((0.215 \mathrm{nm}) .\) (d) Estimate the density of Ba metal at this temperature.

Step-by-Step Solution

Answer

(a) 2 Ba atoms (b) 8 nearest neighbors (c) 0.497 nm (d) 3.50 g/cm³

Step 1: Calculating Ba Atoms per Unit Cell

In a body-centered cubic (bcc) structure, each unit cell contains 2 atoms. This is because it has one atom at each corner, each shared by 8 unit cells, and one atom in the center that is only part of the one unit cell. Therefore, the total number of barium atoms per unit cell is calculated as: \(8 \times \frac{1}{8} + 1 = 2\).

Step 2: Determining Nearest Neighbors

For a body-centered cubic structure, each atom has 8 nearest neighbors. This is due to the atom in the center being surrounded by 8 corner atoms, each equidistant from it.

Step 3: Estimating Length of Unit Cell Edge

In a body-centered cubic structure, the relationship between the atomic radius \(r\) and the cube edge length \(a\) is given by the formula \(a = \frac{4r}{\sqrt{3}}\). Given the atomic radius \(r = 0.215\,\text{nm}\), substitute to find \(a\): \[a = \frac{4 \times 0.215}{\sqrt{3}} = 0.497\,\text{nm}.\]

Step 4: Estimating Density of Barium

To find the density \(\rho\) of barium, use the formula:\[ \rho = \frac{Zn_A}{a^3 \times N_A} \]where \(Z\) is the number of atoms per unit cell (2 for bcc), \(n_A\) is the atomic mass of barium (137.33 g/mol), \(a\) is the edge length (0.497 nm = 4.97 \times 10^{-8} cm), and \(N_A\) is Avogadro's number (6.022 \times 10^{23} mol^{-1}). Substituting these values gives:\[ \rho = \frac{2 \times 137.33}{(4.97 \times 10^{-8})^3 \times 6.022 \times 10^{23}} \approx 3.50\,\text{g/cm}^3.\]

Key Concepts

Understanding the Unit Cell in a Body-Centered Cubic (BCC) Structure

In a crystal, a unit cell is the smallest repeating unit that shows the symmetry and structure of the entire crystal lattice. For barium, which crystallizes in a body-centered cubic (bcc) structure, the unit cell is a cube containing atoms at its corners and one additional atom at the center.

In a bcc lattice:

• Each of the 8 corners contributes 1/8 of an atom to the unit cell because each corner atom is shared by 8 different unit cells.
• The center atom is fully contained within the unit cell.

Thus, the number of atoms per unit cell in a bcc structure is calculated as:\[8 \times \frac{1}{8} + 1 = 2 \text{ atoms per unit cell}.\]This arrangement maximizes packing efficiency while maintaining symmetry.

Exploring the Concept of Atomic Radius

The atomic radius is the distance from the center of an atom's nucleus to the outermost shell of electrons. It determines how tightly atoms can pack together in a crystal lattice. In a bcc structure, the atomic radius plays a crucial role in defining the edge length of the unit cell.

For instance, in barium's bcc structure:

• Each central atom is in direct contact with opposite corner atoms along the body diagonal of the cube.
• The relationship between the atomic radius \(r\) and the cube edge length \(a\) is given by: \[a = \frac{4r}{\sqrt{3}}.\]

Given that barium’s atomic radius is 0.215 nm, we can calculate the edge length as \(0.497\,\text{nm}\), providing a foundation to estimate other physical properties such as density.

Nearest Neighbors in the Body-Centered Cubic Structure

In crystallography, nearest neighbors are the closest atoms surrounding a given central atom. They are crucial for understanding bonding and properties like stability and solidity.

In a bcc lattice:

• Every atom inside the cell wishes to bond with the surrounding atoms.
• For the central barium atom, it is equidistantly surrounded by 8 corner atoms.

This interaction means each atom in a bcc structure like barium has 8 nearest neighbors. These neighboring atoms share faces with the original cube, contributing to a firm and durable crystal matrix.

Calculating the Density of Barium in a BCC Structure

Density is a fundamental property reflecting how much mass is contained in a given volume. Calculating barium's density helps us understand how atoms pack together to form a solid.

To estimate the density of barium in a bcc structure, we use the formula:\[\rho = \frac{Z n_A}{a^3 \times N_A}\]

• Where \(Z=2\) is the number of atoms per unit cell.
• \(n_A\) is the atomic mass (137.33 g/mol for barium).
• \(a\) is the edge length (converted to cm, \(4.97 \times 10^{-8}\,\text{cm}\)).
• \(N_A\) is Avogadro's number \((6.022 \times 10^{23} \text{ mol}^{-1})\).

Using these values, barium's density calculates approximately to \(3.50\,\text{g/cm}^3\), indicating its relatively high compaction and solid state.