Problem 79 - Chemistry An Atoms First Approach

Question

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69 ). Given that the density of cesium chloride is \(3.97 \mathrm{g} / \mathrm{cm}^{3},\) and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm},\) and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

Step-by-Step Solution

Answer

The calculated distance between adjacent Cs+ and Cl- ions in the cesium chloride structure is 3.01 Å, which is relatively close to the expected distance of 3.50 Å based on the sizes of the ions. This result supports the assumption that the ions touch along the body diagonal of the unit cell.

Step 1: Calculate the molar mass of cesium chloride

The molar mass of CsCl is the sum of the molar masses of cesium and chlorine: Molar mass of Cs = 132.9 g/mol (from the periodic table) Molar mass of Cl = 35.45 g/mol (from the periodic table) Molar mass of CsCl = 132.9 g/mol + 35.45 g/mol = 168.35 g/mol

Step 2: Find the volume of one unit cell

To find the volume of one unit cell, we can use the formula: Density = (Mass of one unit cell) / (Volume of one unit cell) We know the density of cesium chloride is 3.97 g/cm³, and there is one formula unit of CsCl per unit cell, so the mass of one unit cell is the molar mass of CsCl. Mass of one unit cell = 168.35 g/mol Using the density formula: 3.97 g/cm³ = (168.35 g/mol) / (Volume of one unit cell) Solving for the volume of one unit cell: Volume of one unit cell = (168.35 g/mol) / (3.97 g/cm³) = 42.42 cm³/mol

Step 3: Calculate the edge length of the unit cell

The unit cell is a cube, so its volume can be expressed as: Volume of one unit cell = (Edge length)³ We have the volume of one unit cell from step 2. Now we can find the edge length: 42.42 cm³/mol = (Edge length)³ Taking the cube root of both sides: Edge length = (42.42 cm³/mol)^(1/3) = 3.48 Angstroms (Å)

Step 4: Calculate the body diagonal of the unit cell

The body diagonal of a cube can be found using the Pythagorean theorem in 3D: Body diagonal = √(Edge length² + Edge length² + Edge length²) Body diagonal = √(3.48 Ų + 3.48 Ų + 3.48 Ų) = 6.02 Å

Step 5: Calculate the distance between adjacent Cs+ and Cl- ions

The distance between adjacent Cs+ and Cl- ions is half the body diagonal of the unit cell: Distance between adjacent Cs+ and Cl- ions = (1/2) × 6.02 Å = 3.01 Å

Step 6: Compare the calculated distance with the expected distance based on ionic sizes

The expected distance between adjacent Cs+ and Cl- ions can be calculated by adding their ionic radii: Expected distance = Ionic radius of Cs+ + Ionic radius of Cl- = 169 pm + 181 pm = 350 pm Converting this value to angstroms, since our calculated distance is in angstroms: Expected distance = 350 pm × (1 Å/100 pm) = 3.50 Å Now we can compare the calculated distance (3.01 Å) and the expected distance (3.50 Å): The difference between the calculated distance and the expected distance is relatively small (0.49 Å), which indicates that our assumption that the ions touch along the body diagonal of the unit cell is reasonable.

Key Concepts

Cesium Chloride

The cesium chloride structure is a classic example of a simple cubic crystal arrangement, often used to illustrate fundamental concepts in solid-state chemistry. In the CsCl arrangement, each cesium ion ( Cs^{+} ) is surrounded by eight chloride ions ( Cl^{-} ), forming a cubic shape. Conversely, each chloride ion is also adjacent to eight cesium ions. This particular setup results in a coordination number of eight for both ion types which is slightly different from more common ionic crystal structures like that of sodium chloride (NaCl), where the coordination number is six. Understanding the CsCl structure helps students comprehend how ions pack in a way that balances forces and optimizes stability. It's a practical example of how three-dimensional geometry impacts the physical properties and density of ionic compounds.

Unit Cell Calculations

Unit cell calculations are essential in determining the physical dimensions and arrangement within a crystalline solid. The unit cell is the smallest repeating structure in the crystal, and its calculation can involve finding edge lengths, volumes, and the body diagonal—vital for understanding how ions like Cs^{+} and Cl^{-} interact spatially.

• Volume Calculation: The volume of the unit cell is inversely proportional to the density, knowledge that marks a crucial step in calculating unit cell dimensions.
• Edge Length: Once the volume is known, the cube root gives the edge length, a measure in nanometers or Angstroms often revealing more about crystal packing.
• Body Diagonal: A key calculation using the Pythagorean Theorem that helps find distances between ions situated along the diagonal of the cube.

This framework is fundamental in solid-state physics and material science, aiding in predicting crystal behavior under various conditions.

Ionic Radius

The ionic radius is a measurement of the size of ions in a crystal lattice, typically measured in picometers (pm) or Angstroms (Å). In the case of the cesium chloride lattice, the ionic radii of cesium ( Cs^{+} ) and chloride ( Cl^{-} ) need to be considered for precise calculations.

• Cs^{+} Ion: It has an ionic radius of 169 pm. This relatively large size reflects its place in the alkaline metals and its readiness to lose an electron and form a cation.
• Cl^{-} Ion: With an ionic radius of 181 pm, chloride ions are large anions, stable due to their full electron shell configuration.

Adding these radii helps in predicting the distance between ions within a crystal, which is crucial for understanding the physical dimensions and characteristics of the material.

Density of CsCl

The density of cesium chloride ( 3.97 ext{ g/cm}^{3} ) is an essential parameter when studying its crystal structure. Density provides insights into how tightly packed the ions are within the structure and has implications for the material's stability and physical properties.

• Mass and Volume Relation: By using the formula ext{Density} = rac{ ext{Mass}}{ ext{Volume}} , one can deduce the volume of a single unit cell when given the density and molecular mass.
• Effective Mass: For these calculations, the effective mass of a unit cell equates to the molar mass as we consider only one formula unit per cell.
• Implications: Higher densities typically indicate tighter packing which could imply stronger ionic attractions or larger effective interactions across the lattice.

By understanding the density, scientists and students can infer many additional properties of the solid, revealing depths of intermolecular forces and material strengths.