Question

In solid KCl the smallest distance between the centers of a potassium ion and a chloride ion is \(314 \mathrm{pm} .\) Calculate the length of the edge of the unit cell and the density of KCl, assuming it has the same structure as sodium chloride.

Short answer

The edge length of the unit cell in solid KCl is approximately 887.99 pm, and the density of KCl is approximately 1.98 g/cm³.

Step-by-step solution

Determine the type of crystal structure for KCl and find the relationship between ion distance and edge length.

Since KCl has the same structure as NaCl, it forms a face-centered cubic (fcc) lattice. In an fcc lattice, the edge length (a) of the unit cell can be determined by using the distance d between the centers of two adjacent ions (a potassium ion and a chloride ion). In this case, d is given to be 314 pm. Since the ions touch along the face diagonal of the cube, the relationship between d and the edge length (a) can be expressed as: \(a\sqrt{2} = 4d\)

Calculate the edge length of the unit cell.

Use the relationship from step 1 to find the edge length (a). The expression to find a is: \(a = \frac{4d}{\sqrt{2}}\) Given the distance d=314 pm, we plug in the value into the expression: \(a = \frac{4 \times 314 \mathrm{pm}}{\sqrt{2}} \approx 887.99 \mathrm{pm}\) The edge length (a) of the unit cell is approximately 887.99 pm.

Calculate the number of KCl formula units in a unit cell.

In an fcc lattice, there are 4 KCl formula units per unit cell. This is because there are 8 corner atoms and 6 face-centered atoms: Corner atoms contribution = \(8 \times \frac{1}{8} = 1\) Face-centered atoms contribution = \(6 \times \frac{1}{2} = 3\) Total KCl formula units per unit cell = 1 + 3 = 4

Calculate the mass of a single KCl formula unit.

To determine the density, we need the mass of a single KCl formula unit. We use the molar masses of potassium (K) and chlorine (Cl) to calculate the mass of one KCl formula unit. The molar masses of K and Cl are approximately 39.10 g/mol and 35.45 g/mol, respectively. Therefore, the molar mass of KCl is: Molar mass of KCl = (39.10 g/mol) + (35.45 g/mol) = 74.55 g/mol The mass of a single KCl formula unit can be determined by dividing the molar mass of KCl by Avogadro's number (6.022 x 10^23 mol^-1): Mass of one KCl formula unit = \(\frac{74.55 \mathrm{\frac{g}{mol}}}{6.022 \times 10^{23} \mathrm{mol}^{-1}} \approx 1.239 \times 10^{-22} \mathrm{g}\)

Calculate the density of KCl.

Now, we will use the edge length, number of KCl formula units per unit cell, and the mass of a single KCl formula unit to calculate the density of KCl. The formula for density is: Density = \(\frac{total \ mass}{total \ volume}\) In this case: Total mass in one unit cell = 4 KCl formula units * (mass of one KCl formula unit) g Total volume = (edge length)^3 Therefore, the density of KCl is: Density = \(\frac{4 \times (1.239 \times 10^{-22} \mathrm{g})}{(8.8799 \times 10^{-8} \mathrm{cm})^3} \approx 1.98 \mathrm{g/cm^3}\) The density of KCl is approximately 1.98 g/cm³.

Key concepts

Understanding Crystal Structure

Crystals are solid materials whose constituent atoms, ions, or molecules are arranged in an ordered pattern extending in all three spatial dimensions. The crystal structure denotes the symmetrical arrangement and spatial organization of atoms within a crystal. Each repeating unit of the structure is known as a unit cell, which acts as the basic building block of the crystal.To understand the properties of a crystalline material such as KCl (potassium chloride), we start by examining its smallest repeating structure. Knowing the crystal structure provides insights into various physical properties, including melting point, hardness, and density. The arrangement of ions in a crystal also affects its electrical conductivity and optical properties. In the given exercise, we attribute a face-centered cubic (fcc) lattice to KCl, a structural type both simple and versatile enough to accommodate numerous compounds, thereby greatly influencing the material's characteristics.

Exploring the Face-Centered Cubic Lattice

The face-centered cubic (fcc) lattice is one of the most common and important crystal structures found in metals and ionic compounds. It is characterized by atoms positioned at each of the eight corners of a cube and a single atom at the center of each of the six faces of the cube.In an fcc lattice, each corner atom is shared by eight adjacent unit cells and each face-centered atom is shared by two. Consequently, the coordination number—which is the number of nearest neighbors to an atom—is 12 in an fcc structure. This packing is known for its high density and efficiency.

Applying the fcc Structure to Ion Distance

In the context of ionic crystals, like in the case of KCl, the fcc arrangement helps to understand the spatial relationships between the ions. As demonstrated in the given exercise, the diagonal passing through the face of the unit cell can be used to express the relationship between the ion distance and the edge length of the cell, enabling the calculation of dimensions that directly contribute to the physical nature of the crystal, such as density.

Density Calculation in Crystals

Density is a fundamental property of materials defined as mass per unit volume. The calculation of the density of crystalline solids like KCl involves knowing the mass of formula units within the crystal's unit cell and dividing it by the unit cell's volume.

Counting the Formula Units

The number of formula units in a unit cell can be ascertained by taking into account the contribution of the atoms situated at corners and faces, as demonstrated in the solution. By identifying that each unit cell in an fcc lattice contains four formula units, we can calculate the total mass within one cell.

Determining the Total Volume

With the edge length of the unit cell known, determining the total volume is a straightforward process of cubing the edge length. By inserting this information into the density formula, we can achieve a numerical representation of the material’s intrinsic mass-to-volume ratio. This final step bridges our conceptual understanding of the crystal structure with practical, quantifiable aspects, ensuring a comprehensive grasp of the crystalline solid's properties.