Question

In the LiCl structure shown in Figure 9.21 , the chloride ions form a face- centered cubic unit cell \(0.513 \mathrm{nm}\) on an edge. The ionic radius of \(\mathrm{Cl}^{-}\) is \(0.181 \mathrm{nm} .\) (a) Along a cell edge, how much space is between the \(\mathrm{Cl}^{-}\) ions? (b) Would an \(\mathrm{Na}^{+}\) ion \((r=0.095 \mathrm{nm})\) fit into this space? a \(\mathrm{K}^{+}\) ion \((r=0.133 \mathrm{nm}) ?\)

Short answer

Answer: The space between Cl⁻ ions along a cell edge is 0.151 nm. Neither Na⁺ (0.095 nm) nor K⁺ (0.133 nm) ions would fit into this space.

Step-by-step solution

Calculate the distance between two Cl⁻ ions along a cell edge

In a face-centered cubic (FCC) structure, the edge length is related to the ionic radius by the following formula: Edge length = 2 * (ionic radius of cation + ionic radius of anion) This is because there are two edge-sharing Cl⁻ ions along a cell edge (corner-edge-corner) Here, the edge length is 0.513 nm and the ionic radius of Cl⁻ is 0.181 nm, so we can set up the following equation: 0.513 = 2 * (ionic radius of cation + 0.181)

Calculate the ionic radius of the cation

We first want to isolate the ionic radius of the cation in the equation. ionic radius of cation = (0.513 - 2 * 0.181) / 2 Now we can calculate the ionic radius of the cation, ionic radius of cation = (0.513 - 0.362) / 2 ionic radius of cation = 0.0755 nm

Calculate the space between Cl⁻ ions along a cell edge

Since the space between Cl⁻ ions in the cell edge = 2 * (ionic radius of cation) Space between Cl⁻ ions = 2 * 0.0755 Space between Cl⁻ ions = 0.151 nm

Check if Na⁺ and K⁺ ions would fit into the space

Now that we found the space between Cl⁻ ions along a cell edge, we can check if Na⁺ and K⁺ ions would fit. For Na⁺ ion (0.095 nm) to fit, the following condition must be satisfied: ionic radius of Na⁺ ≤ ionic radius of cation 0.095 ≤ 0.0755 - which is not true, therefore Na⁺ ion won't fit. For K⁺ ion (0.133 nm) to fit, the following condition must be satisfied: ionic radius of K⁺ ≤ ionic radius of cation 0.133 ≤ 0.0755 - which is also not true, therefore K⁺ ion also won't fit. The calculated space between Cl⁻ ions along a cell edge is 0.151 nm. Neither Na⁺ (0.095 nm) nor K⁺ (0.133 nm) ions would fit into this space.

Key concepts

Ionic Radius

When examining crystalline structures, the ionic radius is a fundamental concept that refers to the effective distance from the nucleus of an ion to the point up to which it has influence on its electron cloud within a crystal lattice. This distance is typically measured in nanometers (nm). In the context of salts like lithium chloride (LiCl), understanding the ionic radius of both the anion (Cl⁻) and the cation (Li⁺) is crucial for determining how these ions pack together in a solid.

As we explore face-centered cubic (FCC) structures, we consider the size of these radii in assessing whether there is a compatible fit between constituent ions. For example, a Cl⁻ ion with a larger ionic radius wouldn't be able to accommodate smaller Na⁺ or K⁺ ions if these ions have larger radii than the calculated spaces in the lattice. This would disrupt the crystal's stability and dictate its physical properties. Proper understanding of ionic radii aids in predicting the potential to form a solid and allows us to understand the nature of ionic bonding within these solids.

Unit Cell Dimensions

The term unit cell dimensions quantifies the size of the smallest repeating unit in a crystal lattice, which represents the entire crystal's symmetry and structure. In our specific case of LiCl, the face-centered cubic unit cell has dimensions defined by its edge lengths. For FCC structures, dimensions are not arbitrary; the edge length can be directly related to the ionic radii of the ions forming the lattice.

The formula tying edge length to the ionic radii of the constituents in a face-centered cubic structure is as follows: \[ \text{Edge length} = 2 \times (\text{ionic radius of cation} + \text{ionic radius of anion}) \]. Hence, by knowing the ionic radius of the anion, we can calculate the length of the cell edge and subsequently determine the entire unit cell's dimensions. Understanding these dimensions is vital for predicting the lattice's physical characteristics, such as density and space available for other ions, as in the exercise example where we determine whether other ions could potentially fit within the LiCl structure.

Cation-Anion Spacing in Crystals

The cation-anion spacing in crystals is another key aspect that influences the structure and properties of ionic compounds. It refers to the distance between the positively charged cations and the negatively charged anions within the crystal lattice. The spacing is pivotal because it affects the electrostatic forces and consequently, the overall stability of the crystal.

In the face-centered cubic structure of LiCl, we examine how closely the ions are packed by calculating the distance between them across the edges and faces of the unit cell. Using the ionic radii and the unit cell dimensions, we can deduce whether the space between Cl⁻ ions would be suitable for other ions of different sizes to occupy, like Na⁺ or K⁺ ions in the exercise example. The spacing between ions is not sufficient for Na⁺ or K⁺, illustrating how the specific arrangement of ions dictates the crystal's ability to incorporate different elements without altering the structure. The understanding of such spacing is essential when it comes to designing new materials and understanding their capacity to host various impurities or dopants.