Problem 115 - General Chemistry: Principles and Modern Applications

Question

A crystalline solid contains three types of ions, \(\mathrm{Na}^{+}, \mathrm{O}^{2-},\) and \(\mathrm{Cl}^{-}\). The solid is made up of cubic unit cells that have \(\mathrm{O}^{2-}\) ions at each corner, \(\mathrm{Na}^{+}\) ions at the center of each face, and \(\mathrm{Cl}^{-}\) ions at the center of the cells. What is the chemical formula of the compound? What are the coordination numbers for the \(\mathrm{O}^{2-}\) and \(\mathrm{Cl}^{-}\) ions? If the length of one edge of the unit cell is \(a,\) what is the shortest distance from the center of a \(\mathrm{Na}^{+}\) ion to the center of an \(\mathrm{O}^{2-}\) ion? Similarly, what is the shortest distance from the center of a \(\mathrm{Cl}^{-}\) ion to the center of an \(\mathrm{O}^{2-}\) ion?

Step-by-Step Solution

Answer

The chemical formula of the compound is \(Na_3OCl\). The coordination numbers for the \(O^{2-}\) and \(Cl^{-}\) ions are 4 and 6, respectively. The shortest distance from the center of a \(Na^{+}\) ion to the center of an \(O^{2-}\) ion is \(a/2\), and from the center of a \(Cl^{-}\) ion to the center of an \(O^{2-}\) ion, it is \(\sqrt{3}/2 * a\).

Step 1: Identify the Ions at Different Positions

Each corner of the unit cell contains an \(O^{2-}\) ion, there is a total of 8 corners in a cubic cell, but since each corner ion is shared by 8 cells, we have totally 1 \(O^{2-}\) ion in a unit cell. Each face of the unit cell contains a \(Na^{+}\) ion, and there are a total of 6 faces, since each face ion is shared by 2 cells, we have totally 3 \(Na^{+}\) ions in a unit cell. The unit cell contains one \(Cl^{-}\) ion at the center which is completely within the cell. Therefore the chemical formula of the compound is \(Na_3OCl\).

Step 2: Determine the Coordination Numbers

The \(Cl^{-}\) in the center is surrounded by \(Na^{+}\) on each face of the cube so coordination number for \(Cl^{-}\) is 6. For the \(O^{2-}\) ions, they are each in the corner of the cube surrounded by four \(Cl^{-}\) in the center of cubes. Hence, the coordination number for \(O^{2-}\) ions is 4.

Step 3: Calculate the Distances between Ions

The shortest distance from the center of a \(Na^{+}\) ion to the center of an \(O^{2-}\) ion is the distance from the center of a face of the cube to one of its corners. This can be calculated using the Pythagorean theorem as one-half of the edge length, or \(a/2\). To calculate the shortest distance from the center of a \(Cl^{-}\) ion to the center of an \(O^{2-}\) ion, we consider the distance from the center of the cube to one of its corners. This is equal to the diagonal of the cube and can be calculated as \(\sqrt{3}/2 * a\).

Key Concepts

Unit Cell

A unit cell is the smallest repeating structure in a crystal lattice that, when stacked together in three-dimensional space, forms the macroscopic crystal structure. It acts like the building block of the crystal, determining its overall symmetry and properties.

In the problem provided, we are looking at a cubic unit cell which contains three types of ions:

• \( \mathrm{O}^{2-} \) ions located at each of the cubic cell's corners
• \( \mathrm{Na}^{+} \) ions found at the center of each face on the cube
• \( \mathrm{Cl}^{-} \) ions residing at the very center of the unit cell

Because of the geometry and fractional occupancy of a cubic unit cell, each corner participates in eight neighboring cells. Consequently, each corner ion counts as only \( \frac{1}{8} \), adding up to a single \( \mathrm{O}^{2-} \) ion per cell. Similarly, face ions share a portion with adjacent cells, counted as \( \frac{1}{2} \) per face, totalling three \( \mathrm{Na}^{+} \) ions. The sole \( \mathrm{Cl}^{-} \) ion at the cube's center claims full ownership within its cell.By determining the total number of each ion in the unit cell, we derive its chemical formula as \( \mathrm{Na}_3 \mathrm{O} \mathrm{Cl} \).Understanding the unit cell arrangement helps explain why this specific formula exists and how ions interact within this miniature framework.

Coordination Number

The coordination number is a critical concept in crystallography, representing the number of nearest neighbor atoms or ions surrounding a given atom or ion in the crystal lattice. Understanding coordination numbers is essential for appreciating how tightly packed or how loose an arrangement in the lattice might be.

In the cubic cell from the question, let's consider both the \(\mathrm{Cl}^{-}\) and \(\mathrm{O}^{2-}\) ions:

• \( \mathrm{Cl}^{-} \) is centralized within the unit cell and encircled by six \( \mathrm{Na}^{+} \) ions, one on each cube face. Therefore, the coordination number for \( \mathrm{Cl}^{-} \) is 6, which indicates an octahedral environment.
• Conversely, an \( \mathrm{O}^{2-} \) ion, situated at a corner, interacts with four \( \mathrm{Cl}^{-} \) ions occupying the centers of adjacent cubes. Thus, its coordination number is 4, forming a tetrahedral setting around the \( \mathrm{O}^{2-} \) ion.

This variation in surrounding ion counts highlights how different ions play distinct roles within the crystal structure, owing to their positions and chemical environment. Knowing the coordination number is helpful for understanding the structural configuration of the crystal and its lattice energy.

Ion Distances

Ion distances are determined based on the geometric arrangement within the crystal lattice, offering insights into the physical and chemical properties of the substance. Calculating these distances allows us to understand how tightly packed ions are in a crystal.

To evaluate ion distances in the cubic unit cell:

• The shortest path from a \( \mathrm{Na}^{+} \) ion at the face center to an \( \mathrm{O}^{2-} \) ion at a cell corner forms one side of an isosceles right triangle. Using Pythagorean theorem, this distance becomes \( \frac{a}{2} \), where \( a \) is the cube's edge length.
• For the \( \mathrm{Cl}^{-} \) ion situated at the cube's center, its shortest trail to any \( \mathrm{O}^{2-} \) ion traverses the cube diagonal; calculated as \( \frac{\sqrt{3}}{2} * a \).

This shows that \( \mathrm{Na}^{+} \) and \( \mathrm{Cl}^{-} \) strategically position to minimize space and maximize stability in the lattice. Understanding these ion distances contributes to grasping ion interaction strength and electrical conductivity within the crystal structure, vital for various applications in material science and chemistry.