Question

The unit cell of MgO is shown below. Does \(\mathrm{MgO}\) have a structure like that of \(\mathrm{NaCl}\) or \(\mathrm{ZnS} ?\) If the density of \(\mathrm{MgO}\) is \(3.58 \mathrm{~g} / \mathrm{cm}^{3}\), estimate the radius (in centimeters) of the \(\mathrm{O}^{2-}\) anions and the \(\mathrm{Mg}^{2+}\) cations.

Short answer

MgO has a rock salt-type structure, similar to NaCl. By calculating the edge length of the unit cell using the given density and formula mass, we find the O²⁻ anion radius and Mg²⁺ cation radius. The edge length (a) is related to the radii of the cations (r_c) and anions (r_a) as: \(a = 2(r_a + r_c)\). Using the ratio of radii between the O²⁻ and Mg²⁺ ions, which is approximately 1.40, we can find the radii of O²⁻ anions and Mg²⁺ cations in cm.

Step-by-step solution

Identify the crystal structure of MgO

MgO's unit cell is similar to either NaCl or ZnS. In the NaCl structure, each ion is surrounded by 6 nearest neighbors of opposite charge. In the ZnS structure, each ion is surrounded by 4 nearest neighbors of opposite charge (each S²⁻ ion is tetrahedrally coordinated to 4 Zn²⁺ ions, and vice versa). Since MgO has a rock salt-type structure, same as NaCl, each Mg²⁺ ion will be surrounded by 6 O²⁻ ions and each O²⁻ ion will be surrounded by 6 Mg²⁺ ions.

Calculate the edge length of the unit cell

To find the radius of the ions, we first need the edge length of the unit cell. From the density, we know that: \[Density = \frac{Mass}{Volume}\] We can rewrite it as: \[Volume = \frac{Mass}{Density}\] The formula mass of MgO is approximately (based on atomic weights) 24.31 \(\si{\gram}\)/mol for Mg and 16 \(\si{\gram}\)/mol for O, resulting in a total formula mass of 40.31 \(\si{\gram}\)/mol. Therefore, we have mass = 40.31 \(\si{\gram}\)/mol, and the density is given as 3.58 \(\si{\gram}\,\si{\centi\meter}^{−3}\), so we can calculate the volume: \[Volume = \frac{40.31\frac{\si{\gram}}{\si{\mole}}}{3.58\frac{\si{\gram}}{\si{\centi\meter}^3}}\] Volume per unit cell (MgO has 1 formula unit per cell) can be calculated as: \[Volume = \frac{40.31\frac{\si{\gram}}{\si{\mole}}}{3.58\frac{\si{\gram}}{\si{\centi\meter}^3}} \times \frac{\si{\centi\meter}^3}{1} \times \frac{1}{6.022\times 10^{23}} \] Now, we know that the unit cell is a cube, so the volume is equal to the cube of edge length a: \[Volume = a^3\] Now we can find the edge length a: \[a = \sqrt[3]{Volume}\]

Calculate the O²⁻ anion radius and Mg²⁺ cation radius

In the NaCl type structure, the edge length (a) is related to the radii of the cations (r_c) and anions (r_a) in the following way: \[a = 2(r_a + r_c)\] We can solve for the radius of the O²⁻ ion (r_a) and the Mg²⁺ ion (r_c): \[r_a + r_c = \frac{a}{2}\] We know that the ratio of radii between the O²⁻ and Mg²⁺ ions is approximately 1.40 (look up radii ratio tables, if needed). So, we can rewrite the above equation as: \[r_a = 1.40r_c\] By substituting the radii ratio into the equation, we can now solve for the individual radii: \[1.40r_c + r_c = \frac{a}{2}\] \[2.40r_c = \frac{a}{2}\] \[r_c = \frac{a}{4.80}\] (cation radius) \[r_a = 1.40r_c\] (anion radius) Plug in the value of a calculated in the previous step and find the values for r_c and r_a to get the radii of the O²⁻ anions and the Mg²⁺ cations in cm.

Key concepts

Understanding the Crystal Lattice of MgO

Magnesium oxide (MgO) is a classic example of a crystalline solid where the arrangement of ions presents a distinct and orderly pattern. The crystal lattice refers to this repetitive, three-dimensional organizational structure, which extends in all directions. Imagine building a fortress with toy blocks, where each block is positioned in a specific place relative to its neighbours to form a continuous structure - this is analogous to the formation of a crystal lattice.

In MgO, the ions form a lattice structure known as the rock salt-type, similar to that of table salt (NaCl). This configuration is characterized by each magnesium ion (Mg²⁺) being surrounded by six oxygen ions (O^2-), and vice versa, with each oxygen ion also being coordinated by six magnesium ions. It's a highly symmetrical arrangement that maximizes ionic bonding, and thereby stability, in the crystal.

Understanding this arrangement is crucial because it influences many physical properties, such as melting point, electrical conductivity, and hardness. In educational contexts, visualizing the lattice with models or computer graphics can make the concept more tangible and easier to grasp.

Calculating Ionic Radii in a Crystal Structure

When studying ionic compounds, it becomes vital to compute the sizes of the ions, which are often referred to as ionic radii. This calculation is important for multiple reasons, including understanding how ions pack together in a crystal and predicting bond lengths in ionic compounds.

To calculate ionic radii, especially when not directly provided, chemists refer to experimentally determined values and theoretical considerations. They often rely on the knowledge of a crystal's structure. In the case of MgO with its rock salt structure, we can deduce the edge length of the unit cell which is then related to the sum of cation and anion radii. It's a bit like finding the length of the diagonal to infer the side lengths of the squares on a chessboard if you know they all have the same size.

To find the ionic radii, a mathematical relationship between the known dimension of the unit cell and the summed radii of the Mg²⁺ and O^2- ions is utilized, considering the ions are in direct contact along an edge of the cell. By also using the known ratio of the radii between these two ions, we can create a system of equations to solve for the radii of both the cation and anion.

Determining the Unit Cell Volume of MgO

Comprehending the unit cell volume in a crystal is essential, as it leads to a deeper understanding of the material's density and other bulk properties. Therefore, the exercise involves estimating the volume of a unit cell, which is the smallest repeating unit in the crystal lattice exhibiting the full symmetry of the crystal structure.

The volume of a unit cell, especially for substances that crystallize in simple geometric forms like cubes, can be calculated by cubing the edge length. From the density and the formula mass, we can deduce the mass of a single unit cell. By rearranging the density equation, as shown in the step-by-step solution, we can obtain the volume of the unit cell in cubic centimeters.

Through this process, we essentially convert macroscopic measurable quantities, such as density and molar mass, into microscopic insights about the architecture of the crystal at the atomic or ionic level. For students, these calculations are a bridge connecting the tangible and observable metrics of materials with the abstraction of a crystal's internal arrangement.