Problem 68 - Chemical Principles: The Quest for Insight

Question

As can be seen in Fig. \(5.33\), not all unit cells are cubic. Other types of unit cells have different restrictions placed on the lattice parameters (edge lengths and angles). Unit cell properties such as cell volume, density, and distances between atoms are calculated just as the calculations are done for cubic unit cells, except the geometry is more complex. (a) With this in mind, calculate the distance between a corner atom and the atom at the body center of a tetragonal unit cell that has \(a=b=549 \mathrm{pm}\) and \(c=769 \mathrm{pm}\). (b) What is the volume of this unit cell?

Step-by-Step Solution

Answer

The distance between a corner atom and the body-centered atom is approximately 989.22 pm, and the volume of the tetragonal unit cell is approximately 231.22 nm^3.

Step 1: Understanding the Tetragonal Unit Cell

A tetragonal unit cell is characterized by three edges: two of them (a and b) are of equal length and the third one (c) is of a different length. The angles between the edges are all 90 degrees. In this cell, the corner atom and the body-centered atom form the endpoints of a body diagonal.

Step 2: Calculating the Body Diagonal Length

To calculate the distance between the corner atom and the body-centered atom, use the Pythagorean theorem in three dimensions, which for the body diagonal (d) gives us the formula: \( d = \sqrt{a^2 + b^2 + c^2} \). Since \(a = b\), it simplifies to \( d = \sqrt{2a^2 + c^2} \).

Step 3: Substitute the Values and Calculate the Distance

Substitute \(a = 549 \text{pm}\) and \(c = 769 \text{pm}\) into the formula: \( d = \sqrt{2(549 \text{pm})^2 + (769 \text{pm})^2} \) and calculate to find the distance.

Step 4: Calculating the Volume of the Unit Cell

The volume of a tetragonal unit cell (V) is given by the product of its edge lengths, so the formula is \( V = a \times b \times c = a^2 \times c \) (since \(a = b\)).

Step 5: Substitute the Values and Calculate the Volume

Substitute \(a = 549 \text{pm}\) and \(c = 769 \text{pm}\) into the volume formula: \( V = (549 \text{pm})^2 \times 769 \text{pm} \) and calculate to find the volume.

Key Concepts

Lattice Parameters

The lattice parameters of a crystal structure include the lengths of the unit cell edges and the angles between them. These parameters define the dimensions of the unit cell, which is the basic repeating unit that constitutes the entire crystal structure. In the case of a tetragonal unit cell, the lattice is defined by two equal edge lengths, denoted as 'a' and 'b', and a third edge length 'c' that is different from 'a' and 'b'. The angles between these edges are all 90 degrees. The specific values of 'a', 'b', and 'c' form the basis for calculating many important properties of the crystal, such as the distance between atoms and the unit cell volume.

For the tetragonal unit cell in our exercise, 'a' and 'b' are equal to 549 picometers (pm), and 'c' is longer at 769 pm. These values are not arbitrary but are derived from experimental measurements and are integral to understanding the crystal's physical attributes and behavior.

Unit Cell Volume

The volume of a unit cell is a direct measure of the three-dimensional space it occupies within a crystal lattice. Calculating the unit cell volume is crucial in determining the density of the material and other macroscopic properties. The volume formula for a tetragonal unit cell is the product of its edge lengths: \( V = a \times b \times c \) where 'a' and 'b' are equal for a tetragonal lattice. Simplifying the formula gives us \( V = a^2 \times c \) since \( a = b \).

In the context of our exercise, with known lattice parameters, the volume calculation becomes a straightforward arithmetic task. By substituting the given edge lengths into the formula, we can derive the unit cell's volume, which then can be utilized in further computations such as determining the material's density.

Crystallography

Crystallography is the scientific study of crystal structures and their properties. It examines the arrangement of atoms within a crystal and how this arrangement affects the crystal's physical properties. The discipline employs various methods, such as X-ray diffraction, to determine the lattice parameters that characterize a unit cell's geometry. Understanding the three-dimensional configuration of unit cells and how they repeat to form a crystal lattice is pivotal in materials science and chemistry.

For students studying crystallography, it's essential to grasp the concept that the crystal structure of a material influences its optical, mechanical, and electrical behaviors. The tetragonal unit cell is just one of the seven crystal systems that crystallographers investigate to understand material properties. Its unique geometry compared to other systems, such as cubic or hexagonal, leads to different physical characteristics.

Pythagorean Theorem in Three Dimensions

The classic Pythagorean theorem is familiar to most students as a method for calculating the hypotenuse of a right triangle in two dimensions. However, in the realm of crystallography, and particularly when dealing with unit cells, we require a three-dimensional version of this theorem.

In the case of our exercise exploring the tetragonal unit cell, the Pythagorean theorem extends to three dimensions to calculate the body diagonal length connecting a corner atom to the body-centered atom. The formula becomes \( d = \sqrt{a^2 + b^2 + c^2} \), which reduces to \( d = \sqrt{2a^2 + c^2} \) for a tetragonal cell due to the equal lengths of 'a' and 'b'. Understanding and applying the Pythagorean theorem in three dimensions is key for students to determine distances within complex crystal structures beyond the simple cubic systems.