Problem 87 - Chemistry: Structure and Properties

Question

A tetrahedral site in a closest-packed lattice is formed by four spheres at the corners of a regular tetrahedron. This is equivalent to placing the spheres at alternate corners of a cube. In such a closest-packed arrangement, the spheres are in contact, and if the spheres have a radius r, the diagonal of the face of the cube is 2r. The tetrahedral hole is inside the middle of the cube. Find first the length of the body diagonal of this cube and then the radius of the tetrahedral hole.

Step-by-Step Solution

Answer

The body diagonal of the cube is \( r\sqrt{6} \), and the radius of the tetrahedral hole is \( \frac{r}{3\sqrt{2}} \).

Step 1: Find the length of the body diagonal of the cube

The body diagonal of a cube can be found using the formula for the diagonal of a cube, which is \( d = \sqrt{3} \cdot a \), where \( a \) is the length of one side of the cube. Given that the face diagonal is \( 2r \), and the face diagonal of a cube is \( \sqrt{2} \cdot a \), we can equate \( 2r = \sqrt{2} \cdot a \) to find \( a = \frac{2r}{\sqrt{2}} = r\sqrt{2} \). We then substitute \( a \) with \( r\sqrt{2} \) in the body diagonal formula to find \( d = \sqrt{3} \cdot r\sqrt{2} = r\sqrt{6} \).

Step 2: Find the height of the tetrahedral hole

The height of the tetrahedral hole (\( h \)) can be found by considering the right-angled triangle formed by the height of the tetrahedron, half the body diagonal of the cube, and the tetrahedron edge (which is the side length of the cube). Using the Pythagorean theorem, we find \( h \) with the relation \( h^2 + (\frac{r\sqrt{6}}{2})^2 = (r\sqrt{2})^2 \), leading to \( h^2 + \frac{3r^2}{2} = 2r^2 \), hence \( h^2 = 2r^2 - \frac{3r^2}{2} = \frac{r^2}{2} \), and therefore \( h = \frac{r}{\sqrt{2}} \).

Step 3: Find the radius of the tetrahedral hole

The radius of the tetrahedral hole (\( r_{\text{hole}} \)) is found by considering another right-angled triangle inside the tetrahedron, formed by the radius of the hole, the height of the tetrahedron, and the radius of one of the spheres (\( r \)). Since the center of the tetrahedral hole is equidistant from all four corners of the tetrahedron, we can divide the height into two segments: the height of the hole (from base to the center of the hole) and the height below the hole to the base. The sum of these two heights is equal to the height of the tetrahedron. Using the Pythagorean theorem, \( r_{\text{hole}}^2 + (\frac{2r}{3})^2 = (\frac{r}{\sqrt{2}})^2 \), we find \( r_{\text{hole}} = \sqrt{\frac{r^2}{2} - \frac{4r^2}{9}} = r\sqrt{\frac{1}{2} - \frac{4}{9}} = r\sqrt{\frac{9 - 8}{18}} = \frac{r}{3\sqrt{2}} \).

Key Concepts

Crystal Lattice Structures

In materials science and crystallography, a crystal lattice structure is a highly ordered arrangement of atoms, ions, or molecules in a crystalline material. It is the geometric pattern of points at which the constituents of the material are positioned.

Closest-packed lattices are a form of crystal structures where particles are packed together as tightly as possible. There are two types of closest-packed structures: hexagonal close-packed (HCP) and face-centered cubic (FCC). These structures achieve the highest packing efficiency – about 74% of the space is filled.

In the context of these structures, a tetrahedral site refers to a space where an atom can fit without disrupting the lattice, and it is surrounded geometrically by four atoms positioned at the vertices of a tetrahedron. Understanding where these sites are located and how they are formed is essential for comprehending various chemical and physical properties of materials.

Geometric Calculations in Chemistry

Geometric calculations are crucial for understanding molecular shapes, crystal structures, and the distances between different parts of a molecule or lattice. In chemistry, geometry helps us visualize and calculate aspects of atomic and molecular structure. One example of geometry applied to chemistry is finding the size of spaces within crystal lattices, such as the tetrahedral hole in closest-packed structures.

To determine the dimensions within lattice structures, chemists often need to calculate distances and sizes using geometric principles. The calculations are guided by formulas that stem from geometric shapes like cubes, spheres, and tetrahedrons. For example, the calculation of the diagonal of a cube helped us understand the spatial arrangement within the closest-packed lattice and further allowed us to calculate the size of the tetrahedral hole where another atom could potentially reside.

Pythagorean Theorem in Chemistry

The Pythagorean theorem is commonly associated with mathematics but finds its importance in chemistry for spatial analysis of molecules and crystal lattices. This theorem, which states that the square of the length of the hypotenuse (the side opposite the right angle) of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides, can be utilized to find unknown distances in three-dimensional space.

In our exercise, the Pythagorean theorem is applied to calculate the height of the tetrahedral hole. By modeling part of the crystal as a right-angled triangle, the theorem allows us to use the known lengths of a cube’s body diagonal and edge to calculate the height of a tetrahedron located within that cube. It beautifully bridges numerical calculations with spatial geometry, vital for predicting the dimensions of unseen spaces in molecular and crystal structures.