
When planning an animation, we always make many calculations. To model a unit cell, for example, we need to determine the Cartesian coordinates (x, y, z) of each ion in order to create it using the Blender^{®}.
An ionic compound is formed by anion and cation layers. If we properly choose a set of anions and join their centers, we will find two different geometric figures, a tetrahedron or an octahedron. The internal space of a tetrahedron is named tetrahedral hole, which can be occupied by a cation. Below, we show the calculations that were made to determine the Cartesian coordinates (x, y, z) of a tetrahedral hole.
We initially determined the center of the base triangle of the tetrahedron on Figure TH1.



The result of TH04 equation and the parameters of the animation on Link TH1, are going to be used to determine the center of the tetrahedron:




With the result of equation TH10, the location of the center of the tetrahedron is solved. We will do some further calculations to determine the cation and anion radius in a tetrahedral hole packing.


The famous angle , shown in Figure TH2, is essential to calculate the ratio between cation and anion radius in a tetrahedral hole packing.


The animation on Link TH2, besides helping us to visualize the parameters of a tetrahedron, also helps us to understand the physical meaning of a tetrahedral hole.


With the parameters of Figure TH2 and the information of the animation on Link TH2, we can calculate the ratio between cation and anion radius.





The solution of equation (TH22), given by (TH23), was used to create the animation on Link TH3. Because the anion and cation are in scale, we can note that a cation is much smaller than an anion. The size difference between the ions that 3D animation allow us to visualize, would be very difficult to be imagined in a different way.


Another important information, which often appears in textbooks, is the occupied volume by ions. We are going to calculate the occupied volume in a cubic closest packing with cations occupying tetrahedral holes, taking as an example, the unit cell of sphalerite, which is shown in the animation on Link TH4 and Figure TH3:



Because this unit cell is cubic, all edges are equal, a = b = c, and the diagonal is equal to 4 times the anion radius. With this information and with the result of equation (TH23), we calculated the ratio between the occupied volume by ions and the unit cell volume. Consider r = radius of the cation and R = radius of the anion.




Equation (TH30) shows that ions occupy 74.9% of the volume of an unit cell in a cubic closest packing with cations located in tetrahedral holes.
Let's now calculate the occupied volume by the ions in an unit cell of a hexagonal closest packing with cations located in tetrahedral holes, taking as an example the wurtzite unit cell, which is shown in the animation on Link TH5.


The volume of the unit cell is the product of its base area by its height; the base area was calculated based on Figure TH4:




The calculation of the height of a wurtzite unit cell was made from Figure TH5. With the base area and height, the volume of the unit cell was calculated. With the unit cell volume and ions volume, the ratio between the volume occupied by ions and the volume occupied by the unit cell was calculated.




The result of equation (TH37) shows that ions occupy 70.6% of the volume of the unit cell in a hexagonal closest packing with cations located in tetrahedral holes. The result of equation (TH30) shows that ions occupy 74.9% of the volume of the cell
unit in a cubic closest packing with cations located in tetrahedral holes.

Several trigonometric calculations were made to obtain these results. This is a typical exercise for college students in their first Chemistry courses. It's amazing how we can know precisely calculate the size of an ion by its unit cell.

