For a cell of the CsCl type (figure 9.18), how is the length of one side of the cell,s, related to the sum of the radii of the ions, r subscript cation + r subscript anion?
Figure 9.18 :
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Another question : Consider the NaCl unit shell in figure 9.18. looking at only the front face, five large cl- ions, 4 small na+ ions, how many cubes share each of the na+ ions in this face? how many cubes share each of the Cl- ions in this face?
To determine the relationship between the length of one side of the CsCl cell (s) and the sum of the radii of the ions (r cation + r anion), we can analyze the arrangement of ions in the crystal structure.
In the CsCl cell, the Cs+ cation occupies the center position and the Cl- anion is located at the corners. Each Cs+ cation is surrounded by eight Cl- anions in a cube. Similarly, each Cl- anion is surrounded by eight Cs+ cations in a cube.
We can consider one side of the cube in this analysis. Taking into account the arrangement of ions in a simple cubic arrangement, the Cs+ cation at the center is in contact with eight Cl- anions at the corners. The distance from the center of the Cs+ cation to the corner of the cube is equal to the sum of the radii of the ions (r cation + r anion).
Therefore, the length of one side of the CsCl cell (s) is equal to twice the ionic radius sum (2(r cation + r anion)).
Answering your second question about the NaCl unit cell in figure 9.18, considering only the front face, let's analyze the arrangement of ions.
For each Na+ ion at the front face, four Na+ ions from the corners of the adjacent cubes are in contact with the central Na+ ion. Thus, each Na+ ion shares its face with four adjacent unit cubes.
For each Cl- ion at the front face, four Cl- ions from the adjacent cubes within the same plane are in contact with it. Therefore, each Cl- ion also shares its face with four adjacent unit cubes.
So, each Na+ ion shares its face with four cubes, while each Cl- ion also shares its face with four cubes in the NaCl unit cell.