Question

A certain metal fluoride crystallizes in such a way that the fluoride ions occupy simple cubic lattice sites, while the metal ions occupy the body centers of half the cubes. What is the formula of the metal fluoride?

Short answer

The formula of the metal fluoride is MF2, based on the given information about the positions of the metal ions and the fluoride ions in the crystal lattice, and the ratio of metal ions to fluoride ions (\(\frac{1M}{2F}\)).

Step-by-step solution

Determine the number of fluoride ions in the lattice

In a simple cubic lattice, there is one ion per lattice point. Therefore, there is one fluoride ion (F-) at each corner of the cube. Since there are 8 corners in a cubic lattice, this gives us a total of 8 fluoride ions associated with the lattice.

Determine the number of metal ions in the lattice

The metal ions occupy the body center of half the cubes in the lattice. In a body-centered cubic (BCC) lattice, there is one ion situated at the center of the cube. As the metal ions occupy the body centers of half the cubes, and there are 8 body centers in a simple cubic lattice, this means that there are 4 metal ions (M) associated with the lattice.

Calculate the ratio of metal ions to fluoride ions

In order to determine the formula of the metal fluoride, we need to find the ratio of metal ions (M) to fluoride ions (F-). From the previous two steps, we have calculated that there are 4 metal ions and 8 fluoride ions. Therefore, the ratio of metal ions to fluoride ions is: \( \frac{4M}{8F} = \frac{1M}{2F} \)

Write the formula of the metal fluoride

Based on the ratio of metal ions to fluoride ions (\( \frac{1M}{2F} \)), the formula of the metal fluoride can be written as MF2.