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 \( MF_2 \).

Step-by-step solution

Understand the Unit Cell_Structure

In this crystal lattice, the fluoride ions occupy simple cubic lattice sites, meaning there is one fluoride ion at each corner of the cube. In a simple cubic lattice, there is one ion on each of the eight corners. Given the metal ions occupy the body centers of half the cubes, it means there is one metal ion at the center of each alternate unit cell.

Calculate the Number of Fluoride Ions in the Unit Cell

In a simple cubic lattice, there is one ion at each of the eight corners of the cube. However, each corner ion is shared with eight other adjacent unit cells. Hence, the contribution of the fluoride ions from the corners to one unit cell is: \( \frac{1}{8} \times 8 = 1 \) fluoride ion per unit cell.

Calculate the Number of Metal Ions in the Unit Cell

Since the metal ion is present at the body center of each alternate unit cell, we need to account for this in our calculation. The body center ion is surrounded by one unit cell and is fully enclosed by it. In half the unit cells, the metal ions are present at the body center, so on average, there will be: \( \frac{1}{2} \times 1 = 0.5 \) metal ion per unit cell.

Find the Formula of the Metal Fluoride

In each unit cell, there is 1 fluoride ion and 0.5 metal ions. To find the formula of the metal fluoride, we will need to find the smallest whole number ratio between the ions. To obtain a whole number ratio, multiply the ions count by 2. \( 1~F^- \times 2 = 2~F^- \) \( 0.5~M \times 2 = 1~M \) So the formula of the metal fluoride is \( MF_2 \).