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

In the given lattice structure, fluoride ions occupy simple cubic lattice sites, while metal ions occupy the body centers of half the cubes. The ratio of metal ions to fluoride ions in the lattice is 1:2. Therefore, the formula of the metal fluoride is \(MF_2\), with M representing the metal ion and F representing the fluoride ion.

Step-by-step solution

1. Understand the Lattice Structure

: First, let's visualize the given structure: - Fluoride ions occupy simple cubic lattice sites. - Metal ions occupy the body centers of half the cubes. In a simple cubic lattice, there is one atom/ion at each corner of the cube. As there are 8 corners, there is a total of 1 ion per unit cell (each corner contributes 1/8 of an ion to the unit cell). So, for the fluoride ions, we have an equivalent of 1 ion per unit cell. The metal ions are in the body center of half the cubes. In a body-centered cubic lattice, there is 1 atom/ion in the center of the cube, which contributes the whole 1 ion to the unit cell. As only half the cubes have metal ions, on average, there is 0.5 metal ion per unit cell.

2. Determine the Ratio between Metal and Fluoride Ions

: Now that we have the number of metal and fluoride ions per unit cell, we can determine the ratio between them: - Metal ions: 0.5 per unit cell - Fluoride ions: 1 per unit cell The ratio of metal ions to fluoride ions in the lattice is 0.5:1, which simplifies to 1:2.

3. Establish the Formula of the Metal Fluoride

: As there is a ratio of 1 metal ion to 2 fluoride ions in this lattice structure, we can represent the metal fluoride's formula with MF2, with M representing the metal ion and F representing the fluoride ion. Therefore, the formula of the metal fluoride is MF2.