Question

A compound consists of three ions X, Y, and Z. The Z ions are arranged in an FCC arrangement. The X ions occupy \(\frac{1}{6}\) of the tetrahedral voids and the Y ions occupy \(\frac{1}{3}\) of the octahedral voids. Which one of the following is the CORRECT chemical formula of the compound?

• A. XY\(_2\)Z\(_4\)
• B. XYZ\(_3\)
• C. XYZ\(_2\)
• D. XYZ\(_4\)

Hint:

In compounds with FCC structures, tetrahedral and octahedral voids are common, and the ions occupy these voids in specific ratios. Calculate the number of ions based on these ratios to get the correct formula.

Answer 1

The Correct Option is B

- In an FCC arrangement, there are 4 unit cells. Each unit cell has 8 octahedral voids and 12 tetrahedral voids.
- Number of Z ions = 4 (since Z occupies all FCC lattice sites).
- Number of X ions = \(\frac{1}{6}\) of the tetrahedral voids = \(\frac{1}{6} \times 12 = 2\).
- Number of Y ions = \(\frac{1}{3}\) of the octahedral voids = \(\frac{1}{3} \times 8 = \frac{8}{3}\).
- The total number of Y ions is 3. Hence, the chemical formula is XYZ\(_3\).