Question

A binary compound has Y-atoms forming FCC unit cell and another type of X-atoms occupying \(\frac{1}{3}\)^rd of tetrahedral voids. Find out the molecular formula of the compound

 

Answer 1

Correct Answer: 23

For 1 unit cell,

 No. of particles

 X = \(\frac{1}{3}\times8\) 

 Y = 4 

 ∴ The formula of Ionic solid = \(X_{\frac{8}{3}}Y_4\) =\(X_2Y_3\)

Answer 2

The formula of the compound can be determined as follows:
In an FCC unit cell, there are 4 atoms located at the corners and one atom at the center of each face. Thus, the total number of Y atoms in the unit cell is 4 from the corners.
In a FCC unit cell, there are tetrahedral voids and octahedral voids. Each void can accommodate one atom. The number of tetrahedral voids in a FCC unit cell is twice the number of atoms per unit cell, which is 4 in this case. Therefore, there are 8 tetrahedral voids in the unit cell. Since X atoms occupy \(\frac{1}{3}\) of the tetrahedral voids, there are \(8\times \frac{1}{3}=\frac{8}{3}\)  X atoms in the unit cell.
The ratio of X atoms to Y atoms in the unit cell is therefore \(\frac{(\frac{8}{3})}{4}=\frac{2}{3}\). This means that the molecular formula of the compound is X₂Y₃. Therefore, option B (X₂Y₃) is the correct answer.