Question

A solid is formed and it has three types of atoms, X, Y, Z. X forms a FCC lattice with Y atoms occupying all the tetrahedral voids and Z atoms occupying half the octrahedral voids. The formula of the solid is:

• A. $X_2{Y}_{4}Z$
• B. $X{Y}_2{Z}_4$
• C. $X_4{Y}_{2}Z$
• D. $X_{4}Y{Z}_2$

Answer 1

The correct option is A $X_2{Y}_{4}Z$

In an fcc unit cell, there are 4 atoms in one unit cell. X forms fcc lattice. Hence, number of X atoms in unit cell = 4. Y atoms occupies all tetrahedral voids. In fcc unit cell, there are 8 tetrahedral voids. Hence, number of Y atoms in unit cell = 8. Z atoms occupy half the octahedral voids. In fcc unit cell, there are 4 octahedral voids. Hence, number of Z atoms $=\frac{4}{2}=2$

Hence, the ratio $X:Y:Z=4:8:2$ or $2:4:1$. Hence, the formula of the solid is $X_2{Y}_{4}Z$