Question

A solid is formed and it has three types of atoms X, Y, Z. X forms an FCC lattice with Y atoms occupying one-forth of tetrahedral viods and Z atoms occupying half of the octahedral voids. The formula of the solid is?

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

Answer 1

The correct option is D $X_{2}YZ$

Since $X$ forms fcc, so effective number of atoms of $X$ in the unit cell $=4$.

Effective number of atoms of $Y=\frac{1}{4}\times numberoftetrahedralvoids$

$=\frac{1}{4}\times 2\times numberofatomsinfcc$

$=\frac{1}{4}\times 2\times 4=2$

Effective number of atoms of $Z=\frac{1}{2}\times numberofoctahedralvoids$

$=\frac{1}{2}\times numberofatomsinfcc$

$=\frac{1}{2}\times 4=2$

So, $X:Y:Z\Rightarrow X:Y:Z$

$4:2:2$ $2:1:1$

So, formula of the compound is $X_{2}YZ$