Question

In a solid ${}^{\prime \prime }A{B}^{\prime \prime }$ having $NaCl$ type structure ${}^{\prime \prime }{A}^{\prime \prime }$ atoms occupy the FCC lattice points of the cubic unit cell. If all the face-centered atoms along one of the axis are removed, then the resultant stoichiometry of the solid is :

• A. $A{B}_2$
• B. $A_{2}B$
• C. $A_4{B}_3$
• D. $A_3{B}_4$

Answer 1

Answer: (D)

$A_3{B}_4$

Since $A$ atoms occupy the corners of the cubic unit cell, the number of atoms occupying the corners is $1$.

If all the face-centred atoms along one of the axes are removed, then only $4$ faces remain, which counts in total to $2$ atoms of $A$.

$B$ is present in octahedral voids, so the total number of atoms of $B$ is $4$.

Hence, the stoichiometry is $A_3{B}_4$.

Hence, option is D.