Question

In a solid $'AB'$ having the $NaCl$ structure, $'A'$ atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

• A. $AB_2$
• B. $A_2 B$
• C. $A_4 B_3$
• D. $A_3 B_4$

Answer 1

Answer: (D)

$A_3 B_4$

Answer 2

Atom A occupies all corner and face center of unit cell and atom B in all octahedral void. As all atoms from face center from one axis is removed, hence number of atom A in unit cell remains 3 and B as 4.

$A=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
$A=4-2\times \frac{1}{2}=3$ after removing atoms

$B=12\times \frac{1}{4}+1\times 1=4$

Formula of compound is $A_{3}B_{4}$