In a solid AB having NaCl structure. "A" atoms occupy the corners of the cubic unit cell. If all the edge centered atoms along one of the axes are removed. Then the resultant stoichiometry of the solid is.

Dear Student,

According to the structure of NaCl (cubic unit cell),

A occupies 8 corners and 6 face centered atoms.
$Total contribution of A = \frac{1}{8} \times 8 + 6 \times \frac{1}{2}= 1 + 3 = 4$
B occupies octahedral voids
$Total contribution of B = 1 + 12 \times \frac{1}{4}= 4$

If we remove all the face centered atoms along one axis, the number of atoms removed from the cube are 2.

So contribution of A becomes :
$A = 8 \times \frac{1}{8} + 4\times \frac{1}{2}= 1 +2 = 3$
Total contribution of A = 3
Total contribution of B = 4

So, stoichiometry of the compound is A₃B₄.


Hope it helps.

Regards