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Q. In a solid 'AB' having rock salt arrangement ( $NaCl$ structure), 'A' atoms occupy the corners and face center of the cubic unit cell. 'B' atoms occupies all octahedral voids of the unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stochiometry of the solid is

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Solution:

There were $6 \,A$ atoms on the face-centres removing face-centred atoms along one of the axes means removal of $2 \,A$ atoms.
Now, number of $A$ atoms per unit cell
Solution
Number of $B$ atoms per unit cell
$=\underset{\left(\text{ edge centred }\right)}{12 \times \frac{1}{4}}+\underset{\left(\text{body centred }\right)}{1 = 4}$
Hence the resultant stoichiometry is $A_{3}B_{4}$