A compound made of particles A,B and C forms a ccp lattice. In the lattice, ions A occupy the lattice points and ions B and C occupy the alternate tetrahedral voids. If all the ions along one of the body diagonals are removed, then formula of the compound is

If all the ions along one of the body diagonals are removed,

Number of A ions removed $=2\times \frac{1}{8}( \mathrm{corner}\mathrm{share} )=\frac{1}{4}$

Number of B ion removed = 1
Number of C ion removed = 1
(Since body diagonal ions are inside the cube so they do not share with other ions) Number of A ions left $=4-\frac{1}{4}=3.75$

Number of B ions left = 4 -1 = 3
Number of C ions left = 4 -1 = 3
Thus, formula is ${\mathrm{A}}_{3.75}{\mathrm{B}}_3{\mathrm{C}}_3={\mathrm{A}}_5{\mathrm{B}}_4{\mathrm{C}}_4$