In a crystalline solid, atoms of element B form a cubic close-packed (ccp) lattice. Atoms of element A occupy all the octahedral voids and all the tetrahedral voids. If all the atoms lying along exactly one body diagonal of the unit cell are removed, the new empirical formula of the solid becomes A _p B _q (where p and q are the simplest whole numbers). The value of p + q is

Step-by-step solution

Initial number of B atoms in ccp = 8 1 8 + 6 1 2 = 4 . Initial number of A atoms = 4 (in octahedral voids) + 8 (in tetrahedral voids) = 12. A body diagonal of a cubic unit cell passes through 2 corners, 2 tetrahedral voids, and 1 octahedral void (at the body center). When atoms along one body diagonal are removed: Number of B atoms removed = 2 corners 1 8 = 1 4 . Remaining B atoms = 4 - 1 4 = 15 4 . Number of A atoms removed = 2 (from tetrahedral voids) + 1 (from body center octahedral void) = 3. Remaining A atoms