In a solid AB having the Nacl lattice the structure A have occupied the corners of cubic unit cell.if all the face centered atoms along one of the axis are removed then the resultant stoichiometry is.pl give me the detailed procedure how it is solved

'AB' has NaCl structure, otherwise known as Rock salt structure. Hence one type of atoms must be arranged in FCC pattern (at all the 8 corners and all the 6 face centers) and other type of atoms must be present on all the 12 edges as well as at the center of the unit cell.

Since, 'A' atoms are occupying the corners of unit cell, they must also be arranged on the face centers. The B atoms can be found at the 12 edges and at the center of the unit cell.

Therefore, before removing face centered atoms:

The number of 'A' atoms = (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 [there are 8 'A' atoms at the corners and 6 'A' atoms at the face centers]

The number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4 [there are 12 'B' atoms at the edges and 1 'B' atom at the center of the unit cell]