In a face centred lattice with all the positions occupied by A atoms the body centred octahedral hole in it is occupied by an atom B of an appropriate size for such a crystal, calculate the void space per unit volume of unit cell. Also predict the formula of the compound.

Step-by-step solution

Let a be the edge of the cube so that aligned & 4 r_A= 2 a & a=2 2 a_A aligned Now since the atoms B has occupied the body centre octaheedral void aligned & 2 r_A+2 r_B=a & 2 r_A+2 r_B=2 2 r_A aligned 2 rB =(2 2 -2) r _ A ; r _ B r _ A = 2 -1=1.414-1=0.414 Volume of the cube a^3=16 2 r_A^3 Volume occupied by A and B = [4 4 3 r_A^3 ]+ [ 4 3 r_B^3 ]= 4 3 (4 r_A^3+r_B^3 ) Volume occupied per unit volume of unit cell = 4 / 3 (4 r_A^3+r_B^3 ) 16 2 r_A^3 = 12 2 [4+ ( r_B r_A )^3 ]= 3.14 1.2 1.414 (4+(0.414)^3 )= 3.14 4.0