Question

In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy corners of the cube of unit cell. If all the face centered atoms along one axes are removed, then, the stiochiometry of the resultant crystal is

• A. $A{B}_2$
• B. $A_{2}B$
• C. $A_4{B}_3$
• D. $A_3{B}_4$

Answer 1

The correct option is D $A_3{B}_4$

There are 6 ‘A’ atoms on the face centers along one axes atoms are removed means 2‘A’ atoms. Now, no. of ‘A’ atoms per unit cell = $=8\times \frac{1}{2}+4\times \frac{1}{2}=3$
(corner) (face centre)
Number of ‘B’ atoms per unit cell.
$=12\times \frac{1}{4}+1=4$
(ed. Cent.) (body cent.)
$\therefore$ formula = $A_3{B}_4$