Question

In a solid AB having the NaCl structure, A atom occupies the cornerns of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid 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 $A_3{B}_4$

AB has NaCl-type structure (fee system).
$\therefore$ Number of A atoms
$=8\left(corner\right)\times \frac{1}{8}\left(percornershare\right)+6\left(faces\right)\times \left(perfacecentershare\right)$
= 1 + 3 = 4/unit cell

Number of B atoms
$=12\left(edges\right)\times \frac{1}{4}\left(peredgecentreshare\right)+1\left(bodycentre\right)$
= 1 + 3 = 4/unit cell

Number of A atoms removed (face centred atoms of one axis)
$=2\left(faces\right)\times \frac{1}{2}\left(perfacecentreshare\right)$
= 1/unit cell

Number of A atoms left = 4 - 1 = 3/unit cell.

Formula = $A_3{B}_4$