Question

In a solid AB having $NaCl$ structure, 'A' atoms occupy the corners of the cubic unit cell.If all the face-centered 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

Answer: (D)

$A_3{B}_4$

According to the structure of NaCl

Here, ${}^{\prime }{A}^{\prime }$ will occupy $8$ corners of cubic unit cell which will contribute $\frac{1}{8}$ each to the cube

therefore $8\times \frac{1}{8}=1$

and 6 face centered atoms will contribute $\frac{1}{2}$ to the cube
Hence $6\times \frac{1}{2}=3$

Therefore total contribution of atom'A' is $3+1=4$
Now B occupies octahedral voids (NaCl structure) which contributes 1/4
therefore $1+12\times \frac{1}{4}=4$
So the formula is A is-
$A_4{B}_{4}orAB$
Now we are removing all face centered atoms along one axis-
so the number of atoms removed from cube are 2
so contribution from 'A' remains $8\times \frac{1}{8}\left(corneratoms\right)+4\times \frac{1}{2}\left(facecenteredatoms\right)=3$
so the resultant becomes$A_3{B}_4$