Question

If atoms along an axis connecting the opposite edge centres on a face are removed from $NaCl$ type solid $AB$ then new empirical formula of the remaining solid would be:

• A. $A_8{B}_5$
• B. $AB$
• C. $A_3{B}_4$
• D. $A_3{B}_8$

Answer 1

Answer: (C)

$A_3{B}_4$

From the structure of $NaCl$, A will occupy $8$ corners of a cube, which will contribute $\frac{1}{8}$ each to the cube.

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

$\Rightarrow 6$ face centered atoms will contribute $\frac{1}{2}$ to the cube.

$\therefore 6\times \frac{1}{2}=3$

Total number of atoms A=$3+1$

Now B occupies voids which is $\frac{1}{4}$

$\therefore 1+12\frac{1}{4}=4\Rightarrow \therefore$ The formula $A_4{B}_4$ (or) $AB$.

Now we are removing atoms along one axis. So, the number of atoms removed is $2$. So, $8\times \frac{1}{8}+4\frac{1}{2}=3$

The resultant $A_3{B}_4$ .