Question

In solid $AB,B^{-}$ forms the fcc lattice and $A^{+}$ occupies all the octahedral voids. If atoms from the edge centers along an axis connecting the opposite edge centers on a same face are removed, what will be the empirical formula of the compound?

• A. $A_8{B}_5$
• B. $A_7{B}_8$
• C. $A_8{B}_5$
• D. $A_3{B}_8$

Answer 1

The correct option is B $A_7{B}_8$

In AB structure, $B^{-}$ ions form fcc and $A^{+}$ ions are present at edge centers and body centers.
$\text{Number of atoms of B per unit cell}=4$
$\text{Number of atoms of A per unit cell}=\frac{1}{4}\times 12+1=3+1=4$
Each atom at the centre of the edge contributes one fourth to the cube.
Now, if atoms from edge centers along an axis joining opposite edges on a same face are removed then,
Effective number of atoms of A removed $=\frac{1}{4}\times 2=\frac{1}{2}$
Effective number of atoms of A present $=4-\frac{1}{2}=\frac{7}{2}$
So, $A:B\Rightarrow \frac{7}{2}:4=7:8$
So, formula of the solid is $A_7{B}_8$