Question

An AB solid has $NaCl$ type of unit cell. If atoms from the edge centers along an axis connecting the opposite edge centers on a face are removed, the empirical formula of the remaining solid would be :

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

Answer 1

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

In $NaCl$ type structure, $C{l}^{-}$ ions form FCC and $N{a}^{+}$ ions are present at edge centers and body centers

So, let $B$ forms FCC and $A$ is present at body center and edge centers.

Number of atoms of $B=4$ (in unit cell)

Number of atoms of $A=\frac{1}{4}\times 12+1=3+1=4$ ( in unit cell)

Now, if atoms from edge centers along an axis joining opposite edges on a face are removed then,

Effective number of atoms of $A$ removed $=\frac{1}{4}\times 2=\frac{1}{2}$

$\frac{1}{4}$ contribution of atoms at edge center

Effective number of atoms of $A$ present $=4-\frac{1}{2}=\frac{7}{2}$

So, $A:B\Rightarrow A:B$

$\frac{7}{2}:4$ $7:8$

So, formula of the solid is $A_7{B}_8$