Question

Consider an ionic solid MX with $NaCl$ structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.
(i) Remove all the anions (X) except the central one
(ii) Replace all the face centered cations (M) by anions (x)
(iii) Remove all the corner cations (M)
(iv) Replace the central anion (X) with cation (M)
The value of $\left(\frac{\text{number of anions}}{\text{number of cations}}\right)$ in Z is.

Answer 1

According to question
$X^{-}$ are present in Octahedral void
$M^{+}$ are present in FCC lattice point

Number of cations and anions left after each sequence are

$\begin{array}{cc}{M}^{+}& X^{-}\\ \left(i\right)4& 4-3=1\\ \left(ii\right)4-6\times \frac{1}{2}& 1+6\times \frac{1}{2}\\ \left(iii\right)1-1& 3+1=4\\ \left(iv\right)0+1& 4-1=3\end{array}$
So, total number of $X^{-}$ in a unit cell = 3
total number of $M^{+}$ in a unit cell = 1
Hence $\frac{anion}{cation}=\frac{3}{1}=3$