Question

From $NaCl$ unit cell, all ions along a diagonal from one edge center to another are absent. The correct expression for packing fraction (P.F.) of that $NaCl$ unit cell (assume ideal) is

• A. $\text{P.F.} = \frac{\frac{4}{3} \text{\pi } \left(\text{r}_{+}^{3} + \text{r}_{-}^{3}\right)}{1 6 \sqrt{2} r_{-}^{3}}$
• B. $\text{P.F.} = \frac{\frac{4}{3} \text{\pi } \left(\frac{5}{2} r_{+}^{3} + 4 \text{r}_{-}^{3}\right)}{1 6 \sqrt{2} r_{-}^{3}}$
• C. $\text{P.F.} = \frac{\frac{4}{3} \text{\pi } \left(\frac{5}{2} r_{+}^{3} + \text{r}_{-}^{3}\right)}{1 6 \sqrt{2} r_{-}^{3}}$
• D. $\text{P.F.} = \frac{\frac{4}{3} \text{\pi } \left(\frac{7}{2} r_{+}^{3} + \text{r}_{-}^{3}\right)}{1 6 \sqrt{2} r_{-}^{3}}$

Answer 1

Answer: (B)

Along edge diagonal only three Na⁺ ions would be absent, 1 body center and two edge centers.
Effective no. of $\left(\text{Na}\right)^{+} = 4 - \left(1 + 2 \times \frac{1}{4}\right) = \frac{5}{2}$
For $\text{Cl}^{-} = \sqrt{2} a = 4 \text{r}^{-}$ $\Rightarrow \text{a} = 2 \sqrt{2} \text{r}_{-}$
$\text{P.F.} = \frac{\text{Vol. of effective no. of cations and anions}}{\text{Vol. of unit cell}}$
$= \frac{\frac{4}{3} \text{\pi } \left(\frac{5}{2} r_{+}^{3} + 4 \text{r}_{-}^{3}\right)}{1 6 \sqrt{2} r_{-}^{3}}$