Select the right expression for determining packing fraction (P.F) of NaCl unit cell (assume ideal), if the ions along an edge diagonal are absent.

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$P . F = \frac{\frac{4}{3} π (r_{+}^{3} + r_{-}^{3})}{16 \sqrt{2} r_{+}^{3}}$

$P . F = \frac{\frac{4}{3} π (r_{+}^{3} - r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

$P . F = \frac{\frac{4}{3} π (\frac{5}{2} r_{+}^{3} + 4 r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

$P . F . = \frac{\frac{4}{3} π (r_{+}^{3} + r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

answer is D.

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Detailed Solution

$r_{+}$ is the radius of cation, sodium ions whereas $r_{-}$ is the radius of anion, chloride ions. Sodium ions have small size and occupy the edge centres whereas chloride ions occupy the FCC lattice. When ions along edge diagonal removed, then two sodium ions are removed from edge centres and one sodium ion is removed at body centre.

There are total four sodium ions and the removed sodium ions:

$\underset{BCC}{1} + \underset{from edge centre}{\underset{⏟}{\frac{1}{4} + \frac{1}{4}}} = \frac{3}{2}$