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.

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

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

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

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P . F = \frac{\frac{4}{3} π (\frac{7}{2} r_{+}^{3} + r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}

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Solution

The correct option is B $P . F = \frac{\frac{4}{3} π (\frac{5}{2} r_{+}^{3} + 4 r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$
An edge diagonal has $1 + \frac{1}{4} + \frac{1}{4} = \frac{3}{2}$ sodium atoms missing.

One unit cell contains $4 - \frac{3}{2} = 2.5$ sodium ions and $4$ chloride ions.

Packing fraction $(P . F) = \frac{Volume occupied}{Total volume}$
Volume occupied $= \frac{4}{3} π (\frac{5}{2} r_{+}^{3} + 4 r_{-}^{3})$
Total volume $= 16 \sqrt{2} r_{-}^{3}$
$P . F = \frac{\frac{4}{3} π (\frac{5}{2} r_{+}^{3} + 4 r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

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