Which of the follwing expression is correct for packing fraction of NaCl if the ions along the face are diagonally removed

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$\frac{\frac{13}{3} {πr}_{-}^{3} + \frac{4}{3} {πr}_{+}^{3}}{8 {(r_{+} + r)}^{3}}$

$\frac{\frac{16}{3} {πr}_{-}^{3} + \frac{13}{3} {πr}_{+}^{3}}{8 {(r_{+} + r_{-})}^{3}}$

$\frac{\frac{13}{3} {πr}_{-}^{3} + \frac{16}{3} {πr}_{+}^{3}}{8 {(r_{+} + r_{-})}^{3}}$

$\frac{\frac{4}{3} {πr}_{-}^{3} + \frac{13}{3} {πr}_{+}^{3}}{8 {(r_{+} + r_{-})}^{3}}$

answer is A.

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Detailed Solution

If the ions along one face diagonal are removed. The ions at the two corners and face centre are removed. Then no. of Cl^– removed are
$1 \times \frac{1}{2} = \frac{1}{2}; 2 \times \frac{1}{8} = \frac{1}{4}$
$Total = \frac{3}{4} . Remaining \frac{1}{4} .$
Na⁺ ions remains are such
no. of Na⁺ ions 4. In NaCl
$\begin{array}{l} 2 r^{+} + 2 r^{-} = a \\ P.F = \frac{4 / 3 \times 13 / 4 {πr}_{-}^{3} + 4 \times 4 / 3 {πr}_{+}^{3}}{{(2 r^{+} + 2 r^{-})}^{3}} \\ = \frac{13 / 8 {πr}_{-}^{3} + 16 / 3 {πr}_{+}^{3}}{8 {(r^{+} + r^{-})}^{3}} \end{array}$