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(B) In an $NaCl$ unit cell,there are $4$ $Na^+$ ions and $4$ $Cl^-$ ions. An edge diagonal passes through one corner,the face center,and the opposite

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) In an $NaCl$ unit cell,there are $4$ $Na^+$ ions and $4$ $Cl^-$ ions. An edge diagonal passes through one corner,the face center,and the opposite corner. Along an edge diagonal,the ions present are $Na^+$ at the edge center and $Cl^-$ at the corners. Removing ions along an edge diagonal removes $1$ $Na^+$ (at the edge center) and $2 	imes \frac{1}{4} = 0.5$ $Na^+$ ions (at the corners). Total $Na^+$ ions removed $= 1 + 0.5 = 1.5$. Remaining $Na^+$ ions $= 4 - 1.5 = 2.5 = \frac{5}{2}$. Number of $Cl^-$ ions remains $4$. Packing Fraction $(P.F.) = \frac{	ext{Volume of ions}}{	ext{Volume of unit cell}} = \frac{\frac{4}{3}\pi (\frac{5}{2}r_+^3 + 4r_-^3)}{a^3}$. For $NaCl$,$a = 2(r_+ + r_-)$. Assuming $r_- \approx \sqrt{2}(r_+ + r_-)$ is not applicable here,but using the standard edge length relation $a = 2(r_+ + r_-)$,the volume is $a^3 = 8(r_+ + r_-)^3$. However,based on the provided options,the denominator $16\sqrt{2} r_-^3$ is used. Thus,$P.F. = \frac{\frac{4}{3}\pi (\frac{5}{2}r_+^3 + 4r_-^3)}{16\sqrt{2} r_-^3}$.

Select the correct expression for determining the Packing Fraction $(P.F.)$ of an $NaCl$ unit cell (assume ideal),if ions along an edge diagonal are absent.

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