Question

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

• A. $\frac{\frac{4}{3}\pi (r_{+}^3\times 4+r_{-}^3\times \frac{15}{4})}{8(r_{+}+r_{-}{)}^3}$
• B. $\frac{\frac{4}{3}\pi (r_{+}^3\times 3+r_{-}^3\times \frac{15}{4})}{8(r_{+}+r_{-}{)}^3}$
• C. $\frac{\frac{4}{3}\pi (r_{+}^3\times 3+r_{-}^3\times \frac{5}{2})}{8(r_{+}+r_{-}{)}^3}$
• D. $\frac{\frac{4}{3}\pi (r_{+}^3\times \frac{5}{2}+r_{-}^3\times \frac{7}{2})}{8(r_{+}+r_{-}{)}^3}$

Answer 1

The correct option is B $\frac{\frac{4}{3}\pi (r_{+}^3\times 3+r_{-}^3\times \frac{15}{4})}{8(r_{+}+r_{-}{)}^3}$

When ions along an body diagonal in $NaCl$ unit cell are absent, then two $C{l}^{-}$ ions at the corner and one $N{a}^{+}$ ion at the body center is removed.
So, effective number of $N{a}^{+}=4-1=3$
Volume of $N{a}^{+}=\frac{4}{3}\pi r_{+}^3\times 3$
Effective number of $C{l}^{-}=4-(2\times \frac{1}{8})=\frac{15}{4}$
Volume of $C{l}^{-}=\frac{4}{3}\pi r_{-}^3\times \frac{15}{4}$
Edge length of unit cell $\left(a\right)=2r_{+}+2r_{-}$
Volume of unit cell $=a^3=(2r_{+}+2r_{-}{)}^3$
So the packing fraction is
$P.F.=\frac{\text{Volume of effective number of cations and anions}}{\text{Volume of unit cell}}=\frac{\frac{4}{3}\pi r_{+}^3\times 3+\frac{4}{3}\pi r_{-}^3\times \frac{15}{4}}{(2r_{+}+2r_{-}{)}^3}=\frac{\frac{4}{3}\pi (r_{+}^3\times 3+r_{-}^3\times \frac{15}{4})}{8(r_{+}+r_{-}{)}^3}$