JEE Advance - Chemistry (2022 - Paper 2 Online - No. 15)
Atom $\mathrm{X}$ occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to
25
35
55
75
Explanation
Atom $X$ occupies FCC lattice sites as well as alternate tetrahedral voids of FCC.
In FCC, tetrahedral voids are 8 (in a unit cell)
Hence
Atom $X$ in a unit cell (FCC lattice sites) $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
Atom $X$ in a unit cell (in T.V.) $=\frac{1}{2} \times 8=4$
Total atom $\mathrm{X}$ in one unit cell $=8$
For relation between a and r, since T.V. forms at $\frac{1}{4}$ th of body diagonal,
$$ \begin{aligned} & \frac{a \sqrt{3}}{4}=2 r \\\\ &\Rightarrow a=\frac{8 r}{\sqrt{3}} \end{aligned} $$
Packing efficiency
$=\frac{8 \times \frac{4}{3} \pi r^3}{a^3} \times 100=\frac{8 \times \frac{4}{3} \pi r^3}{\left(\frac{8 r}{\sqrt{3}}\right)^3} \times 100 \simeq 35 \%$
In FCC, tetrahedral voids are 8 (in a unit cell)
Hence
Atom $X$ in a unit cell (FCC lattice sites) $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
Atom $X$ in a unit cell (in T.V.) $=\frac{1}{2} \times 8=4$
Total atom $\mathrm{X}$ in one unit cell $=8$
For relation between a and r, since T.V. forms at $\frac{1}{4}$ th of body diagonal,
$$ \begin{aligned} & \frac{a \sqrt{3}}{4}=2 r \\\\ &\Rightarrow a=\frac{8 r}{\sqrt{3}} \end{aligned} $$
Packing efficiency
$=\frac{8 \times \frac{4}{3} \pi r^3}{a^3} \times 100=\frac{8 \times \frac{4}{3} \pi r^3}{\left(\frac{8 r}{\sqrt{3}}\right)^3} \times 100 \simeq 35 \%$
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