QJEEAdv21Chf454: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Chf454 Solution: For the given close packed structure of a salt made of cation X …

Radius of cation X = r₊

Number of cation X present = $6 \times {1 \over 2} = 3$

Radius of anion Y = r _$-$

Number of anion Y present = ${1 \over 8} \times 8 = 1$

Side of unit cell = a

As two anions are touching each other, so side of unit cell,

a = 2r _$-$

Volume of unit cell = a³ = (2r _$-$ )³ = 8r $_ - ^3$

Volume of cations X = Number of cation $\times$ Volume of one cation = 3 $\times$ ${4 \over 3}\pi r_ + ^3$

Volume of anions Y = Number of anion $\times$ Volume of one anion = 1 $\times$ ${4 \over 3}\pi r_ - ^3$

Radius ratio for close packed structure, ${{{r_ + }} \over {{r_ - }}} = 0.414$

Packing efficiency = ${{Volume\,of\,cation \times Number\,of\,cations + Volume\,of\,anion \times Number\,of\,anions} \over {Volume\,of\,unit\,cell}} \times 100$

Packing fraction = ${{Packing\,efficiency} \over {100}}$

$= {{1 \times {4 \over 3}\pi r_ - ^3 + 3 \times {4 \over 3}\pi r_ + ^3} \over {8r_ - ^3}}$

$= {{4\pi } \over {3 \times 8}}\left[ {{{r_ - ^3} \over {r_ - ^3}} + {{3r_ + ^3} \over {r_ - ^3}}} \right] = {\pi \over 6}[1 + 3{(0.414)^3}]$

$= {{3.14} \over 6}(1 + 3 \times 0.0710) = {{3.14} \over 6} \times 1.213 = 0.63$