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Solid State

Section: Chemistry

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Question : 5 of 19

Marks: +1, -0

For the given close packed structure of a salt made of cation

X

Y

shown below (ions of only one face are shown for clarity), the packing fraction is approximately

(packing fraction

=\frac{\text{Packing efficiency}}{100}

[JEE Adv 2021 P1]

$0.74$
$0.63$
$0.52$
$0.48$

Solution:

The given system is FCC system.

So, relation between

r

and

a

is

4r = \sqrt{2a} \;\; [\because r = r_{x} + r_{y}]

\therefore \;\; 2(r_{x}+r_{y}) = \sqrt{2a}

\Rightarrow 2 r_{x} + 2 r_{y} = \sqrt{2a}

\Rightarrow 2 r_{x} + a = \sqrt{2a}

\Rightarrow 2 r_{x} + a = \sqrt{2a} \;\; [\because 2 r_{y} = a]

\Rightarrow r_{x} = 0.207 a

or

\;\; r_{y} = 0.5 a

Packing fraction

=\frac{3 \times \text{Vol. of } x + \text{Vol. of } y}{\text{Vol. of unit cell}}

=\frac{3 \times \frac{4}{3} \pi r_{x}^{3} + \frac{4}{3} \pi r_{y}^{3}}{a^{3}} = \frac{4 \pi \times (0.207 a)^{3} + \frac{4}{3} \pi (0.5 a)^{3}}{a^{3}}

=0.63

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