In antifluorite structure,the nearest neighbo | vedclass

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Question

(C) In an antifluorite structure (e.g.,$Na_2O$),the anions $(O^{2-})$ form an $fcc$ lattice,and the cations $(Na^+)$ occupy all the tetrahedral voids.

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) In an antifluorite structure (e.g.,$Na_2O$),the anions $(O^{2-})$ form an $fcc$ lattice,and the cations $(Na^+)$ occupy all the tetrahedral voids.For an $fcc$ unit cell with edge length $a$,the distance between the center of an atom at the corner and the center of an atom in the tetrahedral void (nearest neighbour distance) is given by $d = \frac{\sqrt{3}}{4}a$.Given that the nearest neighbour distance $d = 20\sqrt{3} \ pm$.Equating the two: $\frac{\sqrt{3}}{4}a = 20\sqrt{3}$.Solving for $a$: $a = 20 	imes 4 = 80 \ pm$.

In antifluorite structure,the nearest neighbour distance is $20\sqrt{3} \ pm$. Then,the edge length of the unit cell is ............... $pm$.

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