If AB crystallizes in a zinc blende structure | vedclass

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(C) In a zinc blende $(ZnS)$ structure,the anions $(B^{-})$ form a face-centered cubic $(fcc)$ lattice. The number of atoms per unit cell for an $fcc$

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) In a zinc blende $(ZnS)$ structure,the anions $(B^{-})$ form a face-centered cubic $(fcc)$ lattice. The number of atoms per unit cell for an $fcc$ lattice is $4$. The number of tetrahedral voids in an $fcc$ lattice is $2 	imes (	ext{number of atoms}) = 2 	imes 4 = 8$. Since the stoichiometry of the compound is $AB$,there must be $4$ $A^{+}$ ions per unit cell to maintain electrical neutrality. The fraction of tetrahedral voids occupied by $A^{+}$ ions is $\frac{	ext{Number of } A^{+} 	ext{ ions}}{	ext{Total tetrahedral voids}} = \frac{4}{8} = 0.5$. Expressed as a percentage,this is $0.5 	imes 100 = 50 \%$.

If $AB$ crystallizes in a zinc blende structure with $B^{-}$ ions at lattice points,what fraction of tetrahedral voids is occupied by $A^{+}$ ions? (Express as a percentage)

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