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 $B^{-}$ ions per unit cell is $4$

Vedclass · Accepted Answer

$50$

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(C) In a zinc blende $(ZnS)$ structure,the anions $(B^{-})$ form a face-centered cubic $(fcc)$ lattice.The number of $B^{-}$ ions per unit cell is $4$.The number of tetrahedral voids in an $fcc$ lattice is $2 	imes (	ext{number of atoms}) = 2 	imes 4 = 8$.Since the formula is $AB$,the number of $A^{+}$ ions per unit cell must be $4$.Therefore,the fraction of tetrahedral voids occupied by $A^{+}$ ions is $\frac{4}{8} = 0.5$,which is $50 \%$.

If $AB$ crystallizes in a zinc blende structure with $B^{-}$ ions at lattice points,what fraction of tetrahedral voids is occupied by $A^{+}$ ions? ............... $\%$

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