A binary compound (A^{+} B^{-}) has a zinc bl | vedclass

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Question

(C) In a zinc blende structure,the anions $(B^{-})$ form a face-centered cubic $(FCC)$ lattice.Let the number of $B^{-}$ ions per unit cell be $N = 4$

Vedclass · Accepted Answer

$AB_2$

Vedclass · Answer

(C) In a zinc blende structure,the anions $(B^{-})$ form a face-centered cubic $(FCC)$ lattice.Let the number of $B^{-}$ ions per unit cell be $N = 4$.The number of tetrahedral voids in an $FCC$ lattice is $2N = 2 	imes 4 = 8$.The $A^{+}$ ions occupy $25\%$ of these tetrahedral voids.Number of $A^{+}$ ions $= 25\% 	ext{ of } 8 = 0.25 	imes 8 = 2$.The ratio of $A^{+} : B^{-}$ is $2 : 4$,which simplifies to $1 : 2$.Therefore,the formula of the compound is $AB_2$.

$A$ binary compound $(A^{+} B^{-})$ has a zinc blende structure where $B^{-}$ ions form the lattice and $A^{+}$ ions occupy $25\%$ of the tetrahedral voids. What is the formula of the compound?

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