A binary solid $(A^{+} B^{-})$ has a zinc blende structure with $B^{-}$ ions constituting the lattice and $A^{+}$ ions occupying $25 %$ tetrahedral voids. The formula of solid is

A B

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A_{2} B

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A B_{2}

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A B_{4}

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Solution

The correct option is C $A B_{2}$
Since $A^{+} B^{-}$ has a zinc blende structure, so it has a cubic close packed structure.
Since $B^{-}$ ions constituting the lattice, thus number of $B^{-}$ ions $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$
Again, number of tetrahedral voids would be $8$
So, number of $A^{-}$ ions would be $= \frac{1}{4} \times 8 = 2$
SO the formula of the solid is $A_{2} B_{4} or A B_{2}$

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