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A solid $A^{+} B^{-}$ has the $B^{-}$ Ions arranged as below. If the $A^{+}$ ions occupy half of the tetrahedral sites in the structure, the formula of solid is

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$AB$

${AB}_{2}$

$A_{2} B$

$A_{3} B_{4}$

answer is A.

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Detailed Solution

AB is FCC lattice

Z=4 tetrahedral voids = 8 octahedral voids = 4

$∴ B_{4}, A^{+}$ occupies $\frac{1}{2} \times TV = \frac{1}{2} \times 8 = 4$

$∴ A_{4} B_{4} ≅ AB$

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