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$A$ forms $h c p$ lattice and $B$ are occupying $1 / 3^{\text {rd }}$ of tetrahedral voids, then the formula of compound is
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The correct answer is:
$A_3 B_2$
Let no. of atoms of $A$ used in close packing$=n$
Number of tetrahedral voids $=2 n$
Number of atoms of $B=\frac{1}{3} \times 2 n=\frac{2}{3} n$
$A: B=n: \frac{2}{3} n=3: 2$
Formula of the compound $=A_3 B_2$
Number of tetrahedral voids $=2 n$
Number of atoms of $B=\frac{1}{3} \times 2 n=\frac{2}{3} n$
$A: B=n: \frac{2}{3} n=3: 2$
Formula of the compound $=A_3 B_2$
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