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In a $A_x B_y$ crystal structure, $A^{+y}$ ions occupy all the tetrahedral voids and $\mathrm{B}^{-\mathrm{x}}$ ions make $\mathrm{BCC}$ unit cell: What is the formula of the compound?
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The correct answer is:
$\mathrm{A}_4 \mathrm{~B}_2$
Number of $\mathrm{A}^{+\mathrm{y}}$ ions $=2 \times$ number of atoms per unit cell.
Number of $\mathrm{B}^{-\mathrm{x}}$ ions $=2(\mathrm{BCC})$ per unit cell.
$\Rightarrow$ Number of $\mathrm{A}^{+\mathrm{y}}$ ions $=2 \times 2=4$
$\Rightarrow$ Formula $=\mathrm{A}_4 \mathrm{~B}_2$
Number of $\mathrm{B}^{-\mathrm{x}}$ ions $=2(\mathrm{BCC})$ per unit cell.
$\Rightarrow$ Number of $\mathrm{A}^{+\mathrm{y}}$ ions $=2 \times 2=4$
$\Rightarrow$ Formula $=\mathrm{A}_4 \mathrm{~B}_2$
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