In an A_{x} B_{y} crystal structure,A atoms o | vedclass

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(D) In an $FCC$ unit cell,the number of atoms $(n)$ is $4$. Number of octahedral voids $= n = 4$. Number of tetrahedral voids $= 2n = 8$. Given that $

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$A_3 B$

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(D) In an $FCC$ unit cell,the number of atoms $(n)$ is $4$. Number of octahedral voids $= n = 4$. Number of tetrahedral voids $= 2n = 8$. Given that $A$ atoms occupy all octahedral and all tetrahedral voids: Total number of $A$ atoms $= 4 + 8 = 12$. Given that $B$ atoms are at the $FCC$ lattice points: Total number of $B$ atoms $= 4$. The ratio of $A:B$ is $12:4$,which simplifies to $3:1$. Therefore,the formula of the compound is $A_3 B$.

In an $A_{x} B_{y}$ crystal structure,$A$ atoms occupy all octahedral as well as all tetrahedral voids,and $B$ atoms are at the $FCC$ lattice points. What is the formula of the compound $A_{x} B_{y}$?

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