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(A) In an $fcc$ unit cell,the number of atoms $A$ at the corners is $8 	imes \frac{1}{8} = 1$.Initially,the number of atoms $B$ at the face centers i

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$A_2B_5$

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(A) In an $fcc$ unit cell,the number of atoms $A$ at the corners is $8 	imes \frac{1}{8} = 1$.Initially,the number of atoms $B$ at the face centers is $6 	imes \frac{1}{2} = 3$.If one atom $B$ is removed from a face center,the remaining number of atoms $B$ is $3 - \frac{1}{2} = 2.5 = \frac{5}{2}$.The ratio of $A:B$ is $1 : \frac{5}{2}$,which simplifies to $2 : 5$.Therefore,the formula of the compound is $A_2B_5$.

In an $fcc$ crystal lattice,atoms $A$ are at the corners of the cube and atoms $B$ are at the face centers. If one atom is removed from one of the face centers,find the formula of the compound.

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