In a face-centered cubic (FCC) lattice,A atom | vedclass

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(B) In an $FCC$ unit cell,the number of $A$ atoms at the corners is $8 	imes \frac{1}{8} = 1$.Initially,there are $6$ face centers,so the number of $

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$A_2B_5$

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(B) In an $FCC$ unit cell,the number of $A$ atoms at the corners is $8 	imes \frac{1}{8} = 1$.Initially,there are $6$ face centers,so the number of $B$ atoms is $6 	imes \frac{1}{2} = 3$.After removing one $B$ atom from a face center,the number of $B$ atoms becomes $6 - 1 = 5$ face centers remaining,so the number of $B$ atoms is $5 	imes \frac{1}{2} = \frac{5}{2}$.The ratio of $A:B$ is $1 : \frac{5}{2}$,which simplifies to $2 : 5$.Therefore,the molecular formula is $A_2B_5$.

In a face-centered cubic $(FCC)$ lattice,$A$ atoms are present at the corners and $B$ atoms are present at the face centers. If one $B$ atom is removed from one of the face centers,what will be the molecular formula of the compound?

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