In a face-centered cubic arrangement of atoms | vedclass

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(C) In a face-centered cubic unit cell,there are $8$ corners and $6$ face centers.Number of atoms $A$ at corners $= 8 	imes \frac{1}{8} = 1$.If one a

Vedclass · Accepted Answer

$A_7B_{24}$

Vedclass · Answer

(C) In a face-centered cubic unit cell,there are $8$ corners and $6$ face centers.Number of atoms $A$ at corners $= 8 	imes \frac{1}{8} = 1$.If one atom $A$ is missing from one corner,the number of atoms $A$ per unit cell $= 7 	imes \frac{1}{8} = \frac{7}{8}$.Number of atoms $B$ at face centers $= 6 	imes \frac{1}{2} = 3$.The ratio of $A : B = \frac{7}{8} : 3$.Multiplying both sides by $8$,we get $A : B = 7 : 24$.Therefore,the simplest formula of the compound is $A_7B_{24}$.

In a face-centered cubic arrangement of atoms $A$ and $B$,atoms $A$ are at the corners of the unit cell and atoms $B$ are at the face centers. If one atom $A$ is missing from one corner of the unit cell,what is the simplest formula of the compound?

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