In a face-centered cubic arrangement of A and | vedclass

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(C) In a face-centered cubic $(FCC)$ unit cell,atoms at the corners contribute $8 	imes \frac{1}{8} = 1$ atom per unit cell.There are $6$ faces in a

Vedclass · Accepted Answer

$A_2B_5$

Vedclass · Answer

(C) In a face-centered cubic $(FCC)$ unit cell,atoms at the corners contribute $8 	imes \frac{1}{8} = 1$ atom per unit cell.There are $6$ faces in a unit cell,and each face center atom contributes $\frac{1}{2}$ to the unit cell.Initially,the number of $B$ atoms is $6 	imes \frac{1}{2} = 3$.Since one $B$ atom is missing from one face,the number of $B$ atoms remaining is $3 - \frac{1}{2} = \frac{5}{2}$.The ratio of $A:B$ is $1 : \frac{5}{2}$,which simplifies to $2 : 5$.Therefore,the simplest formula of the compound is $A_2B_5$.

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

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