The FCC unit cell of a compound contains ions | vedclass

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(C) In an $FCC$ unit cell,the number of atoms at the corners is $8 	imes \frac{1}{8} = 1$.Since ions of $A$ are at the corners,the number of $A$ atom

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$AB_3$

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(C) In an $FCC$ unit cell,the number of atoms at the corners is $8 	imes \frac{1}{8} = 1$.Since ions of $A$ are at the corners,the number of $A$ atoms per unit cell is $1$.In an $FCC$ unit cell,the number of atoms at the face centres is $6 	imes \frac{1}{2} = 3$.Since ions of $B$ are at the face centres,the number of $B$ atoms per unit cell is $3$.Therefore,the ratio of $A:B$ is $1:3$.The formula of the compound is $AB_3$.

The $FCC$ unit cell of a compound contains ions of $A$ at the corners and ions of $B$ at the centre of each face. What is the formula of the compound?

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