In a face-centred cubic lattice,atom A occupi | vedclass

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(A) In a face-centred cubic $(FCC)$ lattice:Number of atoms of $A$ at corners = $8 	imes \frac{1}{8} = 1$.Total number of face-centred positions in a

Vedclass · Accepted Answer

$A_2B_5$

Vedclass · Answer

(A) In a face-centred cubic $(FCC)$ lattice:Number of atoms of $A$ at corners = $8 	imes \frac{1}{8} = 1$.Total number of face-centred positions in an $FCC$ unit cell is $6$.Since one atom of $B$ is missing from one of the face-centred positions,the number of atoms of $B$ present = $6 - 1 = 5$.Contribution of each face-centred atom to the unit cell = $\frac{1}{2}$.Number of atoms of $B = 5 	imes \frac{1}{2} = \frac{5}{2}$.The ratio of atoms $A : B = 1 : \frac{5}{2} = 2 : 5$.Therefore,the formula of the compound is $A_2B_5$.

In a face-centred cubic lattice,atom $A$ occupies the corner positions and atom $B$ occupies the face-centred positions. If one atom of $B$ is missing from one of the face-centred points,the formula of the compound is

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