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Question

(A) In an $fcc$ lattice,the number of atoms of $A$ per unit cell is $4$.The number of octahedral voids is equal to the number of atoms in the lattice,

Vedclass · Accepted Answer

$AB_2$

Vedclass · Answer

(A) In an $fcc$ lattice,the number of atoms of $A$ per unit cell is $4$.The number of octahedral voids is equal to the number of atoms in the lattice,which is $4$.The number of tetrahedral voids is twice the number of atoms in the lattice,which is $2 	imes 4 = 8$.Given that $B$ occupies all octahedral voids $(4)$ and half of the tetrahedral voids $(8 	imes \frac{1}{2} = 4)$,the total number of $B$ atoms is $4 + 4 = 8$.Thus,the ratio of $A:B$ is $4:8$,which simplifies to $1:2$.Therefore,the molecular formula is $AB_2$.

In a solid compound,the atoms of $A$ form a face-centered cubic $(fcc)$ lattice,while atoms of $B$ occupy all octahedral voids and half of the tetrahedral voids. The molecular formula of the solid will be ........

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