In a crystal,atoms of A form an FCC lattice,a | vedclass

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(B) 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

$A_2B_2C$

Vedclass · Answer

(B) 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,so the number of atoms of $B$ is $4$.The number of tetrahedral voids is twice the number of atoms in the lattice,which is $2 	imes 4 = 8$.Atoms of $C$ occupy $25\%$ of the tetrahedral voids,so the number of atoms of $C$ is $8 	imes 0.25 = 2$.The ratio of atoms $A:B:C$ is $4:4:2$,which simplifies to $2:2:1$.Therefore,the molecular formula is $A_2B_2C$.

In a crystal,atoms of $A$ form an $FCC$ lattice,atoms of $B$ occupy all octahedral voids,and atoms of $C$ occupy $25\%$ of the tetrahedral voids. The possible molecular formula of the compound is:

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