Atoms A are arranged in an FCC system,and ato | vedclass

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(C) In an $FCC$ unit cell,the number of effective atoms is $n = 4$.Number of octahedral voids $= n = 4$.Number of tetrahedral voids $= 2n = 8$.Atom $A

Vedclass · Accepted Answer

$AB_2$

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(C) In an $FCC$ unit cell,the number of effective atoms is $n = 4$.Number of octahedral voids $= n = 4$.Number of tetrahedral voids $= 2n = 8$.Atom $A$ is in the $FCC$ lattice,so the number of $A$ atoms $= 4$.Atom $B$ occupies all octahedral voids $(4)$ and half of the tetrahedral voids $(8 / 2 = 4)$.Total number of $B$ atoms $= 4 + 4 = 8$.The ratio of $A:B = 4:8 = 1:2$.Therefore,the formula of the compound is $AB_2$.

Atoms $A$ are arranged in an $FCC$ system,and atoms $B$ occupy all the octahedral voids and half of the tetrahedral voids. The formula of the compound is:

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