Question

A' are arranged on FCC system. 'B' are occupied in all the octahedral voids and half of the tetrahedral voids. The formula of the compound is?

Answer 1

No. Of A atoms / unit cell = 4(FCC) No. Of B atoms / unit cell = 4 ( octahedral voids) + 4 ( 1/2 of tetrahedral voids) = 8 Therefore formula is A4B8 = AB2

Answer 2

Answer:

The formula of the compound is: $AB_2$.

Explanation:

Number of sphere sin FCC = n

Number of octahedral voids = n

Number of tetrahedral voids = 2n

Atom A is arranged in FCC . atom B is in all the octahedral voids and half of the tetrahedral voids

Number of A atoms = n

Number of atom B in octahederal void= n

Number of atom B in tetrahedral void = $\frac{2n}{2}=n$

Total atom of B = n + n = 2n

The formula of the compound is:

$A_nB_2n=AB_2$