Question

"a" has fcc arrangement and b is present in 2/3rd of tetrahedral voids. the formula of the compound is

Answer 1

fcc = 4 atoms
in fcc there are 8 tetrahedral voids
therefore b = 2/3×8 = 16/3

formula = a : b
= 4 : 16/3
= 12 : 16 = 6:8 = 3:4
A3B4

Answer 2

Answer:

The formula of the compound is $a_3b_4$.

Explanation:

Number of atomic sphere of in FCC arrangements,n = 4

Number of atoms of 'a' = 4

Number tetrahedral voids in FCC arrangement, =2n=2× 4 = 8

'b' occupies two third of the tetrahedral void.

So, the numbers of b atoms in a cubic crystal=$\frac{2}{3}\times 8=\frac{16}{3}$

the molecular formula of the compound is:

$a_{4}b_{\frac{16}{3}}=a_3b_4$