Question

In a solid oxide ions are arranged in ccp. Cation a occupy one sixth of octahedral void and cation b occupy one third of octahedral voids. What is the formula of the compound

Answer 1

Answer:

$A_{2}B_{4}O_3$

Explanation:

There are four octahedral in a CCP.

Cation occupies 1/6 of octahedral.

So, there are 4*1/6 = 2/3 of the A in unit cell

and 4*1/3 of B in the unit cell of unit cell

So the formula is $A_{2/3}B_{4/3}O$

Hence the formula is

$A_2B_4O_3$