Question

The spinel structure consists of an array of $O^{2 -}$ ions in fcc arrangement. General formula of spinel is $AB_{2}O_{4}$ . Cations of $A$ occupy $\frac{1}{8}^{th}$ the tetrahedral voids and cations of $B$ ions occupy half of the octahedral voids. If oxide ions are replaced by $X^{- 8 / 3}$ ions then number of an ionic vacancy per unit cell will be:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

These are four $O^{2 -}$ ions in the unit cell; let these ions are replaced by n, $X^{- 8 / 3}$ ions.
Charge of $O^{2 -}$ ions = Charge of $X^{- 8 / 3}$ ions
$4(-2) = n (-8/3)$
$n = 3 $(i.e., there will one anionic vacancy)
Thus, there will be one an ionic vacancy per unit cell.