The spinel structure consists of an array of O2 - ions in fcc arrangement. General formula of spinel is AB2O4 . Cations of A occupy (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:

Question

The spinel structure consists of an array of O2 - ions in fcc arrangement. General formula of spinel is AB2O4 . Cations of A occupy (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

Tardigrade Admin · Accepted Answer

These are four O2 - ions in the unit cell; let these ions are replaced by n, X- 8 / 3 ions. Charge of O2 - 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.

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These are four O2− ions in the unit cell; let these ions are replaced by n, X−8/3 ions. Charge of O2− 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.

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.