In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1 / 8 tetrahedral holes are occupied by Z n2 +ions and 50 % of octahedral holes are occupied by F e3 + ions . The formula of the compound is -

Question

In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1 / 8 tetrahedral holes are occupied by Z n2 +ions and 50 % of octahedral holes are occupied by F e3 + ions . The formula of the compound is - (A) Z n2F e2O4 (B) Z n F e2O3 (C) Z n F e2O4 (D) Z n F e2O2

Tardigrade Admin · Accepted Answer

In one unit cell no. of O2 +=4 The no. of Z n2 +=(1/8)× 8=1 The no. of F e3 +=(1/2)× 4=2 ∴ molecular formula of given spinel structure is Z n F e2O4

Q. In a normal spinel types structure, the oxide ions are arranged in $cc p$ whereas $1/8$ tetrahedral holes are occupied by $Z n^{2 +} i o n s$ and $50%$ of octahedral holes are occupied by $F e^{3 +} i o n s$ . The formula of the compound is -

$Z n_{2} F e_{2} O_{4}$

$Z n F e_{2} O_{3}$

$Z n F e_{2} O_{4}$

$Z n F e_{2} O_{2}$

In one unit cell no. of $O^{2 +} = 4$
The no. of $Z n^{2 +} = \frac{1}{8} \times 8 = 1$
The no. of $F e^{3 +} = \frac{1}{2} \times 4 = 2$
$∴$ molecular formula of given spinel structure is $Z n F e_{2} O_{4}$

Q. In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1/8 tetrahedral holes are occupied by Zn2+ions and 50% of octahedral holes are occupied by Fe3+ions . The formula of the compound is -

Zn2​Fe2​O4​

ZnFe2​O3​

ZnFe2​O4​

ZnFe2​O2​