Chemistry - Solid State Question with Solution | TestHub

We have a spinel structure:

O²⁻ ions → ccp (FCC) arrangement

Fe²⁺ → 1/8 tetrahedral voids

Fe³⁺ → 1/2 octahedral voids

The question asks for the minimum internuclear distance between an Fe ion and an Fe ion.

Step 1: Locate Fe ions

In spinel:

Fe²⁺ occupies tetrahedral voids.

Fe³⁺ occupies octahedral voids.

The closest distance between two Fe ions occurs between tetrahedral and octahedral sites.

Step 2: Distance between tetrahedral and octahedral voids in FCC

In an FCC unit cell:

Tetrahedral void is at:

$(\frac{1}{4},\frac{1}{4},\frac{1}{4})$

Nearest octahedral void is at:

$(\frac{1}{2},\frac{1}{2},\frac{1}{2})$

Distance:

$$d=a\sqrt{{\left(\frac{1}{2}-\frac{1}{4}\right)}^2+{\left(\frac{1}{2}-\frac{1}{4}\right)}^2+{\left(\frac{1}{2}-\frac{1}{4}\right)}^2}$$

$$d=a\sqrt{3{\left(\frac{1}{4}\right)}^2}$$

$$d=\frac{\sqrt{3}}{4}a$$

Answer:

$$d_{Fe-Fe}=\frac{\sqrt{3}}{4}a$$