Question

Which of the following compounds represents an inverse 2: 3 spinel structure?
A. $\text{F}{{\text{e}}^{\text{III}}}(\text{F}{{\text{e}}^{\text{II}}}\text{F}{{\text{e}}^{\text{III}}}){{\text{O}}_{4}}$
B. $\text{Pb}{{\text{O}}_{2}}$
C. $\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}$
D. $\text{M}{{\text{n}}_{3}}{{\text{O}}_{4}}$

Answer 1

Hint: Inverse spinel structure consists of the divalent and trivalent cation and a negatively charged element such as oxygen, sulphur, etc. The formula of the inverse spinel unit cell is $\text{B(AB)}{{\text{O}}_{\text{4}}}$. Here A is the divalent cation, B is the trivalent cation whereas O is the negatively charged element.

Complete step by step answer:
- In the given question we have to find out the inverse spinel structure among the given options.
- As we know that inverse spinel is different from that of the normal spinel because in the normal spinel the general formula is $\text{A}{{\text{B}}_{2}}{{\text{X}}_{4}}$ whereas for inverse spinel the general formula is $\text{B(AB)}{{\text{O}}_{4}}$.
-So, we can say that element A has the oxidation state of +2 whereas the oxidation state of B is +3.
- So, now to identify the molecule which has inverse spinel structure we can see that in the first option i.e. $\text{F}{{\text{e}}^{\text{III}}}(\text{F}{{\text{e}}^{\text{II}}}\text{F}{{\text{e}}^{\text{III}}}){{\text{O}}_{4}}$ or $\text{F}{{\text{e}}_{3}}{{\text{O}}_{4}}$.
- The oxidation state of $\text{F}{{\text{e}}^{\text{III}}}$ which is element B is +3 because according to the rules of the oxidation the ion present in the element is its oxidation state.
- Whereas the oxidation state of $\text{F}{{\text{e}}^{\text{II}}}$ which is element A is +2.
- Also, $\text{F}{{\text{e}}^{+3}}$ is present at the $\dfrac{1}{8}$ tetrahedral void and $\dfrac{1}{4}$ octahedral void whereas $\text{F}{{\text{e}}^{+2}}$ is present at the $\dfrac{1}{8}$ tetrahedral void.
- So, the ratio of $\text{F}{{\text{e}}^{+3}}:\ \text{F}{{\text{e}}^{+2}}$ is 2:1 that means for one atom of $\text{F}{{\text{e}}^{+2}}$ two atoms of $\text{F}{{\text{e}}^{+3}}$ is present.
- As we know that oxygen will be present at the FCC or Face centred Cubic lattice, so the atoms are present at the corners and the face edge.
- Total no. of atoms at the corner will be: $\dfrac{1}{8}\text{ }\times \text{ 8 = 1}$ because each atom will contribute to eight units such that per unit contribution of the cell is $\dfrac{1}{8}$.
- Similarly, at the face edge total no. of atoms are 3 because per unit contribution of the cell is $\dfrac{1}{2}$
So, the total no. of atoms present are $\dfrac{1}{2}\text{ }\times \text{ 6 = 3}$.
- So, the total no. of atoms of the oxygen is (1+3)=4.

- Hence, the formula becomes $\text{F}{{\text{e}}^{\text{III}}}(\text{F}{{\text{e}}^{\text{II}}}\text{F}{{\text{e}}^{\text{III}}}){{\text{O}}_{4}}\text{ or F}{{\text{e}}^{+3}}(\text{F}{{\text{e}}^{+2}}\text{F}{{\text{e}}^{+3}}){{\text{O}}_{4}}$ or $\text{F}{{\text{e}}_{3}}{{\text{O}}_{4}}$.
So, the correct answer is “Option A”.

Note: The main basic difference between the octahedral and tetrahedral void is that in tetrahedral, the void is surrounded by the four spheres whereas in octahedral, the void is surrounded by six spheres.