Among the compounds $\mathrm{Fe}_{3} \mathrm{O}_{4}, \mathrm{NiFe}_{2} \mathrm{O}_{4}$ and $\mathrm{Mn}_{3} \mathrm{O}_{4}$
(a)$\mathrm{NiFe}_{2} \mathrm{O}_{4}$ and $\mathrm{Mn}_{3} \mathrm{O}_{4}$ ara normal spinels
(b)$\mathrm{Fe}_{3} \mathrm{O}_{4}$ and $\mathrm{Mn}_{3} \mathrm{O}_{4}$ are normal spinels
(c)$\mathrm{Fe}_{3} \mathrm{O}_{4}$ and $\mathrm{Mn}_{3} \mathrm{O}_{4}$ are inverse spinels
(d)$\mathrm{Fe}_{3}\mathrm{O}_{4}$ and $\mathrm{NiFe}_{2}\mathrm{O}_{4}$ are inverse spinels
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Fe₃O₄ (Fe³⁺ in tetrahedral sites, Fe²⁺/Fe³⁺ in octahedral) and NiFe₂O₄ (Ni²⁺ octahedral, d⁸ CFSE) are inverse spinels; Mn₃O₄ is a normal spinel (Mn²⁺ d⁵ has no CFSE preference, Mn³⁺ octahedral).