Q52 · CSIR-NET Chemistry, December 2013
Paper: CSIR-NET December 2013 · Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Coordination Chemistry – General · Marks: 2 · Difficulty: Easy
$\mathrm{MnCr}_{2} \mathrm{O}_{4}$ is likely to have a normal spinel structure because
(a)$\mathrm{Mn}^{2+}$ will have a LFSE in the octahedral site whereas the Cr will not
(b)Mn is $^{2+}$ oxidation state and both the Cr are in $^{3+}$ oxidation state.
(c)Mn is $^{3+}$ oxidation state and 1 Cr is in $^{2+}$ and the other is in $^{3+}$ state.
(d)$\mathrm{Cr}^{3+}$ will have a LFSE in the octahedral site whereas the $\mathrm{Mn}^{2+}$ ion will not.
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Cr³⁺ gains large octahedral LFSE, Mn²⁺ (d⁵) none — normal spinel (d).
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