AP EAMCET · Chemistry · Coordination Compounds
The correct order about the number of unpaired electrons present in the following complexes is
\[
\left.\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} \underset{\text { II }}{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \underset{\text { III }}{\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6}\right]^{2+}
\]
- A II \(>\) III \(>\) I
- B II \(>\) I \(>\) III
- C I \(>\) II \(>\) III
- D III \(>\) II \(>\) I
Answer & Solution
Correct Answer
(A) II \(>\) III \(>\) I
Step-by-step Solution
Detailed explanation
I. \(\left[\mathrm{Fe}(\mathrm{CN})_6 \mathrm{I}^{4-}\right.\) Atomic number of \(\mathrm{Fe}=26\) Ground state configuration of \(\mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2\) \[ \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6 \] Since, cyanide is a powerful ligand, therefore, all the electrons…
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