Question

The correct order about the number of unpaired electrons present in the following complexes is \[ [Fe(CN)_6]^{4-}\quad (I) \] \[ [Fe(H_2O)_6]^{2+}\quad (II) \] \[ [Co(H_2O)_6]^{2+}\quad (III) \]

• A. II \(\gt \) III \(\gt \) I
• B. II \(\gt \) I \(\gt \) III
• C. I \(\gt \) II \(\gt \) III
• D. III \(\gt \) II \(\gt \) I

Hint:

\(CN^-\) is a strong-field ligand and usually produces low-spin complexes, whereas \(H_2O\) is a weak-field ligand and generally produces high-spin complexes.

Answer 1

The Correct Option is A

Step 1: Identify the metal ions and their $d$-electron counts.
Complex I: $[Fe(CN)_6]^{4-}$ contains $Fe^{2+}$. $Fe = [Ar]\,3d^6\,4s^2$, so $Fe^{2+} = 3d^6$ (6 electrons). Complex II: $[Fe(H_2O)_6]^{2+}$ also has $Fe^{2+}$, so $3d^6$. Complex III: $[Co(H_2O)_6]^{2+}$ has $Co^{2+}$. $Co = [Ar]\,3d^7\,4s^2$, so $Co^{2+} = 3d^7$ (7 electrons).
Step 2: Classify the ligands as strong-field or weak-field.
$CN^-$ is a strong-field ligand (high in spectrochemical series): forces large crystal field splitting, promotes electron pairing (low-spin). $H_2O$ is a weak-field ligand: small splitting, gives high-spin configuration.
Step 3: Electron distribution in Complex I: $[Fe(CN)_6]^{4-}$.
$CN^-$ strong-field with $Fe^{2+}$ ($d^6$): low-spin octahedral configuration: \[ t_{2g}^6\ e_g^0 \] All 6 electrons are paired. Unpaired electrons = 0.
Step 4: Electron distribution in Complex II: $[Fe(H_2O)_6]^{2+}$.
$H_2O$ weak-field with $Fe^{2+}$ ($d^6$): high-spin octahedral: \[ t_{2g}^4\ e_g^2 \] $t_{2g}$: 3 singly occupied + 1 paired = 1 extra unpaired; $e_g$: 2 unpaired. Total unpaired = 4.
Step 5: Electron distribution in Complex III: $[Co(H_2O)_6]^{2+}$.
$H_2O$ weak-field with $Co^{2+}$ ($d^7$): high-spin octahedral: \[ t_{2g}^5\ e_g^2 \] $t_{2g}$: 2 paired + 1 unpaired; $e_g$: 2 unpaired. Total unpaired = 3.
Step 6: Arrange in order of unpaired electrons.
II = 4, III = 3, I = 0. \[ II > III > I \]
Step 7: State the final answer.
\[ \boxed{II > III > I} \]