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: Determine the electronic configuration of \(Fe^{2+}\) and \(Co^{2+}\).
For iron: \[ Fe=[Ar]\,3d^6\,4s^2 \] Therefore, \[ Fe^{2+}=[Ar]\,3d^6 \] For cobalt: \[ Co=[Ar]\,3d^7\,4s^2 \] Therefore, \[ Co^{2+}=[Ar]\,3d^7 \]

Step 2: Analyze complex (I).
\[ [Fe(CN)_6]^{4-} \] \(CN^-\) is a strong field ligand.
Hence, pairing occurs and the complex is low spin: \[ t_{2g}^{6}e_g^{0} \] Number of unpaired electrons: \[ 0 \]

Step 3: Analyze complex (II).
\[ [Fe(H_2O)_6]^{2+} \] \(H_2O\) is a weak field ligand.
The complex is high spin: \[ t_{2g}^{4}e_g^{2} \] Number of unpaired electrons: \[ 4 \]

Step 4: Analyze complex (III).
\[ [Co(H_2O)_6]^{2+} \] \(Co^{2+}\) is \(d^7\) and \(H_2O\) is a weak field ligand.
The complex is high spin: \[ t_{2g}^{5}e_g^{2} \] Number of unpaired electrons: \[ 3 \]

Step 5: Arrange the order.
\[ II=4\ \text{unpaired electrons} \] \[ III=3\ \text{unpaired electrons} \] \[ I=0\ \text{unpaired electrons} \] Therefore, \[ \boxed{II\gt III\gt I} \]

Step 6: Final conclusion.
Hence, the correct option is \[ \boxed{(1)} \]