Question

Explain why $[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic whereas $[Fe(CN)_6]^{3-}$ is weakly paramagnetic. [Atomic number of Fe = 26]

Hint:

Weak ligand = High spin (more unpaired $e^-$). Strong ligand = Low spin (fewer unpaired $e^-$).

Answer 1

Step 1: Concept
Crystal Field Theory (CFT) and ligand strength.

Step 2: Meaning
$Fe^{3+}$ has a $3d^5$ configuration. Unpaired electrons determine paramagnetism.

Step 3: Analysis
Water ($H_2O$) is a weak field ligand, so the crystal field splitting energy is smaller than the electron pairing energy \[ (\Delta_o\lt P). \] Therefore, electrons do not pair in the lower-energy orbitals and instead occupy the higher-energy orbitals according to Hund's rule. The electronic configuration becomes \[ t_{2g}^{3}e_g^{2}, \] containing five unpaired electrons. Cyanide ($CN^-$), on the other hand, is a strong field ligand \[ (\Delta_o\gt P), \] which forces electron pairing. The resulting configuration is \[ t_{2g}^{5}e_g^{0}, \] containing only one unpaired electron.

• Consider a metal ion having the electronic configuration: \[ d^5. \]

• In an octahedral crystal field, the five $d$ orbitals split into: \[ t_{2g} \quad\text{and}\quad e_g. \]

• When the ligand is water ($H_2O$), which is a weak field ligand, \[ \Delta_o\lt P. \]

• Since the splitting energy is smaller than the pairing energy, electrons prefer to occupy all five orbitals singly rather than pair up.

• The electronic configuration therefore becomes: \[ t_{2g}^{3}e_g^{2}. \]

• This arrangement contains: \[ \boxed{5\ \text{unpaired electrons}}, \] making the complex strongly paramagnetic.

• Cyanide ($CN^-$) is a strong field ligand and produces a large crystal field splitting: \[ \Delta_o\gt P. \]

• Consequently, electrons pair in the lower-energy \[ t_{2g} \] orbitals before occupying the higher-energy \[ e_g \] orbitals.

• The configuration becomes: \[ t_{2g}^{5}e_g^{0}, \] containing only \[ \boxed{1\ \text{unpaired electron}}. \]

• Thus, weak field ligands form high-spin complexes, whereas strong field ligands form low-spin complexes.

Step 4: Conclusion
More unpaired electrons result in a stronger paramagnetic character.

Final Answer: $[Fe(H_2O)_6]^{3+}$ has a weak field ligand ($H_2O$), resulting in 5 unpaired electrons. $[Fe(CN)_6]^{3-}$ has a strong field ligand ($CN^-$), forcing pairing and resulting in only 1 unpaired electron. Therefore, the former is strongly paramagnetic while the latter is weakly paramagnetic.