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: Electronic configuration of $Fe^{3+}$.
Iron ($Z = 26$) has the configuration $[Ar]~3d^6~4s^2$. $Fe^{3+}$ loses 3 electrons to give $[Ar]~3d^5$ — five electrons in the d subshell.
Step 2: Weak-field ligand water in $[Fe(H_2O)_6]^{3+}$.
Water is a weak-field ligand. The crystal field splitting energy $\Delta_o$ is smaller than the electron pairing energy $P$ ($\Delta_o < P$). Electrons therefore occupy all five $d$ orbitals singly (Hund's rule), giving a high-spin configuration $t_{2g}^3~e_g^2$ with 5 unpaired electrons. Strong paramagnetism results.
Step 3: Strong-field ligand cyanide in $[Fe(CN)_6]^{3-}$.
Cyanide ($CN^-$) is a strong-field ligand. $\Delta_o > P$, so electrons are forced to pair in the lower $t_{2g}$ orbitals before occupying $e_g$, giving a low-spin configuration $t_{2g}^5~e_g^0$ with only 1 unpaired electron. Only weak paramagnetism results.
Step 4: Summary.
$[Fe(H_2O)_6]^{3+}$: weak-field $H_2O$, high-spin $t_{2g}^3~e_g^2$, 5 unpaired electrons, strongly paramagnetic. $[Fe(CN)_6]^{3-}$: strong-field $CN^-$, low-spin $t_{2g}^5~e_g^0$, 1 unpaired electron, weakly paramagnetic.