Question:medium

\([\mathrm{Fe(H_2O)_6}]^{2+}\) is paramagnetic whereas \([\mathrm{Fe(CO)_5}]\) is diamagnetic. Justify the statement. [Atomic number of Fe = 26]

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Always determine the oxidation state of the metal ion first. Then identify whether the ligand is a weak-field ligand (\(\mathrm{H_2O}, \mathrm{F^-}, \mathrm{Cl^-}\)) or a strong-field ligand (\(\mathrm{CO}, \mathrm{CN^-}, \mathrm{NH_3}\)). Weak-field ligands generally produce high-spin (paramagnetic) complexes, whereas strong-field ligands produce low-spin (often diamagnetic) complexes.
Updated On: Jul 22, 2026
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Solution and Explanation

Step 1: Fix the oxidation state and base configuration of iron in each complex.
In $[Fe(H_2O)_6]^{2+}$ iron is in the $+2$ state, giving the configuration $[Ar]3d^6$. In $[Fe(CO)_5]$ the CO ligands are neutral, so iron stays in the $0$ state, also starting from $[Ar]3d^6 4s^2$ before bonding.
Step 2: Classify the ligand strength in each case.
Water sits low on the spectrochemical series and is a weak field ligand, while carbon monoxide sits at the very strong end of the series.
Step 3: Work out how the electrons arrange themselves in each field.
Because water cannot supply enough splitting energy to force pairing, the six $3d^6$ electrons of $Fe^{2+}$ spread out as far as possible, leaving four orbitals singly occupied, which is why the aqua complex is paramagnetic. Carbon monoxide, being a strong field ligand, provides enough splitting energy that all six d-electrons of iron pair up completely, using $dsp^3$ hybridisation, so no unpaired electron remains.
Step 4: State the conclusion for each complex.
$[Fe(H_2O)_6]^{2+}$ has four unpaired electrons and is paramagnetic, while $[Fe(CO)_5]$ has all electrons paired and is diamagnetic, purely because of the difference in ligand field strength between water and carbon monoxide. \[ \boxed{\text{Weak field } H_2O \Rightarrow \text{paramagnetic; strong field } CO \Rightarrow \text{diamagnetic}} \]
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