Which of the following properties are related to the complex \( [Fe(H_2O)_6]^{3+} \)?
(A) Paramagnetic
(B) number of unpaired electrons are five
(C) \( d^2sp^3 \) hybridisation
(D) \( H_2O \) is a strong field ligand
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
The magnetic behaviour and hybridisation of an octahedral complex follow from the d-electron count of the metal ion and the strength of the ligand field. Iron (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \). In \( \ce{[Fe(H2O)6]^3+} \) the six water molecules are neutral, so iron is +3; losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): \( \mathrm{[Ar]\,3d^{5}} \).
Water is a weak field ligand. The crystal field splitting \( \Delta_o \) is smaller than the pairing energy, so the five electrons occupy all five d orbitals singly: \( t_{2g}^{3}\,e_g^{2} \), with five unpaired electrons (high spin). Since the inner 3d orbitals are all occupied, bonding uses the outer 4d orbitals: \( sp^3d^2 \) hybridisation (outer orbital complex).
(A) Paramagnetic — Correct. With five unpaired electrons the complex is strongly paramagnetic, \( \mu = \sqrt{5(5+2)} \approx 5.92 \) BM.
(B) number of unpaired electrons are five — Correct. \( t_{2g}^{3}\,e_g^{2} \) has five unpaired electrons.
(C) \(d^2sp^3\) hybridisation — Incorrect. \( d^2sp^3 \) needs two empty inner 3d orbitals, which would be possible only if the electrons paired up in a strong field (as in \( \ce{[Fe(CN)6]^3-} \)). Here the hybridisation is \( sp^3d^2 \).
(D) \(H_2O\) is a strong field ligand — Incorrect. In the spectrochemical series water lies well below \( \ce{NH3} \), en and \( \ce{CN-} \); it is a weak field ligand, which is why the complex is high spin.