The d-orbital electron distribution of Fe in the complex [Fe(H2O)6]3+…

The correct answer is Option (1) → \( t_{2g}^{3}e_{g}^{2} \)

In an octahedral complex the five d orbitals split into a lower set t_2g (3 orbitals) and a higher set e_g (2 orbitals), separated by the splitting energy Δ_o. How the electrons fill them depends on the ligand:

• Weak field ligand (Δ_o < P, the pairing energy): electrons occupy e_g before pairing, giving a high spin complex.

• Strong field ligand (Δ_o > P): electrons pair in t_2g first, giving a low spin complex.

Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives Fe³⁺: 3d⁵. In \( \ce{[Fe(H2O)6]^{3+}} \) the oxidation state is x + 6(0) = +3, and H₂O is a weak field ligand. So the five electrons occupy all five orbitals singly:

    \( \displaystyle t_{2g}^{3}e_{g}^{2}\quad (\text{5 unpaired electrons, high spin}) \)

Why the other options are incorrect

• Option (2) \( t_{2g}^{5}e_{g}^{0} \) — Incorrect. This is the low spin d⁵ arrangement, found with a strong field ligand such as \( \ce{CN^{-}} \) in \( \ce{[Fe(CN)6]^{3-}} \).

• Option (3) \( e_{g}^{2}t_{2}^{3} \) — Incorrect. e and t₂ are the labels of tetrahedral splitting, where the lower set is e; they do not describe an octahedral complex.

• Option (4) \( e^{4}t_{2}^{1} \) — Incorrect. This is a low spin tetrahedral arrangement; tetrahedral complexes are almost always high spin, and this complex is octahedral anyway.

NCERT (2026-27 edition):
Crystal Field Theory – “… (i) If Δ_o < P, the fourth electron enters one of the e_g orbitals giving the configuration t_2g³ e_g¹. Ligands for which Δ_o < P are known as weak field ligands and form high spin complexes.”

“(ii) If Δ_o > P, it becomes more energetically favourable for the fourth electron to occupy a t_2g orbital with configuration t_2g⁴e_g⁰. Ligands which produce this effect are known as strong field ligands and form low spin complexes.”