Which of the following hybridisation is correct for [FeF6]3−?

The correct answer is Option (2) → sp³d² with 5 unpaired electrons

According to valence bond theory, an octahedral complex uses six hybrid orbitals of the metal. Which orbitals are used depends on the ligand:

• Weak field ligands (such as \( \ce{F^{-}} \), \( \ce{H2O} \)) do not force the 3d electrons to pair. The inner 3d orbitals stay occupied, so the outer 4d orbitals are used: sp³d², an outer orbital or high spin complex.

• Strong field ligands (such as \( \ce{CN^{-}} \)) pair up the 3d electrons, leaving two inner 3d orbitals empty: d²sp³, an inner orbital or low spin complex.

In \( \ce{[FeF6]^{3-}} \), the oxidation state of iron is:

    \( \displaystyle x+6(-1)=-3 \Rightarrow x=+3 \)

Fe (Z = 26) is [Ar]3d⁶4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Fe^{3+}} \): [Ar]3d⁵. \( \ce{F^{-}} \) is a weak field ligand, so the five 3d electrons stay unpaired, one in each 3d orbital. The six \( \ce{F^{-}} \) ligands use one 4s, three 4p and two 4d orbitals.

    \( \displaystyle \mu=\sqrt{n(n+2)}=\sqrt{5\times7}=\sqrt{35}\approx5.92\ \mathrm{BM} \)

So \( \ce{[FeF6]^{3-}} \) is sp³d² with 5 unpaired electrons.

Why the other options are incorrect

• Option (1) sp³d² with 1 unpaired electrons — Incorrect. Outer orbital hybridisation does not need any pairing, so all five 3d electrons remain unpaired.

• Option (3) d²sp³ with 1 unpaired electrons — Incorrect. This describes \( \ce{[Fe(CN)6]^{3-}} \), where the strong field \( \ce{CN^{-}} \) ligands pair the electrons; \( \ce{F^{-}} \) cannot do this.

• Option (4) d²sp³ with 5 unpaired electrons — Incorrect. To free two 3d orbitals, a d⁵ ion must pair four electrons, leaving only one unpaired. So d²sp³ with 5 unpaired electrons is impossible.

NCERT (2026-27 edition):
Magnetic Properties of Coordination Compounds – …