Select the Correct statement about the nature of bonding in [FeCl6]4−…

The correct answer is Option (1) → sp³d² hybridisation with 4 unpaired electrons

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

• Weak field ligand (\( \ce{F^{-}} \), \( \ce{Cl^{-}} \), \( \ce{H2O} \)): the 3d electrons are not paired, the inner 3d orbitals stay occupied, so the outer 4d orbitals are used: sp³d², an outer orbital (high spin) complex.

• Strong field ligand (\( \ce{CN^{-}} \), CO): the 3d electrons pair up, two inner 3d orbitals become free: d²sp³, an inner orbital (low spin) complex.

For \( \ce{[FeCl6]^{4-}} \), the oxidation state of Fe (each \( \ce{Cl^{-}} \) is −1):

    \( \displaystyle x+6(-1)=-4 \)

    \( \displaystyle x=+2 \)

Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons gives Fe²⁺: 3d⁶. \( \ce{Cl^{-}} \) is a weak field ligand, so the six d electrons spread over the five 3d orbitals with one pair and four single electrons. The 4s, three 4p and two 4d orbitals take the six ligand electron pairs:

    3d: ↑↓ ↑ ↑ ↑ ↑   |   4s 4p³ 4d²: six \( \ce{Cl^{-}} \) pairs

So \( \ce{[FeCl6]^{4-}} \) has sp³d² hybridisation with 4 unpaired electrons (paramagnetic, μ ≈ 4.90 BM).

Why the other options are incorrect

• Option (2) sp³d² hybridisation with 0 unpaired electrons — Incorrect. sp³d² means the 3d electrons were not paired, so a d⁶ ion must keep 4 unpaired electrons.

• Option (3) d²sp³ hybridisation with 4 unpaired electrons — Incorrect. To free two 3d orbitals, all six d electrons must pair up in the other three, which leaves 0 unpaired electrons.

• Option (4) d²sp³ hybridisation with 0 unpaired electrons — Incorrect. This describes the complex with a strong field ligand, \( \ce{[Fe(CN)6]^{4-}} \), not the chloro complex.

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