According to the valence bond theory, [Fe(CN)6]3− is(A) d2sp3(B) sp3d…

The correct answer is Option (3) → (A), (C) and (D) only

In valence bond theory, the metal ion uses empty orbitals of suitable energy, mixes them (hybridisation), and receives one lone pair from each ligand. For an octahedral complex there are two cases:

• Strong field ligand (\( \ce{CN^{-}} \), CO): it forces the 3d electrons to pair up, two inner 3d orbitals become empty, and the hybridisation is d²sp³ (inner orbital, low spin).

• Weak field ligand (\( \ce{F^{-}} \), \( \ce{H2O} \)): no pairing, the outer 4d orbitals are used, and the hybridisation is sp³d² (outer orbital, high spin).

For \( \ce{[Fe(CN)6]^{3-}} \), the charge balance gives x + 6(−1) = −3, so x = +3. Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives Fe³⁺: [Ar]3d⁵.

\( \ce{CN^{-}} \) is a strong field ligand, so the five 3d electrons pair up as far as possible (Hund’s rule within three orbitals):

    3d: [↑↓] [↑↓] [↑] [  ] [  ]   4s: [  ]   4p: [  ] [  ] [  ]

Two empty 3d, one 4s and three 4p orbitals give d²sp³ hybridisation, and one electron stays unpaired.

• (A) d²sp³ — Correct. Inner 3d orbitals are used after pairing.

• (B) sp³d² — Incorrect. This is the outer orbital hybridisation of weak field complexes such as \( \ce{[FeF6]^{3-}} \).

• (C) Paramagnetic — Correct. One unpaired electron, μ = √(n(n + 2)) = √3 ≈ 1.73 BM.

• (D) Low spin complex — Correct. Only one unpaired electron, against five in the high spin case.