Which one of the following complexes is an inner orbital complex?

The correct answer is Option (2) → [Fe(CN)₆]^4–

An inner orbital complex uses d²sp³ hybridisation, so it needs two empty 3d orbitals. For d⁴ to d⁶ ions these are freed only when a strong field ligand (such as \( \ce{CN^-} \)) pairs the 3d electrons.

• Oxidation state of Fe: in \( \ce{[Fe(CN)6]^4-} \), x + 6(−1) = −4, so x = +2.

• Configuration: Iron (Z = 26) is [Ar]3d⁶4s², so \( \ce{Fe^2+} \) is 3d⁶.

Cyanide pairs the six electrons into three 3d orbitals, two 3d orbitals stay empty, and d²sp³ hybridisation follows. It is an inner orbital, low spin, diamagnetic complex.

Why the other options are incorrect

Option (1) [CoF₆]^3- — Incorrect.

• Oxidation state: x + 6(−1) = −3, so x = +3.

• Configuration: Cobalt (Z = 27) is [Ar]3d⁷4s²; losing two 4s and one 3d electron, Co(+3) is 3d⁶.

• Result: fluoride is a weak field ligand, so no pairing occurs and sp³d² (outer orbital) is used.

Option (3) [Ni(NH₃)₆]²⁺ — Incorrect.

• Oxidation state: \( \ce{NH3} \) is neutral, so nickel is +2.

• Configuration: Nickel (Z = 28) is [Ar]3d⁸4s², so \( \ce{Ni^2+} \) is 3d⁸.

• Result: even with pairing, two empty 3d orbitals cannot be obtained, so it is an outer orbital (sp³d²) complex.

Option (4) [Fe(H₂O)₆]³⁺ — Incorrect.

• Oxidation state: \( \ce{H2O} \) is neutral, so iron is +3.

• Configuration: Iron (Z = 26) is [Ar]3d⁶4s²; losing two 4s and one 3d electron, \( \ce{Fe^3+} \) is 3d⁵.

• Result: water is a weak field ligand, so the five electrons stay unpaired and sp³d² is used.

NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In the diamagnetic octahedral complex, [Co(NH₃)₆]³⁺, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d⁶. … Six pairs of electrons, one from each NH₃ molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH₃)₆]³⁺ is called an inner orbital or low spin or spin paired complex.”

5.5.2 Magnetic Properties of Coordination Compounds – “… When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridisation is not directly available (as a consequence of Hund’s rule). Thus, for d⁴ (Cr²⁺, Mn³⁺), d⁵ (Mn²⁺, Fe³⁺), d⁶ (Fe²⁺, Co³⁺) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively.”