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

[Fe(CN)₆]^4–

The correct answer is option 2. [Fe(CN)₆]^4–

The classification depends on which $d$-orbitals are used for hybridization in an octahedral geometry:

• Inner Orbital Complex: Uses inner $(n - 1)d$ orbitals (Hybridization: $d^2 sp^3$). Usually formed with Strong Field Ligands (SFL) that force electron pairing.
• Outer Orbital Complex: Uses outer $nd$ orbitals (Hybridization: $sp^3 d^2$). Usually formed with Weak Field Ligands (WFL).

Option 2: $[Fe(CN)_6]^{4-}$ (The Correct One)

1. Oxidation State: Iron is in the $+2$ state ($Fe^{2+}$).
2. Electronic Configuration: $Fe$ is $[Ar] 3d^6 4s^2$; thus $Fe^{2+}$ is $3d^6$.
3. Ligand Strength: $CN^-$ is a very Strong Field Ligand.
4. Pairing: $CN^-$ forces the 6 electrons in the $3d$ subshell to pair up in the lower $t_{2g}$ orbitals.
5. Result: This leaves two $3d$ orbitals completely empty. These two inner $3d$, one $4s$, and three $4p$ orbitals hybridize to form $d^2 sp^3$.
6. Conclusion: It is an inner orbital complex (Low Spin).

Option 1: $[CoF_{6}]^{3-}$

• $Co^{3+}$ is $3d^{6}$.
• $F^{-}$ is a Weak Field Ligand.
• Type: Outer orbital complex ($sp^{3}d^{2}$).

Option 3: $[Ni(NH_{3})_{6}]^{2+}$

• $Ni^{2+}$ is $3d^{8}$.
• $d^{8}$ systems cannot vacate two inner $d$-orbitals.
• Type: Outer orbital complex ($sp^{3}d^{2}$).

Option 4: $[Fe(H_{2}O)_{6}]^{3+}$

• $Fe^{3+}$ is $3d^{5}$.
• $H_{2}O$ is a Weak Field Ligand.
• Type: Outer orbital complex ($sp^{3}d^{2}$).