Question

Which of the following is not an inner orbital complex?

• A. ${[Fe(CN)6]^4-}$
• B. ${[Co(NH3)6]^3+}$
• C. ${[Ni(CO)4]}$
• D. ${[Cr(NH3)6]^3+}$

Hint:

Inner vs. Outer Orbital Complexes.

• Inner orbital: d$^2$sp$^3$ hybridization, low-spin, strong field ligands.
• Outer orbital: sp$^3$d$^2$ hybridization, high-spin, weak field or tetrahedral geometries.
• Tetrahedral complexes like [Ni(CO)4] are always outer orbital.

Answer 1

The Correct Option is C

Inner orbital complexes involve pairing of electrons in the inner d-orbitals (d$^2$sp$^3$ hybridization). Outer orbital complexes use outer d-orbitals (sp$^3$d$^2$). Let’s analyze:

• {[Fe(CN)6]^4-}: CN$^-$ is a strong field ligand. Causes pairing → uses inner d-orbitals → inner orbital.
• {[Co(NH3)6]^3+}: NH$_3$ is borderline but causes pairing in Co$^{3+}$ → inner orbital.
• {[Ni(CO)4]}: Ni$^0$ (d$^{10}$), CO is strong field ligand. But this is tetrahedral → uses sp$^3$, outer orbital → not inner orbital.
• {[Cr(NH3)6]^3+}: d$^3$ configuration with NH$_3$, forms low spin → inner orbital.

Hence, {[Ni(CO)4]} is not an inner orbital complex.

Answer 2

Step 1: Understanding Inner Orbital Complexes
Inner orbital complexes involve the use of (n-1)d orbitals of the central metal ion for hybridization, typically resulting in low-spin complexes. These complexes often form when ligands cause pairing of electrons in the d-orbitals, allowing hybridization that includes d-orbitals from the inner shell.

Step 2: Analyzing the Complexes
In coordination chemistry, complexes like ${[Ni(CN)_4]}^{2-}$ and ${[Fe(CN)_6]}^{4-}$ usually form inner orbital complexes because strong field ligands such as CN⁻ cause pairing of electrons, allowing d²sp³ or dsp² hybridization involving inner d orbitals.

Step 3: Nature of ${[Ni(CO)_4]}$
${[Ni(CO)_4]}$ is a classic example of an outer orbital complex. The ligand CO is a strong field ligand but in this case, nickel uses its valence 4s and 4p orbitals (not the inner 3d orbitals) for hybridization, leading to an sp³ hybridization. The complex is tetrahedral and high-spin, and does not involve inner d orbitals.

Step 4: Conclusion
Therefore, among the given complexes, ${[Ni(CO)_4]}$ is not an inner orbital complex because it uses outer orbitals (4s and 4p) for bonding, unlike other complexes which use inner (n-1)d orbitals.