Understanding Inner and Outer Orbital Complexes

Coordination complexes are formed when a central metal ion or atom is bonded to a group of molecules or ions called ligands. The nature of bonding and the arrangement of electrons in the metal's d-orbitals determine many properties of the complex, including whether it is an inner orbital or outer orbital complex.

An inner orbital complex (also known as a low spin complex for octahedral geometry) is formed when the central metal ion uses its inner (n-1)d orbitals for hybridization with s and p orbitals to accommodate the ligands. This typically occurs when the ligands are strong field ligands, causing the d-electrons to pair up, leaving inner d-orbitals available for bonding. The hybridization is usually d²sp³ for octahedral complexes or dsp² for square planar complexes.

An outer orbital complex (also known as a high spin complex for octahedral geometry) is formed when the central metal ion uses its outer nd orbitals (from the next higher principal shell) along with ns and np orbitals for hybridization. This usually happens with weak field ligands that do not cause significant pairing of d-electrons, or when the inner d-orbitals are not available for hybridization.

Let's examine each given complex to determine its type:

Analysis of Coordination Complexes

(A) $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$

• Central metal ion: Cobalt (Co).
• Atomic number of Co = 27. Ground state electronic configuration: [Ar] 3d⁷ 4s².
• Oxidation state of Co in $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$: Co + 6(0) = +3. So, it is Co³⁺.
• Electronic configuration of Co³⁺: [Ar] 3d⁶ 4s⁰.
• Ligand: NH₃ (Ammonia). NH₃ is a strong field ligand.
• Due to the strong field ligand NH₃, the 3d electrons in Co³⁺ pair up. The 3d⁶ configuration becomes paired up, leaving two 3d orbitals empty.
• Hybridization: To form bonds with six NH₃ ligands, Co³⁺ needs 6 hybrid orbitals. It uses the two empty 3d orbitals, the 4s orbital, and three 4p orbitals. Hybridization is d²sp³.
• Since (n-1)d orbitals (3d) are used, $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ is an inner orbital complex.

(B) $\text{[CoF}_6\text{]}^{3-}$

• Central metal ion: Cobalt (Co). Co³⁺: [Ar] 3d⁶ 4s⁰.
• Ligand: F^- (Fluoride). F^- is a weak field ligand.
• Due to the weak field ligand F^-, the 3d electrons in Co³⁺ do not pair up. The 3d⁶ configuration remains with unpaired electrons.
• Hybridization: To form bonds with six F^- ligands, Co³⁺ needs 6 hybrid orbitals. Since the inner 3d orbitals are not available (due to unpaired electrons), it uses the 4s orbital, three 4p orbitals, and two 4d orbitals. Hybridization is sp³d².
• Since outer nd orbitals (4d) are used, $\text{[CoF}_6\text{]}^{3-}$ is an outer orbital complex.

(C) $\text{[Ni(CN)}_4\text{]}^{2-}$

• Central metal ion: Nickel (Ni).
• Atomic number of Ni = 28. Ground state electronic configuration: [Ar] 3d⁸ 4s².
• Oxidation state of Ni in $\text{[Ni(CN)}_4\text{]}^{2-}$: Ni + 4(-1) = -2. So, it is Ni²⁺.
• Electronic configuration of Ni²⁺: [Ar] 3d⁸ 4s⁰.
• Ligand: CN^- (Cyanide). CN^- is a strong field ligand.
• Due to the strong field ligand CN^-, the 3d electrons in Ni²⁺ pair up. The 3d⁸ configuration becomes paired up, leaving one 3d orbital empty.
• Hybridization: To form bonds with four CN^- ligands, Ni²⁺ needs 4 hybrid orbitals. It uses the one empty 3d orbital, the 4s orbital, and two 4p orbitals. Hybridization is dsp² (resulting in square planar geometry).
• Since (n-1)d orbitals (3d) are used, $\text{[Ni(CN)}_4\text{]}^{2-}$ is an inner orbital complex.

(D) $\text{[MnCl}_6\text{]}^{3-}$

• Central metal ion: Manganese (Mn).
• Atomic number of Mn = 25. Ground state electronic configuration: [Ar] 3d⁵ 4s².
• Oxidation state of Mn in $\text{[MnCl}_6\text{]}^{3-}$: Mn + 6(-1) = -3. So, it is Mn³⁺.
• Electronic configuration of Mn³⁺: [Ar] 3d⁴ 4s⁰.
• Ligand: Cl^- (Chloride). Cl^- is a weak field ligand.
• Due to the weak field ligand Cl^-, the 3d electrons in Mn³⁺ do not pair up significantly. The 3d⁴ configuration remains with unpaired electrons.
• Hybridization: To form bonds with six Cl^- ligands, Mn³⁺ needs 6 hybrid orbitals. It uses the 4s orbital, three 4p orbitals, and two 4d orbitals. Hybridization is sp³d².
• Since outer nd orbitals (4d) are used, $\text{[MnCl}_6\text{]}^{3-}$ is an outer orbital complex.

(E) $\text{[FeF}_6\text{]}^{3-}$

• Central metal ion: Iron (Fe).
• Atomic number of Fe = 26. Ground state electronic configuration: [Ar] 3d⁶ 4s².
• Oxidation state of Fe in $\text{[FeF}_6\text{]}^{3-}$: Fe + 6(-1) = -3. So, it is Fe³⁺.
• Electronic configuration of Fe³⁺: [Ar] 3d⁵ 4s⁰.
• Ligand: F^- (Fluoride). F^- is a weak field ligand.
• Due to the weak field ligand F^-, the 3d electrons in Fe³⁺ do not pair up. The 3d⁵ configuration remains with all five electrons unpaired.
• Hybridization: To form bonds with six F^- ligands, Fe³⁺ needs 6 hybrid orbitals. Since the inner 3d orbitals are not available (all half-filled), it uses the 4s orbital, three 4p orbitals, and two 4d orbitals. Hybridization is sp³d².
• Since outer nd orbitals (4d) are used, $\text{[FeF}_6\text{]}^{3-}$ is an outer orbital complex.

From the analysis, the inner orbital complexes are $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ (A) and $\text{[Ni(CN)}_4\text{]}^{2-}$ (C).

Revision Table: Inner vs Outer Orbital Complexes

Additional Information: Crystal Field Theory and Ligand Strength

The determination of whether a ligand is strong field or weak field is based on the spectrochemical series. Strong field ligands cause a large splitting of the metal d-orbitals (large crystal field splitting energy, Δ_o or Δ_t), while weak field ligands cause a small splitting.

Spectrochemical series (partial): I^- < Br^- < Cl^- < F^- < OH^- < C₂O₄^2- < H₂O < NCS^- < EDTA^4- < NH₃ < en < CN^- < CO.

Ligands on the left are weak field, and ligands on the right are strong field. For d⁴, d⁵, d⁶, and d⁷ ions in octahedral complexes, the choice between pairing (low spin, inner orbital) and not pairing (high spin, outer orbital) depends on the balance between the pairing energy (P) and the crystal field splitting energy (&Delta;_o).

• If &Delta;_o > P, electrons pair up in the lower energy orbitals (t_2g) before occupying the higher energy orbitals (e_g). This results in low spin complexes, which often use inner d-orbitals for bonding (d²sp³). Strong field ligands cause large &Delta;_o.
• If &Delta;_o < P, electrons occupy the higher energy orbitals (e_g) before pairing up in the lower energy orbitals (t_2g). This results in high spin complexes, which typically use outer d-orbitals for bonding (sp³d²). Weak field ligands cause small &Delta;_o.

For square planar complexes like $\text{[Ni(CN)}_4\text{]}^{2-}$ (d⁸), strong field ligands cause significant splitting, often leading to an empty d-orbital available for dsp² hybridization, making them inner orbital complexes.