\([ \text{Co(NH}_3)_6 ]^{3+}\) and \([ \text{CoF}_6 ]^{3-}\) are respectively known as:
- Spin free Complex, Spin paired Complex
- Spin paired Complex, Spin free Complex
- Outer orbital Complex, Inner orbital Complex
- Inner orbital Complex, Spin paired Complex
The Correct Option is B
Approach Solution - 1
The question asks us about the nature of the complexes \([ \text{Co(NH}_3)_6 ]^{3+}\) and \([ \text{CoF}_6 ]^{3-}\). Let's analyze each complex:
1. Understanding \([ \text{Co(NH}_3)_6 ]^{3+}\)
- Coordination Complex: In this complex, cobalt (Co) is surrounded by six ammonia (NH3) molecules. Ammonia is a strong field ligand.
- Electronic Configuration and Hybridization: Cobalt in this complex has a +3 oxidation state (Co3+), which gives it an electronic configuration of 3d6. Under the influence of a strong field ligand like NH3, it undergoes d2sp3 hybridization, leading to low-spin or spin-paired configurations.
- Conclusion: This is an inner orbital or spin-paired complex.
2. Understanding \([ \text{CoF}_6 ]^{3-}\)
- Coordination Complex: Here, cobalt is surrounded by six fluoride (F-) ions. Fluoride is a weak field ligand.
- Electronic Configuration and Hybridization: With CoF63-, cobalt is also in the +3 oxidation state. Due to fluoride's weak field, it does not cause pairing of electrons, resulting in sp3\d2\ hybridization and a high-spin, or spin-free, complex.
- Conclusion: This is an outer orbital or spin-free complex.
Based on the explanations above, the correct choice matches the respective complex nature: Spin paired Complex, Spin free Complex.
| Complex | Ligand | Ligand Field | Hybridization | Spin Type |
|---|---|---|---|---|
| \([ \text{Co(NH}_3)_6 ]^{3+}\) | NH3 | Strong | d2sp3 | Spin paired |
| \([ \text{CoF}_6 ]^{3-}\) | F- | Weak | sp3d2 | Spin free |
Approach Solution -2
[Co(NH$_3$)$_6$]$^{3+}$:} In this complex, cobalt is in the $+3$ oxidation state with an electronic configuration of [Ar]3$d^6$. Since ammonia (NH$_3$) is a weak field ligand, it causes a small splitting of d-orbitals in the octahedral field, leading to a spin paired (low-spin) configuration. Therefore, [Co(NH$_3$)$_6$]$^{3+}$ is a spin paired complex.
[CoF$_6$]$^{3-}$: In this complex, cobalt is also in the $+3$ oxidation state with an electronic configuration of [Ar]3$d^6$. Fluoride (F$^-$) is a weak field ligand, causing less splitting of d-orbitals in the octahedral field, resulting in a spin free (high-spin) complex configuration. Therefore, [CoF$_6$]$^{3-}$ is a spin free complex.
Conclusion: [Co(NH$_3$)$_6$]$^{3+}$ is a spin paired complex and [CoF$_6$]$^{3-}$ is a spin free complex.
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