Question

\([ \text{Co(NH}_3)_6 ]^{3+}\) and \([ \text{CoF}_6 ]^{3-}\) are respectively known as:

• A. Spin free Complex, Spin paired Complex
• B. Spin paired Complex, Spin free Complex
• C. Outer orbital Complex, Inner orbital Complex
• D. Inner orbital Complex, Spin paired Complex

Answer 1

The Correct Option is B

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 CoF₆^3-, cobalt is also in the +3 oxidation state. Due to fluoride's weak field, it does not cause pairing of electrons, resulting in sp^3\d^2\ 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

Answer 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.