Using crystal field theory,identify from the following complex ions that shows same spin only values (CoF6)-3,(IrCl6)-3,(FeH2O6)+2

To determine which of the given complex ions exhibit the same spin-only values using crystal field theory, we first need to analyze the electronic configurations and the nature of the ligands involved in each complex. The spin-only formula is given by the equation: Spin-only magnetic moment (μ) = √(n(n + 2)) where n is the number of unpaired electrons. Let's break down each complex ion step by step.

1. Analyzing the Complex Ions

(CoF6)³⁻

Cobalt in this complex is in the +3 oxidation state. The electron configuration of Co is [Ar] 3d⁷ 4s². When cobalt loses three electrons to form Co³⁺, the configuration becomes 3d⁶. The fluoride ion (F^-) is a weak field ligand, which means it does not cause significant splitting of the d-orbitals. Therefore, Co³⁺ in (CoF6)³⁻ will have 4 unpaired electrons, leading to:

• n = 4
• Spin-only magnetic moment = √(4(4 + 2)) = √(24) ≈ 4.9 μ_B

(IrCl6)³⁻

Iridium in this complex is also in the +3 oxidation state. The electron configuration of Ir is [Xe] 4f¹⁴ 5d⁷ 6s². For Ir³⁺, the configuration becomes 5d⁶. Chloride (Cl^-) is also a weak field ligand, similar to fluoride. Thus, Ir³⁺ in (IrCl6)³⁻ will also have 4 unpaired electrons:

• n = 4
• Spin-only magnetic moment = √(4(4 + 2)) = √(24) ≈ 4.9 μ_B

(FeH2O6)²⁺

Iron in this complex is in the +2 oxidation state. The electron configuration of Fe is [Ar] 3d⁶ 4s². For Fe²⁺, the configuration becomes 3d⁶. Water (H₂O) is a stronger field ligand than F^- and Cl^-, which can lead to pairing of electrons in the d-orbitals. In an octahedral field, Fe²⁺ can have either 4 unpaired electrons (high-spin state) or 2 unpaired electrons (low-spin state). Assuming high-spin for this analysis:

• n = 4
• Spin-only magnetic moment = √(4(4 + 2)) = √(24) ≈ 4.9 μ_B

Comparing the Results

From our calculations, we find that:

• (CoF6)³⁻ has 4 unpaired electrons and a magnetic moment of approximately 4.9 μ_B.
• (IrCl6)³⁻ also has 4 unpaired electrons and a magnetic moment of approximately 4.9 μ_B.
• (FeH2O6)²⁺, assuming it is in a high-spin state, also has 4 unpaired electrons and a magnetic moment of approximately 4.9 μ_B.

Final Thoughts

All three complex ions—(CoF6)³⁻, (IrCl6)³⁻, and (FeH2O6)²⁺—exhibit the same spin-only magnetic moment of approximately 4.9 μ_B due to having the same number of unpaired electrons in their respective d-orbitals. This analysis highlights how the nature of ligands and the oxidation states of the metal ions influence the electronic configurations and magnetic properties of coordination complexes.