Coordination Compounds

Let's analyze each complex and check their magnetic properties:

1. [VF₆]³⁻:
   V³⁺ has the configuration 3d².
   With a weak field ligand like F⁻, it forms a high-spin complex, so both electrons remain unpaired.
   ✅ Paramagnetic with 2 unpaired electrons — Correct.
2. [CuCl₄]²⁻:
   Cu²⁺ has the configuration 3d⁹.
   One unpaired electron remains.
   ✅ Paramagnetic with 1 unpaired electron — Correct.
3. [Co(NH₃)₆]³⁺:
   Co³⁺ is 3d⁶. NH₃ is a strong field ligand, so it forms a low-spin complex.
   All electrons are paired.
   ✅ Diamagnetic — Correct.
4. [CoF₆]³⁻:
   Co³⁺ is 3d⁶. F⁻ is a weak field ligand, so it forms a high-spin complex.
   This results in 4 unpaired electrons, not 2.
   ❌ Paramagnetic with 2 unpaired electrons — Incorrect.

Therefore, the correct answer is option (D): [CoF₆]³⁻ is paramagnetic with 2 unpaired electrons — which is incorrect because it actually has 4 unpaired electrons.

To determine the magnetic properties of the complexes [Fe(CN)₆]³⁻ and [FeF₆]³⁻, we need to consider:

• Oxidation state of Fe
• Type of ligand (strong field or weak field)
• Electronic configuration and spin pairing

Step 1: Oxidation state of Fe

• In both complexes, Fe is in +3 oxidation state.
• Atomic number of Fe = 26 → Fe³⁺ = 3d⁵

Step 2: Analyze each complex

1) [Fe(CN)₆]³⁻:

• CN⁻ is a strong field ligand (causes pairing of electrons).
• In an octahedral field, the 3d⁵ electrons will pair due to strong field effect.
• Configuration: t_2g⁵ e_g⁰
• Number of unpaired electrons = 1 → Paramagnetic

2) [FeF₆]³⁻:

• F⁻ is a weak field ligand (no pairing).
• Electrons remain unpaired.
• Configuration: t_2g³ e_g²
• Number of unpaired electrons = 5 → Paramagnetic

Conclusion: Both complexes have unpaired electrons → Both are paramagnetic

Correct answer: (A) Both are paramagnetic