The number of paramagnetic complexes among [FeF₆]³⁻,[Fe(CN)₆]³⁻,[Mn(CN)₆]³⁻, [Co(C₂ O₄)₃]³⁻,[MnCl₆]³⁻, and [CoF₆]³⁻, which involved d² sp³ hybridization is _________.

Solution

Chemistry ma’am · 3 steps

Use two filters: inner d orbitals for d²sp³, and at least one unpaired electron left over.

1. What decides it. All six metals are in the +3 state with d⁴, d⁵ or d⁶ configurations. A strong field ligand pairs the d electrons, which frees two inner 3d orbitals, so the complex uses d²sp³. A weak field ligand leaves the electrons unpaired, so the complex uses outer 4d orbitals and is sp³d².

   Complex	Metal ion	Ligand	Hybridisation	Unpaired electrons
   [FeF₆]³⁻	Fe³⁺, d⁵	weak, high spin	sp³d²	5
   [Fe(CN)₆]³⁻	Fe³⁺, d⁵	strong, low spin	d²sp³	1, paramagnetic
   [Mn(CN)₆]³⁻	Mn³⁺, d⁴	strong, low spin	d²sp³	2, paramagnetic
   [Co(C₂O₄)₃]³⁻	Co³⁺, d⁶	low spin with Co³⁺	d²sp³	0, diamagnetic
   [MnCl₆]³⁻	Mn³⁺, d⁴	weak, high spin	sp³d²	4
   [CoF₆]³⁻	Co³⁺, d⁶	weak, high spin	sp³d²	4

2. The cobalt case. Co³⁺ has a high charge, so it pairs up with almost every ligand. Only F⁻ is weak enough to keep it high spin. In the oxalato complex all six electrons fill t_(2g), so it is d²sp³ but has no unpaired electron.
3. So. Only [Fe(CN)₆]³⁻ and [Mn(CN)₆]³⁻ are both d²sp³ and paramagnetic. The answer is 2.

The common slip: 3. Counting every d²sp³ complex, including the diamagnetic [Co(C₂O₄)₃]³⁻, gives 3. The question also asks for unpaired electrons.

The idea to keep. For octahedral d⁴ to d⁶ ions, a strong field ligand pairs the electrons and frees two inner d orbitals for d²sp³. With Co³⁺, almost every ligand except F⁻ does this.

Answer from the NTA final answer key. Solution by PrepWiser.