The number of paramagnetic metal complex species among [Co(NH₃)₆]³⁺,[Co(C₂ O₄)₃]³⁻, [MnCl₆]³⁻,[Mn(CN)₆]³⁻,[CoF₆]³⁻,[Fe(CN)₆]³⁻ and [FeF₆]³⁻ with same number of unpaired electrons is __________.

Solution

Chemistry ma’am · 4 steps

Find the d-count, decide high spin or low spin from the ligand, then count unpaired electrons.

1. Oxidation states. Every metal here is +3. Co³⁺ is d⁶, Mn³⁺ is d⁴ and Fe³⁺ is d⁵.
2. Spin state. CN⁻ is a strong-field ligand and pairs electrons into t_(2g). F⁻ and Cl⁻ are weak-field ligands and give high spin. With the highly charged Co³⁺ ion, both NH₃ and oxalate give a large enough splitting for low spin.

   Complex	Ion, field	Configuration	Unpaired
   [Co(NH₃)₆]³⁺	d⁶, strong	t_(2g)⁶ e_g⁰	0
   [Co(C₂O₄)₃]³⁻	d⁶, strong for Co³⁺	t_(2g)⁶ e_g⁰	0
   [MnCl₆]³⁻	d⁴, weak	t_(2g)³ e_g¹	4
   [Mn(CN)₆]³⁻	d⁴, strong	t_(2g)⁴ e_g⁰	2
   [CoF₆]³⁻	d⁶, weak	t_(2g)⁴ e_g²	4
   [Fe(CN)₆]³⁻	d⁵, strong	t_(2g)⁵ e_g⁰	1
   [FeF₆]³⁻	d⁵, weak	t_(2g)³ e_g²	5

3. Match the counts. Five species are paramagnetic, with 4, 2, 4, 1 and 5 unpaired electrons. Only the value 4 appears twice, in [MnCl₆]³⁻ and [CoF₆]³⁻.
4. So. The number of paramagnetic species that share the same number of unpaired electrons is 2.

The common slip: 3. Treating oxalate as weak field makes [Co(C₂O₄)₃]³⁻ high spin with 4 unpaired electrons, giving 3. With Co³⁺, oxalate pairs all six electrons.

The idea to keep. For octahedral complexes, write the d-count first, then let the ligand decide the spin. A Co³⁺ ion is low spin with almost every ligand except F⁻.

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