The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively :
[Fe(CN)₆]³⁻, [FeF₆]³⁻, [CoF₆]³⁻, [Mn(CN)₆]³⁻

Solution

Chemistry ma’am · 3 steps

Find each d-count, then let the ligand decide: CN⁻ pairs electrons, F⁻ spreads them.

1. Oxidation states. Each complex has six ligands with charge -1 and an overall charge of -3, so every metal is +3. That makes Fe³⁺ d⁵, Co³⁺ d⁶ and Mn³⁺ d⁴.
2. Ligand strength. CN⁻ is a strong-field ligand. It makes Δₒ larger than the pairing energy, so electrons pair up in t_(2g) before entering e_g (low spin). F⁻ is a weak-field ligand, so electrons occupy all five d orbitals singly before any pairing (high spin).

   Complex	Metal ion	Ligand field	Filling	Unpaired
   [Fe(CN)₆]³⁻	Fe³⁺, d⁵	strong, low spin	t_(2g)⁵ e_g⁰	1
   [FeF₆]³⁻	Fe³⁺, d⁵	weak, high spin	t_(2g)³ e_g²	5
   [CoF₆]³⁻	Co³⁺, d⁶	weak, high spin	t_(2g)⁴ e_g²	4
   [Mn(CN)₆]³⁻	Mn³⁺, d⁴	strong, low spin	t_(2g)⁴ e_g⁰	2

3. So. The unpaired electrons are 1, 5, 4 and 2 in that order, which is option (a).

The idea to keep. For an octahedral complex: oxidation state gives the d-count, ligand strength picks low or high spin, then fill t_(2g) and e_g and count the single electrons.

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