Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers : Mn = 25; Fe = 26)
A. [FeF₆]³⁻
B. [Fe(CN)₆]³⁻
C. [MnCl₆]³⁻ (high spin)
D. [Mn(CN)₆]³⁻
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Find the oxidation state, count d electrons, check the ligand, then count unpaired electrons.

1. Oxidation state. In all four complexes the six ligands each carry -1 and the overall charge is -3. So the metal is +3 every time: Fe³⁺ is 3d⁵ and Mn³⁺ is 3d⁴.
2. Ligand strength. CN⁻ is a strong-field ligand, so it forces pairing in the t_(2g) set (low spin). F⁻ and Cl⁻ are weak-field ligands, so the electrons stay spread out (high spin).

   μ = √n(n+2) BM

   Complex	Metal ion	Configuration	Unpaired n	μ (BM)
   B: [Fe(CN)₆]³⁻	Fe³⁺, d⁵	t_(2g)⁵, low spin	1	1.73
   D: [Mn(CN)₆]³⁻	Mn³⁺, d⁴	t_(2g)⁴, low spin	2	2.83
   C: [MnCl₆]³⁻	Mn³⁺, d⁴	t_(2g)³ e_g¹, high spin	4	4.90
   A: [FeF₆]³⁻	Fe³⁺, d⁵	t_(2g)³ e_g², high spin	5	5.92

3. So. The magnetic moment rises with the number of unpaired electrons, 1 < 2 < 4 < 5, so the order is B < D < C < A, option (b).

The idea to keep. Magnetic moment depends only on the number of unpaired electrons. Get the d count from the ion, not the atom, then let the ligand decide high spin or low spin.

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