The correct order of the spin-only magnetic moments of the following complexes is :
(I) [Cr(H₂O)₆]Br₂
(II) Na₄[Fe(CN)₆]
(III) Na₃[Fe(C₂O₄)₃] (Δ₀ > P)
(IV) (Et₄N)₂[CoCl₄]

Solution

Chemistry ma’am · 3 steps

For each complex, find the metal's d count, decide high or low spin, then count unpaired electrons.

1. Spin-only formula. The spin-only moment depends only on the number of unpaired electrons n. More unpaired electrons means a larger moment.

   μ = √n(n+2) BM

2. Pairing decisions. H₂O is a weak-field ligand, so the electrons stay unpaired (high spin). CN⁻ is strong field, so they pair up (low spin). For oxalate, the question tells you Δ₀ > P, so it is low spin too. In a tetrahedral complex the splitting is always small, so it is high spin.

   Complex	Metal ion	Field and spin	Unpaired n	μ (BM)
   (I) [Cr(H₂O)₆]²⁺	Cr²⁺, d⁴	octahedral, high spin: t_(2g)³ e_g¹	4	√24 = 4.90
   (II) [Fe(CN)₆]⁴⁻	Fe²⁺, d⁶	octahedral, low spin: t_(2g)⁶	0	0
   (III) [Fe(C₂O₄)₃]³⁻	Fe³⁺, d⁵	octahedral, Δ₀ > P, low spin: t_(2g)⁵	1	√3 = 1.73
   (IV) [CoCl₄]²⁻	Co²⁺, d⁷	tetrahedral, high spin: e⁴ t₂³	3	√15 = 3.87

3. So. The moments run 4.90 > 3.87 > 1.73 > 0, which is (I) > (IV) > (III) > (II), option (d).

The idea to keep. Oxidation state gives the d count. Ligand strength and geometry decide the pairing. Always check the question for hints like Δ₀ > P before you assume high spin.

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