Question

The correct order of magnetic moment (spin only value in B.m.) is:

विकल्प

• [Fe(CN)₆]⁴⁺ > [MnCl₄]^2– > [CoCl₄]^2–

• [MnCl₄]^2– > [Fe(CN)₆]^4– > [CoCl₄]^2–

• [MnCl₄]^2– > [CoCl₄]^2– > [Fe(CN)₆]^4–

• [Fe(CN)₆]⁴⁺ > [CoCl₄]^2– > [MnCl₄]^2–

Answer 1

[MnCl₄]^2– > [CoCl₄]^2– > [Fe(CN)₆]^4–

Explanation:

[MnCl₄]^2– contains Mn²⁺ (d⁵) with weak-field ligands, giving a high-spin complex with five unpaired electrons (μ ≈ 5.92 B.M.). [CoCl₄]^2– contains Co²⁺ (d⁷) with weak-field ligands, producing a high-spin complex with three unpaired electrons (μ ≈ 3.87 B.M.). [Fe(CN)₆]^4– contains Fe²⁺ (d⁶) with strong-field ligands, resulting in a low-spin complex with no unpaired electrons (μ = 0 B.M.). The correct order of magnetic moments is:

\[\ce{\underset{(Five)}{[MnCl4]^{2–}} > \underset{(Three)}{[CoCl4]^{2–}} > \underset{(Zero)}{[Fe(CN)6]^{4–}}}\]