Question

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

Options

• [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:

The number of unpaired electrons

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