Question

Arrange the following complexes in the increasing order of their spin only magnetic moment (in B.M)
I. \([Fe(CN)₆]⁴⁻\)
II. \([MnCl₄]²⁻\)
III. \([Mn(CN)₆]³⁻\)
IV. \([Cr(NH₃)₆]³⁺\)

• A. II<IV<I<III
• B. III<II<I<IV
• C. I<IV<II<III
• D. I<III<IV<II

Hint:

To solve these problems quickly, remember the spectrochemical series to identify strong-field (CN⁻, CO) and weak-field (halides, H₂O) ligands. Strong ligands cause pairing for \(d^4-d^7\) configurations in octahedral complexes. For tetrahedral complexes, pairing rarely occurs. The magnetic moment increases with the number of unpaired electrons, so you only need to compare \(n\).

Answer 1

The Correct Option is D

Step 1: Calculate Unpaired Electrons (n) for each complex: Formula: \( \mu = \sqrt{n(n+2)} \) B.M. Higher 'n' means higher magnetic moment. I. \( [\text{Fe}(\text{CN})_6]^{3-} \): - Fe is in +3 state (\( 3d^5 \)). - Ligand: \( \text{CN}^- \) (Strong Field). - Pairing occurs. Configuration: \( t_{2g}^5 e_g^0 \). - Unpaired electrons (\( n \)) = 1. II. \( [\text{MnCl}_4]^{2-} \): - Mn is in +2 state (\( 3d^5 \)). - Ligand: \( \text{Cl}^- \) (Weak Field). - No pairing. Tetrahedral geometry (\( e^2 t_2^3 \)). - Unpaired electrons (\( n \)) = 5. III. \( [\text{Mn}(\text{CN})_6]^{3-} \): - Mn is in +3 state (\( 3d^4 \)). - Ligand: \( \text{CN}^- \) (Strong Field). - Pairing occurs. Configuration: \( t_{2g}^4 e_g^0 \). (Wait, usually \( 3d^4 \) strong field is \( t_{2g}^4 \), n=2). - Unpaired electrons (\( n \)) = 2. IV. \( [\text{Cr}(\text{NH}_3)_6]^{3+} \): - Cr is in +3 state (\( 3d^3 \)). - Ligand: \( \text{NH}_3 \) (Strong Field, but for \( d^3 \) it doesn't matter). - Configuration: \( t_{2g}^3 e_g^0 \). - Unpaired electrons (\( n \)) = 3.
Step 2: Order of 'n': I (n=1)<III (n=2)<IV (n=3)<II (n=5).
Step 3: Order of Magnetic Moment: Since \( \mu \) increases with \( n \), the order is: I<III<IV<II.