Question

The spin-only magnetic moments of the complexes \([Mn(CN)_6]^{3-}\) and \([Co(C_2O_4)_3]^{3-}\) are respectively:

• A. \( 2.84 \) BM, \( 0 \) BM
• B. \( 0 \) BM, \( 2.84 \) BM
• C. \( 0 \) BM, \( 3.87 \) BM
• D. \( 5.92 \) BM, \( 2.84 \) BM

Hint:

The number of unpaired electrons determines the spin-only magnetic moment of a coordination complex.

Answer 1

The Correct Option is A

Step 1: Magnetic Moment Formula The spin-only magnetic moment (\(\mu_s\)) is given by: \[ \mu_s = \sqrt{n(n+2)} \text{ BM} \] where \( n \) is the number of unpaired electrons. 

Step 2: Analyzing \([Mn(CN)_6]^{3-}\) - Mn in \([Mn(CN)_6]^{3-}\) is in the +3 oxidation state (\(3d^4\)). - \( CN^- \) is a strong field ligand, causing pairing of electrons, leaving \( n = 2 \). \[ \mu_s = \sqrt{2(2+2)} = \sqrt{8} = 2.84 \text{ BM} \] 

Step 3: Analyzing \([Co(C_2O_4)_3]^{3-}\) - Co in \([Co(C_2O_4)_3]^{3-}\) is in the +3 oxidation state (\(3d^6\)). - \( C_2O_4^{2-} \) is a strong field ligand, leading to full pairing of electrons (\( n = 0 \)). \[ \mu_s = \sqrt{0(0+2)} = 0 \text{ BM} \]