Question

The calculated spin-only magnetic moments of \( K_3[Fe(OH)_6] \) and \( K_4[Fe(OH)_6] \) respectively are:

• A. 4.90 and 4.90 B.M.
• B. 5.92 and 4.90 B.M.
• C. 3.87 and 4.90 B.M.
• D. 4.90 and 5.92 B.M.

Hint:

For calculating the spin-only magnetic moment, determine the number of unpaired electrons using the electronic configuration of the metal in the complex.

Answer 1

The Correct Option is B

The spin-only magnetic moment is given by the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons. Based on the electronic configurations of \( Fe^{3+} \) (in \( K_3[Fe(OH)_6] \)) and \( Fe^{2+} \) (in \( K_4[Fe(OH)_6] \)), the number of unpaired electrons in these complexes is calculated. For \( Fe^{3+} \), there are 5 unpaired electrons and for \( Fe^{2+} \), there are 4 unpaired electrons, leading to magnetic moments of 5.92 and 4.90 B.M., respectively.