Question:
The calculated spin-only magnetic moments of \( K_3[Fe(OH)_6] \) and \( K_4[Fe(OH)_6] \) respectively are:
The calculated spin-only magnetic moments of \( K_3[Fe(OH)_6] \) and \( K_4[Fe(OH)_6] \) respectively are:
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For calculating the spin-only magnetic moment, determine the number of unpaired electrons using the electronic configuration of the metal in the complex.
Updated On: Mar 24, 2025
- 4.90 and 4.90 B.M.
- 5.92 and 4.90 B.M.
- 3.87 and 4.90 B.M.
- 4.90 and 5.92 B.M.
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The Correct Option is B
Solution and Explanation
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.
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