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Multi-Electron Configurations and Orbital Diagrams

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In question (2) magnetic moment is $\sqrt{15}$ where unpaired electrons are 3.

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Solution

(a) The iron is present in +1 oxidation state. Its electronic
confguration in the complex is $[A r] 4 s^{o} 3 d^{7} \begin{matrix} ↑↓ & ↑↓ & ↑↑↑ \end{matrix}$ .
Thus three unpaired electrons are present. This is also confirmed by magnetic moment of 3.87 B.M
$\sqrt{N (N + 2)} = \sqrt{15} = 3.87$

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