Brown coloured ring is formed in nitrate test. Electronic Configuration of iron in the complex is:

[A r] 4 s^{2} 3 d^{5}

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[A r] 4 s^{o} 3 d^{7}

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[A r] 4 s^{1} 3 d^{6}

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[A r] 4 s^{2} 3 d^{7}

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Solution

The correct option is A $[A r] 4 s^{o} 3 d^{7}$
The iron is present in $+ 1$ oxidation state. Its electronic confguration in the complex is:
$[A r] 4 s^{o} 3 d^{7} \begin{matrix} ↑↓ & ↑↓ & ↑ & ↑ & ↑ \end{matrix}$

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