FeSO_4 is a very good absorber for NO. The ne | vedclass

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(D) The reaction between $FeSO_4$ and $NO$ forms the brown ring complex: $[Fe(H_2O)_5(NO)]SO_4$.In this complex,$Fe$ is in the $+1$ oxidation state ($

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$3$

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(D) The reaction between $FeSO_4$ and $NO$ forms the brown ring complex: $[Fe(H_2O)_5(NO)]SO_4$.In this complex,$Fe$ is in the $+1$ oxidation state ($d^7$ configuration).$NO$ acts as a $NO^+$ ligand,making the metal center $Fe^+$,which has a $d^7$ configuration.Due to the strong field nature of $NO^+$,the electrons pair up in the $t_{2g}$ orbitals.The configuration is $t_{2g}^6 e_g^1$,resulting in $1$ unpaired electron in the $e_g$ orbital.However,in the context of the brown ring test,the complex is often represented as $[Fe(H_2O)_5(NO)]^{2+}$ where $Fe$ is $+1$ and $NO$ is neutral,or $Fe$ is $+2$ and $NO$ is $NO^-$. Based on the standard experimental magnetic moment of $3.89 \ BM$,the number of unpaired electrons is $3$.

$FeSO_4$ is a very good absorber for $NO$. The new compound formed by this process is found to contain what number of unpaired electrons?

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