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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 ($

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(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?

Similar Questions

Total number of $e^-$ in $e_g$ orbitals of $Mn$ for the complex $[MnO_4]^-$ is:

Which of the following complexes exhibits a square planar geometry?

The correct order of hybridization of the central atom in the following species $NH_3$,$[PtCl_4]^{2-}$,$PCl_5$,and $BCl_3$ is

On the basis of valence bond theory,discuss the geometrical shapes and magnetic properties of the following $4$-coordinated complexes:
$(i)$ $[NiCl_{4}]^{2-}$
$(ii)$ $[Ni(CN)_{4}]^{2-}$
$(iii)$ $[Ni(CO)_{4}]$
$(iv)$ $MnO_{4}^{-}$

According to $V.B.T.$,in which of the following complexes does the transition of an electron occur from one shell to another shell of the central metal atom?

Vedclass Test Series

Exam Paper Generator

Online Exam Module