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

Which of the following species have a square planar geometry for the central atom?
$(i) XeF_4 \ (ii) SF_4 \ (iii) [NiCl_4]^{2-} \ (iv) [PtCl_4]^{2-}$

Which complex compound possesses $sp^3d^2$ hybridisation?

$d^{2}sp^{3}$ hybridisation of the atomic orbitals gives

For $K_3[Co(CN)_6]$,which of the following statements is not correct?

In solid $CuSO_4 \cdot 5H_2O$,copper is coordinated to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module