The hybridisation and unpaired electrons in [ | vedclass

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

Question

(A) The central metal ion is $Fe^{2+}$,which has an electronic configuration of $[Ar] 3d^6$.$H_2O$ is a weak field ligand,so it does not cause pairing

Vedclass · Accepted Answer

$sp^3d^2 ; 4$

Vedclass · Answer

(A) The central metal ion is $Fe^{2+}$,which has an electronic configuration of $[Ar] 3d^6$.$H_2O$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbital.Thus,the $3d$ electrons are arranged as $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons.The hybridization involves the outer $4d$ orbitals,leading to $sp^3d^2$ hybridization (outer orbital complex).

The hybridisation and unpaired electrons in $[Fe(H_2O)_6]^{2+}$ ion are:

Similar Questions

The hybridisations of $Ni(CO)_4$ and $[Cr(H_2O)_6]^{2+}$ respectively,are

The species having tetrahedral shape is

$Mn^{2+}$ forms a complex with $Br^{-}$ ion. The magnetic moment of the complex is $5.92 \ B.M.$ What could not be the probable formula and geometry of the complex?

The metal $Ni$ $(Z = 28)$ combines with a unidentate ligand $X^-$ to form a paramagnetic complex $[NiX_4]^{2-}$. What are its geometry and the number of unpaired electrons in $Ni$?

The magnetic moment of the high spin complex is $5.92 \ BM$. What is the electronic configuration?

Vedclass Test Series

Exam Paper Generator

Online Exam Module