The species which has four unpaired electrons | vedclass

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

Question

(C) To determine the number of unpaired electrons,we analyze the electronic configuration of the central metal ion in each complex:$1$. In $[Co(CN)_6]

Vedclass · Accepted Answer

$[FeCl_4]^{2-}$

Vedclass · Answer

(C) To determine the number of unpaired electrons,we analyze the electronic configuration of the central metal ion in each complex:$1$. In $[Co(CN)_6]^{4-}$,$Co$ is in $+2$ oxidation state $(d^7)$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^6 e_g^1$,resulting in $1$ unpaired electron.$2$. In $[Cr(H_2O)_6]^{3+}$,$Cr$ is in $+3$ oxidation state $(d^3)$. Configuration: $t_{2g}^3 e_g^0$,resulting in $3$ unpaired electrons.$3$. In $[FeCl_4]^{2-}$,$Fe$ is in $+2$ oxidation state $(d^6)$. $Cl^-$ is a weak field ligand. Configuration: $e^3 t_2^3$,resulting in $4$ unpaired electrons.$4$. In $[Fe(H_2O)_6]^{3+}$,$Fe$ is in $+3$ oxidation state $(d^5)$. $H_2O$ is a weak field ligand. Configuration: $t_{2g}^3 e_g^2$,resulting in $5$ unpaired electrons.Thus,the species with $4$ unpaired electrons is $[FeCl_4]^{2-}$.

The species which has four unpaired electrons is

Similar Questions

How many hybrid orbitals of cobalt ion are involved in the formation of the hexaamminecobalt$(III)$ chloride complex?

Choose the correct statement for the $[Ni(CN)_4]^{2-}$ complex ion (Atomic number of $Ni=28$).

Nickel $(Z = 28)$ forms a paramagnetic complex $[NiX_4]^{2-}$ with a uninegative monodentate ligand $X^-$. The number of unpaired electrons in nickel and the geometry of this complex ion are,respectively:

Which of the following complexes is tetrahedral as well as diamagnetic?

Which of the following has a square planar geometry?
(At. nos.: $Fe=26, Co=27, Ni=28, Pt=78$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module