In which of the following compounds does the | vedclass

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

Question

(B) To determine the number of unpaired electrons,we analyze the electronic configuration of the central metal ion in each complex:$1$. In $[Cr(NH_3)_

Vedclass · Accepted Answer

$[Co(H_2O)_6]^{3+}$

Vedclass · Answer

(B) To determine the number of unpaired electrons,we analyze the electronic configuration of the central metal ion in each complex:$1$. In $[Cr(NH_3)_6]^{3+}$,$Cr$ is in the $+3$ oxidation state. The configuration of $Cr^{3+}$ is $[Ar]3d^3$. Since $NH_3$ is a strong field ligand,but $d^3$ configuration has no choice but to have $3$ unpaired electrons in the $t_{2g}$ orbitals.$2$. In $[Co(H_2O)_6]^{3+}$,$Co$ is in the $+3$ oxidation state. The configuration of $Co^{3+}$ is $[Ar]3d^6$. $H_2O$ is a weak field ligand,so the electrons remain unpaired in the $t_{2g}$ and $e_g$ orbitals,resulting in $4$ unpaired electrons.$3$. In $[Zn(H_2O)_4]^{2+}$,$Zn$ is in the $+2$ oxidation state. The configuration of $Zn^{2+}$ is $[Ar]3d^{10}$. There are $0$ unpaired electrons.$4$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state. The configuration of $Ni^{2+}$ is $[Ar]3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons.Comparing the values,$[Co(H_2O)_6]^{3+}$ has the maximum number of unpaired electrons $(4)$.

In which of the following compounds does the central metal atom have the maximum number of unpaired $e^-$?

Similar Questions

Which of the following is true about the complex $[PtCl_2(H_2O)(NH_3)]$?

Which type of isomerism is exhibited by the isomers $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$?

The most paramagnetic complex is

Complexes $[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$ can be distinguished by

Vedclass Test Series

Exam Paper Generator

Online Exam Module