The electronic configuration of the central a | vedclass

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

Question

(D) $1$. In the complex $[MnO_4]^{2-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -2$,which gives $x = +6$. $2$. The electronic config

Vedclass · Accepted Answer

$e^1 \, t_2^0$

Vedclass · Answer

(D) $1$. In the complex $[MnO_4]^{2-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -2$,which gives $x = +6$. $2$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. $3$. The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$. $4$. In a tetrahedral complex like $[MnO_4]^{2-}$,the $d$-orbitals split into $e$ (lower energy) and $t_2$ (higher energy) sets. $5$. The single $d$-electron occupies the lower energy $e$ orbital. $6$. Therefore,the configuration is $e^1 \, t_2^0$.

The electronic configuration of the central atom of the complex $[MnO_4]^{2-}$ is:

Similar Questions

The $CFSE$ for $[CoCl_6]^{4-}$ complex is $18000 \ cm^{-1}$. The $\Delta_t$ for $[CoCl_4]^{2-}$ will be ........... $cm^{-1}$.

The complex that has the highest crystal field splitting energy $(\Delta)$ is

The electronic spectrum of $[Ti(H_2O)_6]^{3+}$ shows a single broad peak with a maximum at $20,300 \, cm^{-1}$. The crystal field stabilization energy $(CFSE)$ of the complex ion,in $kJ \, mol^{-1}$,is :

The crystal field splitting energy for octahedral $\Delta _0$ and tetrahedral $\Delta _t$ complexes is related as

Which of the following complexes will absorb the maximum wavelength of light?

Vedclass Test Series

Exam Paper Generator

Online Exam Module