The d-electron configuration of [Ru(en)_3]Cl_ | vedclass

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

Question

(C) $1$. For $[Ru(en)_3]Cl_2$: $Ru$ is in the $+2$ oxidation state. $Ru$ belongs to the $4d$ series. For $4d$ and $5d$ series elements,the crystal fie

Vedclass · Accepted Answer

$t_{2g}^6 e_g^0$ and $t_{2g}^4 e_g^2$

Vedclass · Answer

(C) $1$. For $[Ru(en)_3]Cl_2$: $Ru$ is in the $+2$ oxidation state. $Ru$ belongs to the $4d$ series. For $4d$ and $5d$ series elements,the crystal field splitting energy $(\Delta_o)$ is large,even with weak field ligands,leading to low spin complexes. Thus,$Ru^{+2}$ $(d^6)$ will have the configuration $t_{2g}^6 e_g^0$.$2$. For $[Fe(H_2O)_6]Cl_2$: $Fe$ is in the $+2$ oxidation state $(d^6)$. $H_2O$ is a weak field ligand. Therefore,it forms a high spin complex with the configuration $t_{2g}^4 e_g^2$.$3$. Thus,the configurations are $t_{2g}^6 e_g^0$ and $t_{2g}^4 e_g^2$ respectively.

The $d-$electron configuration of $[Ru(en)_3]Cl_2$ and $[Fe(H_2O)_6]Cl_2,$ respectively are :

Similar Questions

Complexes with $CN^{-}$ ligands are generally .......

Which of the following complexes has maximum $CFSE$?

The Crystal Field Stabilisation Energies $(CFSE)$ of high spin and low spin $d^6$ metal complexes in terms of $\Delta_o$,respectively are

Calculate the Crystal Field Stabilization Energy $(CFSE)$ for a high-spin $d^4$ tetrahedral complex.

The tetrahedral crystal field splitting energy is equal to ................. of the octahedral crystal field splitting energy.

Vedclass Test Series

Exam Paper Generator

Online Exam Module