What is the electronic arrangement of a metal | vedclass

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

Question

(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.When the crystal field splitting energy $\Delta_0$ is less than the p

Vedclass · Accepted Answer

$t_{2g}^3 e_g^1$

Vedclass · Answer

(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.When the crystal field splitting energy $\Delta_0$ is less than the pairing energy $P$ (weak field ligand),the electrons prefer to occupy higher energy orbitals rather than pairing up.For a $d^4$ configuration,the first three electrons occupy the $t_{2g}$ orbitals singly.The fourth electron enters the higher energy $e_g$ orbital because $\Delta_0 Therefore,the electronic configuration is $t_{2g}^3 e_g^1$.

What is the electronic arrangement of a metal atom/ion in an octahedral complex with $d^4$ configuration,if $\Delta_0 < P$ (pairing energy)?

Similar Questions

The transition metal complex with the highest value of crystal field splitting $(\Delta_{0})$ is ........ .

The correct order for the wavelength of absorption in the visible region is

The correct order of energies of $d-$ orbitals of a metal ion in a square planar complex is:

According to crystal field theory,for which of the following coordination entities is $\Delta_o$ maximum?

Select the correct order of $\Delta_o$ (Crystal Field Splitting Energy).

Vedclass Test Series

Exam Paper Generator

Online Exam Module