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

Arrange the following complex ions in increasing order of crystal field splitting energy $(\Delta_{0}):$ $[Cr(Cl)_{6}]^{3-}, [Cr(CN)_{6}]^{3-}, [Cr(NH_{3})_{6}]^{3+}$

The $\Delta_0$ of a coordination complex of a metal ion $(3d^1)$ is $1000 \ kJ \ mol^{-1}$. If the energy of $t_{2g}$ orbitals is $-400 \ kJ \ mol^{-1}$,the energy (in $kJ \ mol^{-1}$) of $e_g$ orbitals is

Complexes containing halide ligands are generally .....

The crystal field stabilisation energy $(CFSE)$ and the spin-only magnetic moment in Bohr Magneton $(BM)$ for the complex $K_3[Fe(CN)_6]$ are,respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module