What is the correct order for energy of $d$ orbitals during splitting in the $[NiCl_4]^{2-}$ complex ion?

  • A
    $d_{xy} \cong d_{yz} \cong d_{xz} < d_{x^2-y^2} \cong d_{z^2}$
  • B
    $d_{xy} \cong d_{yz} \cong d_{xz} \cong d_{x^2-y^2} \cong d_{z^2}$
  • C
    $d_{x^2-y^2} \cong d_{z^2} < d_{xy} \cong d_{yz} \cong d_{xz}$
  • D
    $d_{x^2-y^2} > d_{z^2} > d_{xy} \cong d_{yz} \cong d_{xz}$

Explore More

Similar Questions

The wavelengths of light absorbed by the complexes $[Ni(H_2O)_6]^{2+}$,$[Ni(en)_3]^{2+}$,and $[Ni(H_2O)_4en]^{2+}$ are $\lambda_1$,$\lambda_2$,and $\lambda_3$ respectively. The correct order of wavelengths is:

Three complexes,$[CoCl(NH_3)_5]^{2+} (I)$,$[Co(NH_3)_5H_2O]^{3+} (II)$ and $[Co(NH_3)_6]^{3+} (III)$ absorb light in the visible region. The correct order of the wavelength of light absorbed by them is

The crystal field stabilization energy for a high spin $d^4$ octahedral complex is:

The $d-$orbital electronic configuration of the complex among $[Co(en)_3]^{3+}$,$[CoF_6]^{3-}$,$[Mn(H_2O)_6]^{2+}$ and $[Zn(H_2O)_6]^{2+}$ that has the highest $CFSE$ is $:$

$[Pd(F)(Cl)(Br)(I)]^{2-}$ has $n$ number of geometrical isomers. Then,the spin-only magnetic moment and crystal field stabilisation energy $[CFSE]$ of $[Fe(CN)_6]^{n-6}$ are
[Note : Ignore the pairing energy]

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo