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

  • A
    $[Co(NH_3)_6]^{3+} > [Rh(NH_3)_6]^{3+} > [Ir(NH_3)_6]^{3+}$
  • B
    $[NiCl_4]^{2-} > [NiCl_6]^{4-}$
  • C
    $[M(H_2O)_6]^{3+} < [M(NO_2)_6]^{3-} < [M(CN)_6]^{3-}$
  • D
    $[M(CN)_6]^{3-} > [M(CN)_6]^{2-}$

Explore More

Similar Questions

Match List-$I$ with List-$II$.
List-$I$ (Chromium $(III)$ Complexes) - List-$II$ ($\Delta_o$ $(cm^{-1})$)
$A$. $[Cr(CN)_6]^{3-}$ - $I$. $15,060$
$B$. $[CrF_6]^{3-}$ - $II$. $17,400$
$C$. $[Cr(H_2O)_6]^{3+}$ - $III$. $22,300$
$D$. $[Cr(en)_3]^{3+}$ - $IV$. $26,600$
Choose the correct answer from the options given below:

In tetrahedral coordination complexes,what is the order of energy levels of $d$-orbitals during crystal field splitting?

The energies of $d_{xy}$ and $d_{z^2}$-orbitals in octahedral and tetrahedral transition metal complexes are such that

Arrange the following in increasing order of their crystal field splitting energy:
$I. [Co(H_2O)_6]^{3+}$
$II. [Co(NH_3)_6]^{3+}$
$III. [Co(CN)_6]^{3-}$
$IV. [CoF_6]^{3-}$

$A$ metal ion $M^{n+}$ having $d^6$ valence electronic configuration combines with three bidentate ligands to form a complex. Assuming crystal field splitting $(\Delta_0) >$ pairing energy,the $d$-orbital electronic configuration would be

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