The transition metal complex with the highest | vedclass

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

Question

(D) The crystal field splitting energy $(\Delta_{0})$ increases down a group in the periodic table for metal ions with the same charge and same ligand

Vedclass · Accepted Answer

$[Os(H_{2}O)_{6}]^{3+}$

Vedclass · Answer

(D) The crystal field splitting energy $(\Delta_{0})$ increases down a group in the periodic table for metal ions with the same charge and same ligand environment.This is because the $d$-orbitals of the $5d$ series elements are more extended in space compared to the $3d$ and $4d$ series,leading to stronger interaction with the ligands.Among the given complexes,$Cr^{3+}$ $(3d)$,$Mo^{3+}$ $(4d)$,$Fe^{3+}$ $(3d)$,and $Os^{3+}$ $(5d)$ are present.Since $Os$ belongs to the $5d$ series,the complex $[Os(H_{2}O)_{6}]^{3+}$ will have the highest $\Delta_{0}$ value.

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

Similar Questions

For the crystal field splitting in octahedral complexes,

The number of unpaired $d$-electrons in $[Co(H_2O)_6]^{3+}$ is $\qquad$

Consider the metal complexes $[Ni(en)_3]^{2+}$ $(A)$, $[NiCl_4]^{2-}$ $(B)$ and $[Ni(NH_3)_6]^{2+}$ $(C)$. Choose the $\text{CORRECT}$ option by considering the number of unpaired electrons present in $(A)$, $(B)$ and $(C)$ respectively and the order of frequency of absorption.

The electronic spectrum of $[Ti(H_2O)_6]^{3+}$ shows a single broad peak with a maximum at $20,300 \, cm^{-1}$. The crystal field stabilization energy $(CFSE)$ of the complex ion,in $kJ \, mol^{-1}$,is :

Carbon donor ligands are strong ligands and usually form low spin complexes. Among the complexes given below,select the high spin complex.

Vedclass Test Series

Exam Paper Generator

Online Exam Module