Arrange the following Cobalt complexes in the | vedclass

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

Question

(B) The $CFSE$ value depends on two main factors:$1$. Oxidation state of the central metal ion: Higher oxidation state leads to higher $CFSE$.$2$. Str

Vedclass · Accepted Answer

$B < A < C < D$

Vedclass · Answer

(B) The $CFSE$ value depends on two main factors:$1$. Oxidation state of the central metal ion: Higher oxidation state leads to higher $CFSE$.$2$. Strength of the ligand (Spectrochemical series): Stronger ligands cause larger splitting.Let's analyze the complexes:- $[Co(H_{2}O)_{6}]^{2+}$: $Co^{2+}$ $(d^{7})$,weak field ligand $(H_{2}O)$.- $[CoF_{6}]^{3-}$: $Co^{3+}$ $(d^{6})$,weak field ligand $(F^{-})$.- $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ $(d^{6})$,strong field ligand $(NH_{3})$.- $[Co(en)_{3}]^{3+}$: $Co^{3+}$ $(d^{6})$,very strong field ligand $(en)$.Comparing $B$ $(Co^{2+})$ and $A$ $(Co^{3+})$,$B$ has a lower oxidation state,so it has lower $CFSE$. Among the $Co^{3+}$ complexes,the order of ligand strength is $F^{-} Thus,the increasing order of $CFSE$ is: $[Co(H_{2}O)_{6}]^{2+} < [CoF_{6}]^{3-} < [Co(NH_{3})_{6}]^{3+} < [Co(en)_{3}]^{3+}$,which corresponds to $B < A < C < D$.

Similar Questions

$A$ transition metal $(M)$ among $Mn, Cr, Co$ and $Fe$ has the highest standard electrode potential $(M^{3+} / M^{2+})$. It forms a metal complex of the type $[M(CN)_6]^{4-}$. The number of electrons present in the $e_g$ orbital of the complex is $.........$

The correct increasing order for the wavelengths of absorption in the visible region for the complexes of $Co^{3+}$ is:

Calculate the crystal field stabilization energy $(CFSE)$ in $[Co(CN)_6]^{3-}$.

Complexes with $CN^-$ ligands are generally...

The spin magnetic moment of an octahedral complex having $CFSE = -0.8 \Delta_{0}$ and surrounded by weak field ligands can be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module