Crystal field stabilization energy for comple | vedclass

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

Question

(A) In $[Co(CN)_6]^{3-}$,the cobalt ion is in the $+3$ oxidation state,which corresponds to a $3d^6$ configuration.Since $CN^-$ is a strong field liga

Vedclass · Accepted Answer

$- 2.4 \Delta_0 + 3P$

Vedclass · Answer

(A) In $[Co(CN)_6]^{3-}$,the cobalt ion is in the $+3$ oxidation state,which corresponds to a $3d^6$ configuration.Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in a low-spin complex.All $6$ electrons occupy the $t_{2g}$ orbitals,and $0$ electrons occupy the $e_g$ orbitals.The crystal field stabilization energy $(CFSE)$ is calculated as:$CFSE = (n_{t2g} 	imes -0.4 \Delta_0) + (n_{eg} 	imes 0.6 \Delta_0) + mP$Where $n_{t2g} = 6$,$n_{eg} = 0$,and $m$ is the number of extra electron pairs formed due to pairing.For $d^6$ low spin,$3$ pairs are formed,so $m = 3$.$CFSE = (6 	imes -0.4 \Delta_0) + (0 	imes 0.6 \Delta_0) + 3P = -2.4 \Delta_0 + 3P$.

Crystal field stabilization energy for complex $[Co(CN)_6]^{3-}$ will be

Similar Questions

The crystal field stabilization energy $(CFSE)$ of $[CoF_{3}(H_{2}O)_{3}]$ $(\Delta_{0} < P)$ is $:-$

The $3d$ metal ions form coloured compounds because the energy corresponding to the following lies in the visible range of the electromagnetic spectrum:

The conditions and consequence that favour the $t_{2g}^3, e_g^1$ configuration in a metal complex are:

What is crystal field splitting energy? How does the magnitude of $\Delta_{o}$ decide the actual configuration of $d$-orbitals in a coordination entity?

The electronic configuration of the central atom of the complex $[MnO_4]^{2-}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module